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\begin{document}

\begin{titlepage}
\title
{The uniqueness case}
\author{G.~Stroth\\
Martin-Luther-Universit\"at Halle-Wittenberg\\
Institut f\"ur Algebra und Geometrie\\
06099 Halle\\
Germany}
\end{titlepage}
\maketitle

This paper is part of the revision of the classification of
the finite simple groups. The aim is to give a solution of the so
called "uniqueness case problem". Originally this problem was
solved by M.~Aschbacher \cite{Asch1}. We do not follow his proof. The
proof given in this paper uses the amalgam method, which has been
used successfully in many places for dealing with weak closure. Also
our theorem looks different from the one in \cite{Asch1} as definitions
have changed over the time. In fact our hypothesis is weaker as in
\cite{Asch1}. First of all we need some precise definitions. For any group
$X$ let
\begin{center}
$\sigma (X)=\{ p|p\mbox{~odd,~} m_{p}(H)\geq 4$ for some
2-local $H$ of $X$ with $|\,X:H\, |\mbox{~odd} \}\,.$
\end{center}
Furthermore let $Q$ be a $p$-subgroup of $X, k\leq
m(Q )$, let
$$\Gamma_{Q ,k} (X)=\langle N_{X} (R)\, |\, R\leq Q
,\, m(R)\geq k \rangle \, .$$
In this paper we will consider groups $G$ which satisfy the following
conditions
\begin{itemize}
\item[(1)]\begin{itemize}
\item[(i)]If $H$ is a 2-local in $G$ with  $|G:H|\mbox{~odd}$. Fix $S \in $Syl$_2(H)$. Then
$H=\langle \tilde{H} |\, O_{2} (\tilde{H} )=F^{*} (\tilde{H} ), S \leq \tilde{H} \rangle $
(we call $G$ of even type)
\item[(ii)]$\sigma (G)\not= \emptyset$
\end{itemize}
\item[(2)] A group $M$ is called a uniqueness group provided $\sigma (M)\not=
\emptyset$ and one of the following holds
\begin{enumerate}
\item[($\alpha$)] $M$ is a maximal 2-local of $G$ with $F^{*} (M)=O_{2} (M)\, .$
\item[($\beta$)] $F^*(M) = O_2(M)K$, where $K$ is a quasisimple group of Lie type in characteristic 2, $Z(K) = O_2(K)$, and for every $p \in \sigma(M)$ we have $m_p(K) \ge 3$ and $m_p(\bC_M(K)) \le 1.$
\end{enumerate}
\item[(3)]
Let $M$ be a uniqueness subgroup of $G$. For every $p\in\sigma (M)$\, ,\, and $P\in\mbox{~Syl}_{p} (M)$ we have
$\Gamma_{P, 2} (G)\leq M$ and one of the following holds
\begin{itemize}
\item[(i)]If $x\in P, o(x)=p, m_{p}(\bC_{M} (x
) )\geq 3, \mbox{~then~} N_{G}(\langle x \rangle )\leq M\, .$\\
(In particular for $p=3, N_{G}(\langle x \rangle )\leq M$ for any $x\in M, 
o(x)=3$)
\item[(ii)] $F^*(M) = O_2(M)$ and $M/O_{p^{\prime}} (M)=\bar{M} =\bar{X}\bar{Y} ,\bar{X}
\unlhd\bar{M}$ where $m_{p}(\bar{X} )=1$ and $\bar{Q} =O_{p}\, (\bar{Y}
)\unlhd\bar{M}$. Further $\bar{P} =(\bar{P}\cap\bar{X} )\times\bar{Q}$
is elementary abelian of order $p^{\, n}$. Furthermore $\bar{Y}$ is
isomorphic to a Borel subgroup of an automorphism group of $L_{2}
(p^{\, n} )$, and $\Gamma_{Q, 1} (G)\leq M$,
where $Q$ is a preimage of $\bar{Q}$ in $P$.
\end{itemize}
We say $G$ is in the uniqueness case if $G$ is $\cK$-simple and 
satisfies (1) and (3) and the following holds.
\item[(4)]\begin{itemize}
\item[(i)] For every $p \in \sigma(G)$ there is a uniqueness subgroup $M_p$ with
$p\in \sigma(M_p)$.
\item[(ii)] Let $M$ be a uniqueness subgroup of $G$ with $p\in \sigma(M)$. If $H$ is any 2-local subgroup of $G$ such that $H\cap
M\geq E, E\cong E_{p^{\, 2}}, p\in\sigma (M), \Gamma_{E,1} (G)\leq
M$, then $H \le M$.
\item[(iii)] If $M$ is a uniqueness subgroup of $G$, then $N_{G} (S)\leq M$ for
$S\in\mbox{~Syl}_{2} (M)$.
\end{itemize}
\end{itemize}

If $M$ is as in (3) (ii) we call $M$ exceptional. Now we can state our theorem.
\\
\\

{\bf Theorem}. {\it Let $G$ be in the uniqueness case. Let $M$ be a
uniqueness subgroup. If $S\in\mbox{~Syl}_{2} (M)$, then $M$ contains 
every maximal 2-local subgroup of $G$ containing $N_{G} (S)$.}\\

\absa
It remains to explain what $\cK $-simple means. A group is called
$\cK $-simple if it is a minimal counterexample to the
classification theorem, i.e. all simple nonabelian sections of all
proper subgroups are in $\cK $. Here $\cK $ is the set of the
groups of Lie type, the alternating groups and the 26 sporadic
groups.\\
 
\absa
For the proof of the theorem we assume there is a 2-local $N$ such
that $N_{G} (S)\leq N$ but $N\not \leq M$. From this we
construct a pair of subgroups $G_{1}, G_{2}$ such that $G_{2}
\leq N$ and $G_{1}\leq M$. The definition of $G_{2}$ is very simple.
We just choose $G_{2}$ minimal with respect to $|M\, :\, G_{2}\cap M\,
|\mbox{~odd}, F^{*} (G_{2} )=O_{2} (G_{2} )$ and $G_{2}\not\leq M$.
The construction of $G_{1}$ is much more delicate. The definition of
$G_{1}$ is given in chapter 2. The main problem is not to lose the
information given from $p\in\sigma (M)$ but have $G_{1}$ not to
complicated. Then we use the amalgam method for the pair $(G_{1},
G_{2})$, to describe their structure and come down with a
contradiction. Here action of simple groups on $\langle \Omega_{1}
(Z(S))^{G_{i}} \rangle $~~and~~$\langle \Omega_{1}
(Z(S))^{G_{i}} \rangle ^{G_{j}}\, , j,i=1,2\, ,\, i\not= j\, ,\,
S\in\mbox{Syl}_{2} (G_{1}\cap G_{2} )$ plays an important role. In
particular results about quadratic modules \cite{MeiStr1} \cite{MeiStr2},
\cite{Str} and about SC-modules \cite{Coop}, \cite{Asch3} are the key tools. For
convenience of the reader we give proofs of almost all the results
used from \cite{Asch3}, as in this paper are a lot of further references
we would like to avoid. These results usually provide us with Lie
groups over a field of characteristic two and basic modules $V(\lambda
)$. A careful analysis of centralizers of $v\in V(\lambda )$ in the
corresponding Lie group mostly shows that we pick up an $E\cong
E_{p^{\, 2}}, \Gamma_{E,1} (G)\leq M$, which centralize $v$. Now $\bC_{G}
(v)\leq M$, which then at the end forces $G_{2}$ to be in $M$.\\

\absa
For what follows we will always assume that $G$ is in the uniqueness
case and $G$ is a counterexample to the theorem. Furthermore we just
drop the term $\cK $-simple. Whenever the letter $M$ occurs it
will stand for a uniqueness group we will fix from the beginning. In
fact if $3\in\sigma (G)$ we choose $M$ with $3\in\sigma (M)$. With this restriction we choose $M$ of type (2)($\beta$) if possible, 
otherwise any $M$ will be fine.\\
\\
The proof is organized as follows. After the preliminary lemmas we will construct the group  $G_2$ in \S 2. Then in \S 3 we show that for a uniqueness group $M$ we
must have $F^*(M) = O_2(M)$. In \S 4 we construct the group $G_1$. To the groups $G_1$ and $G_2$ we use the amalgam method in \S 6 - \S 8 to arrive at a contradiction. Very important for this is a certain class of modules $V$
for the group $G_2/O_2(G_2)$. These are modules such that $|V : \bC_V(A)| \le |A|^2$ for some nontrivial elementary abelian 2--subgroup of $G_2/O_2(G_2)$.
\\



\begin{center}
\S~1 Preliminaries
\end{center}


\absa
{\bf (1.1)~Lemma.~}{\it Let $X\cong L_{2}(q)\, ,\mbox{~or~} Sz(q)\,
, q\geq 4, q$ even.}
\begin{itemize}
\item[(i)] {\it Let $t\in\Omega_{1} (S), S\in\mbox{Syl}_{2} (X)$. Then
there are conjugates $a, b$ of $t$ such that $X =\langle a,b,t\rangle $.}
\item[(ii)]{\it Let $A\leq\Omega_{1} (S)\, , |A|\geq 4$. Then there
is some $g\in X$ with $X=\langle A, A^{g}\rangle $.}
\end{itemize}

\absa
Proof.~~Let $\langle t,a\rangle \leq A\leq\Omega_{1} (S)\, , |A|\geq
4$. Let
$K=N_{X} (S)\, , |K|=q-1$, and $a,b\in N_{X} (K^{\, g})$, with
$N_{X}(K^{\, g} )=\langle a,b\rangle$ , $g\in X$. Then we get 
$\langle a,b,t\rangle \geq \langle \Omega_{1} (S), 
\Omega_{1} (S)^{\, b}\rangle $. Thus to prove (i) and (ii) it is 
enough to
show $\langle \Omega_{1} (S)\, , \Omega_{1} (S)^{b}\rangle =X$.\\
\indent
We have that $Y=\langle \Omega_{1} (S),\Omega_{1} (S)^{\, g} \rangle $ 
contains at
least $q+1$ conjugates of $\Omega_{1} (S)$. Thus we are done if
$X\cong L_{2}(q)$, as 
$\langle \Omega_{1} (S), \Omega_{1} (S)^{\, b}\rangle $
contains all conjugates.\\
\indent
So let $X\cong Sz(q)$. The number of conjugates of $\Omega_{1} (S)$
in $Y$ is $n q+1$. But then $n q+1\,\Big  |\, q(q^{2}+1)$. Which gives
$n=q$ and so $\Omega_{1} (S)^{X}\leq Y$, hence $X=Y$.

\absa
{\bf (1.2)~Lemma.~}{\it Let $X\cong G(q)$ be a Lie group over a field
of characteristic two, $q > 2$. Let $C$ be the Cartan subgroup and
$m_{p}(C)\leq 3$ for any prime  $p$. Then $X$ is one of the following:
$L_{n}(q), n\leq 4, Sp_{4}(q), Sp_{6}(q), U_{n}(q), n\leq 7,
\Omega_8^{\, -} (q)$, $^2F_{4} (q)$, $^3D_{4} (q)$, $G_{2}(q)$, or 
$Sz(q)$.}

\absa
Proof.~~Let first $X$ be untwisted of Lie rank $r$. Then
$$|\, C\, |=\frac{1}{d} (q-1)^{r}$$
Let $r\geq 4$. Then as $m_{p}(C)\leq 3$, we have $q-1=p=d$.
Furthermore $r =4$. But checking the possible values for $d$ gives
a contradiction. So we have $r\leq  3$ and then 
$X\cong L_{2}(q), L_{3}(q),
L_{4}(q), Sp_{4} (q), Sp_{6}(q)$~~or~~$G_{2}(q)$.\\
\indent
Assume now that $X$ is twisted. Let $X\cong$ $^2E_{6}(q)$.
Then
$$|\, C\, |=\frac{1}{d} (q-1)^{2} (q^{\, 2} -1)^{\, 2}\, , \, 
d=(3,q+1)\, .$$
Let $p\,\Big  |\, q-1$. Then $C$ contains an elementary abelian subgroup
of order $p^{\, 4}$, a contradiction.\\
\indent
Let $X\cong U_{n}(q)$. If $n$ is even then 
$$|\, C\, |=\frac{1}{d} (q-1)(q^{\, 2} -1)^{\,\frac{n}{2} -1}$$
Thus $n\leq 6$.

Let $n$ be odd. Then 
$$|\, C\, |=\frac{1}{d} (q^{\, 2}-1)^{\,\frac{n-1}{2}}$$
This implies $n\leq 7$.\\
\indent
Finally assume that $X\cong\Omega_{2n}^{-} (q)$. Then 
$$|\, C\, |= (q^{\, 2}-1)(q-1)^{n-2}$$
Hence $n-2\leq 2$. We get $n=4$, as $\Omega_6^{-} (q)\cong U_{4}
(q)$.

\absa
{\bf (1.3)~Lemma.~}{\it Let $X/Z(X) \in\cK\,$ , $Z(X)$ a $3$-group, 
then one
of the following holds}
\begin{itemize}
\item[(i)] {\it $m_{3}(X)=0\, ,X\cong Sz(q)$ $,q$ even.}
\item[(ii)] {\it $m_{3}(X)=1\, ,X\cong L_{2}(q), L_{3}(q), U_{3}(q),
J_{1}$.}
\item[(iii)] {\it $m_{3}(X)=2$, $X\cong 3\cdot A_{6}$, $3\cdot A_{7}$, 
$3\cdot M_{22}$, $SL_{3}(q)$, $SU_{3}(q)$, $A_{7}$, $A_{6}$, 
$L_{3}(3)$, $U_{3}(3)$, $L_{3}(q)$, $U_{3}(q)$, $PSp_{4}(q)$, $G_{2}(q)$, 
$^3D_{4}(q)$, $^2F_{4}(q)$, $L_{4}(q)$, $U_{4}(q)$, $L_{5}(q)$, 
$U_{5}(q)$, $M_{11}$, $M_{12}$, $M_{22}$, $M_{23}$, $M_{24}$, 
$J_{2}$, $HiS$, $He$, $Ru$, $J_{4}$.}
\item[(iv)] {\it $m_{3}(X)=3$ and $X\cong 3\cdot O^\prime N$, $A_{9}$, 
$A_{10}$, $A_{11}$, $L_{2}(27)$, 
$PSp_{4}(3)$, $Sp_{6}(q)$, $\Omega^{-}_{8}(q)$,
$L_{4}(q)$, $U_{4}(q)$, $L_{6}(q)$, $U_{6}(q)$, $L_{7}(q)$, $U_{7}(q)$, 
$J_{3}$.}
\end{itemize}

\absa
Proof~. Just inspection of the groups in $\cK$

\absa
{\bf (1.4)~Lemma.~}{\it Let $X/Z(X)\in\cK ,X/Z(X)$ sporadic or
alternating, $Z(X)$ a $3$--group. Suppose $m_{3}(X)\leq 3$, then
$m_{p}(X)\leq 2$ for any odd prime $p\not= 3$.}

\absa
Proof~. By (1.3) we have $X\cong J_{1}$, $M_{11}$, $M_{12}$, $M_{22}$,
$M_{23}$, $M_{24}$, $J_{2}$, $3\cdot M_{22}$, $HiS$, $He$, $Ru$, $J_{4}$, 
$3\cdot O^\prime N$, or $J_{3}$, or $X/Z(X)$ is alternating, i.e. 
$X\cong 3\cdot A_{6}$, $3\cdot A_{7}$, $A_{6}$, $A_{7}$, $A_{8}$, 
$A_{9}$, $A_{10}$, $A_{11}$.\\
\indent
Assume now $p^{\, 3}\,\Big  |\, |X|\, , p\mbox{~odd~} ,p\not= 3$. Then 
we see $X\cong J_{4}\,(p=11)$, $HiS\,(p=5)$, $Ru\,(p=5)$, $He\,(p=7)$ or 
$3\cdot O^\prime N\,(p=7)$. But in all these cases we have $m_{p}(X)=2$ 
by $\cite{CCNPW}$.

\absa
{\bf (1.5)~Lemma.~}{\it Let $K$ be a finite simple group in the list
$\cK , m_{p}(K )\leq 3$ for any odd prime $p$. Then $K$ is one of
the following:}
\begin{itemize}
\item[(i)] {\it $L_{2}(q), Sz(q), L_{3}(q), U_{3}(q), PSp_{4}(q), q$ 
some prime power}
\item[(ii)]{\it $Sp_{4}(2^{n})$, $G_{2}(2^{n})$, $^2F_{4}(2^{n})^\prime$,
$^3D_{4}(2^{n})$, $L_{4}(2^{n})$, $L_{5}(2)$, $L_{6}(2)$, $L_{7}(2)$,
$U_{4}(2^{n})$, $Sp_{6}(2^{n})$, $\Omega^{-}_{8}(2^{n})$}
\item[(iii)] {\it $A_{n},\; 6\leq n\leq 11$}
\item[(iv)]{\it $J_{i}, 1\leq i\leq 4, M_{n}, n\in\{ 11,12,22,23,24\}
,HiS, Ru, He$.}
\end{itemize}
{\it If $K\in\cK , U/Z(U)\cong K, U^\prime =U, 1\not= |\, Z(U)\, |$ odd and
$m_{p}(U)\leq 3$ for every odd prime $p$. Then $U$ is isomorphic to
$3\cdot A_{6}, 3\cdot A_{7}, 3\cdot M_{22}, SL_{3}(q), SU_{3}(q), q$
a prime power, or $3\cdot O^\prime  N$.}

\absa
Proof~. This is easily established by going over the list in (1.3).

\absa
{\bf (1.6)~Lemma.~}{\it Let $X$ be a $p$-group, $p$ odd, cl$(X)\leq
2, X=\Omega_{1} (X)$ and $m_{p}(X)\leq 3$. Then $X$ is elementary
abelian, extraspecial of width 1, a direct product of a cyclic group
of order $p$ with an extraspecial group of width $1$, or an
extraspecial group of width $2$.}

\absa 
Proof~. We may assume cl$(X)=2$. We have $X=\{ x\Big | x^{p} =1\}$.
Let $m_{p}(X)=2$. Then $|Z(X)| =p$. Let $\omega\in X\setminus Z(X)$.
Then $[\omega ,X]=Z(X)$ and so $|\, X:\bC_{X} (\omega )| =p$.
Now $m_{p}(\bC_{X} (\omega ))=2$ and so as 
$|\, Z(\bC_{X} (\omega ))|\geq p^{\, 2}$ we get 
$|\bC_{X} (\omega )| =p^{\, 2}$. This
shows $|X|=p^{\, 3}$ and so $X$ is extraspecial of width 1.\\
\indent
Let now $m_{p}(X)=3$. Assume $|\, Z(X)| =p^{\, 2}$. Again choose
$\omega\in X\setminus Z(X)$. Then $|\, X:\bC_{X} (\omega )| \leq
p^{\, 2}$ and as $|\,\bC_{X} (\omega )| =p^{\, 3}$, we get
$|X|\leq p^{\, 5}$.\\
\indent
Let $|X|=p^{\, 4}$. Then $X^\prime =[X,\omega ]$ is of order $p$. Thus 
$X$ is a direct product of a cyclic group of order $p$ with an
extraspecial group of width 1.\\
\indent
Let $|X|=p^{\, 5}$. Choose $\bC_{X} (\omega )\leq Y < X, |Y|=p^{\, 4}$.
Then as just seen $Y$ is a direct product of a cyclic group of order
$p$ by an extraspecial group of width 1. Now choose $\varphi\in
X\setminus Y$. Then $[\omega ,\varphi ]=t\not\in Y^\prime $. 
Let $\nu\in Y, 1\not= [\omega ,\nu ]=s\in Y^\prime $. We have 
$$[\varphi ,\nu ]=s^{i}\, t^{i}\, ,\mbox{~for some~} 0\leq i, j\leq
p-1\, .$$
But then 
$$[\varphi ,\nu \omega^{\, j} ] =s^{\, i}$$
and so we may assume
$$[\varphi ,\nu ]=s^{\, i}\, .$$
Hence $[X,\nu ]\leq \langle s\rangle $ and then 
$|X:\bC_{X} (\nu )| \leq p$, yielding $|X|=p^{\, 4}$, a contradiction.\\
\indent
So we have $|Z(X)| =p$. Then $Z(X)=\Phi (X)=X^\prime $ and $X$ is
extraspecial. As $m_{p}(X)=3$ we have that $X$ is of width 2.

\absa
{\bf (1.7)~Lemma.~}{\it Let $X$ be some group with $O_{2}(X)=1$ and
$m_{p}(X)\leq 3$ for every odd prime  $p$. Suppose furthermore that for
$S\in\mbox{~Syl}_{2} (X)$ there is exactly one maximal subgroup $Y$
of $X$ containing $S$. Then either $X$ is solvable, or $E(X)$ is one
of the following:}
\begin{itemize}
\item[(i)]{\it $Sz(q)$, $L_{2}(q)$, $(S)L_{3}(q)$, $(S)U_{3}(q)$, 
$Sp_{4}(q)$, $L_{2}(q)\times L_{2}(q)$, $Sz(q)\times Sz(q)$, $q$ even}
\item[(ii)]{\it $L_{2} (p)$, $L_{2}(p^{\, 2})$, $L_{2}(p^{\, 3})$, 
$L_{3}(p)$,
$U_{3}(p)$, $PSp_{4}(p)$, $L_{2}(p)\times L_{2}(p)$, 
$p > 3$ some odd prime, or $L_{2}(27)$ or $L_{3}(3)$}
\item[(iii)]{\it $A_{6}$, $A_{9}$, $3\cdot A_{6}$, 
$3\cdot A_{6}*3\cdot A_{6}$,
$SL_{3}(4)*SL_{3}(4)$, $SU_{3}(8)*SU_{3}(8)$.}
\end{itemize}
{\it or $F(X)$ is a $p$-group with $\Omega_{1} (F(X))$ is elementary
abelian of order $p^{\, 3}$ and $E(X/F(X))\cong L_{2}(p)$ acts
irreducibly on $\Omega_{1} (F(X))$.

\indent
If $E(X)\cong (S)L_{3}(q), Sp_{4}(q), q$ even, $3\cdot A_{6}, 3\cdot
A_{6}*3\cdot A_{6}$, or $SL_{3}(4)*SL_{3}(4)$, there is some $x\in X$
acting nontrivially on the corresponding Dynkin diagram.}

\absa  
Proof~. We may assume that $X$ is nonsolvable. Assume first $E(X)=1$.
Set $F=F(X)$. Let $p\,\Big  |\,|F(X)|$ and $P\in\mbox{~Syl}_{p}
(F(X))$. We may assume that $X/\bC_{X} (P)$ is nonsolvable. Let $C$
be a critical subgroup of $P$ and $U=\Omega_{1} (C)$. Set
$X_{1}=\bC_{X} (U)$.\\
\indent
Suppose $S\cap X_{1}\not= 1$. By the Frattini argument we have
$X=X_{1}N_{X}(S\cap X_{1})$. As $O_{2}(X)=1$ we get $N_{X}(S\cap
X_{1})\not= X$. But then $X_{1}S$ and $N_{X}(S\cap X_{1})$ are
contained in different maximal subgroups containing $S$, a
contradiction.\\
\indent
So we have $S\cap X_{1}=1$. Let $t\in Z(S)^{\sharp}$ such that
$\bC_{X} (t)$ covers $(X/X_{1} )\, /\, O_{p}(X/X_{1})$. As $O_{2}(X)=1$,
we have $\bC_{X} (t)\not= X$. Let $Y$ be the preimage of
$O_{p}(X/X_{1})$. Then $YS\not= X$ and so $\bC_{X} (t)$ and $YS$ are
contained in different maximal subgroups, a contradiction.\\
\indent
Now by (1.6) we know the structure of $U$. The structure of $Sp_{4}
(p)$ \cite{Mi2} shows that $U$ has to be elementary abelian of order
$p^{\, 3}$ and $X/X_{1}$ is a subgroup of $L_{3}(p)$. Now $S\,
X_{1}/X_{1}$ is contained in exactly one maximal subgroup of
$X/X_{1}, p\,\Big  |\, |E(X/X_{1})|$, otherwise $X$ splits over
$P$. So with \cite{Mi1} we get that $E(X/X_{1})\cong L_{2}(p)$. Hence
$X$ acts irreducibly on $U$.\\
\indent
As above the Frattini argument shows that every Sylow subgroup of $X_{1}$
is normal in $X$, so $X_{1} =F$. Suppose $F\not= P$. Then choose
$Q\in\mbox{~Syl}_{q} (F), q\not= p$. We get $X=Q\bC_{X} (Q )$.
So $O^{p^\prime } (X)S\not= X$. Now $O^{p^\prime } (X)S$ and 
$O_{p^\prime } (F)S$
are contained in different maximal subgroups of $X$, a
contradiction.\\
\indent
Suppose now $E(X)\not= 1$. We have $X=E(X)S$, otherwise $N_{X} (S\cap
E(X)), E(X)S$ are contained in different maximal subgroups.
Furthermore this implies that $S$ is transitive on the components of
$E(X)$. In particular $E(X)$ contains at most two components.\\
\indent
Let first $E(X)$ be quasisimple. Then $E(X)$ is described in (1.5).
If $E(X)/Z(E(X))$ is a Lie group over $GF(q)$, then $S\cap
E(X)\leq P, P$ a minimal parabolic. Hence either $E(X)$ is of
rank 1 or $E(X)$ is of rank 2 and $S$ induces a diagram
automorphism. This is (i). Suppose $E(X)/Z(E(X))$ is alternating. If
$n=11$, then $S\cap
E(X)\leq A_{\Omega} ,\Omega =\{ 1,\ldots ,10\}$ and $S \cap E(X)\le
A_{\Omega_{1}}\langle x \rangle , \Omega_{1} =\{ 1,\dots ,9\}, 
x=(1,2)(10,11)$. But
both groups generate $E(X)$. If $n=10$, then $S\cap E(X)$ is in
$A_{\Omega}\langle x\rangle , \Omega =\{ 1,\ldots ,8\}, x=(1,2)(9,10)$ 
and
$N_{E(X)}\Big( \langle (1,2)(3,4),(1,2)(5,6),(1,2)(7,8),(1,2)(9,10)
\rangle \Big)$.
If $n=8$, then $E(X)\cong L_{4} (2)$, a case just done. If $n=7$,
then $S\cap E(X)$ is in $A_{\Omega} ,\Omega =\{ 1,2,\ldots ,6\}$ and 
$A_{\Omega_{1}}\langle x\rangle , \Omega_{1} =\{ 1,\ldots ,5\}, 
x=(1,2)(6,7)$. This finishes the case of an alternating group.\\
\indent
Let now $K/Z(K)$ be sporadic. Application of \cite{RoStr} shows that $K$
is generated by minimal parabolics up to $K\cong J_{1}$ or $M_{11}$.
But the latter contains $M_{10}$ and $GL_{2} (3)$. The group $J_{1}$
contains $Z_{2}\times A_{5}$ and the normalizer of a Sylow 2-subgroup.
Hence in any such group there are at least two maximal subgroups
containing $S\cap E(X)$.\\
\indent
We are left with $K/Z(K)\cong G(r), r\mbox{~odd} , r=p^{\, f}$.
As $m_{p}(K)\leq 3$, we get from (1.5) $K/Z(K)\cong L_{2}(p),
L_{2}(p^{\, 2}),L_{2}(p^{\, 3}), L_{3}(p), U_{3}(p), PSp_{4}(p) $.
This is (ii).\\
\indent
Let now $E(X)=X_{1}X_{2}$. Then $m_{p}(X_{1})=1$ for any prime $p$ with
$p\,\not\!\!\Big |\,\,\,\, |Z(E(X))|$ and $m_{p}(X_{1})=2$ for 
$p\,\Big  |\, |Z(E(X))|$. Application of
(1.3) shows that $X_{1}\cong$ $Sz(q)$, $L_{2}(q)$, $(S)U_{3}(q)$,
$(S)L_{3}(q)$, $J_{1}$, $3\cdot A_{6}$, $3\cdot A_{7}$ or $3\cdot M_{22}$.\\
\indent
If $X_{1}\cong (S)U_{3}(q)$ or $(S)L_{3}(q)$ and $p\,\Big  |\, q+1$ or
$p\,\Big  |\, q-1, p\not=3$, then $X_{1}$ contains an elementary abelian
group of order $p^{\, 2}$ intersecting the center tri\-vially. Then we
get $X_{1}\cong SL_{3}(4), L_{3}(2)$ or $SU_{3}(8)$.\\
\indent
Let $X_{1}\cong J_{1}, 3\cdot A_{7}, 3\cdot M_{22}, 3\cdot A_{6},
SL_{3}(4), L_{3} (2)$. Then there are subgroups $A,B$ of $X_{1}$ such
that $\langle A,B\rangle =X_{1}, S\cap X_{1}\leq A\cap B$. Let $A,B$ 
be normal in $N_{S}(X_{1})$. Choose $g\in X$ with $X_{1}^{\, g}=X_{2}$. 
Then $\langle A,A^{g},S\rangle $ and $\langle B,B^{\, g},S\rangle $ are 
both different from $X$, but $\langle X_{1},S\rangle =X$. This shows 
$X_{1}\cong 3\cdot A_{6}, SL_{3}(4)$ or
$L_{3}(2)$ and there is some $x\in X$ acting nontrivally on the
Dynkin diagram, i.e. $A^{x}=B$.

\absa
{\bf (1.8)~Lemma.~}{\it Let $O_{2}(X)=1$, $S\in\mbox{~Syl}_{2} (X)$, $S$
is contained in exactly one maximal subgroup of $X$. Suppose $X$ is
solvable. Let furthermore $m_{p}(X)\leq 3$ for all odd prime $p$. Then
$X=O_{p}(X)S$ for some prime $p$ and one of the follwing holds}
\begin{itemize}
\item[(i)]{\it $O_{p}(X)$ is cyclic and $S$ is cyclic}
\item[(ii)]{\it $O_{p}(X)\cong Z_{p^{\, 1+2}}$ or abelian of rank two}
\item[(iii)]{\it $O_{p}(X)\cong Z_{p^{\, 1+4}}$}
\item[(iv)] {\it There is $U\cong E_{p^{\, 3}}, U\triangleleft X, 
m_{2}(S)\leq 2$.}
\end{itemize}

Proof~.  Let $F=F(X)$. As $O_{2}(X)=1$, we have that $F$ is of odd
order. Let $S_{1} \le S$ be a preimage of $O_{2}(X/F)$. Now 
$N_{X}(S_{1})$
and $FS$ are in different maximal subgroups of $X$, provided 
$FS\not= X$. So we have $FS=X$. Now $O_{p}(F)S$ and $O_{p^\prime} (F)S$ 
are in different
maximal subgroups or $F=O_{p}(F)$. So we have that $F=O_{p}(X)$.

\indent
If $m_{p}(F)=1$, we get (i). As $S$ is contained in exactly one maximal
subgroup we have that

$(*)$~~$S$ acts irreducibly on $F/\Phi (F)\, .$\\
\\
\indent
Let now $m_{p}(F)=2, C$ be a critical subgroup and $U=\Omega_{1}(C)$. 
If $m_{p}(C)=1$, then $C$ is cyclic and so $X/C$ is abelian. But then 
$|F/\Phi(F)| = p$ by $(\ast)$ and so $F$ is cyclic, i.e. (i) holds.\\
\indent
Let $m_p(U) = 2$. Then $U$ is elementary abelian or of type $p^{1+2}$. 
Hence in any case $X/\bC_{X}(U)U$ is a subgroup of $GL_2(p)$ having a 
normal Sylow p-subgroup. So by $(\ast)$ we get $\bC_{X}(U)U = F$. If 
$U \cong Z_{p^{1+2}}$ we see $U = F$ by $(\ast)$. 
If $U \cong \bZ_{p} \times \bZ_{p}$, we get that $F$ is abelian of rank 
two, this is (ii).\\
\indent
So assume next $m_p(U) = 3$. By (1.6) $U$ is either elementary abelian, 
$\bZ_p \times Z_{p^{1+2}}$, or $Z_{p^{1+4}}$.
Suppose we have $U \cong Z_{p^{1+4}}$. Then $F = U\bC_{F}(U)$. 
By $(\ast)$ we get $\bC_{F}(U) \le U$, so $F = U$.
This is (iii).
\\
\indent
Let $U \cong \bZ_p \times Z_{p^{1+2}}$. Set $Z = Z(U)$. Now 
$|F : \bC_{F}(Z)| \le p$.
By $(\ast)$ we get $Z \le Z(F)$. Furthermore there is 
$W \le Z, W \cap U^\prime = 1$ and 
$W \unlhd X$. Now $F = \bC_{F}(U/W)U$. So by $(\ast) F = U$, a 
contradiction to $(\ast)$ again.\\
\indent
Let finally $U$ be elementary abelian. Then $X/\bC_{X}(U)$ is a 
subgroup of $GL_3(p)$. Suppose 
$m_2(S) = 3$. Then $Z(S)$ contains a fours group $\langle z,t \rangle$. 
Now $F = \langle \bC_{F}(z), \bC_{F}(t), \bC_{F}(zt) \rangle$. We 
may assume $\bC_{F}(z) \not= 1 \not= \bC_{F}(t)$.
Then $\bC_{F}(z)S$ and $\bC_{F}(t)S$ lie in different maximal subgroups, 
a contradiction. So $m_2(S) \le 2$. 

\absa
{\bf (1.9)~Lemma.~}{\it Let $X\cong A_{n}$ or Spor. Suppose 
$P\leq X, P$ contains a Sylow $2$--subgroup of $X$ and $P$ is a 
$\{2,p\}$-group. Then $p=3$ or $X\cong J_{1}$ and $p=7$.} 

\absa 
Proof~.  For $X\cong A_{n}$ this can be found in \cite[(6.1)]{Asch2}. For
the sporadic groups this is in \cite{RoStr}.

\absa
{\bf (1.10)~Lemma.~}{\it Let $K$ be a Lie group in odd
characteristic, $K\not\cong$ $^2G_{2}(3^n)$. Let
$\omega\in\mbox{~Aut} (K), o(\omega )=p > 3, p$ prime, $S$ a Sylow
$2$--subgroup of $K$, $[\omega ,S]\leq S$. Then either $\omega$
induces a field automorphism on $K$ or $\omega$ is a diagonal
automorphism with $[S,\omega ]=1$.}

\absa
Proof~.  This is \cite[(6.3)]{Asch2}
\\
\\
\\
\indent
We are now going to investigate modules for the groups occuring in the
proof of the theorem. Basically there will be two types of modules.
\begin{itemize}
\item[(1)] Modules $V$ over $GF(2)$ admitting some elementary abelian
2-group $A$ such that
$$[V,A,A]=0$$
\item[(2)] Modules $V$ over $GF(2)$ for some group $G$ such that
$|V:\bC_{V} (t)|\leq 2^{\, m_{2}(G)}$ for some involution $t\in G$.
\end{itemize}

\vspace{1cm}

\indent
First we collect some results concerning modules of the first kind.

\absa
{\bf (1.11)~Lemma.~}{\it Let $a$ be an involution in $G$, $L$ be a component
of $G$, $O_{2} (G)=1$ and $A$ a four-subgroup of $G$ with 
$a\in A, [V,a,A]=0$
for a faithful $GF(2)\, G$-module $V$. If $\bC_{A}(L)=1$, then either}
\begin{itemize}
\item[(a)]{\it $[L,a]\leq L$, or}
\item[(b)]{\it $\langle A,L\rangle \cong O_{4}^{\, +}(2^{\, n})$ and 
$[L^{A},V]/\bC_{[L^{A}, V]} (L^{A})$, is a direct sum of natural 
modules for
$\langle L^{A}\rangle \cong \Omega_{4}^{\, +}(2^{\, n})$.}
\end{itemize}

\absa
Proof~. Besides of the assertion that we have a direct sum of natural modules this is \cite[(25.14)]{GoLyS}. But as shown there an element $x\in L$ of order three acts fixed point freely. So we get with 
\cite[(8.2)]{Hi}
that $V$ is a direct sum of natural $L_{2}(2^{\, n})$-modules. As
$\langle a^{\, G}\rangle \cong O^{+}(4, 2^{\, n})$, we get the assertion.

\absa
{\bf (1.12)~Lemma.~}{\it Let $G$ be a group such that $F^{*}(G)$ is
a perfect central extension of a finite simple group. Suppose there
is some elementary abelian subgroup $A$ of $G, |A|\geq 4$, such that
for some irreducible nontrivial faithful module $V$ over $GF(2)$ we
have $[V,A,A]=0$. Then }
\begin{itemize}
\item[(i)]{\it If $F^{*}(G)/Z(F^{*}(G))$ is sporadic, then 
$F^{*}(G)/Z(F^{*}
(G))\cong$ $M_{12}$, $M_{22}$, $M_{24}$, $J_{2}$, $Co_{1}$, $Co_{2}$ or 
$Sz$.
If $|A|\geq 8$, then $F^{*}(G)\cong$ $3\cdot M_{22}$.}
\item[(ii)]{\it If $F^{*}(G)/Z(F^{*}(G))$ is a Lie group in odd
characteristic which is not a Lie group in even characteristic too,
then $E(G)\cong 3\cdot U_{4}(3)$. Furthermore $V$ is the
$12$--dimensional module.}
\item[(iii)]{\it If $F^{*}(G)/Z(F^{*}(G))$ is alternating, then either $V$
is the natural module or a spin module or $F^{*}(G)\cong 3\cdot A_{6}$
and $V$ is the $6$-dimensional module.
If $|A| > 8$, then $V$ is natural or $G\cong A_{8}$ and $|V|=16$.
If $V$ is the spinmodule and $|A| = 4$, then $A$ is conjugate to 
$\langle (12)(34),(13)(24) \rangle$ or 
$\langle (12)(34)(56)(78), (13)(24)(57)(68) \rangle$. 
If $|A| = 8$ the $A$ is conjugate to
$\langle (12)(34)(56)(78), (13)(24)(57)(68), (14)(26)(37)(48)\rangle $ 
under $\Sigma_{n}$.}
\end{itemize}

\absa
Proof~.  (i) This is \cite{MeiStr1}.

(ii) This is \cite{MeiStr2}. For
$F^{*}(G)\cong L_{2}(q), q$ odd it was a quotation to 
\cite[(6.4)]{Asch3}. For convenience of the reader we give a proof 
for the latter. We have $F^{*}(G)\not\cong L_{3}(2), A_{6}, 
3\cdot A_{6}, A_{5}$. If $A\leq F^{*}(G)$, set 
$Y=\langle A^{\, \bC_{G} (a)} \rangle , a\in
A^{\sharp}$. If $A\not\leq F^{*}(G)$ choose $a\in A\setminus
F^{*}(G)$, set $Y=\bC_{G} (a)$. Then in all cases 
$G=\langle Y,Y^{g}\rangle $ for
any $g\in G\setminus\bC (a)$ by \cite[(II.8.27)]{Hu}. Set $W=[V,a]$. Then
by quadratic action $[Y,W]=0$. Now as in the proof of (1.11) we get
that $a^{G}$ is a set of odd transpositions. If $A\leq F^{*}
(G)$, this shows that $8\not\Big |\, |F^{*}(G)|$ and $\bC (a)$ is a
2-group. But then $F^{*}(G)\cong A_{5}$, a contradiction. If $a\not\in
F^{*}(G)$, then $\bC_{Y} (A\cap F^{*}(G))$ is a 2-group. Furthermore
Sylow 2-subgroups are abelian. This shows that $Y\cong L_{2}(5)$ and
$a$ induces a field automorphism. Hence $F^{*}(G)\cong L_{2}(25)$.
Now $G$ is generated by three conjugates of $a$. This shows $|\bC_{V}
(a)/[V,a]|\leq |[V,a]|$. As $[Y,V]\leq \bC_{V} (a), [[V,a],
Y]=0$. We get either $[V,a]\cap [V,b]\not= 0$ for 
$\langle a,b\rangle =A, a\sim b\in G$ or $\bC_{V} (a)=[V,a]\bigoplus 
[V,b]$.\\
\indent
The former is not possible
as $[V,a]\cap [V,b]$ would be centralized by $\langle Y,Y^{g}\rangle =G$. Hence we
have $\bC_{V} (a)=\bC_{V} (b)$ and so $\bC_{V} (a)$ is invariant
under $\langle Y,Y^{g}\rangle $, a contradiction.

(iii)~~This is \cite{MeiStr1}. Suppose $|A| \ge 4$. Let $a\in A^{\sharp}$. Let $k$ be the number of
fixed points of $a$. Then there is 
$K\leq \bC_{G} (a), K\cong \Sigma_{k}$. Furthermore 
$\bC_{\bC_G(a)}(K^\prime)$ is an extension 
of a 2--group by $\Sigma_{m}$, $m = (n-k)/2$. Now choose 
$a \in A$ with $m > 2$ if possible. Suppose first 
$[A, \bC_{\bC_G(a)}(K^\prime)] \not= 1$. 
If $m \ge 5$, then $\Sigma_m$ is nonsolvable and so 
$\bC_{\bC_G(a)}([V,a])$ contains an elementary a
belian subgroup of $O_2(\bC_G(a))$ of order $2^{m-1}$. 
But then this group contains a conjugate of (12)(34) which 
contradicts \cite[(4.3)]{MeiStr1}.\\
\indent
Let $m = 4$. Then $a \sim (12)(34)(56)(78)$. Furthermore as we may 
assume that no $x \sim (12)(34)$ is contained in 
$\langle A^{\bC_G(a)} \rangle$ we see that $A$ is conjugate to a 
subgroup of 
$\langle (12)(34)(56)(78), (13)(24)(57)(68), (15)(26)(37)(48) \rangle$. \\
\indent
Let $m = 3$. Then $\bC(K^\prime) \le \Sigma_6$ and $a \sim (12)(34)(56)$. 
Then $\langle A^{\bC_G(a)} \rangle$ contains some $x \sim (12)(34)$, 
contradicting \cite[(4.3)]{MeiStr1}.\\

So let $[A, O^{2^\prime}(\bC_{\bC_G(a)}(K^\prime))] = 1$. 
If $[A, K^\prime] \not= 1$, then $[K^\prime, [V,a]] = 0$. 
If $k \ge 4$, then $K^\prime$ contains some $
x \sim (12)(34)$. This again contradicts \cite[(4.3)]{MeiStr1}. 
Let $k \le 3$. As $[A, O^{2^\prime}(\bC_{\bC_G(a)}(K^\prime))] = 1$ 
and $m > 2$, there is $x \sim (12)$ in $A$, a contradiction. 
So we are left with 
$[A, K^\prime] = 1 = [A, \bC_{\bC_G(a)}(K^\prime)]$. But this is 
impossible with $m > 2$.\\
\indent
So we have $m \le 2$ for all $a \in A^\sharp$. As there is no fours 
group of transvections we may assume $a = (12)(34) \in A$. Now 
$A \ge \langle a, b \rangle$, $b = (13)(24)$, $(13)(24)$ or $(34)$. Let 
$[b,K^\prime] \not= 1$. Then $b = (12)(56)$ and so $K^\prime$ contains 
no involutions by \cite[(4.3)]{MeiStr1}. This shows $k \le 3$ and so 
$A \le \Sigma_7$. But this group 
$A = \langle (12)(34), (12)(56) \rangle$ does not act quadratically on 
the four dimensional module.\\
\indent
Assume now $b = (34)$. Then $[a, E(\bC_G(b))] \not= 1$. Now 
$E(\bC_G(b)) \cong \Sigma_{n-2}$, which is nonsolvable. But then 
$\langle (34), (12)(56) \rangle$ acts quadratically, a contradiction. 

\absa
{\bf (1.13)~Lemma.~}{\it Let $F^*(G) \cong M_{22}$ or $3\cdot M_{22}$ 
and $V$ be an irreducible faithful $GF(2)G$--module. If $G$ contains $A$ 
with $|A| \ge 4$ and $[V,A,A] = 0$, then one of the following holds}
\begin{enumerate}
\item[(i)]{\it $|A| = 4$, $G \cong {\mbox{Aut}}(M_{22})$ and $V$ is one 
of the two $10$--dimensional mo\-dules.}
\item[(ii)]{\it $4 \le |A| \le 16$ and 
$\langle A^G \rangle \cong 3\cdot M_{22}$, $V$ is the
 $12$--dimensional module for $G$.}
\end{enumerate}

\absa
Proof~. See \cite{MeiStr1}

\absa
{\bf (1.14)~Lemma.~}{\it Let $G=G(q), q=2^{\, n}$, be a Lie group and
$V$ an irreducible $GF(2)$-module. Let $A$ be a fours group with
$[V,A,A]=0$. If $A$ intersects some long root group $R$ nontrivially
but $A\not\leq R$. Then one of the following holds}
\begin{itemize}
\item[(i)]{\it $G\cong (S)L_n(q), (S)U_n(q), Sp_{2n}(q)$ or $F_{4}(q)$ and
$V=V(\lambda )$ for some fundamental weight $\lambda$.}
\item[(ii)]{\it $G=\Omega_{2n}^{\pm} (q)$ and $V$ is the natural or spin
module.}
\item[(iii)]{\it $G=E_{6}(q)$ and $V=V(\lambda_{1} )$ or 
$V(\lambda_{6} )$}
\item[(iv)]{\it $G=E_{7}(q)$ and $V=V(\lambda_{7} )$ }
\item[(v)]{\it $G=$ $^2E_{6}(q)$ and $V=V(\lambda_{4} )$ }
\item[(vi)]{\it $G=G_{2}(q)$ or $^3D_{4}(q)$ and $V$ is the natural
module.}
\end{itemize}

\absa
Proof~. \cite{Str}
\\

\absa
The modules from (1.14) will be called strong quadratic in this paper.
\\

\absa
{\bf (1.15)~Lemma.~}{\it Let $G=G(q), q=2^{\, n}$, be a Lie group, $V$
an irreducible $GF(2)$-module. Let $A$ be an elementary abelian
subgroup of $G$ with $[V,A,A] =0$. Then each of the following implies
that $V$ is one of the modules of (1.14).
\begin{itemize}
\item[$(1)$]$G\cong (S)L_{n}(q),
|A| > q^{\frac{n(n-2)}{4}}$~~or~~$q^{\frac{(n-1)^{\, 2}}{2}}$
according
to $n > 2$ is even or odd
\item[$(2)$]$G\cong (S)U_{n}(q), n\geq 4, |A| > q^{[\frac{n-2}{2}]^{\,
2} +1}, q^{[\frac{n-2}{2}]^{\, 2}}$ for $n > 5$.
\item[$(3)$]$G\cong Sp_{2n}(q), n\geq 3, |A| > q^{{n-1\choose 2} +2}$
\item[$(4)$]$G\cong Sp_4(q), q > 2, |A| > q$
\item[$(5)$]$G\cong \Omega_{2n}^{\pm} (q), n\geq 4,
|A| > q^{\frac{n^2-5n+10}{2}}$
\item[$(6)$]$G\cong G_{2}(q), |A| > q$
\item[$(7)$]$G\cong$ $^3D_{4}(q), |A| > q$
\item[$(8)$]$G\cong$ $^2F_{4}(q), |A| > q$ (i.e. there is no such module)
\item[$(9)$]$G\cong E_{6}(q), |A| > q^{11}$
\item[$(10)$]$G\cong E_{7}(q), |A| > q^{16}$
\item[$(11)$]$G\cong E_{8}(q), |A| > q^{28}$ (i.e. there is no such
module)
\item[$(12)$]$G\cong$ $^2E_{6}(q), |A| > q^{10}$
\item[$(13)$]$G\cong F_{4}(q), |A| > q^{7}$.
\end{itemize}}

\absa

Proof~.  (1): Let $P$ be a parabolic of $G$ with $|O_{2}(P)|=q^{\,
n-1}$ and $O^{\, 2^\prime}(P/O_{2}(P))\cong SL_{n-1}(q)$. We may assume
$A\leq P$. Now the assumption says $A\cap O_{2}(P)\not= 1$.
Hence the assertion follows with (1.14).\\
\indent
(2),(5),(9)-(12): Let $R$ be some long root group, $P=N_{G}(R)$. We
may assume $A\leq P$. Set $E=O_{2}(P)$. Then 
$$|A\cap E|\leq q\, .$$
Otherwise there is  $\langle a,b\rangle \leq A\cap E$ and some $g\in E$ 
with $a^{\, g}=a, b^{\, g}=br, r\in R^{\sharp}$. But then 
$\langle a,r\rangle $ also acts
quadratically on $V$ and we have the assertion with (1.14).\\
\indent
So we have $|A|\leq q\cdot 2^{\, m_{2}(P/E)}$. But this is exactly
the given bound besides the case $U_{n}(q)$ and $\Omega_{2n}^{\pm} (q)
$. Here the bound is $2^{\, m_{2}(P/E)}$. Let $G\cong\Omega_{2n}^{\pm}
(q),O^{2^\prime}(P/E)\cong\Omega_{2n-4}^{\pm} (q)\times L_{2}(q)$. If 
$t\in A$ and
$t$ corresponds to a nontrivial element of the $L_{2}(q)$, then
$|[t,E/R]|=q^{\, 2n-4}$. If $A\cap E\not= 1$ and there is $a\in A\cap
E, b\in\bC_{F} (a)$ with $t^{b}=tc\not= t$, and $\bC_{E} (a)\not=
\bC_{E}(c)$. Now $\langle a,c\rangle $ acts quadratically and as 
$\bC_{E} (a)\not= \bC_{E} (c)$ also $\langle a,r\rangle, r\in R$, 
acts quadratically. The assertion follows with (1.14).\\
\indent
The order of $A$ shows $A\cap E\not= 1$. As $|\bC_{E} (A\cap
E)|=q^{\, 4n-9}$, we see that there is $b\in \bC_{E} (A\cap E)$ with
$t^{b}=tc\not= t, \bC_{E} (c)\not= \bC_{E} (A\cap E)$. Let $G\cong
U_{n}(q)$. Then $A\cap E\not= 1$ and $A$ contains an involution $x$
which does not act as an unitary transvection on $E/R$. But then there
is some $t\in \bC_{E} (A\cap E)$ with $x^{t} =xs, \bC_{E} (A\cap
E)\not= \bC_{E} (s)$. As $\langle a,s\rangle , a\in (A\cap E)^{\sharp}$ 
acts quadratically we now get the same contradiction as before.\\
\indent
(3): Let again $R$ be some short root group, $P=N_{G}(R)$. Then
$O_{2}(P)$ is a direct product of an elementary abelian group of
order $q^{\, 2}$ by a semi\-special group of order $q^{\, 1+2\, (2n-4)}$,
and $O^2(P/O_{2}(P))\cong Sp_{2n-4}(q)\times L_{2}(q)$. Again we have 
$|(A\cap O_{2}(P))Z(O_{2}(P))/Z(O_{2}(P))|\leq q$.
Furthermore $|A\cap Z(O_{2}(P))|\leq q^{\, 2}$. If $|A\cap
Z(O_{2}(P))| > q$ then $A\leq \bC (Z(O_{2}(P)))$ and so
$AO_{2}(P)/O_{2}(P)\leq Sp_{2n-4}(q)$.Hence we have $|A|\leq q^{\,
2}2^{\, m_{2}(Sp_{2n-4}(q))}$, which is exactly the given bound.\\
\indent
(4): Let $S\in \mbox{~Syl}_{2} (G)$. Then $S=FE$, where $E$ and $F$ 
both are
elementary abelian of order $q^{\, 3}$. We may assume $A\leq E$.
As $A$ does intersect trivially any root group, we get $|A\cap E\cap
F|\leq q$. So we may assume (maybe after conjugation in $N_{G}(E)$)
that $A\cap F\not= 1$ and $|A:A\cap F| > 2$. So let $b\in A\cap F$ and
$\langle b_{1},b_{2}\rangle \leq A, \langle b_{1},b_{2}\rangle \cap F=1,
|\langle b_{1},b_{2}\rangle |=4$. Then there are $c_{1},c_{2}\in F$ with
$b_{1}^{\, c_{1}}=b_{1}bt_{1}, b_{2}^{\, c_{2}}=b_{2}bt_{2},
t_{1},t_{2}\in Z(O^{2^\prime}(N_{G}(F)))$. As $q > 2$, $F$ is an
indecomposable $O^{2^\prime}(N_{G}(F))$-module. This proves 
$t_{1}\not= 1$ or $t_{2}\not= 1$. We may assume $t_{1}\not= 1$. Then
$\langle b,bt_{1}\rangle =\langle b,t_{1}\rangle $ acts quadratically on 
$V$. But $t_{1}$ is a
root element. The assertion follows with (1.14).\\
\indent
(6): Let $P$ be the parabolic with $Z(O^{2^\prime}(P))=1, Q=O_{2}(P)$. 
We have $O^{2^\prime}(P)/Q\cong L_{2}(q)$ and so $A\cap Q\not= 1$. 
Let $R$ be a
root group in $Z(Q)$ and $B=\langle R^{P}\rangle $. Then $B$ is the 
natural $O^{2^\prime}(P)/Q$-module. Hence by (1.14) we assume 
$A\cap B=1$. As $G$
does not contain elementary abelian subgroups of order greater than 
$q^{\,3}$,
we get $A\not\leq Q$. Now choose $a\in A\setminus Q,
\tilde{a}\in A\cap Q, s\in B$ with  $a^{s}=at\not= a$. Then
$\langle \tilde{a} ,t\rangle $ acts quadratically and $t$ is a root 
element. The assertion follows with (1.14).\\
\indent
(7): Let $R$ be a root group, $P=O^{2^\prime}(N_{G}(R))$. Then
$P/O_{2}(P)\cong L_{2}(q^{\, 3})$, and $O_{2}(P)$ is semispecial.
Hence by (1.14) $|A\cap O_{2}(P)|\leq q$. If $t\in A\setminus
O_{2}(P)$, then $|[O_{2}(P)/R,t]|=q^{\, 4}$. So if $A\cap
O_{2}(P)\not= 1$, we find $a\in A\cap O_{2}(P), b\in\bC_{O_{2}(P)}
(a), t\in A\setminus O_{2}(P)$, with $t^{b}=tc, \bC_{O_{2}(P)}
(c)\not= \bC_{O_{2}(P)} (a)$, and $\langle a,c\rangle $ acts 
quadratically.\\
\indent
Suppose $A\cap O_{2}(P)=1$. Then $|AO_{2}(P)/O_{2}(P)| > q$. This
implies there are $b,t\in A$ with 
$|[\bC_{O_{2}(P)/R} (b),t]|=q^{\,2}$. As 
$\bC_{O_{2}(P)/R} (b)=[O_{2}(P)/R, b]$. We find 
$\langle a,b\rangle $ acts
quadratically for some $a\in\bC_{O_{2}(P)/R} (b)$. But now as before
we get some $c\in O_{2}(P), \bC_{O_{2}(P)} (a)\not= \bC_{O_{2}(P)}
(c)$ with $\langle a,c\rangle $ acts quadratically.\\
\indent
Now as $\bC_{O_{2}(P)} (a)\not= \bC_{O_{2}(P)} (c)$ there is $x\in
O_{2}(P), a^{x}=a, c^{x}=cr, r\in R^{\sharp}$. Hence 
$\langle a,r\rangle $ acts
quadratically and the assertion follows with (1.14).\\
\indent
(8): Let $R$ be some long root group, $P=O^2(N_{G}(R)),
Q=O_{2}(N_{G}(R))$. We have $A\cap Q\not= 1$. Let $a\in
A\setminus Q$. Then $|Z_{2}(Q):\bC_{Z_{2}(Q)} (a)|=q^{\, 2}$.
Choose $b\in A\cap Q$. Then $|Z_{2}(Q):\bC_{Z_{2}(Q)} (a)|\leq q$.
Hence there is $t\in Z_{2}(Q)$ with $b^{t}=b$ and $a^{b}=ax\not= a$.
This shows that $\langle x,b\rangle $ does act quadratically. 
Furthermore $\bC_{Q}(x)\not= \bC_{Q} (b)$.\\
\indent
If $A\leq Q$, $a,b\in A$, with $\bC_{Z_{2}(Q)} (a)\not=
\bC_{Z_{2}(Q)} (b)$, then there is $t\in Z_{2} (Q)$ with $b^{t}=b,
a^{t}=ax\not= a$ and again $\langle b,x\rangle $ acts quadratically 
$\bC_{Q}(x)\not= \bC_{Q} (b)$.\\
\indent
As $|A| > q$ there are always such pairs $a,b$. Hence we just have to
treat the case of a quadratic fours group 
$\langle b,x\rangle \leq Q, x\in Z_{2}(Q), \bC_{Q} (b)\not= \bC_{Q} (x)$. 
By (1.14) $\langle x,b\rangle \cap 
R=1$. But then there is some $t\in Q$ with $b^{t}=br\not= b,r\in R,
x^{t}=x$. Hence $\langle x,r\rangle $ does act quadratically and the 
assertion follows with (1.14).\\
\indent
(13): Let $R$ be a root group, $P=O_{2}(N_{G}(R)), Q=O_{2}(P)$. Then
$Q$ is a direct product of an elementary abelian group of order
$q^{\, 6}$ by a semispecial group of order $q^{\, 1+8}$. Further
$P/Q\cong Sp_{6} (q)$. Hence $|A:A\cap Q|\leq q^{\, 6}$ and so $|A\cap
Q| > q$. By (1.14) we have $|A\cap Q:A\cap Z(Q)|\leq q$. This now implies
$A\cap Z(Q)\not= 1$. Let $1\not= a\in A\cap Z(Q), b\in A\cap
Q\setminus A\cap Z(Q)$. Then there is some $t\in Q$ with
$b^{t}=br,r\in R^{\sharp}$, and $a^{t}=a$. Hence $\langle a,r\rangle 
$ acts quadratically and the assertion follows with (1.14).\\
\indent
So we may assume $A\cap Q=A\cap Z(Q)$. As $|Z(Q)|=q^{\, 7}$, we see
$A\not\leq Q$. Let now $a\in A\setminus Q, t\in
Q\setminus Z(Q), b\in A\cap Z(Q)\smallsetminus R$. We may choose
$t$ with $a^{t}=ac, c\not\in Z(Q)$. Then $\langle c,b\rangle $ does act
quadratically and there is some $t_{1}\in Q$ with $c^{t_{1}}=cr,r\in
R^{\sharp}$. Now $\langle r,b\rangle $ does act quadratically. The 
assertion follows with (1.14).

\absa
{\bf (1.16)~Lemma.~}{\it Let $G\cong G (q),q=2^{\, n}$, be a Lie
group. Let $G \triangleleft H$ and $V$ be a $GF(2)H$-module. Let
$A\leq N_{H}(G), A$ elementary abelian, $\bC_{A} (G)=1,
[V,A,A]=0$. Let $W\cong V(\lambda ), \lambda$ some fundamental
weight, be a $G$-submodule of $V$. Let $S\in\mbox{~Syl}_{2} (G)$. If
$|A| > |\bC_{W} (S)|$, then $[W,A]\leq W$. (Notice $|\bC_{W} (S)
|=q, q^{\, 2}$ (for $G$ twisted), $q^{\, 3}(\mbox{~for~}$  
$^3D_{4} (q))$).} 

\absa
Proof~.  Let $a\in A^{\sharp}$ with $W_{1} =W^{\, a}\not= W$. Then
$W_{1}$ is an irreducible $G$-module and so $W_{1}+W = W_{1}\bigoplus
W \cong W\bigoplus [W,a]$.\\
\indent
Suppose there is $b\in A^{\sharp}$ with $W^{\, b}=W$. Then by
quadratic action $[[W,b],a]=0$ and so $[W,b]\subseteq W\cap W_{1}=0$.
This shows $W^{b}\not= W$ for all $b\in A^{\sharp}$. We are going to
prove that $W_{1}\bigoplus W$ is $A$-invariant. Suppose there is
$b\in A$ with $W^{b}\not\subseteq W_{1} \bigoplus W$. Then
$W^{b}\bigoplus W_{1}^{\, b}\not= W\bigoplus W_{1}$. Furthermore
$(W_{1}\bigoplus W)^{\, b}\cap (W\bigoplus W_{1})$ is a $G$-module.
As $[[W,a],b]=0$ by quadratic action, we see that $[W,a]\subseteq
(W_{1}\bigoplus W)^{\, b}\cap (W\bigoplus W_{1})$. Furthermore
$|[W,a]|=|W|$ and so we get $(W_{1}\bigoplus W)^{\, b}\cap
(W\bigoplus W_{1})=[W,a]$. Hence $[W,a]$ is a $G$-module. As $A$ is
trivial on $[W,a]$, we see that $[A,G]\leq \bC_{G} ([W,a])=1$, a
contradiction.\\
\indent
This proves $W_{1}\bigoplus W$ is $A$-invariant. As $[[W,a], A]=0$,
we see $[W,a]=[W,A]$. Let $R=\bC_{W} (S), R_{1}=\bC_{W_{1}} (S)$.
Then $[R\times R_{1},a]=[R\times R_{1}, A]$, and for every $a\in
A^{\sharp}, \bC_{R} (a)=1$. This gives $|A|\leq |R|$, the assertion.

\absa
{\bf (1.17)~Lemma.~}{\it Let $X$ be one of the groups in (1.7)(i)
with $q > 2$ or (1.7)(ii) with $E(X)\cong SL_{3} (4)*SL_{3}(4)$.
Let $V$ be an irreducible $GF(2)X$-module, $A$ an elementary abelian
group, $A\leq E(X), [V,A,A]=0$. Then in case of (1.7)(i)
$|A|\leq q^{\, 2}$ while in case of (1.7)(ii) $|A|\leq 16$.}

\absa
Proof~. Suppose false. Then (1.7)(i) implies that $E(X)\cong
Sp_{4}(q), SL_{3}(4)* SL_{3}(4)$. 
Suppose first $E(X)\cong Sp_{4}(q)$. Then by (1.15) and (1.14) $V$
just involves natural or dual natural $Sp_{4}(q)$-modules. As $X$
induces a graph automorphism on $E(X)$, we see that with each natural
module also the dual is involved. But no group of order  greater than 
$q^{\, 2}$ can act quadratically on both modules.\\
\indent
So assume now that we are in (1.7)(ii). Then $E(X)=X_{1}X_{2}$ with
$X_{1}\cong X_{2}, X_{1}\cong SL_{3}(4)$. Furthermore $A\cap
X_{1}\not= 1\not= A\cap X_{2}$. By quadratic action we see that
$[[V,X_{1}], X_{2}]=0$. This shows that $[Z(E(X)),V]=1$. We may
assume that $|A:\bC_{A} (X_{1})|\geq 8$. Now (1.14) implies that
$[V,X_{1}]$ just involves natural $X_{1}$-modules, contradicting
$[V,Z(X_{1})]=0$.

\absa 
{\bf (1.18)~Lemma.~}{\it Let $F^{*}(G)$ be a quasisimple group and
$V$ be an $F$-module over $GF\, (2)$ for $G$. (i.e. there is some
elementary abelian subgroup $1\not= A$ of $G$ with $|V:\bC_{V}
(A)|\leq |A|$). If $V$ is irreducible for $F^{*}(G)$ one of the
following holds:}
\begin{itemize}
\item[(i)]{\it $F^{*}(G)\cong A_{n}$ and $V$ is the natural module.}
\item[(ii)]{\it $F^{*}(G)\cong A_{7}$ and $V$ is the 4-dimensional 
module, $F^{*}(G)\cong 3\cdot A_{6}, |V|=2^{\, 6}$.}
\item[(iii)]{\it $F^{*}(G)\cong (S)L_{n}(q)$, and 
$V\cong V(\lambda_{1}),V(\lambda_{2} ),V(\lambda_{n-2} ),
V(\lambda_{n-1} )$, the natural
module, spin module or their duals.}
\item[(iv)]{\it $F^{*}(G)\cong (S)U_{n}(q)$, $q$ even, $n > 3$ and $V$ is
the natural module.}
\item[(v)]{\it $F^{*}(G)\cong \Omega_{2n}^{\pm} (q),q$ even, $V$ is the
natural module.}
\item[(vi)]{\it $F^{*}(G)\cong \Omega_{10}^{+} (q), q$ even, $V$ is a
spin module.}
\item[(vii)]{\it $F^{*}(G)\cong Sp_{2n}(q), q$ even, $V$ the natural module.}
\item[(viii)]{\it $F^{*}(G)\cong Sp_6(q), q$ even, $V$ the spin module.}
\item[(ix)]{\it $F^{*}(G)\cong G_{2}(q), q$ even, $V$ the natural
6-dimensional $GF(q)$-module.}
\end{itemize}
{\it If $V$ is not irreducible and involves an irreducible module as in
(i),(ii),(iv) with $n=4$, (vi),(viii) or (ix) then $V$ involves 
exactly one nontrivial irreducible module.}

\absa
Proof~. If $F^{*}(G)/Z(F^{*}(G))$ is a Lie group over a field  of
characteristic two the assertion follows with \cite{Coop}. If there are transvections on $V$ we get the assertion with \cite{McL}. Now by
Thompson replacement \cite[(25.2)]{GoLyS} we get a quadratic fours group and we can
apply (1.12). If $E^{*}(G)/Z(F^{*}(G))$ is alternating the result
follows from (1.12)(iii).\\
\indent
As centralizers of involutions in $U_{4}(3)$ are maximal subgroups
and $|V:\bC_{V} (a)|\geq 16$ for $a\in A^{\sharp}; A$ a quadratic 
offending subgroup, we get $[3U_{4}(3), \bC_{V} (A)]\leq\bC_{V}
(A)$, a contradiction. Hence $F^{*}(G)\not\cong 3U_{4}(3)$.\\
\indent
Let $F^{*}(X)/Z(F^{*}(X))$ be sporadic. The modules are given in 
\cite{MeiStr1}. 
Now we see that $|V:\bC_{V} (a)|\geq 8$ for any $a\in
F^{*}(X)^{\,\sharp}, a^{\, 2}=1$. Hence by (1.12)(i) we get that
$F^{*}(X)\cong 3\cdot M_{22}$ and $V$ is the 12-dimensional module.
But now as before we see $|A|=16$ and $\bC_{V} (a)=\bC_{V} (A), a\in
A^{\,\sharp}$. Hence $N_{F^{*}(X)}(A)/A\cong 3\cdot A_{6}$. But for
this group we know $|\bC_{V} (A)|=64$, a contradiction.

\absa
{\bf (1.19)~Lemma.~}{\it Let $X\cong A_{n}, n\geq 5, V$ be a
$GF(2)X$-module with $V/\bC_{V} (X)$ the natural irreducible
permutation module. Assume $[V,X]=V$. Then $|\bC_{V} (X)|\leq 2$, and
$|\bC_{V} (X)|=1$ if $n$ is odd. Furthermore $V$ is a submodule of
the permutation module.}

\absa
Proof~. For simplicity we prove the dual statement.
Let $\bC_{V} (X)=1$ and $[V,X]$ be the natural module.
Then $|V:[V,X]|\leq 2, |V:[V,X]|=1$ for $n$ odd, and $V$ is a factor
of the permutation module.\\
\indent
This will be proved by induction on $n$. For $n=5$ this is well known. 
So let $n > 5, K\cong A_{n-1}, K\leq X$. If $n-1$ is odd, then we have
that $[V,X]=T\bigoplus T_{1}, T_{1}$ the irreducible permutation
module for $K,[T,K]=1$. By induction $V=[V,K]\bigoplus \tilde{T},[V,K]$
the irreducible permutation module. Hence there is $v\in V\setminus 
[V,X], [v,K]=1$, i.e. $\langle v^{\, X}\rangle =V$ is a factor of the 
permutation module.\\
\indent
Let $n-1$ be even. Then we have a $K$-chain. $1 < T <
T_{1} < [V,X] < V$, with $|T|=2, T_{1}/T$ the irreducible
$K$-module and $|[V,X]/T_{1}|=2$. Now by induction $\bC_{V/T}
(K)\not= 1$. As $\bC_{V/T} (K)\not\leq [V,X]/T$, we again get
some $v\in V\setminus [V,X], [v,K]=1$, and so $V$ is a factor of
the permutation module.\\
\\
\indent
For the Lie groups and their modules in (1.18) we also have to know the
nonsplit extension. This is basically contained in \cite{JoPa}. But as it
is not done for all the groups there, we will give a self contained
proof.

\absa
{\bf (1.20)~Lemma.~}(a)~~{\it Let $L$ be one of the following groups
$(S)L_{n}(q)$, $SU_{n}(q )$, $Sp_{2n}(q)$, $\Omega_{2n}^{\pm} (q)$, 
$q$ even, and
$V$ be a module over $GF(2)$ with $[V,L]$ the natural module and 
$\bC_{V}(L)=0$. Then $V=[V,L]$, or one of the following holds:}
\begin{itemize}
\item[(i)]{\it $L\cong L_{2}(q)$, and $[V:[V,L]]\leq q$}
\item[(ii)]{\it $L\cong L_{3}(2)$, and $|V|=16$}
\item[(iii)]{\it $L\cong U_{4}(2)$, and $|V|\leq 2^{\, 10}$}
\item[(iv)]{\it $L\cong Sp_{2n}(q)$, and $|V:[V,L]|\leq q$}
\item[(v)]{\it $L\cong \Omega_6^{+} (2)$, and $|V|=2^{\, 7}\, .$}
\end{itemize}

(b) {\it Let $L\cong\Omega_8^{-} (q)$ or $Sp_{6}(q), q$ even, $V$ be a
module over $GF(2)$ with $\bC_{V} (L)=0$ and $[V,L]\cong V(\lambda_{3}
)$. Then $V\cong V(\lambda_{3} )$.}

(c) {\it Let $L\cong L_{n}(q), n\geq 5$, $\Omega_{10}^{\, +} (q)$ or
$A_{7}$ and $V$ a $GF(2)$-module with $\bC_{V} (L)=0$ and 
$[V,L]\cong V(\lambda_{2} ),V(\lambda_{4} )$ or 
$|[V,L]|=16$, respectively. Then $V=[V,L]$.}

(d) {\it If in (a) or (b) we have $V=[V,L]$ and $V/\bC_{V} (L)$ is the
natural module or $V(\lambda_{3} )$, respectively, then we either get
$\bC_{V} (L)=0$ or we have the exceptions as in (a) and the bound for
$|\bC_{V} (L)|$ is the same as the bound for $|V:[V,L]|$ in (a).}

\absa
Proof~. (d) follows by duality. Hence we just have to prove (a) - (c).
Let us start with (a).\\
\\

\indent
(1)~~Let first $L\cong L_{n}(q)$:\\
\\
\\
\indent
If $L\cong L_{2}(q)$, then for $x\in L, o(x)=2$, we have $|V:\bC_{V}
(x)|=q$. By (1.1) $L$ is generated by three conjugates of $x$. Hence
$|V|\leq q^{\, 3}$.

\indent
Let now $L\cong SL_{3}(q)$. If $q > 4$, then there are three elements 
 $x_{1}, x_{2}, x_{3}$ in $L$ acting fixed point freely\\
\\
\\
$\quad x_{1}=
\left(\begin{array}{ccc}
\omega^{\, -2}\\ 
& \omega &\\
&& \omega
\end{array}\right)$, $x_{2}=\left(\begin{array}{ccc}
\omega\\
& \omega^{\, -2} &\\
&& \omega
\end{array}\right)$, $x_{3}=\left(\begin{array}{ccc}
\omega\\
& \omega &\\
&& \omega^{\, -2}
\end{array}\right)$,
\\
\\
$o(\omega )=q-1$.\\
\\
\\
\indent
 We have $[x_{i},V]=[V,L],i=1,2,3$, and so as
$[x_{i},x_{j}]=1$ for all $i,j$, we get $\bC_{V} (x_{1})=\bC_{V}
(x_{2} )=\bC_{V} (x_{3})$. But $L=\langle \bC_{L} (x_{i})|i=1,2,3\rangle 
$, the assertion.

\indent
If $q=4$, we have $V=[V,Z(L)]=[V,L]$. So let $q=2$. Then there is
$x\in V\setminus [V,L], |x^{L}|=8$. Hence $V$ is a factor module
of the permutation module, which shows $|V|\leq 16$.

\indent
Let next $L\cong L_{4}(2)$. There is 
$\langle \rho\rangle \times A_{5}\leq L, o(\rho)=3, [V,L]=[V,\rho]$ 
and $[V,L]=[V,\gamma ], \gamma\in A_{5},o(\gamma)=3$. Hence 
$\bC_{V} (\rho)=\bC_{V} (\gamma )$. Now as 
$\langle \bC_{L} (\gamma),\bC_{L} (\rho)\rangle =L$, we get $V=[V,L]$.

\indent
Let now $n\geq 4, q > 2$ for $n=4$. Let $P$ be the parabolic in $L$
with $|O_{2} (P)|=q^{\, n-1}$ and $P^\prime /O_{2}(P)\cong SL_{n-1}(q)$.
Assume furthermore $|\bC_{[V,L]} (O_{2}(P))|=q^{\, n-1}$. Then
$[P^\prime ,V]=\bC_{[V,L]} (O_{2}(P))$, as $P^\prime =P^{\prime\prime}$. 
This now shows that $[\bC_{V} (O_{2}(P)),P^\prime  ]$ is the natural 
module. By induction
on $n$ we have that 
$\bC_{V} (O_{2}(P))=\bC_{V}(P^\prime  )\bigoplus\bC_{[V,L]} (O_{2}(P))$. 
Now application of
\cite[(I.17.4)]{Hu} shows $\bC_{V} (P^\prime  )=0$. Hence as 
$|V:\bC_{V}(O_{2}(P))|=q$, we get $V=[V,L]$.

\vspace{4mm}

(2)~~Let next $L\cong SU_{n}(q)$:\\
\\
If $L\cong SU_{3}(q)$. Then there are 
$x_{1},x_{2}\in L, x_{2}\not\in \langle x_{1}\rangle ,
o(x_{1})=q+1=o(x_{2}), [x_{1},x_{2}]=1$ and $[V,L]=[V,x_{i}],i=1,2$.
Hence $\bC_{V} (x_{1})=\bC_{V} (x_{2})$. As $\bC_{V} (x_{1}
)\cong\bZ_{q+1} \times L_{2}(q)$, is a maximal subgroup of $L$, we get
$\langle \bC_{L} (x_{1}),\bC_{L} (x_{2})\rangle =L$, the assertion.\\

\indent
Let now $L=U_{4}(q),q > 2$. There are  
$x_{1},x_{2},x_{3}\in L, (o(\omega) = q + 1)\\
\\
\\
x_{1}=
\left(\begin{array}{cccc}
\omega^{\, -3}\\ 
& \omega &\\
& & \omega &\\
& & & \omega
\end{array}\right),\; x_{2}=
\left(\begin{array}{cccc}
\omega\\
& \omega^{\, -3} &\\
& & \omega\\
& & & \omega
\end{array}\right),\; \\
x_{3}=
\left(\begin{array}{cccc}
\omega\\
& \omega &\\
& & \omega \\
& & & \omega^{\, -3}\\
\end{array}\right).\\$
\\
\\
\indent
As $q > 2$, we have $[V,L]=[V,x_{i}],i=1,2,3$. Now $\bC_{V}
(x_{1})=\bC_{V} (x_{2})=\bC_{V} (x_{3})$. As $\bC_{L}
(x_{i})=\langle x_{i}\rangle SU_{3}(q)$, we get 
$L=\langle \bC_{L} (x_{1}),\bC_{L} (x_{2}),\bC_{L}(x_{3})\rangle $. 
Hence we have the assertion. Let now $q=2$. Let $P$ be the
parabolic with $P/O_{2}(P)\cong L_{2}(4)$. Now $|\bC_{[V,L]}
(O_{2}(P))|=16$. As $\bC_{[V,L]} (O_{2}(P))\not\cong O_{2}(P)$ as
$P/O_{2}(P)$-modules, we get $V=[V,L]\bigoplus\bC_{V} (O_{2}(P))$. We
have $|\bC_{V} (O_{2}(P)):[\bC_{V} (O_{2}(P)),P]|\leq 4$. Hence
$|V:[V,L]|\leq 4$ by \cite[(I.17.4)]{Hu}.

\indent
Let now $L\cong U_{5}(2)$. In this case we have
$x_{1},x_{2},x_{3},x_{4},\\
\\x_{1}=
\left(\begin{array}{ccccc}
\omega^{\, -1}\\
& \omega &\\
& & \omega \\
& & & \omega\\
& & & & \omega
\end{array}\right), x_{2}=
\left(\begin{array}{ccccc}
\omega\\
& \omega^{\, -1} &\\
& & \omega \\
& & & \omega\\
& & & & \omega
\end{array}\right) , \ldots$ \\
\\
and so on, $o(\omega )=3$.\\
\\
Further $[x_i,x_j] = 1$. Now
$[V,L]=[V,x_{j}],i=1,\ldots ,4$ and so 
$\bC_{V} (x_{1})=\bC_{V} (x_{2})= \bC_{V} (x_{3})=\bC_{V} (x_{4})$. 
We have $\bC_{L} (x_{i})\cong \langle x_{i}\rangle \times U_{4}(2)$. 
Hence $L=\langle \bC_{L} (x_{i})|i=1,2,3,4\rangle $ and
then $V=[V,L]$.\\

\indent
If $L\cong SU_{6}(2)$, then $[Z(L),V]=[V,L]$ and we get $V=[V,L]$, as
$|Z(L)|=3$. Let now $L\cong SU_{n}(q), q > 2$ for $n=5$ or 6.

\indent
Let $P$ be the normalizer of a root group $R$ in $L$. We have
$|[[V,L],R]|=q^{\, 2}$. We have $P^\prime  /O_{2}(P)\cong SU_{n-2}(q)$ 
and $\bC_{[V,L]} (R)/[[V,L],R]\cong O_{2}(P)/R$. Now $[V,R]=[[V,L],R]$.
Furthermore as $[P^\prime  ,V]\leq \bC_{[V,L]} (R)$, we see that 
$V/[[V,L],R]=[V,L]/[[V,L],R]\cdot \bC_{V/[[V,L],R]} (O_{2}(P))$. Let
$V_{1}$ be the preimage of $\bC_{V/[[V,L],R]} (O_{2}(P))$. Then
$[V_{1}/[[V,L],R], P^\prime  ] $ is the natural $SU_{n-2} (q)$-module. By
induction $V_{1}/[[V,L],R]=(V_{1}/[[V,L],R])\bC_{V_{1}/[[V,L],R]}
(P^\prime )$.

\indent
Now as $P^\prime =P^{\prime\prime}$, we get that $[V_{2},P^\prime ]=0$ 
for a preimage $V_{2}$
of $\bC_{V_{1}/[[V,L],R]} (P^\prime  )$. Hence $V_{2}=0$ and so 
$V=[V,L]$, the assertion.

\vspace{4mm}

(3)~~Let next $L\cong \Omega_{2n}^{\pm} (q)$:\\
\\
Let first $L\cong\Omega_4^{-} (q)$. Choose $S\in \mbox{Syl}_{2} (L)$.
Then $S$ is normalized by $\omega\in L, o(\omega )=q+1$. Let
$V_{1}=[V,S]$. Chose $x\in S^{\sharp}$, then $[V,x,x]=0$. Hence
$[V,x]\leq [[V,L],x]$. This shows $[V,S]=[V,L,S]$. Furthermore
$V=[V,L]\bC_{V} (x)$. As $[\omega ,V]\leq [V,L]$, we see that
$V=[V,L]\bC_{V} (S)$. By \cite[(I.17.4)]{Hu} we get $V=[V,L]$.

\indent
Let $L\cong\Omega_6^{\, +} (q),q > 2$. Let $P$ be the parabolic with
$P^\prime  /O_{2}(P)\cong L_{2}(q)\times L_{2}(q)$. Then let 
$V_{1}=\bC_{V}(O_{2}(P))$. We have $|V_{1}|=q$, 
$[V,O_{2}(P)]/V_{1}\cong O_{2}(P)$ and $[P^\prime ,V]=[V,O_{2}(P)],$ 
as $P^\prime =P^{\prime\prime}$. 
This shows
$|V/V_{1}:\bC_{V/V_{1}} (O_{2}(P))|=q$. Now 
$|\bC_{V/V_{1}}(O_{2}(P)):\bC_{V} (O_{2}(P))/V_{1}|=q^{\, 4}$. 
Hence $[\bC_{V}(O_{2}(P)),P^\prime ]=0$. This gives with 
\cite[(I.17.4)]{Hu} that $\bC_{V}(O_{2}(P))\leq [V,L]$ and then 
$V=[V,L]$.

\indent
Let $L\cong \Omega_6^{\, +} (2)$. Now $A_{8}\cong\Omega_6^{\, +} (2)$
and $[V,L]$ is the permutation module. Hence the assertion follows
with (1.19).

\indent
Let $L\cong \Omega_{2n}^{\, \pm} (q),n\geq 4$ for $L\cong\Omega_{2n}^{\,
+} (q)$. Let $P$ be the parabolic with 
$P^\prime  /O_{2}(P)\cong\Omega_{2n-2}^{\,\pm} (q)$. Set 
$V_{1}=\bC_{[V/L]} (O_{2}(P)), |V_{1}|=q$. We
have $[P^\prime ,V]=[V,O_{2}(P)]$, as $P^\prime =P^{\prime\prime}$. 
Hence we see $|V/V_{1}:\bC_{V/V_{1}}(O_{2}(P))|=q$. Furthermore 
$\bC_{V/V_{1}} (O_{2}(P))=[\bC_{V/V_{1}}(O_{2}(P)),P^\prime ]
\bC_{V} (O_{2}(P))/V_{1}$. Now we get $[\bC_{V}(O_{2}(P)),P^\prime ]=0$. 
By \cite[(I.17.4)]{Hu} we have $\bC_{V} (O_{2}(P))\leq [V,L]$ 
and so $[V,L]=V$.

\vspace{4mm}

(4)~~Let finally $L\cong Sp_{2n}(q)$:\\
\\
\indent
If $L\cong Sp_4(2)^\prime$, the assertion follows with (1.19). Let now
$L\cong Sp_{2n}(q),q > 2$, for $n=2$. Let $P$ be the parabolic with
$P^\prime /O_{2}(P)\cong Sp_{2n-2}(q), A_{6}$ for $L\cong Sp_{6}(2)$. Now 
we see $V=[V,L]\bC_{V} (Z(O_{2}(P)))$. Set 
$V_{1}=\bC_{V}(Z(O_{2}(P)))$. Now $[V_{1},P^\prime ]=[[V,L],P^\prime ]$. 
Set $V_{2}=[V,Z(O_{2}(P))]$. Then $|V_{2}|=q$. We see that
$|V_{1}/V_{2}:\bC_{V_{1}/V_{2}} (O_{2}(P))|=q$. We have
$V_{1}/V_{2}=([[V,L],P^\prime ]/V_{2})\bC_{V_{1}} (O_{2}(P))/V_{2}$. 
Hence as $P^\prime =P^{\prime\prime}$ we have 
$[\bC_{V_{1}} (O_{2}(P)),P^\prime ]=0$. This shows 
$\bC_{V_{1}}(O_{2}(P))\leq [V,L]$ and so $|V:[V,L]|\leq q$ by 
\cite[(I.17.4)]{Hu}, or $L\cong Sp_{6}(2)$ and 
$[P,\bC_{V_{1}}(O_{2}(P))]\not= 0$. Hence there is some 
$t\in P\setminus P^\prime ,o(t)=2$, with 
$[\bC_{V_{1}} (O_{2}(P)),t]=V_{2}$. We may assume
that $t$ induces a transvection on the natural module. As
$[t,[V,L]/V_{2}]\not= 0$ we now see $|[V,t]|\geq 4$. But 
$\langle t\rangle $ is
conjugate to $Z(O_{2}(P))$ and $[V,Z(O_{2}(P))]=V_{2}$ is of order 2. 
This proves (a).

\vspace{4mm}

To prove (b) let $[V,L]\cong V(\lambda_{3} )$. Let first
$L\cong\Omega_8^{\, -} (2)$. Then there are $x_{1}, x_{2}\in L,
o(x_{1})=o(x_{2})=3, x_{1}\not\in \langle x_{2}\rangle , 
[x_{1},x_{2}]=1,
[V,L]=[V,x_{i}], i=1,2, \bC_{L}(x_{i})=\langle x_{}\rangle 
\times\Omega_6^{\, +}(2)$. This shows 
$\bC_{V} (x_{1})=\bC_{V} (x_{2})$ is invariant
under $L=\langle \bC_{L} (x_{1}),\bC_{L} (x_{2})\rangle $. Hence 
$V=[V,L]$.

\indent
Let now $q > 2$. Let $P$ be the parabolic with
$P^\prime /O_{2}(P)\cong\Omega_6^{\,-} (q)$. Then 
$\bC_{[V,L]} (O_{2}(P))$
is the natural $U_{4}(q)$-module for $P^\prime$ while $O_{2}(P)$ is the
natural $\Omega_6^{\, -} (q)$-module. This shows 
$V=[V,L]\bC_{V} (O_{2}(P))$
and $[\bC_{V} (O_{2}(P)), P^\prime ]=\bC_{[V,L]} (O_{2}(P))$. As 
$q > 2$, we
have $\bC_{V} (O_{2}(P))=\bC_{[V,L]} (O_{2}(P))\bC_{V} (P^\prime )$.
Application of \cite[(I.17.4)]{Hu}  shows $\bC_{V} (P^\prime )=0$ and 
then $V=[V,L]$.

\indent
Let now $L\cong PSp_{6} (q)$. Let $L_{1}\leq L,
L_{1}\cong\Omega_6^{\,+} (q)\cong L_{4}(q)$. Then $[V,L]$ is an
extension of the natural $L_{4}(q)$-module by the natural module.
Hence $V=[V,L]\bigoplus\bC_{V} (L_{1})$. Let $P_{1}\leq L_{1}$
be the parabolic which is the stabilizer of a 2-space in the natural
representation of $L_{4}(q)$. Then $\bC_{V} (P_{1})=\bC_{V} (L_{1})$.
We have $P_{1}\leq P$, $P$ the stabilizer of a 1-space in the
natural representation of $Sp_{6} (q)$. Now $Z(P^\prime )$ centralizes 
$P_{1}$
and so $[Z(P^\prime ), \bC_{V} (P_{1})]=0$. Hence $\bC_{V} (P_{1})$ is
centralized by $\langle L_1,Z(P^\prime )\rangle =L$. This shows 
$\bC_{V} (L_{1})=0$ and then
$V=[V,L]$. This finishes (b).

\indent
For (c) let first $L\cong L_{5} (q)$ and 
$[V,L]\cong V(\lambda_{2})$. Let $P$ be the parabolic with 
$P^\prime /O_{2} (P)\cong SL_{n-1}(q),|\bC_{[V,L]} (O_{2}(P))|=q^{\, n-1}$. 
Now $\bC_{[V,L]} (O_{2}(P))\cong O_{2}(P)^{\, *}$. Furthermore 
$[V,L]/\bC_{[V,L]}(O_{2}(P))\cong V(\lambda_{2} )$ for the group 
$P^\prime /O_{2}(P)\cong SL_{n-1}(q)$. As 
$V(\lambda_{2} )\not\cong O_{2}(P)$, we get
$[V,O_{2}(P)]=\bC_{[V,L]} (O_{2}(P))$.

\indent
If $n=5, q=2$ then let $P_{1}\leq P$ with 
$O_{2}(P)\leq P_{1}$ and $P_{1}/O_{2}(P)\cong A_{7}$. Otherwise set 
$P_{1}=P^\prime  $.
Then by (a), induction or (1.19) we see that 
$V/\bC_{[V,L]}(O_{2}(P))=[V,L]/\bC_{[V,L]} (O_{2}(P))\cdot
\bC_{V/\bC_{[V,L]} (O_{2}(P))}
(P_{1})$. Let $V_{1}$ be the preimage
of $\bC_{V/\bC_{[V,L]} (O_{2}(P))}(P_{1})$. Then as 
$\bC_{[V,L]} (O_{2}(P)) \not\cong O_{2}(P)$ as $P_{1}/O_{2}(P)$-modules, 
we see $[V_{1},O_{2}(P)]=0$. Hence in any case $P^\prime$ acts on 
$V_{1}$. Now by (a) we
have $V_{1}=\bC_{[V,L]} (O_{2}(P))\bC_{V_{1}} (P^\prime )$. 
Application of
\cite[(I.17.4)]{Hu} shows $\bC_{V} (P^\prime )=0$ and so $V=[V,L]$.


\indent
Let now $L\cong \Omega_{10}^{\, +} (q), [V,L]=V(\lambda_{4} )$. Let $P$
be the parabolic with $P^\prime /O_{2}(P)\cong \Omega_8^{\, +} (q)$. Then
$O_{2}(P)$ is the natural $\Omega_8^{\, +} (q)$-module, while $[V,L]/
\bC_{[V,L]} (O_{2}(P))$ and $\bC_{[V,L]} (O_{2}(P))$ both are spin
modules for $\Omega_8^{\, +} (q)$ and so not isomorphic to
$O_{2}(P)$. Hence we get $[V,O_{2}(P)]=\bC_{[V,L]} (O_{2}(P))=V_{1}$.
Now by (a) $V/V_{1}=[V,L]/V_{1}\cdot \bC_{V/V_{1}} (P^\prime )$. Let 
$V_{2}$ be the preimage of $\bC_{V/V_{1}} (P^\prime )$. As 
$\bC_{[V,L]} (O_{2}(P)) \not\cong O_{2}(P)$, we get $[V_{2},O_{2}(P)]=0$. 
Again by (a) $V_{2}=\bC_{[V,L]} (O_{2}(P))\cdot \bC_{V_{2} } (P^\prime )$. 
By \cite[(I.17.4)]{Hu}  $\bC_{V_{2}} (P^\prime )=0$. 
This shows $V=[V,L]$.

\indent
Let finally $L\cong A_{7}, |[V,L]|=16$. Let $L_{1}\cong L_{3}(2),
L_{1}\leq L$, with $|[[V,L],L_{1}]|=8$. Then by (a)(ii)
$V=[V,L]\bC_{V} (L_{1})$. Now by \cite[(I.17.4)]{Hu} we see 
$\bC_{V}(L_{1})=0$ and so $V=[V,L]$.


\absa
{\bf (1.21)~Lemma.~}{\it Let $L=F^{\, *}(G)$ be a quasisimple group
and $V$ be an $F$-module over $GF(2)$ for $G$. Suppose $\bC_{V}
(S)\leq \bC_{V} (F^{\, *}(G))$ for $S\in\mbox{~Syl}_{2} (G)$.
Then one of the following holds}
\begin{itemize}
\item[(i)]{\it $L\cong L_{3}(2)$ and $|[V,L]|=16$.}
\item[(ii)]{\it $L\cong A_{2^{\, m}}$ and 
$|\bC_{V} (L)|=2, [V,L]/\bC_{[V,L]}(L)$ is the natural module.}
\end{itemize}

\absa
Proof~. We may assume $G=LS$ and furthermore $V=[V,L]$. Set
$V_{1}=\bC_{V} (L)$. Assume $\bC_{V} (S)\leq V_{1}$. Let $V_{2}$
be an $L$-submodule of $V, V_{1}\leq V_{2}, V_{2}/V_{1}$
irreducible. Let $T=S\cap N_{G}([V_{2},L])$. If $V_{1}\cap
[V_{2},L]=0$, then $\bC_{\langle [V_{2},L]^{S} \rangle } (L)=0$, but
$\bC_{\langle [V_{2},L]^{\, S}\rangle } (S)\not= 0$.

Hence by (1.19) and (1.20) we are left with $L\cong L_{2}(q)$,
$L_{3}(2)$, $U_{4}(2)$, $Sp_{2n}(q)$, $G_{2}(q)$ or $A_{2x}$. Let $C$ be a
Cartan subgroup of $L$. Then $C$ acts on $\bC_{V_{2}} (S\cap L)$, and
if $\bC_{V_{2}} (S\cap L)\not\leq V_{1}$, then 
$\bC_{V_{2}} (S\cap L)=V_{1}[\bC_{V_{2}} (S\cap L),C]$. As 
$S=(S\cap L)N_{S}(C)$. We see that 
$[\bC_{V_{2}} (S\cap L),C]\cap \bC_{V}(S)\not= 0$. So we may assume 
that either $C=1$ or 
$\bC_{V_{2}} (S\cap L)\leq V_{1}$.

Suppose the latter. We may assume $V_{2}=[V,L]$. If 
$V_{2}/\bC_{V_{2}} (L)$
is the natural  $L_{2}(q)$-module, then as $[x,V_{2}]\bC_{V_{2}}
(L)/\bC_{V_{2}} (L)=[L\cap S, V_{2}]\bC_{V_{2}} (L)/\bC_{V_{2}} (L)$ 
for all $x\in S\cap L$, $x \not= 1$,
we see that $\bC_{V_{2}} (S)\not \leq V_{1}$.

Let $L\cong L_{3}(2)$, then $|V_{2}|=16$ by (1.20) (a). As $V$ is an
$F$-module, we have $V_{2}=V$.\\
\indent
Let $V\cong U_{4}(2)$. By (1.20) (a) $|\bC_{V_{2}} (L)|\leq 4$. Let
$P$ be the parabolic with $P/O_{2}(P)\cong L_{2} (4)$. We have
$\bC_{V_{2}} (O_{2}(P))/\bC_{V_{2}} (L)$ is the natural
$L_{2}(4)$-module. Hence as seen in the $L_{2}(q)$-case this implies 
$\bC_{V} (S\cap P)\not\leq V_{1}$. 

Let $L\cong Sp_{2n}(q)$. By (1.20) (a) $|\bC_{V_{2}} (L)|\leq q$. Let $R$
be a transvection on $V_{2}/\bC_{V_{2}} (L)$. Then
$[R,V_{2}]\bC_{V_{2}} (L)/\bC_{V_{2}} (L)$
is of order $q$ and $[R,[R,V_{2}]]=0$. Let $P=N_{L}(R)$, then
$[P^\prime ,[R,V_{2}]]=0$. Hence we have $S\cap L\not\leq P^\prime$. This
shows $L\cong Sp_6(2)$. But in this case $|[V_{2},R]|=2$ and so
$[S\cap L,[V_{2}, R]]=0$, too.

Let next $L\cong G_{2}(q)$. Let $R$ be a root group, 
$r\in R^{\,\sharp}$. Then $|[V_{2},r]|=q^{\, 2}$ and $\bC_{L} (r)$ 
induces the natural module on $[V_{2},r]$. As 
$[V_{2},r]\cap \bC_{V_{2}} (L)\not= 0$, we have 
$\bC_{V_{2}} (S)\not\leq V_{1}$.

Let finally $L\cong A_{2x}$. Then by (1.19) $V_{2}$ is a submodule of
the permutation module. As $\bC_{V_{2}} (S\cap L)\leq V_{1}$, we have
that a Sylow 2-subgroup has to act transitively and so  $2x=2^{\, m}$
for some $m$. So we have
\begin{center}
$(*)$~\parbox{12cm}{If $\bC_{V_{2}} (S\cap L)\leq V_{1}$, then
$L\cong L_{3}(2), |V_{2}|=16$ or $L\cong A_{2^{\, m}}, V_{2}$ is a
submodule of the permutation module.}
\end{center}

Let now $C=1, \bC_{V_{2}} (S\cap L)\not\leq V_{1}$, but
$\bC_{V_{2}} (L)\not= 0$. Hence we have $L\cong Sp_{2n}(2), G_{2}(2)^\prime$
or $A_{2x}$. Now if $L\not\cong A_{6}$, we have $|G:L|\leq 2$ and so
$V=V_{2}+V_{2}^{\, x}$ for some $x\in G\setminus L$. But then $x$
centralizes $u+u^{\, x}, u\in \bC_{V_{2}} (S\cap L)\setminus
V_{1}$. This leaves us with $G\cong P\Gamma L_{2}(9)$. Furthermore $V$
has at least four composition factors which are  natural
$PSp_{4}(2)$-modules. Now for any $1\not= A\leq S$, $A$ elementary
abelian, $|V:\bC_{V} (A)|\geq 16$. As $|A|\leq 8$, this contradicts
the fact that $V$ is an $F$-module. Hence $(*)$ holds in general. 
As $V$ is an $F$-module we now see that $V=V_{2}$.

\absa
{\bf (1.22)~Lemma.~}{\it Let $X=G(q)$ be a Lie group in
characteristic two. Let $V$ be an irreducible nontrivial
$GF(2)$-module for $X$ and $t\in X$ be an involution. Suppose 
$|V:\bC_{V} (t)|\leq 2^{\, m_{2}(X)}$. Then $X\cong (S)L_{n}(q),
(S)U_{n}(q), Sp_{2n}(q), \Omega^{\,\pm} (2n,q), G_{2}(q), F_{4}(q),
E_{6}(q), E_{7}(q)$, or $^2E_{6}(q)$. Furthermore either $V=V(\lambda )$
for some fundamental weight $\lambda$ or }
\begin{itemize}
\item[(i)]{\it $X\cong L_{n+1} (q)$ and 
$V \cong V(\lambda_{1} )\bigotimes V(\lambda_{1} )^{\, \sigma}$ or 
$V(\lambda_{n} )\bigotimes V(\lambda_{n} )^{\,\sigma}, \sigma$ some 
field automorphism of $X$,
where $n\geq 6$ if $o(\sigma )\geq 3$.}
\item[(ii)]{\it $X\cong L_{n+1} (q), n\geq 6$, and 
$V \cong V(\lambda_{1})\bigotimes V(\lambda_{n} )^{\,\sigma}, 
1\not= \sigma$ some field automorphism of $X$.}
\item[(iii)]{\it $X\cong L_{n+1} (q), V(\lambda_{1} +\lambda_{n} )
\cong V, n\geq 5$}
\item[(iv)]{\it $X \cong$ $^2A_{n}(q)=SU_{n+1}(q)$, $V \cong 
V(\lambda_{1} +\lambda_{n})$, $n\geq 5$, 
$V\cong V(\lambda_{1} )\bigotimes V(\lambda_{1} )^{\,\sigma}$, 
$n\geq 15$, $V\cong V(\lambda_{1} )\bigotimes V(\lambda_{n} )^{\,\sigma}$, 
$n\geq 7$, $\sigma$ a field automorphism of $X, o(\sigma )\geq 3$.}
\item[(v)]{\it $X\cong Sp_{2n}(q),V \cong V(\lambda_{1})\bigotimes 
V(\lambda_{1})^{\,\sigma}, n\geq 7, 1\not= \sigma$ a field automorphism.}
\item[(vi)]{\it $X\cong\Omega_{2n}^{\,\pm} (q), 
V\cong V(\lambda_{1})\bigotimes V(\lambda_{1})^{\,\sigma}, 
n\geq 8, \sigma$ a field automorphism of $X$.}
\end{itemize}

\absa 
Proof~. This is \cite{Coop}.
\\

\absa
{\bf (1.23)~Lemma.~} {\it Let $X\cong G(q), q=2^{\, n}$, be a Lie
group. Let $V$ be an irreducible nontrivial $GF(2)$-module for $X$
and $t\in X$ be an involution with $|V:\bC_{V} (t)|\leq 2^{\,
m_{2}(X)}$. Assume furthermore $V=V(\lambda )$ for some fundamental
weight $\lambda$. Then $m_{2}(X)$ and the minimal value of 
$|V:\bC_{V} (t)|$ are given in the
following table 1.}

\begin{figure}
$\begin{array}{c|c|c|c}
X & m_{2}(X) & \mbox{module} & \mbox{minimal codimension}\\[2.7ex]
A_{\ell }  & (\frac{\ell +1}{2})^{\, 2}n,\ell\; odd & V(\lambda_{1} ) & 1\\
  & (\frac{\ell +2}{4})^{\, 2}n,\ell\; even & V(\lambda_{\ell } ) & 1\\
A_{\ell },\ell\geq 3  & & V(\lambda_{2} ),V(\lambda_{\ell -1} ) & \ell -1\\
A_{5} & & V(\lambda_{3} ) & 6\\
A_{6} & & V(\lambda_{3} ),V(\lambda_{4} ) & 10\\
A_{7} & & V(\lambda_{3} ),V(\lambda_{5} ) & 15\\
\mbox{~} & \mbox{~} & \mbox{~} & \mbox{~}\\
^2A_{\ell } & [\frac{\ell +1}{2} ]^{\, 2}n  & V(\lambda_{1} ) & 2\\
^2A_{3} & & V(\lambda_{2} ) & 2\\
^2A_{\ell }, \ell \geq 4 & & V(\lambda_{2} ) & 2(\ell -1)\\
^2A_{5} & & V(\lambda_{3} ) & 6\\
\mbox{~} & \mbox{~} & \mbox{~} & \mbox{~}\\
B_{\ell } & {\ell +1\choose 2} n & V(\lambda_{1} ) & 1\\
B_{\ell } & & V(\lambda_{2} ) & 2(\ell-1)\\
B_{\ell },3\leq \ell \leq 6 & & V(\lambda_{\ell } ) & 2^{\, \ell -2}\\
\mbox{~} & \mbox{~} & \mbox{~} & \mbox{~}\\
D_{\ell },\ell \geq 4 & {\ell \choose 2} n & V(\lambda_{1} ) & 2\\
D_{\ell },\ell \geq 4 & & V(\lambda_{2} ) & 4\ell -6\\
D_{\ell },4\leq \ell \leq 7 & & V(\lambda_{\ell -1} ),V(\lambda_{\ell } ) & 2^{\, \ell -3}\\
\mbox{~} & \mbox{~} & \mbox{~} & \mbox{~}\\
^2D_{\ell }, \ell \geq 4 & {\ell \choose 2} n  & V(\lambda_{1} ) & 2\\
^2D_{\ell }, \ell \geq 4 & & V(\lambda_{2} ) & 4\ell -6\\
^2D_{\ell }, \ell = 4, 5 & & V(\lambda_{\ell -1} ) & 2^{\, \ell -2}\\
G_{2} & 3n & V(\lambda_{1} ) & 2\\
F_{4} & 11n\leq m_{2}(X)\leq 16n  & V(\lambda_{1} ) & 6\\
E_{6} & 16n & V(\lambda_{1} ),V(\lambda_{6} ) & 6\\
^2E_{6} & 14n\leq m_{2}(X)\leq 20 n & V(\lambda_{1} ) & 12\\
E_{7}& 27n & V(\lambda_{1} ) & 7
\end{array}$
\begin{center} Table 1 \end{center}
\end{figure}

\absa
Proof~. \cite{Coop}

\absa
{\bf (1.24)~Lemma.~}{\it Let $(X,V)$ be as in $(1.22)$, 
$V\not\cong V(\lambda), \lambda$ a basic weight. Then the lower bound 
for the codimension of
$\bC_{V} (t),t$ an involution of $X$ is as in table 2.}

\begin{figure}
\begin{center}
$\begin{array}{c|c|c}
X & V & \mbox{codimension of~~} \bC_{V} (t)\\[2.7ex]
L_{n+1}(q) & V(\lambda_{1} )\bigotimes V(\lambda_{1} )^{\, \sigma} &
2n,n \mbox{~if~}\sigma^{\, 2} =1\\
& V(\lambda_{n} )\bigotimes V(\lambda_{n} )^{\, \sigma} &
2n,n \mbox{~if~}\sigma^{\, 2} =1\\
& V(\lambda_{1} )\bigotimes V(\lambda_{n} )^{\, \sigma} &
2n\\
& V(\lambda_{1} + \lambda_{n} ) &
n-1\\
\mbox{~} & \mbox{~} & \mbox{~}\\
SU_{n+1}(q) & V(\lambda_{1} )\bigotimes V(\lambda_{n} )^{\, \sigma} &
2n\\
& V(\lambda_{1} )\bigotimes V(\lambda_{n} )^{\, \sigma} &
n\\
& V(\lambda_{1} + \lambda_{n} ) &
n-1\\
\mbox{~} & \mbox{~} & \mbox{~}\\
Sp_{2n}(q) & V(\lambda_{1} )\bigotimes V(\lambda_{1} )^{\, \sigma} &
2(2n-1),2n-1\mbox~if~ \sigma^{\, 2}=1\\
\mbox{~} & \mbox{~} & \mbox{~}\\
\Omega_{2n}^{\,\pm} (q) & V(\lambda_{1} )\bigotimes 
V(\lambda_{1} )^{\, \sigma} & 4(n-1)
\end{array}$
\end{center}
\begin{center} Table 2 \end{center}
\end{figure}

\absa
Proof.~ \cite{Coop}


\absa
{\bf (1.25)~Lemma.~}{\it Let $F^{\, *}(G)$ be a perfect central
extension of a sporadic simple group, $V$ a faithful $GF(2)G$-module.
Then $m_{2}(G)$ and the minimal codimension of $\bC_{V} (t)$ in $V$, 
for $t$ an involution in
$G$ are listed in table 3}
\begin{figure}
\begin{center}
$\begin{array}{ccc}
F^{\, *}(G)/Z(F^{\, *}(G)) & m_{2}(G) & {\mbox{codimension of~}} 
\bC_{V}(t)\\[2.7ex]
M_{11}& 2 & 4\\
M_{12}& 4 & 4\\
M_{22}& 5 & 3\\
M_{23}& 4 & 4\\
M_{24}& 6 & 4\\
J_{1} & 3 & 8\\
J_{2} & 4 & 4\\
J_{3} & 4 & 6\\
Mc & 4 & 8\\
Ly & 4 & 33\\
HS & 5 & 6\\
He & 6 & 10\\
Sz & 6 & 8\\
Ru & 6 & 12\\
O^\prime N & 4 & 21\\
Co_{3} & 4 & 8\\
Co_{2} & 10 & 6\\
Co_{1} & 11 & 8\\
M(22) & 10 & 18\\
M(23) & 11 & 18\\
M(24)^\prime & 12 & 18\\
F_{5} & 6 & 40\\
F_{3} &  5 & 9\\
F_{2} & 22 & 54\\
F_{1} & 24 & 54\\
J_{4} & 11 & 50
\end{array}$
\end{center}
\begin{center} Table 3 \end{center}
\end{figure}

\absa
Proof~. This is \cite[(11.1)]{Asch3}.

\absa
{\bf (1.26)~Lemma.} (i) {\it Let $X\cong (S)L_{3}(q), q$ even, $V$ an 
irreducible
$X$-module over $GF(2), t\in X$ an involution and 
$|V:\bC_{V}(t)|\leq q^{\, 2}$. Then $V$ is the natural module or the 
dual, or
$q=t^{\, 2}$ and $V$ is a $9$--dimensional $GF(t)$-module 
($V\bigotimes V^{\,\sigma} ,V$ the natural module, $\sigma$ a 
field automorphism of
order two).}\\
\indent
(ii) {\it Let $X\cong Sp_{4}(q), q$ even, or $3\cdot A_{6}, V$ an
irreducible $X$-module over $GF(2),t\in X$ an involution and
$|V:\bC_{V} (t)|\leq q^{\, 3}$. Then $V$ is the natural module or the
dual, or $[Z(X),V]\not= 0$ and $V$ is a $6$-dimensional module.}\\
\indent
(iii) {\it Let $X\cong (S)U_{3} (q), q$ even, $q\geq 4$. Let $t\in
X,o(t)=2$, and $V$ be an irreducible $GF(2)X$-module with 
$|V:\bC_{V}(t)| < q^{\, 3}$. Then $V$ is the natural module.}

\absa
Proof~. (i),(ii): This follows from (1.22), (1.23) and (1.24).\\
(iii)~~Let $t\in\Omega_{1} (S), S\in\mbox{Syl}_{2} (X)$. If 
$[[t,V],\Omega_{1}(S)]=0$, we get with \cite[Corollary 1]{Str} that 
$V$ is the natural
module or the 8-dimensional module over $GF(q)$. In the latter we
have $|V:\bC_{V} (t)| > q^{\, 3}$, a contradiction.\\
\indent
So we have $[[t,V],\Omega_{1} (S)]\not= 0$. Let 
$\omega\in N(S),o(\omega )=(q+1)/(3,q+1)$. Then $[[V,t],\omega ]\not=0$. 
Let $U\cong L_{2}(q), [U,\omega ]=1$. Then $[[V,\omega ],U]\not= 0$. Now 
$t$
inverts some element $\gamma$ of order $q+1$ in an $U$. As $\omega$
acts freely fixed point freely on $[[[V,\omega ], \gamma ],t]$ we see
that $|[V,t]\cap [V,\omega ]|\geq q^{\, 2}$. This now implies
$[U,\bC_{V} (\omega )]=1$. Hence $[[V,t],\omega ]=[V,t]$. If 
$\langle \omega \rangle $ acts irreducibly on $[V,t]$, we get 
$[[V,t],\Omega_{1} (S)]=0$, a
contradiction. Hence, we are left with $q=8$ and 
$|V:\bC_{V} (t)|\leq 2^{\, 8}$. As 
$[S,\bC_{[V,t]} (\Omega_{1} (S))]\not= 0$, we see that 
$|\bC_{[V,t]} (\Omega_{1}(S))|\geq 2^{\, 4}$. Now we have that 
$[V,\omega ]$ involves at most
two irreducible $U$--modules $W_{1}, W_{2}$.  Let 
$\tau\in N_{U}(\Omega_{1}(S)),o(\tau )=7$. Then $\tau$ acts irreducibly 
on $\bC_{W_{i}} (\Omega_{1}(S))=T_{i}, i=1,2$. Hence $|T_{i}|=2$ or 
$2^{\, 3}$. In both cases
$W_{1}^{\,\omega}=W_{2}$. In the first case 
$[\bC_{[V,t]} (\Omega_{1}(S)), \tau]=0$. As $S$ acts on 
$\bC_{[V,t]} (\Omega_{1} (S))$, we see 
$[S,\bC_{[V,t]} (\Omega_{1} (S))]=0$. In the second case we see that 
$|\bC_{[V,t]} (\Omega_{1} (S))|=2^{\, 6}$ and 
$\langle \omega \rangle \times \langle \tau\rangle $ acts
irreducibly. This again shows $[S,\bC_{[V,t]} (\Omega_{1} (S))]=0$, 
a contradiction.

\absa
{\bf (1.27)~Lemma.~}{\it Let $X\cong Sz(q)$ or $L_{2}(q), q$ even. }

(i){\it If $V$ is some faithful $GF(2)$-module for $X$, then
$|[V,t]|\geq q^{\, 2}, q$, respectively, where $t$ is some involution
in $X$.}

(ii){\it Let $X\cong Sz(q), |V:\bC_{V} (t)|\leq q^{\, 2}$ for
some involution $t$ and $V$ be irreducible. Then $V$ is the natural
module.}

(iii){\it Let $q > 2, V$ be some $GF(2)$-module for 
$X, T\in \mbox{Syl}_{2} (X)$, and $S=\Omega_{1} (T)$. If 
$V=\bC_{V}(S)\bigoplus \bC_{V} (S^{\, g})$ for some $g\in X$, then 
$V$ is a
direct sum of natural modules.}

(iv){\it Let $X\cong L_{2}(q), T\in\mbox{Syl}_{2} (X)$. If $V$ is a
faithful irreducible $GF(2)$-module with $[V,T,T]=0$, then $V$ is the
natural module.}

\absa
Proof~. (i)~~We have that $t$ inverts an element of order 
$q+\sqrt{2q}+1, q+1$, respectively, and so $|[V,t]|\geq q^{\, 2}, q$.\\
\indent
(ii)~~By (1.1) $X$ is generated by three conjugates of $t$. Hence
$|V|\leq q^{\, 6}$. Set $W=V\bigotimes _{GF(2)} GF(q)$. By the tensor
product lemma $W=\bigoplus M\sigma$ where $M$ is a tensor product of
algebraic conjugates of the natural module $T$.  Let $q=2^{\, n}$,
then $\dim W\leq 6n$. But this shows $W=T$.\\
\indent
(iii)~~We have that for $t\in S, t$ inverts some 
$\omega\in X,o(\omega )=3$ for $X\cong L_{2}(q)$ and $o(\omega )=5$ for 
$X\cong Sz(q)$. As $\langle S,S^{\, g}\rangle =X$ by (1.1), we may assume 
$\omega =g$.
Now $\bC_{V} (\omega )=0$. The assertion follows with \cite[(8.2)]{Hi} 
and \cite{Mar}.\\
\indent
(iv)~~By (1.1) there is $g\in X$ with $\langle T,T^{\, g}\rangle =X$. 
Hence
$V=[V,T]+[V,T^{\, g}]$. Furthermore $\bC_{V} (T)\cap\bC_{V} (T^{\, g})=0$. 
As $[V,T,T]=0$, we get $V=\bC_{V} (T)\bigoplus\bC_{V} (T^{\,g})$. 
The assertion follows with (iii).

\absa
{\bf (1.28)~Lemma.~}{\it Let $F^{\, *}(X)=L$ be a quasisimple Lie
group in odd characteristic $p,Z(L)$ a $p^\prime$-group, 
$L\not\cong L_{2}(q)$, $^2G_{2}(q)$, $G_{\, 2}(q)$, $^3D_{4}(q)$ or 
$PSp_{4}(3)$.
Let $V$ be a faithful $GF(2)$-module for $X$ and $t\in X$ be some
involution. Then $m([V,t])\geq (q-1)q^{\, w} d(p)/2\varepsilon p$, where
$d(p)$ is the degree of the smallest nontrivial representation of
$\bZ_{p} ,q^{\, 2w+1}$ is the order of $O_{p}(\bC_{L} (R)), R$ a long 
root group
(see table below) and $\varepsilon =1$ or 
$L\cong PSp_{2n}(q),\varepsilon =2, q > p$.}

\vspace{5mm}
\noindent
$\begin{array}{cccccccccc}
L & L_{n}(q) & U_{n}(q) & \Omega_{n}^{\;\pm} (q) & PSp_{2n}(q) &
F_{4}(q) & E_{6}(q) & ^2E_{6}(q) & E_{7}(q) & E_{8}(q)\\
w & n-2 & n-2 & n-4 & n-1 & 7 & 10 & 10 & 16 & 28
\end{array}$

\absa
Proof~. This is \cite[(10.4)]{Asch3}. We just sketch his proof. Let 
$R$ be 
a long root group in $X$ and $Q=O_{p}(\bC_{L} (R))$. Then $Q$
is a special group of order $q^{\, 1+2w}$, where $w$ is as described
above.\\
\indent
By \cite[(10.1)]{Asch3} we may assume that $t$ inverts some 
$U,|U|=p, U\leq Q, U\not\leq R$, or $X=PSp_{4k}(q)$ and $t$ induces a
field automorphism. In the former we get with \cite[(7.2)]{Asch3} the
assertion.\\
\indent
So assume the latter. Now $E(\bC_{L} (t))\cong PSp_{2k}(q^{\, 2})$
and $t\sim tz,tz$ induces a nontrivial inner automorphism on 
$\bC_{L}(t)$.\\
\indent
We can proceed by induction. First of all by (1.12)  
$[[V,t],E(\bC_{L} (t))]\not= 0$. Set $W=[V,t]$.\\
\indent
Let $k=1$. Then $E(\bC_{L} (t))\cong L_{2}(q^{\, 2})$. Let $R_{1}$ be a
subgroup of order $q^{\, 2}$ in $E(\bC_{L} (t))$. Then 
$N_{E(\bC_{L} (t))}(R_{1})$ has at most two orbits on the hyperplanes of 
$R_{1}$.
Now there is some orbit $\bigtriangleup$ with
%
$$[W,R_{1}]=\bigoplus\limits_{H\in\bigtriangleup} \,\bC_{[W,R_{1}]}(H)\, .$$  
%
This shows~~$m([W,R_{1}])\geq (q^{\, 2}-1)d(p)/2(p-1) > 
(q-1)\, q^{\, w}d(p)/2p$, since $w=1$.\\[2mm]
Let now $k > 1$. Then by induction
%
$$m([V,tz])\geq (q^{\, 2}-1) q^{\, 2u}d(p)/4p\, ,$$ 
%
where $u=k-1$ and $w=2k-1$. Hence $w=2u+1$ and so 
%
$$m([V,t])\geq (q^{\, 2}-1)q^{\, w-1} d(p)/4p > (q-1)\, q^{\,w}d(p)/4p\, .$$

\vspace{5mm}

\absa
{\bf (1.29)~Lemma.~}{\it Let $F^{\, *}(X)=L$ be a quasisimple group
such that $L/Z(L)$ is a Lie group over a field of odd characteristic
which is not a Lie group over a field of characteristic $2$ too. Let
$t\in X$ be an involution and $V$ be an irreducible faithful
$GF(2)$-module for $X$. Then one of the following holds:
\begin{itemize}
\item[$(1)$]$|V:\bC_{V} (t)|\geq 2^{\, 8}$
\item[$(2)$]$L\cong L_{3}(3),\; |V:\bC_{V} (t)|\geq 2^{\, 4} $
\item[$(3)$]$L\cong U_{4}(3), L_{4}(3), G_{2}(3), 3\cdot G_{2}(3),
PSp_{6} (3)$ and $|V:\bC_{V} (t)|\geq 2^{\, 6}$
\item[$(4)$]$L\cong 3\cdot U_{4}(3)$
\item[$(5)$]$L\cong L_{2}(25)$ or $L_{2}(p), p$ prime, 
$|V:\bC_{V} (t)|\geq 2^{\,4}$
\end{itemize}}

\absa
Proof~. (\cite[(10.5)]{Asch3}). Assume $|V : \bC_{V}(t)| \le 2^7$. Let 
first $L\cong L_{2}(q), q$ odd.
Suppose furthermore $q=p^{\, f} > p$. Let $t\in PGL_{2} (q)$. Then
$L\langle t\rangle $ is generated by three conjugates of $t$ and so 
$|V|\leq 2^{\,21}$. Let $P\in\mbox{Syl}_{p} (L)$ and $\triangle$ be one 
orbit of
hyperplanes under $N_{G}(P)$. Then
$V=\bigoplus\limits_{U\in\triangle} \bC_{V} (U)$. We have 
$|\triangle|=(q-1)/(p-1)$ or $G$ does not induce $PGL_{2}(q)$ on $L$ and 
$|\triangle|=(q-1)/2(p-1)$. Let $d\, (p)=|\bC_{V} (U)|$. We get 
$d\, (p)(q-1)\leq 42(p-1)$. As $d\, (p)\geq 2$, we get a rough bound by 
$p+1\leq 21$,
and so $p\leq 19$. If $p > 7$, then $d\, (p)\geq 8$ and so $p+1\leq 5$,
a contradiction. Hence $p=3, 5$ or $7$. Let $p=7$, then $d\, (p)=3$
and so $q-1\leq 14(p-1)$. This shows $L=L_{2}(7^{\, 2})$ and we have
exactly two orbits on the hyperplanes. But now $t$ inverts $P$, a
contradiction.\\
\indent
Let $p=3$. Then $f\geq 3$. Now $L=L_{2}(27)$. But then there is just
one orbit on the hyperplanes and we have $2\cdot 26\leq 21\cdot 2$, a
contradiction. Let finally $p=5$. Then we get $L=L_{2}(25)$. As
$|V|\geq 2^{\, 9}$, we get $|V:\bC_{V} (t)|\geq 2^{\, 4}$.\\
\indent
Suppose next that $t$ induces a field automorphism on $L$. Let
$L_{1}=E(\bC_{L} (t))$ (recall $q > 9$). Suppose $[[V,t], L_{1}]=0$.
Now as in (1.11) we see that $\{ a^{\, L}\}$ is a set of odd
transpositions. This shows that $\bC_{L_{1}} (z)$ is a 2-group for
$z\in L_{1}, z^{\, 2}=1$, and $L_{1}$ has abelian Sylow 2-subgroup.
This shows $L_{1}=L_{2}(5)$ and so $L\cong L_{2}(25)$.\\
\indent
We are going to prove the same result for $[[V,t],L_{1}]\not= 0$. Now
$L_{1}$ acts faithfully on $[V,t]$ and $|[V,t]|\leq 2^{\, 7}$. In
particular $L_{1}\cong L_{2}(p), p\,\Big  |\, |L_{7}(2)|$, or 
$L_{1}=L_{2}(9)$.
But in the latter $PGL_2(9)$ acts on $[V,t]$ and so 
$|[V,t]|\geq 2^{\, 8}$. We are left with $L_{2}(5), L_{2}(7), L_{2}(31)$ 
and
$L_{2}(127)$. But in all cases $PGL_{2}(p)$ is involved and so there
is some involution $s$ which inverts a Sylow $p$-subgroup. This shows
$|[V,t]|\geq 2^{\, 14} (L_{2}(127)), |[V,t]|\geq 2^{\, 10} (L_{2}(31))$. 
Let $\bC_{L} (t)\cong PGL_{2}(7)$. Then 
$|[V,t]|\geq 2^{\, 6}$. There is $z\in L_{1}, o(z)=2$, such that 
$t\sim tz$. Hence we see that $|[V,P]|\leq 2^{\, 18}$ for 
$P\in\mbox{~Syl}_{7}(L_{1})$. As $L_{2}(7^{\, 2})\not\leq L_{18}(2)$, 
we see that 
$\bC_{V} (P)\not= 1$. Let $\triangle$ be the orbit of $P$ in
$N_{G}(E), E\in\mbox{~Syl}_{p} (L), P\leq E$. Then 
$|\triangle|=4$ and $V=\bigoplus\limits_{U\in\triangle} \bC_{V} (U)$. 
Now there is some $U\in\triangle$ with $[U,t]=U$. This shows 
$|\bC_{V} (P)\cap [V,t]|\geq 8$. But then 
$|[V,t]|\geq 2^{\, 9}$, a contradiction.\\
\indent
Let now $L\cong L_{2}(25)$. We may assume $|[V,t]|\leq 8$. Hence
$[[V,t],L_{1}]=1$. As $t$ inverts some elements of order 13, we see
that $L\langle t\rangle $ is generated by three conjugates of $t$. Now 
$|V|\leq 2^{\, 9}$ but $13\not\Big |\, |L_{9}(2)|$.\\
\indent
So we are left with $L\cong L_{2}(p), p$ prime, $p\geq 11$. Suppose
$|[V,t]|\leq 8$, then $|V|\leq 2^{\, 9}$ and so $L\lesssim L_{9}(2)$. 
This
shows $p=31, 127, 73, 17$. As $37\,\Big  |\, |L_{2}(73)|$, but 
$37\,\not\!\!\Big |\,\,\,|L_{9}(2)|, L\not\cong L_{2}(73)$. In $L_{9}(2)$ 
the normalizer of a
Sylow 127-subgroup is of order $2\cdot 3\cdot 7\cdot 127$, hence
$L_{2}(127)\not\leq L_{9}(2)$. On a Sylow 31-subgroup just a group
of order $5$ is induced, as this is true in $L_{5}(2)$, hence
$L_{2}(31)\not\leq L_{9}(2)$. As $t$ cannot invert an element of
order 17, we get $L\langle t\rangle =PGL_{2}(17)$ if $L\cong L_{2}(17)$. 
Now $t$
centralizes a group of order $9$. Hence we get some element of order
$3$ which centralizes $[V,t]$ and $V/\bC_{V}(t)$ as well. But
$|\bC_{V} (t):[V,t]|\leq 8$ and so this element centralizes $V$, a
contradiction.\\
\indent
Let now $[[V,t],E(\bC_{L} (t))]=0$, then we get the assertion with
(1.12). So assume $[[V,t],E(\bC_{L} (t))]\not= 0$.\\ 
\indent
If $L\cong$ $^2G_{2}(q)$, then 
$E(\bC_{L} (t))\cong L_{2}(q), q\geq 27$. But 
$L_{2}(q)\not\leq L_{7}(2)$.\\
\indent
Let $L/Z(L)\cong G_{\, 2}(q)$. If $t\in L$, then 
$\bC_{L} (t)\cong (SL_{2}(q)*SL_{2}(q))\cdot 2$. Hence the structure of 
$L_{7}(2)$
shows $q=3$ or $7$. If $q=3$, then $t$ inverts some element of order
13 and so $|V:\bC_{V} (t)|\geq 2^{\, 6}$. If $q=7$, $t$ inverts some
element of order 817.\\
\indent
Let $t\not\in L$. Suppose $t$ induces a field automorphism. Then
$E(\bC_{L} (t))\cong G_{2}(\sqrt{q} )\not\leq L_{7}(2)$. So
assume that $q=3^{\, f}$ and $E(\bC_{L} (t))\cong$ $^2G_{2}(q)$. This
shows again $q=3$ and $L_{2}(8)$ acts on $[V,t]$, i.e. 
$|[V,t]|\geq 2^{\, 6}$.\\
\indent
Let now $L/Z(L)\cong$ $^3D_{4}(q)$. Then $t$ acts on some $G_{2}(q)$
in $L$. Hence we may assume $q=3$. Now $t$ induces $PGL_2(27)$ on a
subgroup $SL_{2}(27)$ of $L$. Hence we see $|[V,t]|\geq 2^{\, 8}$ as
before.\\
\indent
Suppose now that $L$ is none of these groups but $p \not\!\Big |\,|Z(L)|$.
Then the conclusion follows from (1.28). If $L\cong L_{3}(3)$ then
either $t$ inverts an element of order 13 or an elementary abelian
group $E\cong E_{9}$, with $N_{L}(E)$ transitive on $E^{\,\sharp}$.
In both cases $|[V,t]|\geq 2^{\, 6}$.\\
\indent
So we are left with $p\,\Big  |\, |Z(L)|$. This now leaves us with
$3\cdot\Omega_{\, 7}(3)$.  Now as 
$3\cdot\Omega_{\,7}(3)\not\leq L_{\, 14}(2)$, we get $t\in \bC (Z(L))$.
Furthermore we may assume $[Z(L),V]=V$. Now 
$E(\bC_{L} (t))\leq GL_{3}(4)$, a contradiction. As 
$t\in 3\cdot \Omega_{\, 7} (3)$, we
see $E(\bC_{L} (t))\cong U_{4}(3)$ or $Sp_{4}(3)$, both not in
$GL_{3}(4)$, or we have $\bC_{L} (t)/Z(L)\cong \Sigma_{4}\times
(SL_{2}(3)*SL_{2}(3))\cdot 4\; \cite{CCNPW}$. But now the embedding in
$GL_{3}(4)$ gives some kernel which contains a fours group. Now
we have a quadratic fours group and can apply (1.12) for a
contradiction.

\absa
{\bf (1.30)~Lemma.~}{\it Let $F^{\, *}(X)=L$ be a quasisimple group
and $V$ be an irreducible faithful $GF(2)$-module for $X$. Let $t$ be
some involution in $X$. Then}

(a) {\it If $|V:\bC_{V} (t)|\leq 2$, then $L\cong SL_{n}(2),
Sp_{2n}(2),\Omega_{2n}^{\,\pm} (2)$, $A_{n}$ and $V$ is the natural
module. Furthermore if $L\cong A_{n}$ or $\Omega_{2n}^{\,\pm} (2)$,
then $t\not\in L$.}

(b) {\it If $|V:\bC_{V} (t)|\leq 4$, then either $(L,V)$ is as in
(a) or $L\cong$ $SU_{n}(2)$, $G_{2}(2)^\prime$, $SL_{n}(4)$, $Sp_{2n}(4)$, 
$\Omega_{2n}^{\,\pm}
(4)$ and $V$ is the natural module or one of the following holds:}
\begin{itemize}
\item[(i)]{\it $L\cong 3\cdot A_{6}, |V|=2^{\, 6}$}
\item[(ii)]{\it $L\cong 3\cdot U_{4}(3), |V|=2^{\, 12}$}
\item[(iii)]{\it $L\cong A_{7}, |V|=2^{\, 4}$}
\item[(iv)]{\it $L\cong Sp_{6}(2), |V|=2^{\, 8}$}
\end{itemize}

\vspace{0.1cm}

\indent
(c){\it \,If $|V:\bC_{V} (t)|\leq 8$, then either $(L,V)$ is as in (b)
or (a) or $L\cong SL_{n}(8)$, $Sp_{2n}(8), \Omega_{2n}^{\,\pm} (8), V$ is
the natural module, or one  of the following holds:}
\begin{itemize}
\item[(i)]{\it $L/Z(L)$ is a Lie group in characteristic two}
\item[(ii)]$L\cong M_{22}, |V|=2^{\, 10}\, .$
\end{itemize}

\absa
Proof~. (The proof follows \cite[(13.5)]{Asch3}). Let first $L/Z(L)=G(q),
q$ even. Then if $t\in L$, we get the result with (1.23), (1.22) and
(1.24). So assume $t\not\in L$. Then there is $s\in L$ with $t\sim ts$
in $x$. Hence $|V:\bC_{V} (s)|\leq 16$. Now again we are going to
apply (1.23), (1.22) and (1.24). This implies that we either have the
assertion or $L\cong L_{5}(2), V = V(\lambda_{2} )$ or
$V(\lambda_{3} )$, $L \cong Sp_{8}(2), V=V(\lambda_{4} )$,
$\,L \cong \Omega_{10}^{\, +}(2), V=V(\lambda_{4} )$ or $V(\lambda_{5} )$, 
$\, L \cong \Omega_8^{\, -} (2), V = V(\lambda_{3} )$, 
$L \cong G_{2}(4), V = V(\lambda_{1} )$, 
$L \cong L_{3}(4), |V|=2^{\, 9}$. As $t$ has to act on this module, 
we get $L\cong \Omega_8^{\, -} (2), V=V(\lambda_{3} )$, or 
$L \cong G_{2}(4), V=V(\lambda_{1})$.\\
\indent
Let $L\cong \Omega_8^{\, -} (2)$ and $P$ be the parabolic in $L$ with
$P/O_{2}(P)\cong\Omega_6^{\, -} (2)$. Then $P/O_{2}(P)$ induces the
natural $U_{4}(2)$-module on $\bC_{V} (O_{2}(P))$ and 
$V/\bC_{V}(O_{2}(P))$. Hence $t$ has to induce a transvection on both 
modules. But as these modules are over $GF(4)$, this is not possible.\\
\indent
Let $L\cong G_{2}(4)$, then $t$ has to induce a field automorphism.
But field automorphisms do act freely on $V(\lambda_{1} )$, a
contradiction.\\
\indent
Let next $L/Z(L)\cong A_{n}$. For $n\leq 8$, we just check the modules
to get the result. Let $n\geq 9$. Then $\bC_{L} (t)$ induces a
subgroup of $L_{3}(2)$ on $[V,t]$. Hence there is some involution
$t_{1}\not= t$ such that $[V,t,t_{1}]=0$. This shows that 
$\langle t,t_{1}\rangle $ is a quadratic fours group. Now we apply 
(1.12), which shows 
that $V$ is the natural module, or $V$ is the spin module. In the latter
$\langle t,t_{1}\rangle $ corresponds to 
$\langle (12)(34),(13)(24)\rangle $ 
or $\langle (12)(34)(56)(78), (13)(24)(57)(68)\rangle $ by 
\cite[(4.3)]{MeiStr1} and (1.12). In both
cases $|V:\bC_{V} (t)|\geq 16$, a contradiction.\\
\indent
Let now $L/Z(L)$ be sporadic. Then we quote (1.25) and  get
$L/Z(L)\cong M_{22}$. Furthermore as before we get a quadratic fours 
group. 
Application of (1.13) shows that either $L\cong M_{22}$ and
$|V|=2^{\, 10}$, or $L\cong 3\cdot M_{22}$ and $|V|=2^{\, 12}$. In
the latter as $Z(L)$ acts fixed point freely we get $|[V,t]|\geq 16$
for $t\in L$ and $|[V,t]|=64$ for $t\not\in L$.\\
\indent
Let finally $L/Z(L)\cong G(q), q$ odd, which is not a Lie group over
a field of characteristic two. Now the lemma is a consequence of
(1.29).

\absa
{\bf (1.31)~Lemma.~}{\it Let $X$ be as in (1.7) or (1.8) and $V$ an 
$F$-module
for $X$ over $GF(2)$ which involves a nontrivial irreducible module.
Then}

(i) {\it $E(X)\cong L_{2}(q), q$ even, $V$ involves exactly one
nontrivial irreducible module which is the natural module, or $q=4$
and this module is the ortho\-gonal module. An offending subgroup $A$
is a Sylow 2-subgroup of $E(X)$ or $q=4$ and 
$|A|\leq 4, |A\cap E(X)|\leq 2$.}

(ii) {\it $E(X)=X_{1}\times X_{2}\cong L_{2}(q)\times L_{2}(q), q$
even. Denote by  $W$ the group $[E(X), V]\bC_{V} (E(X))/\bC_{V} (E(X))$. 
Then
$W=W_{1}\oplus W_{2}, [W_{i}, X_{3-i}]=0, i=1,2$. Furthermore $W_{i}$
is the natural $X_{i}$-module, or $q=4$ and $W_{i}$ is the orthogonal
$X_{i}$-module. In the first case an offending subgroup is a Sylow
$2$-subgroup of $X_{1}, X_{2}$ or $E(X)$, while in the second case the
offending subgroup $A$ normalizes $X_{1}$ and $X_{2}$ and 
$|A\cap X_{1}|\leq 2\geq |A\cap X_{2}|\, ,\, |A|\leq 16$.}

(iii) {\it $E(X)\cong A_{9}$ and there is exactly one nontrivial
irreducible module involved which is the natural module.}
 
(iv) {\it $F(X)$ is a 5-group with $E(X/F(X))\cong L_{2}(5)$.
Furthermore $[F(X), V] =0$ and $X/F(X)$ acts as in (i).}

(v) {\it $X$ is a solvable $\{ 2,3\}$-group, $|[V, F(X)]|\leq 16$.
}

\absa
Proof~. Let first $E(X)=X_{1}$ be quasisimple. Then (1.7) and (1.18)
yield $X_{1}\cong L_{2}(q)$, $(S)L_{3}(q), Sp_{4}(q), A_{9}, A_{6}$
or $3\cdot A_{6}$.\\
\indent
Let $X_{1}\cong (S)L_{3}(q), Sp_{4}(q)$ or $A_{6}$, then by (1.18) $V$
involves just trivial and natural modules. By (1.7) there is some
$x\in X$ acting nontrivially on the Dynkin diagram. Thus with the
natural module $W$ also $W^{*}$, the dual module, is involved. But
then $V$ cannot be an $F$-module.\\
\indent
If $X_{1}\cong 3\cdot A_{6}$, then by (1.30) $V$ involves exactly one
nontrivial irreducible module, which is $6$-dimensional. But then again
by (1.7) both $6$-dimensional modules have to be involved, a
contradiction.\\
\indent
Let $X_{1}\cong A_{9}$, then the assertion follows with (1.30) and
(1.18). \\
\indent Assume now $E(X)=X_{1}X_{2}$. Let $A$ be some offending
subgroup. Let $\langle X_{1}^{\, A}\rangle =E(X)$. Then there is some 
$a\in A$ with
$X_{1}^{\, a}=X_{2}$. Set $F=\bC_{E(X)} (a)$. Then $F$ acts on
$\bC_{V} (a)$. Thus as $|A| > 2$ ($a$ cannot induce a transvection), we
see that $\bC_{E(X)} (a)$ induces an $F$-module in $\bC_{V} (a)$,
which shows $X_{1}\cong L_{2}(q)$ as before.\\
\indent
Now $a$ inverts some $\omega\in E(X), o(\omega )=q+1$. Hence
$|V:\bC_{V} (a)|\geq q$ and so $|A|\geq q$. So for $b\in A\cap E(X)$
we have that $b$ inverts an abelian group of order $(q+1)^{\, 2}$.
Hence $|V:\bC_{V} (b)|\geq q^{\, 2}$ and as $|A|\leq 2q, q=2$, a
contradiction.\\
\indent
So we have $A\leq N(X_{1})$. Then by the $A\times B$-Lemma 
$\bC_{V}(\bC_{A} (X_{1}))$ is an $F$-module for $X_{1}A/\bC_{A} (X_{1})$.
This implies $X_{1}\cong L_{2}(q), q$ even, as before.\\
\indent
We now determine the offending subgroups. Let 
$E(X)=X_{1}\cong L_{2}(q), q$ even. By (1.18) $V$ involves exactly one 
irreducible
module. This is the natural one, or $q=4$ and it is the orthogonal
module. Furthermore an offending subgroup is a Sylow 2-subgroup or
$X\cong \Sigma_{5}$ and $|A|=2, 4$ with $|A:A\cap X_{1}|=2$.\\
\indent
Let now $E(X)=X_{1}X_{2}, X_{1}\cong L_{2}(q)\cong X_{2}, q$ even.
Let $[A,X_{1}]\not= 1$. Then $\bC_{V} (\bC_{A} (X_{1}))$ is an
$F$-module for $X_{1}$. So $V$ involves exactly one irreducible
nontrivial $X_{1}$-module, as 
$|V:\bC_{V} (\bC_{A} (X_{1}))|\leq |\bC_{A}(X_{1})|$. Now 
$[[V,X_{1}], X_{2}]=0$. As there is some $x\in X$ with
$X_{1}^{\, x}=X_{2}$, we get the assertion.\\
\indent
Suppose now $E(X)=1$. By (1.7) $F(X)=P$ is a $p$-group and
$E(X/F(X))\cong L_{2}(p), p$ odd. So $|A|\leq 4$. Suppose
$[V,F(X)]\not= 0$. Now $a\in A^{\,\sharp}$ inverts some 
$\omega\in F(X), o(\omega )=p$. As $|V:\bC_{V} (a)|\leq 4$, we get 
$p=5, |V:\bC_{V}(a)|=4=|A|$. But now as $A$ acts on an elementary 
abelian group of
order $5^{\, 3}$, we get $U \cong D_{10}\times D_{10}$ in $PA$ acting
faithfully on $V$. But this is impossible as $|V:\bC (U)|\leq 16$. So
we have $[P,V]=0$. Now as just shown $L_{2}(p)$ has to be isomorphic
to $L_{2}(4)$ and so $p=5$.\\
\indent
Let now $X$ be solvable. By (1.8) $|A|\leq 4$. Furthermore 
$X=PS, S\in\mbox{Syl}_{2}(X), P=O_{p}(X)$. So we get $p\leq 5$.\\
\indent
Suppose $p=5$. Then $|A|=4$. As $P$ is noncyclic and $S$ acts
irreducibly on $P/\Phi (P)$ we get some $D_{10}\times D_{10} =U$
involved in $AP/\Phi (P)$. But now we get $\langle A,A^{\, g}\rangle $ 
covers $U$ for some $g\in P$. As 
$|V:\bC_{V} (\langle A,A^{\, g}\rangle )|\leq 16$, we see
that $U$ cannot act faithfully on $V$ and so $[P,V]=0$, a
contradiction.

\indent
So we have $p=3$. If $P$ is cyclic, we get $|[V,P]|=4$. So let $P$ be
noncyclic. Let $a\in A^{\,\sharp}$ such that $|V:\bC_{V} (a)|=2$.
Then there is some $t\in Z(S)^{\,\sharp}$ such that 
$|V:\bC_{V}(t)|\leq 4$. If $|A|=4$, then $A\cap Z(S)\not= 1$. So in any 
case
$|V:\bC_{V} (t)|\leq 4$ for $t\in Z(S)$. This shows that
$|[P/\Phi (P),t]|\leq 9$. As $S$ acts irreducibly on
$P/\Phi (P)$, we get $|P/\Phi (P)|=9$, and 
$P=\langle \omega,\tilde{\omega} \rangle $ with 
$|V:\bC_{V} (\langle \omega ,\tilde{\omega} \rangle )|\leq 16$.
Hence $[\Phi (P),V]= 0$ and $|[V,P]|=16$.

\absa
{\bf (1.32)~Lemma.~}{\it Let $G\cong L_{2}(q), q$ even. Let $V$ be a
faithful irreducible $GF(2)$-module for $G$. Let $|V| < q^{\, 4}$.}

\indent
(a) {\it One of the following holds:}
\begin{itemize}
\item[(i)]{\it $V$ is the natural module}
\item[(ii)]{\it $q=t^{\, 2}$ and $V$ is the $\Omega_4^{\, -} (t)$-module}
\item[(iii)]{\it $q=t^{\, 3}\, ,\, |V|=t^{\, 8}$.}
\end{itemize}~
\indent
(b) {\it Suppose $|V|=t^{\, 8}, q=t^{\, 3}$. Let $S\in \mbox{Syl}_{2} (G)$.
Then $|\bC_{V} (S)|=t$. If $x\in S^{\,\sharp}$, then $|[V,x]|=t^{\,4}$.}\\

\absa
Proof~. (a) Let $W=GF(q)\bigoplus_{GF(2)} V$. Then we have that
$W=\bigoplus M\sigma, M$ a tensor product of algebraic conjugates of
the natural module $T$. Let $q=2^{\, n}$, then $\dim W < 4n$. This
shows $M=T, T\bigotimes T\tau $ or 
$T\bigotimes T\tau \bigotimes T\tau ^{\, 2}$. Now
in any case just by the dimension formula, we get $W=M$, and so
$q=t^{\,2}, q=t^{\, 3}$, as $GF(q)$ possesses  an automorphism of
order 2 in case of $M=T\bigotimes T\tau $, of order 3 in case 
$M=T\bigotimes T\tau \bigotimes T\tau ^{\, 2}$, respectively.

\indent
(b) $|\bC_{V} (S)|=t$ follows from \cite{Cur}. Now let 
$x\in S^{\,\sharp}$.
By (1.27) (i) $|[V,X]|\geq t^{\, 3}$. Suppose $|\bC_{V} (X)|=t^{\, 5}$. 
Let $\omega\in N(S), o(\omega
)=t^{\, 2}+t+1$. Then $[\omega ,\bC_{V} (S)]=0$. Hence $|[\omega
,V]|=t^{\, 3}$ or $t^{\, 6}$. As there is $y\in S$ with $S\leq
\langle \omega ,\omega^{\, y} \rangle $ and $|\bC_{V} (S)|=t$, we see 
that
$|[\omega ,V]|=t^{\, 6}$, and so $|\bC_{V} (\omega )|=t^{\, 2}$. Let
$z=X^{g}, \omega^{\, z} =\omega^{\, -1}$. Then 
$|[z, [\omega,V]]|=t^{\, 3}$. Hence $[\bC_{V} (\omega ),z]=0$. But 
$\bC_{V}(S)\leq \bC_{V} (\omega )$ and so $\bC_{V} (S)$ is centralized
by $\langle S,y,\omega \rangle =G$, a contradiction. Hence 
$|\bC_{V} (x)|=t^{\,4}$.

\absa
{\bf (1.33)~Lemma.~}{\it Let $G=G(q), q$ even, and $V$ be an
irreducible module for $G$ over $GF(2)$. Let $P$ be some parabolic of
$G$. Then $\bC_{V} (O_{2}(P))$ is an irreducible $P/O_{2}(P)$-module.}

\absa
Proof~. \cite{Ti}

\absa
{\bf (1.34)~Lemma.~}{\it Let $X\cong (S)U_{5}(q), q$ even, $t\in X$
be some involution and $V$ be some irreducible nontrivial
$GF(2)X$-module.}
\begin{itemize}
\item[(i)]{\it If $|[V,t]|\leq q^{\, 4}$, then $V$ is the natural module}
\item[(ii)]{\it Let $P=O^{\, 2}(N_{X}(R)), R$ a transvection group in the
natural representation. If $|[V,t]| < q^{\, 6}$, then 
$\bC_{V} (P)\not= 0$.}
\end{itemize}

\absa
Proof~. (i) follows from (1.22), (1.23) and (1.24)\\
\indent
(ii) Suppose $\bC_{V} (P)=0$. We may assume 
$t\in P\setminus O_{2}(P)$. Let $r\in R^{\,\sharp}$. Then 
$[P,[V,r]]\not= 0$. As
$P/O_{2}(P)\cong SU_{3}(q)$, we get with (1.26) (iii) that $[V,r]$ is
the natural module and $V/\bC_{V} (r)$ is the natural module.
Furthermore $[P^\prime , \bC_{V} (r)] =[V,r]$. As $\bC_{V} (P^\prime )=0$, 
we get that $|\bC_{V} (r):[V,r]|\leq q^{\, 2}$. Now 
$[O_{2}(P),\bC_{V}(r)]=[V,r]$, as $\bC_{V} (O_{2}(P))$ is an irreducible 
$P$-module by
(1.33). As  $[V,r]\cong O_{2}(P)/R$ as $P$-modules, we see that 
$\bC_{\bC_{V} (r)}(x)=[V,r]$ for every $x\in O_{2}(P)\setminus R$. 
Let now $r^{\, g}\in O_{2}(P)\setminus R, g\in X$. Then we see that 
$[V,r^{\, g}]=[V,r]$
as $[V,r^{\, g}]\leq \bC_{V} (r)\cap\bC_{V} (r^{\, g})$. But now
$[V,r]\lhd \langle \bC_{X} (r), \bC_{X} (r^{\, g})\rangle =X$, a  
contradiction.



\absa
{\bf (1.35)~Lemma.~}{\it Let $X$ be one of the group in (1.7).
Suppose $X$ possesses a nontrivial $GF(2)$-module $V$ with 
$\bC_{V} (F^{\, *}(X))=0, [V,F^{\, *}(X)]=V$ such that for some 
elementary abelian subgroup $1\not= A$ of $X$ we have 
$|V:\bC_{V} (A)|\leq |A|^{\, 2}$. Then one of the following holds:}
\begin{itemize}
\item[(1)] {\it $V$ is an $F$-module for $X$.}
\item[(2)] {\it $E(X)\cong SL_{3}(q)$ and $V$ is a direct sum of the 
natural and the dual module.}
\item[(3)] {\it $E(X)\cong Sz(q), V$ is the natural module, or
$E(X)=X_{1}\times X_{2}, X_{1}\cong X_{2}\cong Sz(q)$ and
$V=V_{1}\bigoplus V_{2}, [V_{i}, X_{3-i}]=0$ and $V_{i}$ is the
natural module for $X_{i}, i=1,2$.}
\item[(4)]{\it $E(X)\cong L_{2}(q), V$ is a direct sum of two natural 
modules.}
\item[(5)] {\it $E(X)\cong L_{2}(q), q=t^{\, 2}, V$ is the orthogonal 
module.}
\item[(6)]{\it $E(X)\cong A_{5}\cong L_{2}(4), V$ is a direct sum of two
orthogonal modules. }
\item[(7)] {\it $E(X)\cong SU_{3}(q), V$ is the natural module.}
\item[(8)]{\it $E(X)\cong X_{1}X_{2}, X_{1}\cong X_{2}\cong L_{2}(q),
V=V_{1}\bigoplus V_{2}, [V_{i},X_{3-i}]=0, [V_{i},X_{i}]=V_{i},
i=1,2$, and $V_{i}$ is orthogonal, a direct sum of two natural
modules, or $q=4$ and $V_{i}$ is a direct sum of two orthogonal
modules.}
\item[(9)]{\it $E(X)=X_{1}X_{2}, X_{1}\cong X_{2}\cong L_{2}(q)$ and 
$V$ is the natural $O_4^{\, +}(q)$-module.}
\item[(10)] {\it $E(X)=X_{1}X_{2}, X_{1}\cong X_{2}\cong L_{3}(2),
V=V_{1}\bigoplus V_{2}, [V_{i}, X_{3-i}]=0, [V_{i},X_{i}]=V_{i},
i=1,2$, and $V_{i}$ is a direct sum of the natural and the dual
module.}
\item[(11)]{\it $E(X)\cong Sp_{4}(q), V$ is a direct sum of the natural 
and the dual module.}
\item[(12)] {\it $E(X)\cong 3\cdot A_{6}, [V,Z(E(X))]=0$ and $V$ is a 
direct sum of the natural $4$--dimensional module and its dual.}
\item[(13)] {\it $E(X)=X_{1}* X_{2}$, $X_{1}\cong X_{2}\cong 3\cdot A_{6}$,
$[V,Z(E(X))]=0$, $V=V_{1}\bigoplus V_{2}$, $[V_{i},X_{3-i}]=0$, $[V_{i},
X_{i}]=V_{i}, i=1,2$ and $V_{i}$ is a direct sum of the $4$--dimensional
module and its dual.}
\item[(14)] {\it $E(X)\cong A_{9}$}
\item[(15)] {\it $X$ is a solvable $\{ 2,5\}$-group. }
\item[(16)] {\it $X$ is a $\{ 2,3\}$-group, $X=PS, P=O_{3}(X)$, and one 
of the following holds}
\begin{itemize}
\item[(i)] {\it $P$ is cyclic and $|V|=16$}
\item[(ii)] {\it $m_{3}(P)=2, S\cong \bZ_{\, 4}$ and $|V|=16$}
\item[(iii)] {\it $m_{3}(P)=2, S\cong D_{\, 8}$ and $|V|=2^{\, 8}$}
\item[(iv)]$P\cong 3^{\, 1+2}, |V|=2^{\, 6}$
\item[(v)]$P\cong 3^{\, 1+4}, |V|=2^{\, 8}, [Z(P),V]=0$
\item[(vi)]{\it $m_{3}(P)=3, X$ is as in (1.8) (iv), $|V|\leq 2^{\, 8}$.}
\end{itemize}
\end{itemize}

\absa
Proof~. Let us go over the list of possibilities for $X$. Suppose
first that $X$ is nonsolvable. Let $E(X)$ be quasisimple.

\indent
(i) $E(X)\cong Sz(q)$: Then for any involution $t\in E(X)$ we have
$|V:\bC_{V} (t)|\geq q^{\, 2}$, so we get $|A|=q$ and then by (1.27)
$V$ is the natural module. This is (3).

\indent
(ii) $E(X)\cong L_{2}(q)$: Let first $A\leq E(X)$. If $|A| > 2$
then by (1.1) $E(X)\langle A,A^{\, g}\rangle $ for suitable $g\in E(X)$. 
As
$|A|\leq q$, we get $|V|\leq q^{\, 4}$. If $|V|=q^{\, 4}$, then
$A\in\mbox{Syl}_{2}(E(X))$ and by (1.27) 
$V=V_{1}\bigoplus V_{2}, V_{i}$ the natural module for $E(X)$. 
This is (4).\\
\indent
So assume $|V| < q^{\, 4}$. Then by (1.32) $V$ is either the natural
module, $q=t^{\, 2}$ and $V$ is the orthogonal module, or $q=t^{\, 3}$
and $V$ is the eight dimensional $GF(t)$-module.\\
\indent
Suppose we have the last case. Then $|V:\bC_{V} (x)|=t^{\, 4}$ for
$x\in A^{\,\sharp}$. Hence $|A|\geq t^{\, 2}$. But for $B\leq
L_{2}(q),\, |B| > t$, we have $|V:\bC_{V} (B)|\geq t^{\, 6}$. This
shows $A\in\mbox{Syl}_{2} (E(X))$. Then $|\bC_{V} (A)|=t$, a
contradiction. So we have (1) or (5).\\
\indent
Suppose now $|A|=2$. Then $|V:\bC_{V} (A)|\leq 4$. As $A$ inverts
some element of order $q+1$, we get $q=4$ and $E(X)\cong L_{2}(4)$.
So we now have (1) or (5) again.\\
\indent
Assume next $A\not\leq E(X)$. Then $q=t^{\, 2}$ and $|A|\leq
2t$. If $|A\cap E(X)|\not= 1$, then as before 
$|V|\leq 16t^{\, 4}=16q^{\, 2}$. If $|V| < q^{\, 4}$ we may argue 
as before. So assume $q=4$
and $|V|=q^{\, 4}$. Then $|A|=4$ and so $V$ is  a direct sum of two
orthogonal modules, which is (6).\\
\indent
Let finally $A\cap E(X)=1$. Then $|A|=2$ and $|V:\bC_{V} (A)|\leq 4$.
As $A$ inverts some element of order $t+1$, we see $t=2$ or $4$. If $t=2$,
then as before we have (1) or (6). Let $t=4$, i.e. 
$E(X)\cong L_{2}(16)$. As $X$ is generated by four conjugates of $A$, 
we get
$|V|=2^{\, 8}$, and so $V$ is the orthogonal module. This is (5).

\indent
(iii) $E(X)\cong (S)L_{3}(q)$:~~Suppose first $A\leq E(X)$. If
$V$ is not irreducible, we get some submodule $W$ of $V$ such that
$|W:\bC_{W} (t)|\leq |A|$, for some $t\in A$. By (1.26) $W$ is a
natural module or a tensor product of a natural and an algebraically
conjugate module. But now by (1.7) also $W^{\, *}$ is involved. This
shows that $W$ is the natural module and we are in (2).

\indent
So assume that $V$ is an irreducible $(S)L_{3}(q)$-module. Suppose
first  $|A|\leq q$. Then (1.26) implies that $V$ is the natural
module or a tensor product of the natural module with an
algebraically conjugate module. As $X$ contains some $x$  inducing a
Dynkin automorphism on $E(X)$ this is impossible.\\
\indent
So we may assume $|A| > q$. Let $P$ be the maximal parabolic in $E(X)$
with $A\leq O_{2}(P)$. Now there is $g\in G, h\in P$ such that
$\langle A^{\, g}, A^{\, gh}\rangle =O^{\, 2^\prime }(P)=P_{1}$. 
Furthermore $A^{\, g}\cap O_{2}(P)\not= 1$. Set $V_{1}=\bC_{V} (O_{2}(P))$. 
Now $|V_{1}:\bC_{V_{1}} (A^{\, g})|\leq 
|A^{\, g}:A^{\, g}\cap O_{2}(P)|^{\, 2}$. By (1.33) $V_{1}$ is an 
irreducible
$P/O_{2}(P)$-module. So $V_{1}$ is a natural or orthogonal
$L_{2}(q)$-module, whence $|V_{1}|=q^{\, 2}$.\\
\indent
Furthermore $O_{2}(P)=\langle A,A^{\, r}\rangle $ for some 
$r\in P, A\cap A^{\,r}\not= 1$, as $|A| > q$ and $|O_{2}(P)|=q^{\, 2}$. 
Now 
$$|V|\,|V:\bC_{V} (A\cap A^{\, r})|\leq 
|V|\, |V:\bC_{V} (A)\bC_{V} (A^{\,r})|=
\frac{|V|^{\, 2}}{|\bC_{V} (A)\bC_{V} (A^{\,r})|} = $$
$$|V:\bC_{V} (A)|^{\, 2}|\bC_{V} (A)\cap\bC_{V} (A^{\, r})|=
q^{\, 2}|V:\bC_{V}(A)|^{\, 2}\leq q^{\, 2}|A|^{\, 4}=
q^{\, 2}(|A|\, |A^{\, r}|)^{\,2}=$$
$$q^{\, 2}\; |AA^{\, r}|^{\, 2}\, |A\cap A^{\, r}|^{\,2}=
q^{6}|A\cap A^{\, r}|^{\, 2}.$$
  
As we may assume $|V:\bC_{V}(A\cap A^{\, r})| > |A\cap A^{\, r}|^{\, 2}$, 
we get $|V| < q^{\, 6}$. In
particular $q > 2$.\\
\indent
Hence we may assume $A=O_{2}(P)$. Application of (1.10) shows
$[O_{2}(P), V]\not\leq V_{1}$. Let $V_{3}\leq V$ such that
$V_{3}/V_{1}$ is a nontrivial irreducible submodule of $V/V_{1}$.
Then $[V,O^{\, 2^\prime }(P)]=V_{3}$. As $[V,O_{2}(P)]\not\leq V_{1}$,
we have $V > V_{3}$ and so $V_{3}/V_{1}\cong O_{2}(P)$ as
$L_{2}(q)$-module. Choose $\omega\in P, o(\omega )=q+1$. There is
some $t$ such that $t^{\, g}\in A$ for some $g\in G$ with 
$\omega^{\,t} =\omega^{\, -1} $. As $[t, V]\leq V_{3}$ and 
$V_{3}=[V,\omega ]$, we get $|[V,t]|=q^{\, 2}$, contradicting (1.26).\\
\indent
So there is no nontrivial irreducible submodule in $V/V_{1}$. Let
$V_{2}$ be maximal such that $[V_{2}, O^{\, 2^\prime }(P)]=V_{1}$. 
We have $\bC_{V_{2}} (t)=V_{1}$ for any $t\in O_{2}(P)^{\, \sharp}$.\\
\indent
Let now $V_{2}\leq V_{3}\leq V, V_{3}/V_{2}$ a nontrivial
irreducible $L_{2}(q)$-module. Then $[O_{2}(P),V_{3}]=V_{1}$, and so
$V_{3}\not= V$. Furthermore there is some $v\in V\setminus V_{3}$
such that 
$O^{\, 2^\prime }(P)=O_{2}(P)\bC_{O^{\, 2^\prime }(P)} (vV_{1})$ and
$O_{2}(P)\cap\bC_{O^{\, 2^\prime }(P)} (vV_{1})=1$. But then
$[O_{2}(P),v]\cong O_{2}(P)$ as $P/O_{2}(P)$-module, and then we have
a natural submodule for $L_{2}(q)$ in $V/V_{1}$, a contradiction.\\
\indent
So assume next $A\not\leq E(X)$. Then $|A\cap E(X)|\leq q$ and
$|A:A\cap E(X)|\leq 4$. Let first $A\cap E(X)\not= 1$. Then by (1.9)
we may assume $|V:\bC_{V} (t)| > q^{\, 2}$ for any $t\in (A\cap
E(X))^{\,\sharp}$.

\indent
Suppose first $|A:A\cap E(X)|=4$. Then $q=t^{\, 2}$ and 
$|A\cap E(X)|\leq t$. So $|V:\bC_{V} (A)|\leq 16t^{\, 2}=16q$. As 
$|V:\bC_{V}(A)| > q^{\, 2}$, we get $q=4, |A|=8$. There is some $x\in A$ 
such that 
$\bC_{X} (x)$ contains $U\cong L_{3}(2)\cdot 2$. There is 
$t\in U^\prime $ such
that $x\sim xt$. As $|V:\bC_{V} (t)|\geq 32$, we get 
$|V:\bC_{V}(x)|\geq 8$. But there is some 
$y\in U\setminus U^\prime , y\in A$.
Hence 
$|\bC_{V} (x):\bC_{\bC_{V} (x)} (\langle y,t\rangle )|\geq 2^{\, 4}$. 
This implies $|V:\bC_{V} (A)|\geq 2^{\, 7}$, a contradiction.\\
\indent
So we have $|A:A\cap E(X)|=2$. Suppose $A$ contains a field $\times$
diagram automorphism. Then $q=t^{\, 2}$ and $|A|\leq 2t$. So
$|V:\bC_{V} (A)|\leq 4t^{\, 2}=4q$. As $|V:\bC_{V} (A)| > q^{\, 2}$, we
get a contradiction. Let $p$ be a primitive prime divisor of 
$q^{\,3}-1, p=9$ if $q=4$. Then there is some $a\in A\setminus E(X)$ 
inverting some $\omega ,o(\omega )=p$. Now we get 
$|V:\bC_{V} (a)|\geq t^{\,3}$, where $q=t^{\, 2}$ or $q=t$, if $q$ is 
not a square.\\
\indent
Let $a$ be the diagram automorphism. Then 
$\bC_{E(X)} (a)\cong L_{2}(q)$. Hence 
$|V:\bC_{V} (A)|\geq qt^{\, 3}=tq^{\, 2}$. So
$4\geq t$. This shows $E(X)\cong L_{3}(2),(S)L_{3}(4)$ or
$(S)L_{3}(16)$. In case $(S)L_{3}(16)$ we have 
$|V:\bC_{V} (a)|=4^{\,3}$ and so $[[V,\omega ],a]=1$. Hence 
$|\bC_{V} (\omega )\cap\bC_{V}(A)|=1$. But this implies 
$|\bC_{V} (\omega )|\leq 16$, i.e. $|V|\leq q^{\, 4}$, a contradiction.\\
\indent
Let $E(X)\cong (S)L_{3}(4)$. Now 
$|\bC_{V} (\omega ):\bC_{\bC_{V}(\omega )} (a)|\leq 2$. Hence 
$|\bC_{V} (\omega )|\leq 2\cdot 4=8$.
This shows $|V|\leq 2^{\, 9}$, again $|V:\bC_{V} (t)|\leq 2^{\, 4}$
for $t\in E(X), t^{\, 2}=1$, a contradiction.\\
\indent
Let $E(X)\cong L_{3}(2)$. Then $|\bC_{V} (\omega )|\leq 4$, so
$|V|\leq 2^{\, 5}$, a contradiction.\\
\indent
So we have $q=t^{\, 2}$ and $A$ contains $a$, a field automorphism.
Hence $\bC_{X} (a)$ contains $U\cong (S)L_{3}(t)\cdot 2$. Again
$|V:\bC_{V} (a)|\geq t^{\, 3}$. As 
$|V:\bC_{V} (A)| > q^{\, 2}=t^{\,4}$, we get $|A\cap U| > t$. Now 
$|\bC_{V} (a):\bC_{V} (A)| > t^{\, 2}$.
Hence $|V:\bC_{V} (A)| > t^{\, 5}=q^{\, 2}t$ so $4 > t$, i.e. $q=4$. Now 
$|V:\bC_{V}(a)|=8$. So $|[V,\omega ]|=2^{\, 6}$ and 
$\bC_{\bC_{V} (\omega )}(A)=1$. This shows 
$|\bC_{V} (\omega )|\leq 2^{\, 3}$. So $|V|\leq 2^{\, 9}$, a 
contradiction.\\
\indent
So we are left with $A\cap E(X)=1$. Let $|A|=2$. Then 
$|V:\bC_{V}(A)|\leq 4$ and so for $t\in E(X), t^{\, 2}=1$, we have 
$|V:\bC_{V}(t)|\leq 16,$ i.e. $\; E(X)\cong L_{3}(2)$ by (1.26). But as 
$a\in A^{\,\sharp}$ inverts an element of order 7, $|V:\bC_{V} (A)| > 4$, 
a contradiction. So we have $|A|=4$ and $q=t^{\, 2}$. There is some
$a\in A^{\,\sharp}$ with $|V:\bC_{V} (a)|\geq t^{\, 3}$. This shows
$t=2$. But then for this $a$ we may assume 
$\bC_{X} (a)\geq U\cong L_{3}(2)\cdot 2$ and 
$|\bC_{V} (a):\bC_{V} (A)|=2$. But there are now
$a\in (U\cap A)^{\,\sharp}$ inducing transvections on 
$\bC_{V}(a)$.

\indent
(iv)~~$E(X)\cong (S)U_{3}(q)$: Then $|A|\leq q$. So 
$|V:\bC_{V}(a)|\leq q^{\, 2}$ for any $a\in A^{\,\sharp}$. Now (1.26) 
yields (7).

\indent
(v)~~$E(X)=Sp_{4}(q)$, including $E(X)\cong A_{6}$: Suppose first
$A\leq E(X)$. If $V$ is reducible, then we have some submodule
$W$ with $|V:\bC_{V} (t)|\leq q^{\, 3}$ for some $t\in A^{\,\sharp}$.
Now by (1.26) we may assume that $W$ is the natural module. As $X$
contains some diagram automorphism also the dual module is involved.
Now we have (11).\\
\indent
So we may assume  that $V$ is irreducible as a $Sp_{4}(q)$-module
and $|V:\bC_{V} (a)| > q^{\, 3}$ for any $a\in A^{\,\sharp}$. In
particular $|A| > q$. Suppose $|A|\leq q^{\, 2}$. Let $P$ be the
parabolic with $A\not\leq O_{2}(P)$. As 
$[O^{\, 2^\prime }(P),\bC_{V}(O_{2}(P))]\not= 0$, we see that 
$|\bC_{V} (A\cap O_{2}(P)):\bC_{V}(A)|\geq q$. Now 
$|V:\bC_{V} (A\cap O_{2}(P))| > q^{\, 3}$ and so 
$|V:\bC_{V}(A)| > q^{\, 4}\geq |A|^{\, 2}$, a contradiction.\\
\indent
So we have $|A| > q^{\, 2}$. Let now $P_{1}$ be the parabolic with
$A\leq O_{2}(P_{1})$. Then for some  $g\in P_{1},
O_{2}(P_{1})=\langle A,A^{\, g}\rangle $. Now choose $A$ with $|A|$ 
mini\-mal. As in (iii) we see 
$|V|\, |V:\bC_{V} (A\cap A^{\, g})|\leq q^{\, 8}|A\cap A^{\, g}|^{\, 2}$. 
Hence $|V| < q^{\, 8}$. So also $|V:\bC_{V}(O_{2}(P_{1}))|\leq 
|O_{2}(P_{1})|^{\, 2}$. Choose $a\in Z(O_{2}(P_{1}))^{\,\sharp}$. Then 
$|[V,a]| > q^{\, 3}$. As $\bC_{V}(P_{1})=0$ we get that $[V,a]$ contains 
a sub\-module $[[V,a],O^{\,2^\prime }(P_{1})]=W$. Now as 
$[O_{2}(P_{1}),[V,a]]\not= 0$ by (1.10) we
have $[O_{2}(P_{1}),[V,a]]=W$ and so 
$W\cong O_{2} (P_{1})/Z(O^{\,2^\prime }(P_{1}))$, i.e. $W$ is the 
natural module. As $W=\bC_{[V,a]}(O_{2}(P_{1}))$, we get 
$|[V,a]|\leq q^{\, 3}$, a contradiction.\\
\indent
So we have $A\not\leq E(X)$. Assume first $A\cap E(X)\not= 1$.
Recall $|V:\bC_{V} (t)| > q^{\, 3}$ for any 
$a\in A\cap E(X), a\not= 1$. Suppose $|A\cap E(X)|\leq q$. Then 
$|V:\bC_{V} (A)|\leq 16 q^{\,2}, 4q^{\, 2}$, if $q$ is not a square. 
Hence $q\leq 4$.\\
\indent
Let $q=4$, then $|A|\leq 8$, so $|V:\bC_{V} (A)|\leq 64=q^{\, 3}$, a
contradiction. So let $q=2, |A|=4$. But now inspection of the
$A_{6}$-modules just implies (11).\\
\indent
So assume $|A\cap E(X)| > q$. Then $q=t^{\, 2}$ and $A$ contains a
field automorphism. So $|A|\leq 2t^{\, 3}$. Hence 
$|V:\bC_{V}(A)|\leq 4t^{\, 6}=4q^{\, 3}$. Now as there is some 
$x\in E(X)^{\,\sharp}$ such that $a\sim ax, a$ the field automorphism, 
we get $|V:\bC_{V} (a)|\geq 2t^{\, 3}$. So 
$|\bC_{V} (a):\bC_{V}(A)|\leq 2t^{\, 3}$ and by induction just the 
natural and its dual
is involved. As $|A\cap E(X)|=t^{\, 3}$, we then get 
$|\bC_{V} (a):\bC_{V}(A)|\geq t^{\, 5}$, a contradiction.\\
\indent
So we are left with $A\cap E(X)=1$. Hence $|A|\leq 4$. For 
$u\in E(X)^{\,\sharp}, u^{\, 2}=1$, we have 
$|V:\bC_{V} (u)|\leq 2^{\, 8}$.
This implies $q\leq 4$. If $q=2$, we have $E(X)\cong A_{6}, |A|=2$.
Now inspection of the $A_{6}$-modules gives a contradiction.\\
\indent
So let $q=4, |A|=4$. Now $|V:\bC_{V} (a)|=16$ for any 
$a\in A^{\,\sharp}$. This yields that 
$E(X)A=\langle \bC_{E(X)} (a)|a\in A^{\,\sharp} \rangle $ acts on 
$\bC_{V} (A)$, a contradiction.

\indent
(vi)~~$E(X)\cong L_{2}(r), r\geq 11, (S)U_{3}(r), (S)L_{3}(r), r$
odd: Then in any case $|A|\leq 8$ and so $|V:\bC_{V} (A)|\leq 64$.
Application of (1.29) shows $E(X)\cong L_{2}(p), p$ prime,
$L_{2}(25)$, or $L_{3}(3)$. Let $E(X)\cong L_{2}(25)$ or $L_{3}(3)$
and suppose there is some $a\in A^{\,\sharp}$ inverting some element
of order 13. Then $|V:\bC_{V} (a)|=64$ and so $[\bC_{V} (a),A]=0$.
Hence there is a quadratic fours group in $A$, contradicting (1.12).
So in any case we have $|A|=4$. Now $E(X)\not\cong L_{2}(25)$. As $A$
cannot act quadratically we get $|V:\bC_{V} (a)|\leq 8$ for 
$a\in A^{\,\sharp}$, which contradicts (1.30). Let finally 
$E(X)\cong PSp_{4}(p),p\geq 5$. Then $|A|\leq 16$. Hence 
$|V:\bC_{V} (A)|\leq 2^{\, 8}$. By (1.29) $\bC_{V} (A)=\bC_{V} (t)$ for 
any $t\in A^{\,\sharp}$. Hence $[V,A,A]=0$. But this contradicts (1.12).

\indent
(vii)~~$E(X)\cong 3\cdot A_{6}$ and $[V,Z(E(X))]\not= 0$: Hence the
irreducible modules involved are of dimension 1, 6, 6 or 18. If one
6-dimensional module is involved than also the other 6-dimensional
module is involved, as there is a diagram automorphism $x$ in $X$.
Hence $|V:\bC_{V} (a)|\geq 16$ for any $a\in A^{\,\sharp}$, i.e.
$|A|\geq 4$. But for a fours group we have $|V:\bC_{V} (A)|\geq 64$.
So we have $|A|=8$. Now some $a\in A^{\,\sharp}$ inverts $Z(E(X))$.
Hence $|V:\bC_{V} (a)|=64$ and $|V|=2^{\, 12}$ is a direct sum of the
two 6-dimensional modules. But now $\bC_{V} (a)=\bC_{V} (A)$, a
contradiction as $Z(E(X))$ acts on $\bC_{V} (A\cap E(X))$.

\indent
(viii)~~From now on let $E(X)=X_{1}X_{2}$. Let $X_{1}\cong L_{2}(q),
q$ even. Suppose first $[A,X_{1}]=X_{1}$. If $\bC_{A} (X_{1})=1$, then
we have that as a $X_{1}$-module $V$ involves just modules which are
as in (1), (4), (5) or (6). Suppose $[[V,X_{1}],X_{2}]=0$. Then we
get $V=V_{1}\bigoplus V_{2}, [V_{i},X_{i}]=V_{i}$ and
$[V_{i},X_{3-i}]=0, i=1,2$. So we have (8).\\
\indent
Assume now $[[V,X_{1}], X_{2}]\not= 0$. Then there are at least two
nontrivial irreducible $X_{1}$-modules involved. 
This implies that both are natural and so 
$[V,X_{1}]=V_{1}\bigoplus V_{2}, V_{i}$ natural $X_{i}$-modules, 
$i=1,2$. Hence $V$ is the $O_4^{\, +}(q)$-module (9).\\
\indent
So assume now $B=\bC_{A} (X_{1})\not= 1$. Then $X_{1}$ acts on
$\bC_{V} (B)$. By the $A\times B$-Lemma we have 
$[X_{1},\bC_{V}(B)]\not= 0$. If $|V:\bC_{V} (B)|\leq |B|^{\, 2}$, 
we may argue as before.\\
\indent
So suppose $|V:\bC_{V} (B)| > |B|^{\, 2}$. Then 
$|\bC_{V} (B):\bC_{V}(A)| < |A/B|^{\, 2}$ and so the action of $X_{1}$ 
on $\bC_{V} (B)$ is
as in (1), (4), (5) or (6). If $[X_{2},[V,X_{1}]]=0$, we have the
assertion. Now there is just one irreducible nontrivial
$X_{1}$-module in $\bC_{V} (B)$. As $|V:\bC_{V} (B)| < q^{\, 4}$, we
get that $[V,X_{1}]$ involves at most two nontrivial irreducible
$X_{1}$-modules and so again $V$ is as in (9).\\
\indent
So let now $a\in A$ with $X_{1}^{\, a}=X_{2}$. This shows 
$|A|\leq 2q$. As $a$ inverts some $\omega \in X_{1}, o(\omega )=q+1$, we 
get
$|V:\bC_{V} (a)|\geq q$. Let $|A|=2$, then we get $q=4$, and for
$1\not= y\in\bC_{E(X)} (a), y^{\, 2}=1$ we get 
$|V:\bC_{V} (y)|\leq 16=q^{\,2}$. \\
\indent
Assume that $[V,X_{1}]$ involves at least 3 nontrivial irreducible
submodules for $X_{1}$. Then for 
$1\not= y\in\bC_{E(X)} (a), y^{\,2}=1$, we get 
$|V:\bC_{V} (y)|\geq q^{\, 3}$. Suppose 
$A\cap E(X)\not= 1$, then $|V:\bC_{V} (y)|\leq 4q^{\, 2}$ and so $q=4$. 
Suppose $A\cap E(X)=1$, then $|A|\leq 4$ and so 
$|V:\bC_{V} (y)|\leq 2^{\, 8}$ and again $q=4$.\\
\indent
So we have: If $[V,X_{1}]$ involves at least 3 nontrivial irreducible
modules, then $q=4, |A|\geq 4$.\\
\indent
Now suppose $[[V_{1},X_{1}],X_{2}]=0$. Then as 
$[V_{1},X_{1}]^{\,a}=[V_{2},X_{2}]$, we get that there are at most two 
$X_{1}$-modules
involved. Hence if there are at least 3 $X_{1}$-modules involved
$[[V_{1},X_{1}],X_{2}]\not= 0$. Suppose these modules are natural
modules. Then $|A|=8$ and $A$ acts quadratically. But now by (1.11)
we have a contradiction. So these modules are all orthogonal modules 
and then we have four of them. Now again $A$ acts quadratically,
contradicting (1.11) again.\\
\indent
So we have that there are at most two nontrivial irreducible
$X_{1}$-modules involved and $[[V_{1},X_{1}],X_{2}]\not= 0$. But then
there are exactly two such modules involved and so again we have (9).
So let finally $[[V_{1},X_{1}],X_{2}]= 0$. Then there is exactly one
nontrivial irreducible $X_{1}$-module involved. As 
$[V,X_{1}]^{\,a}=[V,X_{2}]$, we get $|V:\bC_{V} (a)|\geq q^{\, 2}$.\\
\indent
Furthermore for $b\in A\cap N(X_{1})$, we have 
$|V:\bC_{V} (b)|\geq t, t^{\, 2}=q$ or $t=q$. So we have 
$|V:\bC_{V} (b)|\geq tq^{\, 2}$
and then $t\leq 4$. We get $t=q=4$. As $\bC_{V} (E(X))=0$, we see
$|[V,X_{1}]|=16$ and so we have (8).

\indent
(ix)~~Let $E(X)=X_{1}X_{2}, X_{i}\cong Sz(q), i=1,2$. Let 
$[A,X_{1}]=X_{1}$. 
Then as in (viii) we get that 
$V=V_{1}\bigoplus V_{2}, [V_{i}, X_{i}]=V_{i}, [V_{i},X_{3-i}]=0$. 
So we have (3). If
$X_{1}^{\, a}=X_{2}$, for some $a\in A$ we get $|A|\leq 2q$. So 
$|V:\bC_{V}(b)|\leq 4q^{\, 2} < q^{\, 3}$ for $b\in A\cap E(X)$. 
Suppose $A\cap E(X)\not= 1$, then $V$ involves exactly one irreducible
$X_{1}$-module which is nontrivial. So again we have (3).\\
\indent
Let $A\cap E(X)=1$, then $|A|=2$ and $|V:\bC_{V} (a)|\leq 4$. But $a$
inverts some $\omega ,o(\omega )=q-1$, a contradiction.

\indent
(x)~~Let $E(X)=X_{1}X_{2}, X_{i}\cong L_{2}(r), r$ odd, $r\geq 11,
i=1,2$. Suppose $[A,X_{1}]=X_{1}$. If $[[V,X_{1}], X_{2}]=0$ we get
the same contradiction as in (vi). So assume 
$[[V,X_{1}],X_{2}]\not= 0$. Now we get that there are at least 5 
nontrivial irreducible
$X_{1}$-modules involved. This shows that $A$ has to induce
transvections on these modules, contradicting (1.30).\\
\indent
So let $a\in A, X_{1}^{\, a}=X_{2}$. Then $|A|\leq 16$. Now
$|[V,a]|\geq 8$. Let $B=A\cap N(X_{1})$. Then 
$|\bC_{V} (a):\bC_{V}(B)| < |B|^{\, 2}$, a contradiction.

\indent
(xi)~~Let now $E(X)=X_{1}X_{2}, X_{1}\cong X_{2}\cong 3\cdot A_{6},
L_{3}(2), SL_{3}(4)$ or $SU_{3}(8)$. Suppose first $[A,X_{1}]=X_{1}$.
Let $B=\bC_{A} (X_{1})=1$. Then as an $X_{1}$-module we get that $V$
is as in (2), (7) or (12). In any case $[X_{2},[V,X_{1}]]=0$. So we
get that $X_{1}\cong 3\cdot A_{6}$ and we have (12) or 
$X_{1}\cong L_{3}(2) $ and we have (2). Furthermore we get (13) or (10).\\
\indent
So let $B\not= 1$. Then $[X_{1},\bC_{V} (B)]\not= 0$. We may assume 
$|V:\bC_{V}(B)| > |B|^{\, 2}$ and so 
$|\bC_{V} (B):\bC_{\bC_{V} (B)}(A/B)| < |A/B|^{\, 2}$. Now it is not 
possible to apply our results just
established as there is no reason why the diagram
automorphism should act on $\bC_{V} (B)$. We just get 
$X_{1} \not\cong SU_{3}(8)$.\\
\indent
Now in any case $[[V,X_{1}], X_{2}]\not= 0$. So $[V,X_{1}]$ involves
more than one irreducible nontrivial $X_{1}$-module. Suppose that
$[V,X_{1}]$ involves just the modules from (1.26) i.e. natural
modules or a 6-dimensional module in case $X_{1}\cong 3\cdot
A_{6}$.\\
\indent
Let $X_{1}\cong L_{3}(2)$, then we have at least 3 natural and 3 dual
modules. So $|V:\bC_{V} (b)|\geq 2^{\, 6}$. This shows 
$|A|=2^{\,4}$. But then $|V:\bC_{V} (A)|\geq 2^{\, 9}$, a contradiction. 
\\
\indent
Let
$X_{1}\cong SL_{3}(4)$. Then again there are six modules involved and
so $|V:\bC_{V} (b)|\geq 2^{\, 12}$ and then $|A|\geq 2^{\, 6}$. Hence
$|A\cap X_{1}|\geq 4$. If $|A\cap X_{1}| > 4$, we get 
$|V:\bC_{V}(A\cap X_{1})|\geq 2^{\, 18}$, a contradiction. But now 
$|A\cap E(X)|\leq 16$ and so we get a contradiction again.\\
\indent
Let $X_{1}\cong 3\cdot A_{6}$. Then we must have two 6-dimension
modules and their duals or four natural modules and their duals. In both
cases $|V:\bC_{V} (b)|\geq 2^{\, 8}$. So $|A|\geq 2^{\, 4}$. Now
$|A\cap E(X)|\not= 1$ and so $|V:\bC_{V} (b)|\geq 2^{\, 16}$ in the
second case, a contradiction. Hence we have the first case. Now
$[Z(E(X)),V]=V$ and so there are three 6-dimensional modules and their
duals involved. Hence $|V:\bC_{V} (b)|\geq 2^{\, 12}$ and so 
$|A\cap X_{1}|\geq 4$. But now $|V:\bC_{V} (A\cap X_{1})|\geq 2^{\, 18}$, 
a contradiction.\\
\indent
So we have that $[V,X_{1}]$ also involves modules not from (1.26). In
particular for $b\in X_{2}\cap A$, we have 
$|V:\bC_{V} (b)| > 2^{\,2m_{2}(X_{2})}$. Suppose $X_{2}\cap A\not= 1$. 
Then there is some
$b\in X_{2}\cap A$ centralized by $X_{1}$ and the diagram
automorphism. Now we see that $\bC_{V} (b)$ just involves modules from
(1.26). But then $V$ just involves modules from (1.26) as $X_{1}$
acts on $[V,b]$ as on $V/\bC_{V} (b)$. So we have $A\cap X_{2}=1$. 
Now the same applies for $X_{1}$.\\
\indent
Let $X_{1}\cong L_{3}(2)$. Then $|A|\leq 4$. So $|A|=4$. But then
$A$ contains some $a$ inverting a group of order $7^{\, 2}$, i.e.
$|V:\bC_{V} (a)|\geq 64$, a contradiction.\\
\indent
Let $X_{1}\cong 3\cdot A_{6}$, then $|A|\leq 4$. Furthermore $|A|=4$,
otherwise for some $b\in X_{1},o(b)=2$, we have 
$|V:\bC_{V} (b)|\leq 16$, a contradiction. So there is some 
$a\in A^{\,\sharp}$ inverting
a group of order 9 in $E(X)$. Hence there is some 
$\zeta\in X_{1},o(\zeta )=3, |[V,\zeta ]|\leq 16$. But now inspection 
of the
$A_{6}$--modules and $3\cdot A_{6}$--modules gives that there are just
$X_{1}$-modules from (1.26) involved. \\
\indent
Let $X_{1}\cong SL_{3}(4)$. Now $|A|\leq 16$. By (1.26) for 
$b\in X_{1},o(b)=2$, we have $|V:\bC_{V} (b)|\geq 2^{\, 15}$. This gives
$|A|=16$ and $|V:\bC_{V} (a)|=2^{\, 8}$ for any $a\in A^{\, \sharp}$. In
particular $\bC_{V} (a)=\bC_{V} (A)$ for any $a\in A^{\,\sharp}$. But
there is some fours group in $A$ acting nontrivially on a group of
order $7^{\, 2}$, a contradiction.\\
\indent
Now we have some $a\in A$ with $X_{1}^{\, a}=X_{2}$. Let 
$X_{1}\cong L_{3}(2)$, then for some $b\in E(X),b^{\, 2}=1$, we have 
$|V:\bC_{V}(b)|\leq 2^{\, 6}$. This shows that $[V,X_{1}]$ either is 
irreducible or involves at most three natural and dual modules.\\
\indent
Suppose $[[V,X_{1}],X_{2}]=0$. Then $|[V,X_{1}]|\leq 2^{\, 6}$ and so
we have (10). Assume $[[V,X_{1}],X_{2}]\not= 0$. Then we have that
$[V,X_{1}]$ involves three natural and three dual modules. As
$N_{A}(X_{2})$ induces a fours group on $[V,X_{1}]$, we see that 
$|V:\bC_{V}(A\cap X_{1})|\geq 2^{\, 9}$, a contradiction.\\
\indent
Let $X_{1}\cong 3\cdot A_{6}$. Then for $b$ as above 
$|V:\bC_{V}(b)|\leq 2^{\, 8}$. If $[V,X_{1},X_{2}]=0$, then as 
$|V:\bC_{V}(a)|\leq 2^{\, 8}$ we have (12). Let 
$[[V,X_{1}],X_{2}]\not= 0$. Then
inspection of the $A_{6}$--modules and $3\cdot A_{6}$--modules implies 
that either
natural $A_{6}$--modules and their duals, or 6-dimensional 
$3\cdot A_{6}$-modules and their duals are involved. In any case 
$A\cap E(X)=1$ and so $|A|=4$. But then for $b\in N(X_{1})\cap A$, we 
have $|V:\bC_{V} (b)|\geq 2^{\, 6}$, a contradiction.\\
\indent
Let $X_{1}\cong SL_{3}(4)$. Then $|V:\bC_{V} (b)|\leq 2^{\, 10}$. Let
$[V,X_{1},X_{2}]=0$. Then $|[V,X_{1}]|\leq 2^{\, 10}$. But now by
(1.26) and (1.14) we get a contradiction.\\
\indent
So we have $[V,X_{1},X_{2}]\not= 0$. Now $[V,X_{1}]$ involves more
than one nontrivial irreducible $(S)L_{3}(4)$-module. Suppose that
for any $b$ we have $|V:\bC_{V} (b)|=2^{\, 10}$. Hence $|A\cap
N(X_{1})|=16$. But now $A\cap N(X_{1})$ acts quadratically on $V$ and
so by (1.10) we just get natural modules involved. So in any case we
just have the modules from (1.26) involved in $V$. Now we have three
natural and their duals involved. But then 
$|V:\bC_{V} (b)|\geq 2^{\,12}$, a contradiction.\\
\indent
Let finally  $X_{1}\cong SU_{3}(8)$. Now $|V:\bC_{V} (b)|\leq 2^{\, 8}$
for $b\in E(X),o(b)=2$. So by (1.26) there is just one nontrivial
irreducible module involved in $[V,X_{1}]$ which is the natural
module. But now $[V,X_{1},X_{2}]=0$ contra\-dic\-ting $[Z(E(X)),V]=V$.

\indent
(xii)~~~Let now $F(X)$ be a $p$-group and still $X$ be nonsolvable.
Now $|A|\leq 4$ and then $|V:\bC_{V} (A)|\leq 16$. As $A$ acts on an
elementary abelian group of order $p^{\, 3}$, we get $p=5$. Now
$|[V,F(X)]|=2^{\, 12}$ and so $X\leq GL_{12}(2)$. But 
$5^{\,4}\not\Big |\, |GL_{12}(2)|$.

\indent
(xiii)~~~Now let $X$ be a solvable $\{ 2,p\}$-group. As there is
$U\cong D_{2p}\times\ldots\times D_{2p}$ in $X$ with 
$A\in\mbox{Syl}_{2}(U)$ by ~\cite[(24.1)]{GoLyS}, we see that $p\in\{ 3,5\}$. 

\indent
(xiv)~~~Let $X$ be a $\{ 2,3\}$-group. Let $O_{3}(X)$ be
cyclic. Then $|A|=2$ and so $|V:\bC_{V} (A)|\leq 4$, hence
$|[V,O_{3}(X)]|\leq 16$. This is (16) (i). Let next
$m_{3}(O_{3}(X))=2$. Then $|A|\leq 4$. By (1.8) in any case $O_{3}(X)$
is generated by two elements. So $|[V,O_{3}(X)]|\leq 2^{\, 8}$.
Suppose $|[V,O_{3}(X)]|=16$. Then $X/\bC_{X} (V)\lesssim A_{8}$. So
either $V$ is an $F$-module or we have (16) (ii).\\
\indent
Let $|[V,O_{3}(X)]|=2^{\, 6}$. Let $|A|=4$. Now again 
$X/\bC_{X}(V)\lesssim GL_{6}(2)$. Suppose 
$O_{3}(X)/\bC_{O_{3}(X)} (V)\cong \bZ_{3}\times \bZ_{3}$. Further 
$S\cong D_{8}$ or $S_{16}$. Now we have
that for 3 of the 4 subgroups $U$ of $O_{3}(X)/\bC_{O_{3}(X)} (V)$ of
order 3, $|\bC_{V}(U)|=4$ while for the forth $\bC_{V} (U)=1$. But
$D_{8}$ induces 2 orbits of length 2 on these subgroups, while $S_{16}$
induces just one orbit of length 4.\\
\indent
Let $|A|=2$ and $O_{3}(X)$ be abelian. Now as before we see that we
cannot  have the proper number of orbits.\\
\indent
So assume $|V|=2^{\, 8}$. Let again 
$O_{3}(X)/\bC_{O_{3}(X)} (V)\cong \bZ_{3}\times \bZ_{3}$. Now we either 
have $|\bC_{V} (u)|=4$ for any $u$ which
induces a group of order 3 or we have two such subgroups with 
$|\bC_{V}(u)|=16$ and two with $\bC_{V} (u)=1$. Assume the latter. Then
$S\cong D_{8}$ or $\bZ_{4}$ and $|A|=2$. But $A$ inverts
some fixed point free element of order 3, a contradiction. So we may
assume that we have the former. Furthermore $|A|=4$ and
$S\cong S_{16}$. Then for some $a\in A^{\,\sharp}$ we have 
$\bC_{V} (a)=\bC_{V}(A)$. But now $A=\langle a,b\rangle $ and $b$ 
inverts some element $c$ of 
order 8. So $[c^{\, 2},\bC_{V} (a)]=0$, a contradiction. So we have (16)
(iii).

\indent
(xv)~~~Let now $O_{3}(X)\cong 3^{\, 1+4}$. Let
$V=[V,Z(O_{3}(X))]$. Then we get $|V|\geq 2^{\, 18}$. As
$m_{2}(X)\leq 3$ (a Sylow 2-subgroup of $X$ is contained in
Aut$(Sp_{4}(3))$), we get that $[A,Z(X)]=1$. So $|A|\leq 4$. Let
$a\in A,\rho\in O_{3}(X)\setminus Z(X),\rho^{\, a} =\rho^{\,-1}$. Then 
$|V:\bC_{V} (\rho )|\leq 2^{\, 8}$. So  
$|V:\bC_{V} (\rho)\cap \bC_{V} (Z(O_{3}(X)))|\leq 2^{\, 16}$, a 
contradiction.\\
\indent
So we have $[V,Z(O_{3}(X))]=0$. Now by irreducible action of $S$ we
get $|V|=2^{\, 8}$ or $|V|\geq 2^{\, 16}$. The former is (16) (v). So
assume the latter. Let $t\in A$ such that 
$\bC_{O_{3}(X)}(t)\leq Z(O_{3}(X))$. Then 
$|V:\bC_{V} (t)|\geq 2^{\, 8}$. But
as $|A|\leq 2^{\, 3}$ and so $|V:\bC_{V} (A)|\leq 64$, this is
impossible. So we have $|A|\leq 4$ and then $|V:\bC_{V} (A)|\leq 16$.
Let $a\in A^{\,\sharp}$, then $a\sim at$, where 
$\bC_{O_{3}(X)}(t)\leq Z(O_{3}(X))$. This shows 
$\bC_{V} (a)=\bC_{V} (A)$, for all
$a\in A^{\,\sharp}$. But as there is some 
$\Sigma_{3}\times\Sigma_{3} \cong U$
induced on $V$, with $A\in\mbox{Syl}_{2} (U)$, this is impossible.

\indent
(xvi)~~~Let $m_{3}(O_{3}(X))=3$. By (1.8) $m_{2}(S)\leq 2$. Then
for $t\in Z(S)$, we get $|V:\bC_{V} (t)|\leq 16$ and so 
$|V|\leq 2^{\, 8}$.

\absa
{\bf (1.36)~Corollary.~}{\it Let $X,V$ be as in (1.35). But assume
now additionally that $|V:\bC_{V} (A)| < |A|^{\, 2}$. Then one of the
following holds:
\begin{itemize}
\item[$(1)$]$V$ is an $F$-module.
\item[$(2)$]$E(X)\cong (S)L_{3}(q),q$ even, and $V$ is a direct sum of
the natural and the dual module.
\item[$(3)$]$E(X)\cong L_{2}(q), q=t^{\, 2}, q$ even, $V$ is the
orthogonal module.
\item[$(4)$]$E(X)\cong Sp_{4}(q), q$ even, or $3\cdot A_{6}, V$ is a
direct sum of the natural module and the conjugate $4$-dimensional
module. 
\item[$(5)$]$E(X)\cong A_{9}$.
\item[$(6)$]$E(X)=X_{1}X_{2}\cong L_{2}(q)\times L_{2}(q), q=t^{\, 2}, q$ even,
$V=V_{1}\bigoplus V_{2}, [X_{i},V]=V_{i}, [X_{3-i}, V_{i}]=0, i=1,2,
V_{i}$ is the orthogonal $X_{i}$-module, $i=1,2$.
\item[$(7)$]$E(X)\cong L_{2}(q)\times L_{2}(q), q$ even, $V$ is the
natural $O^{\, +}(4,q)$-module.
\item[$(8)$]$E(X)=X_{1}X_{2}, X_{i}\cong L_{3}(2), V=V_{1}\bigoplus
V_{2}, [V_{i},X_{i}]=V_{i}, [V_{i},X_{3-i}]=0, V_{i}$ is a direct sum
of the natural and the dual $X_{i}$-module, $i=1,2$.
\item[$(9)$]$E(X)=X_{1}X_{2}, X_{i}\cong 3\cdot A_{6}, [Z(E(X)),V]=0,
V=V_{1}\bigoplus V_{2}, [V_{i},X_{i}]=V_{i}, [V_{i},X_{3-i}]=0, V_{i}$
is a direct sum of the two $4$-dimensional $A_{6}$-modules, $i=1,2$.
\item[$(10)$]$X$ is solvable, $F(X)\cong 3^{\, 1+4}, |V|=2^{\, 8}$.
\end{itemize}}

\absa
Proof~. Inspection of the list in (1.35) gives the assertion up to
the case in (1.35) (16) (vi). Suppose that there is some $A$ with
$|A|=4$ and $|V:\bC_{V} (A)|=8$. Then $A\cap Z(S)\not= 1$. Choose
$t\in A\cap Z(S), t\not= 1$. Then 
$\bC_{O_{3}(X) /\Phi (O_{3}(X))} (t)=1$. As $|V:\bC_{V} (t)|\leq 8$, 
we get $|O_{3}(X):\Phi (O_{3}(X)|\leq 27$. As a 2-group cannot act
irreducibly on an elementary abelian group of order 27, we get
$|O_{3}(x):\Phi (O_{3}(X))|=9$. Now $|[V,O_{3}(X)]|\leq 2^{\, 6}$
and so $O_{3}(X)/\bC_{O_{3}(X)} (V)\cong \bZ_{3}\times \bZ_{3}$ or 
$3^{\,1+2}$. But in both cases we do not have $A$ with 
$|V:\bC_{V} (A)|=8$ and $V$ is not an $F$-module.

\absa
{\bf (1.37)~Lemma.~}{\it Let $G=G_{2}(q), q$ even, $V$ be an
irreducible faithful $F$-module, $A\leq G$ be an offending
subgroup. Then $|A|=q^{\, 3}$.}

\absa 
Proof~. By (1.18) $V$ is the natural module, $V=V(\lambda_{1} )$. By
(1.23) the minimal codimension is 2. Assume $|A|=q^{\, 2}$. Then 
$\bC_{V}(a)=\bC_{V} (A)$ for all $a\in A^{\,\sharp}$.

Let $P$ be the minimal parabolic with $|\bC_{V} (O^{\, 2^\prime }(P))|=q$.
We have that there is no $g\in P$ with 
$\langle A,A^{\, g}\rangle =O^{\, 2^\prime }(P)$. Hence by (1.1) 
$A\cap Z(O_{2}(P))\not= 1$. Let $g\in O_{2}(P)$. Then 
$A\cap A^{\, g}\not= 1$. Hence 
$\bC_{V} (A^{\, g})=\bC_{V} (A\cap A^{\, g})=\bC_{V}(A)$. 
Now either $A=Z(O_{2}(P))$, or there is some 
$x\in \langle A^{\,O_{2}(P)}\rangle $, with $ x^{\, 2}\not= 1$. Then 
$|\bC_{V} (x)|=q^{\, 4}$ and 
so $\bC_{V} (x)=\bC_{V} (x^{\, 2})$, a contradiction. This now implies 
that $A=Z(O_{2}(P))$. But
$$
G=\langle \bC_{G} (a)|a\in Z(O_{2}(P))^{\,\sharp} \rangle 
$$ 
and $\bC_{G} (a)$ acts on $\bC_{V} (a)$. This implies that 
$\bC_{V}(a)$ is $G$-invariant, a contradiction.

\absa 
{\bf (1.38)~Lemma.~}{\it Let $G=G(q)$ be a Lie group in even
characteristic, $V$ be a faithful $GF(2)G$-module, and $t\in G$ be an
involution with $|V:\bC_{V} (t)|\leq 2^{\, m_{2}(G)}$. Let $\ell$ be the
Lie rank of $G$ and suppose that $V$ involves all $V(\lambda )$,
$\lambda$
a fundamental weight, but maybe not $V(\lambda_{1} )$ and 
$V(\lambda_{2} )$.
Furthermore assume that there is some irreducible nontrivial module
involved in $V$ which is not of type $V(\lambda ),\lambda$ a 
fundamental weight. Then $G\cong (S)L_{2}(q)$ or $(S)L_{3}(q)$.}

\absa
Proof~. By (1.22) $G\cong (S)L_{n}(q), (S)U_{n}(q), n\geq 6, Sp_{2n}(q),
n\geq 7$ or $\Omega_{2n}^{\,\pm} (q), n\geq 8$. Now application of
(1.23) shows $G\cong SL_{n}(q), n\leq 7$ or $(S)U_{6}(q)$.\\
\indent
If $G\cong (S)U_{6}(q)$, then $V(\lambda_{2} )$ is involved. But by
(1.23) 
$|V(\lambda_{2} ):\bC_{V(\lambda_{2} )} (t)|\geq q^{\, 8}$. 
Hence $V$ just involves $V(\lambda_{2} )$, as $G$ contains no
elementary abelian subgroup of order greater than $q^{\, 9}$ by (1.23).\\
\indent
Let $G\cong (S)L_{7}(q)$. Then 
$V(\lambda_{2} ),V(\lambda_{3}),V(\lambda_{4} ),V(\lambda_{5} )$ 
are involved. But by (1.23)
$|V(\lambda_{3} ):\bC_{V(\lambda_{3} )} (t)|\geq q^{\, 10}\leq
|V(\lambda_{4} ):\bC_{V(\lambda_{4} )} (t)|$ and there are no
elementary abelian subgroups of order greater than $q^{\, 15}$.\\
\indent
Let $G\cong (S)L_{6}(q)$. Then 
$V(\lambda_{2} ), V(\lambda_{3}),V(\lambda_{4} )$ are involved. Now 
$|V(\lambda_{2}):\bC_{V(\lambda_{2} )} (t)|\geq q^{\, 4}\leq 
|V(\lambda_{4}):\bC_{V(\lambda_{4} )} (t)|$ and 
$|V(\lambda_{3}):\bC_{V(\lambda_{3} )} (t)|\geq q^{\, 6}$ by (1.23). 
But by (1.23) there are 
no elementary abelian subgroups of order greater than $q^{\, 9}$.\\
\indent
Let $G\cong (S)L_{5}(q)$. Then $V(\lambda_{2} )$ and $V(\lambda_{3} )$
are involved. By (1.23) 
$|V(\lambda_{2} ):\bC_{V(\lambda_{2} )}(t)|\geq q^{\, 3}\leq 
|V(\lambda_{3} ):\bC_{V(\lambda_{3} )} (t)|$.
As there are no elementary abelian subgroups of order greater than 
$q^{\, 6}$ in $G$ by (1.23), we see that just $V(\lambda_{2} )$ and 
$V(\lambda_{3} )$
are involved, a contradiction.\\
\indent
Let finally $G\cong (S)L_{4}(q)$. Then $V(\lambda_{2} )$ is involved
and $|V(\lambda_{2} ):\bC_{V(\lambda_{2} )} (t)|$ $\geq q^{\, 2}$.
Furthermore by (1.22) either 
$V(\lambda_{1} )\bigotimes V(\lambda_{1})^{\,\sigma}$ or 
$V(\lambda_{3} )\bigotimes V(\lambda_{3} )^{\,\sigma}$
is involved. Now by (1.23)
$$
|V(\lambda_{1} )\bigotimes V(\lambda_{1} )^{\,\sigma}
:\bC_{V(\lambda_{1} )\bigotimes 
V(\lambda_{1} )^{\,\sigma}} (t)|\geq q^{\, 3}\, .
$$
But $(S)L_{4}(q)$ does not possess elementary abelian subgroups of
order $q^{\, 5}$ by (1.23).

\absa
{\bf (1.39)~Lemma.~}{\it Let $F^{\, *}(G)\cong\Omega_6^{\,\pm} (r), r$
even. Let $V$ be the natural module for $G$. Suppose there is
$A\leq G, A$ an elementary abelian $2$--group with $[V,A,A]=0$. If
$|A| > r^{\, 3}$ then the following holds}
\begin{itemize}
\item[(i)]{\it $|A|=2\cdot r^{\, 3}, A\not\leq F^{\, *}(G)$}
\item[(ii)] {\it There is some $t\in A$ with $|[t,V]|=r^{\, 3}$}
\item[(iii)]{\it If $F^{\, *}(G)\cong\Omega_6^{\, +} (r)\cong L_{4}(r)$,
then $A$ does not act quadratically on the natural $L_{4}(r)$-module}
\item[(iv)]{\it If $F^{\, *}(G)\cong\Omega_6^{\, -} (r)\cong U_{4}(r)$,
then $A$ does not act quadratically on the natural $U_{4}(r)$-module.}
\end{itemize}

Proof~. Suppose first $A\leq L=F^{\, *} (G)$. Let $P$ be a
subgroup of $L, P/O_{2}(P)\cong \Omega_4^{\,\pm} (r),
|O_{2}(P)|=r^{\, 4}$. Now $O_{2}(P)$ is the natural module for
$P/O_{2}(P)$. As we may assume $A\leq P$ a moments thought shows
$A\leq O_{2}(P)$. Now we have that for some 
$v\in V, [A,\langle v\rangle ]=0,
[v^{\,\perp},A]\subseteq \langle v\rangle , \bC_{v^{\,\perp}} (A)=
\langle v\rangle ,
[V,A]\subseteq v^{\,\perp}$. By (1.27) $(V\cong V(\lambda_{2} ))$ we
see that $[V,A]\not\subseteq \langle v\rangle $. Hence $[V,A,A]\not= 0$, 
a contradiction.\\
\indent
We have $A\not\leq L$. Then we get $|A|=2\cdot r^{\, 3}$ and
$AL\cong O_6^{\,\pm} (r)$. This is (i). Furthermore $A$ contains all
types of involutions in $O_6^{\,\pm} (r)$, which proves (ii). As
$O_6^{\,\pm} (r)$ does not act on the natural $L_{4}(r)$-module we
also have (iii).\\
\indent
Let $L\cong U_{4}(r), x\in A\setminus L$. Then $x$ induces a
field automorphism on $L$ and on $W$, the natural $U_{4}(r)$-module.
We have $\bC_{L} (x)\cong Sp_{4}(r)$ and so 
$[[V,x],\bC_{L} (x)]\not= 0$. But this contradicts $[[V,x],A]=0$. 
This proves (iv).

\absa
{\bf (1.40)~Lemma.~}{\it Let $X\cong$ $^3D_{4}(q)$, $^2F_{4}(q)$, $q$
even, $V$ be a nontrivial irreducible $GF(2)$-module for $X$,
$t\in X$ be an involution. Then $|[V,t]| > q^{\, 5}$.}

\absa
Proof~. We have $|A|\leq q^{\, 5}$ for any elementary abelian
subgroup of $X$. By \cite{Coop} $|[V,t]| > q^{\, 5}$.

\absa
{\bf (1.41)~Lemma.~}{\it Let $F^{\, *}(X)\cong G(r)$ be a Lie group
over a field of odd characteristic which is not a Lie group in characteristic two too. Let $V$ be a faithful irreducible
$GF(2)$-module for $X$ and $t$ be an involution in $X$ with
$|[V,t]|\leq 2^{\, m_{2}(X)+1}$. Then  $F^{\, *}(X)$ $\cong$ 
$3\cdot U_{4}(3)$, $L_{3}(3)$, $U_{4}(3)$, $L_{4}(3)$, where we have 
equality
in the last three cases and $m_{2}(F^{\, *}(X)) < m_{2}(X)$.}

\absa
Proof~. (The proof follows \cite[(10.9)]{Asch3}). Let first 
$F^{\, *}(X)\cong L_{2}(r)$. Then $m_{2}(X)\leq 3$ or $m_{2}(X)=2$ for 
$r$ prime . Now the 
assertion follows with (1.30) and (1.29). If $F^{\, *}(X)\cong$
$^2G_{2}(r)$, $(S)L_{3}(r)$, $(S)U_{3}(r)$, $(S)L_{4}(r)$, $(S)U_{4}(r)$, 
$3\cdot U_{4}(3)$, $PSp_{4}(r)$, $G_{2}(r)$, $3\cdot G_{2}(3)$ or $^3D_{4}(r)$, then 
$m_{2}(X)\leq 6$ and the assertion follows with (1.29), unless 
$F^{\, *}(X) \cong L_{3}(3), U_{4}(3), L_{4}(3)$ or $3\cdot U_{4}(3)$, 
where we have equality in the first three cases.\\
\indent
Let now 
$T\in\mbox{Syl}_{2}(X),\triangle = {\mbox{Fun}}(T),k=|\triangle|,
Y=\langle \triangle \rangle $, and $YT^{\, *}=YT/\bC_{YT} (Y)$. 
Then $Y^{\, *}$
is a direct product of $k$ copies of $L_{2}(r)$ permuted by $T$. Let
$B^{\, *}$ be an elementary abelian subgroup of $T^{\, *}$ of maximal
order, $K\in\triangle ,D^{\, *}$ a complement to 
$E^{\,*}=N_{B}(K)^{\, *}$ in $B^{\, *},t=m_{2}(D^{\, *})$, and 
$S=\langle K^{\, D} \rangle \cap\, T$. Then $S^{\, *}$ is the direct 
product of $2^{\, t}$ copies 
of $L_{2}(r)$ and $\varepsilon =m_{2}(\bC_{K^{\, *}} (E^{\, *}))=1$ or
$2$. Hence $m_{2}(\bC_{S^{\, *}} (E^{\, *}))=\varepsilon 2^{\, t}$
and 
$m_{2}(S^{\, *}\cap B^{\, *})=m_{2}(\bC_{S^{\, *}} (B^{\,*}))=
\varepsilon$, so $m_{2}(E^{\, *}\bC_{S^{\, *}} (B))\geq m_{2}(E^{\, *})+
\varepsilon 2^{\, t}-\varepsilon\geq m_{2}(E^{\,*})+t=m_{2}(B^{\, *})$, 
so that $E^{\, *}\bC_{S^{\, *}} (E^{\, *})$
is also elementary abelian of maximal order. Thus we may choose $B$
to fix each member of $\triangle$. Let $C^{\, *}$ be the subgroup of
$B^{\, *}$ inducing inner automorphisms in $PGL_{2}(r)$ on each
member of $\triangle^{\, *}$. Then $m_{2}(B/C)\leq 1$ and
$m_{2}(C^{\, *})\leq 2k$, so $m(B^{\, *})\leq 2k+1$. Set 
$Z=T\cap Z(Y)$. Then $m_{2}(Z)\leq k$ and 
$m_{2}(\bC_{T} (Y))\leq m_{2}(Z)+i(F^{\, *}(X))$, where $i(F^{\, *}(X))$ 
can be found in \cite[(10.8)]{Asch3}. So we have 
$$
m_{2}(X)\leq 3k+1+i(F^{\, *}(X))\, ,i(F^{\, *}(X))\leq 3\, .
$$
Let $m$ be the lower bound for $[V,t]$ supplied by (1.28). Then
$$m\geq (p-1)p^{\, w-1}d(p)/2\varepsilon\geq 2(3^{\,w-1)}\varepsilon,$$ 
where $r=p^{\, s},w$ is given by 
$|O_{p}(\bC_{F^{\, *}(X)} (R))|=r^{\, 1+2w},R$ some root group in 
$F^{\, *}(X),\varepsilon =1$
or $F^{\, *}(X)=PSp_{2n}(r), r > p$, where $\varepsilon =2$. Furthermore
$k$ is listed in \cite[Theorem 2]{Asch2}. In particular $k\leq w$ unless 
$F^{\,*}(X)=PSp_{\, 2n}(r)$, where $k=n$ and $w=n-1$.\\
\indent
So assume $F^{\, *}(X)\not= PSp_{\, 2n}(r),n\geq 3$. Then 
$$2(3^{\, w-1}) > 3w+5\geq 3k+2+i(F^{\, *}(X))\,.$$ 
Hence the lemma holds.\\
\indent
So let $F^{\, *}(X)\cong PSp_{\, 2n}(r)$. Then $i(F^{\, *}(X))=0,
k=n, w=n-1$. Hence
$$(p-1)r^{\, n-1}d(p)/2\varepsilon p > 3n+2\geq m_{2}(X)+1\, ,$$ 
unless $F^{\, *}(X)\cong PSp_{\, 6}(3)$. But then $|[V,t]|\geq 2^{\, 6}$. As
$m_{2}(\mbox{Aut~}(PSp_6(3)))=4$, the lemma holds.

\absa
{\bf (1.42)~Lemma.~}{\it Let $F^{\, *}(X)$ be a quasisimple group, $V$
be a faithful $F_{1}$-module for $X$. Then either}
\begin{itemize}
\item[(i)] {\it $F^{\, *}(X)$ is a Lie group in characteristic two}
\item[(ii)]{\it $F^{\, *}(X)\cong 3\cdot U_{4}(3)$}
\item[(iii)]{\it $F^{\, *}(X)/Z(F^{\, *}(X))\cong
M_{22},M_{23},M_{24},M_{12}$ or $J_{2}$ (in fact $M_{12}$ or $J_{2}$
do not possess $F_{1}$-modules but we will not need this result)}
\item[(iv)]{\it $F^{\, *}(X)\cong 3\cdot A_{6}$ and $|[V,X]|=2^{\, 6}$}
\item[(v)]{\it $F^{\, *}(X)\cong A_{n}$ and $V$ involves just permutation
modules, or $n=8,7,5$ and also $4$--dimensional modules are involved, or
$n=9$ and $V$ has just one nontrivial module involved, a spin module.}
\end{itemize}

\absa
Proof~. (See also \cite{Asch3}). Let first $F^{\, *}(X)$ be a Lie group in
odd characteristic. By (1.41) either (i) or (ii) holds, or 
$F^{\,*}(X)\cong L_{3}(3), U_{4}(3)$ or $L_{4}(3)$.\\
\indent
Now let $A$ be some offending subgroup, i.e. 
$|V:\bC_{V} (A)|\leq 2|A|$. Suppose $F^{\, *}(X)\cong U_{4}(3), L_{4}(3)$ 
or $L_{3}(3)$.
Then by (1.41) $|V:\bC_{V} (A)|=2|A|=|V:\bC_{V} (t)|$ for any 
$t\in A^{\,\sharp}$. Furthermore $|A| > 2$. Hence $A$ acts quadratically 
on $V$, contradicting (1.12).\\
\indent
Let next $F^{\, *}(X)/Z(F^{\, *}(X))$ be sporadic. Application of
(1.25) shows that either (iii) holds or $F^{\, *}(X)\cong HS, Co_{2}$
or $Co_{1}$. In case of $HS$ we get $|V:\bC_{V} (A)|=|V:\bC_{V} (t)|$
for any $t\in A^{\,\sharp}$, and so we get a quadratic fours group,
contradicting (1.12).\\
\indent
Let $F^{\, *} (X)\cong Co_{1}$. For centralizers of involutions in
$Co_{1}$ and $Co_{2}$ see \cite{CCNPW}. Let $z$ be a 2-central involution in
$X$. Then $\bC_{X} (z)\cong 2^{\, 1+8}\Omega^{\, +}_{\, 8} (2)$. Set
$Q=O_{2}(\bC_{G} (z))$. Suppose the 24-dimensional module is not
involved in $V$. Then by \cite{MeiStr1} there is no quadratic fours group. 
Hence $\bC_{\bC_{X} (z)} ([V,z])=\langle z\rangle $. This shows 
$|[V,z]|\geq 2^{\, 9}$. Let $A$ be an offending subgroup. If $z\in A$, 
then we
have $|A|\geq 2^{\,8}$ and so $|A:A\cap Q|\geq 8$. But then
$|[V,z]:\bC_{[V,z]} (A)|\geq 4$. This now shows $|A|\geq 2^{\, 10}$ and
so $|A:A\cap Q|\geq 2^{\, 5}$. Furthermore 
$|[V,z]:\bC_{[V,z]} (A\cap Q)|\geq 2$ and so we now see 
$|[V,z]:\bC_{[V,z]} (A)|\geq 16$. Hence
$|A|\geq 2^{\, 12}$, a contradiction. We have $z\not\in A$. Then we
may assume $A\cap Q=1$ as all involutions in $Q$ are conjugated to
$z$. Now $|A|\leq 2^{\, 6}$. But this contradicts (1.25).\\
\indent
So we are left with the 24-dimensional module, $z\in A$. Now
$|[V,z]|=2^{\, 8}$, and $A\not\leq Q$. Hence $|\bC_{V} (z):
\bC_{\bC_{V} (z)} (A)|\geq 16$. This gives $|A|=2^{\, 11}$. But there 
is just one elementary abelian subgroup of order $2^{\, 11}$ in a Sylow
2-subgroup. By \cite{MeiStr1} we have $|\bC_{V} (A)|=2$, a
contradiction.\\
\indent
Let now $X\cong Co_{2}$. Let us again assume that there is no
quadratic fours group. Let $z$ be 2-central, $Q=O_{2}(\bC_{G} (z))$.
As before $|[V,z]|\geq 2^{\, 9}$. Suppose $z\in A$. Then
$A\not\leq Q$. Furthermore $A\cap Q\not\leq \langle z\rangle $. Hence
$|[V,z]:\bC_{[V,z]} (A)|\geq 8$. This would imply $|A|\geq 2^{\, 11}$, a
contradiction.\\
\indent
Hence $z\not\in A$. Let $a\in A$ with 
$O_{2}(\bC_{X} (a))\cong 2^{\,1+6}\cdot 2^{4}, 
\bC_{X} (a)/O_{2}(\bC_{X} (a))\cong A_{\, 8}$. As
there is no quadratic fours group we see 
$|[V,a]|\geq 2^{\,8}$. Furthermore as 
$z\not\in A, |A\cap O_{2}(\bC_{X} (a))|\leq 16$.
Now $|A:A\cap O_{2}(\bC_{X} (a))|\geq 8$. But 
$|[V,a]:\bC_{[V,a]} (A)|\geq 4$, and so $|A|\geq 2^{\, 9}$. This shows 
$|A:A\cap O_{2}(\bC_{X}(a))|\geq 2^{\, 5}$, a contradiction to 
$m_{2}(A_{8})=4$. \\
\indent
So neither $z$ nor $a$ is in $A$. Hence $A\cap Q=1$. Now 
$|A|\leq 2^{\, 6}$. 
Let $b\in A$, then $\bC_{X} (b)\cong 2^{\, 10}
P\Gamma L_{2}(9)$. Now we get $|[V,b]|\geq 2^{\, 8}$, a
contradiction.\\
\indent
Hence we have quadratic fours groups. By \cite{MeiStr2} this gives
$|V|=2^{\, 22}$. We show that $V$ is not an $F_{1}$-module.\\
\indent
We have $|\bC_{V} (z)|=2^{\, 16}$. Furthermore 
$|\bC_{V} (z):\bC_{\bC_{V} (z)}(A\cap Q)|\geq |A\cap Q|/2$. Hence 
$A\not\leq Q$. Let $z\in A$. Now for
$B\leq A, B\cap Q=1$, we have 
$|[V,z]:\bC_{[V,z]} (B)|\leq |B|/2^{\, 5}$. But this is impossible. 
Hence $z\not\in A$. Let $a\in A^{\,\sharp}, a$ as before. Then 
$|[V,a]|=2^{\, 8}$ and $|A|\leq 2^{\, 8}$ as before. This shows 
$|A:A\cap O_{2}(\bC_{X} (a))|\geq 8$.
But then $[A,[V,a]]\not= 0$ and so $|A:A\cap O_{2}(\bC_{X} (a))|=16$
and $A$ has to induce transvections on $[V,a]$, a contradiction.\\
\indent
So $a^{\, X}\cap A=\emptyset$. This shows $A\subseteq b^{\, G}$. But
$|[V,b]|\geq 2^{\, 8}$, and $|A|\leq 2^{\, 6}$, as $A\cap Q=1$, we
have a contradiction. This proves (iii).\\
\indent
(iv)~~For $n\leq 7$ we get the assertion by checking the irreducible
modules. Let $n=8$ or $9$. By (1.30) we have $|V:\bC_{V} (a)|\geq 16$ 
for any
$a\in X$. Hence $|A|\geq 8$. If $|A|=8$, then 
$\bC_{V} (a)=\bC_{V}(A)$ for any $a\in A^{\,\sharp}$. Hence $A$ acts 
quadratically and
the assertion follows with (1.12). So assume $|A|=16$. Now 
$|V:\bC_{V}(a)|\leq 32$ for any $a\in A^{\,\sharp}$. If $(12)\in A$, then
$\bC_{X} (x)\cong \langle x\rangle \times\Sigma_{6}$ or 
$\langle x\rangle \times\Sigma_{7}$. As $A\leq\bC_{X} (x)$ and 
$|[V,x]|\geq 16$, we see 
that $|[V,x]:\bC_{[V,x]} (A)|\leq 2$. This shows 
$[E(\bC_{X} (x)),[V,a]]=0$. Now there is a
quadratic fours group on $V$. This gives the assertion, or $n=9$ and
we have the spin module. But there is no quadratic fours group of
this type on the spin module by \cite[(4.3)]{MeiStr1}.\\
\indent
Hence we have $A\leq A_{n}$. In particular 
$$A=\langle (12)(34),(13)(24),
(56)(78),(57)(68)\rangle. $$ 
Let $b=(12)(34)(56)(78)$.
Then $A\not\leq O_{2}(\bC_{V} (b))$. As 
$|[V,b]:\bC_{[V,b]}(A)|\leq 2$, we see again that we have a quadratic 
fours group
containing $b$. By (1.12) we have the assertion, or $n=9$ and we
have a spin module. Now 
$A_{1}=\langle (12)(34),(13)(24)\rangle\,\leq A_{5}\leq A_{9}$. As 
$g=(123)$ acts fixed point freely on $V$,
we see with \cite[(8.2)]{Hi} that $V$ is a direct sum of two natural
$L_{2}(4)$-modules. Hence $|V:\bC_{V} (A_{1})|=16$. We have 
$\bC_{X}(A_{1})\leq K, K\cong A_{ \{ 5,6,7,8,9\} }$ with
$A_{2}=\langle (56)(78),(57)(68)\rangle\;\leq K$. Now 
$(567)=\gamma\in K$ and $[V,\gamma]=V$. Hence $\bC_{V} (A_{1})$ is the 
natural $K$-module.
This shows $|V:\bC_{V} (A)| = 2^{\, 6}$, a contradiction.\\
\indent
Let now $n > 9$. Let 
$\Gamma \cup\!\!\!\!^\cdot\,\,\Gamma^{\, \prime } =\Omega,\Omega =
\{ 1,\ldots ,n\}$ be a partition with  $\Gamma^{\, a} =\Gamma$
or $\Gamma^{\, \prime }$ for $a\in A$. Let $|\Gamma |=m$, and if
$A\not\leq N(\Gamma )$ then $n$ is a power of two. Suppose first
$\Gamma^{\, a} =\Gamma^{\, \prime }$ for some $a\in A$. Then $2m=n$. Let
$E=E(X_{\Gamma^{\, \prime }} )$. Then $\bC_{E\, E^{\, a}} (a)\cong E$.
Furthermore $\bC_{X} (a)=\langle a\rangle\times K, K\cong \Sigma_{m}$. 
Let $A=\langle a\rangle \times B, B\leq K$. We have 
$[\bC_{V} (a), E(K)]\not= 0$. Hence as 
$|V:\bC_{V} (a)|\geq 16$ by (1.30) we get 
$|\bC_{V} (a):\bC_{\bC_{V} (a)}(B)|\leq |B|/4$. By induction 
$\bC_{V} (a)$ involves just natural or
4-dimensional modules. But, as can be easily seen, on the natural
module $W$ we get $|W:\bC_{W} (B)|\geq |B|/2$, a contradiction. Hence
$m\leq 8$ and 4-dimensional modules are involved. Furthermore 
$m\geq 7$. As $K\cong \Sigma_{m}$, we get that both 4-dimensional 
modules are
involved and so again $|\bC_{V} (a):\bC_{\bC_{V} (a)} (B)|\geq |B|$,
a contradiction.\\
\indent
So let now $A\leq N(\Gamma )$. Choose $\Gamma$ with $m$ maximal.
Then either $A$ has a fixed point and $m=n-1$ or $m=n-2$.\\
\indent
Let $m=n-1$. Then $A\leq X_{\Gamma^{\, \prime }}\cong  A_{n}$ or
$\Sigma_{n-1}$. By induction $V$ just involves natural 
$X_{\Gamma^\prime }$-modules, or $m=9$. As for a natural module $W$ we 
have $|W:\bC_{W}(A)|\geq |A|/2$, we get that just one natural module is 
involved or $|A|\leq 8$.\\
\indent
Suppose the latter. By (1.30) we may assume $|V:\bC_{V} (a)|\geq 16$
for any $a\in A$. Hence we have $|A|=8$ and $\bC_{V} (a)=\bC_{V} (A)$
for any $a\in A$. In particular $A$ acts quadratically. Now we get the 
assertion with (1.12), as on the spin module $\rho =(1,2,3)$ acts fixed 
point freely but
$\rho\in A_{m-1}$.\\
\indent
Assume $V$ involves exactly one natural $X_{\Gamma^\prime }$-module. If
$\bC_{V} (X_{\Gamma^\prime } )\not= 1$. Then choose 
$v\in \bC_{V} (X_{\Gamma^\prime })^{\,\sharp}$. Now we get that some 
permutation module for $X$ is
involved in $V$. Then induction on $\dim V$ yields the assertion.
Hence we may assume $\bC_{V} (X_{\Gamma^\prime})=1$. Now (1.19) shows
$|V:[V,E(X_{\Gamma^\prime} )]|\leq 2$, and $V$ is a factor module of the
permutation module for $X_{\Gamma^\prime}$. Hence there is a quadratic 
group
of order eight $\langle (12)(34), (12)(56), (12)(78) \rangle$. 
By (1.12) $V$ is a permutation module for $X$. We are
left with $m=9$ and $V$ involves a spin module just once. Now $|A|=8$
and $|V:\bC_{V} (a)|=16$ for any $a\in A^{\,\sharp}$. Hence $A$ acts
quadratically on $V$ and so by (1.12) $V$ is the natural module or a
spin module for $A_{\, 10}$. But $|\bC_{V} (a)|\geq 2^{\, 8}$ for the
spin module for $A_{\, 10}$. We have the assertion.\\
\indent
Assume now $m=n-2$. Let $Q=O_{2}(X_{\Gamma} ))$. Then $|Q|\leq 2$. Let
first $A\cap Q=1$. Now $[E(X_{\Gamma^\prime } ), \bC_{V} (Q)]\not= 0$. 
Hence
by induction just natural, 4-dimensional or a spin module are
involved.  The last two cases just occur for $n=10, 11$, respectively.
So assume that just natural modules are involved. If there is more than 
one 
natural module then $|A|\leq 8$. Now $A$ acts quadratically on $V$ and 
we get 
the assertion as above. Hence we may assume that there is just one 
natural module involved. This shows $[[V,Q],E(X_{\Gamma^\prime } )]=1$, 
as otherwise 
$V$ involves at least two natural $E(X_{\Gamma^\prime } )$-modules and 
we get again $|A|\leq 8$.\\
\indent
Now by (1.19) we have 
$V=\bC_{V} (E(X_{\Gamma^\prime } ))V_{1}, V_{1}/V_{1}\cap
\bC_{V} (E(X_{\Gamma^\prime } ))$ is a factor module of the
permutation module. As $m-2\geq 8$, we get a quadratic group of order
eight in $E(X_{\Gamma^\prime } )$. Hence the assertion follows with
(1.12).\\
\indent
So let next $n=10, m-2=8$. Furthermore there are the  natural
$L_{4}(2)$-module and its dual involved in $\bC_{V} (Q)$. Now
$|\bC_{V} (Q):\bC_{\bC_{V} (Q)} (A)|\geq |A|$. In particular there
are no further nontrivial modules involved. This shows with (1.20)
that $|[E(X_{\Gamma^\prime } ),\bC_{V} (Q)]|=2^{\, 8}$. Furthermore
$[[V,Q],E(X_{\Gamma^\prime } )]=0$. Hence 
$|[E(X_{\Gamma^\prime } ),V]|=2^{\, 8}$
and so there is a quadratic group of order 16 ($O_{2}$ of the
stabilizer of a 2-space in the $L_{4}(2)$-representation). By (1.12)
$V$ just involves natural modules. Assume now $n=11$ and $\bC_{V} (Q)$
involves the spin module for $A_{9}$. As $\Sigma_{\, 9}$ is involved, both
spin modules are in $\bC_{V} (Q)$. But then $|V:\bC_{V} (A)| > 2|A|$,
as the spin module is not an $F$-module for $A_{9}$ by (1.18).\\
\indent
So we are left with $Q\not= 1,Q\leq A$. Let $B$ be a complement
to $Q$ in $A$. First of all $|V:\bC_{V} (Q)|\geq 16$ by (1.30). Hence 
$|\bC_{V}(Q):\bC_{\bC_{V} (Q)} (B)|\leq |B|/4$. But as $\bC_{V} (Q)$ 
just
involves natural modules or $n=10$ and $\bC_{V} (Q)$ involves both
4-dimensional modules or $n=11$ and $V$ involves a spin module we see 
$|\bC_{V}(Q):\bC_{\bC_{V} (Q)} (B)|\geq |B|/2$, a contradiction.

\absa
{\bf (1.43)~Lemma.~}{\it Let $X$ be as in (1.7), $X$ nonsolvable, and
$V$ be an $F_{1}$-module. If $\bC_{V} (X)=0,V=[V,X]$, then one of the
following holds:
\begin{itemize}
\item[$(1)$]$V$ is an $F$-module
\item[$(2)$]$E(X)\cong A_{\, 5}$ or $L_{2}(16)$ and $V$ is the
orthogonal module. In case of $L_{2}(16)$ the offending subgroup $A$
is of order eight or two and not contained in $E(X)$. 
\item[$(3)$]$E(X)\cong L_{3}(2)$ and $V$ is a direct sum of the natural
module and the dual module.
\item[$(4)$]$E(X)\cong A_{9}, V$ is the natural module or a spin module.
\item[$(5)$]$E(X)=X_{1}X_{2}$, $X_{1}\cong X_{2}\cong A_{\, 5}$, 
$L_{2}(16)$
or $L_{3}(2)$ and $V=V_{1}\bigoplus V_{2}$, $[V_{i},X_{i}]=V_{i}$,
$[V_{i},X_{3-i}]=0$, $V_{i}$ is an $F_{1}$-module for $X_{i},i=1,2$.
\item[$(6)$]$E(X)\cong A_{\, 5}\times A_{\, 5}$ and $V$ is the 
$O^{\,+}(4,4)$-module.
\item[$(7)$]$E(X)\cong A_{\, 5}$ or $A_{9}$ and $V=V_{1}\bigoplus
V_{2}, V_{i}$ permutation modules for $E(X),i=1,2$, and the offending
subgroup $A$ is of order 2.
\item[$(8)$]$F\, (X)$ is a $p$-group, $p\in\{ 5,7\} ,E(X/F(X))\cong
L_{2}(p), [V,F(X)]=0$ and $V$ is an $F_{1}$-module for $X/F\, (X)$, as
in $(2)$, $(3)$ or $(7)$.
\end{itemize}}

\absa 
Proof~. We have $|V:\bC_{V} (A)|\leq 2|A|\leq |A|^{\, 2}$. So all
modules will be found in (1.36). Inspection gives the assertion up to
the result about $A_{9}$. Now (4) and (7) follow from (1.42).

\absa
{\bf (1.44)~Lemma.~}{\it Let $X$ be one of the groups in (1.7) (i),
$q > 2$, or (1.7) (iii) with $E(X)\cong SL_{3}(4)* SL_{3}(4)$ or
$E(X)\cong SU_{3}(8)* SU_{3}(8)$. Let $V$ be an irreducible
$X$-module over $GF\, (2), t\in E(X)$ be some involution and
$|[V,t]|\leq 2q$. Then }
\begin{itemize}
\item[(i)]{\it $E(X)\cong L_{2}(q)$ or $L_{2}(q)\times L_{2}(q)$}
\item[(ii)]{\it Let $E(X)\cong L_{2}(q)\times L_{2}(q)$, 
$A=\langle t| 1 = t^2, t\in E(X), |[V,t]|\leq 2q\rangle $. 
Suppose $A\cap E(X)$ is not
contained in either of the two components, then 
$|V:\bC_{V} (A)|\geq q^{\, 2}$.}
\end{itemize}

\absa
Proof~. By (1.26) and (1.27) we get the assertion (i) as $q > 2$. 
We have $|[V,t]| < q^{\, 2}$ and in case of $E(X)\cong (S)L_{3}(q),
Sp_{4}(q),SL_{3}(4)* SL_{3}(4)$ we have that the natural module
and its dual is involved and so $q^{\, 2}\leq |[V,t]|$.\\
\indent
To prove (ii) assume $|V:\bC_{V} (A)| < q^{\, 2}$. Let
$E(X)=X_{1}\times X_{2}, X_{i}\cong L_{2}(q)$. Then we see that
$[V,X_{1}]$ involves exactly one irreducible nontrivial module $T$.
Hence $[[V,X_{1}],X_{2}]=0$. Now $A$ has to be nontrivially  on
$[V,X_{1}]$ and $[V,X_{2}]$. This shows $|T:\bC_{V} (A)| < q$. But this
contradicts (1.27) (i).


\absa
{\bf (1.45)~Lemma.~}{\it Let $X=(S)U_{n}(q)$ or $\Omega_{2n}^{\, -}(q),q$
even, and $V$ be the natural module for $X$. Let $A\leq X$ be an
elementary abelian $2$--group with $|V:\bC_{V} (A)|\leq q^{\, 2}$. Then
$|A|\leq q$.}

\absa
Proof~. We may assume $|A|\not= 1$. Then $|V:\bC_{V} (A)|=q^{\, 2}$.
If $X\cong (S)U_{n}(q)$, then $A$ is a group of transvections to the
hyperplane $\bC_{V} (A)$ and so $|A|\leq q$. \\
\indent
Let $X\cong \Omega_{2n}^{\, -} (q)$. Now we have some $a\in A^{\,\sharp}$
such that $[a,V]^{\,\perp} \geq\bC_{V} (A)$. Let $v$ be some singular
vector in $[a,V]$. Then $\bC_{X} (v)$ acts on $v^{\,\perp}$. We have
$O_{2}(\bC_{X} (v))$ is elementary abelian and 
$\bC_{X} (v)/O_{2}(\bC_{X}(v))\cong \Omega_{2n-2}^{\, -} (q)$. 
Furthermore
$v^{\,\perp}/v^{\,\perp\perp}\cong O_{2}(\bC_{X} (v))$ as a module for 
$\bC_{X}(v)/O_{2}(\bC_{X} (v))$. As there are no transvections on the 
natural
module in $\Omega_{2n-2}^{\, -} (q)$ and $|v^{\,\perp}:\bC_{V} (A)|=q$,
we get $A\leq B\leq O_{2}(\bC_{X} (v))$ where 
$B\cong\bC_{V}(A)^{\, \perp} /v^{\,\perp\perp}$. So $|B|=q$ and we are 
done.

\absa
{\bf (1.46)~Lemma.~}{\it Let $X\cong Sp_6(r), r=2^{\, n},V$ be the
natural module. Let $A$ be an elementary abelian $2$--subgroup of $X$.
If $|V:\bC_{V} (A)|\leq r^{\, 2}$, then $|A|\leq r^{\, 3}$.}

\absa
Proof~. Let $R$ be a transvection group. We may assume
$A\leq\bC_{X} (R)$. Let $V_{1}=\bC_{V} (R)$ and $V_{2}=[V,R]$.
Then for $B=O_{2} (\bC (R))$, we have $[V_{1},B]=V_{2}$ and
$[V/V_{2},B]=V_{1}/V_{2}$. Hence for 
$B_{1}\leq B,B_{1}\cap R=1$, we get $|V:\bC_{V} (B_{1})|\geq |B_{1}|r$. 
This now implies
$|B\cap A:B\cap R|\leq r$. Hence $|B\cap A|\leq r^{\, 2}$. Let 
$a\in A\setminus B$. Then $|[V_{1}/V_{2},a]|\geq r$. Suppose
$|V_{1}/V_{2}:\bC_{V_{1}/V_{2}} (A)|\geq r^{\, 2}$. Then $B\cap A=1$.
As $Sp_4(r)$ does not contain elementary abelian subgroups of order
greater than $r^{\,3}$, we have the assertion.\\
\indent
So let $R\cap A\not= 1$. Then 
$|V_{1}/V_{2}:\bC_{V_{1}/V_{2}}(A)|\leq r$ and so $|A:A\cap B|\leq r$, 
whence $|A|\leq r^{\, 3}$.

\absa 
{\bf (1.47)~Lemma.~}{\it Let $X\cong (S)L_{n}(r),r=2^{\, m},V$ an
irreducible faithful $X$-module over $GF\, (2)$. Any one of the
following implies that $V$ is the natural module.}
\begin{itemize}
\item[(a)]{\it $n=6,7,\quad |V:\bC_{V} (t)|\leq r^{\, 3},\; t$ some
involution in $X$}
\item[(b)]{\it $n=5,\quad |V:\bC_{V} (t)|\leq r^{\, 2},\; t$ some
involution in $X$}
\item[(c)]{\it $n=4,\quad |V:\bC_{V} (t)|< r^{\, 2},\; t$ some
involution in $X$.}
\end{itemize}

\absa
Proof~. We have that $V$ is an $SC$-module. By (1.22) 
$V\cong V(\lambda ), \lambda$ a fundamental weight or 
$V\cong V(\lambda_{1})\bigotimes V(\lambda_{1} )^{\, \sigma}$, 
$V(\lambda_{n-1} )\bigotimes V(\lambda_{n-1} )^{\, \sigma}$, 
$V(\lambda_{1} )\bigotimes V(\lambda_{n-1} )^{\, \sigma}$, $\sigma$ a 
field automorphism, or we have (a) and 
$V\cong V(\lambda_{1} +\lambda_{n-1} )$.\\
\indent
Suppose $V\not\cong V(\lambda )$. Then application of (1.24) shows
$|[V,t]|\geq r^{\, (n-1)}$ or $V\cong V(\lambda +\lambda_{n-1} )$ and
$|[V,t]|\geq r^{\, 4}$, a contradiction\\
\indent
So we have $V\cong V(\lambda ),\lambda$ a fundamental weight. Now
(1.23) yields the assertion.

\absa 
{\bf (1.48)~Lemma.~}{\it Let $X\cong G\, (r)$ be a Lie group over a field
of characteristic $2$, $X\not\cong$ $^3D_{4}(r)$. Let $V$ be some
irreducible $GF\, (2)$-module for $X$ and $t$ be an involution in
$X$. Suppose $|V:\bC_{V} (t)|\leq r^{\, 2}$ and 
$|V:\bC_{V} (t)|\leq r^{\, 4}$ if $X$ is twisted. Then 
$X\cong (S)L_{n}(r),Sp_{2n}(r),(S)U_{n}(r),\Omega_{2n}^{\,\pm} (r),G_{2}(r)$ 
or $Sz(r)$.}

\absa
Proof~. Suppose first that $X$ contains elementary abelian subgroups
of order $r^{\, 2},r^{\, 4}$, respectively. 
Then $V$ is an
$SC$-module. By (1.22) $X\cong SL_n(r),Sp_{2n}(r),SU_n(r),
\Omega_{2n}^{\,\pm}(r),G_{2}(r),F_{4}(r),E_{6}(r)$, $^2E_{6}(r)$, 
$E_{7}(r)$.\\
\indent
In the last four cases $V=V(\lambda ),\lambda$ a fundamental weight,
by (1.22). Now the assertion follows with (1.23).\\
\indent
If $X$ is untwisted and does not contain elementary abelian subgroups
of order $r^{\, 2}$, we see $X\cong L_{2}(r)$.\\
\indent
If $X$ is untwisted and does not contain elementary abelian subgroups
of order $r^{\, 4}$, we see $X\cong (S)U_{3}(r)$ or $Sz(r)$.

\absa
{\bf (1.49)~Lemma.~}{\it Let $X=Sp_{4}(q),q$ even, $V$ a 
$GF\,(2)X$-module such that $V/\bC_{V} (X)$ is the natural module. Let
$S\in \mbox{Syl}_{2} (X)$ and $Z=Z(S)$. If $[Z,V,Z]=0$, then there is
some $A\leq V,|A|=q,A\cap \bC_{V} (X)=0$ and 
$|Z:\bC_{Z} (A)|=q$.}

\absa
Proof~. Let $U\leq Z, |U|=q$, be the group of transvections on
$V/\bC_{V} (X)$. Let $V_{1}$ be the preimage of 
$\bC_{V/\bC_{V}(X)} (U)$. Let $A\leq V_{1},|A|=q$, with
$\bC_{V/\bC_{V}(X)} (Z)A\bC_{V} (X)/\bC_{V} (X)=V_{1}/\bC_{V} (X)$. We 
are going to show
that $[A,U]=0$. Let $P=\bC_{X} (U)$. Then $P/O_{2}(P)\cong L_{2}(q)$
and $P/O_{2}(P)$ acts irreducibly on 
$V_{2}=(V_{1}/\bC_{V} (X))/([U,V]\bC_{V}(X)/\bC_{V} (X))$. As 
$[Z,V,Z]=0$, we have $[[U,V]\bC_{V} (X),U]=0$.
Hence $[V_{1},U]$ is trivial or isomorphic to $V_{2}$ as a
$P/O_{2}(P)$-module. As $[V_{1},U]\leq \bC_{V} (X)$, we get
$[V_{1},U]=0$ and so $[A,U]=0$.

\absa
{\bf (1.50)~Lemma.~}(a) {\it Let $X\cong U_{4}(r),r=2^{\, n}$, and $V$ be
the natural module. If $v\in V^{\,\sharp},A\leq X,A$ an elementary
abelian $2$--group, $[V,A,A]=0$, then $|[v,A]|\not= r^{\, 4}$.}

(b){\it Let $X\cong Sp_6(r),r=2^{\, n}$, and $V$ be the spin module.
If $v\in V^{\,\sharp}, A\leq X, A$ elementary abelian,
$[V,A,A]=0$, then $|[v,A]|\not= r^{\, 4}$.}

(c) {\it Let 
$X_{1},X_{2}\leq X, T_{i}\lhd X_{i},X_{i}/\bC_{X_{i}}(T_{i}) \cong 
Sp_6(r),T_{i}$ the spin module for 
$X_{i}/\bC_{X_{i}}(T_{i}),i=1,2$. Let 
$\langle T_{1},T_{2}\rangle \leq X_{1}\cap X_{2}$. Then
$[T_{1},T_{2}]=1$.
}

\absa
Proof~. (a) and (b):~~Suppose $|[v,A]|=r^{\, 4}$ for some 
$v\in V^{\,\sharp}$. Then as $[V,A,A]=0,[v,A]\leq \bC_{V} (A)$. Let
$W\leq V$ be a 4-dimensional subspace such that  $N_{X}(W)$
induces $L_{2}(r^{\, 2})$ or $Sp_{4}(r)$ on $W$. Then $N_{X}(W)$
also induces $L_{2}(r^{\, 2})$ and $Sp_{4}(r)$ on $V/W$. Hence all
cosets of $W$ in $V$ are conjugated under $N_{X}(W)$. We may assume
$A\leq N_{X}(W)$. As $|[v,A]|=r^{\, 4}$, we have $v\not\in W$.
Now we may assume $v^{\, A}=v+W$. This implies that $X$ acts      
transitively on $V^{\,\sharp}$, a contradiction.\\
\indent
(c):~~Suppose $[T_{1},T_{2}]\not= 1$. We have
$[T_{1},T_{2},T_{2}]=1=[T_{2},T_{1},T_{1}]$. By (b) there is no 
$t\in T_{2}$ with $|[T_{1},t]|=r^{\, 4}$. Let 
$W\leq T_{1}, |W|=r^{\,4}$, such that $N_{X_{1}}(W)$ induces $Sp_{4}(r)$ 
on $W$ and
$T_{1}/W$. Set $Z(O^{\, 2^\prime }(N_{X_{1}}(W))/\bC_{X_{1}}(T_{1}))=Z$. 
Then $[Z,T_{1}]=W$. We may assume that 
$[T_{2}\bC_{X_{1}} (T_{1})/\bC_{X_{1}}(T_{1}),Z]=1$. Hence $T_{2}$ 
induces transvections on either $W$ or $T_{1}/W$. Set 
$U_{2}=T_{2}\bC_{X_{1}} (T_{1})/\bC_{X_{1}} (T_{1}) \cap 
O_{2}(N_{X_{1}} (W)/\bC_{X_{1}} (T_{1}))$. Then 
$|U_{2}|\leq r^{\, 3}$. Hence 
$|T_{2}\bC_{X_{1}} (T_{1})/\bC_{X_{1}} (T_{1})|\leq r^{\, 4}$.
Furthermore the same holds for $T_{1}$, i.e. 
$|T_{1}\bC_{X_{2}}(T_{2})/\bC_{X_{2}} (T_{2})|\leq r^{\, 4}$. By (1.23) 
we see that $|T_{2}:\bC_{X_{2}} (T_{1})|\geq r^{\, 3}$. Hence 
$|U_{2}|\geq r^{\,2}$. \\
\indent
Suppose $u\in U_{2},u$ is an element of type $c_{2}$ in the Suzuki 
notation. Then there is some $g\in X_{1}$ with  
$u^{\, g}\in N_{X_{1}}(W)\setminus\bC_{X_{1}} (W)$ and $u^{\, g}$ is an 
element of type $c_{2}$ in $Sp_{4}(r)$. Hence 
$|[W,u^{\, g}]|=r^{\,2}=|[T_{1}/W,u^{\, g}]|$, a contradiction.\\
\indent
So we have that all elements in $U_{2}$ are of type $a_{2}$. Hence
$|U_{2}|=r^{\, 2}$ and $|T_{1}:\bC_{T_{1}} (U_{2})|=r^{\, 2}$. But
now $\bC_{T_{1}} (u)=\bC_{T_{1}} (U_{2})$ for all 
$u\in U_{2}^{\,\sharp}$. Let $R$ be a root group of type $a_{2}$, then
$N_{X_{1}} (R)$ is a maximal subgroup of $X_{1}$ and so we have
$\langle \bC_{X_{1}/\bC_{X_{1} (T_{1})}} (u)|u\in U^{\,\sharp} \rangle =
X_{1}/\bC_{X_{1}} (T_{1})$, a contradiction.

\absa
{\bf (1.51)~Lemma.~}{\it Let $O_{r}(X)$ be elementary abelian of order
$r^{\, 3}$, and $X/O_{r}(X)\cong L_{2}(r)$, $r$ an odd prime, $r > 3$. Let
$\bC_{X} (O_{r}(X))=O_{r}(X)$ and $V$ be a $GF\, (2)X$-module. If for
some involution $t\in X$ we have $|[V,t]|\leq 32$, then
$[O_{r}(X),V]=0$.}

\absa
Proof~. Suppose $[O_{r}(X),V]\not= 0$. Now 
$|[O_{r}(X),t]|\geq r^{\,2}$. This implies $r=5$ and there is some 
$\omega\in O_{5}(X), |[V,\omega ]|=16$. As $|O_{5}(X)|=5^{\, 3}$, we get
$|[V,O_{5}(X)]|\leq 2^{\, 12}$. But $5^{\, 4}\not\Big |\,  |GL_{12}(2)|$,
a contradiction.

\absa 
{\bf (1.52)~Lemma.~}{\it Let $X=O_{p}\, (X)F, O_{p}\, (X)$
elementary abelian of order $p^{\, n}, n\geq 3$ and $X$ a subgroup 
contained in the Borel subgroup of
$\mbox{Aut~} (L_{2}(p^{\, n}))$,
containing the Borel subgroup of $L_{2}(p^{\, n})$. Let $V$ be some 
$GF\,(2)X$-module with $\bC_{V}(O_{p}\, (X))=0$. If 
$\omega\in O_{p}\, (X), o(\omega )=p$, then $|[\omega,V]| > 4$. In 
particular for $x\in F,o(x)=2$, then $|V:\bC_{V} (x)| > 2$.}

\absa
Proof~. Suppose $|V:\bC_{V} (x)|=2$. Then $p=3$. Let 
$\omega\in O_{p}\, (X)$
with $\omega^{\, x} =\omega^{\, -1}$. Then $|[V,\omega ]|=4$. So
we get $|[V,O_{p}\, (X)]|\leq 2^{\, 2n}$. Now $F$ contains a group of 
order $(3^{\, n}-1)/2$ acting fixed
point freely on $O_{p}\, (X)$. This shows that
$n\geq (3^{\, n}-1)/4$, which gives $n=2$, a contradiction.


\absa 
{\bf (1.53)~Lemma.~}{\it Let $X=G\, (r)$ be a Lie group, 
$r=2^{\, n}, X\not\cong Sz(r), L_{2}(r), U_{3}(r)$. Let 
$S\in \mbox{Syl}_{2} (X), A\triangleleft S, A$ elementary abelian. Then 
there is some parabolic $P$ of 
$X, O^{\, 2^\prime } (P/O_{2}(P))\cong L_{2}(r), L_{2}(r^{\, 2})$ or 
$U_{3}(r)$, such that $A\leq O_{2}(P)$. If
$X\cong$ $^2F_{4}(r), A\leq  O_{2}(P)$ for both minimal
parabolics. If $X\cong$ $^3D_{4}(r)$, 
$O^{\, 2^\prime }(P/O_{2}(P))\cong L_{2}(r^{\, 3})$.}

\absa
Proof~. By way of induction it is enough to prove the assertion for
Lie groups of rank two, i.e. $X\cong L_{3}(r)$, $Sp_{4}(r)$, $U_{4}(r)$,
$U_{5}(r)$, $G_{2}(r)$, $^3D_{4}(r)$, $^2F_{4}(r)$.\\
\indent
Let $P_{1},P_{2}$ be the two minimal parabolics. In case of $X\cong$
$U_{4}(r)$, $U_{5}(r)$, $^3D_{4}(r)$, $^2F_{4}(r)$ choose notation such
that $O^{\, 2^\prime } (P_{1}/O_{2}(P_{1}))\cong $ $L_{2}(r)$, $SU_{3}(r)$,
$L_{2}(r^{\, 3})$, $Sz(r)$, respectively.\\
\indent
If $X\cong L_{3}(r)$ or $Sp_{4}(r), O_{2}(P_{1})$ and $O_{2}(P_{2})$
are the only maximal elementary abelian subgroups of $X$. Hence
$A\leq O_{2}(P_{1})$ or $O_{2}(P_{2})$.\\
\indent
Let $X\cong U_{4}(r)$. Let $A\not\leq O_{2}(P_{1})$. We have
$O_{2}(P_{1})/Z(O_{2}(P_{1}))$ is elementary abelian and for 
$a\in A\setminus O_{2}(P_{1}),|[O_{2}(P_{1})/Z(O_{2}(P_{1})),a]|=
r^{\, 2}$. This implies 
$|\langle Z(O_{2}(P_{1})),A\rangle | > r^{\, 3}$. \\
\indent
Let now $A\not\leq O_{2}(P_{2})$. We have 
$\Omega_{1}(O_{2}(P_{2}))$ is elementary abelian of order $r^{\, 4}$ and 
$O^{\,2^\prime } (P_{2}/O_{2}(P_{2}))\cong L_{2}(r^{\, 2})$.
Furthermore $\Omega_{1} (O_{2}(P_{2}))$ is the 
$\Omega^{\, -}(4,r)$-module for $L_{2}(r^{\, 2})$. As $A$ acts 
quadratically on
$\Omega_{1} (O_{2}(P_{2}))$, we see $|A:A\cap O_{2}(P_{2})|\leq r$.
Now $|\langle Z(O_{2}(P_{1})), A\rangle \cap\,\Omega_{1} (O_{2}(P_{2}))| > 
r^{\, 2}$. But this contradicts (1.27).\\
\indent
Let $X\cong G_{2}(r)$. Let $O_{2}(P_{1})/Z(O_{2}(P_{1}))$ be
elementary abelian of order $r^{\, 4}$. If 
$a\in A\setminus O_{2}(P_{1})$, then 
$|[a,O_{2}(P_{1})/Z(O_{2}(P_{1}))]|=r^{\, 2}$.
Hence $|\langle A,Z(O_{2}(P_{1}))\rangle | > r^{\, 3}$, contradicting 
the fact that
$A$ contains no elementary abelian subgroup of order greater than 
$r^{\, 3}$ (see (1.23)).\\
\indent
Let $X\cong$ $^3D_{4}(r)$. Then $V=O_{2}(P_{1})/Z(O_{2}(P_{1}))$ is the 
8-dimensional $GF\,(r)$-module for $L_{2}(r^{\, 3})$.  Suppose that 
$A\not\leq O_{2}(P_{1})$. Now by (1.32)
$|[a,O_{2}(P_{1})/Z(O_{2}(P_{1}))]|=r^{\, 4}$. Then
$[a,O_{2}(P_{1})]=\bC_{O_{2}(P_{1})} (a)$. Let 
$\gamma\in N(S),o(\gamma)=r^{\, 2}+r+1$. Then 
$[\gamma ,\bC_{V} (S)]=0$ by (1.32). Hence 
$\bC_{O_{2}(P_{1})}(\gamma )$ is a Sylow 2-subgroup of $L_{3}(r)$. 
But we may assume there is $a^{\, g}=b\in L_{2}(r^{\, 3})$ with 
$\gamma^{\, b} =\gamma^{\, -1}$. But there is no automorphism $b$ of a 
Sylow 2-subgroup $T$ of $L_{3}(r)$ such that $[T,b]$ is elementary 
abelian of order $r^{\, 2}$.\\
\indent
Let $X\cong$ $^2F_{4}(r)$. We have 
$Z_{2}(O_{2}(P_{1}))=\Omega_{1}(O_{2}(P_{1}))$. Now 
$[a,O_{2}(P_{1})]\leq Z_{2}(O_{2}(P_{1}))$ for
$a\in A$, and so $A\leq O_{2}(P_{1})$. Hence 
$A\leq Z_{2}(O_{2}(P_{1}))\leq \bC (Z(O_{2}(P_{1})))\leq O_{2}(P_{2})$.

\absa
{\bf (1.54)~Lemma.~}{\it Let $X\cong U_{\, 5}(q), q$ even, 
$A\leq X$ be an elementary abelian $2$--group, $|A| > q^{\, 2}$. Then 
$A$ contains some $t$
which induces a transvection on the natural module.}

\absa
Proof~. Let $P$ be the parabolic with $Z(O_{2}(P))=1$. Then
$|\Omega_{\, 1} (O_{2}(P))|=q^{\, 4}$ and 
$O^{2^\prime}(P/O_{2}(P))\cong L_{2}(q^{\, 2}),\Omega_{\, 1} (O_{2}(P))$ 
is the natural $\Omega^{\, -} (4,q)$-module.\\
\indent
Let $R=Z(S),S\in\mbox{Syl}_{2} (P)$. We may assume $A\leq P$ and
$R\cap A\not= 1$. Hence $|A\, R| > q^{\, 3}$. If 
$a\in S\setminus O_{\,2}(P)$, then 
$|\bC_{\,\Omega_{\, 1} (O_{\, 2}(P))} (a)|=q^{\, 2}$.
Hence $A\, R\leq \Omega_{\, 1} (O_{\, 2}(P))$. Now 
$A\leq \Omega_{\, 1} (O_{\, 2}(P))$. As the conjugates of $t$ correspond 
to the singular vectors and the largest subgroup of the orthogonal
$4$-dimensional module containing no singular vector is of order
$q^{\, 2}$, we get the assertion.

\absa
{\bf (1.55)~Lemma.~}{\it Let $E(X)\cong 3\cdot U_{4}(3)$ and $u\in X$
be an involution. Then there is an elementary abelian subgroup
$E,|E|=27$ of $E(X)$ such that $u\in N(E),|\bC_{E} (u)|\geq 9$. If
$u\in \bC (Z(E(X)))$, even $[E,u]=1$.}

\absa
Proof~. According to $\cite{CCNPW}$ any involution in 
$\mbox{Aut~}(U_{4}(3))$ centralizes an elementary abelian subgroup $F$ 
of order 9 in $U_{4}(3)$. An easy inspection shows that this group has an
elementary abelian preimage $E$.

\absa
{\bf (1.56)~Lemma.~}{\it Let $G = G(q)$ be a group of Lie type over 
$GF(q)$,
$q = 2^n$, $G \not\cong L_2(q)$, $L_3(q)$, $U_3(q)$, $Sz(q)$, $G_2(q)$ or 
$^2F_4(q)$. Let $R$ be a long root group, $Q = O_2(N_G(R)/R)$ and $L$ 
be a Levi complement in $N_G(R)$. Then $Q$ has the following $L$--module 
structure}
\begin{enumerate}
\item[(i)]{\it $G \cong L_n(q)$, $O^{2^\prime}(L) \cong L_{n-2}(q)$, 
$Q = V_1 \oplus V_2$, $V_1$ is the natural $L$--module and $V_2$ its 
dual.}
\item[(ii)]{\it $G \cong Sp_{2n}(q)$, 
$O^{2^\prime}(L) = Sp_{2n-4}(q) \times L_2(q) = L_1 \times L_2$, 
$Q = V_1 \oplus V_2$, $[V_2, L_1] = 1$, $V_1$ is the natural
 $L_2$--module, $V_1 = V_1^{(1)} \oplus V_1^{(2)}$, $V_1^{(i)}$, 
$i = 1,2$, are natural $L_1$--modules and $[L_2,V_1] = V_1$.}
\item[(iii)]{\it $G \cong \Omega_{2n}^{\pm}(q)$, 
$O^{2^\prime}(L) = \Omega_{2n-4}^{\pm}(q) \times L_2(q) = L_1 \times L_2$, 
$Q = V_1 \oplus V_2$,
$V_i$, $i = 1,2$, are natural $L_1$--modules and $[Q,L_2] = Q$.}
\item[(iv)]{\it $G \cong U_n(q)$, $O^{2^\prime}(L) = U_{n-2}(q)$, $Q$ is 
the natural module.}
\item[(v)] $G \cong E_6(q)$, $O^{2^\prime}(L) \cong L_6(q)$, 
$Q \cong V(\lambda_3)$.
\item[(vi)] $G \cong$ $^2E_6(q)$, $O^{2^\prime}(L) \cong U_6(q)$, 
$Q \cong V(\lambda_3)$.
\item[(vii)] $G \cong E_7(q)$, $O^{2^\prime}(L) \cong \Omega^{+}_{12}(q)$, 
$Q \cong V(\lambda_6)$.
\item[(viii)] $G \cong E_8(q)$, $O^{2^\prime}(L) \cong E_7(q)$, 
$Q \cong V(\lambda_1)$.
\item[(ix)] {\it $G \cong F_4(q)$, $O^{2^\prime}(L) \cong Sp_6(q)$, 
$Q$ is an extension of the natural module by a spin module.}
\item[(x)]{\it $G \cong$ $^3D_4(q)$, $O^{2^\prime}(L) \cong L_2(q^3)$, 
$Q$ is the $8$--dimensional $GF(q)$--module for $L$.}
\end{enumerate}

\absa
Proof~. This can easily be checked using the Chevalley commutator 
formula (see also \cite{AschSe}).

\absa
{\bf (1.57)~Lemma.~}{\it Let $G = G(q) \not\cong$ $^2F_4(q)$, 
$ q = 2^m$, be a group of Lie type and $R$ be a long root group. Set 
$Q = O_2(\bC_G(R))$. Let $A \le Q$ be elementary abelian with 
$[A,Q] \not= 1$. Then there is $U \le Q$, $|U| = q$, with 
$|A : \bC_A(U)| \le q$.}

\absa
Proof~. We have $R = Q^\prime$ is of order $q$. Hence $|[A,Q]| \le q$. 
Furthermore $Q$ is generated by subgroups $R_i$, $|R_i| = q$, and 
$R_i$ is a TI--set in $Q$. Let $x \in A$ with $[x,r] = 1$ for some 
$r \in R_i$, $r \not= 1$.
Then $[x,R_i] = 1$, as $R_i$ is a TI--set. This now implies that 
$R_1 \cap \bC(A) = 1$ for some $R_1$ and 
$|A : \bC_A(R_1)| = |A : \bC_A(r)| \le q$, for $r \in R_1$, $r \not= 1$.

\absa
{\bf (1.58)~Lemma.~}{\it Let $G = F_4(q)$, $q = 2^m$, $R$ be a 
root group, $A$ be
an elementary abelian subgroup of $N_G(R)$. Let $Z = Z(O_2(N_G(R)))$ and 
$S \in {\mbox{Syl}}_2(N_G(R))$. If $A \not\le O_2(N_G(R))$ but 
$A \le O_2(\bC_{N_G(R)}(Z(S)))$, then 
$|Z : \bC_Z(A)| \ge |A : A \cap O_2(N_G(R))|$.}

\absa
Proof~. We have that $Z/R$ is the natural $Sp_6(q)$--module by (1.56). 
Furthermore by assumption about $A$ we have that there is a subgroup 
$Z_1$
of $Z$, $|Z : Z_1| = q$, with $Z_1 \ge [Z,A]$ and $[Z_1,A] \le Z(S)$. 
Now
obviously $|Z_1 : \bC_{Z_1}(A)| \ge |A : A \cap O_2(N_G(R))|$.

\absa
{\bf (1.59)~Lemma.~}{\it Let $G$ be a group of Lie type over $GF(q)$, 
$q = 2^n$, $G \not\cong$ $^2F_4(q)$. Let $R$ be a long root group, 
$Q = O_2(N_G(R)/R)$. If $t$ is an involution in 
$\bC(R) \setminus O_2(N_G(R))$ such that 
$|[Q,t]| \le q^2$, then $G \cong L_n(q)$, $U_n(q)$ or $Sp_{2n}(q)$ and 
$|[Q,t]| = q^2$, or $G \cong G_2(q)$.}

\absa
Proof~. We have that $Q$ is a $GF(q)$--module (see (1.56)) and so 
$|[Q,t]| = q$ or $q^2$. In the former $t$ induces a 
$GF(q)$--transvection and so by (1.22), 
(1.23) and (1.24) we get a contradiction to (1.56).\\
\indent
So let $|[Q,t]| = q^2$. If $G \cong \Omega_{2n}^{\pm}(q)$, then by (1.56) 
$Q$ is
a sum of two modules. On neither of them $t$ can induce a transvection. 
If $G \cong F_4(q)$, then again by (1.56) there are two modules in $Q$ 
and $t$ cannot induce a transvection on the spin module by (1.23).\\
\indent
If $G \cong E_6(q)$, $E_7(q)$, $E_8(q)$, or $^2E_6(q)$, then by (1.23) 
and (1.56) the minimal codimension for centralizers of involutions in 
$Q$ is greater than $2$.\\
\indent
So we are left with $G \cong$ $^3D_4(q)$. But in this case the minimal 
codimension is four by (1.32).

\absa
{\bf (1.60)~Lemma.~}{\it Let $G = G(2)$ be a group of Lie type over 
$GF(2)$.}

(a){\it \, Assume $m_3(G) \ge 4$. Let $x$ be a long root element, 
$t \in O_2(\bC_G(x))$ be an involution, $S \in {\mbox{Syl}}_2(\bC_G(x))$ 
and $[t,S] \le \langle x \rangle$. Then $3 \Big | |\bC_G(t)|$.}

(b){\it \, Assume $m_3(G) \ge 3$. If $S \in {\mbox{Syl}}_2(G)$ then either 
$G \cong Sp_6(2)$ or $m_3(\bC_G(Z(S))) \ge 2$.}

\absa
Proof~. (a) Set $H = \bC_G(x)$. If $G \not\cong L_n(2)$, then 
$O_2(H)/Z(O_2(H))$ is
an irreducible module for $H/O_2(H)$ and so $tZ(O_2(H))$ is centralized 
by some 
parabolic $P$ in $H/O_2(H)$ by (1.56). Hence $3 \Big | |P|$ or 
$H/O_2(H)$ is solvable. The latter just occurs for 
$G \cong \Omega_8^+(2)$. But in this group any involution is centralized 
by a $3$--element.\\
\indent
Assume now $G \cong L_n(2)$. Then by (1.56) 
$O_2(H)/Z(O_2(H)) = H_1 \oplus H_2$,
where $H_1$ is the natural $L_{n-2}(2)$--module and $H_2$ its dual. So 
$tZ(O_2(H))$ is centralized by a $3$--element as $n \ge 8$.\\
\indent
Hence we may assume that $tZ(O_2(H))$ is centralized by a $3$--element 
in $H$. 
If $\langle x \rangle = Z(O_2(H))$ the same applies to $t$. So assume 
$\langle x \rangle < Z(O_2(H))$. This means $G \cong Sp_{2n}(2)$ or 
$F_4(2)$.\\
\indent
Let $G \cong Sp_{2n}(2)$ and set $\tilde{H} = \bC_H(Z(O_2(H)))$. Then 
$O_2(H)/Z(O_2(H)) \cong H_1 \oplus H_2$, where $H_i$, $i = 1,2$, are 
natural
modules for $\tilde{H}/O_2(H) \cong Sp_{2n-4}(2)$ by (1.56). Now as 
$2n-4 \ge 4$, we have that $tZ(O_2(H))$ is centralized by a 
$3$--element in 
$\tilde{H}$ and then also $t$ is centralized by a $3$--element.\\
\indent
So we hare left with $G \cong F_4(2)$. Now $tZ(O_2(H))$ is centralized 
by a subgroup $U \cong 2^6L_3(2)$ in $H/O_2(H)$. As 
$[O_2(U),t] \le \langle x \rangle$, we see that 
$|t^U\langle x \rangle| = 1$ or $7$. In both cases $t$ is
centralized by a $3$--element.

(b) Let $H$ be as in (a). If $|\Omega_1(Z(S))| = 2$ this is the same as 
saying $m_3(H/O_2(H)) \ge 2$. Hence the assumption holds or 
$H/O_2(H) \cong \Sigma_3$
or $L_3(2)$ by (1.3). But as $m_3(G) \ge 3$, an easy inspection shows 
that this is not possible. So we have $|\Omega_1(Z(S))| = 4$ and 
$G \cong Sp_{2n}(2)$ or
$F_4(2)$. In the latter $\Omega_1(Z(S))$ is centralized by $Sp_4(2)$ and 
in the former by $Sp_{2n-4}(2)$ (see (1.56)). Hence $m_3(\bC(Z(S))) \ge 2$ 
or $G \cong Sp_6(2)$.

\absa
{\bf (1.61)~Lemma.~}{\it Let $G \cong U_n(q)$, $\Omega_{2n}^-(q)$ or 
$^2E_6(q)$, $q = 2^n$. Suppose no parabolic subgroup contains an 
elementary abelian subgroup of order $27$. Then $G \cong U_n(q)$, 
$n \le 7$, or $\Omega_8^-(q)$.}

\absa
Proof~. In $^2E_6(q)$ we have a parabolic which involves 
$L_2(q) \times L_3(q^2)$ and so it has $3$--rank at least 3.\\
\indent
In $\Omega_{2n}^-(q)$ we have a parabolic which involves 
$L_2(q) \times \Omega_{2n-4}^-(q)$. As $m_3(\Omega_6^-(q)) \ge 2$, 
we have the assertion (remember $\Omega_6^-(q) \cong U_4(q)$).\\
\indent
In $U_n(q)$ we have a parabolic which involves 
$L_2(q) \times U_{n-4}(q)$. As $m_3(U_4(q)) \ge 2$, we have the 
assertion. 
 
\absa
{\bf (1.62)~Lemma.~}{\it Let $G$ be a group with 
$F^{*}(G) = O_2(G) \not= 1$ and $A \le G$, $A$ elementary abelian, 
$A \not\le O_2(G)$, but $A \unlhd S \in {\mbox{Syl}}_2(G)$. Then there 
is some $g \in G$ such that for $X = \langle A, A^g \rangle$ the 
following hold}
\begin{enumerate}
\item[(1)]{\it $X/O_2(X) \cong D_{2u}$, $u$ odd, $L_2(q)$, $Sz(q)$ or $PSp(4,q)$, 
$q$ even.}
\item[(2)] $Y = (A \cap O_2(X))(A^g \cap O_2(X)) \unlhd X$
\item[(3)] $Y \not= A \cap O_2(X)$
\item[(4)] $|A : \bC_A(Y/\bC_Y(A))| \le |Y : \bC_Y(A)|^2$,
\end{enumerate}
{\it  or there is
some $B \le S$, $B$ elementary abelian, with $[B,A] \not= 1$ and 
$|A : \bC_A(B)| \le |B : \bC_B(A)|$.}

\absa
Proof~. Set $\bar{G} = G/O_2(G)$. Let $[O_r(\bar{G}), \bar{A}] \not= 1$ 
for some odd prime $r$. Then we get some $g \in O_r(\bar{G})$ with 
$X/O_2(X) \cong D_{2r}$, $X = \langle A, A^g \rangle$.\\
\indent
Let now $L$ be some component of $\bar{G}$ with 
$1 \not= [L, \bar{A}] \le L$. If $|\bar{A} : \bC_{\bar{A}}(L)| = 2$ we 
again get some $g \in L$ with $X/O_2(X) \cong D_{2u}$, $u$ odd. As 
$O_2(X) \le AO_2(G) \le S$ in both cases 
we see $\langle A \cap O_2(X), A^g \cap O_2(X) \rangle = Y \unlhd X$. 
As $[X, O_2(G)] \le Y$ and $\bC_G(O_2(G)) \le O_2(G)$, we see 
$A \cap O_2(G) \not= Y$. 
We have $A \cap A^g = \bC_A(Y)$ and $A \cap Y = \bC_Y(A)$. Now 
$$|A : \bC_A(Y)| = 2|A \cap Y : A \cap A^g| \le 
|A^g \cap Y : A \cap A^g||A \cap Y : A \cap A^g|$$
$$ = |Y : A \cap A^g| = |Y : Y \cap A||A \cap Y : A \cap A^g| = 
|Y : A \cap Y|^2.$$
We have that $A$ acts quadratically on $O_2(G)$. From now on assume 
$1 \not= [\bar{A}, L] \le L$, but $|\bar{A} : \bC_{\bar{A}}(L)| \ge 4$. 
If $L$ is a group of Lie type in odd characteristic which is not a group 
of Lie type in even
characteristic too. Then by (1.12) $L/Z(L) \cong U_4(3)$. Let $i$ be a 
$2$--central involution in $\bar{S} \cap L$. If 
$\bar{A} \not\le O_2(\bC_L(i))$ we get for suitable $g \in \bC_L(i)$ 
that $X/O_2(X) \cong \Sigma_3$, where
$X = \langle A, A^g \rangle$. Now the assertion follows as above. So let 
$\bar{A} \le O_2(\bC_L(i))$. As $\bC_L(i)$ does not contain an 
elementary
abelian normal subgroup of order at least four, we get some 
$g \in \bC_L(i)$ with $\bar{A} \not= \bar{A}^g$. Now 
$1 \not= [A, A^g] \le A \cap A^g$. Hence the second alternative holds.\\
\indent
Let now $L \cong A_n$, $ n \ge 5$. Let $m_1, \ldots , m_r$ be the dyadic 
decomposition of $n$. Then $S \cap A_n$ is in 
$\Sigma_{m_1} \times \ldots \times \Sigma_{m_r}$. We may assume 
$m_{i} \ge 4$ for all $i$. If $m_i \ge 8$ for all $i$, we have 
induction. So we may assume $m_r = 4$. Then we may suppose 
$\bar{A}/\bC_{\bar{A}}(L) \le A_4 \le A_5$. Hence for 
$g \in L$ either 
$\langle A, A^g \rangle /O_2(\langle A, A^g \rangle ) \cong D_{2u}$, $u$ 
odd, or $A_5$. In the first case we get the assertion. If 
$X/O_2(X) \cong A_5$, we get $|Y : A \cap A^g| \ge 16$, as $A_5$ has to 
act nontrivially on $Y/A \cap A^g$. But now again 
$|A : \bC_A(Y)| \le |Y : \bC_Y(A)|^2$.\\
\indent
Assume that $L$ is sporadic. With (1.12) we get 
$L/Z(L) \cong M_{12}$, $M_{22}$, $M_{24}$, $J_2$, $Co_1$, $Co_2$, or 
$Suz$. Again choose $i \in Z(S \cap L)^\sharp$. If 
$L \cap \bar{A} \not\le O_2(\bC_L(i))$, then we have induction. So 
assume $L \cap \bar{A} \le O_2(\bC_L(i))$. If there is some 
$g \in \bC_L(i)$ with $[\bar{A}, \bar{A}^g] \not= 1$, 
then as before we get the second alternative of the statement.
 So assume that $\langle \bar{A}^{\bC_L(i)} \rangle$ is abelian. 
Then we have $L/Z(L) \cong M_{12}$, $M_{22}$, or $M_{24}$. In the first case $\bar{A} \cap L \le P\Gamma L_2(9)$ and we have the assertion by
induction. If $L \cong M_{24}$, we get $\bar{A} \cap L \le M_{23}$, 
a contradiction to (1.12). If $L/Z(L) \cong M_{22}$, then 
$\bar{A} \cap L \le (S)L_3(4)$, a case we will handle next.\\
\indent
Let now $L \cong G(r)$ be a group of Lie type in even characteristic. 
Let $R$ be a root group in $Z(\bar{S} \cap L)$. If 
$\bar{A} \not\le O_2(N_L(R))$, we
have induction again. Hence assume $\bar{A} \le O_2(N_L(R))$. 
Furthermore we may assume that $\langle A^{N_L(R)} \rangle$ is 
abelian. If $\bar{A} \le R$, then $\bar{A} \le \tilde{L} \le L$, where 
$\tilde{L} \cong L_2(r)$ or $Sz(r)$, and we have induction again. So we 
are left with $L \cong L_n(r)$, $Sp_{2n}(r)$, $F_4(r)$, $^2F_4(r)$, 
$Sz(r)$, or $U_3(r)$.\\
\indent
Let $L \cong L_2(r)$. Then we may assume $X/O_2(X) \cong L_2(r)$. So 
$|A : \bC_A(Y)| \le |Y : \bC_Y(A)|^2$, as $|Y : A \cap A^g| \ge r^2$.\\
\indent
Let $L \cong L_n(r)$, $n \ge 3$. Then we have two parabolics $P_1$ and 
$P_2$ which involve $L_{n-1}(r)$. If $\bar{A} \not\le O_2(P_i)$, for 
some $i$, we
have induction again. This shows 
$\bar{A} \le O_2(P_1) \cap O_2(P_2) = R$. Now
$\bar{A} \le L_2(r)$ and we are done.\\
\indent
Let $L \cong Sp_{2n}(r)$. We have $\bar{A} \cap L \le Z(O_2(N_L(R)))$. 
Hence
$\bar{A} \cap L \le Sp_4(r)$. So it is enough to prove the assertion 
for
$L \cong Sp_4(r)$. Then we have two parabolics $P_1$ and $P_2$, which 
involve $L_2(r)$. If $\bar{A} \not\le O_2(P_1) \cap O_2(P_2)$, we have 
induction. So assume $\bar{A} \le O_2(P_1) \cap O_2(P_2)$. 
Then either $X/O_2(X) \cong L_2(r)$ or $PSp_4(r)$. In the latter 
$|Y : A \cap A^g| \ge r^4$. As $|A : A \cap Y| \le r^2$, we again get 
$|A : \bC_A(Y)| \le |Y : \bC_Y(A)|^2$. In the former we get the 
assertion by induction.\\
\indent
Let now $L \cong F_4(r)$. We have two root subgroups $R_1$ and $R_2$ and 
$\bar{A} \cap L \le Z(O_2(N_L(R_1))) \cap Z(O_2(N_L(R_2))) = 
Z(\bar{S} \cap L)$. But this group is contained in some $PSp_4(r)$, and 
we may argue by induction.\\
\indent
Let $L \cong$ $^2F_4(r)$. As $A$ acts quadratically, we get with (1.14) 
that $\bar{A} \cap L \le R$ and
so $X/O_2(X) \le L_2(r)$, and induction applies. The same is true for 
$L \cong U_3(r)$.\\
\indent
Let $L \cong Sz(r)$. Then $|Y : A \cap A^g| \ge r^4$ and so 
$|A : \bC_A(Y)| \le |Y : \bC_Y(A)|^2$.\\
\indent
Let now finally $[A,L] \not\le L$. As $\bar{A} \unlhd \bar{S}$, we have 
that $L$ possesses abelian Sylow 2--subgroups. Furthermore we have 
$|A : \bC_A(L)| \ge 4$. By (1.11) we get 
$L \cong L_2(r)$, $r$ even. Now 
$N(\bar{S} \cap L)/\bar{S} \cap L \cong  \bZ_{r-1} \times \bZ_{r-1}$ and 
so some $\bar{a} \in \bar{A}^\sharp$ acts nontrivially on this group. 
Hence $g$ can be chosen in such a way that $X/O_2(X)$ is dihedral. 


\absa
{\bf (1.63)~Lemma.~}{\it Let $G$ be in the uniqueness case. Let 
$R\leq G,E\leq R\, \cap\, M, E$ elementary
abelian $p$-subgroup, $M$ not exceptional. Suppose one of the
following holds:}
\begin{itemize}
\item[(a)]{\it $p\in\sigma (M)$, $|E|\geq
p^{\, 2}$ and $\Gamma_{\, E,1} (G)\leq M$}
\item[(b)]{\it $p\not\in\sigma (M)$, $F^*(M) = O_2(M)$, $|E|=p^{\, 3}$, 
$\Gamma_{\, E,1}(G)\leq M$ and there is $H\trianglelefteq M$ such that
$EO_{2}(H)\triangleleft H, H/O_{2,p}(H)\cong L_{2}(p),
(H/O_{2,p}(H))^\prime =H/O_{2,p}(H)$.}
\end{itemize} 
{\it Then $R\leq M$ or (a) holds and $E(R/O_{p^\prime }(R))=L$ is simple 
and 
we have one of the following  $(P\in\mbox{Syl}_{p} (R\cap M))$:}
\begin{itemize}
\item[(i)]{\it $L\cong L_{2}(p^{\, n}), n > 1,U_{3}(p^{\, n})$ or
$^2G_{2}(3^{\, n}),n > 1; M\cap L=N_{L}(P\cap L)$}
\item[(ii)]{\it $p=3\, , L\cong L_{2}(8)\, , L_{3}(4), M_{11},A_{6}; 
M\cap L=N_{L} (P\cap L)$}
\item[(iii)]{\it $p=5\, , L\cong Sz(32), McL$, $^2F_{4}(2)^\prime ; 
M\cap L=N_{L}(P\cap L)$}
\item[(iv)]{\it $p=5$ or $7$\, and $L\cong HS$, $Ru$, $He$, $O^\prime  N$, or $M(24)^\prime$ ;
$m_{p}(L)=2$ and
$\Gamma_{\, P\cap L,1} (G)\not\leq M$}
\item[(v)]$p=5\, , L\cong A_{10}$
\item[(vi)]$p=11\, , L\cong J_{4}; M\cap L=N_{L}(P\cap L)$.
\item[(vii)] {\it $p = 5$, $L \cong M(22)$ and $E(M) = F^*(M).$}
\end{itemize}
{\it In any case $O_{p^\prime } (R)=O_{2^\prime }(R)$ and 
$O_{2}(R\cap M)=1$ or $L\cong L_{3}(4)$ and some graph $\times$ 
field automorphism is involved.}

\absa
Proof~. Let $R\not\leq  M$. Set $K=O_{p^\prime  }(R)$. As 
$K=\langle \bC_{K}(x)|x\in E^{\,\sharp}\rangle $ we get $K\leq M$. 
Set $U=O_{p^\prime ,p}(R)$. 
Let $P\in \mbox{Syl}_{p} (R),E\leq P$ and $1 \not= x\in Z(P)\cap U$. Then
$[x,E]=1$ and so $x\in M$. If $x\in E$, then $P\leq M$. Assume
$x\not\in E$. Then as $m_{p}(M)=3$ in case (b) we have (a). Now
$m_{p}(\langle x,E\rangle )\geq 3$ and so 
$P\leq N_{G}(\langle x\rangle )\leq M$.\\
\indent
So $U\leq M$ in any case. Suppose first $U=K$. Let now $W$ be
the preimage of $E(R/U)$. Then $W > K$ and $p\,\Big  |\, |W|$. We first 
show
that $W/K$ is simple. Let $E\leq P\in\mbox{Syl}_{p} (R)$, then as
before $P\leq M$. Let $W_{1}/K\cdot W_{2}/K\cdots W_{r}/K=W/K$. Suppose
there is $\omega\in E$ with $(W_{1}/K)^{\,\omega}\not= W_{1}/K$. Then
we see that $r\geq p$. As $p\geq 3$, we get in case (a) that
$W_{2}/K\cdots W_{r}/K\leq N_{W/K}((P\cap W_{1})K/K)\leq M$
and $W_{1}/K\cdots W_{r-1}/K\leq N_{W/K}((P\cap W_{r})K/K)\leq M$.\\
\indent
Hence $W\leq M$ and as $R=WN_{R}(W\cap P)$ we see $R\leq M$.
Suppose (b), then $r=3$. Now $|N_{E}(W_{1}/K)|=3^{\, 2}$. As
$m_{3}(M)=3$, we get $N_{E}(W_{1}/K)\leq W$. But now 
$\bC_{\Omega_{\,1} (P\cap W)} (\omega )\geq N_{E}(W_{1}/K)$ which 
contradicts
$(W_{1}/K)^{\,\omega} =W_{2}/K,(W_{2}/K)^{\,\omega} =W_{3}/K$.\\
\indent
So we have $E\leq N(W_{1}/K)$. There is $x\in Z(P\cap W_{1})$ with
$N_{G}(\langle x\rangle )\leq M$, so $W_{2}W_{3}\cdots W_{r}\leq M$. But 
the
same is true for $W_{r}$, i.e. $W_{1}W_{2}\cdots W_{r-1}\leq M$.
Hence we have $W\leq M$. If we are in case (a), we get 
$R\leq M$. So assume (b). As $m_{p}(M)=3$, we get $r\leq 3$. 
We have  $E\cap W_{i}\not= 1,i=1,\ldots ,r$. So at least one $W_{i}$ has 
a cyclic Sylow
$p$-subgroup, say $W_{1}$. Now if $W_{1}\triangleleft R$, we get
$R=W\, N_{R}(E\cap W_{1})\leq M$. Hence we have that
$R/N_{R}(W_{1})\lesssim \Sigma_{3}$, and $N_{R}(W_{1})\leq M$.
Furthermore $R=N_{R}(W_{1})\, N_{R}(E\cap W)$. Where 
$N_{R}(E\cap W)=\bC_{R} (E\cap W)\langle x|\bC_{\, E\cap W} (x)\not= 
1\rangle \leq M$.\\
\indent
So we have that $W/K$ is simple. Now we may apply \cite[(24-9)]{GoLy} if
$p\in\sigma (M)$, \cite[(24-12)]{GoLy} in case (b). We get a list of
possibilities for $W/K$. \\
\indent
Suppose first case (a): As $W\not\leq \Gamma_{\, P,2} (G)$ we get
$W/K\not\cong PSp_{4}(p),L_{3}(p)$ or $A_{3p}$. Let $W/K$ not be as in
(i). Suppose furthermore $p=3$. Then $\Gamma_{\, P,1} (G)\leq M$. Now
we get $W/K\cong L_{2}(8),L_{3}(4)$ or $M_{11}$. In any case 
$M\cap W/K=N_{M\cap W}(P)K/K$.\\
\indent
So let $p\geq 5$. If $W/K\cong M(22)$, then $M\cap R$ involves
$D_{4}(2)$. If $F^*(M) = O_2(M)$ then $3\in\sigma (M)$ and so 
$m_{3}(M\cap R)\geq 4$. But
then $R\leq M$.\\
\indent
Let $W/K\cong$ $^2F_{4}(32)$. Then $m_{p}(P)\geq 3$ and so also $W$
satisfies the assumption for some $\widetilde{E}\leq P\cap W$. But
this contradicts \cite[(24-9)]{GoLy}.\\
\indent
Let $W/K\cong A_{\, 2p}$, then $A_{p}\times A_{p}$ is in $M\cap R$.
If $p > 5$, then $A_{p}$ contains a 2--group which is normalized by an 
elementary abelian group of order 9, a contradiction as before.\\
\indent
So assume (b). As $|E|=p^{\, 3}$ and $m_{p}(P)=3$, we get 
$W/K\cong A_{3p}$,$L_{2}(p^{\, 3})$, $G_{2}(p)$, $PSp_{4}(p)$ or $p=3$ 
and $W/K\cong Sp_{6}(2)$, $A_{10}$, $A_{11}$, $J_{3}$, $Ly$, $F_{3}$, 
$F_{5}$, $Co_{1}\,(p=5)$, or $F_{1}\,(p=7)$. \\
\indent
As $E\cap W\triangleleft M\cap W/O_{2}(M\cap W)$ we see by
inspection of \cite{CCNPW} that $W/K\cong L_{2}(p^{\, 3})$. Now there is
some $H\leq M\cap W$ with $|H/H\cap K|=(p^{\, 3}-1)/2$ and $H$ acts
irreducibly on $E$. Hence $M$ induces $L_{3}(p)$ on $E$, a
contradiction.\\
\indent
So we just have to show that $U=K$. Suppose $P\cap U\not= 1$. Then
$R=UN_{R}(Z(U\cap P))$. So in case (a) we have 
$N_{R}(Z(U\cap P))\leq M$. Thus we are in (b). If 
$[E,\Omega_{\, 1} (Z(U\cap P))]\not= 1$,
then there is some $\omega\in Z(U\cap P)$ inducing a transvection on
$E$. But now $M$ induces $L_{3}(p)$ on $E$, a contradiction. So
$\Omega_{\, 1} (Z(U\cap P))\leq E$. So all we have to show is:\\

$(*)\,\hfill \mbox{Let~} 1\not= F\leq E,\mbox{~then~} N_{G}(F)\leq 
M\, .\hfill$
\\
\\
As $EO_{2}(M)\triangleleft M$, we get some
$\widetilde{E},|\widetilde{E}|=r^{\, 2},r\in\sigma (M)$ with
$\Gamma_{\, E,1} (G)\leq M$ and $\widetilde{E}\leq N_{G}(F)$. Now
the just proved part (a) tells us that either $N_{G}(F)\leq M$
or $O_{r^\prime }(N_{G}(F))\leq M$ and let $X$ be a preimage of
$E(N_{G}(F)/O_{p^\prime  }(N_{G}(F)))$ then $N_{G}(F)=N_{M}(F)X$. As
$\Gamma_{F,1} (G)\leq M$, we see that 
$\bC_{X} (F)\leq O_{p^\prime }(N_{G}(F))$. This gives 
$X/O_{p^\prime  }(N_{G}(F))\leq L_{3}(p)$. But none of the groups in 
the assertion has this property.
So we have $N_{G}(F)\leq M$.\\
\indent
In case (a) we just have to prove the additional assertion about
$M\cap R$. Let $2\,\Big  |\, |K|$. Then let $T\in\mbox{Syl}_{2} (K)$. So
$R=K\, N_{R}(T)$. As we may assume $E\leq N_{R}(T)$, we get 
$N_{R}(T)\leq M$, a contradiction. So $2\not\Big |\, |K|$.\\
\indent
We see that $O_{2}(N_{L}(P\cap L))=1$. So assume there is some
$\omega\in\mbox{Aut~} (L)$ with $[\omega ,P\cap L]=1, o(\omega )=2$.
Then we just have $L\cong L_{3}(4)$. 

\absa
{\bf (1.64)~Lemma.~}{\it Let $X=G(q)$ be a Lie group, $q$ even,
$q > 2$. Let $r=2^{\, n}$ and $x$ be a primitive prime divisor of $r-1,
x=9$ in case $r=64$. Suppose $r > q$. Let 
$\omega\in \mbox{Aut} (X),o(\omega )=x$, $\omega$ normalizes a 
Borel subgroup $B$ of $X$. Then
one of the following holds}
\begin{itemize}
\item[(i)]{\it $r=q^{\, 2}$ and $X\cong U_{n}(q), \Omega_{2n}^{-} (q)$ or
$^2E_{6}(q)$.}
\item[(ii)]{\it $r=q^{\, 3}$ or $r^2 =q^{\, 3}$ and $X\cong$ $^3D_{4}(q)$.}
\item[(iii)]{\it $x=3$ or $9$ and $X\cong$ $^3D_{4}(q)$, $D_4(q)$, 
$q\leq 32$.}
\end{itemize}

Proof~.  Suppose first that $\omega$ induces a graph automorphism on
$G(q)$. Then $X\cong$ $^3D_{4}(q)$ or $D_4(q)$ and $x=3$ or $9$. If $x=3$, 
we get
$r=4$ and so by assumption $q=2$, a contradiction.
Let $x=9$, then $r=64$. So $q\leq 32$ and (iii) holds.\\
\indent
So assume now that $\omega$ induces a field automorphism. Then
$q=t^{x}$ or $x=9$ and $q=t^3$. Suppose the former. As
$x\,\Big  |\, t^{x-1}-1 < q-1$, we see that $r\leq q$, which contradicts 
the
choice of $x$. So we have $q=t^3$ and $x=9$. Hence $r=64$ and so
$q=8$. This shows $r=q^{\, 2}$. Now $\omega^{\, 3}$ induces an
inner $\times$ diagonal automorphism. As $3\not\!\Big |\, q-1$, we see 
that $X$ is
a 2-fold twisted group and so we have (i).\\
\indent
So assume finally that $\omega$ induces an inner $\times$ diagonal
automorphism. Let $x=p$, prime. Then $x\not\Big |\, q-1$. This now 
shows
$x\,\Big |\, q^{\, 2}-1$ or $X\cong$ $^3D_{4}(q)$ and
 $x\,\Big  |\, q^{\,3}-1$.\\
\indent
In the former by the choice of $x$ we have $r=q^{\, 2}$ and this is
(i). In the latter we get $r=q^{\, 3}$ or $r^2=q^{\, 3}$ and so we
have (ii).\\
\indent
So we are left with $x=9, r=64, q\leq 32$. As the torus of $B$ is an
abelian group, we see that $9\,\Big  |\, q-1, 9\,\Big  |\, q^{\, 2}-1$ or
$9\,\Big  |\, q^{\, 3}-1$. This gives $q=8, q=4$ and $X\cong$ 
$^3D_{4}(q)$, or
$q=16$ and $X\cong$ $^3D_{4}(q)$. Hence we have (i) or (ii). \\


\absa
{\bf (1.65)~Lemma.~}{\it Let $X=G(q)$ be a Lie group, $q$ even. Let
$r=2^{\, n}$ and $x$ be a primitive prime divisor of $r-1, x=9$ for
$r=64$. Suppose $r > q$. Let 
$\omega\in\mbox{Aut} (X),o(\omega)=x$. Let $S\in\mbox{Syl}_{2} (X)$. 
Suppose that $\langle S,\omega \rangle $ is a
$\{ 2,x\}$-group. Then one of the following holds:}
\begin{itemize}
\item[(i)]{\it $q > 2, \omega$ normalizes a Borel subgroup of $X$}
\item[(ii)]{\it $q=2, o(\omega )=3$ or $9$ and $\omega\in X$ if 
$\langle S,\omega \rangle =S\langle \omega \rangle ,
S\not\!\triangleleft \langle S,\omega \rangle $}
\item[(iii)]{\it $X\cong U_{n}(2),\Omega_{2n}^{\, -}(2)$, $^2E_{6}(2)$, $^3D_{4}(2)$, $o(\omega )=3$ or $9$ and
$\omega$ normalizes a Borel subgroup}
\item[(iv)]{\it $X\cong$ $^2F_{4}(2)^\prime$, 
$o(\omega )=5, \omega\in X$, if
$\langle S,\omega \rangle =S\langle \omega \rangle ,
S\not\!\triangleleft\, S\langle \omega \rangle .$}
\item[(v)]{\it $X\cong$ $^3D_{4}(2)$, $o(\omega )=7,\omega$ normalizes a
Borel subgroup}
\item[(vi)]{\it $X\cong D_{4}(2), o(\omega )=3$ or $9$, $\omega$ 
induces a diagram automorphism on $X$.}
\end{itemize}

\vspace{6mm}

\absa
Proof~. As $\langle S,\omega \rangle $ is a $\{ 2,x\}$-group we either 
have that
$\omega$ induces an automorphism which normalizes  a Borel subgroup
or $X$ possesses a solvable parabolic.\\
\indent
Suppose the former. Then if $q > 2$ we have (i). Let $q=2$. Then
$\omega$ induces a graph or diagonal  automorphism. In the first case
we have (vi). So assume that $\omega$ induces a diagonal
automorphism. Then $X\cong U_{n}(2),\Omega_{2n}^{\, -} (2)$, $^2E_{6}(2)$
or $^3D_{4}(2)$. This is (iii) or (v).\\
\indent
So assume that $\omega$ does not normalize a Borel subgroup of $X$. We
then have a solvable minimal parabolic $P$. This first shows $q=2$.
Furthermore $\langle S,\omega \rangle \cap\, X\geq P$. We have 
$P/O_{2}(P)\cong \Sigma_{3} ,SU_3(2)$ or $F_{20}$. Which shows $x=3$ or 
$9$, or
$X\cong$ $^2F_{4}(2)$ and $o(\omega )=5$.\\
\indent
Suppose now that we have 
$\langle S,\omega \rangle =S\langle \omega \rangle $. Then $\omega$
induces an inner automorphism. As 
$S\not\!\triangleleft\, S\langle \omega \rangle $, we see that 
$\omega\in \langle S,\omega \rangle^\prime\leq X$. \\

\absa
{\bf (1.66)~Lemma.~}{\it Let $X$ be one of the groups of (1.7) or
(1.8). Let $M$ be the maximal subgroup of $X$ containing $S$. Suppose
there is some $\omega\in M,o(\omega )=p, p > 2$, a prime such that 
$N_{X}(\langle \omega \rangle )\leq M$, then $X$ is one of the 
following: $L_{2}(r),
L_{3}(r), U_{3}(r), r$ odd, $PSp_{4}(r), L_{2}(r)\times L_{2}(r), r$
prime, $(S)L_3(q)$, $Sp_4(q)$, $q$ even , or $F\, (X)$ is elementary abelian of order $r^{\, 3}$ and
$E(X/F\, (X))\cong L_{2}(r),L_{2}(r)\not\cong A_{\, 5}$ or $X$ is
solvable and (1.8) (i) or (iv) holds. If $E(X) \cong $$(S)L_3(q)$ or $Sp_4(q)$, $q$ even, then there is $E \leq M$, $E$ elementary abelian of order $p^2$ such that $N_X(E) \not\leq M$.}

\absa
Proof~. Suppose first that $E(X)$ is as in (1.7) (i). Then 
$M\cap E(X)=N_{E(X)}(S\cap E(X))$. In any case there is some 
$x\in E(X)$
with $\omega^{\, x} =\omega^{\, -1} ,x\not\in M$.\\
\indent
Suppose now that $E(X)$ is as in (1.7) (iii). Then we have
$E(X)\cong A_{9}$. Now obviously $p=5$ or $7$ as $M\cong A_{8}$. But
in both cases $N_{A_{\, 9}}(\langle \omega \rangle )\not\leq A_{\, 8}$.\\
\indent
So we have $F\, (X)\not= 1$. Then $X$ is solvable. By (1.8) we have
(1.8) (i) or (iv).\\

\absa
{\bf (1.67)~Lemma.~}{\it Let $X$ be quasisimple $1\leq m_{p}(X)\leq 2$
for some odd prime $p,m_{r}(X)\leq 3$ for all odd prime. Suppose $X$
admits an outer automorphism $\omega$ of order $p$ such that
$m_{p}(X\langle \omega \rangle ) > m_{p}(X)$, then 
$X\cong L_{2}(r^{\,3}),SL_{3}(r),SU_{3}(r),3\,\Big  |\, |Z(X)|, r$ 
prime and $p=3$, or $X\cong G(q)$, a Lie group over a field of 
characteristic $2$.}

\absa 
Proof~. As no sporadic or alternating group does admit an outer
automophism of order $p$, we have that $X$ is a Lie group, we may
assume $X=G(q), q$ odd. \\
\indent
Let $q=r^{\, t}, r$ prime. Now $m_{r}(X)\leq 3\geq m_{3}(X)$. This
shows $X/Z(X)\in \{ L_{2}(r), L_{2}(r^{\, 2}), L_{2}(r^{\, 3}),
L_{3}(r), U_{3}(r), Sp_{4}(r),G_{\, 2}(r)\}$.\\
\indent
As $X$ has to admit an outer automorphism of order $p$ we get
$X/Z(X)= L_{2}(r^{\, 3}),p=3,L_{3}(r),p=3,p\,\Big  |\, r-1,U_{3}(r),p=3,
p\,\Big  |\, r+1$. 
Let $p\not\Big |\, |Z(X)|$. In case $L_{3}(r)$ and $U_{3}(r)$, we 
have $m_{3}(X)=2$. Furthermore
$\omega$ is a diagonal automorphism, i.e. 
$\bC_{X} (\omega )\cong SL_{2}(r)$. 
So $m_{3}(\bC_{X} (\omega ))=1$, hence 
$m_{3}(X\langle \omega \rangle )=2$, a 
contradiction. So we have $p\,\Big  |\, |Z(X)|$.

\absa
{\bf (1.68)~Lemma.~}{\it Let $G$ be in the uniqueness case. There is no 
normal subgroup $M_{1}$ of $M$ with $|M:M_{1}|=p, p\in \sigma (M)$.}

\absa 
Proof~. Let $P\in\mbox{Syl}_{p} (M)$. By Alperin's lemma \cite[chapter
7, Theorem (2.6)]{Go}  we have 
$P\cap G^\prime =\langle N_{G}(X)^\prime 
|X\leq P,X=O_{p}(N_{G}(X))\rangle $. Now choose 
$X\leq P,X=O_{p}(N_{G}(X))$. As
$m_{p}(P)\geq 2$, we see that $m_{p}(X)\geq 2$. Hence $N_{G}(X)\leq M$.
This shows $P\cap G^\prime \leq P\cap M^\prime\leq M_{1}$, 
contradicting $P\leq G^\prime$, as $G$ is simple.\\
\\


\begin{center}
\S 2 The group $G_2$
\end{center}

\noindent
For the proof of the main theorem we need two groups $G_1$ and $G_2$ on
which we are going to apply the amalgam method. In this chapter we will
to give the construction of the group $G_2$. This group is
dependent of the uniqueness group $M$ and a Sylow 2-subgroup $S$ of $M$.\\

We fix some uniqueness group $M$. All we assume is that $3\in \sigma (M)$ if
$3\in \sigma (G)$. Furthermore we fix some Sylow 2-subgroup $S$ of $M$.\\

\noindent Let
\[{\cal{N}}_1 = \{\tilde{N} \mid \tilde{N} \mbox{ is a maximal 2-local subgroup
of~} G, N_G(S)\le \tilde{N},\tilde{N} \not\le M\} \]
and
\[{\cal {N}} = \{N\mid S\le N, F^\ast (N) = O_2(N), N\le \tilde N\in {\cal {N}}_1,
N\not\le M \}\]
and
\[{\cal{N}}_{min} = \{ N\mid N\in {\cal{N}},S \mbox{ is in exactly one
maximal subgroup of~} N\} \] \\
By assumption ${\cal{N}}_{min}\neq \emptyset$. Now choose $G_2\in {\cal{N}}_{min}$
such that the maximal subgroup containing $S$ is contained in $M$. If
possible choose $G_2$ to be solvable.\\[3ex]
\indent
{\bf (2.1)  Lemma.} 
{\it (a) If $\sigma (G_2) = \emptyset$, then $G_2 / O_2(G_2)$ is as 
in (1.7) and (1.8)\\
\hspace*{0.3cm} (b) If $\sigma (G_2)\neq \emptyset$, then $3\not\in \sigma (G_2)$ and
either $G_2/O_2(G_2) = ES$, $E$ extraspecial $p$ - group or elementary
abelian, or $G_2/O_2(G_2) = E(G_2/O_2(G_2))S$, $E(G_2/O_2(G_2))\cong L_2(r)
\times L_2(r)$, or $L_2(r), r$ odd.\\
\hspace*{0.3cm} (c) If $G_2$ is one of the groups in (b), then $G_2$ does not possess an 
$F$-module $V$ with $O_2(G_2/\bC_{G_2}(V)) = 1$.}\\[2ex]
\indent Proof. (a) If $\sigma (G_2) = \emptyset$, then $m_p(G_2)\le 3$ for any odd
prime $p$. So the assertion follows with (1.7) and (1.8).\\
\hspace*{0.3cm} (b) Let $p\in \sigma (G_2)$. Then by the choice of $G_2$ and $M$, we have
$p\neq 3$. In particular $m_3(G_2)\le 3$. Let X = $G_2/O_2(G_2)$. Assume
first $E(X) = 1$. Then $\bC_X(F(X))\subseteq F(X)$. Let $R\in Syl_r (F(X))$.
Suppose $m_r(R)\le 3$. Then as in (1.7) we get a contradiction. So we have
$m_r(R)>3$, $r\in \sigma (G_2)$. Now $R\not \le M$, and so $X=RS$, in
particular $X$ is solvable.

Let $C$ be a critical subgroup of $R$  and $C_1=\Omega_1(C)$. Suppose first
$m_r(C_1)\ge 3$. Then $C_1S = X$, hence $m_r(C_1)\ge 4$. Let $D$ be an
elementary abelian subgroup of order $r^2, D\unlhd X$. Then $D\le M$. So
$\bC_{C_1}(D) = D$, but this is impossible. So we have $C_1$ is elementary
abelian or extraspecial.

So suppose now $m_r(C_1)\le 2$. Then $C_1S\le M$. But $\Gamma_{C_1,1}
(G)\le \tilde N$, where $N\le \tilde N, \tilde N$ a uniqueness group
for the prime $r$, a contradiction. 

If $O_2(M) = F^*(M)$ we get a contradiction by definition
of the uniqueness group. So assume $E(M)=K$ is quasisimple. Application
of (1.63) shows $M\subseteq \tilde N$. Now another application of (1.63)
gives $\tilde N\subseteq M$ and so $G_2\subseteq \tilde N = M$, a 
contradiction.

So assume now $E(X)\neq 1.$ Then $X = E(X) S$. Suppose that 
$E(X)$ contains at least four components. Then these
components are of type $Sz(q)$. But then $N_X(E(X)\cap S)$ is covered by
$(G_2\cap M)/O_2(G_2)$ and $m_r(N_X(E(X)\cap S))\ge 4$ for some odd $r$, a 
contradiction.

So we have that $E(X)$ contains at most two components. Suppose that $E(X)$ 
is quasisimple. Then $m_3(E(X))\le 3$ and so (1.3) applies. As $m_p(E(X))
\ge 4$, we get by (1.4) that $E(X)/Z(E(X))$ is not sporadic or alternating.
If $E(X)/Z(E(X))$ is a Lie group in characteristic two we have by the 
minimal choice of $G_2$ that the Lie rank is at most two. But then $m_p
(E(X))\le 3$. So $E(X)/Z(E(X))$ is a Lie group in odd characteristic. As
$M\cap G_2$ cannot contain an $E\cong E_{p^2}$ with $m_p(\bC_{G_2}(E))
\ge 3$, we see that $E(X)\cong L_2(p^n)$.

So assume $E(X) = E_1\ast E_2$. Then $m_3(E_1) = 1$ if $3\not\,\mid$ $| Z(E_1)\mid$ 
and $m_3(E_1) = 2$ if $3\mid \mid Z(E_1)\mid$.
Again with (1.3) and (1.4) we get that $E_1\cong L_2(r), (S)L_3(r)$ or
$(S)U_3(r)$. Suppose $r$ to be odd. In the last two cases for $r\in \sigma
(G_2)$ we get $SL_2(r)\ast SL_2(r)$ is involved in $G_2\cap M$. But this
is impossible as before. So $E(X)\cong L_2(r)\times L_2(r), r$ odd. If
$r$ is even, then $E_1\cong (S)L_3(r)$ or $(S)U_3(r)$ and for $p\in \sigma
(G_2)$ we get $p\mid r-1$ or $r+1$, respectively. But again $G_2\cap M$
contains $E\cong E_{p^2}$ with $\Gamma_{E,1}(G)\le \tilde N$, where $\tilde N$
is a uniqueness group containing $G_2$, a contradiction as before.\\
\hspace*{0.3cm} (c) If $G_2$ is nonsolvable the assertion follows with (1.18). So let $G_2$
be solvable. Let $A$ be some offending subgroup on $V$. By \cite[(24.1)]{GoLyS} $G_2/\bC_{G_2}(V)$ contains some $U\cong D_{2p}\times \ldots D_{2p}, \mid U\mid =
2^np^n$, where $\mid A\mid = 2^n$. Now $\mid V: \bC_V(A)\mid \le \mid
A\mid$ and so $\mid V:\bC_V(U)\mid \le 2^{2n}$. But as $p>3$ and
$GL_{2n}(2)$ does not contain an elementary abelian subgroup of order $p^n,
p>3$, we have a contradiction.

\absa
{\bf (2.2) Lemma.} {\it Let $E(L_2/Q_2) \cong$ $L_2(r) \times L_2(r)$, $Sz(r) \times Sz(r)$, $Sp_4(r)$, $(S)L_3(r)$, $r$ even, or $SU_3(8)*SU_3(8)$, $SL_3(4)*SL_3(4)$. Let $\Z_p \times \Z_p \cong E \leq L_2 \cap M$. Then $p \not\in \sigma (M)$.}

\absa
Proof. ~ The assertion follows from the fact that $L_2 = \langle L_2 \cap M, N_{L_2}(E) \rangle$.

 
























\begin{center}
\S 3 The nonconstrained case
\end{center}

The purpose of this chapter is to prove that a uniqueness group $M$ has to
satisfy $F^*(M)=O_2(M)$. For this we assume that $F^*(M)=KO_2(M)$,where 
$O_2(M)$ might be trivial.


\absa
{\bf(3.1) Lemma.} 
{\it If $3\in \sigma (G)$, then $3\in \sigma (M)$}.

\absa
Proof. Let $\tilde{M}$ be a uniqueness group for the prime $3$ such that 
$\tilde{M}\cap M$ contains a Sylow 3-subgroup of M. By (1.3) $m_3(M)>1$.
Now application of (1.63) gives the assertion.\\[2ex]
Next we choose a long root group $R$ in $K/Z(K)$. Let $\tilde{R}$ be the
preimage in $K$ and $G_1=N_M (\tilde{R})$. Then $G_1\cap K$ is 2-constrained.
(in case of $K\cong F_4(q)$ extended by a field automorphism we choose
$G_1$ such that $G_1\cap K$ is the parabolic with $Sp_4(q)$ on top.)

\absa
{\bf(3.2) Lemma.} 
{\it There ist $E\subseteq G_1, E\cong E_{p^2}$,
$\Gamma_{E,1} (G)\le M$}.

\absa
Proof. Let first $K\cong G(q), q>2$. Then  application of (1.2) shows that
we have the assertion or $K/Z(K)\cong L_4(q), Sp_6(q), U_n(q), n\le 7,
\Omega_8^-(q)$. If $m_p(K)\le 2$, then there is a diagonal automorphism of 
$K$ of order $p$ contradicting (1.61). 

So in case $K/Z(K)\cong 
L_4(q),Sp_6(q), U_4(q),\Omega_8^-(q)$, we have $m_p(K)\ge 3$. Hence $p\mid q-1, q^2-1,
q+1,q^2-1$ respectively, and we see again that the assertion holds. We are
left with $K/Z(K)\cong U_n(q),5\le n\le 7$, and $p=5,3,7$, respectively.
Now in any case $p\mid q+1$ and then again the assertion holds.

Let now $K=G(2)$ be defined over $G F(2)$. If $3\not\in \sigma (M)$ then by (3.1)
$K/Z(K)\cong L_n(2), 4\le n\le 7, Sp_6(2), U_4(2), \Omega_8^-(2), G_2(2)$. 
But $m_p(K)\ge 3$, and so we get a contradiction. Hence $3\in \sigma (G)$.
Now it is enough to find $Z_3\times Z_3$ in $G_1\cap K$. Inspection shows that we
may assume $K\cong L_4(2), L_5(2), U_4(2), G_2(2)$. But now $m_3(K)\le 3$
and so $m_3 (\bC(K))=1$. Hence it is enough to have $3\mid\ \ \mid G_1\cap 
K\mid$ which is the case.\\

Now we choose $G_2$ as in chapter 2 as a $2-$local with $S\subseteq G_2$, where $S\in 
Syl_2 (G_1\cap K)$ and $G_2$ is minimal with respect to $S\subseteq G_2$
and $G_2\not\subseteq M$.

\absa
{\bf(3.3) Lemma.} {\it $O_2(\langle G_1,G_2\rangle)=1$}
 
\absa
Proof. This follows from (3.2) and the definition of the uniqueness case.\\

Set $Z_2=\langle \Omega_1 (Z(S))^{G_2}\rangle$. We have $Z_2\subseteq  Q_2 = 
O_2(G_2)$. Set $Q_1=O_2(G_1)$.


\absa
{\bf(3.4) Lemma.}  {\it $[Z_2,O_2(G_1\cap K)]\not= 1$}.

\absa
Proof. Suppose false
\\ \\
\hspace*{1.5cm} $(\alpha) \ \ \ \ {\bC}_G(Z_2)\subseteq M$.\\

Suppose $\bC_G(Z_2)\not\subseteq M$. As $m_p(K)\ge 2$, we get 
$Z_2\cap \bC(K)=1$. In particular $O_2(M)=1$. Now (3.2) shows that
$K=G(q)$ and $p\mid q-1$. Furthermore $m_p(G_1)=2$. Hence 
$m_p(K)\ge 4$. Application of (1.2) shows $K\cong L_4(q), Sp_6(q),U_n(q), 
5\le n\le 7$, or $\Omega^-(8,q)$. As $m_p(K)\ge 4$ and $p\mid q-1$, we get
a contradiction.
\\ \\
\hspace*{1.5cm} $(\beta)$ \ \ \ \ If $P\in Syl_p(\bC_M(Z_2))$, then
$N_G(P)\not\subseteq M$, in particular $m_p(P)\le 1$.\\

Otherwise $N_G(Z_2)=\bC_G(Z_2)N_G(P)\subseteq M$ by $(\alpha)$ and
the Frattiniargument.\\

Let $Z(O_2(G_1\cap K))/Z(K)$ be a root group $R$. Then by $(\beta)$ we
have $m_p(G_1)=2$ and $p\mid q-1$. Application of (1.2) and (1.61) yield
$K\cong \Omega_8^-(q)$. But then $\bC_K(R)$ involves $L_2(q^2)\times
L_2(q)$, a contradiction to $(\beta)$.

So let now $K\cong Sp_{2n}(q)$
or $F_4(q)$ and $Z_2\cap K / Z(K)$ not be contained in $R / Z(K)$.
If $K\cong Sp_{2n}(q)$, then $Z_2$ is centralized by $Sp_{2n-4}(q)$. If
$n\ge 4$, then $p\mid q^2-1$, a contradiction. Hence we have $K\cong Sp_6(q)$.
But in this case there is a $p$-element $\omega \in \bC(Z_2)$ with $m_p
(\bC_M(\omega))\ge 3$, hence $N_G(\langle \omega \rangle) \subseteq M$,
contradicting $(\beta)$.\\

Now $Z_2\subseteq Z(N(R))$. But then any $x\in Z_2^{\sharp}$ is centralized by
$Sp_4(q)$ in $K$. Hence $\bC_G(x)\subseteq M$ for every 
$x\in Z_2^{\sharp}$. This shows $N_G(Z_2)\subseteq M$, a contradiction.

\absa
{\bf(3.5) Lemma.} 
{\it We have $[Z_2, K\cap G_1] \not\subseteq O_2(G_1)$.}

\absa
Proof. Suppose false. By (3.4) we get $[Z_2, O_2(G_1\cap K)]\neq 1$. Suppose
that in case of $K\cong F_4(q)$ we do not have a diagram automorphism in M.
Now there is a group $U$ of order $q$ in $O_2(G_1\cap K)$ with $U\cap \bC
(Z_2)=1$ and $\mid Z_2:\bC_{Z_2}(U)\mid \le q$ (see (1.57)). Hence $Z_2$ is an
$F$-module. If $K\cong F_4(q)$ and $[Z_2, N_K(R)]\not\subseteq O_2(N_K(R)))$, 
we see with (1.58) that $\mid Z_2:\bC_{Z_2}(Z(O_2(N_K(R))))\mid$ $\le \mid Z
(O_2(N_K(R))):\bC_{Z(O_2(N_K(R)))}(Z_2)\mid$.
Hence also in this case $Z_2$ is an $F$ - module. Now by (2.1) and (1.31) 
we have that $G_2$ is solvable or $E((G_2/Q_2) /F(G_2/Q_2))$ $\cong L_2(r),
L_2(r)\times L_2(r), r$ even, or $A_9.$\\ 

Assume first that in case of $F_4(q)$ there is no diagram automorphism
involved.\\

Let first $K=G(q), q>2$. Then for every $u\in U^\sharp$ we have $\bC_{Z_2}
(u)=\bC_{Z_2}(U)$ and so we get $E(G_2/Q_2)\cong L_2(r)$ or $L_2(r)\times
L_2(r)$, and $\mid U\mid =r$. Now we see $q=r$. Let $\omega \in M\cap G_2, o 
(\omega) = p\mid q-1$. Then $N_G(\langle \omega \rangle) \not\le M$. 
By (1.2) we see $K\cong L_n(q), n\le 4, Sp_6(q)$, or $U_n(q), n\le 7$.

If $K\cong L_n(q)$ or $Sp_6(q)$, then every $t\in \sigma (M)$ divides $q-1$, 
a contradiction.
Let $K\cong U_n(q), n\le 7$. Now there is $t\in \sigma (M), t\mid q+1$. Then 
there is $\nu \in K, o(\nu)=t, N_G(\langle \nu\rangle) \le M,[\nu,a]=1,a\in 
Z_2, a\sim x \in R^\sharp$. But $\bC_G(x)\subseteq M$ and so $\bC_G(a)
\subseteq M^g$ for suitable $g\in G_2\setminus M$. Hence $\nu \in M^g$. 
But then $M=M^g$, a contradiction.

So we are left with $K=G(2)$. Now $U$ induces transvections and so $G_2$ is
solvable or $E((G_2/Q_2)/F(G_2/Q_2))\cong L_2(4)$ or $A_9$. For some prime 
$p$ we have $m_p(K)\ge 3$. This shows with (1.3) $p=3$ and so $G_2/Q_2\cong  
\Sigma_3$. As before we see that $t\in Z_2$ with $[t,O_2(G_1)]\neq 1$ is
not centralized by a $3-$element in $K$. By (1.60) we see that $m_3(K)=3$ and 
so $O_2(K)\neq 1$. Hence $\Omega_1(Z(S))\cap O_2(K)\neq 1$. This gives
$Z(G_2)\neq 1$ and so $m_3(\bC_K(Z(S)\cap K))\le 1$. Application of (1.60)
again implies that no $2$-local in $K$ contains an elementary abelian group
of order $27$, contradicting the definition of an uniqueness group.

So we are left with $K\cong F_4(q)$ and some diagram automorphism is involved.
Furthermore we may assume that $Z_2\not\subseteq O_2(N_K(R)), R$ a root group.
Now choose $x\in Z(O_2(N_K(R))), [x, Z_2]\neq 1$. Then $\mid [x,Z_2]\mid \ge q$. 
Let 
$y\in O_2(N_K(R)) \backslash Z(O_2(N_K(R)))$ with $[y,Z_2]\neq 1$, then $\mid 
[y,Z_2]\mid \ge  q^2$.

Suppose first $q>2$. Then we get $\mid [y,Z_2] \mid \ge 16$ and so we see that
$Z_2 / \bC_{Z_2}(G_2)$ is the natural module and $E(G_2/\bC_{G_2}(Z_2))
\cong L_2(r), r\ge q^2$. As $[x,Z_2]\neq [y,Z_2]$, we see that one of
$x$ or $y$ has to induce a field automorphism on $L_2(r)$. As $\langle x,y
\rangle$ acts quadratically on $Z_2$ this is impossible.

So we have $q=2$. Now $3\in \sigma (G)$ and $O_2 (M)\neq 1$. Hence again
$Z_2 / \bC_{Z_2}(G_2)$ is the natural module. As $\mid [y,Z_2] \mid  \ge 4$,
we get $E(G_2 / \bC_{G_2}(Z_2)) \cong L_2(r), r\ge 4$. As there is a field
automorphism involved, we get that $3\mid r-1$. But then for some $\omega \in
G_2\cap M, o(\omega) =3$, we have $N_G (\langle \omega \rangle)\subseteq M$, 
a contradiction.\\

Now we have $Z_2\not\subseteq O_2(G_1)$. By (1.62) we get that there is some
$g\in G_1$ such that for $X=\langle Z_2,Z_2^g\rangle$ we have\\[1ex]
\begin{itemize}
\item[(1)]   $X / O_2(X)\cong D_{2u}$ ($u$ odd), $L_2(q_1)$ or
      $PSp_4(q_1),q_1$ even.
\vspace{2ex}  
\item[(2)]   $Y=(Z_2\cap O_2(X)) (Z_2^g \cap O_2(X) \unlhd X$
\vspace{2ex}
\item[(3)]   $Y\neq Z_2\cap O_2(X)$
\vspace{2ex}
\item[(4)]   $\mid Z_2:\bC_{Z_2}(Y/\bC_Y(Z_2))\mid
 \le \mid Y:Y\cap Q_2\mid^2$
\end{itemize} 
\vspace{3ex}
Suppose equality in (4). Then we have $\mid Y:Z_2\cap Z_2^g\mid=4,q_1^2,q_1^
4$, respectively. In all cases $X$ acts transitively on 
$(Y/Z_2\cap Z_2^g)^\sharp$. Hence $Y'=1$. But then $\mid Z_2:\bC_
{Z_2}(Y)\mid =2,q_1,q_1^2$, a contradiction. So we have

\absa
{\bf(3.6) Lemma.}  {\it $\mid Z_2:\bC_{Z_{2}}(Y)\mid <\mid Y:Y\cap Q_2\mid ^2$}.
\\

Now application of (1.36) implies that we have one of the following 
possibilities for the structure of $G_2$ and $\bar{Z}_2=Z_2/\bC_{Z_{2}}
(E(G_2/Q_2))$.\\[1ex]
\begin{itemize}
\item[(1)] $Z_2$ is an F-module
\vspace{2ex}
\item[(2)] $E(G_2/Q_2)\cong (S)L_3(r)$ and $\bar{Z}_2$ is the natural module
extended by its dual
\vspace{2ex}
\item[(3)] $E(G_2/Q_2)\cong L_2(r), r=t^2, \bar{Z}_2$ is the orthogonal 
module.
\vspace{2ex}
\item[(4)]$E(G_2/Q_2)\cong PSp_4(q)$ or $3\cdot A_6, \bar{Z}_2$ is a
direct sum of the natural and the dual module.
\vspace{2ex} 
\item[(5)] $E(G_2/Q_2)\cong A_9$
\vspace{2ex}
\item[(6)]$E(G_2/Q_2) = X_1\times X_2\cong L_2(t^2)\times L_2(t^2),\bar{Z}_2 =
\bar{Z}_2^{(1)}\bigoplus \bar{Z}_2^{(2)}, \bar{Z}_2^{(i)}$ is the ortho\-gonal 
$X_i$-module and $[X_i,\bar{Z}_2^{(j)}]=1$ for $i\neq j, i,j = 1,2$.
\vspace{2ex}
\item[(7)]$E(G_2/Q_2)\cong L_2(r)\times L_2(r)$ and $Z_2$ ist the natural
$O_4^+(r)$-module.
\vspace{2ex}
\item[(8)]$E(G_2/Q_2)=X_1\times X_2\cong L_3(2)\times L_3(2), \bar{Z}_2=
\bar{Z}_2^{(1)}
\bigoplus \bar{Z}_2^{(2)}, \bar{Z}_2^{(i)}$ is the direct sum of the natural
module and its dual for $X_i$ and $[X_i,\bar{Z}_2^{(j)}]=1$ for $i\neq j,
i, j=1,2$.
\vspace{2ex}
\item[(9)]$E(G_2/Q_2)=X_1X_2\cong 3A_6\ast 3A_6, \bar{Z}_2=\bar{Z}_2^{(1)}
\bigoplus \bar{Z}^{(2)},\bar{Z}_2^{(i)}$ is the direct sum of the natural and
dual module for $X_i$ and $[X_i,\bar{Z}_2^{(j)}]=1$ for $i\neq j, i,j=1,2$.
\vspace{2ex}
\item[(10)]$G_2$ is solvable, $F(G_2/Q_2)\cong 3^{1+4},\mid \bar{Z}_2 \mid =
2^8$.
\end{itemize}

We now fix the following notation. Let $t$ be a Zsigmondy prime dividing 
$r-1, t=9$ in case of $r=64$, and $\omega \in G_2\cap K, o(\omega)=t$.


\absa
{\bf(3.7) Lemma}. {\it $[K,\omega] \neq 1$.}

\absa
Proof. Otherwise by (1.63) we have $N_G(\langle \omega \rangle) \subseteq M$. But 
this contradicts the structure of $G_2$. 

\absa
{\bf(3.8) Lemma}. {\it $\omega$ normalizes a Borel subgroup of $K.$}

\absa
Proof. Suppose false. As $\langle S,\omega \rangle$ is a $\{2,t\}$ - group,
we have that $K$ has to have a solvable minimal parabolic. This now implies
$K=G(2)$ and $t=3$ or $5$. In particular $r=4,64$ or $16$ and $t=3$, while
$t=5$ implies $K\cong$ $^2F_4(2)$, a contradiction. As $m_p(K)\ge3$ for
$p\in \sigma (M)$, we get $p=3$, contradicting $o(\omega)=3$ or $9$ and
$N_G(\langle \omega \rangle)\not\subseteq M$.

\absa
{\bf (3.9) Lemma}. {\it $G_2$ is solvable or $E(G_2 / Q_2)\cong L_3(2),
A_6, 3\cdot A_6, A_9,L_3(2)\times L_3(2)$ or $3A_6 \ast 3A_6$}.\\

Proof. Let $r\le q$. By (3.8) $\omega$ normalizes a Borel subgroup of $K$.
This implies $O_2(G_1\cap K)Q_2 / Q_2\subseteq E(G_2 / Q_2)$. Now we see
$\mid O_2(G_1\cap K) / R:\bC_{O_2(G_1\cap K) /R}(t)\mid \le r^2\le 
q^2$, for $t\in Z_2$, and $\mid O_2(N_K(R)/R):{\bC}_{O_2(N_K(R)/R)}(t)
\mid \le^2\le q^2$ for $K\cong F_4(q)$ and $M$ involves a diagram 
automorphism.

This now implies with (1.59) that $q=r$ and $K\cong (S)L_n(q)$, $(S)U_n(q)$,
 $Sp_{2n}(q)$. Furthermore $\mid O_2(G_1\cap K)/R:\bC_{O_2(G_1\cap K)/R}
(t)\mid=q^2$. Inspection of the groups in (1) - (7) shows that we have
$E(G_2/Q_2)\cong (S)L_3(q),Sp_4(q)$ or $L_2(q)\times L_2(q)$, where $\bar{Z}_2
$ is the $O^+(4,q)$ - module in the latter.

Let $E(G_2/Q_2)\cong (S)L_3(q)$ or $Sp_4(q)$. Then we may assume that $Z_2
^{(1)}\not\subseteq O_2(G_1\cap K)$, where $\bar{Z}_2^{(1)}$ is the natural
module. Hence $R\cap Z_2^{(1)}\neq 1, R$ a root group. But ${\bC}_G(u)
\subseteq M$, for $u\in R^\sharp$, a contradiction.
We are left with the $O^+(4,q)$ - module. Now $p\not\,\mid  q-1$ for $p\in \sigma
(M)$. This implies $K\cong Sp_6(q)$ or $(S)U_n(q),n\le 7$, by (1.2). As
$M\cap G_2$ has a factorgroup isomorphic to $\Bbb{Z}_{q-1}\wr \Bbb{Z}_2$, we 
see that in case of $K\cong Sp_6(q)$ there is some $\omega_1 \in M\cap K,
o(\omega_1)=t,[\omega_1, K]=1$. Now we get with (1.63) $N_G(\langle 
\omega_1 \rangle) \subseteq M$, a contradiction.

Hence we have $K\cong (S)U_n(q)$. Then there is $p\in \sigma (M)$ such that
$\Omega_1 (Z(S))$ is centralized by an elementary abelian group of order
$p^2$ in $K$. This implies $Z(G_2)=1$ and so $\mid Z_2\mid =q^4$. Let
$R$ be the root group in $K$, then we see $R\subseteq [Z_2,O_2(G_1\cap K)]$
and so $\Omega_1(Z(S))\subseteq R$. Hence $F^*(M)=K$. Now we see
$\mid O_2(G_1):{\bC}_{O_2(G_1)}(Z_2)\mid = q^2$. Hence ${\bC}_{O_2(G_1)
Z_2}(Z_2)=Z_2\cdot T$, where $T$ is a special group of order $q^{1+2(n-2)}$.
In particular $Z_2$ contains a conjugate $R^g, g\in K, R^g\neq R$. As
$N_G(R)\subseteq M$, we have that $N_G(R^g)\subseteq M$. But $\langle 
N_{G_2}(R), N_{G_2}(R^g)\rangle =G_2$, a contradiction.

So assume now $r > q$. By (1.64) $K\cong (S)U_n(q)$,$^2E_6(q),\Omega_{2n}^-(q),r=q^2$. Now $3\mid r-1$ and so in any case $3\not\in \sigma(M)$. This shows $K\cong (S)U_n(q),n\le 7,\Omega_8^-(q)$ by (1.61). Now
in any case there is $p\mid q^2-1,p\in \sigma (M)$. This implies
that there is some $\omega_1 \in G_2 \cap M,o(\omega_1)=p$. If 
$m_p({\bC}(K)) \neq 1$, we see $m_p({\bC}_M(\omega_1)) \ge 3$. Hence
$N_G(\langle \omega_1 \rangle) \subseteq M$, a contradiction.

So we have $m_p(Aut(K))\ge 4$. As there is no field automorphism of order
$p$ involved in $M$ by (1.68), we see $K\cong (S)U_n(q), 5\le n\le 7,
p\mid q+1$. Now $\omega_1$ normalizes a parabolic $P$ in $Aut (K)$ with
$P/O_2(P)$ contains $(S)U_3(q)\times \Bbb{Z}_{q+1}, L_2(q)\times L_2(q^2)
\times \Bbb{Z}_{q+1}$, or $L_2(q)\times (S)U_3(q)$, respectively. Hence in
all cases $m_p ({\bC}_M(\omega_1)) \ge 3$, the same contradiction as 
before.

\absa
{\bf (3.10)  Proposition.} {\it If $M$ is a uniqueness group in $G$ then
$F^*(M)=O_2(M)$}.

\absa
Proof. Suppose first $E(G_2/Q_2)\not\cong A_9$. Then (1) - (10) shows that
there is some $t\in Z_2\backslash O_2(G_1\cap K)$ such that $\mid O_2(G_1
\cap K):\bC_{O_2(G_1\cap K)}(t)\mid \le 4$. Now (1.59) yields
$K\cong L_n(2),(S)U_n(2)$ or $Sp_{2n}(2)$. As $m_p(Aut(K))\ge 3$, we see
$3\in \sigma (M)$. Hence we get that $E(G_2/Q_2)\cong L_3(2)$ or $G_2/
{\bC}_{G_2}(Z_2)\cong \Sigma_3$. In the latter $\mid O_2(G_1\cap K):
{\bC}_{O_2(G_1\cap K)}(t)\mid =2$, a contradiction.

So we are left with $E(G_2/Q_2)\cong L_3(2)$. It is $O_2(G_1\cap K)Q_2/Q_2$
an elementary abelian normal subgroup in $S/Q_2$. But then $\mid O_2(G_1
\cap K): O_2(G_1\cap K)\cap Q_2\mid = 2$ and so $\mid O_2(G_1\cap K)
:{\bC}_{O_2(G_1\cap K)}(t)\mid \le 2$, a contradiction.

So we are left with $E(G_2/Q_2)\cong A_9$. We have $3\not\in \sigma (M)$.
Further $M\cap G_2/Q_2\ge A_8$. If $(M\cap G_2)^{(\infty)} \subseteq
{\bC}(K)$, then $m_3(K)=1$. But then we have a contradiction to (1.3)
(i),(ii). Hence we have that $K$ possesses a parabolic $P$ with $O^{2'}
(P/O_2(P))\cong A_8$. Then $K\cong L_n(2), Sp_{2n}(2),\Omega_{2n}^+(2),E_n(2)$,
or $^2E_6(2)$. As $3\not\in \sigma (M)$, we get $K\cong L_n(2),n\le 7,
Sp_8(2)$, or $\Omega_8^+(2)$. As $m_p(K)\ge 3$ for some prime $p\neq 3$,
we have a contradiction.

















\begin{center}
\S~4 The group $G_{1}$ 
\end{center}

\vspace{1cm}

In this chapter we construct the group $G_{1}$. To the pair $G_{1}$,
$G_{2}$ we are going to apply the amalgam method.

\vspace{5mm}

By (3.10) we have $F^*(M) = O_2(M)$ for the uniqueness group $M$. 
If $3\in\sigma (G)$, we
choose $M$ such that $m_{3}(M)\geq 4$.\\

\absa
{\bf (4.1)~Lemma.~}{\it Let $S\in\mbox{Syl}_{2} (M)$ and $N$ be any
$2$-local of $G$ with $S\leq M\cap N$. If $m_{3}(N)\geq 4$, then
$N\leq M$. Furthermore if $m_{3}(M)\geq 4$, and $m_{3}(N)\geq
2$, then $N\leq M$.}

\absa
Proof~. In both cases by choice of $M$ we have $3\in\sigma (G)$ and
$m_{3}(M)\geq 4$. Let $P\in\mbox{Syl}_{3} (N)$. Then $P$ contains
some $E$, $E$ elementary abelian of order 9. Let $P\leq
Q\in\mbox{Syl}_{3} (G)$. Then there is some $g\in G$ such that
$Q\leq M^{\, g}$. As $\Gamma_{E,1} (G)\leq M^{\, g}$, we now get
$N\leq M^{\, g}$. So $S\leq M^{\, g}$. But then we may
assume that $g\in N_{G}(S)\leq M$, i.e. $M=M^{\, g}\geq N$.

\absa
{\bf (4.2)~Lemma.~}{\it Let $p \in \sigma(M)$ and $E$ be an elementary ablian group of
order $p^{\, 3}$ in $M$ such that $E\cap M^{\, g}\not= 1$ for some
$g\in G$, where $E\leq O_{p^\prime ,p}(M)$ if $M$ is exceptional.
Then $M=M^{\, g}$.}

\absa
Proof~. Let $\nu\in E\cap M^{\, g}, o(\nu)=p$. Then
$N_{G}(\langle\nu\rangle )\leq M$. Let $P\in\mbox{Syl}_{p}
(\bC_{M^g}(\langle\nu\rangle ))$. Then $m_{p}(P)\geq 2$. Now
$N_{G}(P)\leq M^{\, g}$. Using this argument several times, we get
that $M$ and $M^{\, g}$ share a Sylow p-subgroup. But then $M=M^{\,
g}$, as $M^{\, g}$ contains $E$ and $\Gamma_{E,1} (G)\leq M$. 

\absa
{\bf (4.3)~Lemma.~}{\it Denote by ~$\bar{\,}$~ the natural homomorphism onto
$M/O_2(M)$. Let $\overline{K}\lhd\overline{M} ,m_{r}(\overline{K} )\leq
3$, for all odd primes $r$. Let $\omega\in K,o(\omega )=r, r$ an odd
prime. Then $N_{G}(\omega )\leq M$ or $M$ is exceptional or
$L=O^{\, p^\prime } (M/O_{p^\prime }(M))$ is a Sylow $p$-subgroup of
$L_{2}(p^{\, n}), U_{3}(p^{\, n})$ or $^2G_{2}(3^{\, n})$ and $M$
induces a Borel subgroup on $L$. Furthermore in the latter
$\bC_{O_{2}(M)} (\omega )=1$.}

\absa
Proof~. Let $\omega\in R\in\mbox{Syl}_{r}(\overline{K} )$. Then $\overline{M}
=\overline{K} N_{\overline{M}} (R)$. Set $Z=\Omega_{\, 1} (Z(R))$. Then
$m_{p}(N_{G}(Z))\geq 4$ and so application of (1.63) shows that 
$E(N_{G} (Z)/O_{p^\prime} (N_{G}(Z)))\cong L_{2}(p^{\, n}
),U_{3}(p^{\, n})$, $^2G_{2}(3^{\, n})$, or $M$ is exceptional, or $N_{G}(Z)\leq
M$.

Suppose that $N_{G}(Z)\leq M$. Let
$H=N_{G}(\langle\omega\rangle )$. Then $Z\leq H$. Suppose
furthermore $H\not\leq M$. Suppose finally that
$N_{M}(\langle\omega\rangle )$ contains an elementary abelian
$p$ - group $E$ such that $\Gamma_{E,1} (G)\leq M$. Then the structure
of $H$ is given in (1.63). We have $R\cap H\lhd H\cap M/H\cap
O_{2}(M)$. Now we get that $R\cap H\leq O_{p^\prime} (H)$. Let
$R_{1}\in\mbox{Syl}_{r} (O_{p^\prime} (H)), R\cap H\leq R_{1}$.
Then $H=O_{p^\prime} (H)N_{H} (R_{1})$. In particular
$N_{H}(R_{1})\not\leq M$. As $Z\leq R_{1}$, we get that $\bC_{H}
(R_{1})\leq O_{p^\prime} (H)$. As $O_{p^\prime}(H) = 
\langle \bC_{O_{p^\prime}(H)}(x) | x \in E^\sharp \rangle $ $\leq M$,
we have $R_1 \leq M$. If $m_{r}(R_{1})\geq 4$, then $r$ is an
uniqueness prime, contradicting $H \not\leq M$. 
So we have $m_{r}(R_{1})\leq 3$. Let $C$ be a
critical subgroup of $R_{1}, C_{1}=\Omega_{\, 1} (C)$. Then
$C_{1}\cong \bZ_{r},\bZ_{r}\times \bZ_{r}, Z_{r^{1+2}},\bZ_{r}\times Z_{r^{\,
1+2}}, \bZ_{r}\times \bZ_{r}\times \bZ_{r}$ or $Z_{r^{\, 1+4}}$. In any
case $C_{H}(C_{1})\leq O_{p^\prime }(H)$. So $E(H/O_{p^\prime} (H))\lesssim
L_{3}(r)$ or  $Sp_{4}(r)$. Application of (1.63) and \cite{Mi1},\cite{Mi2}
shows $E(H/O_{p^\prime }(H))\cong A_{6}$. But either $2\cdot A_{6}$ is
induced, contradicting $2\not\Big |\, |O_{p^\prime} (H)|$ by (1.63) or
$3\cdot A_{6}$ is induced, but $p=3$, as $m_{p}(H/O_{p^\prime}
(H))=2$, and now $3\,\Big |\, | O_{p^\prime} (H)|$, a contradiction.

So we have $N_{G}(\langle\omega\rangle )\leq M$ if $N_{G}(Z)\leq M$ and
$N_{G}(\langle\omega\rangle )$ contains $E$ with $\Gamma_{E,1}
(G)\leq M$.

So let now $F\leq M, F$ elementary abelian of order $p^{\, 4}, F\leq
N(R)$. Suppose furthermore $|\bC_{F} (\omega )|\leq p$. 

Let $D$ be a critical subgroup of $R$ and $D_{1}=\Omega_{1} (D)$.
Then $|F : \bC_F(D_{1})| \leq p$. This immediately shows that $D_{1}\cong
\bZ_{r}\times\bZ_{r}\times\bZ_{r}$ , or $D_{1} \cong Z_{p^{\,
1+4}}$ and $p\,\Big |\, r-1$. 

Suppose the latter. Then we get that $R\cong D_{1}*D_{2},D_{2}$
cyclic. As $\omega\in R$, we get $\omega\in D_{1}$ and so
$\langle\omega\rangle$ is normalized by an elementary abelian group
of order $p^{\, 2}$ in $F$.

So we have $D_{1}$ is elementary abelian of order $r^{\, 3}$. But now
$F$ induces an elementary abelian group of order $p^{\, 3}$ in
$GL_3(r)$. This now shows that $M$ possesses a normal subgroup of
index $p$ contradicting (1.68).
So we have that $N_G(\langle \omega \rangle )$ does contain some 
$E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \leq M$. Hence 
$N_{G}(\langle\omega\rangle )\leq M$, if
$N_{G}(Z)\leq M$.

Suppose finally $N_G(Z) \not\leq M$. Then a Borel subgroup $Y$ of $L_2(p^n)$,
$U_3(p^n)$ or $^2G_2(3^n)$ acts on $R$. Now let again $D_1 = \Omega_1(D)$,
$D$ a critical subgroup of $R$. As $m_r(D_1) \leq 3$, we see that $O_p(Y)$
acts trivially on $D_1$ and so it centralizes $R$. Hence by (1.63) we have
$\bC_{O_2(M)}(\omega) = 1$.\\

The choice of $G_{1}$ is a little bit complicated. The aim is to
have $G_{2}\cap M\leq G_{1}$ and $m_{p}(G_{1})$ large enough for some
$p\in\sigma (M)$. Furthermore $G_{1}$ should have a more or less
simple structure.

Denote by $\overline{M} =M/O_{2}(M)=K_{0}$. Fix some $p_{\, 1}\in\sigma (M)$.
Let $K_{1}=O_{{p_{1}}\!^\prime} (\overline{M} )$. Suppose there is some
$p_{\, 2}\in\sigma (M)$ with $m_{p_{\, 2}}(K_{1})\geq 4$. Then set
$K_{2}=O_{{p_{2}}\!^\prime} (K_{1})$. After finitely many steps we
end with some $K=K_{i}$ such that $m_{r}(K)\leq 3$ for any prime $r,r\,\Big |\,
|K|$. Furthermore $m_{p}(K_{i-1})\geq 4$ for some $p\in\sigma (M)$.
Set $H=K_{i-1}(\overline{G_{2}\cap M} )$. Now $F(H/K)$ is a $p$-group. We
construct $G_{1}$ in a preimage of $H$. 

\absa
{\bf (4.4)~Lemma.~}{\it One of the following holds}
\begin{itemize}
\item[(a)] {\it For suitable choice of $G_1$ we have that $G_2 \cap M \leq G_1$ 
and one of the following holds for $Z_1 = \langle \Omega_1(Z(S))^{G_1} \rangle$}
\begin{itemize}
\item[(i)]{\it $G_1 = \bC_{G_1}(Z_1)(G_2 \cap M)$, $m_p(\bC_{G_1}(Z_1)) \geq 3$.
If $G_2/O_2(G_2) \cong Sz(q)$ (including $q = 2$, i.e. $F_{20}$),
$\langle \Omega_1(Z(S))^{G_2} \rangle = Z_2$ is the natural module and for
$Z = \bC_{Z_2}(\Omega_1(S/O_2(G_2)))$ we have $T_1 = \langle Z^{G_1} \rangle$
$\not= Z$ then $m_p(\bC_{G_1}(T_1)) \leq 1$.}   
\item[(ii)]{\it $G_1 \not= \bC_{G_1}(Z_1)(G_2 \cap M)$, $m_p(G_1) \geq 2$. If 
$m_p(G_1) \leq 4$ set $m_p(G_1) = 4 - r$. If $r \not= 0$, then we have $m_p(\bC_M(Z_1)) \geq 1$ and $m_p(\bC_M(Z_1)) \geq 2$ if not $r=1$ and $m_p(\bC_{G_1}(Z_1)) = 0$.  Furthermore
if $x \in \bC_{G_1}(Z_1)$ is a 2--element then $x \in G_2 \cap M$. If 
$r = 1$, any $E \cong E_{p^2}$, $E \leq G_1$ is contained in some $F \cong E_{p^3}$ with
$|F \cap \bC_{M}(Z_1)| \geq p$. If $r = 2$ any $x \in Z_1$ is centralized 
by some $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \leq M$.}
\end{itemize}
\item[(b)]{\it $G_2/O_2(G_2) \cong Sz(q)$, $\langle \Omega_1(Z(S))^{G_2} \rangle = Z_2$
is the natural module and for $Z = \bC_{Z_2}(\Omega_1(S/O_2(G_2)))$ we
have $T_1 = \langle Z^{G_1} \rangle \not= Z$ and $m_p(\bC_M(T_1)) \geq 2$.
If $y \in G_1$, $y$ a 2--element with $[T_1,y] = 1$, then 
$y \in O_2(G_1)$. If $x \in T_1$ and $x \not= 1$, then $M$ is the unique maximal
2--local of $G$ containing $\bC_G(x)$.}
\end{itemize}
{\it Furthermore in (a) and (b) if $F(H/K)$ is a Sylow $p$--subgroup of
$L_2(p^n)$, $U_3(p^n)$, or $^2G_2(3^n)$ with $m_p(F(H/K)) \geq 3$ and $M$ induces a Borel
subgroup $B$ of the corresponding simple group on $F(H/K)$, then $G_1K/K \geq B$.}

\absa
Proof~. We will prove (4.4) in several steps according to the structure of
$H/K$.

\absa
{\bf (4.4.1)~}{\it If $m_p(F(H/K)) \geq 4$ the lemma holds.}

\absa
Let $H_1 \leq H$ be minimal with respect to $G_2 \cap M \le H_1$ and
$H_1K/K \geq F(H/K)$. Set $Z_1 = \langle \Omega_1(Z(S))^{H_1} \rangle$ and
$T_1 = \langle Z^{H_1} \rangle$ if $G_2/O_2(G_2) \cong Sz(q)$ and 
$Z_2$ is the natural module. 

Let first $m_p(\bC_{H_1}(Z_1)) \ge 3$. Let $G_1$ be minimal with $G_2 \cap M \leq G_1$ and 
$G_1K/K \geq \bC_{H_1}(Z_1)K/K$. Hence (a) holds or $G_2/O_2(G_2) \cong Sz(q)$,
$Z_2$ is the natural module, $T_1 \not= Z$ and $m_p(\bC_{G_1}(T_1)) \geq 2$.
Suppose the latter. If $m_p(\bC_{H_1}(T_1)) \geq 3$ set $G_1$ minimal with
respect to $G_1K/K \geq \bC_{H_1}(T_1)K/K$ and $G_2 \cap M \leq G_1$
hence again (a)(i) holds.

So assume next that $m_p(\bC_{H_1}(T_1)) = 2$. Suppose 
$\bC_{H_1}(T_1)K/K \leq F(H/K)$. Then set $G_1 = \bC_{H_1}(Z_1)(G_2 \cap M)$.
Hence either (a)(i) or (b) holds. In fact we have that any 2--element
$y \in \bC(T_1)$ is contained in the preimage of $F(H/K)$ and so is
contained in $O_2(G_1)$. Finally any $x \in T_1$ is centralized by
some $E \cong E_{p^2}$, $E \leq H_1$, with $\Gamma_{E,1}(G) \leq M$, 
as $m_p(F(H/K)) \geq 4$.

So assume now that $\bC_{H_1}(T_1) \not\leq F(H/K)$. Set
$E = F(H/K)$ and $U = [\bC_{H_1}(T_1), E/\Phi(E)]$. As
$O_p((H/K)/F(H/K)) = 1$ we see that $(H_1K/K)/E$ is
a subgroup of $GL_2(p)$. Set $\tilde{S} = O_2(\bC_{H_1}(T_1))$. Then
$\tilde{S}K/K$ is nontrivial. Set $\tilde{U} = \bC_E(\tilde{S})$. As
$m_p([\tilde{S}, E]) \leq 2$, we see that $m_p(\tilde{U}) \geq 2$. 
Let $m_p(\bC_E(T_1)) \geq 2$, then let $G_1$ be minimal with
$G_2 \cap M \leq G_1$ and $\tilde{U} \leq G_1K/K$. Then we have (b).
If $y$ is a 2--element centralizing $T_1$, then $y \in \tilde{S}$ and so
$[y, \tilde{U}] = 1$, i.e. $y \in O_2(G_1)$.

So let $m_p(\bC_E(T_1)) = 1$ and then also $m_p([\tilde{S}, E]) = 1$.
In this case set $G_1 = (G_2 \cap M)\bC_{H_1}(Z_1)$ and so (a)(i) holds.


So assume now $m_p(\bC_{H_1}(Z_1)) \leq 2$. If $\bC_{H_1}(Z_1)K/K \leq
F(H/K)$, then set $G_1 = H_1$ and we have (a)(ii).

So assume $\bC_{H_1}(Z_1)K/K \not\leq F(H/K)$. Set again
$E = F(H/K)$ and let $U = [\bC_{H_1}(Z_1), E/\Phi(E)]$. Let $\tilde{S} = O_2((\bC_{H_1}(Z_1)K/K)/E)$. Then
again $\tilde{S}$ is nontrivial and $(H_1K/K)/E$ is isomorphic to a subgroup of
$GL_2(p)$. Set $\tilde{U} = \bC_E(\tilde{S})$. Suppose first $[\bC_{G_2 \cap M}(Z_1), \tilde{U}] = 1$.
As $m_p([\tilde{S}, E]) \leq 2$, we see $m_p(\tilde{U}) \geq 2$. Now
let $G_1$ be minimal with $G_2 \cap M \leq G_1$ and $G_1K/K \geq \tilde{U}$. Then we
have (a)(ii). Obviously any 2--element in $G_1$ centralizing $Z_1$ is
contained in $G_2 \cap M$. Let $m_p(\tilde{U}) = 3$ and let $\hat{U} \leq \tilde{U}$ with $\hat{U} \cong E_{p^2}$.
Set $U_1 = [\tilde{S}, E]$. As $m_p(U_1) \leq 2$ a critical subgroup $V$ of $U_1$ is of
rank at most two. As $\tilde{S}$ acts on $V$ we get some $W \leq \Omega_1(V)$,
$|W| = p$, $W \cap \hat{U} = 1$ and $[W, \hat{U}] = 1$. So
$\langle \hat{U}, W \rangle \cong E_{p^3}$.

Suppose now $m_p(\tilde{U}) = 2$, then $m_p([\tilde{S},E]) = 2$ and so
$m_p(\bC_{H_1}(Z_1)) \geq 2$. Furthermore any $x \in Z_1$ is
centralized by some $F \cong E_{p^2}$, $F \leq [\tilde{S},E]$ with 
$\Gamma_{F,1}(G) \leq M$, as $m_p(F(H/K)) \geq 4$ and so 
$m_p(\bC_{F(H/K)}(F)) \geq 3$. 

So we are left with $[\bC_{G_2 \cap M}(Z_1), \tilde{U}] \not= 1$.
But then the argument above yield some $\tilde{S}^{(1)}$, a nontrivial 
2--subgroup with $[\tilde{S}^{(1)}, \tilde{U}] \not= 1$ and 
$\tilde{S}^{(1)} \unlhd \bC_{H_1}(Z_1)K/K$, a contradiction to the
choice of $\tilde{S}$.

\absa
{\bf (4.4.2)~}{\it If $m_p(F^*(H/K)) \geq 4$ the lemma holds.}

\absa
By (4.4.1) we may assume $m_p(F(H/K)) \leq 3$. As before choose $H_1$ 
minimal with respect to $G_2 \cap M \leq H_1$ and $H_1K/K \geq F^*(H/K)$. 
Define $Z_1$ and $T_1$ as before. 

If $m_p(\bC_{H_1}(Z_1)) \geq 3$ set $G_1 = \bC_{H_1}(Z_1)(G_2 \cap M)$ or
if $m_p(\bC_{H_1}(T_1)) \geq 3$ set $G_1 = \bC_{H_1}(T_1)(G_2 \cap M)$.
Now as in (4.4.1) we get (a)(i).

So we may assume $m_p(\bC_{H_1}(Z_1)) \leq 2$ or $m_p(\bC_{H_1}(Z_1)) \geq 3$ 
and $m_p(\bC_{H_1}(T_1)) \leq 2$. Set $U_1 = Z_1$ in the first case and 
$U_1 = T_1$ in the second case. We have 
$m_{p}(\bC_{H_{1}} (U_{1}))\leq 2$. Let $E_{1}$ be some
component with $[E_{1},U_{1}]=1$. Then $m_{p}(E_{1})\leq 2$. If
$m_{p}(E_{1})=2$, then choose $G_{1}$ minimal with respect to $G_{\,
2}\cap M\leq G_1$ and $G_{1}K/K\geq\bC_{H/K} (E_{1})$. Let $t$ be a
2-element with $[Z_{1},t]=1$ but $t\not\in G_{\, 2}\cap M$. Then
$Z(E_{1})\not= 1$ and $[Z(E_{1}),t]\not= 1$. We then let $G_{1}$ be
minimal with respect to $G_{\, 2}\cap M\leq G_{1}$ and $\bC_{\bC_{H/K}
(E_{1})} (t)\leq G_{1}$. Now any 2-element centralizing $Z_{1}$ is in
$G_{\, 2}\cap M$. So we have (a)(ii) as we have $m_p(G_1) \geq 2$.

So let next $m_{p}(E_{1})=1$. If there is a second component $E_{2}$
with $[U_{1},E_{2}]=1,m_{p}(E_{2})=1$, then choose $G_{1}$ minimal
with respect to $G_{\, 2}\cap M\leq G_{1}$ and $G_{1}K/K\geq
\bC_{H/K} (E_{1}E_{2})$ and we get (a)(ii) as before.

So we have
\begin{quote}
$(*)$~~There is at most one component $E_{1}$ of $F^{*}(H/K)$ with
$[E_{1},U_{1}]=1$. Furthermore $m_{p}(E_{1})=1$ if it exists.
\end{quote}

\indent
We may assume $m_{p}(\bC_{H_{1}} (Z_{1}))\geq 1$, otherwise choose
$H_{1}=G_{1}$. So assume first $m_{p}(\bC_{H_{1}} (Z_{1}))\geq 3,m_{p}(\bC_{H_{1}}
(T_{1}))\leq 2$. Further we assume there is some 2--element 
$x\in \bC_{H_{1}}(T_{1})$ 
with $[x,F(H/K)]\not= 1$. As in (4.4.1) there is a nontrivial 2--group
$\tilde{S} \leq \bC_{H_1}(T_1)$ with $\tilde{S} \unlhd (H/K)/F^*(H/K)$.
Now $m_p([F(H/K), \tilde{S}]) \leq 2$. Let again 
$\tilde{U} = \bC_{F(H/K)}(\tilde{S})$. If $m_{p}([F(H/K),\tilde{S}])=2$,
we choose $G_{1}$ minimal with $G_{\, 2}\cap M\leq G_{1}$ and 
$$
G_{1}K/K\geq \langle\bC_{E(H/K)} (\bC_{E(H/K)}
(T_{1})),\tilde{U}\rangle\, .
$$
Then we have (b).

Suppose $m_{p}([F(H/K),\tilde{S} ])=1$. If $\bC_{E(H/K)}
(T_{1})\not= 1$, we just choose $G_{1}$ as before. Hence assume that $\bC_{E(H/K)}
(T_{1})=1$. Now choose $G_{1}$ minimal with respect to $G_{\, 2}\cap
M\leq G_{1}$ and $\bC_{H_{1}} (Z_{1})\leq G_{1}K/K$, and we have (a)(i). 

So we may assume that $[\bC_{H_{1}} (T_{1}),F(H/K)]=1$. Let $E_{1}$
be some component with $[E_{1},T_{1}]=1$. If there is no such
component we may choose $H_{1}=G_{1}$ and so any 2-element
centralizing $T_{1}$ is contained in $O_2(G_{\, 1})$ and we have (b).

Now $m_{p}(E_{1})=1$ by $(*)$ or there is a component $E_1$ with 
$[T_1,E_1] = 1$.
We
choose $G_{1}K/K \geq \bC_{F^*(H_1/K)}(E_1)$ and $G_2 \cap M \leq G_1$.
Then we have (b).

Assume now $m_{p}(\bC_{H_{1}} (Z_{1}))\leq 2$.
Suppose there is some 2-element $x\in G_{\, 2}\cap M, [x,Z_{1}]=1$
and $[x,F(H/K)]\not= 1$. As before there is some $\tilde{S}\leq\bC
(Z_{1}),\tilde{S}\lhd (H/K)/F^{*}(H/K)$. Hence
$m_{p}([F(H/K),\tilde{S} ])\leq 2$. If $m_{p}([F(H/K),\tilde{S}
])<m_{p}(F(H/K))$ we may choose $\tilde{U}$ as before. Then choose
$G_{1}$ minimal with $G_{\, 2}\cap M\leq G_{1}$.
$$G_{1}K/K\geq\langle
E(H/K),\tilde{U}\rangle\mbox{~if~}\bC_{E(H/K)} (Z_{1})=1
$$
$$G_{1}K/K\geq\langle
\bC_{E(H/K)} (E_{1}),\tilde{U}\rangle\mbox{~if~} E_{1}=\bC_{E(H/K)}
(Z_{1})\, .$$
Hence we have (a)(ii).

So assume $m_{p}([F(H/K),\tilde{S} ])=m_{p}(F(H/K))$. In particular we have
$m_{p}(F(H/K))\leq 2$. If $\bC_{E(H/K)} (Z_{1})=1$, we let $G_{1}$ be
minimal with respect to $G_{\, 2}\cap M\leq G_{1}$ and
$G_{1}K/K\geq E(H/K)$. In the other case we choose $G_{1}$
minimal with $G_{\, 2}\cap M\leq G_{1}$ and
$G_{1}K/K\geq\bC_{E(H/K)} (E_{1}), E_{1}=\bC_{E(H/K)}
(Z_{1})$. We get (a)(ii) again.

So we may assume $[\bC_{H_1}(Z_1), F(H/K)] = 1$. If 
$\bC_{E(H/K)}(Z_1) \leq F(H/K)$, then choose $G_1 = H_1$ and
we have (a)(ii). So assume there is a component $E_1$ with
$[E_1,Z_1] = 1$. Then by $(*)$ $m_p(E_1) = 1$ and
$E_1$ is the only component with this property. Now choose $G_1$
with $G_2 \cap M \leq G_1$ and $G_1K/K \geq \bC_{F^*(H/K)}(E_1)$. Again we
have (a)(ii).

Assume now that we have the exceptional case or $F(H/K)$ is a Sylow
$p$-subgroup of $L_{2}(p^{\, n}),U_{3}(p^{\, n})$ or $^2G_{\,
2}(3^{\, n})$, where $H$ induces a Borel subgroup of the
corresponding group on $F(H/K)$.\\
\indent
Then choose $G_{1}$ minimal with $G_{\, 2}\cap M\leq G_{1}$ and
$G_{1}K/K$ covers the Borel subgroup of $L_{2}(p^{\, n}),U_{3}(p^{\,
n})$ or $^2G_{\, 2}(3^{\, n})$. In these cases we always have
$m_{p}(G_{1})\geq 3$, where $m_{p}(G_{1})=3$ just happens if we have
the situation of $L_{2}(p^{\, 3})$. Hence we get the assertion.
\\
\\

So assume from now on $m_{p}(F^{*}(H/K))\leq 3$.

\absa
{\bf (4.4.3)~}{\it If there is a component $E_{1}$ of
$H/K$ with $m_{p}(E_{1})\geq 3$, then the lemma holds.}

\absa
By (4.4.2) we may assume that $E_{1}=F^{*}(H/K)$. Let $H_{1}=E_{1}$. 
Define $Z_1$ and $T_1$ as before. Assume
$[E_{1},Z_{1}]\not= 1$. Choose $U > E_{1},m_{p}(U)=4,(G_{\, 2}\cap
M)K/K\leq U$. Set $H_{1}=U$. Now $\bC_{U} (Z_{1})\leq G_{\, 2}\cap M$
and so we may choose $G_{1}$ minimal with $G_{\, 2}\cap M\leq G_{1}$
and $G_{1}K/K\geq U$. So assume now $[E_1, Z_1] =1$. Let $G_1$ be minimal
with $G_2 \cap M \leq G_1$ and $G_1K/K \geq E_1$. Then we have (a)(i) or we have the
Suzuki case and $[E_1,T_1] \not= 1$. 

\absa
{\bf (4.4.4)~}{\it If there is some component $E_{1}$ of
$H/K$ with $m_{p}(E_{1})=2$, then the lemma holds.}

\absa
First of all $E_{1}\not= F^{*}(H/K)$. Otherwise $E_{1}$ admits an
outer automorphism group $E\cong E_{p^{\, 2}}$. By (1.67) $E_{1}$ is
a Lie group and $E$ contains some field automorphism. By construction
$K\lhd \overline{M}$ and $E(H/K)\lhd\overline{M} /K$ besides possibly for
$A_{8}\leq\overline{G_{\, 2}\cap M} ,A_{8}$ a component of $H/K$. As $A_{8}$
does not possess an outer automorphism group of order $p^{\, 2}$, we
have that $E_{1}\lhd\overline{M} /K$. In particular $M$
possesses a normal subgroup of index $p$. This contradicts (1.68).

Now we have $F^{*}(H/K)=E_{1}E_{2}, m_{p}(E_{2})\leq 2$ and
$m_{p}(E_{1}E_{2})=3$. Again there is some $U >
E_{1}E_{2},m_{p}(U)=4,\overline{G_{\, 2}\cap M}\leq U$. Set $H_{1}=U$. Let
$U_{1}=Z_{1}$ or $T_{1}$. If $[E_{1}E_{2},U_{1}]=1$, let $G_{1}$ be
minimal with $G_{\, 2}\cap M\leq G_{1}$ and $E_{1}E_{2} \leq G_{1}K/K$.
Assume $[E_{1}E_{2},U_{1}]\not= 1$. Let $[E_{1},U_{1}]=1$. If
$m_{p}(\bC_{U} (U_{1}))=3$ let $G_{1}$ be minimal with respect to $G_{\, 2}\cap
M\leq G_{1}$ and $\bC_{U} (U_{1})\leq G_{1}K/K$.

So assume $m_{p}(\bC_{U} (U_{1}))=2$. As $m_p(U) = 4$ we see that there is
a $p$--element $\omega \in U \setminus E_1E_2$ such that 
$m_p(E_1E_2\langle \omega \rangle) = 4$. Suppose first $[\omega, E_1] = 1$.
Then let $G_1$ be minimal with $G_1 \geq G_2 \cap M$ and $G_1K/K \geq 
E_2\langle \omega \rangle$. If $E_2$ is nonsolvable or $E_2$ is cyclic
we have that $\bC_{E_2}(U_1) = 1$ and so we get (a)(ii) or (b). So let
$E_2$ be solvable of rank two. Then $p \Big | |Z(E_1)|$. By (1.5) we see $p = 3$.
Now $E_2 = F(\bC_H(E_1))$ and so as $\omega \not\in E_2$, we
get $\bC_H(E_1)/E_2 \cong SL_2(3)$ or $GL_2(3)$. Hence for a
critical subgroup $C$ of $E_2$ we have $\Omega_1(C) \cong Z_{3^{1+2}}$.
This shows that any 2--element in $\bC_{G_1}(U_1)$ is contained in $\bC(E_2)$
and so we have (a)(ii) or (b).

Assume now $[\omega,E_1] \not= 1$. If $p \not\!\!\Big |~ |Z(E_1)|$ we get
$m_p(E_2) = 1$ and $m_p(E_2\langle \omega \rangle) = 2$. Furtermore
$\bC_{E_2\langle \omega \rangle}(U_1) = 1$ and so let $G_1$ be as 
before we get (a)(ii) or (b) provided $\omega$ normalizes 
$\overline{G_2 \cap M} \cap E_1$.
As $m_p(E_1\langle \omega \rangle) = 3$ we get with (1.67) that 
$E_1 \cong SL_3(r), SU_3(r)$ or a Lie group in even characteristic. 
If we have a Lie group in even characteristic $\overline{G_2 \cap M} \cap E_1$
is a parabolic in $E_1$ and so $\omega$ normalizes 
$\overline{G_2 \cap M} \cap E_1$. If $E_1 \cong SL_3(r)$ or $SU_3(r)$, 
$r$ odd, then $r$ has to be a prime and $\omega$ induces a 
diagram automorphism on $E_1$. But then we get a 
contradiction with (1.68).

So assume now that $p \Big| |Z(E_1)|$. Then again $p = 3$ and
$E_1 \cong SL_3(2^n)$ or $SU_3(2^n)$. Hence $\omega$ normalizes
$\overline{G_2 \cap M} \cap E_1$. If $E_2$ is nonsolvable we let
$G_1$ be as before. So let $E_2$ be solvable. Then $m_3(E_2) = 2$. Now
suppose there is $x \in U$, $o(x) = 2$ with $[x,U_1] = 1$ and
$[x,E_2] \not= 1$. Let $C$ be a critical subgroup of $E_2$ and
$C_1 = \Omega_1(C)$. Then we see that $C_1 \cong E_9$ or $\bZ_3$. As
$C_1 \geq Z(E_1)$, we get that $U$ induces $\bZ_2$ or $\Sigma_3$ on
$C_1$. In the latter we have that 
$U = E_1[x,U]\bC_{E_2}(x)\langle x \rangle$. But $\bC_{E_2}(x)$ is
cyclic and so $m_p(\langle E_1, [x,U] \rangle) \geq 3$. But 
$[\langle E_1, [x,U] \rangle,U_1] = 1$, a contradiction. So we
have the former and then we may assume that $\omega$ normalizes 
$\bC_{E_2}(x)$ and then let $G_1$ be minimal with $G_2 \cap M \leq G_1$ and
$G_1K/K \geq \langle \bC_{E_2}(x), \omega \rangle K/K$. 
Hence we get (a)(ii) or (b) again.

So we may assume $[E_1,U_1] \not= 1$. Suppose $[E_2,U_1] = 1$. Then
either $E_2$ is solvable or $m_p(E_2) = 1$. Choose $\omega$ as before. If
$[\omega, U_1] = 1$ choose $G_1$ minimal with $G_2 \cap M \leq G_1$ and
$G_1K/K \geq E_1$. Now we have (a)(ii) or (b).

So let $[\omega ,U_{1}]\not= 1$. Suppose $[E_{1},Z_{1}]=1$. Then
$m_{p}(\bC_{E_{1}E_{2}} (Z_{1}))=3$. If $m_{p}(E_{2})=2$, we choose
$G_{1}$ minimal with $G_{\, 2}\cap M\leq G_{1}$ and $E_{1}\leq
G_{1}K/K$ and (b) holds. If $m_{p}(E_{2})=1$ we
choose $G_{1}$ minimal with $G_{\, 2}\cap M\leq G_{1}$ and
$E_{1}E_{2}\leq G_{1}K/K$ and we have (a)(i). 

So assume $[E_{1},Z_{1}]\not= 1$. If $[\omega,Z_1] = 1$, then 
$[\omega, E_1] = 1$ and so for $G_1$ minimal with $G_2 \cap M \leq G_1$ 
and $E_1 \leq G_1K/K$ we get (a)(ii). So assume
$[\omega ,Z_{1}]\not= 1$. If $m_p(E_2) = 2$ then choose $G_1$ 
minimal with $G_2 \cap M \leq G_1$ and $E_1 \leq G_1K/K$. Hence
we have (a)(ii). If $m_p(E_2) = 1$ and $[\omega, E_2] = 1$,
then choose $G_1$ minimal with respect to $G_2 \cap M \leq G_1$
and $E_1\langle \omega \rangle \leq G_1K/K$. hence we have 
(a)(ii) again.

So let $[E_2, \langle \omega \rangle ] \not= 1$. If $E_2$ is 
nonsolvable then by (1.67) we get (a)(ii) for $G_1$ minimal
with $G_2 \cap M \leq G_1$ and $E_1\langle \omega \rangle \leq G_1$, 
or $E_2 \cong L_2(r^3)$ and $p = 3$. Furthermore $\omega$ induces
a field automorphism on $E_2$. But now we get a normal subgroup of
index $p$ in $M$, contradicting (1.68). So assume that $E_2$ is a
cyclic $p$--group. Let $x \in U$ be a 2--element with $[x,U_1] = 1$
but $[x,E_1] \not= 1$. Then we may assume that $[\omega, x] = 1$. Let
$\langle x \rangle \bC_U(E_2)/\bC_U(E_2) \in Syl_2(U/\bC_U(E_2))$. Then
let $G_1$ be minimal with $G_2 \cap M \leq G_1$ and $E_1\bC_U(x) \leq G_1K/K$.
Again we have (a)(ii).

So assume now also $[E_{2},U_{1}]\not= 1$. If $E_{2}$ is nonsolvable
or any 2-element of $U$ in $\bC (U_{1})$ is contained in $G_{\, 2}\cap
M$, let $G_{1}$ be minimal with respect to $G_{\, 2}\cap M\leq G_{1}$
and $G_{1}K/K=U$ and we have (a)(ii).

So let $x$ be a 2-element, $[x,U_{1}]= 1, x\not\in G_{\, 2}\cap M$. Again
we may assume $[x,U]\leq E_{2}$. As $[E_{2},U_{1}]\not= 1$, we have
$m_{p}(\bC_{E_{2}} (x))\geq 1$ and also $m_{p}(\bC_{E_{2}}
(U_{1}))\geq 1$, as $[[x,E_{2}],U_{1}]=1$. We have $m_{p}(E_{2})=2$
and as $m_{p}(E_{1}E_{2})=3$, we see $p\,\Big |\, | Z(E_{1})|, [E_{2},x]$ is
cyclic and $[E_{2},x]\cap E_{1}=1$, as $[E_1,x] = 1$. 
Furthermore $\bC_{E_{2}} (x)\cap
Z(E_{1})\not= 1$. If $[Z_{1},E_{1}]=1$, we choose $G_{1}$ minimal
with $G_{\, 2}\cap M\leq G_{1}$ and $\langle
E_{1},[E_{2},x]\rangle\leq G_{1}K/K$ and we have (a)(i).

So assume $[Z_{1},E_{1}]\not= 1$. Let $[Z_{1}, E_{2}]=1$. Then we
choose $G_{1}$ minimal with $G_{\, 2}\cap M\leq G_{1}$ and $E_{1}\leq
G_{1}K/K$ and we have (a)(ii).

So we may assume $[Z_{1},E_{2}]\not= 1$. 
If $m_{p}(\langle E_{1},\bC_{E_{2}} (x)\rangle )\geq 3$, we
let $G_{1}$ be minimal with $G_{\, 2}\cap M\leq G_{1}$ and $\langle
E_{1},\bC_{E_{2}} (x)\rangle \leq G_{1}K/K$. Now any $E\leq G_{1},
E\cong E_{p^{\, 2}}$ centralizes some $\nu\in [E_{2},x]^\sharp$, and we
have (a)(ii).

It remains the case that $\bC_{E_{2}} (x)$ is cyclic and
$m_{p}(E_{1}\bC_{E_{2}} (x))=2$. Now let $\omega$ be as before. We
may assume $[\omega ,x]=1$. Hence $\omega$ acts on $E_{1}\bC_{E_{2}}
(x)$. If $[\omega ,U_{1}]\not= 1$, we let $G_{1}$ be minimal with
respect to $G_{\, 2}\cap M \leq G_{1}$ and $\langle E_{1},\bC_{E_{2}}
(x),\omega\rangle\leq G_{1}K/K$. As $[\omega ,\Omega_{1}
([E_{2},x])]=1$, we have (a)(ii).

Let $[\omega ,U_{1}]=1$. In this case $[\omega ,E_{1}]=1$. As
$E_{2}=O_{p}(U)$ and $[x,U] \leq E_2$, we get $m_{p}([x,E_{2}])\geq 2$, 
a contradiction, as $[x,E_2] \cap E_1 = 1$ and $m_p(E_1E_2) =3$. This
proves (4.4.4).

\absa
So we have $m_{p}(E_{1})=1$ for any component of $H/K$.

\absa
{\bf (4.4.5)~}{\it If $E_{1},E_{2}$ are two components of $H/K$.
Then the lemma holds.}

\absa
We have $F^{*}(H/K)=E_{1}\times E_{2}\times E_{3},\, E_{3}$ cyclic or
nonsolvable of $p$-rank 1. Let
$E_{1}E_{2}E_{3} < U,m_{p}(U)=4$. Now there is some $\omega$
with $[E_{1},\omega ]\not= 1,m_{p}(E_{1}\langle\omega\rangle )=2$. By
(1.67) $\omega$ induces a field automorphism on $E_{1}$. Now (1.68)
implies that there is some $t\in S$ with $E_{1}^{\,
t}=E_{2},\omega^{\, t} =\omega^{\,-1}$. Now we may consider $\langle
E_{1},E_{2}\rangle$ as a rank 2 component and we get the
assertion as in (4.4.4).

\absa
{\bf (4.4.6)~}{\it If $m_{p}(E(H/K))=1$, the lemma holds.}

\absa
Now we have $F^{*}(H/K)=E_{1}E_{2}, m_{p}(E_{1})=1, E_{2}$ is
solvable with $1\leq m_{p}(E_{2})\leq 2$. Suppose first $m_{p}(E_{2})=1$. Then
we get some $\langle\omega_{1} ,\omega_{2}\rangle\leq$ Out $(E_{1}), 
\langle\omega_{1} ,\omega_{2}\rangle$ an elementary abelian
$p$-subgroup. By (1.67)  we may assume that $\omega_{1}$ is a field
automorphism. But this contradicts (1.68).

So we have $m_{p}(E_{2})=2$. Now choose again $U >
E_{1}E_{2},m_{p}(U)=4,\overline{G_{2}\cap M} \leq U$. Again by (1.67) 
and (1.68)
for some $\omega\in U\setminus E_{1}E_{2},o(\omega )=p$, we have
$[\omega ,E_{1}]=1$. Hence $U=E_{1}\bC_{U} (E_{1})(\overline{G_{\, 2}\cap M} )$.

If $\bC_{E_{1}E_{2}} (Z_{1})=1$ then choose $G_{1}$ minimal with $G_{\, 2}\cap
M\leq G_{1}$ and $G_{1}K/K=U$. We have (a)(ii).

Assume next $[E_{1},Z_{1}]=1$. If $\bC_{E_{2}} (Z_{1})=1$. The
structure of $U$ tells us $[\omega ,Z_{1}]\not= 1$. Hence choose
$G_{1}$ minimal with respect to $E_{2}\langle\omega\rangle\leq
G_{1}K/K$ and $G_{\, 2}\cap M\leq G_{1}$. Then as any $E\cong E_{p^{\,
2}}$ in $G_{1}$ centralizes some $p$-element in $E_{1}$, we have
(a)(ii) again.

We may assume $\bC_{E_{2}} (Z_{1})\not= 1$. Let $m_p(\bC_{E_2}(Z_1)) = 1$.
Then choose $G_1$ minimal with $G_2 \cap M \leq G_1$ and $G_1K/K \geq 
\langle E_2 , \omega \rangle$. As $O_p(U) = E_2$, we have that 
$SL_2(p)$ is induced on $E_2$ by $\langle \omega, \overline{G_2 \cap M} \rangle$.
If $x$ is a 2--element in $G_1$ centralizing $Z_1$ not contained in
$G_2 \cap M$ then we see that there is also a 2--element $y$ with
$m_p([E_2,y]) = 2$ and $[y,Z_1] = 1$, a contradiction. Hence we have
(a)(ii).

Suppose $m_{p}(\bC_{E_{2}}
(Z_{1}))=2$. Then we choose $G_{1}$ minimal with $G_{\, 2}\cap M\leq
G_{1}$ and $G_{1}K/K\geq\bC_{U} (Z_{1})$. We have (a)(i) as long as
we do not have the $Sz(q)$-case with $\bC_{E_2}(T_1) = 1$. So let assume that we have the $Sz(q)$--case. Let $[T_{1},E_{1}]=1$. If
$m_{p}(\bC_{E_{2}} (T_{1}))=2$, we choose $G_{1}$ minimal with $G_{\, 2}\cap
M\leq G_{1}$ and $\bC_{U} (T_{1})\leq G_{1}K/K$ and we have (a)(i).
So assume next $m_{p}(\bC_{E_{2}} (T_{1}))=1$. As in the case of 
$m_{p}(\bC_{E_{2}} (Z_{1}))=1$, we choose $G_{1}$
minimal with $G_{\, 2}\cap M\leq G_{1}$ and $G_{1}K/K\geq\langle
E_{2},\omega\rangle$ and we have (b).

So assume now $[T_1, E_1] \not= 1$. If $\omega \in \bC_U(T_1) \not\leq E_2$,
then $m_p(\bC_{E_2}(T_1)) = 2$. Now we just choose $G_1$ minimal with
respect to $G_2 \cap M \leq G_1$ and $\bC_U(T_1) \leq G_1K/K$.
So let $\omega \not\in \bC_U(T_1)$. Again we have $m_p(\bC_{E_2}(T_1)) = 2$, otherwise we have (a)(ii).
If $x$ is a 2--element with $[x,T_1] = 1$ and $x \not\in G_2 \cap M$,
then we see that $m_p(\bC_U(x)) \geq 2$ and so we choose $G_1$ minimal
with respect to $G_2 \cap M \leq G_1$ and $\bC_U(x) \leq G_1K/K$.
Hence we have (b) in any case.

So assume now $[E_1,Z_1] \not= 1$. If $\omega \in \bC_U(Z_1)$, then $m_p(\bC_{E_2}(Z_1)) = 2$ and we choose $G_1$ minimal with respect to $G_2 \cap M \leq G_1$ and $\bC_U(Z_1)K/K \leq G_1K/K$. So we have (a)(i). So assume $\omega\not\in \bC_{U} (Z_{1})$. Then we are going 
to prove that we have (a)(ii). For this we may assume
that there is some 2-element $x\not\in G_{\, 2}\cap M,[x,Z_{1}]=1$. Then
$m_{p}(\bC_{E_{2}} (Z_{1}))=2=m_{p}(\bC_{U} (Z_{1}))$. Now let $G_{1}$
be minimal with respect to $G_{\, 2}\cap M\leq G_{1}$ and $\bC_{U}
(x)\leq G_{1}K/K$. As $m_{p}(\bC (Z_{1}))=2$, we have (a)(ii).

\absa
To prove the lemma we just have to show 

\absa
{\bf (4.4.7)~}{\it If $F^{*}(H/K)=F(H/K)$ and $m_{p}(F(H/K))\leq 3$,
then the lemma holds.} 

\absa
Let $C$ be a critical subgroup of $E=F(H/K)$ and $C_{1}=\Omega_{1}
(C)$. As $H/F(H/K)$ acts faithfully on $C_{1}$, and $m_{p}(\mbox{Out}
(C_{1}))\leq 1$ for $m_{p}(C_{1})\leq 2$, we get $m_{p}(E)= m_p(C_1) = 3$. As $E\geq\bC_{H/K} (C_{1}/\Phi (C_{1}))$, we see that $H/E$
is involved in $\mbox{Aut~} (C_{1})$. So $H/E$ is involved in
$GL_{2}(p),GL_{3}(p)$ or $Sp_{4}(p)$. Now choose $X$ minimal with
respect to $G_{\, 2}\cap M\leq X,m_{p}(X)\geq 4$ and $XK/K\geq
C_{1}$. Set $Z_{1}=\langle\Omega_{1} (Z(S))^{\, X}\rangle
,T_{1}=\langle Z^{\, X}\rangle$.

Suppose $m_{p}(\bC_X(Z_{1})) \geq 3$. Then choose $G_{1}$ minimal with
respect to $G_{\, 2}\cap M\leq G_{1}$ and $G_{1}K/K\geq\bC
(Z_{1})$. If we do not have the $Sz(q)$-case then we get (a)(i).
The same holds for $m_p(\bC_X(T_1)) \leq 1$. So assume  we have the
$Sz(q)$-case. If $m_{p}(\bC_{X} (T_{1}))\geq 3$, we choose $G_{1}$
minimal with respect to $G_{\, 2}\cap M\leq G_{1}$ and $\bC_{X}
(T_{1})\leq G_{1}K/K$. Again we have (a)(i). So for the remainder set $U_1 = Z_1$
or $T_1$. We just have $m_p(\bC_X(U_1)) \leq 2$. If any 2--element centralizing
$U_1$ is in $E$ we may choose $G_1 = X$ and so we have (a)(ii) or (b).

 Suppose next
$|\bC_{X}(U_{1})/O_{2}(X)|$ is even. Set $U=C_{1}/\Phi (C_{1})$.
Now $[\bC_{\overline{X}}(U_{1}),U]\leq\bC_{U}(U_{1})$. Let 
$|[\bC_X(U_{1}),U]|=p$.
Then $\bC_{\overline{X}} (U_{1})/O_{p}(\bC_{\overline{X}} (U_{1}))$ is cyclic.
Let $T=S\cap\bC (U_{1}),X_{1}=\langle N_{X}(T/O_{2}(X)),G_{\, 2}\cap
M\rangle$. As $m_{p}(\bC_{E} (T/O_{2}(X)))\geq 2, m_{p}(X)=4$ and
$X_{1}KE/KE=XKE/KE$, we get $m_{p}(X_{1})\geq 2$ and
$m_{p}(X_{1})\geq 3$ if $m_{p}(\bC_{C_{1}} (U_{1}))=1$. Furthermore
any 2-element in $X_{1}$ centralizing $U_{1}$ is in $G_{\, 2}\cap M$.
We may choose $G_{1}$ minimal in $X_{1}$ such that 
$G_{\, 2}\cap M\leq G_{1}$
and $G_{1}K/K\geq N_{X} (T/O_{2}(X))$. Now we have (a)(ii) or (b) 
provided we can prove that any $\tilde{E} \cong E_{p^2}$, 
$\tilde{E} \leq G_1$
is contained in some $F \cong E_{p^3}$, $|F \cap \bC(U_1)| \geq p$.
But $|\bC_U(\tilde{E})| \geq p^2$ and so $\bC_{C_1}(\tilde{E})$ 
contains some $F_1 \cong E_{p^2}$
and then $F = F_1\tilde{E}$ does the job or $\tilde{E} \leq C_1$ and
$\bC_{C_1}(\tilde{E}) = \tilde{E}$. But then $C_1 \cong Z_{p^{1+2}}$ and so
$X$ induces $SL_2(p)$ on $C_1$ and all elementary abelian subgroups
of order $p^2$ in $C_1$ are conjugate, hence they are contained in some 
$F \cong E_{p^3}$. 

Suppose now $|[U,\bC (U_{1})]|=p^{\, 2}$. Then $m_{p}(\bC_{C_{1}}
(U_{1}))=2$. Now $W=O^{\, p}(\bC (U_{1})/O_{p}(\bC (U_{1})))\lesssim
GL_{2}(p)$. Let $T = S \cap O_2(W)$. Let $C_{1}$ be elementary abelian of order $p^{\, 3}$.
Then $(H/K)/E\lesssim GL_{3}(p)$. As $O_{p}((H/K)/E)=1$ and
$m_{p}(E)=3$, we now get $p^{\, 2}GL_{2}(p) \lesssim (H/K)/\bC_{H/K} (C_1)$. But $\bC_{H/K} (C_{1})\leq E$ and so
$|[E,C_{1}]|=p$, and $\bC_{E} (U_{1})\bC_{E} (C_{1})=E$. This finally
gives $[\bC (U_{1}),C_{1}]=C_{1}$, a contradiction.

Assume next $C_{1}\cong Z_{p^{\, 1+4}}$. Then
$\bC_{C_{1}} (T)\cong Z_{p^{\, 1+2}}$. Set $X_{1}=\bC_{H/K} (T)$. Then
$m_{p}(X_{1})\geq 2$. So let $G_{1}$ be minimal with respect to $G_{\, 2}\cap M\leq G_{1}$
and $G_{1}K/K\geq \bC_{C_{1}} (T)$. Then any 2--element of
$G_{1}$ centralizing $U_{1}$ is contained in $G_{\, 2}\cap M$.
So we have (a)(ii) or (b).

Suppose now $C_{1}\cong \bZ_{p}\times Z_{p^{\, 1+2}}$ or $C_{1}\cong Z_{p^{\,
1+2}}$. Let $F\leq (H/K)/E,F\cong SL_{2}(p)$. Suppose that $\bC
(U_{1})$ does not cover $F$. Then let $G_{1}$ be minimal with 
$G_{\, 2}\cap M\leq G_{1}$ and $G_{1}K/K\geq\bC_{H/K} (T)$. Now
$m_{p}(G_{1})\geq 2$ and any 2-element centralizing $U_{1}$ is in $G_{\, 2}\cap
M$. Again we have (a)(ii) or (b) provided we can prove that 
$\tilde{E} \cong E_{p^2}$, $\tilde{E} \leq G_1$ is contained in some 
$F \cong E_{p^3}$. This is clear as long $C_1 \not\cong Z_{p^{1+2}}$ as before. 
In the latter we see that $C \cong \bZ_{p^n}*Z_{p^{1+2}}$ and 
$E \cong C_1\bC_E(C_1)$. Either there is a normal $Z_{p^{1+2}}$ in
$\bC_E(C_1)$ and we are done, or $F$ covers $\bC_X(\bC_E(C_1))$
as we may assume $p > 3$. But now we have no $\tilde{E} \leq G_1$,
$\tilde{E} \cong E_{p^2}$ and $\tilde{E} = \Omega_1(\bC_{G_1}(\tilde{E}))$.  

So assume that $\bC (U_{1})$ does cover $F$. Now if $p>3$, then $\bC
(U_{1})$ does contain a subgroup $F_{1}\cong SL_{2}(p)$ and $\bC
(U_{1})$ does contain $C_{2}\leq C_{1}, C_{2}\cong Z_{p^{\, 1+2}}$. Now
$m_{p}(C_{2} F_{1})=3$, a case handled before.

So we have $p=3$ and $(H/K)/E\cong SL_{2}(3)$ or $GL_{2}(3)$. If
$H/K\not= E\bC_{H/K} (U_{1})$, then choose $G_{1}$ as before. So
assume first $\bC_{H/K} (U_{1})/\bC_{E} (U_{1} )\cong GL_{2}(3)$. Let
$P\in\mbox{Syl}_{3} (H/K),T_{1}=T\cap N(P)$. Then $T=\langle
x_{1},x_{2}\rangle$ is a fours group. Furthermore
$[x_{1},P]=C_{2}\cong Z_{3^{\, 1+2}}$ and $[\bC_{P} (x_{1}),x_{2}]$ is
cyclic. Hence $[P,T_{1}]\cong \bZ_{3^n}*Z_{3^{\, 1+2}}$. As
$m_{3}(P)\geq 4$, we get $m_{3}(\bC_{P} (T_{1}))\geq 2$. But $\bC_{P}
(T_{1})\leq E$ and  so as $[\bC_{P} (T_{1}),C_{2}]=1$ and $\bC_{P}
(T_{1})\cap C_{2}=1$, we see $m_{3}(E)\geq 4$, a contradiction.

So we have $\bC_{H/K} (Z_{1})/\bC_{E} (Z_{1})\cong SL_{2}(3)$. Let $Q$
be the quaternion group in $SL_{2}(3)$. Then we see that $[Q,E]=C_{2}$
and $E=C_{2}\bC_{E} (Q)= C_{2} *\bC_{E} (Q)$. Now as we may assume
that $\bC (U_{1})$ does not contain $SL_{2}(3)$, otherwise we get a
contradiction as before, we get $m_{3}(\bC_{E} (Q))\geq 2$. Now let
$G_{1}$ be minimal with $G_{\, 2}\cap M\leq G_{1}$ and $N_{H/K}(Q)\leq
G_{1}K/K$. As $Z_{1}=\langle\Omega_{1} (Z(S))^{\, G_{1}}\rangle$, we
see that any 2-element centralizing $U_{1}$ is in $O_{2}(G_{1})$.
Furthermore $U_{1}$ is centralized by $E_{9}$ and we have (a)(ii) or (b).

So assume finally $C_{1}\cong \bZ_{p}\times \bZ_{p}$ and
$[C_{1},U_{1}]=1$. Now $(H/K)/E$ acts as a subgroup of $GL_{2}(p)$ on
$C_{1}$. Let $\tilde{E}\leq H/K,\tilde{E}$ be elementary abelian of
order $p^{\, 4}$. Then $|\tilde{E} :\tilde{E}\cap E|=p$. But
$|C_{1}:\bC_{C_{1}} (\tilde{E} )|=p$. This implies $|(\tilde{E}\cap
E)C_{1}|=p^{\, 4}$, a contradiction. This proves the lemma.
\\
\\

In (5.8) we will see that (4.4)(b) does not exist.
\\
\\

If $G_{2}$ is solvable or $E(G_{\, 2}/O_{2}(G_2))\cong L_{2}(r)$,
$L_{2}(r)\times L_{2}(r), r$ odd, $r\geq 11$ or $PSp_{4}(r),r\geq 7$, then
we change the choice of $G_{1}$ in (4.4) a little bit. We are just 
interested in having $S$ in $G_1 \cap G_2$ not the whole of $G_2 \cap M$.


\absa
{\bf (4.5)~Lemma.~}{\it Let $G_{\, 2}$ solvable or $E(G_{\,
2}/O_{2}(G_2))\cong L_{2}(r)$ or $L_{2}(r)\times L_{2}(r), r$ odd, $r\geq
11$ or $PSp_{4}(r),r\geq 7$, then choose $G_{1}$ minimal such that
$S\leq G_{1}$ and $G_{1}/K$ satisfies (4.4)(a).}
  
\absa
{\bf (4.6)~Lemma.~}{\it We may choose $G_1$ such that (4.4) holds and  
$\overline{G_{2}\cap G_{1}}\cap K\leq
O_{2}(\overline{G}_{1} )$, or  $F(H/K)$ is a Sylow $p$--subgroup of
$L_2(p^n)$, $U_3(p^n)$ or $^2G_2(p^n)$ with $m_p(F(H/K)) \geq 3$ 
and  $[\Omega_{1}(Z(S)), G_1 \cap G_2] \not= 1$.}

\absa
Proof~. Suppose $\overline{G_{2}\cap G_{1}}\cap K$ is a 2-group $T$. Then
$\overline{G}_{1} K=KN_{\overline{G}_{1}} (T)$. So we  may assume 
$T\lhd\overline{G}_{1}$.

Suppose $\overline{G_{2}\cap G_{1}}\cap K$ is not a 2--group
and  furthermore $N_{G} (\langle \nu\rangle )\leq M$, for all
$\nu\in\overline{G_{2}\cap G_{1}}\cap K,o(\nu )$ odd. Then by (1.66)
$G_{\, 2}$ is solvable, $E(G_{\, 2}/O_2(G_2))\cong L_{2}(r)$,
$L_{2}(r)\times L_{2}(r)$, $(S)L_{3}(r)$, $(S)U_{3}(r)$, $PSp_{4}(r)$ 
or $F(G_{\, 2}/O_{2}(G_2))$ is elementary abelian of order $r^{\, 3}$
and $E((G_{\, 2}/O_{2}(G_2))/F(G_{\, 2}/O_{2}(G_2)))$ $\cong L_{2}(r)$. Let $T\in
\mbox{Syl}_{2} (K)$. Then $\overline{G}_{1} K=KN_{\overline{G}_{1}} (T)$. 
So by choice of $G_{1}$ in (4.5), we get that $E(G_{\, 2}/O_{2}(G_2))\cong (S)L_{3}
(r)$ or $(S)U_{3}(r),r$ odd, or $|F(G_{\, 2}/O_{2}(G_2))|=r^{\, 3}$ and
$G_{\, 2}$ is nonsolvable.

In all cases $r\,\Big |\, | K\cap \overline{G_{\, 2}\cap G_{1}}|$ and so even
$r^{\, 2}\,\Big |\, | K\cap\overline{G_{\, 2}\cap G_{1}}|$. This implies that
we have $|F(G_{\, 2}/O_{2}(G_2))|=r^{\, 3}$, as in the other cases
$((G_{1}\cap G_{\, 2})/O_{2}(G_2))\cap E(G_{\, 2}/O_{2}(G_2))\cong GL_{2}(r)$.
Now $r^{\, 3}\,\Big |\, | K\cap\overline{G_{\, 2}\cap G_{1}} |$. But then $G_{\,
2}\leq M$, a contradiction.

So we have that there is some $\nu\in\overline{G_{1}\cap G_{\, 2}}\cap
K,o(\nu )$ odd, such that $N_{G}(\langle \nu\rangle )\not\leq M$.\\
\indent
The assertion follows with (4.3). As $\bC_{O_2(M)}(\nu) = 1$, we 
also see that $[\nu, \Omega_1(Z(S))] \not= 1$.
\\
\\
\indent
If we have the exceptional case of (4.6) we also change the definition
of $G_{1}$ a little bit.


If $\Omega_{1} (Z(S))\not\!\!\lhd G_{\, 2}$ and $G_{1}$ is as in
the exceptional case in (4.6) then replace $G_{1}$ by some group 
minimal with respect to
containing the normalizer of the corresponding Sylow $p$--subgroup..


Now in any case we have 

\absa
{\bf (4.7)~Lemma.~}{\it $\overline{G_{2}\cap G_{1}}\cap K\leq
O_{2}(\overline{G}_{1} )$.}

\absa
{\bf (4.8)~Lemma.~}{\it The group $G_{1}$ has the following
structure
\begin{center}
$G_{1}/O_{2}(G_{1})=(X_{1}* X_{2}*\ldots
*X_{t}*Y_{1}*Y_{2}*\ldots *Y_{f})U$,
\end{center}
where
$X_{1},\ldots ,X_{t}$ are the quasisimple groups, $Y_{i}\,^\prime
=Y_{i},Y_{i}/O_{r_{i}}(Y_{i})\cong L_{2}(r_{i})$ or $L_{3}(r_{i})$, and
$U$ is $p$-constrained. If $\nu$ is a $p$-element with $[\nu
,Y_{i}]\leq O_{r_i} (Y_{i})$ then $\nu\in Y_{i}\bC (Y_{i})$.}

\absa
Proof~. Let $XK/K$ be a component of $G_{1}K/K$. Let $F=F(K)$ and
$R\in\mbox{Syl}_{r} (K)$. We have $m_{r}(R)\leq 3$. Suppose that
$\bC_{XK} (R)\leq K$. Then $XK/K\lesssim L_{3}(r)$ or $Sp_{4}(r)$.
As $2\not\Big |\, |K|$, we get $XK/K\lesssim L_{3}(r)$. So by \cite{Mi1} we
get $XK/K\cong L_{2}(r),A_{5},L_{3}(r), L_{2}(7)$ or $A_{6}$. In case of $A_6$ 
the group $3\cdot A_6$ will be induced. 
So we get $G_{1}/O_{2}(G_{1})=(X_{1}\times\ldots\times X_{t}
*H_{1})F,\, X_{1}, \ldots ,X_{t}$ quasisimple groups, $E(H_{1}K/K)$
has just components $L_{3}(r)$, $L_{2}(r)$, $A_{5}$, $L_2(7)$ or $A_6$, 
$F$ is an
automorphism group of $X_{1}\times\ldots\times X_{t}$.

Let now $Y_{1}K/K\cong L_{3}(r_{1}),L_{2}(r_{1})$, 
$Y_{2}K/K\cong L_{3}(r_{2})$, $L_{2}(r_{2})$,
$A_{5}$, $L_{2}(7)$, or $3\cdot A_{6}$ and $Y_{1}Y_{2}K/K\leq H_{1}K/K$.

Set $U_{1}=(Y_{1}K)^{\, \infty},U_{2}=(Y_{2}K)^{\,\infty}$. Then
$[U_{1},U_{2}]\leq U_{1}\cap U_{2}\leq K$. 
Let $U_{1}=A_{0}>A_{1}>A_{2}>\ldots >A_{f}=1, A_{i}$
characteristic in $U_{1}$. $A_{0}/A_{1}\cong
L_{2}(r_{1}),L_{3}(r_{1}), A_{5},L_{2}(7)$, or $3\cdot A_{6}$ and
$A_{i}/A_{i+1}$ groups of prime power order $s_{i}^{\, a_{i}},s_{i}\not=
s_{i+1}$. Let $[U_{2},A_{i}]\not\leq A_{i+1}$ for some $i$. Then
Aut$(A_{i})$ either involves $U_{1}K/K\times U_{2}K/K$ or
$U_{1}=A_{1}\bC_{U_{1}} (A_{i}/A_{i+1})$.

Suppose the former. As $|A_{i}/A_{i+1}|=s_{i}^{\, a_{i}}$ for some
prime $s_i$, we get a $s_{i}$-subgroup $W$ of $A_{i}$ such that Aut$(W)$
involves $U_{1}K/K\times U_{2}K/K$. But we have $m_{s_i}(W)\leq 3$, so
Aut $(\Omega_{1} (C)), C$ a critical group in $W$, has a subgroup
which is an extension of a group of odd order by a direct product of
two groups in $\{
L_{3}(r_{1}),L_{2}(r_{2}),L_{2}(r_{1}),L_{2}(r_{2}),A_{5},
A_{6},L_{2}(7)\}$. But we have that this group is involved in 
$L_{3} (s_i)$ or $Sp_{4}(s_i)$,
a contradiction to \cite{Mi1} and \cite{Mi2}.

So we have the latter. Choose $i$ minimal with respect to $\bC_{U_{1}}
(A_{i}/A_{i+1})\not\leq A_{1}$. Now $A_{i}/A_{i+1}\leq Z(U_{1}/A_{i+1})$
as $U_{1}=U_{1}^\prime$. Let $i>1$. Then $A_{i-1}/A_{i+1}\cong
A_{i}/A_{i+1}\times A_{i-1}/A_{i}$.
So we produce a new series
$A_{0}>A_{1}=B_{1}>B_{2}>\ldots >B_{f}$, 
where $\bC_{A_{0}} (B_{j}/B_{j+1})\not\leq A_{1}$~~for~~$j<i\, .$ 

This shows that we may assume $A_{1}/A_{2}\leq Z(A_{0}/A_{2})$. Now
the Schur multipliers yield $A_{0}/A_{1}\cong L_{3}(r_{1})$ and
$|A_{1}/A_{2}|=3$ so $A_{0}/A_{2}\cong SL_{3}(r_{1})$.

Now there is an elementary abelian subgroup $C$ of
$A_{2}, |C|=r_{1}^{\, 3}$, such that $A_{0}/A_{2}$ induces
$SL_{3}(r_{1})$ on $C$. As $\bC_{C} (Z(A_{0}/A_{1}))=1$, we see that
$A_{0}$ contains a subgroup $A\cong SL_{3}(r_{1})$. Now
$m_{r_{1}}(C\, A)=4$, a contradiction.

So we have that $\bC_{A_{0}} (A_{i}/A_{i+1}) \leq A_1$ for all $i$. In
particular $[U_{1},U_{2}]=1$. Furthermore the same argument implies
that $A_{1}/A_{3}\cong A_{1}/A_{2}\times A_{2}/A_{3}$ and
so $A_{1}$ is nilpotent.

Let now $\nu$ be a $p$-element with $[Y_{1},\nu ]\leq A_{1}$. Suppose
$[\nu ,A_{1}]\not= 1$. Then $\nu$ acts on some $E\cong E_{r^{3}}$. As
$Y_{1}/A_{1}$ acts irreducibly on $E$, we see that $\nu$ acts as an
element in $Z(GL_3(r))$. We have $\overline{M} = KN_{\overline{M}}(\overline{E})$.
Hence by (1.68) we know that $\nu\in
Y_{1}\,^\prime$. Now as seen before we have $Y_{1}/A_{1}\cong 3\cdot
A_{6},o(\nu )=3$, and so $\nu\in Y_{1}\bC (Y_{1})$.

Suppose $gcd(|Y_{1}K/K|,|K|)=1$. Then by Zassenhaus' lemma \cite[(I.18.1),(I.18.2)]{Hu} we get a subgroup isomorphic to $Y_{1}/A_{1}$ which is
normalized by $(\overline{G_{\, 2}\cap G_{1}} )$. By the minimal choice of
$G_{1}$, (4.7) and the fact that any $p$-element normalizing $Y_{1}$
is in $Y_{1}\bC (Y_{1})$, we have $A_{1}=1$.

If $Y_{1}/A_{1}\cong L_{3}(r_{1}),L_{2}(r_{1}),r_{1}\,\Big |\, | A_{1}|$. Then
by the same argument, minimal choice  and (4.7) again, we have that $A_{1}$ is an
$r_{1}$-group.


Now we make the following choice for $L_{1},L_{2}$. Set $L_{2}=G_{\,
2}$. Suppose $f\geq 1$, $Y_{1}/O(Y_{1})\cong L_{2}(r)$. Let
$\hat{Y}_{1}$ be the preimage of $Y_{1}$. Set $L_{1}=\hat{Y}_{1}
(G_{1}\cap G_{\, 2})$.

So we have the following:

\absa
{\bf (4.9)~Lemma.~}{\it $L_1$ is of one of the following types}
\begin{itemize}
\item[(i)]{\it $L_{1}/O_{2}(L_{1})=(X_{1}*\ldots *X_{t}*Y_{1}*\ldots
*Y_{f}*U)A, X_{1},\ldots ,X_{t}$ quasisimple groups, \linebreak[4]
$Y_{i}/O_{r_{i}}(Y_{i})\cong L_{3}(r_{i}), r_{i}$ prime,
$U$ is $p$-constrained,
$A$ is an automorphism group of $(X_{1}*\ldots *X_{t}*Y_{1}*\ldots*Y_{f}*U)$.
Furthermore if $m_{p}(X_{1}*\ldots *X_{t}*Y_{1}*\ldots *Y_{f}*U)\geq
4,A\leq G_{\, 2}$.}
\item[(ii)]$L_{1}/O_{2}(L_{1})\cong A\cdot B, B/O_{r}(B)\cong
L_{2}(r)$,
$A=G_{\, 2}\cap L_{1}/O_{2}(L_{1}),r>3$.
\end{itemize}

\absa
{\bf (4.10)~Lemma.~}{\it Let $L_{1}$ be as in (4.9)(ii). Then $M$ is the
only maximal subgroup of $G$ containing an elementary abelian
subgroup of order $r^{\, 3}$ from $O_{r}(B)$.}

\absa
Proof~. By (4.3) we have $N_{G}(\langle\omega\rangle )\leq M$ for any
$\omega\in O_{r}(B),o(\omega )=r$. Now the assertion follows with
(1.63). \\

If $L_{2}$ is solvable with $\sigma (L_{2})\not= \emptyset$
we are going to change notation a little bit in order to get more
symmetry. This will become important in (6.35).

Let $N$ be the uniqueness group with $L_{2}\leq N$. If we go over the
construction of $L_{2}$ in (2.1) we see that either
$L_{2}/O_{2}(N)\leq F(N/O_{2}(N))S$ or $L_2/O_2(N) \leq 
\langle E,S/O_2(N)\rangle, E\leq
E(N/O_{2}(N)),E\cong Sz(q)$. Hence in any case we get that $\langle
L_{1}\cap N,L_{2}\rangle$ is of the same shape, i.e. $\langle
L_{1}\cap N,L_{2}\rangle /O_{2}(\langle L_{1}\cap N,L_{2}\rangle )$
is an extension of a $p$-group by $G_{1}\cap N/O_{2}(\langle
G_{1}\cap N,L_{2}\rangle )$.

Now we are going to choose $H_{1}\leq L_{1}$ minimal with respect to
$H_{1}\not\leq N$ and $S\leq H_{1}$. If $\sigma (H_{1})=\emptyset$, we
replace $L_{1}$ by $\langle H_{1}\cap N, L_{2}\rangle$ and $L_{2}$ by
$H_{1}$. Hence we just interchange the roles of $L_{1}$ and $L_{2}$.\\

So from now on assume that we are not able to choose $H_{1}$ with $\sigma
(H_{1})=\emptyset$.

\absa
{\bf (4.11)~Lemma.~}{\it Let $L_{2}$ be solvable with $\sigma
(L_{2})\neq\emptyset$. Suppose there is no $H_{1}\leq G_{1},
H_{1}\not\leq N,S\leq H_{1},\sigma (H_{1})=\emptyset$. Then if $H_{1}$
is minimal with respect to $H_{1}\not\leq N,S\leq H_{1}$ we have
either\\}
\indent
(i)~~{\it$H_{1}$ is a $\{ 2,p\}$-group, $m_{p}(H_{1})\geq
4$, or\\}
\indent
(ii)~~{\it$E(H_{1}/O_{2}(H_{1}))=K_{1}\times\ldots\times
K_{t}, K_{i}$ are Lie groups over fields of odd characteristic, not
$U_{3}(3)$, $PSp_{4}(3)$, $^2G_{\, 2}(3)$ or $3\cdot U_{4}(3)$.
Furthermore $S$ acts transitively on $\{ K_{1},\ldots ,K_{t}\}$}.

\absa
Proof~. Choose notation as in (4.8). Let $X_{1}$ be a quasisimple
component of $G_{1}/O_{2}(G_{1})$. Suppose $X_{1}\not\leq (G_{1}\cap
N)/O_{2}(G_{1})$.

Let first $X_{1}\cong A_{n}$ or $3A_{7}, n\geq 7$. As 
$m_p(\langle X_1^{G_1} \rangle) \geq 4$ we have $p=3$. There
is a subgroup $U_{1}$ of $X_{1}, U_{1}/O_2(U_{1}) \cong \Sigma_{\lfloor\frac{n}{2}\rfloor}$.
We may assume $U_{1}\leq N$, otherwise we argue by induction. Hence
$m_{3}(\Sigma_{\lfloor\frac{n}{2}\rfloor} )=1$. In particular
$\lfloor\frac{n}{2}\rfloor\leq 5$ and so $n\leq 11$. If $n$ is odd
there is $A_{n-1}\leq X_{1}$. Hence we may assume $n$ to be even. Now
$n=8$ or $10$. As $\Sigma_{10}$ contains $\bZ_{2}\times\Sigma_{8}$, which
has the same Sylow 2-subgroup, it is enough to treat $X_{1}\cong
A_{8}$. But $\Sigma_{8}$ is generated by $S$-invariant $\{
2,3\}$-groups, the assertion.

Let $X_{1}$ be a sporadic group. By (1.4) $p=3$. Let $X_{1}\not\cong
M_{11}$. Then $X_{1}$ is generated by minimal parabolics $P$ and in
case $X_{1}\cong J_{1}$ we have $X_{1}$ is generated by
$N_{X_{1}}(S\cap X_{1})$ and $\bC_{X_{1}} (t)$ for some involution
$t\in S\cap X_{1}$.

Suppose $X_{1}$ is normalized by $S$. Then we see from \cite{RoStr} that
$1\leq m_{3}(P)\leq 3$, for any such $P$, a contradiction. As
$m_{3}(M_{11})=2$, we get $X_{1}$ is not normalized by $S$ if
$X_{1}\cong M_{11}$.

So assume that $[X_{1},S]\not\leq X_{1}$. Now as in the $A_{n}$-case
we may assume that there is no solvable parabolic. As $GL_{2}(3)\leq
M_{11}$, \cite{RoStr} provides us with a contradiction.

So let $X_{1}$ be a Lie group $G(q),q=2^{\, n}$. As $X_{1}$ is
generated by minimal parabolics we may assume $\hat{H}_{1} \leq
G_{1},\hat{H}_{1}\not\leq N,S\leq\hat{H}_{1}$ with:
$$
F^{*}(\hat{H}_{1} /O_{2}(\hat{H}_{1} ))\cong K_{1}\times\ldots\times
K_{t}, K_{1}\cong K_{2}\cong\ldots\cong K_{t},m_{p}(\hat{H}_{1} )\geq
4\, .
$$
$$K_{i}\cong L_{2}(r),U_{3}(r),Sz(r),L_{3}(r),Sp_{4}(r),r \mbox{~even.}$$

If $K_{i}\cong L_{2}(q),Sz(q)$, then $t\geq 4$, as $m_{p}(\hat{H}_{1}
)\geq 4$. In the remaining cases $t\geq 2$. Let $r>2$. Then we get
that $m_{p}(C)\geq 2$, where $C$ is a Cartan subgroup of
$E(\hat{H}_{1}/O_{2}(\hat{H}_{1} ))$. Hence $C\not\leq N$ and as $C$
is solvable we get the assertion.

If $r=2$, then either $\hat{H}_{1}$ is solvable or $K_{1}\cong
L_{3}(2)\cong L_{2}(7)$ or $K_{1}\cong Sp_{4}(2)^\prime\cong
L_{2}(9)$, which are Lie groups in odd characteristic, the
assertion.

As $U_{3}(3)\cong G_{\, 2}(2)^\prime ,PSp_{4}(3)\cong\Omega_6^-(2)$ and 
$^2G_{\, 2}(3)^\prime\cong L_{2}(8)$, it just remains to handle
$X_{1}\cong 3\cdot U_{4}(3)$. Hence $p=3$. But $3\cdot U_{4}(3)$ is
generated by subgroups $T$ with $T/O_{2}(T)\cong \Sigma_{3}$. Now we get
that $H_{1}$ is solvable as before.
So we have shown that:
\\
\\
\noindent{\it{The assertion holds or any quasisimple
component of $G_{1}/O_{2}(G_{1})$ is covered by $N$.}}
\\

Now we consider $Y_{1}$. Suppose $G_{1}\cap
N/O_{2}(G_{1})\not\geq Y_{1}$. In any case $S$ normalizes
$Y_{1}$. But $m_{p}(Y_{1})\leq 3$ for any prime $p$, a contradiction.
So we are left with $U$. But $U$ is $p$-constrained and so we are
done if $O_{p}(U)S\not\leq N$. Suppose $O_{p}(U)S\leq N$. Then
$m_{p}(O_{p}(U))\leq 1$. But $m_{p}(X_{1}*\ldots *X_{t}*Y_{1}*\ldots
*Y_{f})\leq 1$, as $X_{1}*\ldots *X_{t}*Y_{1}*\ldots
*Y_{f}$ is covered by $N$. This is impossible as
$m_{p}(G_{1})\geq 4$. The lemma is proved.\\
\\

Now we choose $H_{1}$ according to (4.11) and set $L_{1}=H_{1}$. In
particular we have symmetry between $L_{1}$ and $L_{2}$.

\absa
{\bf (4.12)~Lemma.~}{\it Let $L_1$ be as in (4.9). Suppose there is
some $\omega\in L_{1}, o(\omega )=p^{\, a}, \omega$ induces a field
automorphism on some of the $X_{i}$ or $1\not=[\omega ,Y_{i}]\leq
O_{r}(Y_{i})$ for some $i$. Then one of the following holds}
\begin{itemize}
\item[(1)] {\it There is some $s\in S, X_{1}^{\, s}=X_{2}, \omega$ induces
a field automorphism on both of $X_{1}$ and $X_{2}, \omega^{\, s}
=\omega^{\, -1}$, and $X_{1}\cong X_{2}\cong L_{2}(q)$ or $Sz(q),q$
even}
\item[(2)] {\it $X\cong X_{1}\times X_{2}$ or $X_{1}\times
X_{2}\times X_{3}, X\lhd L_{1}/O_{2}(L_{1}), X_{1}\cong X_{2}\cong
X_{3}\cong L_{2}(q)$ or $Sz(q),\bC_{L_{1}/O_{2}(L_{1})} (X)\leq
(L_{1}\cap L_{2})/O_{2}(L_{1})$.} 
\end{itemize}

\absa
Proof~. Suppose first that $\omega$ induces a field automorphism on
$X_{1}$ (notations are as in (4.9)). We recall the construction given
in (4.4). This shows that $X_{1}$ is a component of $(H/K)/F(H/K)$.
Furthermore by (1.68) we see that there is some $g\in M$ with
$X_{1}^{\, g}\not= X_{1}$ and $\omega$ induces a field automorphism
on $X_{1}^{\, g}$.  If $m_p(F^*(H/K)) \geq 4$, the construction given 
in (4.4) shows that we may assume $\omega \in G_1 \cap G_2$. But as
$m_p(\bC_M(\omega)) \geq 3$, we see with (1.66) and the construction
of $L_1$ in (4.5) that $E(G_2/O_2(G_2)) \cong L_3(r) $ or $U_3(r)$ 
or $E((G_2/O_2(G_2))/F(G_2/O_2(G_2))) \cong L_2(r)$ and $F(G_2/O_2(G_2))$
is an elementary abelian group of order $r^3$. As 
$N_G(\langle \omega \rangle) \leq M$ and $m_p(G_1 \cap G_2) \leq 1$, we see that
$E(G_2/O_2(G_2)) \cong L_3(r)$ or $U_3(r)$ and 
$\omega O_2(G_2) \in (G_1 \cap G_2/O_2(G_2))^{(\infty)}$.
But this is impossible as $\omega$ has to induce a field automorphism on 
$X_1$. Hence $m_p(F^{*}(H/K))\leq 3$. If
there are just two conjugates of $X_{1}$ in $E(H/K)$ we have (1), as
$m_{x}(X_{1})=1$ for any odd prime $x$ and (1.3). If $|X_{1}^{\,
M}|=3$, we have $F^{*}(H/K)=X_{1}\times X_{2}\times X_{3}$, and we
get (2).

That $1\not=[Y_{i},\omega ]\leq O_{r}(Y_{i})$ is not possible has been
shown in (4.8).\\

Let $Z_{1}=\langle\Omega_{1} (Z(S))^{\, L_{1}}\rangle$ and
$Z_{2}=\langle\Omega_{1} (Z(S))^{\, L_{2}}\rangle, Q_{i} = O_{2}(L{i}), i = 1,2$.
Because of $C_{L_{i}}(Q_{i})\leq Q_{i}$, we have $Z_{i}\leq Q_{i}, i = 1,2$.\\

There are a few special cases to be treated in (7.3) and chapter 8, which we will investigate now.

\absa
{\bf (4.13)~Lemma.~}{\it Suppose $Z_{1}\leq Z_{2},L_{1}=
\bC_{L_{1}} (Z_{1})(L_{1}\cap L_{2})$ and
$L_{1}$ is not of type (ii) in (4.9). Let $\tilde{Z}_{2}=\bC_{Z_{2}/Z_{1}}
(S)\bmod Z_{1}$. Assume $\tilde{Z}_{2}\leq O_{2}(L_{1})$. Set
$V_{1}=\langle \tilde{Z}_{2}^{\, L_{1}}\rangle$ and suppose $O^{\,
2^\prime} (\bC_{L_{1}} (V_{1}))\not\leq O_{2}(L_{1})$. Let 
$(L_{2},Z_{2})$ be one of the following:}
\begin{itemize}
\item[(i)]{\it $E(L_{2}/Q_{2})\cong L_{2}(q),q$ even, or
$E((L_{2}/Q_{2})/F(L_{2}/Q_{2}))\cong L_{2}(4)$ with
$[F(L_{2}/Q_{2}),Z_{2}]=1$. Further assume that $Z_{2}$ is the natural module, a direct
sum of two natural modules or the natural $O^{-}(4,t)$-module $t=2$,
or $q=t^{\, 2}$. }
\item[(ii)]{\it $E(L_{2}/Q_{2})\cong L_{2}(q)\times L_{2}(q),q$ even,
$Z_{2}$ the natural $O^{+}(4,q)$-module.}
\item[(iii)]{\it $E(L_{2}/Q_{2})\cong SU_{3}(q),q$ even, $Z_{2}$ the
natural module.}
\item[(iv)]{\it $E(L_{2}/Q_{2})\cong Sz(q), Z_{2}$ the natural module,
$\tilde{Z}_{2}\not\!\!\lhd L_{1}$, or $\tilde{Z}_{2}\lhd L_{1}$ and we
replace $\tilde{Z}_{2}$ by $\bC_{Z_{2}/\tilde{Z}_{2}} (S)\bmod
\tilde{Z}_{2}$, $m_p(\bC_{L_1}(\tilde{Z}_2)) \leq 1$.}
\item[(v)]{\it $L_{2}$ is solvable and one of the following holds}
\begin{itemize}
\item[($\alpha$)]$|Z_{2}|=4,L_{2}/\bC_{L_{2}} (Z_{2})\cong \Sigma_{3}$
\item[($\beta$)]{\it $|Z_{2}|=16$ and one of the following holds}
\begin{enumerate}
\item[(a)] {\it $ L_{2}/\bC_{L_{2}} (Z_{2})\cong F_{10}\mbox{~or~} F_{20}$,
$m_p(\bC_{L_1}(\tilde{Z}_2)) \leq 1$.}
\item[(b)] $ L_{2}/\bC_{L_{2}} (Z_{2})\cong \Sigma_{3}$
\item[(c)] $ L_{2}/\bC_{L_{2}} (Z_{2})\cong E_{9}\bZ_{4}$
\item[(d)] $ L_{2}/\bC_{L_{2}} (Z_{2})\cong \Sigma_{3} \wr\bZ_{2}$
\end{enumerate}
\item[($\gamma$)]{\it $|Z_{2}|=64$ and $ O_{3}(L_{2}/\bC_{L_{2}}
(Z_{2}))\cong Z_{3^{\, 1+2}}$}
\end{itemize}
\item[(vi)]{\it $E(L_{2}/Q_{2})\cong A_{9},Z_{2}$ is the natural module.}
\end{itemize}
{\it Then there is
$\hat{L}_{1}\leq L_{1},L_{2}\cap \hat{L}_{1} = L_{2} \cap L_{1}$ with}
\begin{enumerate}
\item[(1)]$m_{p}(\hat{L}_{1} )\geq 2$
\item[(2)]{\it If $\hat{V}_{1}
=\langle\tilde{Z}_{2}^{\,\hat{L}_{1}}\rangle$, then $O^{\, 2^\prime}
(\bC_{\hat{L}_{1}} (\hat{V}_{1} ))\leq O_{2}(\hat{L}_{1} )$}
\item[(3)]{\it $\hat{L}_{1}$ is of the same general
shape as $G_{1}$ in (4.9)}
\item[(4)]{\it Suppose $m_{p}(\hat{L}_{1} )=2$. If
$\omega\in\hat{L}_{1},o(\omega )=p$, there is $\nu\in M,o(\nu
)=p,[\nu ,\hat{V}_{1} ]=1$ and for $E=\langle\omega, \nu\rangle$ we have
$\Gamma_{E,1} (G)\leq M$.}
\end{enumerate}

\absa
Proof~. We choose $L_{1}^{\, (1)}\leq L_{1}$ minimal with respect to\\
\\
\hspace*{1cm} (1)~~$L_{1}^{\, (1)}$ is of the same
general shape as $L_{1}$ in (4.9).\\
\hspace*{1cm} (2)~~$m_{p}(L_{1}^{\, (1)})\geq 3$.\\
\hspace*{1cm} (3)~~$L_{1}\cap L_{2}\leq L_{1}^{\, (1)}$.

Hence we may assume $L_{1}^{\, (1)}=L_{1}$. Set
$U_{1}=X_{1}\times\ldots\times X_{t},W_{1}=Y_{1}\times\ldots\times
Y_{f}$ (notation as in (4.9)). Let $I\sbq\{ 1,\ldots ,t\},
[X_{i},V_{1}]\not= 1$ for $i\in I, J\sbq\{ 1,\ldots ,f\}
,[Y_{j},V_{1}]\not= 1,j\in J$. If $m_{p}(\prod_{i\in I}
X_{i}\times\prod_{j\in J} Y_{j})\geq 2$, then let $\hat{L}_{1}^{\, (1)}$ be
the preimage of $(\prod_{i\in I} X_{i}\times\prod_{j\in J}
Y_{j})(L_{1}\cap L_{2})$.
Now $O^{\, 2^\prime} (\bC_{L_{1}}
(V_{1})O_{2}(L_{1})/O_{2}(L_{1}))\cap(\prod_{i\in I}
X_{i}\times\prod_{j\in J} Y_{j})=1$.
If $O^{\, 2^\prime} (\bC_{\hat{L}_{1}} (\hat{V}_{1} ))\not\leq O_{\,
2}(\hat{L}_{1})$, then there is $E\cong E_{p^{\, 2}}$ in
$L_{1},\Gamma_{E,1} (G)\leq M$ such that $[\tilde{Z}_{2} ,E]=1$. But
in all cases except $L_{2}/Q_{2}\cong Sz(q)$, we have
$L_{2}=\langle\bC_{L_{2}} (x)|x\in\tilde{Z}_{2}^{\,\sharp}\rangle$.
So we are left with $L_{2}/Q_{2}\cong Sz(q)$. Now by assumption we
have $\tilde{Z}_{2}\not\!\!\!\lhd L_{1}$ and 
$m_{p}(\bC_{L_{2}} (\tilde{Z}_{2} ))\leq 1$, a
contradiction. Obviously (4) is satisfied, so we have found 
$\hat{L}_{1}$. 

Let us
now assume $m_{p}(U_{1}\times W_{1})\leq 1$. If $
U_{1}\times W_{1}=1$, then $m_{p}(X_{1}\times\ldots\times X_{t}\times
Y_{1}\times\ldots\times Y_{f})\leq 2$. In case of equality we have
some $E\cong E_{p^{\, 2}}\leq X_{1}\times\ldots\times X_{t}\times
Y_{1}\times\ldots\times Y_{f}$ with $\Gamma_{E,1} (G)\leq M$, a
contradiction as before. So we have 
$m_{p}(X_{1}\times\ldots X_{t}\times Y_{1}\times\ldots
\times Y_{f})\leq 1$. By (4.12) we see that $[O^{\, p^\prime} (U),X_{1}\times\ldots X_{t}\times Y_{1}\times\ldots
\times Y_{f}]=1$, and so $m_{p}(O_{p}(U))\geq 2$.

Suppose $m_{p}(U_{1}\times W_{1})=1$. Then $m_{p}(X_{1}\times\ldots X_{t}\times Y_{1}\times\ldots
\times Y_{f})\leq 2$. Suppose that $U$ contains some $\omega$ which
induces a field automorphism on one of the components. Then by (4.12) 
$\bC_{L_{1}/Q_{1}} (X_{1}\times\ldots\times X_{t}\times Y_{1}\times\ldots\times
Y_{f})\leq X_{1}\times\ldots X_{t}\times Y_{1}\times\ldots
\times Y_{f}$. Now we may choose $X_{1}=U_{1}\times W_{1}$ and
$X_{2}\not= X_{1}$. Let $\hat{L}_{1} =\langle (X_{2}\bmod
Q_{1}),\omega ,L_{1}\cap L_{2}\rangle$. Then (1) - (4) holds. Hence
in any case we may assume $[O_{p}(U),X_{1}\times\ldots X_{t}\times Y_{1}\times\ldots
\times Y_{f}]=1$.

Suppose $O_{p} (U)$ contains an elementary abelian characteristic
subgroup $C$ of order $\geq p^{\, 3}$. Then by choice of $L_{1}$ we
have $L_{1}/O_{2}(L_{1}) =$ $C\cdot (L_{1}\cap L_{2})/O_{2}(L_{1})$. Now
as before $|\bC_{C} (\tilde{Z}_{2} )|\leq p$. We have $[C,Q_{2}]\leq
\bC_{C} (\tilde{Z}_{2} )$. Let $D=\bC_{C} (Q_{2})$. Suppose $D\not=
C$. Let $\hat{L}_{1}$ be the preimage of $D\cdot (L_{1}\cap
L_{2})/O_{2}(L_{1})$. Obviously $\hat{L}_{1}$ satisfies (1) - (4).

Assume $D=C$. Then $Q_{2}\leq Q_{1}$. Hence $Z_{2}$ is an $F$-module
with offending subgroup in $Q_{1}$. As $Q_{2}\not= Q_{1}$, (vi) is
not possible. Hence with (1.31) we see that $Z_{2}$ is the natural
$L_{2}(q)$-module, or $L_{2}/\bC_{L_{2}} (Z_{2})\cong\Sigma_{3}
\wr \bZ_{2}$. In the former we get a contradiction with (4.2). 
So assume the latter.
Let now $D$ be an $S$-invariant complement of $\bC_{C} (\tilde{Z}_{2}
)$. Let $\hat{L}_{1}$ be the preimage of $D\cdot S/O_{2}(L_{1})$.
Then $\hat{L}_{1}$ satisfies (1) - (4).

Let now $C$ be a characteristic subgroup of $O_{p}(U),C\cong
E_{p^{\, 2}}$. Then $m_{p}(\bC_{L_{1}} (C))\geq 3$. If $\bC_{C}
(\tilde{Z}_{2} )=1$, we let $\hat{L}_{1}$ be the preimage of
$C(L_{1}\cap L_{2})/O_{2}(L_{1})$. Then (1) - (4) holds. (Recall
$Q_{2}\leq\hat{Q}_{1}$ and so $Z_{2}$ is an $F$-module, hence $p\not\Big |\, 
|L_{1}\cap L_{2}|$).

So assume $|\bC_{C} (\tilde{Z}_{2} )|=p$. If $U_{1}\times W_{1}\not=
1$, set $D=\bC_{C} (Q_{2})$. Suppose $D\not= C$. Let $\hat{L}_{1}$ be
a preimage of $(U_{1}\times W_{1}\times D)\, (L_{1}\cap L_{2})/O_{2}(L_{1})$.  
Then (1) - (4) holds.

Assume $[C,Q_{2}]=1$. Now $C\leq N_M(J(Q_{2}Q_{1}))$ and so we have
that $Z_{2 }$ is an $F$-module with offending subgroup in
$Q_{2}Q_{1}$. Now as before $L_{2}/\bC_{L_{2}} (Z_{2})\cong\Sigma_{3}
\wr \bZ_{2}$. Let $D$ be an $S$-invariant complement of $\bC_{C}
(\tilde{Z}_{2} )$ in $C$, and $\hat{L}_{1}$ be a preimage of
$(U_{1}\times W_{1}\times D)S/O_{2}(L_{1})$. Again (1) - (4) holds.

We may assume $U=L_{1}/O_{2}(L_{1})$. Suppose $\Omega_{1}
(O_{p}(U))=C$. Now $\bC_{U} (C)\leq O_{p}(U)$ and we see
$U=O_{p}(U)(L_{1}\cap L_{2})/O_{2}(L_{1})$, as any $p$-element
stabilizes  $1<\bC_{C} (\tilde{Z}_{2} ) < C$. But now there is
$\omega\in L_{1}\cap L_{2},o(\omega )=p$ with $[\omega ,C]=1$. This
contradicts $N_{G}(\langle\omega\rangle )\not\leq M$ by (1.66).

We have $m_{p}(O_{p}(U))\geq 3$ and so $U=O_{p}(U)(L_{1}\cap
L_{2})/O_{2}(L_{1})$ again. Let $[Q_{2},O_{p}(U)]\not= 1$. Then
$m_{p}([Q_{2},O_{p}(U)])=1$. Hence $m_{p}(\bC_{O_{p}(U)} (Q_{2}))\geq
2$. The same holds if $[Q_{2},O_{p}(U)]=1$. Hence in any case
$N_{G}(J(Q_{1}Q_{2}))$ contains $E\cong E_{p^{\, 2}} ,\Gamma_{E,1}
(G)\leq M$. This shows that $Z_{2}$ is an $F$-module with offending
subgroup in $Q_{1}Q_{2}$. Now as before this implies $L_{2}/\bC_{L_{2}}
(Z_{2})\cong\Sigma_{3} \wr \bZ_{2}$. Then $L_{1}\cap L_{2}=S$. Let $T=\bC_{S}
(\tilde{Z}_{2} )$ and let $\hat{L}_{1}$ be the preimage of
$\bC_{O_{p}(U)} (T)S$. As $m_{p}(\bC_{O_{p}(U)} (T))\geq 2$, we have
(1) - (4).

So we are left with $O_{p}(U)$ is extraspecial, $\Omega_{1}
(O_{p}(U))=O_{p}(U)$. Furthermore we may assume $\bC_{O_{p}(U)}
(\tilde{Z}_{2} )\not\leq \Phi (O_{p}(U))$. As $\bC_{O_{p}(U)} (V_{1})\lhd
L_{1}/O_{2}(L_{1})$, we now have $m_{p}(\bC_{O_{p}(U)} (V_{1}))=2$.
We know that for $E\leq\bC_{O_{p}(U)}
(V_{1}),m_{p}(E)=2,\Gamma_{E,1} (G)\not\leq M$. This implies
$U_{1}\times W_{1}=1$ and $|O_{p}(U)|=p^{\, 3}$ and
$U/O_{p}(U)\lesssim GL_{2}(p)$. As $m_{p}(L_{1})=3$, we see that $p\,\Big |\,
|U:O_{p}(U)|$. But now again there is some $E\leq O_{p}(U),E\cong
E_{p^{\, 2}},\Gamma_{E,1} (G)\leq M$, as $E\leq F,F\cong E_{p^{\,
3}}$. This is a contradiction, proving the lemma.\\
\\

Hence if we are in the situation of (4.13) we replace $L_{1}$ by
$\hat{L}_{1}$.

\absa
{\bf (4.14)~Lemma.~}{\it $O_{2}(\langle L_{1},L_{2}\rangle )=1$.}

\absa
Proof~. If $L_{1}$ is as in (4.9)(ii) the assertion follows with (4.10)
and $L_{2}\not\leq M$. Otherwise $m_{p}(L_{1})\geq 2$ and $L_{1}$
contains an elementary abelian group $E$ of order $p^{\, 2}$ with
$\Gamma_{E,1} (G)\leq M$. Hence $\langle L_{1},L_{2}\rangle\leq M$ if
$O_{2}(\langle L_{1},L_{2}\rangle )\not= 1$, a contradiction.

\vspace{1cm}


Now for the remainder of this paper let us fix some notations:
\\
\\
Let $\Gamma =\Gamma (L_{1},L_{2})$ be the coset graph.
Furthermore let\\
\\
 $b_{i}=\min\{ d(i,\alpha )|\alpha\in \Gamma$ such
that $Z_{i}\leq L_{\alpha}$ but $Z_{i}\not\leq L_{\beta}$ for some
$\beta\in\triangle (\alpha )\},i=1,2 $,  where $\triangle (\alpha )$
denotes the set of neighbours of $\alpha$ in $\Gamma$. Let 
\begin{center}
$b=\min\{ b_{1},b_{2}\}$.
\end{center}
A critical pair $(i,\alpha )$ is a pair of vertices in $\Gamma$ such
that $b=b_{i},d(i,\alpha )=b$ and $Z_{i}\not\leq L_{\beta}$ for some
$\beta\in\triangle (\alpha )$. 

We recall some basic facts about $\Gamma$, which might be found in
\cite{Gol}.

\begin{itemize}
\item[(1)]$\Gamma$ is connected.
\item[(2)]$L=\langle L_{1},L_{2}\rangle$ acts edge transitively on
$\Gamma$.
\item[(3)]Let $\alpha\in \Gamma$, then $L_\alpha$ is either conjugate to
$L_{1}$ or to $L_{2}$ (this will be denoted by $\alpha\sim 1$ or
$\alpha\sim 2$). 
\item[(4)]Let $B=L_{1}\cap L_{2},U\leq B$ such that
$L_{i}=N_{L_{i}}(U)B,i=1,2$, then $U=1$. 
\end{itemize}

We fix the following notation:
\begin{itemize}
\item[(a)]There are vertices $1,2\in\Gamma$ whose stabilizers in $L$
which are $L_{1}$ and $L_{2}$, respectively.
\item[(b)]Let $N_{i}\unlhd G_{i},i=1,2 ,\alpha\in \Gamma$ and $g\in L$
with $i^{\, g}=\alpha$. Then set $N_{\alpha}=N_{i}^{\, g}$. As $g$ is
uniquely  determined up to $h\in G_{i}$, we have that $N_{\alpha}$
is well defined.
\end{itemize}

There is some unsymmetry between $L_{1}$ and $L_{2}$. While the
structure of $L_{2}$ is restricted, all we know about $L_{1}$ is some
information about $p$-elements. The proof of the theorem is devided
into several steps.
\begin{itemize}
\item[(i)] We show that the exceptional $Sz(q)$--situation from (4.4)
does not occur (\S 5)
\item[(ii)]We prove that $\Omega_{1} (Z(S))\lhd L_{1}$. ((5.1),\S 6)
\item[(iii)]We show that $\Omega_{1} (Z(S))\lhd L_{2}$. (\S 5,7,8)
\end{itemize}
Obviously (i) and (ii) yield a contradiction to (4).\\

In both cases (i) and (ii) there is a basic subdivision $b>2$ and
$b\leq 2$, where arguments are different. The so called amalgam
method, i.e. investigation of $\Gamma$, just works for $b>2$.























\begin{center}
\S~5 Action of $L_{1}/Q_{1}$ and $L_{2}/Q_{2}$ on some modules
\end{center}

\vspace{1cm}

\absa
{\bf~(5.1) Lemma~}{\it Let $V \le Z_1$ be some module for 
$L_{1}/Q_{1}$. Let $K$ be a quasisimple component of $L_{1}/Q_{1}$.
Suppose $V$ is an $F$-module for $K$. Then there is a submodule $U$
of $V$, $[U,K]=U$, such that any $x\in U^{\sharp}$ is centralized by
some $E_{x}\cong E_{p^{2}}$ with $\Gamma_{E_{x},1}(G)\leq M$. In particular
$\bC_{G}(x)\le M$, or }

(i) {\it $K\cong L_{2}(q)$, $q$ even, $U=[V,K]$ and $U/\bC_U(K)$ is the natural
module. Furthermore $\bC_{U}(K)\neq 1$ and $U$ is centralized by $\langle
\rho\rangle$ with  $o(\rho)=p$ and $N_{G}(\langle \rho\rangle)\le M$. }

(ii) {\it $K\cong U_{4}(2)$, $U=[V,K]$ and $U/\bC_U(K)$ is the natural module.
Further\-more $|\bC_{U}(K)|\leq 4$ and every $x\in U^{\sharp}$ is centralized by some
$\rho\in L_{1}$ with $o(\rho)=3=p$.}

\absa
Proof.~ By (1.18) $K/Z(K)\cong A_{n}$, $L_{n}(q)$, $Sp_{2n}(q)$,
$\Omega_{2n}^{\pm}(q)$, $U_{n}(q)$ or $G_{2}(q)$, $q$ even. Further by
(4.4) we may assume that $m_p(G_1) \ge 4$.

\bigskip


(1)~~\parbox[t]{14cm}{$K/Z(K)\cong A_{n}$}

\bigskip

Let $U$ be a submodule with $[U,K]=U$. Suppose that $U/\bC_{U}(K)$ is
the permutation module. If $p=3$ and $n\geq 8$, then every $x\in U$ is
centralized by $E_{x}\cong E_{9}$ in $K$.

So assume next $p\neq 3$. Then $n\leq 11$. Furthermore $U=[V,K]$. We
have $m_{p}(K)\leq 2$. Hence by (4.4) some $E\cong E_{p^{2}}$ centralizes $U$,
the assertion. This also applies for $n\leq 7$, $p=3$.

Hence by (1.18) we may assume that $n=8$ or $7$ and
$|U/\bC_{U}(K)|=16$, $n=6$ and $|U/\bC_{U}(K)|=2^{6}$, or $n=5$ and
$U/\bC_U(K)$ is the $L_{2}(4)$-module. If $n\neq 8$, then $U=[V,K]$ and
we can argue as above. 

So we are left with $n=8$, $U\neq [V,K]$. As $V$ is an $F$-module, we
have that $[V,K]$ has at least three nontrivial composition factors.
Hence for $p>3$, we get with (4.4) some $E\cong E_{p^{2}}$ with $[E,[V,K]]=1$, the
assertion. Let $p=3$. Then by (4.4) there is some $3$-element $\rho$ with
$[\rho,[V,K]]=1$. But every $x\in U^{\sharp}$ is centralized by some
$3$-element in $K$, the assertion.

\bigskip

(2)~~\parbox[t]{14cm}{$K/Z(K)\cong L_{n}(q)$, $q$ even}

\bigskip

Suppose $U/\bC_{U}(K)$ is the natural module. Then every $x\in
U^{\sharp}$ is centralized by $L_{n-1}(q)$. So we may assume that
there is no $E\cong E_{p^{2}}$, $E \le L_{n-1}(q)$ with
$\Gamma_{E,1}(G)\leq M$. This shows $K/Z(K)\cong L_{2}(q)$,
$L_{3}(q)$, $L_{4}(q)$, $L_{5}(2)$, $L_{6}(2)$ or $L_{7}(2)$.

Suppose next $p\, \Big |\, q^{2}-1$. Then we have $K/Z(K)\cong L_{n}(q)$, $n\leq
4$. Furthermore $m_{p}(K)\leq 2$. If $m_{p}(\bC(K))\geq 3$, we get $U$
with $[E,U]=1$ for some $E\cong E_{p^{2}}$ with $m_{p}(\bC(E))\geq 3$. 

So we may assume $m_{p}(\bC(K))\leq 2$. Hence by (4.12) either
$K\cong L_{2}(q)$ and field automorphisms are involved, or
$m_{p}(K)=2$. In the latter $U$ is centralized by $\rho$ with
$o(\rho)=p$. Now for every $x\in U^{\sharp}$, $p\, \Big |\, |\bC_{U}(x)|$ as
$p\, \Big |\, q^{2}-1$ and so we get $E_{x}\cong E_{p^{2}}$, $[E_{x},x]=1$ and
$\Gamma_{E_{x},1}(G)\leq M$.

Suppose $K\cong L_{2}(q)$. Then just one module is involved. Now
$[U,\rho]=1$, for some $\rho\in\bC(U)$. This is (i).

So assume now $p \not\!\Big |\, q^{2}-1$. Let $m_{p}(K)\geq 2$. Then we get
$K\cong L_{6}(2)$ or $L_{7}(2)$ and $p=7$. Now $p\, \Big |\, |L_{n-1}(2)|$.
This shows $K\cong L_{6}(2)$. By the construction given in \S~4 there is $E_{7^{2}}\cong E\le
\bC(K)$. Now we may choose $U$ such that $U$ is centralized by
$\rho\in E$, $o(\rho)=7$, and now every $x\in U^{\sharp}$ is
centralized by $E_{x}\cong E_{7^{2}}$ with $\Gamma_{E_{x},1}(G)\leq M$.


We may assume that $U/\bC_{U}(K)$ is not natural. By (1.18) 
$U/\bC_{U}(K)\cong V(\lambda_{2})$, or $K\cong L_{2}(r^{2})$,
$r^{2}=q$, and $U/\bC_{U}(K)$ is the orthogonal
$\Omega_4^{-}(q)$-module. In the latter by (1.20) $\bC_{U}(K)=1$ and
so every $x\in U^{\sharp}$ is centralized by $E_{x}\cong E_{p^{2}}$ with 
$\Gamma_{E_{x},1}(G)\leq M$.

Suppose now $U/\bC_{U}(K)\cong V(\lambda_{2})$. Then $V$ is the set
of forms on $V^{\ast}$. Hence for any $x\in U$, we get $\bC_{K}(x)$
picks up $L_{2}(q)\times L_{n-2}(q)$ or $Sp_{4}(q)$. So we may assume
that $p \not\!\Big |\, q^{2}-1$. In this case $n\leq 7$. Furthermore $q=2$ for
$n>4$. Finally we have $m_{p}(K)\geq 2$, otherwise we may choose $U$
with $[E,U]=1$ for some $E\cong E_{p^{2}}$ with $\Gamma_{E,1}(G)\leq M$. This
shows $K\cong L_{6}(2)$ or $L_{7}(2)$ and $p=7$. By (1.23) we see in
both cases that there are at most two composition factors which are
isomorphic to $V(\lambda_{2})$. Hence some $E\cong E_{p^{2}}$ with
$\Gamma_{E,1}(G)\leq M$ centralizes $U$.

\bigskip

(3)~~\parbox[t]{14cm}{$K/Z(K)\cong Sp_{2n}(q)$, $q$ even}

\bigskip


Suppose first $U/\bC_{U}(K)$ is the natural module. Then every $x\in
U^{\sharp}$ is centralized by $Sp_{2n-2}(q)$. Hence we have the
assertion or $K/Z(K)\cong Sp_4(q)$ or $Sp_6(q)$. In the latter we may
assume $m_{p}(K)=1$. Anyway, if $m_{p}(K)=1$ by (4.12) and (4.4) there is some
$E\cong E_{p^{2}}$ with $\Gamma_{E,1}(G)\leq M$ and $[E,U]=1$. 

So let $K/Z(K)\cong Sp_4(q)$, $m_p(K)=2$. Now by (4.12) and (4.4) there is
$\rho\in\bC(K)$, $o(\rho)=p$, and we may assume $[\rho,U]=1$. As any
$x\in U^{\sharp}$ is centralized by $L_{2}(q)$ in $K$ and $p\, \Big | \,
q^{2}-1$ (as $m_p(K)=2$), we get the assertion.

Hence by (1.18) we may assume $K\cong Sp_6(q)$ and $U/\bC_{U}(K)$ is
the spin module. Furthermore we may assume that $m_{p}(K)\geq 2$. Hence
$p\, \Big |\, q^{2}-1$. As elements in $U$ are centralized by $L_{3}(q)$ or
$G_{2}(q)$, we see that $p\, \not\!\Big |\, q-1$. Hence $p\, \Big |\, q+1$. As
$m_{p}(\bC(K))\geq 1$ by (4.12) we get that there are at least two
composition factors which are spin modules. Hence $|A|\geq q^{4}$ by
(1.23), $A$ an offending subgroup. Furthermore $|A|\neq q^{6}$, as
$|[U,t]|\geq q^{4}$ for a transvection $t$ (in the natural
representation). Now $|U:\bC_{U}(A)|=q^{2}$ and so by (1.23) again
$\bC_{U}(A)=\bC_{U}(a)$ for every $a\in A^{\sharp}$. But $\langle
\bC_{K}(a)\, \Big | \, a\in A^{\sharp} \rangle = K$, a contradiction.



 \bigskip

(4)~~\parbox[t]{14cm}{$K/Z(K)\cong \Omega_{2n}^{\pm}(q), n\geq 4,q$ even}

\bigskip


Let $U/\bC_{U}(K)$ be the natural module. Then every $x\in U^{\sharp}$
is centralized by $\Omega_{2n-2}^{\pm}(q)$. Hence we may assume
$m_{p}(\Omega_{2n-2}^{\pm}(q))\leq 1$. This shows $K\cong \Omega_8^{-}(q)$
and $m_{p}(K)=1$. Now by (4.12) we may assume that there is $E\cong
E_{p^{2}}$ with $\Gamma_{E,1}(G)\leq M$, and $[E,U]=1$. Now by (1.18) we may assume that $K\cong \Omega_{10}^{+}(q)$ and
$U/\bC_{U}(K)$ is the spinmodule. In this module every element is
centralized by $L_{4}(q)$ or $Sp_{6}(q)$. As $p \,\Big |\, q^{2}-1$, we get
the assertion. 


\bigskip

(5)~~\parbox[t]{14cm}{$K/Z(K)\cong U_{n}(q)$, $n\geq 4$, $q$ even}

\bigskip

If $U/\bC_{U}(K)$ is the natural module, then as every $x\in U^{\sharp}$
is centralized by $U_{n-1}(q)$  or $U_{n-2}(q)$, we get $n\leq 5$.
Furthermore in case $n=5$ we have $q=4$. Let first $n=5$. Then $p\neq
5$ as $m_{5}(U_{3}(4))=2=m_{5}(U_{4}(4))$. Hence $m_{p}(K)\leq 2$. If
$m_{p}(K)=2$, then $p=3$. Now there is $\rho\in\bC(K)$, $[\langle
\rho\rangle,U]=1$. As $3\, \Big |\, |U_{4}(4)|$ and $|U_{3}(4)|$, we have the
assertion.

So let $m_{p}(K)=1$. Then we may assume that there is $E\cong
E_{p^{2}}$ with $\Gamma_{E,1}(G)\leq M$ and $[U,E]=1$.

Assume $n=4$. If $m_{p}(K)=1$, we may argue as before. Hence
$m_{p}(K)\geq 2$. Now $p\, \Big |\, |U_{3}(q)|$ and $p \,\Big |\,  |U_{2}(q)|$. The
same argument as before shows $m_{p}(K)\geq 3$, i.e.\ $p\, \Big |\,  q+1$. By
(1.20) $\bC_{U}(K)=1$ or $q=2$ and $|\bC_{U}(K)|\leq 4$. The latter
just is (ii). In the former by (1.45), we see that $U$ is invariant
under $L_{1}$. But now every $x\in U^{\sharp}$ is centralized by
$E_{x}\cong E_{p^{2}}$ with $\Gamma_{E_{x},1}(G)\leq M$.


So assume that $n=4$, $U_{n}(q)\cong\Omega_6^{-}(q)$ and
$U/\bC_{U}(K)$ is the natural orthogonal module. Again by (1.20)
$\bC_{U}(K)=1$. Furthermore by (1.45) $U\triangleleft G_{1}$. Hence
$p \not\:\Bigm | \, |L_{2}(q^{2})|$ or $p \not\,\Bigm |\, |Sp_{4}(q)|$. In particular
$m_{p}(K)=1$. But now by (4.12) we may assume that there is $E\cong
E_{p^{2}}$ with $\Gamma_{E,1}(G)\leq M$ and $[U,E]=1$.


\bigskip

(6)~~\parbox[t]{14cm}{$K/Z(K)\cong G_{2}(q),q$ even.}

\bigskip

By (1.18) $U/\bC_{U}(K)$ is the natural $6$-dimensional module. As
before we may assume $m_{p}(K)=2$, i.e.\ $p \,\Big |\,  q^{2}-1$. Now every
$x\in U^{\sharp}$ is centralized by some $p$-element in $K$. By
(4.12) we may assume there is some $\rho\in\bC(K)$, $o(\rho)=p$ and
$[\rho,U]=1$, the assertion.

\absa
{\bf (5.2)~Hypothesis.~} (a)~~$Z_{1}\le Z_{2}$,
$m_{p}(\bC_{L_{1}}(Z_{1}))\geq 3$. \\
\indent
(b)~~$b=b_{2}>2$. 

\absa
Set  $V_{1}=\langle Z^{G_{1}}_{2}\rangle$. We have $V_{1}\le
Q_{1}$ if $b>2$.


\absa
{\bf (5.3) Lemma.~}{\it Assume (5.2). Let $J(S)\not\le Q_{1}$ and $J(S) \le Q_{2}$. 
There is  $U_{1}\le L_{1}$, $S\le U_{1}$ such that for
$\tilde{Q_{1}}=O_{2}(U_{1})$, we either have that
$E(U_{1}/\tilde{Q_{1}})$, is a direct product
$L^{(1)}\times\ldots\times L^{(n)}$, $L^{(1)}\cong L^{(2)}\cong \ldots \cong \L^{(n)}$,
$L^{(1)}\cong L_{2}(r)$, $U_{3}(r)$, $Sz(r)$, $^{2}F_{4}(r)$, $r$ even,
sporadic, a group of Lie type in odd characteristic, a
solvable $\{2,r\}$-group, or $L_{1}$ is isomorphic to $r^{3}L_{2}(r)$, $r$ an
odd prime. Furthermore $J(S)\not\le \tilde{Q}_{1}$ and 
$Z_{2}\ntriangleleft U_{1}$.}


\absa
Proof.~Let $L/Q_1$ be a component of $L_{1}/Q_{1}$ with $[J(S),L]\not\subseteq
Q_1$. Suppose $L/Q_1$ is alternating or a group of Lie type 
$G(r)\not\cong\,^{2}F_{4}(r)$ in even characteristic. Then there is some minimal
parabolic $P$ such that $J(S)\not\le O_{2}(\langle P^{J(S)} \rangle J(S))$. If $P\not\le
N_{L_{1}}(Z_{2})$, we choose $P$ as $L^{(1)}$. So assume $P\le
N_{L_{1}}(Z_{2})$ for every $P$ with $J(S)\not\le O_{2}(\langle P^{J(S)} \rangle J(S))$. Then
$[O^{2}(P),Z_{2}]=1$. Let $p\in\sigma(M)$ with $p~\mid~|P|$. By the
Frattini argument we have $m_p(\bC_G(Z_2)) \le 1$. Hence we see that for some $x \in P$ with $o(x) = p$ we get $N_G(Z_2) = \bC_G(Z_2)N_G(\langle x \rangle)$.
As $N_G(Z_2) \not\le M$, we have $N_G(\langle x \rangle) \not\le M$ and
so $m_p(\bC_M(x)) \le 2$ for all $x \in P$ with $o(x) = p$. By the construction in chapter 4 we may assume $m_p(L) \geq 4$. But then we easily see 
$m_p(\bC_M(x)) > 2$ for some $x \in P$, with $o(x) = p \in \sigma(M)$. So we have $p \not\:\Bigm | \,
|P|$. In particular $p\neq 3$ and $p \not\Big |\,  r^{2}-1$ if $L/Q_1 \cong G(r)$. This implies
$m_{p}(L)\leq 2$.

Suppose first $m_{p}(L)=2$. Then we see $L \unlhd L_1/Q_1$ and $L/Q_1 \cong A_{10}$, $A_{11}$,
$L_{6}(2)$, $L_{7}(2)$. In all cases $p \,\Big | \, |\bC_{L_{1}/Q_{1}}(L/Q_1)|$.
We first suppose $[J(S)Q_{1}/Q_{1},\bC_{L_{1}/Q_{1}}(L/Q_1)]\neq 1$.
As $m_{3}(L)=3$, $p\neq 3$, we have that 
$\bC_{L_{1}/Q_{1}}(L/Q_1)$ is solvable or has components of type $Sz(r)$. Now choose $L^{(1)}$
as such a component. Suppose now
$[J(S)Q_{1}/Q_{1},\bC_{L_{1}/Q_{1}}(L/Q_1)]=1$. Let $L/Q_1 \cong A_{10}$, $A_{11}$.  Then $p = 5$ and there is some
parabolic $P$ in $L$ with $P/O_{2}(P)\cong S_{5}$ and $J(S)\le
O_{2}(\langle P^{J(S)} \rangle J(S))$. Now $J(S)$ is normalized by $E\cong E_{25}$ with
$\Gamma_{E,1}(G)\le M$, a contradiction to $L_{2}\not\le M$.


Let next $L/Q_1 \cong L_{6}(2)$ or $L_{7}(2)$. Then $p = 7$ . Now  there is more than one
parabolic  $P$ with $J(S) \not\le O_2(\langle P^{J(S)} \rangle J(S))$ and so we get $\Sigma_{3}\times\Sigma_{3}$ which is
involved in $\bC(Z_{2})$. Call this group $U_2$. We are going to prove a similar result for
$m_{p}(L)=1$. 

Let now $m_{p}(L)=1$. If $m_r(L) \leq 1$ for all primes $r$, we may choose $L$ as $L^{(1)}$. So we may assume that there is some prime $r$ with $m_r(L) > 1$ and so $L \unlhd L_1/Q_1$.   Then
$[J(S)Q_{1}/Q_{1},\bC_{L_{1}/Q_{1}}(L/Q_1)]\neq 1$. Now we may assume that there is a subgroup $\tilde{L}/Q_1$ of $\bC_{L_1/Q_1}(L/Q_1)$ with $[J(S), \tilde{L}] \not\le Q_1$ too. We may assume
that $\tilde{L}$ is alternating or of type $G(\tilde{r})$. Finally as we may interchange the r\^oles of $L$ and $\tilde{L}$ we have $m_p(\tilde{L}) = 1$ too. As $3\not\in\sigma(M)$,
we may assume $m_{3}(L)=1$. Hence $L\cong (S)L_{3}(r)$. Now
$m_{3}(\tilde{L})\leq 2$. Then $\tilde{L}\cong L_{3}(\tilde{r})$,
$Sp_{4}(\tilde{r})$, $G_{2}(\tilde{r})$, $^{3}D_{4}(\tilde{r})$,
$L_{4}(\tilde{r})$, $U_{4}(\tilde{r})$, $U_{5}(\tilde{r})$,
$L_{5}(\tilde{r})$, or $A_{7}$ by (1.4). Now we get $L_{2}(r)\times
L_{2}(\tilde{r})$ or $L_{2}(r)\times U_{3}(\tilde{r})\lesssim 
\bC(Z_{2})$. Hence we have that there is $U_{2}\le L_{1}$,
$[U_{2},Z_{2}]=1$, $U_{2}/O_{2}(U_{2})\cong K_{1}\times K_{2}$,
$K_{1}\cong L_{2}(r)$, $K_{2}\cong L_{2}(\tilde{r})$ or
$SU_{3}(\tilde{r})$ including $\tilde{r}=2$, $r=2$. For each $\omega
\in K_{i}$, $o(\omega)$ prime, $o(\omega)>2$, we have that $\omega$
is centralized by some $E\cong E_{p^{2}}$ with $\Gamma_{E,1}(G)\le M$.
As $p>3$ and $\bC_{O_{2}(M)}(\omega)\neq 1$, we get with (1.63)
$N_{G}(\langle \omega\rangle)\le M$.

But we have $U_2 \le M_{3}$, $M_{3}\neq M_{1}$. Now by
construction $L/O_{p^\prime}(L)$ was a component of $M/O_{p'}(M)$. Let $\hat{L}$ be the
corresponding group in $M_3$. So assume first
$1\neq[\hat{L}O_{p'}(M_{3})/O_{p'}(M_{3}),U_{2}]\le
\hat{L}O_{p'}(M_{3})/O_{p'}(M_{3})$. Then as $L\cong L_{6}(2)$,
$L_{7}(2)$ or $L_{3}(r)$, we see
$\bC_{U_{2}}(\hat{L}O_{p'}(M_{3})/O_{p'}(M_{3}))\not\le O_{2}(U_{2})$
or $L\cong L_{6}(2)$, $L_{7}(2)$ and $(\hat{L}O_{p'}(M_{3})\cap
U_{2})O_{2}(U_{2}) = U_{2}$.  But in the latter $3 \not\,\Bigm | 
|O_{p'}(M_{3})|$ as $3\not\in \sigma(M)$ and so $\omega\in K_{1}$ is
centralized by $\nu \in M_{3}$, $o(\nu)=7$, $\nu\in E\cong E_{p^{3}}$,
$E\le M_{3}$. As $\nu \in N_{G}(\langle \omega \rangle)\le
M$, we have a contradiction to (4.2).

So we have that
$\bC_{U_{2}}(\hat{L}O_{p'}(M_{3})/O_{p'}(M_{3}))O_{2}(U_{2})\geq
K_{1}$ or $K_{2}$. Now choose
$\omega\in\bC_{U_{2}}(\hat{L}O_{p'}(M_{3})/O_{p'}(M_{3}))$,
$o(\omega)=3$. As $m_{3}(O_{p'}(M_{3}))\leq 3$, and $L/Q_1 \not\lesssim
L_{3}(3)$ or $Sp_{4}(3)$, we see that for $U_{3}\in
Syl_{3}(\bC_{M_{3}}(\hat{L}O_{p'}(M_{3})/O_{p'}(M_{3})))$ we have
$\bC_{M_{3}}(U_{3})O_{p'}(M_{3})/O_{p'}(M_{3})\geq
\hat{L}O_{p'}(M_{3})/O_{p'}(M_{3})$. Hence there is some
$\nu\in M_{3}$, $o(\nu)=p$ with $[\nu,\omega]=1$ and $\nu\in E\cong E_{p^{3}}$,
contradicting (4.2).

\absa
{\bf (5.4)~Lemma}~{\it Assume (5.2)(a). \\
\\
(a)~Let $J(S)\not\le Q_{1}$ and $[J(S),Z_1] = 1$. Then
one of the following holds }\\
\\
\indent
(i) $[\langle J(S)^{L_{1}}\rangle,Z_{2}\cap Z(O_{2}(\langle
J(S)^{L_{1}}\rangle))]=1$.\\
\\
\indent
(ii) {\it There is a component $L_{3}(2)$ of $L_{1}/Q_{1}$ which
induces an
indecomposable submodule $U$ of dimension 4 in $\Omega_{1}(Z(O_{2}(\langle J(S)^{L_{1}}\rangle)))$ .}\\
\\
\indent
(iii) {\it There is a component $A_{2^{m}}$ of $L_{1}/Q_{1}$ which
induces a
submodule $U$ in $\Omega_{1}(Z(O_{2}(\langle J(S)^{L_{1}}\rangle)))$   which is an extension of the trivial module by the
permutation module.}\\
 \\
(b) ~ {\it In (a)(ii) and (a)(iii) for every $1\neq x\in Z_{2}\cap U$ we have
$\bC_{G}(x)\le M$.}

\absa
Proof.~Set $W=\langle J(S)^{L_{1}}\rangle$ and $\overline{W} = W/O_2(W)$. As $[J(S),Z_1] = 1$, we have $[W, \Omega_1(Z(S))] = 1$. Hence $[O(\overline{W}), Z(O_2(W))] = 1$. Assume now that we do not have (i). Then by construction in chapter 4 we have that there are components of $L_1/Q_1$ which act nontrivially on $\Omega_1(Z(O_2(W))) = V$. Then $V$
is an $F$-module.  Let $\tilde{L}$ be a component of
$\overline{W}$, such that $\tilde{L}$ induces an $F$-module on $V$, which exists by \cite[(26.24)]{GoLyS} Then by (1.18), $\tilde{L}\cong SL_{n}(r)$, $SU_{n}(r)$,
$Sp_{2n}(r)$, $\Omega_{2n}^{\pm}(r)$,$G_{2}(r)$, $r$ even, or $A_{n}$.
Now we have some submodule $U$ such that $U$ is a nonsplit extension
of a trivial module by an irreducible $F$-module for $\tilde{L}$.
Furthermore $\bC_{U}(S\cap N_{L_{1}}(\tilde{L}))=\bC_{U}(\tilde{L})$.
Application of (1.21) shows $\tilde{L}\cong L_{3}(2)$ or $A_{2^m}$.

In case $\tilde{L}=L_{3}(2)$ we have that $U$ is the indecomposable
$4$-dimensional module.

If $m_{p}(\tilde{L})=1$, there is $E\cong
E_{p^{2}}$ with  $[E,\tilde{L}]=1$ and $\Gamma_{E,1}(G)\leq M$. Now $[E,U]=1$,
and we have the assertion with (1.63).

So assume $\tilde{L}\cong A_{2^{m}}$ and $p=3$,
$m_{p}(\tilde{L})\geq 2$. As every $v\in U$ is centralized by $E_9$
in $\tilde{L}$, we again get the assertion with (1.63).

\absa
{\bf (5.5) Lemma.}~{\it Assume (5.2). Then one of the following holds }

(i){\it $Z_2$ is an $F$--module}

(ii) {\it $J(S) \le Q_2$ and there is $U_{1}$ such that $U_1$ is as in (5.3) or $U_1/O_2(U_1)$ is
a component of $L_1/Q_1$ with $m_{p}(U_{1})\geq
2$, $p\in\sigma(M)$. Furthermore $\langle
Z^{U_{1}}_{2}\rangle/Z_{1}$ involves at least two nontrivial
irreducible $U_{1}$-modules}

(iii) {\it $J(S) \leq Q_2$ and there is $U_1$ such that $U_{1}/O_2(U_1)\cong A_{2^{m}}$ is a component of 
$L_1/Q_1$. Furthermore there is
$U\le Z(Q_{1})$, $U$ is an extension of the natural $U_1$--module by a
trivial module, $|Z_{2}\cap U|\geq 8$ and for every $x\in
(Z_{2}\cap U)^{\sharp}$, $\bC_{G}(x)\le M$.}

\absa
Proof.~Suppose $Z_2$ is not an $F$--module. Then $J(S)\le Q_{2}$ and
$J(S)\not\le Q_{1}$. By (5.3) we have that there is $U_1$ as in (ii) or (iii) which does not normalize $Z_2$. Set $W_{1}=\langle
Z^{U_{1}}_{2}\rangle$. Let $\tilde{W}_{1}$ be the largest submodule
in $W_{1}$ which possesses just trivial $U_{1}$-composition factors.
Let $W\geq \tilde{W}_{1}$ such that $W/\tilde{W}_{1}$ is a
nontrivial irreducible $U_{1}$-module. Suppose $W_{1}=WZ_{2}$. Set
$X=\bC_{O_2(U_1)}(\tilde{W}_{1})$. Suppose first
$[X,W_{1}]\tilde{W}_{1}\geq W$. As $X\le L_{2}$ we have
$W_{1}=\tilde{W}_{1}Z_{2}$. Now as $J(S)\le Q_{2}$,
$[Z_{2},J(S)]=1$. This implies $J(S)\leq O_2(U_1)$, a
contradiction.

So we have $[W_{1},X]\le \tilde{W}_{1}$. Suppose next
$[W,X]\neq 1$. Let $\tilde{W}$ be a hyperplane in $\tilde{W}_{1}$
such that $[W,X]\not\le \tilde{W}$. We have
$\bC_{W_{1}/\tilde{W}}(U_{1})=\tilde{W}_{1}/\tilde{W}$. Now
$\bC_{W_{1}/\tilde{W}}(X)=\tilde{W}_{1}/\tilde{W}$. Hence $X$ induces
an offending subgroup on $Z_{2}$, a contradiction.

So we have $[W,X]=1$. As $[O_2(U_1),U_{1}]\le X$, we get
$W\le \tilde{W}_{1}Z(O_2(U_1))$. Now we apply (5.4). As $[U_1, Z(O_2(U_1))] \not=1$, we do not have (5.4)(i). Hence we 
get some component $\tilde{L} \cong L_{3}(2)$ or $A_{2^{m}}$. But we know that $Z_{2}$
is not an $F$-module. Let first $\tilde{L} \cong L_3(2)$. As $[J(S),Z_{2}]=1$, we see that
$|\bC_{U}(J(S))|=4$, where $U$ is as in (5.4). Now there is $x\in\tilde{S}\setminus Q_{1}$
with $|Z_{2}:\bC_{Z_{2}}(x)|=2$, a contradiction.

Let $\tilde{L}\cong A_{2^m}$. Then $|Z_{2}\cap U|\geq 8$, otherwise
again $Z_{2}$ is an $F$-module. Now the assertion follows with (5.4)

\absa
{\bf (5.6)~Proposition.~}{\it Assume (5.2). Then one of the following
holds}


\begin{itemize}
\item[(i)] {\it $Z_{2}$ is an $F$-module.}

\item[(ii)] {\it $E(L_{2}/Q_{2})\cong L_{2}(t^2)$, $t$ even and
$Z_{2}$ is the natural $O_4^{-}(t)$-module, or $F(L_{2}/Q_{2})\cong
E_{r^{3}}$ and $E((L_{2}/Q_{2})/F(L_{2}/Q_{2}))\cong L_{2}(4)$  and we have
the orthogonal module.}

\item[(iii)] {\it $E(L_{2}/Q_{2})\cong L_{2}(q)\times L_{2}(q)$, $q$ even,
$Z_{2}$ is the natural $O_4^{+}(q)$-module.}


\item[(iv)] {\it $E(L_{2}/Q_{2})\cong L_{2}(q)$, $q$ even, $Z_{2}$ is a
direct sum of two natural modules.}

\item[(v)] {\it $E(L_{2}/Q_{2})\cong SU_{3}(q)$, $q$ even, $Z_{2}$ is the
natural module.}

\item[(vi)] {\it $E(L_{2}/Q_{2})\cong Sz(q)$, $q$ even, $Z_{2}$ is the
natural module.}

\item[(vii)] {\it $L_{2}$ is solvable and one of the following holds

$(\alpha)$  ~~~~$L_{2}/\bC_{L_{2}}(Z_{2})\cong F_{10}$, $F_{20}$,
$|Z_{2}|=16$.

$(\beta)$~~~~$L_{2}/\bC_{L_{2}}(Z_{2})\cong \Sigma_{3}$, $|Z_{2}|=16$.

$(\gamma)$~~~~$L_{2}/\bC_{L_{2}}(Z_{2})\cong E_{9}\bZ_{4}$,
$|Z_{2}|=16$.

$(\delta)$~~~~$O_{3}(L_{2}/\bC_{L_{2}}(Z_{2}))\cong 3^{1+2}$,
$|Z_{2}|=64$.}
\end{itemize}

{\it (Notice that $(\alpha)$ is the solvable version of (vi), $(\beta)$
the solvable version of (iv), $(\gamma)$ the solvable version of (iv)
and $(\delta)$ the solvable version of (v))}


\absa
Proof.~Suppose $Z_{2}$ is not an $F$-module. Then $J(S)\le
Q_{2}$. Now $J(S)\not\le Q_{1}$. Hence we may apply (5.3). Let the notation be as in (5.3). Let $A\in
\cal A(S)$, $A\not\le \tilde{Q_{1}}$. We have $A\cap
\tilde{Q_{1}}=(A\cap \bC_{\tilde{Q_{1}}}(V_{1}))C$  with $C\cap
\bC_{\tilde{Q_{1}}}(V_{1})=1$.

Assume first $C=1$. Then $|A:A\cap
\tilde{Q_{1}}|\geq|V_{1}:\bC_{V_{1}}(A)|$. Hence $V_{1}$ is an
$F$-module for $U_{1}$. This now implies with (1.18) and \cite[(26.24)]{GoLyS} that
$L^{(1)}\cong L_{2}(r)$ or solvable and $|A:A\cap
\tilde{Q_{1}}|=|V_{1}:\bC_{V_{1}}(A)|$. But now by (1.27) there is
just one nontrivial irreducible module involved which contradicts
(5.5).

So we have $C\neq 1$. Then there is some $g\in U_{1}$ such that
$[Z_{2}^{g},C]\neq 1$. Set $C_{1}=\bC_{C}(Z^{g}_{2})$. Suppose there is no elementary abelian subgroup $A_{1}$ of $L_{2}$ with $A_1 \not\leq Q_2$ 
such that

\begin{center}
$|Z_{2}:\bC_{Z_{2}}(A_{1})|\leq|A_{1}:A_{1}\cap Q_{2}|^{2}$
\end{center}

\noindent
Then we get

\begin{center}
$|Z_{2}^{g}:\bC_{Z_{2}^{g}}(C)|>|C:C_{1}|^{2}$
\end{center}


\noindent
Now repeating this argument implies

\begin{center}
$|V_{1}:\bC_{V_{1}}(C)|>|C|^{2}$
\end{center}

\noindent
As $|V_{1}:\bC_{V_{1}}(A)|\leq |C||A:A\cap\tilde{Q}_{1}|$ we get

\begin{center}
$|C|<|A:A\cap \tilde{Q}_{1}|$
\end{center}

\noindent
and then 
$$|V_{1}:\bC_{V_{1}}(A)|<|A:A\cap \tilde{Q}_{1}|^{2}$$.

\bigskip

Now with (5.5) we get that $V_{1}$ involves an $F$-module for
$U_{1}$. Hence we get $L^{(1)}\cong L_{2}(r)$ or solvable again. Now 

\begin{center}
$|A:A\cap \tilde{Q}_{1}|^{2}\leq |V_{1}:\bC_{V_{1}}(A)|$
\end{center}

\noindent
by (1.27). This gives a contradiction. So we have shown: There is
some $A_{1}\le L_{2}$ with $A_1 \not\leq Q_2$ and

\begin{center}
$|Z_{2}:\bC_{Z_{2}}(A_{1})|\leq |A_{1}:A_{1}\cap Q_{2}|^{2}$
\end{center}


Now we may apply (1.35). By (5.2)(a) we have that $G_2 \not= \langle \bC_{G_2}(x) ~|~ x \in Z_1^\sharp \rangle$. Hence we see that
$E(L_{2}/Q_{2})\cong L_{3}(2)$ in (1.35)(2), $E(L_{2}/Q_{2})\cong
Sz(q)$ in (1.35)(3), $q = 4$ and the $V_{i}$
are direct sums of two orthogonal modules in (1.35)(8) and
$E(L_{2}/Q_{2})\cong A_{6}$ in (1.35)(11).

\bigskip

Suppose first that $L_{2}$ is nonsolvable. Let $E(L_{2}/Q_{2})$ be as in
(1.35)(2), (6), (8), (10), (11), (12), (13). Then in all cases $|Z_{1}| \leq 4$.
Furthermore in (1.35)(6) we get that $V_{1}$ involves exactly two
$F$-modules for $L_{2}(r)$. In all other cases even
$|Z_{2}:\bC_{Z_{2}}(A_{1})|\leq|A_{1}:A_{1}\cap
Q_{2}|^{\frac{3}{2}}$.

The same argument as before now shows that

\begin{center}
$|V_{1}:\bC_{V_{1}}(A)|\leq|A:A\cap\tilde{Q}_{1}|^{3}$.
\end{center}

\noindent
Hence in any case there are at most three modules involved.

Set $\hat{L}_{1}=\langle J(S)^{L_{1}}\rangle$. Then
$m_{p}(\hat{L}_{1})\geq 2$. As $U_{1}\le \hat{L}_{1}$, we have
that also $\hat{L}_{1}$ induces at most three nontrivial irreducible
modules and at most two of them in case (1.35)(6). Let
$\hat{Q}_{1}=O_{2}(\hat{L}_{1})$.

Set $\tilde{Z}_{2}=\bC_{Z_{2}/Z_{1}}(\hat{Q}_{1})(\mbox{mod}~Z_{1})$.
Then by (5.4)(b) $[\tilde{Z}_{2}\cap Z(\hat{Q}_{1}),\hat{L}_{1}]=1$ and so
$\hat{Z}_{2}\cap Z(\hat{Q}_{1})=Z_{1}$ or of order $4$ in case (6).
Furthermore for $V^{(1)}_{1}=\langle
\tilde{Z}_{2}^{\hat{L}_{1}}\rangle$ we have $V^{(1)}_{1}>Z_{1}$ and
$V_{1}^{(1)}/Z_{1}$ is a nontrivial module. As
$[V^{(1)}_{1},\hat{Q}_{1}]=Z_{1}$ we get that $\tilde{Q}_{1}$ acts
nontrivially on the set of irreducible $E(L_{2}/Q_{2})$-submodules or
$E(L_{2}/Q_{2})\cong L_{2}(4)$. Set
$Z^{(1)}_{2}=\bC_{Z_{2}/\tilde{Z}_{2}}(\hat{Q}_{1})$. Then
$Z^{(1)}_{2}\not\le V^{(1)}_{1}$ and as
$[\hat{Q}_{1},Z^{(1)}_{2}]\le \tilde{Z}_{2}$, we see that
$\langle {Z^{(1)}_2}^{\hat{L_1}}\rangle= V^{(2)}_{1} > V^{(1)}_{1}$ and
$V^{(2)}_{1}/V^{(1)}_{1}$ involves a nontrivial module. But even in
(1.35)(6) we have $Z_{2}\neq Z_{2}^{(1)}$. So $\langle
Z_{2}^{\hat{L}_{1}}\rangle > V^{(2)}_{1}$, a contradiction. Now in
all other cases set
$Z^{(2)}_{2}=\bC_{Z_{2}/Z_{2}^{(1)}}(\hat{Q}_{1})$. We have
$Z^{(2)}_{2}\neq Z_{2}$ and so $V^{(3)}_{1}=\langle
Z^{(2)\hat{L}_{1}}_{2}\rangle > V^{(2)}_{1}$ and
$V^{(3)}_{1}/V_{1}^{(2)}$ and $V_{1}/V_{1}^{(3)}$ are nontrivial
modules, again a contradiction.

Suppose now $E(L_{2}/Q_{2})\cong A_{9}$. Then $Q_{1}\le Q_{2}$
and so $Z_{2}\le Z(\hat{Q}_{1})$. But this contradicts (5.4). 

So it remains $L_{2}$ to be solvable. Let $L_{2}$ be a
$\{2,5\}$-group. Then again there are exactly two modules involved in $V_1$.
Suppose we do not have (vii)($\alpha$). By
(1.8) $O_{5}(L_{2}/Q_{2})\cong 5^{1+2}$ or $5^{1+4}$.
Suppose $[\Phi(O_{5}(L_{2}/Q_{2})),Z_{2}]\neq 1$ and $[A_{1},Z(O_{5}(L_{2}/Q_{2}))]\neq 1$. As
$|Z_{2}:\bC_{Z_{2}}(A_{1})|\leq 2^{6}$, we get that
$|[Z(O_{5}(L_{2}/Q_{2})),Z_{2}]|\leq 2^{12}$. But Sylow $5$-subgroups
of $GL_{12}(2)$ are abelian. Hence $[A_{1},Z(O_{5}(L_{2}/Q_{2}))]=1$.
Then we have that $|A_{1}:A_{1}\cap Q_{1}|=2$ in the first case and
$|A_{2}:A_{1}\cap Q_{1}|\leq 4$ in the second. Now $x\in
A^{\sharp}_{1}\setminus Q_{1}$ inverts a group of order $25$ in
$O_{5}(L_{2}/Q_{2})/\Phi(O_{5}(L_{2}/Q_{2}))$. But then
$|Z_{2}:\bC_{Z_{2}}(A_{1})|>4$, $16$, respectively. 

So $[\Phi(O_{5}(L_{2}/Q_{2})),Z_{2}])=1$. But now
$|Z_{2}|=2^{8}$, $|Z_{2}|=2^{16}$ respectively and
$Z_{2}=Z^{(1)}_{2}\oplus Z^{(1)}_{2}$, $Z_{2}=
Z^{(1)}_{2}\oplus Z^{(2)}_{2}\oplus Z^{(3)}_{2}\oplus Z^{(4)}_{2}$,
$Z^{(i)}_{2}$ $O_{5}(L_{2}/Q_{2})$-modules. Now the same argument as
above shows that we have more than two modules involved in $V_1$.

So let $L_{2}$ be a $\{2,3\}$-group. By (1.35) we have to handle
$O_{3}(L_{2}/Q_{2})\cong \bZ_{3}\times \bZ_{3}$ and $S\cong D_{8}$,
$|Z_{2}|=2^{8}$, or $O_{3}(L_{2}/Q_{2})\cong 3^{1+4}$ and
$|Z_{2}|=2^{8}$, or $m_{3}(O_3(L_2/Q_2))=3$ and $|Z_{2}|\leq 2^{8}$. 

If $O_{3}(L_{2}/Q_{2})\cong \bZ_{3}\times \bZ_{3}$, then as 
$O_{3}(L_{2}/Q_{2})$-module we have $Z_{2}=Z^{(1)}_{2}\oplus Z^{(2)}_{2}$,
$|Z_{2}^{(i)}|=16$ and so $Q_{1}Q_{2}=S$. Now we see that
$|Z_{1}|=4$, $|\bC_{Z_{2}/Z_{1}}(S)(\mbox{mod}~Z_{1})|=16$,
$|[S,Z_{2}]|=64$ and so again three modules are in $V_{1}$, a
contradiction.

Let $O_{3}(L_{2}/Q_{2})\cong 3^{1+4}$. Then $|Z_{1}|=2$. We have as
$O_{3}(L_{2}/Q_{2})$-module that $Z_{2}=Z^{(1)}_{2}\oplus 
Z^{(2)}_{2}\oplus Z^{(3)}_{2}\oplus Z^{(4)}_{2}$ and $S$ induces a
transitiv subgroup of $S_{4}$ on these modules. Hence $Q_{1}$
contains either $\langle (12)(34),(13)(24)\rangle$ or $\langle
(1234)\rangle$. Now we easily see that there are at least four
modules involved in $V_1$.

So let finally $m_{3}(O_{3}(L_{2}/Q_{2}))=3$ and $m_{2}(S/Q_{2})=2$.
Then $|Z_{2}|\leq 2^{8}$ and so $O_{3}(L_{2}/\bC_{L_{2}}(Z_{2}))\cong
3^{1+2},3^{2}$ or $3^{4}$. But now we may argue as before.

\absa
{\bf (5.7) Lemma.~}{\it  Assume
$\bC_{G}(x)\le M$ for every $x\in Z^{\sharp}_{1}$. If $E(L_2/Q_2) \not\cong Sz(q)$ or $L_2/\bC_{L_2}(Z_2) \not\cong F_{20}$ then assume additionally $Z_1 \leq Z_2$. Let $Z_{2}$ be
one of the modules in (5.6) or $E(L_{2}/Q_{2})\cong L_{2}(4)\times
L_{2}(4)=X_{1}\times X_{2}$, $Z_{2}=Z^{(1)}_{2}\oplus Z^{(2)}_{2}$,
$[Z^{(i)}_{2},X_{j}]=1$ for $i\neq j$, $Z^{(i)}_{2}$ the orthogonal
$X_{i}$-module, $i=1,2$. Then $Z_{2}\le O_{2}(M)$.}

\absa
Proof.~Assume $Z_{2}\not\le O_{2}(M)$. Then
$O_{2}(M)\not\le Q_{2}$. Suppose first $E(L_{2}/Q_{2})\cong U_{3}(q)$ or $Sz(q)$ including the
solvable case, i.e.\ $O_{3}(L_{2}/Q_{2})\cong 3^{1+2}$, or
$L_{2}/Q_{2}\cong F_{20}$.


\bigskip


(1)~~\parbox[t]{14cm}{$\Omega_{1}(O_{2}(M))\not\le Q_{2}$.}

\bigskip

Otherwise $[\Omega_{1}(O_{2}(M)),Z_{2}]=1$. As
$[O_{2}(M),Z_{2}]\le O_{2}(M)\cap Z_{2}\le
\Omega_{1}(O_{2}(M))$, we get $x \in \bC_M(O_2(M))$ with $o(x)$ odd and $x^tO_2(M) = x^{-1}O_2(M)$ for some $t \in Z_2$. Hence $\bC_{M}(O_{2}(M))\not\le O_{2}(M)$,
a contradiction.

\absa
Now in any case $|\Omega_{1}(O_{2}(M)):\Omega_{1}(O_{2}(M))\cap
Q_{2}|=q$.


\bigskip


(2)~~\parbox[t]{14cm}{$\Omega_{1}(O_{2}(M))'=1$.}

\bigskip


Suppose $Z(\Omega_{1}(O_{2}(M)))\neq \Omega_{1}(O_{2}(M))$. We have
$Z_{1}\le Z(\Omega_{1}(O_{2}(M)))$. Now $Z_{1}\le$ $
\Omega_{1}(O_{2}(M))'$, as $N_{M \cap G_2}(\Omega_1(Z(S)))$ acts transitively on $\Omega_1(Z(S))^\sharp$.  Choose $t\in Z_{2}\setminus
\bC_{Z_{2}}(\Omega_{1}(O_{2}(M)))$ with
$[t,\Omega_{1}(O_{2}(M))]\le Z_{1}$. Then $t\in
\Omega_{1}(O_{2}(M))$. If $E(L_{2}/Q_{2})\cong U_{3}(q)$, then
$[\Omega_{1}(O_{2}(M)),Z_{2}]=Z_{1}$, a contradiction.  If $E(L_{2}/Q_{2})\cong Sz(q)$, then
$Z(\Omega_{1}(O_{2}(M)))\le Q_{2}$ and so
$[Z_{2},Z(\Omega_{1}(O_{2}(M)))]=1$. But
$[Z_{2},\Omega_{1}(O_{2}(M))]\le Z(\Omega_{1}(O_{2}(M)))$ and
$[Z_{2},O_{2}(M)]\le \Omega_{1}(O_{2}(M))$. This
contradicts $\bC_{M}(O_{2}(M))\le O_{2}(M)$.

\bigskip


(3)~~\parbox[t]{14cm}{If $E(L_{2}/Q_{2})\cong Sz(q)$, then
$O_{2}(M)'=1$.}

\bigskip


As $[Z_2,O_2(M)] \leq \Omega_1(O_2(M))$, we have $[Z_2, \Omega_1(Z(O_2(M)))] \not= 1$. By (2) $|Z_{2}\cap O_{2}(M)|=q^{2}$. Now as $[O_{2}(M),Z_{2}]\le
Z_{2}\cap O_{2}(M)$ we have $|O_{2}(M):O_{2}(M)\cap Q_{2}|=q$. But
then there is $t\in Z_{2}\setminus O_{2}(M)$ with $[t,O_{2}(M)]\leq
Z_{1}$ and so $O_{2}(M)'=1$, as otherwise $Z_1 \leq O_2(M)'$.

\bigskip


(4)~~\parbox[t]{14cm}{If $E(L_{2}/Q_{2})\cong SU_{3}(q)$, then
$|Z_{2}\cap O_{2}(M)|=q^{4}$.}

\bigskip

As $[Z_{2},O_{2}(M)]\le \Omega_{1}(O_{2}(M))$, we have
$\bC_{Z_{2}}(\Omega_{1}(O_{2}(M)))\le O_{2}(M)$, and so
$|Z_{2}\cap O_{2}(M)|=q^{4}$.

\bigskip


(5)~~\parbox[t]{14cm}{If $E(L_{2}/Q_{2})\cong SU_{3}(q)$, then
$O_{2}(M)'=1$.}

\bigskip

We have $Z_{1}\le O_{2}(M)'$. Now there is some $t\in Z_{2}\setminus O_{2}(M)$ with $[\Omega_{1}(O_{2}(M))O_{2}(M)'/O_{2}(M)',t]=1$. But
$[t,O_{2}(M)]\le \Omega_{1}(O_{2}(M))$, contradicting
$\bC_{M}(O_{2}(M))\le O_{2}(M)$.

\bigskip

So we have shown :

(*) If $E(L_2/Q_2) \cong SU_3(q)$ or $Sz(q)$ (including $q = 2$), then $O_2(M)' = 1$.

\absa
Assume next $E(L_{2}/Q_{2})\cong L_{2}(q)\times L_{2}(q)$, $q>2$, $Z_{2}$ the orthogonal
$O^{+}(4,q)$-module, $q=4$, $Z_{2}=Z^{(1)}_{2}\oplus Z^{(2)}_{2}$, $Z^{(i)}_{2}$
orthogonal $L_{2}(4)$-modules, or $|Z_{2}|=16$, $F(L_{2}/\bC_{L_2}(Z_2))\cong
\bZ_{3}\times \bZ_{3}$, $S/Q_{2}\cong D_{8}$ or $\bZ_{4}$. As $\bC_G(x) \leq M$ for any $x \in Z_1^\sharp$, we see that for $q > 2$ just the orthogonal module is possible. 

Suppose $q>2$. Then $|O_{2}(M):O_{2}(M)\cap Q_{2}|=q^{2}$. Hence in
any case $|Z_{2}\cap O_{2}(M)|=q^{3}$ and $|Z_{2}:Z_{2}\cap
O_{2}(M)|=q$. 

Set $W=\langle (Z_{2}\cap O_{2}(M))^{M}\rangle$. Then as
$\bC_{M}(O_{2}(M))\le O_{2}(M)$, we have $[W,Z_{2}]\neq 1$, and
so $|W:W \cap Q_{2}|=q^{2}$. As $O_{2}(M)'\neq 1$, we get $W\not\le
O_{2}(M)'$. As $[O_{2}(M)',Z_{2}]=1$, we see that we have
$A\le Z_{2}$, $|A|=q$, $|[W/W \cap O_{2}(M)^\prime ,A]|=q^{2}$, if
$Z_{2}$ is the $O_4^{+}(q)$-module and  $|A|=t$,
$|[W/W \cap O_{2}(M)',A]|=t^{2}$, $t^{2}=q$, if $Z_{2}$ is a sum of
two $O_4^{-}(t)$-modules. Furthermore in that case $[W\cap
O_{2}(M)',Z_{2}]=1$. So we have


\bigskip


(6)~~\parbox[t]{12.5cm}{If $E(L_2/Q_2) \cong L_2(q) \times L_2(q)$, $q>2$, then for $W = \langle (Z_2 \cap O_2(M))^M \rangle$ we have $[W\cap O_{2}(M)',Z_{2}]=1$.
There is $A\le Z_{2}$, $A\cap O_{2}(M)=1$, $|A|=q,t$,
respectively and $|[W/W \cap O_{2}(M)',A]|=|A|^{2}$.}

\bigskip

Suppose now that $L_{2}$ is solvable, i.e.\ $|Z_{2}|=16$. If
$|\Omega_{1}(O_{2}(M)):Q_{2}\cap \Omega_{1}(O_{2}(M))|=2$, we get as
before that $O_{2}(M)$ is abelian and $|Z_{2}:Z_{2}\cap O_{2}(M)|=4$,
$|O_{2}(M):O_{2}(M)\cap Q_{2}|=2$.

Let now $|\Omega_{1}(O_{2}(M)):\Omega_{1}(O_{2}(M))\cap Q_{2}|>2$,
then $S/Q_{2}\cong D_{8}$. Let
$[\Omega_{1}(O_{2}(M)),Z_{2},\Omega_{1}(O_{2}(M))]=1$. Then we see
that $O_{2}(M)'=1$. Furthermore there is $x\in Z_{2}\setminus
O_{2}(M)$ with $|[x,O_{2}(M)]|=2$.

If $\Omega_{1}(O_{2}(M))$ does not act quadratically. We see
$|O_{2}(M)\cap Z_{2}|=8$. Define $W$ as before. Then $O_{2}(M)'\neq 1$
and $|[Z_{2},W/W \cap O_{2}(M)']|=4$. Furthermore $[W\cap
O_{2}(M)',Z_{2}]=1$.

Let now $E(L_{2}/Q_{2})\cong L_{2}(q)$, $Z_{2}$ the natural or
orthogonal module, or the direct sum of two natural modules,
including $q=2$, or $L_{2}/\bC_{L_2}(Z_2)\cong F_{10}$.

If we have the natural module or a sum of two natural modules or
$F_{10}$, then $[O_{2}(M),Z_{2}]\leq Z(O_{2}(M))$. We have
$Z(O_{2}(M))\cap Z_{2}\le O_{2}(M)'$, if $O_{2}(M)'\neq 1$, a
contradiction to $\bC_M(O_2(M)) \leq O_2(M)$. So we have $O_{2}(M)'=1$.

\bigskip

(7)~~\parbox[t]{12.5cm}{If $E(L_{2}/Q_{2})\cong L_{2}(q)$, or $L_2/\bC_{L_2}(Z_2) \cong F_{10}$, then
$O_{2}(M)'=1$, or $E(L_2/Q_2) \cong L_2(q)$ and $Z_{2}$ is the orthogonal module.}

\bigskip

Let $Z_{2}$ be the orthogonal module. Then $|Z_{2}\cap
O_{2}(M)|=t^{3}$, $t^{2}=q$. Now set $W=\langle (Z_{2}\cap
O_{2}(M))^{M}\rangle$. Again $[W,Z_{2}]\neq 1$ and $W\cap
O_{2}(M)'\neq 1$. Hence there is $A\le Z_{2}$, $|A|=t$ with $A\cap
O_{2}(M)=1$ and $|[W/W\cap O_{2}(M)',A]|=q$.

Now we have all possible cases from (5.6) besides
$E(L_{2}/Q_{2}) \cong A_{9}$. But in that case $O_{2}(M)\le Q_{2}$, contradicting $Z_2 \not\leq O_2(M)$. We collect

\bigskip


(8)~~One of the following holds:  
\begin{enumerate}
\item[($\alpha$)] $E(L_{2}/Q_{2})\cong L_{2}(q)$, $Sz(q)$, including $q=2$,
or $L_2/\bC_{L_2}(Z_2) \cong 
F_{10}$,  and $O_{2}(M)'=1$.

\item[($\beta$)] $E(L_{2}/Q_{2})\cong SU_{3}(q)$, $O_{2}(M)'=1$.
$|Z_{2}:Z_{2}\cap O_{2}(M)|=q^{2}$ and $|[Z_{2},O_{2}(M)]|=q^{2}$,
$\bC_{O_{2}(M)}(z)=\bC_{O_{2}(M)}(Z_{2})$ for every $z\in
Z_{2}\setminus O_{2}(M)$. 

\item[($\gamma$)] $E(L_2/Q_2) \cong L_2(q) \times L_2(q)$ or $L_2/\bC_{L_2}(Z_2)$ is a solvable $\{2,3\}$--group with $m_3(L_2/\bC_{L_2}(Z_2)) = 2$. In the latter set $q = 2$. For $W=\langle (Z_{2}\cap O_{2}(M))^{M}\rangle$ we
have $[W\cap O_{2}(M)',Z_{2}]=1$ and there is some $A\le Z_{2}$, $A\cap O_{2}(M)=1$ with  $|A|=q$ or $q=t^{2}$ and orthogonal
$L_{2}(t^{2})$-modules are involved in which case $|A| = t$, such that
$|[W/W \cap O_{2}(M)',A]|=|A|^{2}$ and
$\bC_{W/W\cap O_{2}(M)'}(a)=\bC_{W/W\cap O_{2}(M)'}(A)$ for every $a\in
A^{\sharp}$ we have Furthermore for $u\in[W,Z_{2}]^\sharp$,
we have $\bC_{G}(u)\le M$ iff $u\in Z_{1}$.


\item[($\delta$)] $L_{2}$ is a solvable $\{2,3\}$ - group with $m_3(L_2) > 1$, $O_{2}(M)'=1$. There is $x\in
Z_{2}$, $|[O_{2}(M),x]|=2$, $\bC_{G}(x)\not\le M$.

\end{enumerate}
\noindent
Now we show

\bigskip

(9)~~\parbox[t]{14cm}{$(8)(\alpha)$ holds. }

\bigskip


Suppose we have ($\beta$), ($\gamma$) or ($\delta$). Set $A=Z_{2}/Z_{2}\cap O_{2}(M)$ in case $(8)(\beta)$ and
$A=\langle x\rangle$ in case $(\delta)$. Set $V=O_{2}(M)$ in case
$(8)(\beta)$, or $(\delta)$, $V=W/W\cap O_{2}(M)'$ in case
$(8)(\gamma)$.

Let $K$ be a component of $E(\bar{M}),\bar{M}=M/O_2(M)$ with $1\neq
[\bar{A},K]\le K$. As $\bC_V(A) = \bC_V(a)$ for all $a \in A^\sharp$ we have $\bC_A(K) =1$. Furthermore $|[V,A]| = |A|$ or $|A|^2$. In case $(\gamma) \, [K,O_{2}(M)'\cap W]=1$, as
$[Z_{2},O_{2}(M)'\cap W]=1$. Hence we have

\bigskip

(9.1)~~\parbox[t]{14cm}{If ($\gamma$) holds, then $[K,Z_{1}]=1$.}

\bigskip

If $|A| > 2$, then by quadratic action (1.12) we have $K/Z(K) \cong$ $A_n$, $M_{12}$, $M_{22}$, $M_{24}$, $J_{2}$, $Co_{1}$,
$Co_{2}$, $Sz$, $U_{4}(3)$ or $K/Z(K)\cong G(r)$, a group of Lie type in even
characteristic. If $|A|=2$, then $|V:\bC_{V}(A)|\leq 4$ and by (1.30) the same holds.

We first are going to show that $|A| = 2$. So until further notice we assume $|A| > 2$. Then there is some $\omega \in L_1 \cap L_2$, $o(\omega)$ is odd and $\langle \omega \rangle$ acts transitively on $A^\sharp$. 

Suppose that $K$ is not a group of Lie type in even characteristic. Then by (1.12) and (1.25) $o(\omega) = 3$ and so $3 \not\in \sigma(M)$. This shows $m_p(K) \le 2$ and there is $E \cong E_{p^2}$ with $[E,K] = 1$. Furthermore by (1.25) $V$ involves exactly one nontrivial irreducible $K$--module, i.e. $[[K,V],E] = 1$. This implies that there is some $\rho \in K$, $o(\rho)$ odd, with $\rho^a = \rho^{-1}$ for some $a \in A^\sharp$ and $[[V,\rho],E] = 1$. As $[[V,\rho],a] \leq Z_2$ but $[V,\rho] \cap Z_1 = 1$, we see that there is some $x \in Z_2 \cap O_2(M) \setminus Z_1$ with $\bC_G(x) \le M$. But $L_2 = \langle L_2 \cap L_1 , \bC_{L_2}(x) \rangle$, a contradiction. So we have


\bigskip

(9.2)~~\parbox[t]{14cm}{If $|A| > 2$ then $K$ is a group of Lie type in even characteristic}

\bigskip

Next we show

\bigskip

(9.3)~~\parbox[t]{14cm}{$[K,\omega]\le K$. }

\bigskip

Suppose $[K,\omega] \not\leq K$. Then $[K^\omega, K] = 1$ and $K^{\langle \omega \rangle}$ contains at least three components $K$, $K^\omega$, $K^{\omega^2}$.  Assume first $[[K,V],K^\omega] = 1$. Then we may assume $[[K,V],K^\omega * K^{\omega^2}] = 1$ too. Now there is $x \in Z_2 \cap M \setminus Z_1$ with $[x,K^\omega * K^{\omega^2}] = 1$. This shows that $\bC_G(x) \le M$ and so $L_2 = \langle L_1 \cap L_2, \bC_{L_2}(x) \rangle \leq M$, a contradiction. So we have $[K^\omega, [K,V]] \not= 1$ and so $[K,V]$ involves at least two modules which then are $F$--modules with $A$ offending.

We have that $m_p(\bC_W(S \cap K)) \leq 1$ for any prime $p~|~|K|$, where $W$ is some nontrivial irreducible $K$--submodule in $V$. Hence we see with (1.18) that $K$ induces on $V$ the group $A_7$, $3\cdot A_6$, $L_4(2)$, $L_3(r)$, $L_2(r)$, $SU_4(r)$, $L_2(r^2)$, $Sp_4(r)$, or  $G_2(r)$. Furthermore we have some irreducible module $W$ invariant under $A$. Now the action of $K^\omega * K^{\omega^2}$ on $[V,K]$ shows that there are at least 8,8,8,6,4,16,8,8, or 12 such modules involved, respectively. Hence we have $|A| \geq r^4$, $r^4$, $r^4$, $r^3$, $r^2$, $r^8$, $r^4$, $r^4$, $r^6$, respectively. This shows that $|A| > 2^{m_2(\mbox{\tiny{Aut}}(K))}$, a contradiction. So we have (9.3).
\
\bigskip

(9.4)~~\parbox[t]{14cm}{$[K,\omega]\neq
1$.}

\bigskip

This is clear, as $A = [A,\omega]$. 
\\

By (9.2) we have that $K$ is a group of Lie type in even characteristic. Let first
$K\cong G(r)$ with $|A|\leq r$. Then either $K\cong Sz(r)$, $U_{3}(r)$ or
$L_{2}(r)$ or $V$ is an $SC$-module for $K$ with offending subgroup
$A$ and so by (1.22) and (1.23) $K\cong (S)L_{n}(r)$, $(S)U_{n}(r)$,
$Sp_{2n}(r)$, $\Omega_{2n}^{\pm}(r)$ or $G_{2}(r)$. Hence $|A|=r$ or
$K\cong (S)L_{n}(r)$ or $Sp_{2n}(r)$ and $|A|=\sqrt{r}$. 

As $|A|\leq r$, we do not have case $(\beta)$. 
Suppose first some $p \,\Big |\,  r-1$ for some $p\in\sigma(M)$. If there is some $x \in Z_2 \cap M \setminus Z_1$ centralized by $Sp_4(r)$, $L_3(r)$, $U_4(r)$ or $\Omega^+(4,r)$, then $x$ is centralized by some $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \leq M$ and so $\bC_G(x) \leq M$. But $L_2 = \langle L_2 \cap L_1, \bC_{L_2}(x) \rangle$. Hence we get $K \cong SL_3(r)$, $L_2(r)$, $U_3(r)$, $U_4(r)$, $U_5(r)$, $G_2(r)$ or $Sz(r)$. Now $m_p(K) \le 2$ and so $m_p(\bC(K)) \geq 1$. As $|A| = r$, there is $\nu \in \langle \omega \rangle$, $o(\nu) = p$ and so $m_p(\bC_M(\nu)) \geq 3$. Hence $N_G(\langle \nu \rangle ) \leq M$, but $L_2 = \langle L_1 \cap L_2, N_{L_2}(\langle \nu \rangle) \rangle$. So we have $|A| = \sqrt{r}$ and $K \cong SL_3(r)$ or $L_2(r)$. Furthermore $V$ involves just one nontrivial irreducible module, the natural one. Finally $A \leq K$. Hence there is some $\rho \in K$, $o(\rho)$ odd, and some $E \cong E_{p^3}$, with $[E,\langle \rho, a \rangle] = 1$. But then
some $F\le E$, $|F|=p^{2}$ centralizes some $z\in Z_{2}\cap
O_{2}(M)\setminus Z_{1}$, a contradiction.

So we have if $p \in \sigma(M)$ then $p \not\!\Big |\,  r-1$. Let $p\, \Big |\,  r^{2}-1$ and $K\cong
(S)U_{n}(r)$. Again no element $x \in (V/Z_1)^\sharp$ is centralized by $U_3(r)$ or $U_4(r)$ in $K$. This shows $K \cong SU_4(r)$ or $SU_3(r)$ and $m_p(K) \leq 3$. If $K \cong SU_3(r)$ then there is $\rho \in K$, $o(\rho)$ odd, with $\rho^a = \rho^{-1}$ for some $a \in A^\sharp$ and $[E, \langle a, \rho \rangle ] = 1$ for some $E \cong E_{p^3}$. We get the same contradiction as above. So we have $K\cong
SU_{4}(r)$. In any case $E(L_{2}/Q_{2})\cong L_{2}(|A|^{2})$ or $L_{2}(|A|)\times
L_{2}(|A|)$. In the former there is some $\nu\in L_{1}\cap L_{2}$,
$o(\nu)=p$. Now $m_{p}(\bC_{M}(\nu))\geq 3$, a contradiction. So we
have the latter. Then $T=\bZ_{r-1}\times \bZ_{r-1}\le
L_{1}\cap L_{2}$. This implies $T\cap K\neq 1$, otherwise any prime $u$ with $u ~|~r-1$ is in $\sigma(M)$.  As $[T\cap
K,Z_{1}]=1$, we have $[T\cap K,A]=1$. Now $m_{p}(N_{M}(T\cap K))\geq
2$ and there is $E\le N_{M}(T\cap K)$, $E\cong
E_{p^{2}}$ with $\Gamma_{E,1}(G)\leq M$. This implies with (1.63) that
$N_{G}(T\cap K)\le M$, a contradiction.

Suppose now first that $[K,V]$ involves exactly one irreducible
nontrivial module. As for $p \in \sigma(M)$ we have $p \not\!\Big |\,  r-1$, and $p ~\not|~r+1$ in case of $K \cong (S)U_n(r)$, we get that any
$p$-element $\mu$, with $[K,\mu]=1$ centralizes $[K,V]$. Hence
$m_{p}(\bC_{\bar{M}}(K))\leq 1$. Now $m_{p}(Aut(K))\geq 3$. This implies
$m_{p}(K)\geq 2$ and so $K\cong L_{4}(r)$, $Sp_{4}(r)$, $Sp_{6}(r)$,
$\Omega^{-}_{8}(r)$, $G_{2}(r)$. In any case $p\, \Big |\,  r^{2}-1$. But
there is some $a\in A^\sharp$ such that $[a,V]$ is centralized by $L_{2}(r)$ in
$K$. Hence $m_{p}(\bC_{\bar{M}}(K))=0$ and so $m_{p}(K)\geq 3$. So we
are left with  $Sp_{6}(r)$, $\Omega^{-}_{8}(r)$. Now $[a,V]$ is
centralized by $Sp_{4}(r)$, $\Omega^{-}_{6}(r)$, and so we again see
that $\bC_{G}([a,V])\leq M$, a contradiction. 

Hence we have $K\cong SL_n(r)$ or $Sp_{2n}(r)$, $|A|=r$ and $[K,V]$
involves exactly two natural modules. Again $p \not\!\Big |\,  r-1$ for $p \in \sigma(M)$. So we
have $K\cong L_{2}(r)$, $SL_{3}(r)$, $SL_{4}(r)$, $Sp_{4}(r)$ or
$Sp_{6}(r)$. If $K\cong Sp_{6}(r)$, then even $p \not\Big |\, \, r^{2}-1$, as otherwise some $x \in Z_2 \cap M \setminus Z_1$ is centralized by $Sp_4(r)$ and so $\bC_G(x) \leq M$, a contradiction. But
now in the case of $Sp_6(r)$ we have $m_{p}(K)=1$ and so $m_{p}(\bC_{\bar{M}}(K))\geq 2$. But
$[[K,V],\bC_{\bar{M}}(K)]=1$, a contradiction. So in any case
$m_{p}(K)\leq 2$, $m_{p}(\bC_{\bar{M}}(K))=1$ and
$[\bC_{\bar{M}}(K),[V,K]]\neq 1$. Hence we have a subgroup $K_{1}$
of $\bC_{\bar{M}}(K)$, $K_1$ isomorphic to a subgroup of $L_2(r)$. Furthermore $A$ cannot
centralize this group otherwise we may argue as before. Hence
$K_{1}\cong L_{2}(r)$ and $[V,K_{1}]$ is a sum of two natural
modules, i.e.\ $|[V,K_{1}]|=|A|^{4}$ and so $|[V,K]|=|A|^{4}$. This
shows $K\cong L_{2}(r)$ too, $[V,K\times K_{1}]$ is the
$O^{-}(4,r)$-module. As above $T=\bZ_{r-1}\times \bZ_{r-1}\le L_{1}\cap L_{2}$. As $L_{2}(r)\times L_{2}(r)$ does
not contain $\bZ_{r-1}\times \bZ_{r-1}$ normalizing $A$, we get
that $T\cap \bC(K\times K_{1})\neq 1$, a contradiction to (1.63).

So we are left with $|A|>r$. By (1.64) and (1.65) we get $K\cong SU_{n}(r)$,
$\Omega_{2n}^{-}(r)$, $^2E_6(r)$, $^{3}D_{4}(r)$, or $^2F_4(2)^\prime$, or $r=2$, $o(\omega)=3$,
$|A|=4$. In the latter we get a contradiction as in the sporadic and
alternating case. As $|A|$ acts quadratically $K\not\cong$
$^{3}D_{4}(r)$ nor $^2F_4(2)^\prime$ by (1.15) and (1.14). Hence we have $|A|=r^{2}$ and
$K\cong SU_{n}(r)$ or $\Omega_{2n}^{-}(r)$ by (1.23).  In any case by (1.45)
$|V : \bC_{V}(A)| > |A|$ and so $(8)(\beta)$ is not possible. Now
$A\cap Z(S/O_{2}(M))\neq 1$. As always $\omega$ normalizes a Borel subgroup we get $|Z(S\cap K)|\geq |A|$, a
contradiction.

\noindent
So we have shown

\bigskip

(9.5)~~\parbox[t]{14cm}{$|A|=2$. }


\bigskip

Now we have $(8)(\gamma)$ or $(\delta)$ and  $|V:\bC_{V}(A)|\leq 4$, and
so by (1.30) $K/Z(K)\cong L_{n}(2)$, $SL_{n}(4)$, $SU_{n}(2)$,
$Sp_{2n}(2)$, $Sp_{2n}(4)$, $\Omega_{2n}^{\pm}(2)$, $\Omega_{2n}^{\pm}(4)$,
$G_{2}(2)$, $A_{n}$ or $U_{4}(3)$.

In any case either $3 \,\Big |\, |L_{1}\cap L_{2}|$ or $m_{3}(L_{2})\geq 2$ as we assume that ($\alpha$) does not hold.
Hence we have $3\not\in \sigma(M)$. This now implies $K/Z(K)\cong
L_{n}(2)$, $n\leq 7$, $SL_{n}(4)$, $n\leq 4$, $SU_{4}(2)$, $Sp_4(2)$,
$Sp_6(2)$, $Sp_4(4)$, $Sp_6(4)$, $\Omega_8^{-}(2)$,
$\Omega_8^{-}(4)$, $G_{2}(2)$, $A_{n}$, $n\leq 11$. Now any
$p$-element $\mu\in\bC_{\bar{M}}(K)$ centralizes $[V,K]$. This
implies $m_{p}(Aut(K))\geq 3$, and so $m_{p}(K)\geq 3$. This implies
$K\cong Sp_6(4)$ or $\Omega_8^{-}(4)$, $p=5$. Now $[V,A]$ is
centralized by $Sp_4(4)$, or $Sp_6(4)$, and so by some $E\cong
E_{p^{2}}$ with $\Gamma_{E,1}(G)\leq M$, a contradiction.

\noindent
So we have shown


\bigskip

(9.6)~~\parbox[t]{14cm}{If $[K,A]\neq 1$ and $[K,V] \not= 1$, then $[K,A]\not\le K$. }


\bigskip

Next we show

\bigskip

(9.7)~~\parbox[t]{14cm}{$[[A,E(\bar{M})],V]=1$. }

\bigskip

Let $[K,A] \not= 1$ and $[K,V] \not= 1$. If $|A|>2$ then by (1.11) $K\cong L_{2}(r)$, $r$ even, and $[V,K^{A}]$
just involves $O_4^{-}(r)$-modules. But now there is $T\cong
D_{2(r+1)}\times D_{2(r+1)}\le \langle K^{A}\rangle$, $A\cap
T\in Syl_{2}(T)$. As $\bC_{V}(a)=\bC_{V}(A)$ for every $a\in A^{\sharp}$
this is impossible.

Hence $|A|=2$. As $3\not\in \sigma(M)$, $K$ possesses cyclic Sylow
$3$-subgroups. Hence $K\cong L_{2}(r)$, $U_{3}(r)$, $L_{3}(r)$,
$Sz(r)$ or $J_{1}$. Furthermore as $|[V,A]|\leq 4$, we have that
$|K|=2^{x}3^{y}5^{z}$, with cyclic Sylow $5$ subgroups as $A$
cannot invert an elementary abelian group of order $5^{2}$. Hence
$K\cong L_{2}(4)$. Now we see that for $P\in Syl_{p}(K\times K^{a})$,
$|[V,P]|=2^{8}$. Then there is $E\cong E_{p^{2}}$ with
$\Gamma_{E,1}(G)\leq M$ and $[[V,P],E]=1$. This shows that for some
$v\in O_{2}(M)\cap Z_{2}\setminus Z_{1}$, we have
$\bC_{G}(v)\le M$, a contradiction. 


\bigskip

(9.8)~~\parbox[t]{12cm}{Let $\bar{P} \le \bar{M}$ be
a $u$-group, $u$ an odd prime, and $[\bar{P}, A] \leq \bar{P}$. If $|A|>2$, then $[A,\bar{P}] \leq \bC_{\bar{P}}(V)$.}

\bigskip


Let $[A,\bar{P}]\neq 1$. Then as $\bC_{V}(a)=\bC_{V}(A)$ for $a\in
A^{\sharp}$, we have $\bC_{A}([A,\bar{P}])=1$. Now by \cite[(24.1)]{GoLyS} there is
$U_{1}\leq \bar{P}A$, $U_{1} \cong D_{2 u}\times\ldots \times D_{2 u}$,
$A\in Syl_{2}(U_{1})$. But then $[O_{u}(U_{1}),V]=1$. As
$[A,O_{2}(M)]\leq V$, we see $[O_{u}(U_{1}),O_{2}(M)]=1$, a
contradiction. 

\bigskip

Now we come to the final contradiction. Let $\bar{P}\in Syl_{u}(F(\bar{M}))$, $u$ an odd prime with $[A,\bar{P}]\neq 1$ and $[V, \bar{P}] \not= 1$.
Then by (9.8) $|A|=2$. Furthermore $u\in\{3,5\}$ . Suppose $u \not\in \sigma(M)$. Then $m_u(\bar{P}) \leq 3$. Hence $m_p(\mbox{Aut}(\bar{P})) \leq 1$. This gives some $E \cong E_{p^3}$ with $\Gamma_{E,1}(G) \leq M$ and $[E,[V,A]] = 1$, a contradiction.

So we have $u = 5 \in \sigma(M)$. If $m_5(\bar{P}) \geq 3$, then by \cite[(11.18)]{GoLyS} $A$ normalizes some $E \cong E_{5^3}$ and so some $F \le E$, $|F| = 25$, centralizes $[V,A]$, a contradiction again. So we have $m_5(\bar{P}) \leq 2$ and $\sigma(M) = \{5\}$. Now there is a critical subgroup $\bar{P_1}$ of $\bar{P}$ with $\Omega_1(\bar{P_1}) \cong \bZ_5$, $\bZ_5 \times \bZ_5$ or $5^{1+2}$. In the last two cases $\bC_G(x) \leq M$ for all $x \in \Omega_1(\bar{P_1})^\sharp$. Hence $[V,A]$ is centralized by some $E \cong E_{25}$ with $\Gamma_{E,1}(G) \leq M$, a contradiction. This shows that $\bar{P_1}$ is cyclic with $[\bar{P}, A] \leq \bar{P_1}$ and $[\bar{P_1}, A] = \bar{P_1}$. Furthermore $|[\bar{P_1}, V]| = 16$ and so $\bC_M([\bar{P_1}, A])$ contains an elementary abelian group of order $5^3$, a contradiction.

So we have shown that $[F^*(\bar{M}), A] \leq \bC_{F^*(\bar{M})}(V)$. This now implies $[O_2(M), [F^*(\bar{M}), A]]$ $ \leq V$ and so $[[F^*(\bar{M}),A], O_2(M)] = 1$, a contradiction to $\bC_M(O_2(M)) \leq O_2(M)$. This proves (9). 

\bigskip

By $(8)$ and $(9)$ we know that $O_{2}(M)'=1$. Set
$\tilde{L}_{2}=\langle O_{2}(M_{1}),O_{2}(M_{3})\rangle$. Then 
$\tilde{L}_{2}/O_{2}(\tilde{L}_{2})\cong L_{2}(q)$, $Sz(q)$ or
$F_{10}$. 

Now $[O_{2}(M_{1})\cap O_{2}(M_{3}),\tilde{L}_{2}]=1$. As
$[\tilde{L}_{2},Q_{2}]\le O_{2}(\tilde{L}_{2})$, and
$Z(L_{2})=1$, we get $O_{2}(M_{1})\cap O_{2}(M_{3})=1$ or
$\tilde{L_2}/\bC_{\tilde{L}_{2}}(Z_2)\cong F_{10}$ and $L_{2}/\bC_{L_2}(Z_2)\cong
F_{20}$. In the latter $\bC_{Q_{2}}(\tilde{L}_{2}Q_{2})\neq 1$ and so also
$\bC_{Q_{2}}(L_{2})\neq 1$, a contradiction. So we have 


\bigskip

(10)~~\parbox[t]{11.5cm}{$O_{2}(M_{1})\cap O_{2}(M_{3})=1$ and
$\tilde{W}= \langle O_{2}(M_{1})\cap Q_{2},O_{2}(M_{3})\cap Q_{2} \rangle$ is a
direct sum of natural modules.}


\bigskip


\noindent
The latter follows from (1.27).

\bigskip

So we have that $O_{2}(M)$ is an $F$-module for $M$ with offending
subgroup $O_{2}(M_{3})\cap Q_{2}$. As $|O_{2}(M)|>16$, we have
$|O_{2}(M_{3})\cap Q_{2}|>4$. Furthermore $\bC_{O_{2}(M)}(x)=
\bC_{O_{2}(M)}(O_{2}(M_{3})\cap Q_{2})$ for every $1\neq x\in
O_{2}(M_{3})\cap Q_{2}$. Hence there is some component $K$ of $\bar{M}$
with $1\neq [K,O_{2}(M)\cap Q_{2}]\le K$. Finally
$\bC_{O_{2}(M_{3})\cap Q_{2}}(K)=1$. If $|O_{2}(M):O_{2}(M)\cap
Q_{2}|=2$, then $O_{2}(M_{3})\cap Q_{2}$ induces the full group of
transvections on $O_{2}(M)$ i.e.\ $K\cong L_{n}(2)$,
$|O_{2}(M)|=2^{n}$.

Let $q>2$. We know that $O_{2}(M)$ involves exactly one nontrivial 
irreducible module. Hence let $\omega\in L_{1}\cap L_{2}$,
$o(\omega)=q-1$. Then $[\omega,K]\le K$. As
$[\omega,O_{2}(M_{3})\cap Q_{2}]=O_{2}(M_{3})\cap Q_{2}$, we have
$[\omega,K]\neq 1$. By (1.18) we know the structure of $K$. By (1.65) and (1.64) we see that $K$ is a Lie group over $GF(q)$ or $K \cong \Omega_{2n}^-(\sqrt{q})$ or $U_n(\sqrt{q})$. But as $|O_2(M_3) \cap Q_2| > q$, we get with (1.45) that $K$ is a Lie group over $GF(q)$ and then $|O_2(M) : \bC_{O_2(M)}(O_2(M) \cap Q_2)| = q$. Hence by (1.23) we see that $K\cong SL_{n}(q)$
and $|O_{2}(M)|=q^{n}$. Now in any case $M/O_{2}(M)\lesssim \Gamma
L_{n}(q)$. Hence $(O_{2}(M_{3})\cap Q_{2})O_{2}(M)$ is normalized by
$SL_{n-1}(q)$ in $M$. This implies that there is $E\cong E_{p^{2}}$ with
$\Gamma_{E,1}(G)\leq M$ and $E\le N_{M}((O_{2}(M_{3})\cap
Q_{2})O_{2}(M))$. But then also $E\le N_{M}(\tilde{W})$, and so
$L_{2}\le N_{G}(\tilde{W})\le M$, a contradiction. 

\absa
{\bf~(5.8) Lemma.~}{\it The configuration in (4.4)(b) does not occur.}

\absa
Proof.~We consider the graph $\Gamma(G_1,G_2) = \Gamma$. Let $\alpha \in \Gamma$ be of minimal distance from 1 such that $T_1 \le G_\alpha$ but there is some $\beta \in \Delta(\alpha)$ with $T_1 \not\le G_\beta$.

\bigskip
(1)~~~~ $\alpha \not\sim 1$

\bigskip

Suppose $\alpha \sim 1$. Then $T_\alpha \leq G_1$ and so $[T_1, T_\alpha] \leq T_1 \cap T_\alpha$. As $T_1 \not\le O_2(G_\alpha)$ we have $[T_1,T_\alpha] \not= 1$ by (4.4). Furthermore $M$ is the unique maximal 2--local containing $\bC_G(x)$ for $x \in [T_1,T_\alpha]^\sharp$. Hence $M = M_\alpha$. By (5.7) $Z_2 \leq O_2(M)$ but $T_1 \not\leq O_2(M_\alpha)$, a contradiction.

\bigskip

(2)~~~~~ $Z_\alpha \not\leq G_1$

\bigskip

Suppose $Z_\alpha \leq G_1$. Then $\bC_G(x) \leq M$ for every $x \in [T_1, Z_\alpha]^\sharp$. We have $|[T_1, Z_\alpha]| = q^2$ and there is some $x \in [T_1, Z_\alpha]^\sharp$ such that $M_{\alpha - 1}$ is the unique maximal 2--local containing $\bC_G(x)$. This again proves $M = M_{\alpha - 1}$. By (5.7) $Z_\alpha \leq O_2(M_{\alpha - 1}) = O_2(M)$. We have $[T_1, \Omega_1(O_2(M))] = 1$ and so $[Z_\alpha, T_1] = 1$, a contradiction.

Now we have $Z_\alpha = \langle T_\beta ~|~ \beta \in \Delta(\alpha) \rangle$, and so $Z_\alpha \leq G_2$. As $Z_\alpha \not\leq G_1$ we have $[Z_2,Z_\alpha] \not= 1$. Furthermore $Z_2 = \langle T_\gamma~|~\gamma \in \Delta(2) \rangle$ and so $Z_2 \leq G_\alpha$. But this implies that $Z_2$ is an $F$--module, a contradiction to (1.18).










 \begin{center}
\S~6 The case $b=b_{1}$
\end{center}

\vspace{1cm}

In this chapter we are going to investigate the case $b=b_{1}$. The
notation is as introduced in chapter 4.

\absa
{\bf (6.1)~Lemma.~}{\it Let $\alpha\in\Gamma$ with
$Z_{1}\le L_{\alpha}$ and $Z_{\alpha}\le L_{1}$. If
$[Z_{1},Z_{\alpha} ]\not= 1$, then $\alpha\sim 2$.}

\absa
Proof.~ Suppose false, i.e. $\alpha\sim 1$. We have
\begin{center}
$[Z_{1},Z_{\alpha}]\le Z_{1}\cap Z_{\alpha}$
\end{center}

Hence $Z_{1}$ is an F-module with offending subgroup $Z_{\alpha}
/Z_{\alpha}\cap\bC (Z_{1})$. Let $L_1$ not be in the exceptional case. Suppose $[E(L_1/Q_1), Z_\alpha] \not= 1$. By (1.11) $Z_{\alpha}$ induces orbits of
length at most two on the set of components of $L_{1}/Q_{1}$.
Furthermore either for a component $K$ we have $[Z_{\alpha} ,K]\leq
K$, or $ |Z_{\alpha} :N_{Z_{\alpha}} (K)|=2$ and $K\cong L_{2}(2^{\, n})$,
or $|Z_{\alpha} :\bC_{Z_{\alpha}} (K)|=2$ and $N_{Z_{\alpha}}
(K)=\bC_{Z_{\alpha}} (K)$. Hence in the latter two cases by (1.11)
$Z_{1}$ is not an F-module for $\langle K^{\, Z_{\alpha}}\rangle Z_\alpha$
with offending group $Z_{\alpha} /\bC_{Z_{\alpha}} (K)$. Hence there
is a component $K$ of $L_{1}/Q_{1}$ with $[Z_{\alpha} ,K]\le K$
such that $Z_{1}$ is an F-module for $KZ_\alpha$ with offending subgroup
$Z_{\alpha} /\bC_{Z_{\alpha}} (K)$.

Now we may apply (5.1). Let $U$ be the
K-module given by (5.1). Let $K^{\, (\alpha)}$ be the corresponding
component to $K$ in $L_{\alpha}/Q_{\alpha}$. 

We will assume that we do not have case (5.1) (i) or (ii). Choose
$u\in U^{\, \sharp}$. We have $x\in [Z_{\alpha} ,u]^{\,\sharp}$ is
centralized by $E_{x}\cong E_{p^{\, 2}}$  with $\Gamma_{E_{x},1} (G)\leq M$. 

Assume $[u,K^{\, (\alpha )}]=1$. Let $[[u,Z_{\alpha} ],K^{\, (\alpha
)}] = 1$ . Then $K^{\, (\alpha )}$ is covered by $M\cap
M_{\alpha}$. Now (4.2) implies $M=M_{\alpha}$. But $Z_{1}\le
Z(O_{\, 2}(M))=Z(O_{\, 2}(M_{\alpha} ))\le Q_{\alpha}$, a
contradiction. So $[[u,Z_{\alpha} ],K^{\, (\alpha
)}] \not= 1$. Moreover we have that $m_p(K^{\, (\alpha )})=1$. But in
this case we have by (1.18) that $U$ is centralized by $E\cong
E_{p^{\, 2}}$ with $\Gamma_{E,1} (G)\leq M$. Now again $K^{\, (\alpha )}$ is
covered by $M\cap M_{\alpha}$, a contradiction.

So we have $[u,K^{\, (\alpha )}]\not= 1$. Now there is $z\in
Z_{\alpha}$ such that $[u,z]$ is centralized by $E\cong E_{p^{\, 2}}$
in $M$ and $M_{\alpha}$ as well. This gives $M=M_{\alpha}$, a
contradiction.

Assume now (5.1) (i). In this case $U=[Z_{1},K]$ and $[Z_{\alpha} ,U]$
is normalized by $E\cong E_{p^{\, 3}}$. For $u\in U$, we have
$|[u,Z_{\alpha} ]|\leq q$. Hence for $[u,K^{\, (\alpha )} ]=1$, we
have $[[u,Z_{\alpha} ],K^{\, (\alpha )} ]=1$. By (5.1) (i) we see
that we may choose $u$ with $[u,Z_{\alpha} ]$ is normalized by
$E\cong E_{p^{\, 2}}$ with $\Gamma_{E,1} (G)\leq M$. Hence for such $u$ we
have $[u,K^{\, (\alpha )} ]\not= 1$. This now shows that $Z_{\alpha}$
is an F-module for $K^{\, (\alpha )}$ with offending subgroup $U$.
But then $[U,Z_{\alpha}]$ is normalized by $E\cong E_{p^{\, 3}}$ in
$M$ and $M_{\alpha}$, contradicting $M\not= M_{\alpha}$.

So we are left with (5.1) (ii). Then $U=[Z_{1},K]$ and $[Z_{\alpha}
,U]$ is normalized by $E\cong E_{9}$. But
$[Z_{\alpha} ,U]$ is normalized by $E\cong E_{9}$ in $M_{\alpha}$ as
well, which is clear for $[U,K_\alpha] = 1$ and in case $[U,K_\alpha] \not=1$ we get this as $U$ offends on $U_\alpha$. Hence we have
$M=M_{\alpha}$, a contradiction. 

So we have $[E(L_1/Q_1), Z_\alpha] = 1$. By (4.7) $O_{p'}(L_{1})\cap L_{2} \le Q_{\, 1}$. So we get a faithfull $\{ 2,p\}$-group
$K_{1} \leq L_1/Q_1$ on $Z_{1}$ with $O_{\, 2}(K_{1})=1$ and having $Z_\alpha Q_1/Q_1$ as a Sylow 2--subgroup.. This implies $p=3$, and
by symmetry we may assume that there are $x\in Z_{1}, y\in
Z_{\,\alpha}$ with $[x,y]=[x,Z_{\,\alpha}]=[y,Z_{1}]\le
[Z_{1},\rho ], \rho$ of order 3, $|[Z_{1},\rho ]|=4$. By (4.4) we get some $E\cong E_{9}$, with $[[x,y],E]=1$ and
$\Gamma_{E,1} (G)\leq M$. As the same holds in $M_{\alpha}$, we get a
contradiction.

So we are left with the exceptional case: $L_{\, 1}/O_{\, 2}(L_{\,
1})=BA$. $A=L_{\, 1}\cap L_{\, 2}/O_{\, 2}(L_{\, 1})$ and $B$ is an
extension of an $r$-group by $L_{\, 2}(r), r$ odd, $r>3$. Then $[Z_{\,
1},O_{\, r}(B)]=1$. Hence we get a contradiction with (4.10). 

\absa
{\bf (6.2)~Lemma.~}{\it Let $b=b_{\, 1}$ and $[Z_{1},Z_{\,\alpha}
]\not= 1$. Then $[Z_{1},Z_{2}]\not= 1$.}

\absa
Proof~. Suppose $Z_{1}\le Q_{\, 2}$. By (6.1) we have that
$\alpha\sim 2$. So either $Z_{2}$ or $Z_{1}$ is an F-module. Suppose
that $Z_{2}$ is an F-module. Then by (1.31) and (2.1) either $E(L_{\,
2}/Q_{\, 2})\cong L_{2}(q),L_{2}(q)\times L_{2}(q), q$ even, $A_{9}$,
or $E((L_{2}/Q_{\, 2})/F(L_{2}/Q_{\, 2}))\cong A_{5}$, or $L_{2}$ is
a solvable $\{ 2,3\}$-group. So in any case by inspection we get that
$Z_{1}$ is an F-module too. So we may assume that $Z_{1}$ is an
F-module wright from the begining. To prove the lemma we now are going to argue as in (6.1).

Let $K$ be some component of $L_{1}$ such that $Z_{\,\alpha}
/\bC_{Z_{\,\alpha}} (K)$ offends on $Z_{1}$. Then (5.1) provides us
with some module $U$. Suppose $\bC_{G} (u)\le M$ for any $u\in
U^{\,\sharp}$. Let now $\alpha +1$ be some neighbor of $\alpha$ such
that $U\not\le M_{\alpha +1}$. Let $U_{\alpha +1}$ be the
corresponding module in $L_{\alpha +1}$. Then $U_{\alpha +1}\le
G_{\, 2}$. 

Suppose $[U_{\alpha +1},Z_{2}]\not= 1$. Then $Z_{1}\le \bC_{G}
([U_{\alpha +1},Z_{2}])\le M_{\alpha +1}$, by (5.1), a
contradiction. So we have $[U_{\alpha +1},Z_{2}]=1$, i.e. $U_{\alpha +1}\le
L_{1}$. Now $\bC_{U_{\alpha +1}} (Z_{1})=1$, i.e. $U_{\alpha +1}\cap
Q_{\, 1}=1$ otherwise $Z_1 \leq M_{\alpha+1}$. As $b > 1$, we see that $\langle U_{\alpha+1}^{L_\alpha} \rangle$ is abelian. hence $U_{\alpha+1}$ acts quadratically on $Z_1$. This together with (1.11) shows that $[U_{\alpha+1}, K] \leq K$. As $[U_{\alpha +1},Z_{\,\alpha} ]=1$, we see that $[U_{\alpha
+1},U]\le U$, and so for $u\in U\setminus M_{\alpha +1}$,
we get $[u,U_{\alpha +1} ]\cong U$, a contradiction.

Now by (5.1) $K\cong L_{\,
2}(q)$ or $U_{4}(2)$. Then $Z_{\alpha}$ is an F-module too. In the first case we get $E(L_{\,
2}/K_{2})\cong L_{\, 2}(q)$ or $L_{\, 2}(q)\times L_{\, 2}(q)$. Furthermore $p~|~q-1$. But now there is $\omega\in L_{1}\cap L_{\, 2}$ with
$o(\omega )=p$. This shows $m_{p}(\bC_{L_{1}} (\omega ))\geq 3$. But
then $N_{G}(\langle\omega\rangle )\le M$, a contradiction to (1.66).

If $K\cong U_{4}(2)$, then $|Z_{1}:Z_{1}\cap Q_{\,\alpha} |=16$.
Furthermore $3\in\sigma (M)$. Now by (1.31) $E(L_{\, 2}/Q_{\,
2})\cong L_{\, 2}(r)$. As $|Z_{1}:Z_{1}\cap Q_{\,\alpha} |=16$, we
get $r=16$, but then $3\, \Big |\, |L_{1}\cap L_{\, 2}|$, contradicting (1.66).

So we have $[E(L_1/Q_1), Z_\alpha] = 1$. Let $K$ be a $t$--group on which $Z_\alpha$ acts nontrivially. Suppose first $t = p$. Then we get $p=3$.
Furthermore, as $Z_{1}$ is an F-module, there is some $t\in Z_{\alpha}$
inducing a transvection on $Z_{1}$. By (1.52) we do not have the exceptional case.  By quadratic action there is also
some $u\in Z_{1}$ inducing a transvection on $Z_{\alpha}$, \cite[(25.12)]{GoLyS},
\cite[(24.1)]{GoLyS}. So by (1.31) $E(L_{\, 2}/Q_{\, 2})\cong A_{5}, A_{5}\times
A_{5}, A_{9}$, or $E((L_{\, 2}/Q_{\, 2})/F(L_{\, 2}/Q_{\, 2}))\cong
A_{5}$, or $L_{\, 2}$ is a solvable $\{ 2,3\}$-group. By (2.1) $m_{3}(L_{\, 2})=1$. 

Furthermore by (1.66) $m_{3}(L_{1}\cap L_{\, 2})=0$. This implies
that $L_{\, 2}/Q_{\, 2}\cong\Sigma_{3}$ and $Z_{2}=Z(L_{\, 2})\times
\tilde{Z} _{2}, |\tilde{Z} _{2}|=4$. Now let $Z_{1}\not\le M_{\alpha
+1}$, for some $\alpha +1\in\bigtriangleup (\alpha )$. We have
$|Z_{\alpha +1}:Z_{\alpha +1}\cap L_{1}|\le 2$. As $\langle Z_{\alpha
+1}^{L_\alpha}\rangle =V_{\alpha}$ is abelian by (6.1) we see that $V_{\alpha}\cap L_{1}$ 
acts quadratically on $Z_{1}$. Now there is a $\{ 2,3\}$-group $P(V_{\alpha}\cap
L_{1})Q_{1}/Q_{1}$ where $P$ is normal in this group. Let
$A=\bC_{V_{\alpha}\cap L_{1}} (P)$ and $C$ be a complement of $A$ in
$(V_{\alpha}\cap L_{1})Q_{1}/Q_{1}$. We have that $PC$ acts
nontrivially on $\bC_{Z_{1}} (A)$ by the $A\times B$-Lemma. Now by
\cite[(24.1)]{GoLyS} and \cite[(25.12)]{GoLyS} there is some $u\in\bC_{Z_{1}} (A)\setminus
Q_{\, 2}$ such that $|V_{\alpha}\cap L_{1}:\bC_{V_{\alpha}\cap L_{1}}
(u)|=2$. Hence $|V_{\alpha}:\bC_{V_{\alpha}} (u)|\leq 4$, i.e. $|Z_{\alpha
+1}:\bC_{Z_{\alpha +1}} (u)|\leq 4$. 

By (4.4) every subgroup of index 4 in $Z_{\alpha +1}$
contains  some $v\not= 1$, such that $v$ is centralized by an
elementary abelian group $E_{\alpha +1}\le L_{\alpha +1}$ such
that $\Gamma_{E_{\alpha +1}, 1} (G)\leq M_{\alpha +1}$.
So $\bC_{G} (v)\le M_{\alpha +1}$ and then $u\in M_{\alpha
+1}$. But $u\not\in Q_{\,\alpha}$, a contradiction.

So let $t = p$. Then we have the exceptional case $L_{1}/Q_{1}=BA,
B/O_{r}(B)\cong L_{\, 2}(r)$. Again $[O_{r}(B),Z_{1}]=1$. 

Let $\alpha +1\in\bigtriangleup (\alpha )$ with $Z_{1}\not\leq M_{\alpha
+1}$. Then by (4.10) we get that $\bC_{G} (v)\le M_{\alpha +1}$
for every $v\in Z_{\alpha +1}^{\,\sharp}$. As $Z_1 \le \bC([Z_2, Z_\alpha])$, we see that $[Z_{\alpha
+1},Z_{2}]=1$ and so $Z_{\alpha +1}\le L_{1}$. Now $Z_{\alpha
+1}\cap Q_{1}=1$. Choose $u\in Z_{1},u\not\in M_{\alpha +1}$. Then $|[Z_{\alpha
+1},u]|=|Z_{\alpha +1}|=|Z_{1}|$, a contradiction as $|u^{\,
L_{1}}|\leq |Z_{1}|-1$. \\

\absa
Suppose $b=b_{1}$ and $[Z_{1},Z_{\alpha}
]\not=1$.
Then by (6.2) $\alpha =2$. Furthermore as in (6.2) we
see that $Z_{1}$ is an F-module with $Z_{2}/Z_{2}\cap Q_{1}$
offending. 

\absa
{\bf (6.3)~Hypothesis.~} $b=b_{1}$ and $[Z_{1},Z_{\alpha}
]\not=1$. Suppose furthermore there is some component $K$ such that $Z_{1}$ is an
F-module for $K$ with $Z_{2}/\bC_{Z_{2}}(K)$ offending. 


\absa
Under (6.3) we have 

\absa
{\bf (6.4)~Hypothesis.~} $L_{2}$ is nonsolvable with $E(L_{\,
2}/Q_{\, 2})\cong L_{\, 2}(r)$, $(S)L_{3}(r)$, $(S)U_{3}(r)$,
$Sz(r)$, $Sp_{4}(r)$, $L_{\, 2}(r)\times L_{\, 2}(r)$, $Sz(r)\times
Sz(r)$, $r>2$ even, $SL_{3}(r)*SL_{3}(r), r=4$,
$SU_{3}(r)*SU_{3}(r)$, $r=8$, or $F(L_{\, 2}/Q_{\, 2})\not= 1$ and
$E((L_{\, 2}/Q_{\, 2})/F(L_{\, 2}/Q_{\, 2}))\cong L_{\, 2}(r)$,
$r=4$.

\absa
As by construction $L_{\, 2}\cap M=L_{\, 2}\cap L_{1}$ we see that
$|Z_{1}:Z_{1}\cap Q_{\, 2}|=r$ or $E(L_{\, 2}/Q_{\, 2})\cong
Sp_{4}(r)$, $L_{\, 2}(r)\times L_{\, 2}(r)$, $Sz(r)\times Sz(r)$,
$SL_{3}(r)*SL_{3}(r)$ or $SU_{3}(r)*SU_{3}(r)$ in which cases $|Z_{1}:Z_{1}\cap
Q_{\, 2}|=r^{\, 2}$.

\absa
Now let $x$ be a primitive prime divisor of $r-1$, which exists by
\cite{Zy}, or $x=9$ in case $r=64$. We fix the following notation
\begin{center}
$\omega\in L_{1}\cap L_{\, 2},\; o(\omega )=x\, .$
\end{center}

\absa 
By (1.66) there is some $\omega$ with $N_G(\langle \omega \rangle) \not\le M$. Furthermore $x \not\in \sigma(M)$ for $E(L_2/Q_2) \cong (S)L_3(r)$ or $Sp_4(r)$, $r$ even.

\absa
{\bf (6.5)~Lemma.~}{\it Suppose (6.3) and (6.4). If $N_G(\langle \omega \rangle) \not\le M$, then we have $[K,\omega ]\le K$.}

\absa
Proof~. Suppose false. Then let $K_{1}K_{2}\ldots K_{y}$ be some
orbit of $\langle\omega\rangle$, i.e. $y=x$ or $y=3$ and $x=9$.

Let $x=y$. Then $\bC_{M} (\omega )$ involves some group isomorphic to
$K$. Application of (1.63) yields that $m_{p}(K)=1$ for every prime $p\, \Big |\, |K|$. So we get
with (1.5) that $K\cong L_{2}(q)$, $Sz(q)$ or $J_{1}$. As $K$ has to
admit an F-module, we get with (1.18) $K\cong L_{2}(q), q$ even or
$K\cong L_{2}(7)$.

If $y=3$ and $x=9$, then $3\not\in \sigma (M)$ and so $m_{3}(K)=1$. Hence
by (1.5) we see $K\cong L_{2}(q)$, $Sz(q)$, $J_{1}$, $(S)L_{3}(q)$ or
$(S)U_{3}(q)$. Again (1.18) implies $K\cong L_{2}(q)$ or
$(S)L_{3}(q)$, $q$ even.

So in any case we have $K\cong L_{2}(q)$ or
$(S)L_{3}(q)$, $q$ even. Now we see $|Z_{1}:Z_{1}\cap Q_{\, 2}|\geq
q^{\, y}$. This shows $r\geq q^{\, y}$ or $r^{\, 2}\geq q^{\, y}$.

Suppose $y=x$. Then $x|2^{\, y-1}-1$. Hence $2^{\, y-1}\geq r$. This
shows $q=2$. So $K\cong L_{3}(2) $ and $|Z_{1}:Z_{1}\cap Q_{\,
2}|=r^{\, 2}$.

Suppose first $3\in\sigma (M)$. Then by (4.1) $E(L_{2}/Q_{\, 2})\cong
Sz(r)\times Sz(r)$. Now $4^{\, y}\geq |Z_{2}:\bC_{Z_{2}} (\langle
K^{\langle \omega \rangle}\rangle )|\geq r^{\, 4}$. Hence as $r^{\,
2}\geq 2^{\, y}$, we see that $r^{\, 2}=2^{\, y}$. But then $y$ is
even, a contradiction. So we have $3\not\in \sigma (M)$. So $y=3=x$ and
then $r=4$. But we have $[K^{\langle \omega\rangle} ,Z_{1}]$ is a direct
sum of three natural modules for the three components. Hence we see
that $|[Z_{1},K^{\langle\omega\rangle} ]/\bC_{[Z_{1},K^{\langle\omega\rangle} ]}
(Z_{2})|=8$, a contradiction as $L_{1}\cap L_{2}$ acts on this group
and so the order has to be power of $r$.

As $\bC_G(u) \leq M$ for any $u \in \bC_{Z_2}(\langle K^{\langle \omega \rangle} \rangle)^\sharp$, we see  $|Z_2 : \bC_{Z_2}(\langle K^{\langle \omega \rangle} \rangle )| \geq r^4$.

We are left with $x \not= y$, i.e. $y=3,x=9$. Then $r=64$. Furthermore $q\leq 16$. Now
choose $\nu\in L_{1}\cap L_{2},o(\nu )=7$. Then $[K,\nu ]\le
K$ as $3 \not\in \sigma(M)$. As $N_{G}(\langle\nu\rangle )\not\le M$, we get
$m_{7}(K)\leq 1$. This shows that $K\cong L_{2}(q)$ or $K \cong (S)L_3(q)$ and $[\nu ,K]=1$.
In the latter we get $m_{p}(K)=1$ for $p\in\sigma (M)$ by (1.63).
Hence $K\cong L_{2}(7)$. 

So in any case $K\cong L_{2}(q), q\leq 16$, or $L_{2}(7)$. Suppose
$7 \! \not\!\!\Big | \,|K|$, then $[\nu ,K]=1$ and so $[\nu ,K^{\langle \omega \rangle}
]=1$. Again by (1.63) we get a contradiction. So we have $K\cong
L_{2}(8)$ or $L_{2}(7)$. So in both cases $7\in\sigma (M)$ and
$\nu\in E, E$ elementary abelian of order $7^{\, 3}$, so
$N_{G}(\langle\nu\rangle )\le M$, a contradiction. This proves
(6.5).

\absa
{\bf (6.6)~Lemma.~}{\it Assume  (6.3) and (6.4). There is some $\omega \in L_1 \cap L_2$ with $N_{L_2}(\langle \omega \rangle) \not\leq M$ and $[K,\omega ]\not= 1, [K,\omega^{\, 3} ]\not= 1$ for
$o(\omega )=9$. }

\absa
Proof~. Suppose $[K,\omega ]=1$ or $[K, \omega^3] = 1$ for all $\omega \in L_1 \cap L_2$, $o(\omega) = x$, with $N_{L_2}(\langle \omega \rangle) \not\leq M$.  Then by (1.63)
$m_{p}(K)=1$ for $p\in\sigma (M)$. By (1.18) $K\cong (S)L_{n}(q)$, $n\leq 4$ for $q>2$, $n\leq 7$
for $q=2$, $Sp_{4}(q)$, $Sp_{6}(q)$, $\Omega_8^{\, -} (q)$,
$U_{4}(q)$, $G_{2}(q)$, $A_{n}$, $n\leq 11$. 

Set $\Omega = \langle \omega \rangle$ if $(r-1)^2 ~\not|~|L_1 \cap L_2|$ and
$\Omega = \langle \omega , \omega_1 \rangle$, where $|\Omega| = x^2$ if $(r-1)^2 ~|~ |L_1 \cap L_2|$.Furthermore we choose notation such that $N_G(\langle \omega_1 \rangle) \not\leq M$, too. So we have $[\Omega, K] = 1$ or $[\Omega^3, K] = 1$, where $\Omega^3 = \langle \omega^3, \omega_1^3 \rangle$.
We have that $|\bC_{S} (\Omega ):\bC_{Q_{1}Q_{2}} (\Omega )|\leq 2$.
Hence we get $Z_{2}/Z_{2}\cap Q_{1}$ projects into $Z(S/Q_{1}\cap K)$
or $q=2$ and $|Z_{2}/\bC_{Z_{2}} (K)|\leq 4$, or 8 if $K\cong
Sp_{4}(2)$ or $Sp_{6}(2)$. 

In any case we may choose $\omega$ such that $[Z_1, \omega] \not\leq Q_2$ and then $K$ acts nontrivially on $V_1 = [Z_1, \omega]$ or $V_1 = [Z_1, \omega^3]$, respectively, which is a module over $GF(r)$ or $GF(4)$.

Suppose first $|Z_{2}/\bC_{Z_{2}} (K)|\leq q$, then we have $K\cong
(S)L_{n}(q)$, $Sp_{4}(q)$ or $Sp_{6}(q)$, $K\cong A_{11}$, and in $V_1$ there
is just one natural module involved (1.18). This shows $[Z_2,V_1] \leq V_1$. We have $|V_1 : V_1 \cap Q_2| = r$ or $r^2$ or $4^t$. On the other hand $|V_1 : \bC_{V_1}(Z_2)| = q$. So we have $q = r$ or $q = r^2$, or $q$ is a power of 4. So $K\cong SL_{n}(q)$, $q>2$, $Sp_{4}(q)$ or
$Sp_{6}(q)$. As $N_{G}(\langle\omega\rangle )\not\le M$, we get
$m_{x}(K)\leq 2$, $m_{3}(K)\leq 2$ for $x=9$. This shows $K\cong
L_{2}(q)$, $SL_{3}(q)$ or $Sp_{4}(q)$. Now also $Z_{2}$ is an
F-module.

Suppose next that $K\cong Sp_{4}(q)$ or $Sp_{6}(q)$ and
$|Z_{2}/\bC_{Z_{2}} (K)|>q$. Now $|Z_{1}:Z_{1}\cap Q_{\, 2}|=q^{\,
2}$. Hence again $V_{1}$ is irreducible and we get $r=q$ again. Again
we get that $Z_{2}$ is an F-module.

So assume $|Z_{2}/\bC_{Z_{2}} (K)|\leq 4$, 8 respectively. As $|[V_{1},u]| \geq 4$ for $u \in Z_2$ with $[u,K] \not= 1$ we see that $V_1$ involves just one nontrivial irreducible module again. So either $Z_2$ is an $F$--module or $|Z_2/\bC_{Z_2}(K)| = 8$ and $|V_1 : \bC_{V_1}(Z_2)| = 4$. Now $m_3(K) \leq 2$ and so $K \cong Sp_4(2)'$ and the natural module is involved, which is not a module over $GF(4)$.

So in any case $Z_2$ is an $F$--module with $Z_1$ offending. Now as $[\Omega ,Z_{2}/\bC_{Z_{2}} (K)]=1$, we get a contradiction to
(1.31) and the action of an offending group on an F-module for $
L_{2}$. Now the assertion follows with (6.5).

\absa
We now fix the $\omega$ from (6.6) for the remainder.

\absa
{\bf (6.7)~Lemma.~}{\it Assume (6.3) and (6.4). If $K\cong G(q)$, a
group of Lie type in even characteristic, then $r\not= q$.}

\absa
Proof~. So suppose $r=q$. Let $\langle\nu\rangle\le L_{1}\cap
L_{2}, o(\nu )=q-1,\omega\in\langle\nu\rangle$. Let $\langle\nu_1\rangle
=N_{\langle\nu\rangle} (K)$. By (6.5) $\omega \in \langle \nu_1 \rangle$. We are going to show
\begin{itemize}
\item[(1)] $p  \not\Big | \,q-1\, .$
\end{itemize}

Suppose false. As $r=q$, we see that by (6.6) $\omega$ induces an inner-diagonal
automorphism on $K$. As $N_{L_2}(\langle \omega \rangle) \not\leq M$  we get $m_x(K) \leq 3$ and so that $K\cong L_{n}(q)$, $n\leq 4$,
$Sp_{2n}(q)$, $n\leq 3$, $U_{n}(q)$, $n\leq 7$, $\Omega^{\, -}_{\, 8}
(q)$ or $G_{2}(q)$.

On the other hand let $\mu\in\langle\nu\rangle, o(\mu )=p$. As $N_{L_2}(\langle \omega \rangle) \not\leq M$ also $N_{L_2}(\langle \mu \rangle) \not\leq M$ . If
$\mu\in\langle\nu_{1}\rangle$, then $m_{p}(\bC_{M} (\mu
))\leq 2$. If furthermore $p~|~|\bC_{M/O_{p'}(M)}(KO_{p'}(M)/O_{p'}(M))|$, then $m_p(\bC_M(\mu)) \geq 3$. Hence we have $m_{p}(K)\geq 4$. As $p\, \Big | \,q-1$, this is
impossible.

So we have $\langle\nu\rangle\not= \langle\nu_{1}\rangle$. As $\omega$
normalizes $K^{\, \nu^{\, i}}$ for any $i$ and $\bC_{K^{\nu^i}}(\omega)$ contains an element of order $q-1$ for all $i$, we get that there are at most three components $K^{\nu^i}$, otherwise $N_G(\langle \omega \rangle) \leq M$. So $|\langle\nu\rangle
:\langle\nu_{1}\rangle |=3$. As $\mu\not\in \langle\nu_{1}\rangle$, we
get $p=3$. But now $3\, \Big | \,|\langle\nu\rangle |$ and so by (1.66), we get
again $L_{2}\le M$, a contradiction.

By (1) we get 
\begin{itemize}
\item[(2)]$K\cong L_{n}(q)$, $n\leq 4$, $Sp_{2n}(q)$, $n\leq 3$,
$U_{n}(q)$, $n\leq 7$, $\Omega^{\, -}_{\, 8} (q)$ or $G_{2}(q)$.
\item[(3)]$E(L_{2}/Q_{\, 2})\not\cong Sp_{4}(q)$
\end{itemize}

Suppose $E(L_{2}/Q_{\, 2})\cong Sp_{4}(q)$. Then $|Z_{1}:Z_{1}\cap
Q_{\, 2}|=q^{\, 2}$. Furthermore by quadratic action and (1.14) $Z_{2}$ just involves
natural, dual of natural and trivial modules. 

As $L_{2}$ contains some element inducing the diagram automorphism on
$E(L_{2}/Q_{2})$, we see that with any natural module also its dual
is involved. So $|Z_{2}:Z_{2}\cap Q_{\, 1}|\geq q^{\, 4}$. As before we have that $K^{L_1 \cap L_2}$ contains at most three components. As $p ~\not\!|~ q-1$ by (1), we get $K \cong L_2(q)$ or $U_3(q)$. But then $|Z_2/\bC_{Z_2}(\langle K^{L_1 \cap L_2} \rangle)| \leq q^2$, a contradiction. So we have that $L_1 \cap L_2$ normalizes $K$. As $|Z_{1}:\bC_{Z_{1}} (Z_{2})|=q^{\, 2}$, we see with (1.45) and
(1.18) that $K\cong L_{4}(q),Sp_{6}(q)$.  Now $L_{1}\cap L_{2}$ contains $(\bZ_{q-1}\times
\bZ_{q-1})\cdot 2$. Hence $m_{p}(K(L_{1}\cap L_{2}))\geq 4$ for
$p\, \Big | \,q-1$, if $K\cong Sp_{6}(q)$. This implies $K\cong L_{4}(q)$.
Furthermore $Z_{2}$ acts as the uniquely determined subgroup of order
$q^{\, 4}$ on $Z_{1}$. This shows with (1.18) that $[K,Z_{1}] = U_{1}$
is the natural module. But now for $t\in U_{1}$ we have
$|Z_{2}:\bC_{Z_{2}} (t)|\leq q^{\, 2}$. But this is impossible as
$Z_{2}$ involves the natural module and its dual and no element can
induce transvections on both modules. This contradiction proves (3).

\absa
By (3) in any case we now have
\begin{itemize}
\item[(4)] $Z_{2}=\langle t \,\Big | \,|Z_{1}:\bC_{Z_{1}} (t)|\leq q\rangle\, .$
\end{itemize}

Hence $Z_{2}$ induces transvections on $Z_{1}$. So we have 
\begin{itemize}
\item[(5)] $Z_{1}$ involves exactly one irreducible nontrivial
module.
\end{itemize}

Now by (5.1) and (1) and $q>2$ we get that for $[K,Z_{1}]=U_{1}$, every
$x\in U_{1}$ is centralized by some elementary abelian group $E$ of
order $p^{\, 2}$ with $\Gamma_{E,1} (G)\leq M$. So let $g\in L_{2}$ with
$U_{1}^{\, g}\not\le M$. The structure of $E(L_{2}/Q_{\, 2})$
tells us that for $X=\langle U_{1},U_{1}^{\, g}\rangle$ we may assume
$X/O_{2}(X)\cong L_{2}(q)$ or $Sz(q)$ and $O_{2}(X)\le L_{1}$.  Now $U_{1}\cap U_{1}^{\, g}\le Z(X)$. So $U_{1}\cap U_{1}^{\,
g}=1$. 

As $U_{1}\cap O_{2}(X)\unlhd O_{2}(X)$, we get $\langle U_{1}\cap
O_{2}(X),U_{1}^{\, g}\cap O_{2}(X)\rangle =(U_{1}\cap O_{2}(X))\times
(U_{1}^{\,g}\cap O_{2}(X))$. So $U_{1}^{\, g}\cap O_{2}(X)$ induces the
full group of transvections of $U_{1}$ to the hyperplane $U_{1}\cap
O_{2}(X)$. This shows that 
\begin{itemize}
\item[(6)] $K\cong L_{n}(q), n\leq 4, U_{1}=[Z_{1},K]$ is the natural
module.
\end{itemize}

\begin{itemize}
\item[(7)] $Z_{1}=Q_{\, 1} = O_2(M)$
\end{itemize}

By (6) we have $Z_{1}=U_{1}\bigoplus \bC_{Z_{1}} (K)$. Suppose
$U_{1}\not= Z_{1}$. We are going to show that there is some $V_1$ with $V_1 \leq Z_1$ and $U_1 < V_1$, such that for any $u \in V_1^\sharp$ we have $\bC_G(u) \leq M$. By (4.4) we may assume $m_p(L_1) \geq 3$ and $m_p(\bC_{L_1}(Z_1)) = 0$ for $m_p(L_1) = 3$. We have $m_{p}(K)\leq 2$, and $m_{p}(K)=1$ for
$n\leq 3$. So we may assume $m_{p}(\bC (K))\geq 1$, and $m_{p}(\bC (K))\geq 2$
for $n\leq 3$. If $m_{p}(\bC (K))\geq 3$, there is some subgroup $U$ of order $p^2$ in $\bC(K)$ such that $\bC_{Z_1}(U) > V_1$ and we are done.

Let $m_{p}(\bC (K))=2$. And assume there is no such $V_{1}$. There
is some $\rho\in\bC (K),o(\rho )=p$ and some $Z_{1} > \tilde{V}
_{1} > U_{1}$, such that $[\rho ,\tilde{V_{1}}]=1$. If every $z\in Z_{1}$
is centralized by some $p$-element in $K$, we get that any element in
$V_{1}=\tilde{V}_{1}$ is centralized by some elementary abelian
$p$-group $E$ of order $p^{\, 2}$ with $\Gamma_{E,1} (G)\leq M$. 

Assume now that there is some $z\in U_{1}$ which is not centralized
by some $p$-element in $K$. Then we see $m_{p}(K)=1$. If $m_p(L_1) > 3$ there
is some $x\in L_{1},o(x)=p, x$ induces a field automorphism on $K$.
Furthermore by (4.12) we may assume $[x,\langle\rho ,\rho_{1}\rangle
]=1, \langle\rho ,\rho_{1}\rangle
\le\bC (K)$, elementary abelian of order $p^{\, 2}$. Now we may
replace $x$ by $x\rho_{1}^{\, i}$, for suitable $i$, such that
$V_{1}=U_{1}\bC_{\tilde{V}_{1}} (x\rho_{1}^{\, i} )>U_{1}$. But now
as every $z\in U_{1}$ is centralized by a conjugate of $x\rho_{1}^{\,
i}$, we see that every $z\in V_{1}$ is centralized by some $E,E$
elementary abelian, $|E|=p^{\, 2}, \Gamma_{E,1} (G)\leq M$. 

So let $m_p(L_1) = 3$. Then $m_p(\bC_{L_1}(Z_1)) = 0$ and so by (4.4) we have some $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \leq M$, such that $\rho \in E$ and $\bC_E(Z_1) \not= 1$. Set $V_1 = \tilde{V_1}$.

Let finally $m_p(\bC(K)) = 1$. Now $m_p(K) = 2$. By (4.12) no field automorphisms are involved. So we have $m_p(L_1) = 3$. Now any $u \in Z_1^\sharp$ is centralized by some $p$--element in $K$ and so by (6.4) any $u \in Z_1^\sharp$ is centralized by some $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \leq M$.

So we have found our $V_1$. But now we replace $U_{1}$ by $V_{1}$ and see that $V_{1}^{\, g}\cap
L_{1}$ has to induce all transvections on the hyperplane $V_{1}\cap
L_{1}^{\, g}$, a contradiction, as this hyperplane is of order greater than
$q^{\, n-1}$.

So we have $Z_{1}=U_{1}$. If $|Z_{2}:Z_{2}\cap Q_{1}|=q$, then we see
that $Z_{2}$ is an F-module, so $E(L_{2}/Q_{\, 2})\cong L_{2}(q)$ or
$L_{2}(q)\times L_{2}(q)$ by (1.31). Hence $Z_{1}\not\le
Q_{1}^{\,\prime}$. So assume $|Z_{2}:Z_{2}\cap Q_{\, 1}|>q$. Then $Q_{1}Q_{\, 2}$ does
not contain a Sylow 2-subgroup of $K$. This again shows that
$Q_{1}Q_{\, 2}=Z_{1}Q_{\, 2}$, so $Z_{1}\not\le Q_{1}^{\,
\prime}$. As $Z_{1}$ is irreducible, we get (7).

By (1) we have $m_{p}(K)\leq 2$. As $\bC_M(O_2(M)) \leq O_2(M)$ we get with (1) and (6.4) that $m_p(L_1) \geq 4$. Hence we have
that $m_{p}(\bC (K))=0$. But as $p  \not\!\!\Big |  \,q-1$ this would force some $p$-element in $M$ to
centralize $O_{2}(M)=Z_{1}$, a contradiction again.

\absa
{\bf (6.8)~Lemma.~}{\it Assume (6.3) and (6.4). If $K\cong G(q)$ is a group of Lie type, then $r=4$, $q=2$, $\langle
S,\omega\rangle\cap K/O_{2}(\langle S,\omega\rangle \cap K)\cong
\Sigma_{3}$, $\omega\in K$.}

\absa
Proof~. By (6.7) we know $q\not= r$. First we show 
\begin{itemize}
\item[(1)] $r>q$.
\end{itemize}

Suppose false. By quadratic action, (1.14) and (1.49) in case
$E(L_{2}/Q_{\, 2})\cong Sp_{4}(r)$, we get some $t\in Z_{2}$ with
$|Z_{1}:\bC_{Z_{1}} (t)|\leq r$, a contradiction to (1.18). 

Suppose first $q =2$. By (1.18) we have $K \cong L_n(2)$, $SU_n(2)$, $\Omega_{2n}^{\pm}(2)$, $Sp_{2n}(2)$ or $G_2(2)'$. Now (1.65) tells us that $o(\omega) = 3$ or $9$. Hence we have $3 \not\in \sigma(M)$. This shows that $K \cong L_n(2)$, $n \leq 7$, $SU_n(2)$, $n \leq 4$, $Sp_{2n}(2)$, $n \leq 3$, $\Omega_8^-(2)$ or $G_2(2)'$. As these groups do not posses automorphisms of order 9 interchangeable with a Sylow 2-subgroup, we see that $r = 4$.

 If $K \cong Sp_{2n}(2)$, $G_2(2)'$ or $L_n(2)$, we have the assertion as long as $\langle S, \omega \rangle = S\langle \omega \rangle$. So suppose false. Then we have $9 ~|~|L_1 \cap L_2|$ and so $E(L_2/Q_2) \cong (S)L_3(4)$, $Sp_4(4)$, $L_2(4) \times L_2(4)$ or $SL_3(4)*SL_3(4)$. But in all cases any element of order 3 in $E(L_2/Q_2)$ has its normalizer not contained in $M$. So as $\langle S, \omega \rangle \cap K/O_2(\langle S, \omega \rangle \cap K) \cong \Sigma_3$, we see that there is some $\omega_1 \in L_1 \cap L_2$, $o(\omega_1) = 3$, which is in $K$. Hence we may assume $\omega \in K$. 

So assume now $q > 2$. Then by (1.65) and (1.64) we get $r=q^2$ and $K \cong U_n(q)$ or $\Omega_{2n}^-(q)$, or $o(\omega) = 3$ or $9$ and $K \cong O_8^+(q)$, $q \leq 32$. But in the latter $3 \in \sigma(M)$, a contradiction. So we have\\

\begin{itemize}
\item[(2)] $r=q^{\, 2}$ and $K\cong U_{n}(q)$ or $\Omega_{2n}^{\, -}(q)$.
\end{itemize}


By (5.1) there is $U_{1}\le Z_{1}$, such that every $x\in U_{1}$
is centralized by an elementary abelian subgroup $E$ of order $p^{\,
2}$ in $M$, such that $\Gamma_{E,1} (G)\leq M$. Recall that (5.1)
(ii) cannot occur as in this case $ 3 \in\sigma (M)$ and $r = 4$, i.e. $o(\omega) = 3$. Let $g\in L_{2}$ such that $\langle U_{1},U_{1}^{\, g}\rangle
=X\not\leq M$. Suppose $|U_{1}:U_{1}\cap O_{2}(X)|=r$, then $U_{1}^{\, g}\cap
O_{2}(X)$ induces the full group of transvections to the subgroup
$U_{1}\cap O_{2}(X)$. As $|U_{1}:U_{1}\cap O_{2}(X)|=q^{\, 2}$, we
get with (1.45) that $|U_{1}^{\, g}\cap O_{2}(X)|\leq q$. But then
$|U_{1}|\leq q^{\, 3}$, a contradiction.

So we have $|U_1 : U_1 \cap O_2(X)| > r$. This shows $E(L_{2}/Q_{\, 2})\cong Sp_{4}(r)$. But now again
by (1.49) there is some $\tilde{Z}_{2}\le Z_{2}, |\tilde{Z}_{2}| =r$, $\tilde{Z}_{2}\cap Q_{\, 1}=1$ but $|Z_{1}:\bC_{Z_{1}}
(\tilde{Z}_{2} )|=r=q^{\, 2}$. But this contradicts (1.45).

\absa
{\bf (6.9)~Lemma.~}{\it Assume (6.3) and (6.4). Then $\langle S,\omega\rangle /O_{2}(\langle S,\omega\rangle
) \cong \Sigma_{3}$, $\omega\in K$ and $r=4$. }

\absa
Proof~. By (6.8) and (1.18) it is enough to treat the case $K\cong A_{n}, n\geq 6$.
Now by (1.9) we get $o(\omega )=3$, so $r=4$. Furthermore $
\langle S,\omega\rangle /O_{2}(\langle S,\omega\rangle )\cong \Sigma_{3}$ as in (6.8).\\

\absa
By (6.9) we have $3\not\in \sigma (M)$. Hence $K\cong L_{n}(2)$,
$n\leq 7$, $Sp_{6}(2)$, $U_{4}(2)$, $\Omega_8^{\, -} (2)$,
$G_{2}(2)^{\prime}$ or $A_{n}, n\leq 11$. 

If $m_p(\bC_{L_1/Q_1}(K)) \geq 2$, we have that $\omega$ is centralized by $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \leq M$. Suppose $m_p(\bC_{L_1/Q_1}(K))\leq 1$. If $K \leq E(M/O_{p'}(M))$, then we see that $m_p(\bC_M(KO_{p'}(M)/O_{p'}(M)) \geq 2$, as $m_p(K) \leq 2$ and $K$ does not allow outer $p$--automorphisms. Hence in this case too $\omega$ is centralized by $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \leq M$. If $K \not\leq E(M/O_{p'}(M))$, then by the construction in (4.4) we have that $[E(M/O_{p'}(M)), K] = 1$, so we may assume $m_p(E(M/O_{p'}(M)) \leq 1$ and so we have $m_p(F(M/O_{p'}(M)) \leq 3$. Hence as $K$ acts faithfully on $F(M/O_{p'}(M))$, we see $K \cong L_2(7)$ or $K \cong A_6$ and $Z_1$ is centralized by $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \leq M$. 

Suppose we do not have the latter. Then by (1.63) $\bC_{Z_1}(\omega) =1$. We have that $\omega$ is in some minimal parabolic.  If $K \cong L_n(2)$ then $\omega$ centralizes some nontrivial vector in the natural module. Hence the same is true for the module $V(\lambda_2)$ as the maximal parabolic with $L_{n-1}(2)$ on top induces the natural module. Hence we have that $K \not\cong L_n(2)$. If $K \cong Sp_6(2)$ then by (1.18) $Z_1$ involves the natural or spin module. Again in the natural module $\omega$ has a fixed point. So we have that $Z_1$ is the spin module. If $K \cong U_4(2)$ we must have the natural $O_6^-(2)$--module. If $K \cong \Omega_8^-(2)$ then $Z_1$ is the natural module by (1.18) and so $\omega$ has a fixed point. If $K\cong A_n$ and $Z_1$ is the permutation module, we get a fixed point for $\omega$ or $n = 6$.

\absa
So we summarize. We have $K \cong L_3(2)$, $A_6$, $A_7$, $U_4(2)$ (and $Z_1$ is the $O_6^-(2)$--module), $G_2(2)'$, or $Sp_6(2)$ (and $Z_1$ is the spin module). In any case for all $u \in Z_1^\sharp$ we have $\bC_G(u) \leq M$.

Now choose $g \in Z_2$ with $Z_1^g \not\leq M$. Set $X = \langle Z_1, Z_1^g \rangle$. Suppose first $|Z_1 : Z_1^g \cap O_2(X)| = 4$. As we have $Z_1^g \cap O_2(X) \cap Z_1 = 1$, we see that $Z_1^g \cap O_2(X) \cap Q_1 = 1$. Hence $|Z_1^g \cap O_2(X)| \leq 2^{m_2(\mbox{Aut}(K))}$. Further $Z_1^g \cap O_2(X)$ centralizes a subgroup of index 4 in $Z_1$. As $|Z_1^g \cap O_2(X)| \leq 8$ for $K \cong G_2(2)'$ we see that $|Z_1| \leq 2^5$, a contradiction. If $K \cong Sp_6(2)$, we get $|Z_1^g \cap O_2(X)| = 2^6$ and so $Z_1^g \cap O_2(X)$ is the uniquely determined subgroup of order 64 in a Sylow 2-subgroup of $Sp_6(2)$. But the centralizer of this group in the spin module is of order two. Let $K\cong U_4(2)$. Then $|Z_1^g \cap O_2(X)| = 2^4$. But a subgroup of Aut$(K)$ centralizing a subgroup of order 4 in the natural $O_6^-(2)$--module is never of order 16. 

So we are left with $K \cong L_3(2)$, $A_6$ or $A_7$ and $|Z_1| = 2^3$ or $2^4$. Now there is $t \in Z_1 \setminus Q_2$ with $|Z_2 : \bC_{Z_2}(t)| \leq 4$. Application of (1.43) shows $E(L_2/Q_2) \cong A_5$ or $A_5 \times A_5$ and so $Z_1 \not\leq Q_1'$. If $Z_1$ is irreducible then $Q_1' = 1$. We show that $Q_1$ is abelian in general. So we may assume $|Z_1| = 16$ and $K \cong L_3(2)$. but as $|Z_1 : \bC_{Z_1}(u)| = 4$ for any $u \in Z_2 \setminus Q_1$, we see that $Z_1$ is an indecomposabe module having the natural module $U_1$ as a submodule. Now $|U_1 : U_1 \cap Q_2| = 4$ and so $U_1 \not\leq Q_1'$. This again shows $Q_1' = 1$.

Suppose now that $|Z_1^g : Z_1^g \cap O_2(X)| = 16$ in which case $E(L_2/Q_2) \cong Sp_4(4)$. Then $X/O_2(X) \cong Sp_4(4)$ and again $Z_1 \cap O_2(X) \cap Z_1^g = 1$. Furthermore $|Z_1 \cap O_2(X)| \geq 16$. This shows that $K \cong U_4(2)$ or $Sp_6(2)$. Furthermore $|Z_1| \geq 2^8$ and so $K \cong Sp_6(2)$ and $Z_1$ is the spin module. Hence $|Z_1 \cap O_2(X)| = 16$. As $Z_2$ centralizes a subgroup of index 16 in $Z_1$ we see that $|Z_2 : \bC_{Z_2}(Z_1)| \leq 2^5$ and so $Z_1$ offends as an $F_1$--module on $Z_2$. But this contradicts (1.43).

So we have $Q_1' = 1$ and so $Q_1 = O_2(M)$ is elementary abelian. Furthermore $|Z_1| \leq 16$ and $K \cong L_3(2)$, $A_6$ or $A_7$ and $E(L_2/Q_2) \cong L_2(4)$. Hence $O_2(M)$ is an $F$--module for $K$ with $Z_2/\bC_{Z_2}(O_2(M))$ offending. So we see that $[K,O_2(M)] \leq Z_1$. As $\Omega_1(Z(S)) \leq Z_1$, we get with (1.20) that $|O_2(M)| \leq 2^5$. But this contradicts $m_p(M) \geq 4$ and $\bC_M(O_2(M)) \leq O_2(M)$.


\absa
So we have with (2.1)

\absa
{\bf (6.10)~Lemma.~}{\it Assume (6.3). Then
$L_{2}$ is solvable or $E(L_{2}/Q_{2})\cong L_{2}(r)$, $L_{3}(r)$,
$U_{3}(r)$, $PSp_{4}(r)$, $L_{2}(r)\times L_{2}(r)$, $r$ odd,
$A_{6}$, $A_{9}$, $3\cdot A_{6}$, $3A_{6} * 3A_{6}$.}

\absa
As $Z_{1}\not\le Q_{2}$, we have $E(L_{2}/Q_{2})\not\cong
A_{9}$. Next we show

\absa
{\bf (6.11)~Lemma.~}{\it Assume (4.3). Then
$|Z_{1}:Z_{1}\cap Q_{\, 2}|>2$. }

\absa
Proof~. For this assume $|Z_{1}:Z_{1}\cap Q_{\, 2}|=2$. Then $Z_{2}$
induces transvections on $Z_{1}$ and so $K\cong L_{n}(2)$,
$Sp_{2n}(2)$, $\Omega_{2n}^{\, -} (2)$ or $A_{n}$. By (5.1) there is
$U_{1}\le Z_{1}$, such that any $z\in U_{1}$ is centralized by
some elementary abelian group $E$ of order $p^{\, 2}$ in $M$ with
$\Gamma_{E,1} (G)\leq M$. So let $g\in L_{2}$ such that $\langle
U_{1},U_{1}^{\, g}\rangle /O_{2}(\langle U_{1},U_{1}^{\, g}\rangle )$
is dihedral of order $2t, t$ odd. Then we see that $U_{1}^{\,
g}\cap Q_{\, 2}$ induces the full group of transvection on $U_{1}$ to
the hyperplane $U_{1}\cap Q_{\, 2}$. This implies $K\cong L_{n}(2)$
and $U_{1}$ is the natural module. If $Z_{1}>U_{1}$, then we see as in
(6.7) (3) that there is some $V_{1} > U_{1}$ such that every $z\in
V_{1}$ is centralized by some elementary abelian $E$ of order $p^{\,
2}$ with $\Gamma_{E,1} (G)\leq M$, a contradiction. So we have $Z_{1}$
is the natural module. In particular 
\setcounter{equation}{0}
\begin{itemize}
\item[(1)]~~~~$|\Omega_{1} (Z(S))|=2.$
\end{itemize}


Now let $\nu\in\langle U_{1},U_{1}^{\, g}\rangle ,o(\nu )=t$. Then
$(U_{1}\cap Q_{2})\times (U_{1}^{\, g}\cap Q_{2})=[Q_{2},\nu ]$ and
$\Omega_{1} (Z(S))\le [Q_{2},\nu ]$. As $\Phi
(Q_{2})\le \bC_{Q_{\, 2}} (\nu )$ we see that
$\Phi (Q_{\, 2})=1$. As in a linear group the full transvection
group is its own centralizer we see that 

\begin{itemize}
\item[(2)] $S/Q_{1}Q_{\, 2}$
involves a Sylow 2-subgroup of $L_{n-1}(2)$.
\end{itemize}

Next suppose that $z\in Z_{1}\setminus Q_{\, 2}$ inverts an
elementary abelian group of order $t^{\, 2}, t$ a prime. Then there
is some $\omega\in L_{2}$, $\omega^{\, z} =\omega^{\, -1}$,
$\bC_{[Q_{\, 2},z]} (\omega )\not= 1$. But $[Q_{\, 2},z]\le
Z_{1}$ and every $x\in Z_{1}$ is centralized by some elementary
abelian group $E$ of order $p^{\, 2}$ with $\Gamma_{E,1} (G)\leq M$. So we have
$\omega\in M$, a contradiction as $z\in O_{2}(M)$. As $z\in Z(S/Q_{\,
2})$, this now implies together with (2.1)
\begin{itemize}
\item[(3)]  $E(L_{2}/Q_{\, 2})\cong L_{2}(r)$, $L_{3}(r)$,
$U_{3}(r)$, $3\cdot A_{6}$, $PSp_{4}(r)$, or $L_{2}$ is a solvable
$\{ 2,t\}$-group and $\sigma (L_{2})=\emptyset$.
\end{itemize} 

Suppose now $n\geq 5$ and  $n\geq 6$ in case of $E(L_{2}/Q_{\, 2})\cong
PSp_{4}(r)$. Then (3) implies that $L_{2}$ is solvable. Let $C$ be a
critical subgroup of $O_{t}(L_{2}/Q_{\, 2})$, $C_1=\Omega_{1} (C)$. As
Aut$(C_1)$ involves a Sylow 2-subgroup of type $A_{8}$, and
$m_{t}(C_1)\leq 3$, we get a contradiction with (1.6). So (3) and (2) imply
\begin{itemize}
\item[(4)] $n\leq 4$ and $n\leq 5$ in case of $E(L_{2}/Q_{\, 2})\cong
PSp_{4}(r)$.
\end{itemize}

Suppose now that $L_{2}$ is solvable and $O_{t}(L_{2}/Q_{\, 2})$ is
not cyclic or $E(L_{2}/Q_{\, 2})\cong 3\cdot A_{6}$. Then by (1.6)
and (2) $z\in Z_{1}\setminus Q_{\, 2}$ acts on some extraspecial
group $T$ of order $t^{3}$ such that $\bC_{T}(z)=Z(T)$. As
$|[z,Q_{\, 2}]|\leq 8$, we see $t=3$. Let $u\in T\setminus
Z(T)$. We may assume $[u,Q_{\, 2}]=(Z_{1} \cap Q_{\, 2})\times (Z_{1}^{\,
g} \cap Q_{\, 2})$. Now $Z(T)$ acts on $[u,Q_{\, 2}]$. As $T\cap
M\le Z(T)$ and every element in $Z_{1} \cap Q_{\, 2}$ is centralized by
an elementary abelian group $E$ of order $p^{\, 2}$  with
$\Gamma_{E,1} (G)\leq M$, we see $\bC_{Q_{\, 2}} (Z(T))\cap ((Z_{1} \cap Q_{\,
2})\times (Z_{1}^{\, g} \cap Q_{\, 2}))\not= 1$. But then $T$ acts as an
elementary abelian group of order $9$ on $\bC_{Q_{\, 2}} (Z(T))$,
which gives that some $\tilde{u}\in T\setminus Z(T)$ centralizes
some $x\in Z_{1} \cap Q_{\, 2}$, $x\not= 1$, a contradiction. So we have 
\begin{itemize}
\item[(5)] (i) If $L_{2}$ is solvable, then $O_{t}(L_{2}/Q_{\, 2})$
is cyclic.\\
(ii) $E(L_{2}/Q_{\, 2})\not\cong 3\cdot A_{6}$.
\end{itemize}

Let $L_{2}$ be nonsolvable. Then $|Q_{\, 2}:\bC_{Q_{\, 2}}
(Z_{1})|\leq 8$ and $|Q_{\, 2}:\bC_{Q_{\, 2}} (Z_{1})|\leq 16$ in case of
$E(L_{2}/Q_{\, 2})\cong PSp_{4}(r)$. With (1.30) we see
$E(L_{2}/Q_{\, 2})\cong L_{2}(7)\cong L_{3}(2)$. 

Now $E(L_{2}/Q_{\, 2})$ is generated by four conjugates of $Z_{1}$. This
shows that $|Q_{\, 2}:\bC_{Q_{\, 2}} (E(L_{2}/Q_{\, 2}))|\leq
|[Z_{1},Q_{\, 2}]|^{\, 4}\leq 2^{\, 12}$. As $\Omega_{1}
(Z(S))\le Z_{1}\cap Q_{\,  2}$, we get with (1) $\bC_{Q_{\, 2}}
(E(L_{2}/Q_{\, 2}))=1$. So we now see $|Q_{1}|\leq 2^{\, 9}$.

If $O_{t}(L_{2}/Q_{\, 2}) \not= 1$ we get $n=3$ and $|Q_{\, 2}:\bC_{Q_{\, 2}}
(O_{t}(L_{2}/Q_{\, 2}))|\leq 16$. Hence $|Q_{\, 2}|\leq 16$ and so
$t=3$ or $5$, and $|Q_{1}|\leq 2^{\, 3}$. So we have 
\begin{itemize}
\item[(6)] $|Q_{1}|\leq 2^{\, 9}$ and $|Q_{1}|\leq 2^{\, 3}$ for
$L_{2}$ solvable.
\end{itemize}

As $O_{2}(M)\le Q_{1}$ and some elementary abelian subgroup of
order $p^{\, 4}, p$ odd, acts faithfully on $O_{2}(M)/\Phi
(O_{2}(M))$, we see $|O_{2}(M)/\Phi (O_{2}(M))|\geq 2^{\, 8}$.

This shows $2^{\, 8}\leq |Q_{1}|\leq 2^{\, 9}$, $n=4$, and $p=3$. As
$|Z(Q_{1})|\geq 2^{\, 4}$, we also get that $Q_{1}$ is elementary
abelian. But as $|\Omega_{1} (Z(S))|=2$ by (1), we now have
$\bC_{Q_{1}} (K)=1$, a contradiction to $|[Q_{1},K]|=2^{\, 4}$. This proves (6.11).

\absa
{\bf (6.12)~Lemma.~}{\it Assume (6.3). Set
$U_{1}=[K,Z_{1}]$. Suppose there is some $x\in Z_{2}$ with $[x,K]\not=1$
such that $|U_{1}:\bC_{U_{1}} (x)|=2$. Then for $z\in
[Z_{2},U_{1}]$ with  $z\not= 1$, we have $\bC_{L_{2}} (z)\le M$}

\absa
Proof~. By assumption $K\cong L_{n}(2)$, $Sp_{2n}(2)$, $\Omega_{2n}^{\,\pm}
(2)$ or $A_{n}$ and $U_{1}/\bC_{U_{1}} (K)$ is the natural module.
Now by (5.1) every $z\in U_{1}$ is centralized by some 
elementary abelian subgroup $E$ of order $p^{\, 2}$ in $M$ with $\Gamma_{E,1} (G)\leq M$.
Hence $\bC_{G} (z)\le M$, if $z\not= 1$.

\absa
{\bf (6.13)~Lemma.~}{\it Assume (6.3). Then
$L_{2}$ is solvable.}

\absa
Proof~. Let $E(L_{2}/Q_{\, 2})=X_{1}$ or $X_{1}X_{2}$, $X_{1}$
quasisimple. If $|Z_{1}:\bC_{Z_{1}} (X_{1})|\leq 2$, we see that
$[E(L_{2}/Q_{\, 2}),Z_{2}]$ is a sum of two modules $Z_{2}^{\, (1)}$,
$Z_{2}^{\, (2)}$ with $[Z_{2}^{\, (i)}, X_{3-i}] = 1$. Furthermore there is some $x \in Z_2^{(1)}$ with $[x, U_1] \not= 1$ and $[x,K] \not= 1$, where $U_1 = [K,Z_1]$. Now
$|U_{1}:\bC_{U_{1}} (x)|=2$ and $[U_{1},Z_{2}^{\, (1)}]$ is
centralized by $X_{2}$. But this contradicts (6.12).

So we have $|Z_{1}:\bC_{Z_{1}} (X_{1})|> 2$. Application of (1.12)
shows $X_{1}\cong L_{3}(2)$, $A_{6}$ or $3A_{6}$. As $Z_{1}\unlhd S$
and there is no normal fours group in a Sylow 2-subgroup of ${\mbox{Aut}}(L_{3}(2))$, we get $X_{1}\cong A_{6}$ or $3 A_{6}$ and
$N_{L_{2}/Q_{\, 2}}(X_{1})/\bC_{L_{2}/Q_{\, 2}}(X_{1})\cong P\Gamma
L_{2}(9)$. Hence $|Z_{1}:\bC_{Z_{1}} (X_{1})|=4$ and $Z_{1}/\bC_{Z_{1}}
(X_{1})$ corresponds to the center of a Sylow 2-subgroup of $\Sigma_{6}$.
By (1.14) $Z_{2}$ involves natural and dual $Sp_{4}(2)$-modules. But
now again there is some $x\in Z_{2}$ (in some natural module) such
that $|U_{1}:\bC_{U_{1}} (x)|=2$. If $[x,K]=1$, we would get
$[x,U_{1}]\leq \bC_{U_{1}} (K)$ and so $[x,U_{1}]=1$. Thus we have
$[x,K]\not= 1$. But in the natural module every element is centralized
by some 3-element in $\Sigma_{6}$. So $[U_{1},Z_{2}]^{\,\sharp}$ contains
an element $z$ with $\bC_{L_{2}} (z)\not\le M$, contradicting
(6.12).

\absa
{\bf (6.14)~Lemma.~}{\it We have that (6.3) does not hold.}

\absa
Proof~. Suppose false. Then by (6.13) $L_{2}$ is solvable.
Now by \cite[(24.1)]{GoLyS} there is some subgroup $\hat{L}_{2}$ of $L_{2}/\bC_{L_{2}}
(Z_{2})$ such that $\hat{L}_{2} =D_{1}\times\ldots\times D_{n}$ with $D_{i}\cong
D_{2t}$ where $|Z_{1}:Z_{1}\cap \bC_{L_{2}} (Z_{2})|=2^{\, n}$ and
$Z_{1}\bC_{L_{2}} (Z_{2})/\bC_{L_{2}} (Z_{2})$ is a Sylow 2-subgroup
of $\hat{L}_{2}$.

Set $U_1 = [K,Z_1]$. By \cite[(25.12)]{GoLyS} there is some $z\in Z_{2}$ with $|[U_{1},z]|=2$ and
$[K,z]\not= 1$. By (6.11) $n>2$. But now by quadratic action there
is some $1\not= x\in [U_{1},Z_{2}]$ which is centralized by some
$t$-element in $\hat{L}_{2}$. As $Z_{1}\le O_{2}(M)$, we have
$\hat{L}_{2}\cap (L_{2}\cap M)/\bC_{L_{2}} (Z_{2})=Z_{1}\bC_{L_{2}}
(Z_{2})/\bC_{L_{2}} (Z_{2})$, a contradiction to (6.12).

\absa
{\bf (6.15)~Proposition.~}{\it If $b=b_{1}$, then $[Z_{1},Z_{\alpha}
]=1$.}

\absa
Proof~. Suppose $b = b_1$ and $[Z_1,Z_\alpha] \not= 1$. Let first $K$ be $p$-constraint such that $Z_1$ is an $F$--module for $KZ_2$ with $Z_2/\bC_{Z_2}(K)$ offending. Then we get  $p=3$.
Furthermore by quadratic action there is some $z\in Z_{1}$ inducing a
transvection on $Z_{2}$. This implies that $L_{2}$ is a solvable $\{
2,3\}$-group, $E(L_{2}/Q_{\, 2})\cong A_{5}$, $A_{5}\times A_{5}$ or
$A_{9}$, or $E(L_{2}/Q_{\, 2})/F(L_{2}/Q_{\, 2})\cong L_{2}(4)$. As
$p=3$, we get $m_{p}(L_{2}\cap M)= 0$. This implies that $L_{2}/Q_{\,
2}\cong \Sigma_{3}$. Hence $Z_{2}$ induces transvections on $Z_{1}$ to
the hyperplane $Z_{1}\cap Q_{\, 2}$. We get that
$[Z_{2},O_{3}(K/\bC_{K} (Z_{1}))]=N$ is of order three and
$|[N,Z_{1}]|=4$. As $Z_{2}Q_{1}/Q_{1}\lhd S/Q_{1}$, we see that
$[N,Z_{1}]\unlhd S$. Now $|[[N,Z_{1}],Q_{\, 2}]|=2$. So let $\omega\in
L_{2},o(\omega )=3$. Then $|Q_{\, 2}:\bC_{Q_{\, 2}} (\omega )|=4$.
Hence $\omega$ centralizes a subgroup of index 8 in $Z_{1}$. Let $F$
be an elementary abelian 3-subgroup of maximal rank in $L_{1}$. Then
$|F:\bC_{F} (Z_{1})|\geq 9$. So every subgroup of $Z_{1}$ of index 8
contains some $1\not= z$ such that $|\bC_{F} (z)|\geq 9$. As
$\Gamma_{\bC_{F} (z),1} (G)\leq M$, we get $\bC_{G} (z)\leq M$. But then $\omega\in M$, a contradiction.

So by (6.14) we may assume that we have the exceptional case $K_{1}/Q_{1}=BA$,
$B/O_{r}(B)\cong L_{2}(r)$. Now again $[O_{r}(B),Z_{1}]=1$. As $Z_{1}$
is an F-module we have $r=7$ or $r=5$. By (4.10) $\bC_{G}
(z)\le M$ for any $z\in Z_{1}^\sharp$. Hence
\begin{itemize}
\item[(1)]  $\bC_{Z_{2}} (L_{2})=1$. If $\omega\in L_{2}$,
$\bC_{Z_{1}\cap Z_{2}} (\omega )\not= 1$, then $\omega\in M$.
\item[(2)] If $z\in Z_{2}$ and $[B,z]\not= 1$, then $[Q_{1},z]\le
Z_{1}$.
\end{itemize}

Otherwise $[Q_{1},z]\le Z_{1}$ and so
$[O_{r}(B),Q_{1}]\le Z_{1}$. But then $[O_{r}(B),Q_{1}]=1$, a
contradiction.
\begin{itemize}
\item[(3)] $| Z_{1}:Z_{1}\cap Q_{\, 2}|=2$
\end{itemize}

Otherwise $Z_{2}$ is an F-module with $Z_{1}$ offending. As
$[Z_{1},Z_{2}]\le Z_{1}$, we see with (1.31) that $L_{2}$ is
solvable or $L_{2}/\bC_{L_{2}} (Z_{2})\cong A_{5}$ or $S_{5}$ and
$Z_{2}$ is the natural $L_{2}(4)$-module.

In the latter $Z_{1}\cap Z_{2}=\bC_{Z_{2}} (Z_{1})$. So let $z\in
Z_{2}\setminus Z_{1}\cap Z_{2}$ with $[z,S]\le Z_{1}
\cap Z_{2}$. Then $[z,Q_{1}]=[Z_{1},z]$, a contradicting (2).

So we have that $L_{2}$ is solvable. Now by \cite[(24.1)]{GoLyS} there is a subgroup
$\hat{L}_{2}$ in $L_{2}/\bC_{L_{2}} (Z_{2})$ such that
$\hat{L}_{2}=D_{1}\times\ldots\times D_{n}$ with $D_{i}\cong\Sigma_{3}$ and
$Z_{1}\bC_{L_{2}} (Z_{2}) /\bC_{L_{2}} (Z_{2})\in\mbox{~Syl}_{2}
(\hat{L}_{2})$. Furthermore $n\geq 2$ by assumption. By quadratic
action $[D_{2},[Z_{1}\cap D_{1},Z_{2}]]=1$. But this contradicts (1) and $Z_{1}\le O_{2}(M)$.

By (2) we have that $Z_{2}$ induces transvections and so $r=7$. Hence
$p=3$, Sylow 3-subgroups of $L_{2}$ are cyclic and $m_{3}(L_{2} \cap
M)=0$ by (4.1) and (1.66). This together with (1.23) and (1.43) shows that $E(L_{2}/Q_{\,
2})\cong L_{3}(2)$, $L_{2}/Q_{\, 2}\cong \Sigma_{3}$ or $L_{2}$ is a
solvable $\{ 2,5\}$-group. 

In the first case we have by (1.43) a submodule in $Z_{2}$ which is
the natural one. But then every element in this submodule is
centralized by a 3-element in $L_{2}$, contradicting (1).

So $L_{2}$ is solvable. Let $5 \,\Big | \,|L_{2}|$.  As $Z_{1}Q_{\, 2}/Q_{\,
2}\le Z(S/Q_{\, 2})$ we have that $Z_{1}Q_{\, 2}/Q_{\,
2}$ acts fixed point freely on $O_{5}(L_{2}/Q_{\, 2})/\Phi (O_{5}(L_{2}/Q_{\,
2}))$. But $|Z_{2}:\bC_{Z_{2}} (Z_{1})|\leq 4$ and so we have that $O_{5}(L_{2}/Q_{\,
2})$ is cyclic. As $Z_{1}\le O_{2}(M)$ we have $|O_{5}(L_{2}/Q_{\,
2})|=5$ and $|Z_{2}|=16$. So $|Z_{2}:Z_{2}\cap Q_{1}|=4$. As
$B/O_{7}(B)\cong L_{2}(7)$ we see that $Q_{1}Q_{\, 2}\not= S$ and so
$L_{2}/\bC_{L_2}(Z_2)\cong F_{20}$.

But now in both cases $L_2/\bC_{L_2}(Z_2) \cong F_{20}$ and $L_2/Q_2 \cong \Sigma_3$ there is some $z\in Z_{2}\setminus
Z_{1}\cap Z_{2}$ with $[S,z]\le Z_{1}\cap Z_{2}$. Hence
$[S,z]\le Z_{1}$, contradicting (2).

\absa
{\bf (6.16)~Proposition.~}{\it Suppose $b=b_{1}>1$. Set
$V_{2}=\langle Z_{1}^{\, L_{2}}\rangle$. Then $V_{2}$ involves at
least two nontrivial irreducible $L_{2}/Q_{\, 2}$-modules.}

\absa
Proof~. Suppose $b=b_{1}>1$. By (6.15) $[Z_{1},Z_{\alpha} ]=1$ hence
$\alpha\sim 2$. This shows $\Omega_{1} (Z(S))\lhd L_{2}$. Let $(1,\alpha )$
be a critical pair and $(1,2\ldots ,\alpha )$ be a path of length
$b$. Then $V_{2}\le L_{\alpha}$ and $V_{\alpha}\le
L_{2}$. We have 
$$[V_{2},V_{\alpha} ]\le V_{2}\cap V_{\alpha} ,V_{2}^{\prime}
=1\mbox{~and~} [Z_{1},V_{\alpha} ]\not= 1\, .$$
\indent This shows
$$[Z_{1},Z_{\alpha +1}]\not= 1\mbox{~for some~}\alpha +1\in\triangle
(\alpha )\, .$$
\indent Choose $\hat{V}_{2}$ maximal in $V_{2}$ such that
$\hat{V}_{2}\trianglelefteq L_{2}$ and $[\hat{V}_{2},O^{\, 2^\prime}
(L_{2})]=1$. Let $\hat{V}_2 \le A\le V_{2}$ such that $A/\hat{V}_{2}$ is an
irreducible $L_{2}/Q_{2}$-module. Suppose $[V_{2},O^{\, 2^\prime}
(L_{2})]\le A$. We show
$$ V_{2}\not= AZ_{1}$$


Suppose false. Assume $V_{2}=\hat{V}_{2} Z_{1}$. As $Z_{1}\not\unlhd
L_{2}$, we get $Q_{1}\le Q_{\, 2}$. But this contradicts
$Z_{1}\le Q_{\alpha -1}$, $Z_{1}\not\le Q_{\, \alpha}$. So we have $\hat{V}_2Z_1 \not= V_2$.
 We have $[Q_{\, 2},Z_{1}]\le Z_{1}$, and so
$[V_{2}/\hat{V}_{2} ,Q_{\, 2}]\le Z_{1}\hat{V}_2/\hat{V}_2$, as $[A,Q_2] \le \hat{V}_2$. As $\hat{V}_{2}
Z_{1}\not= V_{2}$, we have $\hat{V}_{2} Z_{1}\not\geq A$. This
now shows that $[V_{2},Q_{\, 2}]\cap A\le \hat{V}_{2}$. Hence
we have $[V_{2},Q_{\, 2}]\le \hat{V}_{2}$.

Set $X=\bC_{Q_{\, 2}} (\hat{V}_{2} )$. Assume $[X,A]\not= 1$. Choose
$1\not= v\in\hat{V}_{2} \cap [A,X]$. Let $F$ be a hyperplane of $\hat{V}_{2}$
not containing $v$. Then $[A/F,X]=\hat{V}_{2}/F$.

Now we get $\bC_{V_{2}/F} (X)=\hat{V}_{2} /F$ as $[V_2,O^{2'}(L_2)] \le A$. Hence $|X:\bC_{X}
(Z_{1})|\geq |Z_{1}:Z_{1}\cap\hat{V}_{2}|$ and so $Z_{1}$ is an
F-module with offending subgroup $X/\bC_{X} (Z_{1})$. We have
$[X,Z_{1}]\le \hat{V}_{2}$. Hence $[Z_{1},X,X]=1$. This shows
that $X$ acts quadratically. By (1.11) $X/\bC_{X} (Z_{1})$ induces
orbits of lenght $\leq 2$ on the components of $L_{1}/Q_{1}$. Suppose $[X, E(L_1/Q_1)] \not= 1$.  Hence
again by (1.11) and as $X \unlhd S$ there is some component $K$ normalized by $X$ such
that $X/\bC_{X} (K)$ offends on $Z_{1}$. Now (5.1) provides us with
some submodule $U_{1}$ of $Z_{1}$ such that every $u\in U_{1}^{\,\sharp}$
is centralized by some $E\le M$, $E$ elementary abelian of order
$p^{\, 2}$ and $\Gamma_{E,1}(G) \leq M$, or $L_{1}$ possesses a
component $K\cong L_{2}(q)$, or $U_{4}(2)$ and $U_{1}$ the natural
module. 

As $\bC_{U_{1}} (X)\not= 1$, we get $U_{1}\cap \hat{V}_{2}\not= 1$,
and then $O_{{2}^{\prime}} (L_{2})\le\bC (U_{1}\cap\hat{V}_{2}
)\le M$ in the first case, a contradiction. Hence we just have the latter. By
(5.1) we have $\bC_{U_{1}} (K) \not= 1$. But $\bC_{U_{1}}
(K) = 1$, as $\bC_{U_{1}} (K)\cap\hat{V}_{2}\not= 1$, we have a
contradiction.

So we have $[E(L_1/Q_1), X] = 1$. We get $p=3$. By (4.10) we have $p \in \sigma(M)$. As
$\bC_{Z_{1}} (X)\le\hat{V}_{2}$, we get with \cite[(24.1)]{GoLyS} that
$|X/\bC_{X} (K)|\leq 4$. Now there is a $p$--group $K$ $\in Syl_2(F(L_1/Q_1))$ with $[K,X] \not= 1$. There is some element $\rho$ of order 3  in $K$ with $|[Z_1, \rho]| = 4$ and $|[Z_1, \rho] \cap \hat{V_2} \not= 1$. Then there is some $E$ elementary abelian of
order 9 with $\Gamma_{E,1} (G)\leq M$ and $[[Z_1,\rho] \cap \hat{V}_2, E] = 1$, contradicting 
$L_{2}\not\leq M$.

We have shown
\\
$$X=\bC_{Q_{\, 2}} (A)\, .$$
\\
\indent
As $[O^{\, 2^\prime} (L_{2}/Q_{\, 2}),Q_{\, 2}]\leq X$, we see that
$A/\hat{V}_{2}\le Z(Q_{\, 2})\cap V_{2}/\hat{V}_{2}$. In
particular $Z(Q_{\, 2})$ is a nontrivial $O^{2'}(L_{2}/Q_{\, 2})$-module. But
now we may argue as in (6.15).


So we have $V_{2}\not= AZ_{1}$. As $V_{2}=\langle Z_{1}^{\,
L_{2}}\rangle$, there is some nontrivial irreducible module in
$V_{2}/A$. This proves the lemma.
\\

\absa
Let $[V_{\alpha}\cap Q_{\, 2},Z_{\beta} ]=1$ for all
$\beta\in\triangle (2)$ with $Z_{\beta}\not\leq Q_{\,\alpha}$.
Suppose furthermore $[V_{2}\cap Q_{\alpha} ,Z_{\delta} ]=1$ for all
$\delta\in\triangle (\alpha )$ with $Z_{\delta}\not\le Q_{\,
2}$. Let $\gamma\in\triangle (2)$ with $[V_{\alpha} \cap Q_{\,
2},Z_{\gamma} ]\not= 1$. Then $Z_{\gamma}\le Q_{\,\alpha}$ and
so there is some $\varepsilon\in\triangle (\alpha )$ with $[Z_{\gamma}
,Z_{\varepsilon} ]\not= 1$. Now $Z_{\gamma}\leq Q_{\,\alpha}\leq
L_{\varepsilon}$ and $Z_{\varepsilon}\leq Q_{\, 2}\leq L_{\gamma}$, a
contradiction to (6.1). So we have $[V_{2},V_{\alpha}\cap Q_{\,
2}]=1=[V_{2}\cap Q_{\,\alpha} ,V_{\alpha}]$. Hence $V_{2}$ is an
F-module for $L_{2}/Q_{\, 2}$ with offending subgroup $V_{\alpha}
/V_{\alpha}\cap Q_{\, 2}$. By (1.31) $V_{2}$ involves exactly one
irreducible nontrivial $O^{\, 2^\prime} (L_{2}/Q_{\, 2})$-module,
contradicting (6.16). So we have 

\absa
{\bf (6.17)~Lemma.~}{\it Let $b=b_{1}>1$. Then we may choose notation
such that $[V_{\alpha}\cap Q_{\, 2},Z_{1}]\not= 1$.}

\absa
For what follows it is very important to show
$V_{\alpha}\le M$. For this it is enough to show that there is
some $1\not= W\le V_{\alpha}\cap L_{1}$ such that $W$ is
normalized by $E$, $E$ elementary abelian of order $p^{\, 2}$ with
$\Gamma_{E,1} (G)\leq M$. In this case we get
$V_{\alpha}\le\bC_{G} (W)\le M$.

\absa
If (6.4) holds then as before we choose $\omega\in L_{1}\cap L_{2}$,
$o(\omega )=x$, where $x$ is minimal with $x \, \Big | \,r-1$, i.e. it does not divide any smaller power of two minus one, and $x=9$ for
$r=64$. According to (1.66) we may choose $\omega$ with $N_G(\langle \omega \rangle) \not\leq M$. Fix $\alpha + 1 \in \Delta(\alpha)$ with $[Z_1, Z_{\alpha+1}] \not= 1$, $Z_1 \not\leq M_{\alpha+1}$.

\absa
{\bf (6.18)~Lemma.~}{\it Assume (6.4), $b=b_{1}>1$. Let $K$ be
a component of $L_{1}/Q_{1}$. Then $[\omega ,K]\le
K$ or $[(V_{\alpha}\cap L_{1})Q_1/Q_1, K^{\langle\omega\rangle} ] = 1$.}

\absa
Proof~. Suppose false. Let $K_{1},\ldots ,K_{y}$ be some orbit under
$\langle\omega\rangle$, i.e. $y=x$ or $y=3$ and $x=9$.

Let $y=x$. Then by (1.63) we get $m_{p}(K)=1$ for every prime $p \,\Big | \,|K|$.
So with (1.5) we have $K\cong L_{2}(q)$, $Sz(q)$ or $J_{1}$.

If $y=3$ and $x=9$, then $3\not\in\sigma (M)$, so $m_{3}(K)=1$. Hence by (1.3) we see $K\cong L_{2}(q)$, $Sz(q)$, 
$(S)L_{3}(q)$ or $(S)U_{3}(q)$. Let $\hat{K}_{1} =\bC_{K_{1}}
(\omega^{\, 3} )$. If $p \,\Big | \,|\hat{K}_{1} |$, then $p\not\in\sigma
(M)$, $p$ prime. 

Suppose $K_{1}=(S)L_{3}(q)$ or $(S)U_{3}(q)$. If
$K_{1}\cong (S)U_{3}(q)$, then $\omega^{\, 3}$ centralizes some
$\gamma\in K_{1}, o(\gamma ) \,\Big | \,q+1/3$. This implies $q=8$, a
contradiction. Suppose $
K_{1}\cong (S)L_{3}(q)$. Then $\omega^{\, 3}$ is not inner $\times$
diagonal, so $\omega^{\, 3}$ induces a field automorphism. Now it
centralizes $(S)L_{3})(\sqrt[3]{q} )$ and so $\sqrt[3]{q} =2$, i.e.
$q=8$. But $m_{7}((S)L_{3}(8))\geq 2$ where $7 \,\Big | \,|SL_{3}(2)|$, a
contradiction. So in any case we have $K\cong L_{2}(q)$, $Sz(q)$ or $J_{1}$.
Further $K^{\langle \omega \rangle} \unlhd L_1/Q_1$.

Let $x\in V_{\alpha}\cap L_{1}$, $[x,K^{\langle\omega\rangle} ]=1$ and
$[x,Z_{1}]\not= 1$. But now $\bC_{L_{1}/Q_{1}} (K^{\langle\omega\rangle} )\not= 1$ and as $[(V_\alpha \cap L_1)Q_1/Q_1, K^{\langle \omega \rangle}] \not= 1$ we get by quadratic action 
$\bC_{Z_{1}} (K^{\langle\omega\rangle} )\not= 1$. As $\bC_{Z_1}(K^{\langle \omega \rangle}) \unlhd S$ there is
some $1\not= t\in\Omega_{1} (Z(S))$ such that $t$ is centralized by
some elementary abelian subgroup $E$ of order $p^{\, 2}$ in $L_{1}$ with
$\Gamma_{E,1} (G)\leq M$,  a contradiction to $L_2 \not\subseteq M$. This gives $\bC_{(V_{\alpha}\cap L_{1})Q_1/Q_1} (K^{\langle\omega\rangle}
)=1$. 

Let $y>3$. Then every $p$ with $p \,\Big | \,|K|$
is in $\sigma (M)$. Let first $3\in\sigma (M)$. By (6.4) we get $Z_{1}=\langle t \,\Big |  \,|V_{\alpha}
:\bC_{V_{\alpha}} (t)|\leq sr^3\rangle$, where $s = q$, if $q$ is even, $s = 2$ for $K \cong J_1$ and $s = 4$ for $q$ odd .   Now we get some $E$ with $E$ elementary abelian of
order $p^{\, 2}$ and $\Gamma_{E,1} (G)\leq M_{\alpha +1}$ such that
$|\bC_{Z_{\alpha +1}} (E)|\geq q^{2(y-3)}$. As $Z_{1}\not\le
M_{\alpha +1}$, we see $q^{2(y-3)}\leq qr^3$ and so $q^{2y-5}\leq r^3$.
As $r\leq 2^{\, y-1}$ we get $q^{2y-5}\leq 2^{3(y-1)}$. As
$q\geq 4$ we have $4y-10\leq 3y-3$, i.e. $y\leq 7$. Hence $y=5$ or $y = 7$.
Furthermore we have that $r^3$ divides $|E(L_2/Q_2)|$ which shows $E(L_2/Q_2) \cong Sp_4(r)$. Then by (1.3) $3 \not\in \sigma(M)$. Hence $K \cong Sz(q)$ and so $q \geq 8$. This now shows even $6y - 15 \leq 3y -3$ and so $y \leq 4$, a contradiction.

 So we have  $y=3$ and then  $3\not\in\sigma (M)$. This implies $x=9$ or
$x=3$, so $r=64$ or $r=4$.

We have $K^{L_1 \cap L_2} = K^{\langle \omega \rangle}$ otherwise we may argue as before. Let  $\bZ_{9}\times \bZ_{9} \cong F\le L_{1}\cap L_{2}$.  As $3 \not\in \sigma(M)$, we see $m_{3}(\langle K , (L_{1}\cap L_{2})/Q_1\rangle )\leq 3$. Hence $F$ acts faithfully on $K^{\langle \omega \rangle}$. But then $\bC_{K_{1}^{\langle\omega\rangle}}
(\omega )$ contains an elementary abelian group of order 9 and
 we get the
contradiction $m_3(K^{\langle \omega \rangle}) \geq 6$. . The same argument shows that  $E(L_{2}/Q_{\, 2}) \not\cong
SL_{3}(4) * SL_{3}(4)$ as $L_{1}\cap L_{2}$ does not contain an
elementary abelian group of order 27.

So we have $|V_{2}:V_{2}\cap Q_{\, \alpha} |\leq r^{\, 2}=16$ or
$r=64$ and $|V_{2}:V_{2}\cap Q_{\alpha} |\leq r=64$. 

Let $[Z_{\alpha +1}\cap L_{1},Z_{1}]=1$. There is some $E\leq
M_{\alpha +1}$, $E$ elementary abelian of order $p^{\, 2}$ with
$\Gamma_{E,1} (G)\leq M_{\alpha +1}$, and $|\bC_{Z_{\alpha +1}}
(E)|\geq 64$, as $p\geq5$. This shows $p=7$ ,$r=64$ and $|Z_{\alpha+1} : Z_{\alpha+1} \cap L_1| = 64$. But now there
is $\nu\in L_{1}\cap L_{2}$, $o(\nu )=7$, $[K,\nu ]\le
K$. So $m_{7}(K)=1$ and $\nu$ centralizes some elementary 
abelian group of order $7^{\, 3}$ in $L_{1}$, which shows
$N_{G}(\langle\nu\rangle )\le M$. But as $L_{2}=\langle
L_{2}\cap M,N_{L_{2}}(\langle\nu\rangle )\rangle\le M$, we
have a contradiction.

Thus $[Z_{\alpha +1}\cap L_{1},Z_{1}]\not= 1$. Suppose next that
$[[K_{1},Z_{1}],K_{2}\times K_{3}]=1$. Then as we may assume $[Z_{\alpha +1}\cap
L_{1},[K ,Z_{1}]]\not= 1$, we get $V_{\alpha}\le L_{1}$. Now
$Z_{1}=\langle t \,\Big | \,|Z_{\alpha +1}:\bC_{Z_{\alpha +1}} (t)|\leq
2^{m_{2}(K)}\rangle$. Hence $|[K,Z_{1}]|\leq 2^{\,
m_{2}(K)}$, which is not possible. 

So we have that $[[K_{1},Z_{1}],K_{2}\times K_{3}]\not= 1$. In
particular there are at least 4 nontrivial irreducible
$K_{1}$-modules involved. By symmetry we have some $s\in Z_{\alpha +1}\cap L_{1}$, $[s,K_1] \not= 1$ with
$1 \not= |Z_{1}:\bC_{Z_{1}} (s)|\leq 2^{\, m_{2}(K_{1})}\cdot 64$. Hence
there is some irreducible module $V$ for $K_{1}$ such that for some
involution $s\in\mbox{Aut~} (K_{1})$ we have $|V:\bC_{V} (s)|\leq
2^{\, m_{2}(K_{1})/4}\cdot\sqrt[4]{64}$. 

Let $K_{1}\cong L_{2}(p)$ or $J_{1}$, $p$ odd. Then $|V:\bC_{V}
(s)|\leq 4$. This implies with (1.30)(a) $K_{1}\cong L_{2}(5)$ or $L_{2}(7)$. Suppose $K_{1}\cong L_{2}(q)$, $q$ even. Then $|V:\bC_{V} (s)|\geq
q$. So $q^{\, 3}\leq 64$, i.e. $q\leq 4$. If $K_{1}\cong Sz(q)$, then $|V:\bC_{V} (S)|\geq q^{\, 2}$ and so
$q^{\, 7}\leq 64$, a contradiction.

Let $K_{1}\cong L_{2}(5)$. Then $r=64$. But now some $\nu\in
L_{1}\cap L_{2}$, $o(\nu )=7$, centralizes
$K_{1}^{\,\langle\omega\rangle}$, and so $N_{G}(\langle\nu\rangle
)\le M$ by (1.63) as $O_{2}(M)\cap N_{G}(\langle\nu\rangle
)\not= 1$, a contradiction.

So we have $K_{1}\cong L_{2}(7)$. But in this case at least 6
nontrivial modules are involved. Hence also $r=64$ and then again 
$N_{G}(\langle\nu\rangle )\le M$.

\absa
{\bf (6.19)~Lemma.~}{\it Assume (6.4), $b=b_{1}>1$. Let $K/Z(K)\cong
A_{n}$, with $m_p(K) \ge 2$, be a component of $L_1/Q_1$. Then $[(V_{\alpha}\cap L_{1})Q_1/Q_1,K] = 1$.}

\absa
Proof~. Suppose false. By (6.18) $[\omega ,K]\le K$. By (1.63) we have $[\omega ,K]\not= 1$. Hence by (1.9) $o(\omega )=3$
and so $r=4$. Furthermore $3\not\in\sigma (M)$ and then $n\leq 11$.
As $m_{p}(K)\geq 2$, we get $n=10$ or $11$ and $p=5$. Now we see that $KO_{3'}(M)/O_{3'}(M)$ $\leq E(M/O_{3'}(M))$. The construction of $G_1$ in \S 4 tells us that $KO_{3'}(M)/O_{3'}(M) = E(M/O_{3'}(M))$. Hence we have that a Sylow 3--subgroup of $M$ is centralized by some $E \cong E_{5^2}$ with $\Gamma_{E,1}(G) \leq M$. As $[\omega, Z_2] = 1$, we see with (1.63) that $N_G(\langle \omega \rangle) \leq M$, a contradiction.


\absa
{\bf (6.20)~Lemma.~}{\it Assume (6.4) with $b=b_{1}>1$. Let $K/Z(K)$
be a component of $L_{1}/Q_{1}$. If $K/Z(K)$ is a sporadic simple
group with $m_{p}(K)\geq 2$, then $[(V_{\alpha}\cap L_{1})Q_1/Q_1,K] = 1$.}

\absa
Proof~. Suppose false. By (6.18) $[\omega, K] \leq K$ and then $[\omega ,K]\not= 1$ by (1.63). By (1.9)
$o(\omega )=3$ or $o(\omega )=7$ and $K\cong J_{1}$. The latter is
not possible as $m_{p}(K)\geq 2$. So we have $r=4$ and $3\not\in\sigma
(M)$.

As $m_3(K) \leq 3$ we get with (1.3) $K/Z(K) \cong$ $J_1$, $M_{11}$, $M_{12}$, $M_{22}$, $M_{23}$, $M_{24}$, $J_2$, $HiS$, $He$, $Ru$, $J_4$, $O'N$, $J_3$. As $m_p(K) \geq 2$, we get $K/Z(K) \cong$ $J_2$, $J_3$, $J_4$, $HiS$, $He$, $Ru$, or $O'N$. Finally $p = 5, 7$ or $11$. 

As $K$ cannot centralize any element in $\bC_{Z_1}(S)^\sharp$ we get that $(V_\alpha \cap L_1)Q_1/Q_1$ acts faithfully on $K$. Let first $|Z_{\alpha +1}\cap L_{1}:Z_{\alpha +1}\cap Q_{1}|\leq 2$.
 As there is $E\le M_{\alpha +1}$ and $|\bC_{Z_{\alpha +1}}
(E)|\geq 2^{\, 6}$ we get $|Z_{\alpha +1}:\bC_{Z_{\alpha +1}} (Z_{1})|\geq 2^{\, 6}$. But this contradicts (1.17).

So we have $|Z_{\alpha +1}\cap L_{1}:Z_{\alpha +1}\cap Q_{1}|\geq 4$.
As $\bC_{Z_{\alpha +1}\cap L_{1}/Z_{\alpha +1}\cap Q_{1}} (K)=1$, we
get with (1.12)
$$K\cong J_{2} \mbox{~and~} p=5\, .
$$

Now $|Z_{\alpha +1}:\bC_{Z_{\alpha +1}} (Z_{1})|\leq 2^{\, 6}$ by
(1.12)(i) and so every $t\in Z_{1}$ centralizes some $x\in Z_{\alpha +1}$
such that $x$ is centralized by some elementary abelian $E,
E\le M_{\alpha +1}$, $|E|=25$ and $\Gamma_{E,1} (G)\leq M$. Hence
$t\in\bC_{G} (x)\le M_{\alpha +1}$. This implies the
contradiction $Z_{1}\le M_{\alpha +1}$.

\absa
{\bf (6.21)~Lemma.~}{\it Assume (6.4) with $b=b_{1}>1$. Let $K/Z(K)$
be a component of $L_{1}/Q_{1}$, $K/Z(K)=G(q)$ be a group of Lie type over a
field $GF(q)$ of odd characteristic. If $m_{p}(K)\geq 2$, then
$[(V_{\alpha} \cap L_{1})Q_1/Q_1,K] = 1$.}

\absa
Proof~. Suppose false. Let $|V_\alpha \cap L_1 : \bC_{V_\alpha \cap L_1}(K)| \geq 4$. Then by (1.12) $K/Z(K) \cong$ $U_4(3)$, $A_6$ or $U_3(3)$, and so $3 \in \sigma(M)$. As $[\omega, K] = 1$, we get a contradiction. So we have $|V_{\alpha}\cap
L_{1}:\bC_{V_{\alpha}\cap L_{1}} (K)|=2$. Let $K_1 * \cdots * K_u = K^S$. Then as $L_2 \not\subseteq M$, we have $V_\alpha \cap L_1/V_\alpha \cap Q_1$ acts fatihfully on $K^S$. Hence we get that $Z_1 = \langle t~\Big |~|V_\alpha \cap L_1 : \bC_{V_\alpha \cap L_1}(t)| \leq 2 \rangle$. If $V_\alpha \leq L_1$ then even $Z_1 = \langle t ~\Big | ~|[V_\alpha,t]| \leq 2 \rangle$. By (6.16) then $Z_1 \leq Q_\alpha$, a contradiction. So we have $V_\alpha \not\subseteq L_1$ and $\bC_{V_\alpha \cap L_1}(K) = V_\alpha \cap Q_1$. Hence $|V_{\alpha} \cap L_{1}:V_{\alpha} \cap Q_{1}|=2$. Now $|V_{\alpha}
:\bC_{V_{\alpha} } (Z_{1})|\leq 2r^{\, 2}$ by (1.17). By (6.16) and (1.44) $E(L_{2}/Q_{\, 2})\cong L_{2}(r)\times L_{2}(r)$ or $L_2(r)$ and
$|Z_{1}:Z_{1}\cap Q_{\alpha} |\leq r$. But in the latter by (6.16) we would get $|V_\alpha : \bC_{V_\alpha}(Z_1)| \geq r^2$, which contradicts $|V_\alpha : V_\alpha \cap L_1| \leq r$.  As $V_2 \not\leq M_{\alpha = 1}$ we see that $|V_2 : V_2 \cap Q_{\alpha+1}| \leq 2$ too and so $|V_{2}:\bC_{V_{2}}
(Z_{\alpha +1})|\leq 2r^{\, 2}$, we get $|V_{2}:V_{2}\cap Q
_{\alpha }|\geq \frac{1}{2} r^{\, 2}\leq |V_{\alpha} :V_{\alpha} \cap
Q_{\, 2}|$. Hence there is $\beta\in\triangle (2)$ such that $V_{\alpha}
\le G_{\beta}$, and $Z_{\beta} Q_{\alpha}\not= Z_{1}Q_{\alpha}
,Z_{\beta}\not\le Q_{\alpha}$. Now we get $|V_{\alpha} :\bC_{V_{\alpha} }
(Z_{1}Z_{\beta}) |\leq 4r^{\, 2}$. By (6.16) and (1.44) we may choose $Z_\beta$ such that  $|V_{\alpha} :\bC_{V_{\alpha}}
(Z_{1}Z_{\beta} )|\geq r^{\, 4}$, a contradiction.

\absa
{\bf (6.22)~Lemma.~}{\it Assume (6.4) with $b=b_{1}>1$. Let $K$ be a
component of $L_{1}/Q_{1}$ with $K/Z(K)=G(q)$ a group of Lie type over a
field of characteristic two, $q>2$, If $m_{p}(K)\geq 2$ then $[(V_{\alpha}
\cap L_{1})Q_1/Q_1,K] = 1$.}

\absa
Proof~. Suppose first that $\omega$ induces a field automorphism on
$K$. Then $q=t^{\, x}$ and $\omega$ centralizes $\tilde{K}\cong
G(t)$. By (1.63) $m_{p}(\tilde{K} )\leq 1$. Then $K\cong L_{n}(q)$,
$n\leq 7$, where $t=2$ for $n>4$ , $Sp_4(q)$ or $Sp_{6}(q)$, $G_{\,
2}(q)$, $U_{n}(q)$, $n\leq 7$, $\Omega_8^{\, -} (q)$, $^{2}F_{4}(q)$
or $^3D_{4}(q)$. 

Suppose that $\omega$ induces an inner $\times$ diagonal automorphism.
Then $\omega$ centralizes some Cartan subgroup $C$.
Hence $m_{x}(C)\leq 3$ and $m_{x}(C)\leq 2$ for $m_{x}(K)\geq 4$. This
shows with (1.2) that $K$ is one of the groups above.

Suppose that $K\not\!\!\unlhd L_{1}/Q_{1}$. Then in the first case
$m_{\ell}(\tilde{K} )\leq 1$ for all prime $\ell$, so $\tilde{K}\cong L_{3}(2)$ and $K\cong
L_{3}(2^{\, x})$. Replace $K$ by $\langle K^{\, L_{1}}\rangle$.
We have that $\langle K^{L_{1}}\rangle$ contains at most three
components.

In the second case we get $m_{s}(C)\leq 1$ for any prime $s$. But this implies that $K/Z(K)\cong
U_{3}(q)$ or $L_{3}(4)$. If $K/Z(K) \cong L_3(4)$ we have $3\in\sigma (M)$, a
contradiction as $o(\omega )=3$. So we have the former. But now $x$
centralizes some $E,E\cong E_{p^2}$, $p \,| \,q+1$, a contradiction too.

So we have
\begin{itemize}
\item[(1)] $K\lhd L_{1}/Q_{1}$ or $K\cong L_{3}(2^{\, x})$ and there
are at most three conjugates.
\end{itemize}

Next we show

\begin{itemize}
\item[(2)] $[Z_{1}\cap Q_{\alpha} ,V_{\alpha} ]\not= 1\, .$
\end{itemize}

Otherwise $V_{\alpha}$ centralizes a subgroup of index $r^{\, 2}$ in
$Z_{1}$ by (1.17). Application of (1.22) shows that $Z_{1}$ has a
submodule $T\cong V(\lambda )$, $\lambda$ some weight, or $r>q$ and
$K\cong U_{3}(q)$. 

Suppose $K\unlhd L_{1}/Q_{1}$. Let first $K \not\cong U_{3}(q)$, if $r>q$ and $T \not\cong V(\lambda)$.
Suppose furthermore $T\unlhd L_{1}$. Then $\Omega_{1} (Z(S))\cap T\not=
1$. Hence there is no $E,E\cong E_{p^2}$ with $\Gamma_{E,1} (G)\leq M$,
such that $\bC_{\bC_{T} (S)} (E)\not= 1$ as $L_2 \not\leq M$.

As $p \,| \,q^{\, 2}-1$, we see that $K\cong L_{4}(q)$, $L_{3}(q)$,
$Sp_{4}(q)$, $Sp_{6}(q)$, $G_2(q)$, $U_{n}(q)$, $n\leq 5$ or $\Omega_8^{\, -}
(q)$.

In any case there is some $\mu\in K,o(\mu )=p$, such that $[\bC_{T}
(S),\mu ]=1$. By (4.4) we get $m_p(L_1) \geq 4$. Further we see that on $\bC_T(S \cap K)$ there is just a cyclic group induced. But this now implies that we 
have $K\cong U_{5}(q)$, $p \,| \,q+1$. Hence $T$ is not the natural module. As
$r^{\, 2}\leq q^{\, 4}$ this contradicts (1.34)(i).

So we have $T_{1}\bigoplus T_{2},T_{i}\cong T$ or $T^{\,\star}$ as $GF(2)$-modules,
involved in $Z_{1}$. Now $V_{\alpha}\cap L_{1}$ induces $GF(q)$--transvections
on $T_{i}$, i.e. $T$ is the natural module for $L_{n}(q)$,
$Sp_{2n}(q)$ and $q=r$, $\Omega_6^{\, -}(q)$, $\Omega_8^{\, -} (q)$, or $U_{n}(q)$ and  $q=r^{\,
2}$. 

Now as before we get $K\cong L_{3}(q)$, $L_{4}(q)$,
$Sp_{4}(q)$ or $U_{4}(q)$. Furthermore as $m_p(L_1) \geq 4$ we get that either $K\cong
U_{4}(q)$ and $p \,| \,q+1$, $K\cong L_{4}(q)$ or $L_{3}(q)$,
$T_{1}\bigoplus T_{2}\cong T\bigoplus T^{\,\star}$. In the latter we
also get that $p \,| \,q-1 = r-1$. Now $\omega$ centralizes some $F\cong
E_{p^2}$ with $\Gamma_{F,1}(G)\leq M$. By (1.63) we have
$N_{G}(\langle\omega\rangle )\le M$. 

So we are left with $K\cong U_{4}(q)$, $r=q^{\, 2}$, $p\,| \,q+1$.
Furthermore $T$ is the natural $U_{4}(q)$-- or $\Omega_6^{\, -}
(q)$--module. Now $L_{1}\cap L_{2}$ contains $\tilde{E}
=\langle\tilde{\omega}\rangle\times\langle\tilde{\omega}_1\rangle\cong
E_{p^2}$, contradicting (2.2). 

Assume now $K\cong U_{3}(q)$, $r>q$. Then $\omega$ induces a diagonal
automorphism and so $r=q^{\, 2}$. 
We get $p \,| \,q+1$ and $L_{1}\cap L_{2}$ contains an elementary abelian
group of order $p^{\, 2}$, a contradiction as before.

So we are left with $K\not\!\!\unlhd L_{1}/Q_{1}$ and  $K\cong L_{3}(2^{\, x})$. Now $r<q$
and so by (1.26)(i) $[K,Z_{1}]$ is the natural module. In particular
$[[K,Z_{1}],K^{\, g}]=1$ for $K^{\, g}\not= K,g\in L_{1}/Q_1$. But
then $[K^{\, L_{1}/Q_{1}},Z_{1}]\cap\Omega_{1} (Z(S))$ is centralized
by $E,E\cong E_{p^2}$ with $\Gamma_{E,1} (G)\leq M$, a contradiction.
This proves (2).


Now choose $\beta\in\triangle (\alpha )$ such that $[Z_{1}\cap Q_{\,
\alpha},Z_{\beta} ]\not= 1$.
\begin{itemize}
\item[(3)] Let $K \unlhd L_1/Q_1$ and $K_{\beta}$ be the component in $L_{\beta}$
corresponding to $K$ in $L_{1}$. If there is $t\in Z_{1}\cap Q_{\alpha}$ such
that $[t,K_{\beta} ]=1$, then $[t,Z_{\beta}] =1$.
\end{itemize}


Suppose false. We have $[K_{\beta} ,[t,Z_{\beta} ]]\le
[t,Z_{\beta} ]$ and so $V_2 \leq L_\beta$.  If $[K_{\beta}
,[t,Z_{\beta} ]]=1$, then $K$ centralizes some $x\in\Omega_{1}
(Z(S))^{\,\sharp}$, a contradiction. So we have 
$$[K_{\beta}
,[t,Z_{\beta} ]]\not= 1 \mbox{~and~} [K_{\beta}
,Z_{1}]=1.$$

 As $[V_{\alpha}\cap L_{1},K]\not= 1$, we get $\bC_{V_{\alpha}\cap
L_{1}/V_{\alpha}\cap Q_{1}} (K)=1$. Let $|V_{\alpha}\cap L_{1}:V_{\alpha}\cap Q_{1}|=q^{\, x}$. Then by
(1.17) $|[Z_{\beta} ,t]|\leq q^{\, x}r^{\, 2}$. If $K \cong$ $^3D_4(q)$, then $x \leq 5$ and by (1.64) and (1.65) $r \leq q^3$. This gives $|[Z_\beta,t]| \leq q^{11}$, contradicting (1.23). If $K \cong$ $^2F_4(q)$, then $r \leq q^2$ and so we have $|[Z_\beta,t]| \leq q^{10}$ a contradiction to (1.23) again. If $K \cong G_2(q)$, we have $r \leq q$ and so $[Z_\beta,t]| \leq q^5$, a contradiction.

So we are left with $K/Z(K) \cong$ $L_n(q)$, $Sp_4(q)$, $Sp_6(q)$, $U_n(q)$ or $\Omega_8^-(q)$. Let $K/Z(K) \cong \Omega_8^-(q)$, then $r \leq q^2$ and so $|[Z_\beta,t]| \leq q^xq^4$. We have $|[Z_\beta,t]| \geq q^8$ and so $x \geq 4$. By (1.15) and (1.14) there is some natural or spin module involved in $[Z_\beta,t]$. But for the latter the dimension is 16 and $x < 12$, a contradiction. So just one natural module is involved. But now $\bC_K(\bC_T(S))$ contains  some $E$, $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \leq M$, a contradiction.

If $K/Z(K) \cong U_6(q)$ or $U_7(q)$ then $x \leq 9$. Then we see as before that $x = 8$ and $K/Z(K) \cong U_6(q)$. But now again with (1.14) and (1.15) we get a contradiction.

By the same argument we get $T$ is the spin module for $K\cong
Sp_{6}(q)$. But then $|V_{\alpha}\cap L_{1}:V_{\alpha}\cap
Q_{1}|=q^{\, 6}$. The spin module does not admit a quadratic group of
order $q^{\, 6}$. We see that $K/Z(K) \not\cong U_5(q)$.  

If $K\cong U_{4}(q)$, and $T$ is the natural
module, then $r=q^{\, 2}$ and so $L_{1}\cap L_{2}$ contains an
elementary abelian group of order $p^{\, 2}$, a contradiction to (2.2). If $T$ is the orthogonal module, then $|V_{\alpha}\cap
L_{1}:V_{\alpha}\cap Q_{1}|\geq q^{\, 2}$ and so there is some $v\in
V_{\alpha}\cap L_{1}$ such that $[Z_{1},v]$ contains some $1\not= x$
with $\bC_{K} (x)$ involves $Sp_{4}(q)$. So $x$ is centralized by
$E\cong E_{p^2}$ with $\Gamma_{E,1} (G)\leq M$, which gives
$V_{\alpha}\le \bC_{G} (x)\le M$. But this contradicts
(1.44) and (6.16), as $|V_{\alpha}\cap L_{1}:V_{\alpha}\cap
Q_{1}|\leq q^{\, 3}<r^{\, 2}$. 

Let $K\cong Sp_{4}(q)$. Then $T$ is the natural module and
$|V_{\alpha}\cap L_{1}:V_{\alpha}\cap Q_{1}|\geq q^{\, 2}$. By (1.16)
$V_{\alpha}\cap L_{1}$ normalizes $T^{\, g}$ for every $g\in L_{1}$.
As $|T| = q^4$, we see that $V_\alpha\not\leq M$. Hence we have with (4.4) that $m_p(L_1) \geq 4$. Let $U$ be the kernel of the action of $L_1$ on $\{T^g ~\Big |~g \in L_1\}$. Then we see that $m_p(U) = 2$ as $V_\alpha \cap L_1 \leq U$. Now we also see that $\bC_U(T)$ is solvable. Let $U_1$ be the largest solvable normal subgroup of $U$. Then $[O^{2'}(U_1), T] = 1$ and so $O^{2'}(U_1)$ centralizes $T^g$ for all $g$ in question. As $p \not\!\Big|~|U_1|$ this implies that there is some $t \in V_\alpha \cap L_1 \setminus Q_1$ such that $tO^{2'}(U_1)$ is centralized by some $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \leq M$, and so $[\langle T^g \Big |~g \in L_1 \rangle, t]$ is normalized by $E$, i.e. $V_\alpha \leq L_1$, a contradiction.


So we have $K\cong L_{n}(q)$. If $p \,| \,q-1$, we see that $n\leq 4$,
otherwise for some $1\not= x\in \Omega_{1} (Z(S))$ we have $\bC_{G}
(x)\le M$. Furthermore for $n\geq 5$, we have $p \,| \,q-1$, so we
have $n\leq 4$ anyway.

Let $K\cong L_{4}(q)$, then $T$ is the natural or dual or the
orthogonal module. In the latter some $1\not=x\in \bC_{T} (S)$ is
centralized by $E,E\cong E_{p^2}$ with $\Gamma_{E,1}(G) \leq M$ and  $E\le K$, a contradiction. So we have that $T$ is the natural module. Furthermore as
$V_{\alpha} \not\le L_{1}$ by (6.1), we get $p \,| \,q+1$, i.e. $m_{p}(K)=2$. Now by (1.16)
we see that $(V_{\alpha}\cap L_{1})Q_{1}/Q_{1}$ normalizes some conjugate of $T$ in $L_1$  as
$|V_{\alpha}\cap L_{1}:V_{\alpha}\cap Q_{1}|\geq q^{\, 2}$. But now we may argue as in the case of $Sp_4(q)$. 

So we have $K\cong L_{3}(q)$, $p \,| \,q-1$ and $T$ is the natural module. If
$|V_{\alpha}\cap L_{1}:V_{\alpha}\cap Q_{1}|>q$ we may argue as
before. So assume $|V_{\alpha}\cap L_{1}:V_{\alpha}\cap Q_{1}|\leq q$. Thus we must have  
$r=q$. Now $L_{1}\cap L_{2}$ contains an elementary abelian group $E$ of
order $p^{\, 2}$. As $m_p(L_1) = 4$, we see that $\Gamma_{E,1}(G) \leq M$, a contradiction.

%So we are left with $K\not\!\!\unlhd L_{1}/Q_{1}$, i.e. $K\cong L_{3}(2^{\,
%x})$. Furthermore $p \,| \,2^{\, x}-1$. Suppose
%$\langle\omega\rangle\times\langle\omega_{1}\rangle\le
%L_{1}\cap L_{2}$, $o(\omega )=x=o(\omega_{1} )$. Then we may assume
%that $\omega_{1}$ induces an inner $\times$ diagonal automorphism on $K$. But
%then $\omega_{1}$ centralizes some $E\cong E_{p^2}$, $E\le
%K$, a contradiction.

%So we get that $|V_{\alpha}:V_{\alpha}\cap L_{1}|\leq r$. Again
%$Z_{1}\le L_{\beta}$ and by (4.16) $V_{\alpha}\le L_{1}$,
%so $\bC_{V_{\alpha}\cap L_{1}/V_{\alpha}\cap Q_{1}} (K)=1$. Furthermore %$|V_{\alpha}\cap L_{1}:V_{\alpha}\cap Q_{1}|\leq q^{\, 2}$
%and so $|[Z_{\beta} ,t]|\leq q^{\, 2}r<q^{\, 3}$. Hence
%$[[t,Z_{\beta} ],K_{\beta} ]=1$. Let $K_{\beta}^{\, g}\not=
%K_{\beta}$, then we may assume $[t,K_{\beta}^{\, g} ]\not= 1$,
%otherwise $[t,Z_{\beta} ]$ is centralized by $\langle K_{\beta}^{\,
%L_{\beta} }\rangle$, a contradiction to $\bC_{G} (x)\not\leq M$ for
%$x\in\Omega_{1} (Z(S))$. 

%Now we have $[[K_{\beta} ,Z_{1}],K_{\beta}^{\, g} ]=1$. As
%$V_{\alpha}\cap L_{1}$ normalizes $[K,Z_{1}]$ and $m_{p}(K)=2$, we
%get $V_{\alpha}\le M$, a contradiction.

\begin{itemize}
\item[(4)] Let $K\lhd L_{1}/Q_{1}$. Then $Z_{1}$ is either a strong
quadratic module or an SC-module, or $Z_{1}$ is the natural module
for $K\cong U_{3}(q)$.
\end{itemize}

\indent
Suppose false. Let $t \in Z_1 \cap Q_\alpha$ with $[t, Z_\beta] \not= 1$. By (3) $[t, K_\beta] \not= 1$.  If $K\cong L_{n}(q)$, then we get $|[Z_{\beta} ,t]|>
q^{\, (\frac{n}{2} )^{\, 2}}$ or $q^{\,\frac{(n^{\, 2}-1)}{4}}$,
according to $n$ is even or odd. This shows $|V_{\alpha}\cap
L_{1}:V_{\alpha}\cap Q_{1}|>q^\frac{n^{\, 2}-8}{4}$ or $q^\frac{n^{\,
2}-9}{4}$. Hence (1.15) implies $n=3$ and $|V_{\alpha}\cap
L_{1}:V_{\alpha}\cap Q_{1}|\leq q$. As $r \leq q$ we see that $|Z_\beta : Z_\beta \cap Q_2| > r$. Now we have $\langle \omega \rangle \times \langle \omega_1 \rangle$ involved in $L_1 \cap L_2$. Hence we may assume that $\omega_1$ induces an inner $\times$ diagonal automorphism on $K$. As $p \Big |~q-1$, we get $N_G(\langle \omega_1 \rangle) \leq M$. Application of (1.66) shows $E(L_2/Q_2) \cong (S)L_3(r)$ or $Sp_4(r)$. We have $|V_2 : \bC_{V_2}(Z_\beta)| \geq r^6$ by (6.16). In particular $|V_2 \cap L_\beta : V_2 \cap Q_\beta| \geq r^4$. As $r^2q > q^2$, i.e. $r > q$, we see that $(V_2 \cap L_\beta)Q_\beta/Q_\beta \cap \bC(K_\beta) \not= 1$. Hence $\bC_{Z_\beta}(K_\beta) \not= 1$ and so as $K \unlhd L_1/Q_1$ we have that $\bC_{\Omega_1(Z(S))}(K) \not= 1$, contradicting $L_2 \not\leq M$. 


Let $K\cong U_{n}(q)$, $n\leq 7$. Now $|[Z_{\beta} ,t]|>q^{\,
(\lfloor\frac{n}{2} \rfloor)^{\, 2}}$. Hence $|V_{\alpha}\cap L_{1}:V_{\alpha}\cap
Q_{1}|>q^{\, (\lfloor \frac{n}{2} \rfloor )^{\, 2}-4}$. By (1.15) we get $n\leq 5$.
Suppose $n>3$. Then by (1.15) $|V_{\alpha}\cap L_{1}:V_{\alpha}\cap
Q_{1}|\leq q^{\, 2}$. This now implies $r>q$. Hence $r=q^{\, 2}$ and
$\omega$ induces some diagonal automorphism. We have that $p \Big |~r-1$. So there is $\Z_p \times \Z_p \cong E \leq L_1 \cap L_2$, which contradicts (2.2). So $|V_{\alpha} :V_{\alpha}\cap L_{1}|\leq
r=q^{\, 2}$, a contradiction too.

Let $K\cong U_{3}(q)$.  If
$r>q$, then $\omega$ induces a diagonal automorphism. So we have
$r=q^{\, 2}$. As $p \,| \,q+1$, we have some $\nu\in L_{1}\cap L_{2}$,
$o(\nu )=p$. But $\nu$ centralizes some $E\cong E_{p^ 2}$ with
$\Gamma_{E,1} (G)\leq M$ and $E\le K$. This gives $N_{G}(\langle\nu\rangle
)\le M$, a contradiction. So $r\leq q$. As before we see that $|V_\alpha : V_\alpha \cap Q_2| \leq r$ and now $|[Z_{\beta}
,t]|\leq q^{\, 2}$. By (1.26)(iii) $Z_{\beta}$ is the natural module.

Let $K\cong Sp_{4}(q)$ or $Sp_{6}(q)$. Then $|[Z_{\beta} ,t]|>q^{\,
3},q^{\, 6}$ respectively. Hence $|V_{\alpha}\cap
L_{1}:V_{\alpha}\cap Q_{1}|>q^{\, 4}$ in the latter, contradicting
(1.15). So we have $K\cong Sp_{4}(q)$ and $|V_{\alpha}\cap
L_{1}:V_{\alpha}\cap Q_{1}|>q$, which contradicts (1.15) too.

Let $K\cong \Omega^{\, -} (8,q)$, then $|[Z_{\beta} ,t]|>q^{\, 6}$.
Hence $|V_{\alpha}\cap L_{1}:V_{\alpha}\cap Q_{1}|>q^{\, 2}$. By (1.15) 
we have $|V_{\alpha}\cap L_{1}:V_{\alpha}\cap Q_{1}|\leq q^{\, 3}$.
So we get $r=q^{\, 2}$ and $|V_{\alpha} :V_{\alpha}\cap L_{1}|>r$.
But as $\omega$ induces a diagonal automorphism and $p \Big |~r-1$ we get a contradiction.

Let $K\cong G_{2}(q)$. Then
$|V_{\alpha}\cap L_1 : V_{\alpha}\cap Q_{1}|>q$, in which case $Z_{\beta}$ is strong quadratic by (1.15).

Let $K\cong$ $^{3}D_{4}(q)$. As $Z_{\beta}$ is not strong quadratic
we get with (1.15) that $|V_{\alpha}\cap L_{1}:V_{\alpha}\cap
Q_{1}|\leq q$. Then as $|[Z_{\beta} ,t]|>q^{\, 5}$, we get $r=q^{\, 3}$
and $\langle\omega\rangle\times\langle\omega_{1}\rangle\le
L_{1}\cap L_{2}$, $o(\omega )=o(\omega_{1} )$.
So we again get $N_G(\langle \omega_1 \rangle) \leq M$ and then $|V_2 : \bC_{V_2}(Z_\beta)| \geq r^6$. Hence $|V_2 \cap L_\beta : V_2 \cap Q_\beta| \geq r^4 = q^{12}$. This implies that $(V_2 \cap L_\beta)Q_\beta/Q_\beta \cap \bC(K_\beta) \not= 1$. Hence we have $\bC_{Z_\beta}(K_\beta) \not= 1$, a contradiction.

Let $K\cong$ $^{2}F_{4}(q)$. By (1.15) $|V_{\alpha}\cap L_{1}:V_{\alpha}\cap
Q_{1}|\leq q^{\, 2}$. So $|[Z_{\beta} ,t]|\leq q^{\, 4}$, a
contradiction. This proves (4).

As no element $1\not= x\in \Omega_{1} (Z(S))$ is centralized by
$E\cong E_{p^2}$, we now get one of the following for $K \unlhd L_1/Q_1$
\begin{itemize}
\item[(a)] $K\cong L_{n}(q),n\leq 4,Z_{1}$ contains the natural
module or the dual or $V(\lambda )\bigoplus V(\lambda ^{\,\sigma} )$ 
\item[(b)] $K\cong Sp_{4}(q),Z_{1}$ just involves the natural module
\item[(c)] $K\cong Sp_{6}(q),Z_{1}$ just involves spin modules
\item[(d)] $K\cong U_{n}(q),n\leq 5$. In case $n=5$ there is no
natural submodule involved
\item[(e)] $K\cong G_{2}(q)$ and the natural module is involved
\item[(f)] $K\cong$ $^{3}D_{4}(q)$ and the natural module is involved
\item[(g)] $K\cong\Omega_8^{\, -} (q)$ and just the spin module is
involved.
\end{itemize} 
\begin{itemize}
\item[(5)]Let $K\lhd L_{1}/Q_{1}$. If $|V_{\alpha}\cap L_{1}:V_{\alpha}\cap
Q_{1}|>q$, $q^{\, 2}$ in case (d) and (g) and $q^{\, 3}$ in case (f), then
$V_{\alpha}\le L_{1}$. This is also true if $\alpha$ is replaced by 2 and 1 is replaced by $\beta$.
\end{itemize}

By (1.16) $V_{\alpha}\cap L_{1}$ normalizes every $K$-submodule $T$. In
any case above we see that there is some $x \in V_{\alpha} \cap L_1$ such that  some $1 \not= U\le [x,Z_{1}]$ is normalized
by some $\mu\in K,o(\mu )=p$. If there is some $\mu_{1}\in M$ with
$[T,\mu_{1} ]=1$, then $V_{\alpha}\le L_{1}$.

So we may assume there is no such $\mu_{1}$. In particular $\langle
L_{1}\cap V_{\alpha} ,K\rangle Q_{1}/Q_{1}\cap \bC (K)=1$. Now let again $U_1$ be the kernel of the action of $L_1$ on $\{T^g \Big |~g \in L_1 \}$. Then $V_\alpha \cap L_1 \leq U_1$. Furthermore let $U_2 = \bC_{U_1}(T)$. Then $p \not\!|~|U_2|$ and $m_p(\mbox{Aut}(KO^{2'}(U_2)/O^{2'}(U_2))) \geq 4$. 

If $m_{p}(K)=2$, this implies that $U$ is normalized by $E_{p^2}$,
a contradiction. So $m_{p}(K)\geq 3$. By (4.12) no field
automorphism is induced.

This now implies $K\cong U_{5}(q),p \,| \,q+1$. By (1.54) $V_{\alpha}\cap
L_{1}/V_{\alpha}\cap Q_{1}$ contains some $u\in K$, $u$ a
transvection in the natural representation. But $[u,Z_{1}]$ is
normalized by $U_{3}(q)$ and so $N_{G}([u,Z_{1}])\le M$, i.e.
$V_{\alpha}\le L_{1}$.
\begin{itemize}
\item[(6)] $K\not\!\!\unlhd L_{1}/Q_{1}$
\end{itemize}

Suppose false. By (6.1) either $Z_1 \not\subseteq L_\beta$ or $Z_\beta \not\subseteq L_1$. As in then proof of (6) we will never use $V_2 \not\subseteq Q_\alpha$ we now by (5) have symmetry between 1 and $\beta$. So we may assume $Z_\beta \not\subseteq L_1$.  So we may
assume that $|V_{\alpha}\cap L_{1}:V_{\alpha}\cap Q_{1}|\leq q$,
$q^{\, 2}$ in case (d) and $q^{\, 3}$ in case (f). Hence for $t\in
Z_{1}\cap L_{\beta}$ we have 
\begin{center}
$|[Z_{\beta} ,t]|\leq qr^{\, 2},\; q^{\, 2}r^{\,2}\mbox{~~or~~} q^{\, 3}r^{\, 2}$
\end{center}

Suppose first that $Z_{\beta}$ and also $Z_{1}$ involve exactly one
irreducible nontrivial module $T_{1}$. Let first $K\cong$
$^{3}D_{4}(q)$. As $|[t,Z_{\beta} ]|\geq q^{\, 6}$, we see that
$r>q$. Suppose first that
$W = \langle\omega\rangle\times\langle\omega_1 \rangle\le L_{1}\cap
L_{2}$, $o(\omega )=o(\omega_{1} )$. As $r>q$, we see that $\bC_{W}
(K)\not= 1$ if all elements in $W$ induce inner $\times$ diagonal
automorphism on $K$. But then for some $1\not= \omega_{2} \in W$ we have
$N_{G} (\langle\omega_{2} \rangle )\le M$, by (1.63). This implies with (1.66) that $E(L_2/Q_2) \cong $ $(S)L_3(r)$ or $Sp_4(r)$ and so by (6.16) $|V_2 : V_2 \cap Q_\beta| \geq r^6$. Hence $|V_2 \cap L_\beta : V_2 \cap Q_\beta| \geq r^4$. As $r^2 \geq q^3$, we have $|V_2 \cap L_\beta : V_2 \cap Q_\beta| \geq q^6$ and then $(V_2 \cap L_\beta)Q_\beta/Q_\beta \cap \bC(K_\beta) \not= 1$. This shows $\bC_{Z_1}(K) \not= 1$, contradicting $L_2 \not\leq M$.

So we may assume that $\omega$ induces a field automorphism on $K$.
By (1.64) and (1.65) this implies $x=3$ or $9$, and $\omega_{1} $ induces a diagonal $\times$
inner automorphism. Let $U_1$ be the parabolic with $L_2(q^3)$ on top. Then $\bC_{T_1}(O_2(U_1))$ is the natural $L_2(q^3)$--module and $U_1/O_2(U_1) \cong L_2(q^3) \times Z_{q-1}$. As $x \not|~q-1$ and as $r > q$, we see that $x \not= 3$. If $3 \not|~q^3-1$, we have that $\bC_{\langle K,\omega_1\rangle}(T_1) \not= 1$. Hence we have $[\omega_1,K] = 1$ and so $N_G(\langle \omega_1 \rangle) \leq M$. Now we may argue as above. Hence $3 ~|~ q^3 -1$, i.e. $3~|~q-1$. Hence as $r^2 \geq q^3$, we get $q = 4$ or $q = 16$. But then $m_3(\langle K, W\rangle) = 4$, a contradiction. 

So we may assume $|[Z_{\beta} ,t]|\leq q^{\, 3}r$. Hence $r\geq q^{\,
3}$. Now by (1.64) and (1.65) either $r=q^{\, 3}$ and $\omega$ induce
a diagonal $\times$ inner automorphism or $x=9$ and $q=4$, and so $r = q^3$ too. If $p \not|~q^3-1$, then any $p$--element $\mu \in \bC(K)$ centralizes $[Z_1,K]$. As by (4.12) no field automorphism of order $p$ is involved we see that $[Z_1,K]$ is centralized by $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \leq M$, or $p = 3$. The former implies $L_2 \leq M$, a contradiction. So we have $p = 3$ and $m_3(K) = 3$. But then $[Z_1,K]$ is centralized by a cyclic group of order 3. As $\bC_{[Z_1,K]}(S)$ is centralized by a 3--element in $K$, we get that $N_G(\bC_{[Z_1,K]}(S)) \leq M$, a contradiction. So $p \Big |~q^3-1$.
Now choose $\nu\in L_{1}\cap L_{2}$, $o(\nu )=p$. By (4.4)(ii) we get $m_p(L_1) \geq 3$. By (4.12) no field automorphisms of order $p$ are involved. Hence $p ~|~|\bC_{L_1/Q_1}(K)|$ and so $m_p(\bC_{L_1}(\nu)) \geq 3$, which gives $N_G(\langle \nu \rangle) \leq M$, a contradiction to (1.66) and (2.2).

So we may assume $K\not\cong$ $^{3}D_{4}(q)$. In this case there is some
$\mu\in K$, $o(\mu )=p$ such that $[\mu ,\bC_{T_{1}} (S)]=1$. Hence
we see that $m_{p}(K)\geq 4$. This implies $K\cong U_{5}(q)$,
$p \,| \,q+1$. By (1.34) (i) we have $r>q$. Hence by (1.64) and (1.65)
$r=q^{\, 2}$. Further $L_1 \cap L_2$ contains a subgroup of order $(r-1)^2$. As $p~|~r-1$ we get a contradiction with (2.2).

So assume next that $T_{1}\bigoplus T_{2}$ is involved. Then we have
$K\cong L_{n}(q)$, $n\leq 4$, $T_{1}$ is the natural module, $K\cong
U_{4}(q)$ or $\Omega_6^{\, -} (q)$ and $T_{1}$ is the natural module.
Suppose $K\cong U_{4}(q)\cong\Omega_6^{\, -} (q)$. If $r = q^2$ then $p~|~|L_1 \cap L_2|$ and so again by (4.4) and (4.12) $N_G(\langle \mu \rangle) \leq M$ for $\mu \in L_1 \cap L_2$, $o(\mu) = p$, a contradiction. So we have $r = q$ by (1.64) and (1.65). Now $\omega$ induces an inner $\times$ diagonal automorphism and $p~|~q+1$. But now
$\langle\omega\rangle\times\langle\omega_{1}\rangle\le
L_{1}\cap L_{2}$ and so $\bC_{\langle\omega\rangle\times\langle\omega_{1}\rangle}
(K)\not= 1$, as  $[\langle\omega\rangle\times\langle\omega_{1}\rangle
,\Omega_{1} (Z(S))]=1$.  So there is $\omega_2 \in \langle \omega \rangle \times \langle \omega_1 \rangle$ with $N_G(\omega_2 \rangle) \leq M$. By (1.66) we get $E(L_2/Q_2) \cong $ $(S)L_3(q)$ or $Sp_4(q)$ and so by (6.16) $|V_2 : V_2 \cap Q_\beta| \geq r^6$. Hence $|V_2 \cap L_\beta : V_2 \cap Q_\beta| \geq r^4 = q^4$. If $(V_2 \cap L_\beta)Q_\beta/Q_\beta \cap \bC(K_\beta) \not= 1$ we get $\bC_{Z_1}(K) \not= 1$, contradicting $L_2 \not\leq M$. So we have $(V_2 \cap L_\beta)Q_\beta/Q_\beta \cap \bC(K_\beta) = 1$ and then $|V_2 \cap L_\beta : V_2 \cap Q_\beta| = q^4$,i.e. $T_1$ and $T_2$ are natural $U_4(q)$--modules as $V_2 \cap L_\beta$ acts quadratically on $Z_\beta$. But now as $|Z_\beta \cap L_1 : Z_\beta \cap Q_1| = q^2$, we see that $|Z_1 : \bC_{Z_1}(Z_\beta \cap L_1)| \geq q^8 = r^8$ and then $|Z_1 \cap L_\beta : Z_1 \cap Q_\beta| \geq r^6$, a contradiction.

So we have $K \cong$ $(S)L_3(q)$ or $L_4(q)$. Suppose first that $p~|~q+1$. Then $K \cong L_4(q)$ and $m_p(K) = 2$. Further by (4.4) and (4.12) we have that $m_p(\bC_{L_1/Q_1}(K)) \geq 2$. As $Z_1$ involves at most three natural modules, we get that $\bC_{L_1/Q_1}(K)$ induces on $\langle T_1^g~|~g \in L_1 \rangle$ a subgroup of $SL_3(q)$ and so as $p~|~q+1$ we have that $p~|~|\bC_{L_1}(\langle T_1^g~|~g \in L_1 \rangle)|$. But then as $p~|~|\bC_{K}(s)|$ for $s \in V_\alpha \cap T_1$, we see that $V_\alpha \leq L_1$, a contradiction. So we have that $p~|~q-1$. As $V_\alpha \not\leq L_1$, no element in $T_1^\sharp$ is centralized by an elementary abelian subgroup of order $p^2$ in $K$. Hence $n = 3$. As $\omega$ cannot centralize a subgroup of order $(q-1)^2$ in $K$, we see that $\omega$ induces a field automorphism on $K$. This implies $2r = q$ or $x = 9$, $r = 64$ and $q = 2^9$. Hence in any case we get that $(r-1)^2~|~|L_1 \cap L_2|$ and so there is some $\omega_2 \in L_1 \cap L_2$, $o(\omega_2) = x$ and $\bC_G(\omega_2) \leq M$. Now by (1.66) $E(L_2/Q_2) \cong$ $(S)L_3(r)$ or $Sp_4(r)$ and so by (6.16) $|V_2 : V_2 \cap Q_\beta| \geq r^6$. Hence $|V_2 \cap L_\beta : V_2 \cap Q_\beta| \geq r^4$. On the other hand we have $|V_2 \cap L_\beta : V_2 \cap Q_\beta| \leq q^2$, so $r^2 \leq q$. As $r \geq 4$ this is impossible.

\absa
For proving the lemma we just have to treat the case $K\cong
L_{3}(2^{\, x})$ and $K\not\!\!\unlhd L_{1}/Q_{1}$. 

\absa
Let $t \in Z_1 \cap Q_\alpha$ with $[t, Z_\beta] \not= 1$. Assume first $[K_\beta,t] = 1$. Then $V_2 \subseteq L_\beta$ and so $V_\alpha \not\subseteq L_1$ by (6.1). Hence by quadratic action $|V_\alpha \cap L_1 : V_\alpha \cap Q_1 | \leq q^2$ and so $|[Z_\beta,t]| \leq r^2q^2$. As $q = 2^x$ we see that $2r = q$ or $q = 2^9$ and $r = 64$, i.e. $x = 9$. If $[[Z_\beta,t],K_\beta]\not=1$, then we have $(r-1)^2~|~|L_1 \cap L_2|$. But then there is some $\omega_1 \in L_1 \cap L_2$, $o(\omega_1) = 3$ or $x$ with $[\omega_1,K] = 1$. This implies $N_G(\langle \omega_1 \rangle) \leq M$ and so by (1.66) $E(L_2/Q_2) \cong (S)L_3(r)$ or $Sp_4(r)$. Now $|V_\alpha : \bC_{V_\alpha}(Z_1)| \geq r^6$ by (6.16) and then $|V_\alpha \cap L_1 : V_\alpha \cap L_2| \geq r^4$. As $(V_\alpha \cap L_1)Q_1/Q_1 \cap \bC(K) = 1$, we see that $|V_\alpha \cap L_1 : V_\alpha \cap Q_1| \leq q^2$. But as $r \geq 4$ this is not possible. Hence we have $[[Z_\beta,t],K_\beta] = 1$. If $[\langle K_\beta^{L_\beta} \rangle,t] = 1$, we get $[[Z_\beta,t],\langle K_\beta^{L_\beta}\rangle] = 1$ and so there is $1 \not= u \in \Omega_1(Z(S))$ with $[u, \langle K^{L_1}\rangle] = 1$, i.e. $\bC_G(u) \leq M$, contradicting $L_2 \not\leq M$. So we get a conjugate $K_\beta^g$ with $[t,K_\beta^g] \not= 1$.

So we may assume $[t,K_\beta] \not= 1$ from the beginning. Assume $V_\alpha \not\leq L_1$ again and so $|[Z_\beta,t]| \leq r^2q^2$. Let $g \in L_1$, $K^g \not= K$. Then $K^g$ acts nontrivially on the set of irreducible $K$--modules involved in $Z_1$. In particular $|[Z_\beta,t]| \geq q^3$. But now we have $(r-1)^2 ~|~|L_1 \cap L_2|$ and we may argue as above to get a contradiction.
So we have $V_\alpha \leq L_1$. By (6.1) $V_2 \not\leq L_\beta$. Now by quadratic action for $s \in Z_\beta$ we have $|[Z_1,s]| \leq q^2r^2$. As before $|[Z_1,s]| \geq q^3$ and so $(r-1)^2 ~\Big|~|L_1 \cap L_2|$ and $|V_\alpha : \bC_{V_\alpha}(Z_1)| \geq r^6$. If $V_\alpha \cap \bC(K) \not=1$ then we have there is some $u \in [t,Z_\beta]^\sharp$ which is centralized by $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \leq M_\beta$ and so $V_2 \leq L_\beta$, a contradiction. Hence we have $|V_\alpha : V_\alpha \cap Q_1| \leq q^2$. So $r^6 \leq q^2$. But $q \leq r^2$ and then $r^6 \leq r^4$, a contradiction.


\absa
{\bf (6.23)~Lemma.~}{\it Assume (6.4) with $b=b_{1}>1$. Let $K$ be a
component of $L_{1}/Q_{1}$ with $K/Z(K)\cong G(2)$ and $m_{p}(K)\geq 2$.
Then $[V_{\alpha}\cap L_{1},K]\le Q_{1}$.}

\absa
Proof~. Suppose false. As $[\omega ,K]\not= 1$, we see with (1.65) that $o(\omega
)=3$ or $o(\omega )=5$ and $K\cong$ $^{2}F_{4}(2)$ or $o(\omega )=7$
and $K\cong$ $^{3}D_{4}(2)$.

Suppose $E(L_2/Q_2) \cong $ $(S)L_3(r)$ or $Sp_4(r)$. Then $|V_\alpha : \bC_{V_\alpha}(Z_1)| \geq r^6$. Hence $|V_\alpha \cap L_1 : V_\alpha \cap Q_1| \geq r^4 \geq 2^8$. As $K \unlhd L_1/Q_1$ we see that $\bC_{(V_\alpha \cap L_1)Q_1/Q_1}(K) = 1$. Hence $K \cong L_n(2)$, $o(\omega) = 3$. But now $m_3(K) = 3$ and so $\omega \in K$. As $m_p(K) = 2$, we see that $\omega$ is centralized by $E$, $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \leq M$ and so $N_G(\langle \omega \rangle) \leq M$ by (1.63), a contradiction. So we have by (1.66) that $N_G(\langle \omega_1 \rangle) \not\leq M$ for any $\omega_1 \in L_1 \cap L_2$, $o(\omega_1) = y$. If $\omega \in K$ then we argue as before. So let $\omega \not\in K$. Let first $y \not= 7$. then $\langle K,\omega \rangle \cong K \times Z_y$ and $\omega_1 \in \bC_{\langle K,\omega \rangle}(K) \leq L_1 \cap L_2$, as elements of order $y$ in parabolics are inverted. But then $N_G(\langle \omega_1 \rangle) \leq M$ by (1.63), a contradiction.

So we are left with $o(\omega) = 7$ and $K \cong$ $^3D_4(2)$. Further $p = 3$. Finally $49 \not|~|L_1 \cap L_2|$. So $|V_\alpha : V_\alpha \cap L_1| \leq 8$. This shows that $[Z_1 \cap Q_\alpha, V_\alpha] \not= 1$ and so we may assume $[Z_1 \cap Q_\alpha, Z_\beta] \not= 1$ for some $\beta \in \Delta(\alpha)$. Choose $t \in Z_1 \cap Q_\alpha$ with $[t,Z_\beta] \not= 1$. Suppose $[[t,Z_\beta],K_\beta] = 1$. Then $\bC_{Z_\beta}(K_\beta) \not= 1$ and so $\bC_{\Omega_1(Z(S))}(K) \not= 1$, a contradiction. Hence as $|[t, Z_\beta]| \leq 2^8$, we see $[t, K_\beta] \not= 1$ and by (1.40) $|[Z_{\beta} ,t]|\geq 2^{\, 6}$.
This gives that $|V_{\alpha}\cap L_{1}:V_{\alpha}\cap Q_{1}|\geq
2^{\, 3}$, as $|V_{\alpha} :V_{\alpha}\cap Q_{1}|\leq r=8$. Hence by
(1.15) $Z_{\beta}$ is a strong quadratic module, i.e. it involves
just one nontrivial irreducible module $T$, the natural one. But as
$3 \,\Big | \,|\bC_{M} (T)|$, we see that every $x\in T^{\,\sharp}$ is centralized
by an elementary abelian group of order $9$, which gives $\bC_{G}
(\Omega_{1} (Z(S)))\le M$, a contradiction.

\absa
{\bf (6.24)~Lemma.~}{\it Assume (6.4) with $b=b_{1}>1$. Let $K$ be $p$--constrained with $m_p(O_p(K)) \geq 2$ or 
$O_{t}(K)\not= 1$ and $K/O_{t}(K)\cong L_{3}(t)$ or $L_{2}(t)$, then $[V_{\alpha}\cap
L_{1},K]\le Q_{1}$. (In particular the exceptional case in
(4.9) does not occur.)}

\absa
Proof~. Assume first $V_{\alpha}\le L_{1}$. Then there is some
$t\in Z_{1}$ with $|V_{\alpha} :\bC_{V_{\alpha}} (t)|\leq 4$. By
(6.16) $t\in Q_{\alpha}$. Let $\beta\in\triangle (\alpha )$ with
$[Z_{\beta} ,t]\not= 1$. Then $Z_{1}\le L_{\beta}$. But this
contradicts (6.1).

So we have $V_{\alpha}\not\le L_{1}$. In particular either
$\bC_{V_{\alpha}\cap L_{1}/V_{\alpha}\cap Q_{1}} (K)=1$ or $\bC_{V_{\alpha}\cap L_{1}/V_{\alpha}\cap Q_{1}} (O_p(K))=1$ . Furthermore
by \cite[(24.1)]{GoLyS} and \cite[(25.12)]{GoLyS} we see that $|V_{\alpha}\cap L_{1}:V_{\alpha}\cap
Q_{1}|\leq 4$. Let $t\in Z_{1}\setminus Q_{\alpha}$. Then $|V_{\alpha}
:\bC_{V_{\alpha}} (t)|\leq 4|V_{\alpha} :V_{\alpha}\cap L_{1}|$. By
(6.16) and (1.44) we get $E(L_{2}/Q_{\, 2})\cong L_{2}(4)$ or
$L_{2}(r)\times L_{2}(r)$. Suppose the latter with $r>4$. Then as $|V_\alpha : \bC_{V_\alpha}(Z_1)| \leq 4r^2$ we get with (6.16) that 
$|Z_{1}:Z_{1}\cap Q_{\alpha} | \leq r$. Choose $\alpha +1\in\triangle
(\alpha )$ such that $Z_{1}\not\le L_{\alpha +1}$. Then also
$Z_{\alpha +1}\not\le L_{1}$.

Hence $|V_{2}:\bC_{V_{2}} (Z_{\alpha +1})|\geq r^{\, 2}$. This shows
$V_{2}Q_{\alpha} >Z_{1}Q_{\alpha}$. Let $\lambda\in\triangle (2)$
with $Z_{\lambda} Q_{\alpha} \not\le Z_{1}Q_{\alpha}$. Then
also $|V_{\alpha} :\bC_{V_{\alpha}} (Z_{\lambda} )|\leq 4r^{\,
2}$ and so $|V_{\alpha} :\bC_{V_{\alpha}} (Z_{1}Z_{\lambda} )|\leq
16r^{\, 2}$. On the other hand by (1.44) and (6.16) $|V_{\alpha}
:\bC_{V_{\alpha}} (Z_{1}Z_{\lambda} )|\geq r^{\, 4}$, a
contradiction. This shows $r=4$. 

Choose $\alpha +1\in\triangle (\alpha )$ as before. Now $|Z_{\alpha
+1} :\bC_{Z_{\alpha +1}} (Z_{1})|\leq 64$. This implies $p=7$ and
$|Z_{\alpha +1} :\bC_{Z_{\alpha +1}} (Z_{1})|=64$, hence
$E(L_{2}/Q_{\, 2})\cong L_{2}(4)\times L_{2}(4)$ and $|Z_{\alpha +1}
:Z_{\alpha +1}\cap L_{1}|=16=|Z_{1}:Z_{1}\cap Q_{\alpha} |$. But then
by (6.16) $|V_{\alpha} :\bC_{V_{\alpha}} (Z_{1})|\geq 2^{\, 8}$, a
contradiction.

\absa
{\bf (6.25)~Lemma.~}{\it Assume (6.4) with $b=b_{1}>1$. Suppose we
have components $K_{1},K_{2},K_{3}$ of $L_{1}/Q_{1}$,
$m_{p}(K_{i})=1$, where the $K_i$ might be cyclic $p$-groups. Then there is some $i$ with $[V_{\alpha}\cap
L_{1},K_{i}]\le Q_{1}$.}

\absa
Proof~. Suppose false. As in the proof of (6.24) we may assume that
$|V_{\alpha}\cap L_{1}:\bC_{V_{\alpha}\cap L_{1}} (K_{i})|\geq 8$ for
every $i$, in particular the $K_i$ are all nonsolvable.


We have $K_i \cong $ $L_2(q_i)$, $(S)L_3(q_i)$, $(S)U_3(q_i)$ or $Sz(q_i)$, $i = 1,2$, where we may assume $q_1 \leq q_2$. Suppose first $r \leq q_1$, $r \leq q_1^2$ in case of $K_1 \cong (S)U_3(q_1)$. 
\begin{itemize}
\item[(1)]~~~~~~~~~If $Z_1 \not\leq L_\beta$ then $[Z_\beta, Z_1 \cap Q_\alpha] = 1$.
\end{itemize}

Suppose false. Then let $[K_{1\beta}, Z_1 \cap Q_\alpha] \not= 1$. 

Never $[[K_i,Z_1],K_j \times K_k] = 1$ for $\{i,j,k\} = \{1,2,3\}$. Otherwise we have $[K_{i\beta}, Z_1 \cap L_\beta] = 1$. But then even $[K_{i\beta}^{L_\beta}, Z_1 \cap L_\beta] = 1$. This implies $K_{i\beta} \unlhd L_\beta/Q_\beta$ and so there is some $x \in \Omega_1(Z(S)) \cap [Z_1, K_i]$ which is centralized by $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \leq M$, contradicting $L_2 \not\leq M$. 

If now $\bC_{V_\alpha \cap L_1}(K_k) \not\leq Q_1$ for some $k \in \{1,2,3\}$ then $[[K_k, Z_1], K_j] = 1$ for some $j \in \{1,2,3\} \setminus \{k\}$. As $V_2 \not\leq L_\beta$ we have $[[K_k, Z_1], K_i] \not=1$ for $\{i,j,k\} = \{1,2,3\}$. In particular $[K_i, \bC_{V_\alpha \cap L_1}(K_k)] = 1$. But then $[[K_j, Z_1], K_i] = 1 = [[K_j,Z_1], K_k]$, a contradiction. So we have $\bC_{V_\alpha \cap L_1}(K_i) \leq Q_1$ for all $i$.

There is $X \leq Z_\beta$, $|X| \geq q_1^2$ ($q_1^4$ for $K_1 \cong (S)U_3(q_1)$) with $X \cap \bC(Z_1) = 1$. Suppose $X \cap L_1 \leq Q_1$. Then $q_1 = r$ ($q_1^2 = r$ respectively) and $|V_\alpha : V_\alpha \cap L_1| = r^2$. As $|V_2 : \bC_{V_2}(X)| \geq r^4$ by (6.16) we get $K_1 \cong (S)L_3(r)$ and $[Z_1, K_1]$ involves exactly two natural $K_1$--modules. As $L_1 \cap L_2$ centralizes $\Omega_1(Z(S))$ we get $m_x(K_1(L_1 \cap L_2)) \geq 4$, a contradiction. So we have $[X \cap L_1, Z_1] \not= 1$. Now let $t \in X \cap L_1$, then $|[t,Z_1]| \leq q_1^3$ ($q_1^5$ in case of $K_1 \cong (S)U_3(q_1)$) or $q_1^4$ in case of $K_1 \cong (S)L_3(q_1)$ and $|V_2 \cap Q_\alpha : \bC_{V_2 \cap Q_\alpha}(K_{1\beta})| = q_1^2$. As $\bC_{V_\alpha \cap L_1}(K_1) \leq Q_1$ we may assume $[t,K_1] \not= 1$. If there are four nontrivial irreducible modules involved in $[Z_1,K_1]$ then we again see $q_1 = r$ and as before we get a contradiction. Suppose there are exactly three nontrivial irreducible modules involved. Then $K_1 \not\cong (S)U_3(q_1)$ and $[K_1,Z_1]$ is centralized by $K_2$ or $K_3$. In particular $|[t,[K_1,Z_1]]| \geq q_1^3$. Suppose $[K_2, [K_1,Z_1]] = 1$. We have $|[t,[K_2,Z_1]]| \geq q_2^2 \geq q_1^2$ as $\bC_{V_\alpha \cap L_1}(K_1) \leq Q_1$, a contradiction. So $[K_2, [K_1,Z_1]] \not= 1$. As $[t, [K_3,Z_1]] \not= 1$, we even have $|[t,Z_1]| > q_1^3$ and so $K_1 \cong SL_3(q_1)$. We have $|[K_2, [K_1,Z_1], t]| \geq q_2^3$. Now $K_2$ is isomorphic to a subgroup of $SL_3(q_1)$. As $q_2 \geq q_1$, we have $K_2 \cong L_2(q_2)$.  We have $|[[K_2,[Z_1,K_1]],t]| \geq q_2^3$. As $[K_3,[Z_1,K_1]] = 1$, we see that $[K_2,[K_3,Z_1]] \not= 1$ , so $|[K_2, [K_1,Z_1],t]| \geq q_2$, and so $|[Z_1,t]| \geq q_2^4$. But then $q_1 = q_2 = r$, a contradiction as $(r-1)^2~\Big|~|L_1 \cap L_2|$. 

So there are exactly two nontrivial irreducible modules involved. Again either $[[K_1,Z_1],K_2] = 1$ or $[[K_1,Z_1],K_3] = 1$. In particular $|[Z_1,t]| > q_1^2$ ($q_1^4$ for $K_1 \cong (S)U_3(q_1)$). Let $[K_2,[K_1,Z_1]] = 1$. Then we have $|[Z_1,t]| \geq q_1^2q_2^2 \geq q_1^4$ (or for $K_1 \cong (S)U_3(q_1)$ we have $|[Z_1,t]| \geq q_1^4q_2^2 \geq q_1^6$). So $K_1 \not\cong (S)U_3(q_1)$ and we have $q_1 = q_2 = r$, a contradiction again. 

Hence $[K_2, [K_1,Z_1]] \not= 1$ and so $K_2$ is isomorphic to a subgroup of $L_2(q_1)$ or $L_2(q_1^2)$ in case of $K_1 \cong (S)U_3(q_1)$. Let first $K_1 \cong (S)U_3(q_1)$. As $q_2 \geq q_1$, we have $K_2 \cong L_2(q_1)$ or $L_2(q_1^2)$. We have further that there are at least six nontrivial irreducible $K_2$--modules involved if $K_2 \cong L_2(q_1)$ and at least three nontrivial irreducible modules involved if $K_2 \cong L_2(q_1^2)$ , so $|[t,Z_1]| \geq q_1^6$, a contradiction. Hence we have $K_1 \not\cong (S)U_3(q_1)$. This gives $q_2 = q_1$ and $K_2 \cong L_2(q_1)$. Suppose $K_1 \cong Sz(q_1)$. Then we have $[Z_1,t]| \leq q_1^3$. But on the other hand $|[t,Z_1]| \geq q_1^4$, a contradiction. So we have $K_1 \cong L_2(q_1)$ or $(S)L_3(q_1)$ and then $K_3 \cong L_2(q_3)$, $(S)L_3(q_3)$, $(S)U_3(q_3)$ or $Sz(q_3)$. Hence $q_3 < q_1$. If $K_1 \cong L_2(q_1)$, then $|[Z_1,t]| \leq q_1^3$ and so there are at most three nontrivial irreducible $L_2(q_1)$--modules involved. As $[K_1,[K_2,Z_1]] \not= 1$, we see $[K_3, [K_2,Z_1]] = 1$. But then $[K_3, [K_1 \times K_2, Z_1]] = 1$ and so $[K_1 \times K_2, [K_3, Z_1]] = 1$, a contradiction. So we have $K_1 \cong (S)L_3(q_1)$ and $|[Z_1,t]| > q_1^3$. Furthermore $[[K_2,Z_1],K_3] \not= 1$. But then $[K_2,Z_1]$ has to involve at least five nontrivial irreducible modules, a contradiction.

So we have $[Z_1 \cap Q_\alpha, K_{1\beta}] = 1$, and we may assume $[Z_1 \cap Q_\alpha , K_{1\beta}^{L_\beta}] = 1$. Furthermore we see $K_{1\beta} \unlhd L_\beta/Q_\beta$ otherwise $V_2 \leq L_\beta$, a contradiction. Now we have $[Z_1 \cap Q_\alpha, K_{2\beta}] \not= 1 \not= [Z_1 \cap Q_\alpha, K_{3\beta}]$ as otherwise again $V_2 \leq L_\beta$. In particular $q_1 < q_2$. 

Let $K_1 \cong (S)U_3(q_1)$. Then either $[K_1, [Z_1,K_2]] \not=1 $ or $[K_1, [Z_1, K_3]] \not= 1$. In any case there is $X \leq Z_\beta$, $|X| \geq q_1^6$ and $\bC_X(Z_1) = 1$. As $|X \cap L_1 : X \cap Q_1| \leq q_1$, we get $r > q_1^2$, a contradiction. 

Now we have $r \leq q_1$. There is $X \leq Z_\beta$, $|X| \geq q_2^2$ with $\bC_X(Z_1) = 1$. If $X \cap L_1 \leq Q_1$ then $r \geq q_2$ but $r \leq q_1$, a contradiction. So we have $X \cap L_1 \not\leq Q_1$. Now for $t \in X \cap L_1 \setminus Q_1$ we have $|Z_1 : \bC_{Z_1}(t)| \leq q_2^2r^2$ or $q_2r^2$ (for $K_2 \not\cong (S)L_3(q_2)$). As $r < q_2$ there are at most three nontrivial irreducible modules involved (if $K_2 \not\cong (S)L_3(q_2)$ there are exactly two). So we have $[K_2,Z_1]$ is centralized by $K_1$ or $K_3$. If $[K_1,[K_2,Z_1]] = 1$, then as $r \leq q_1$ and $r < q_2$, and there are at least two nontrivial irreducible modules involved, we see that $|Z_1 : \bC_{Z_1}(t)| \geq q_2^2q_1^2 \geq q_2^2r^2$. Hence we have $K_2 \cong L_3(q_2)$ and $r = q_1$. Finally both $[K_2,Z_1]$ and $[K_1,Z_1]$ involve exactly two nontrivial irreducible modules. This shows $K_3$ is isomorphic to a subgroup of $L_2(q_1)$ and $L_2(q_2)$ as well. So $K_3 \cong L_2(q_3)$ for $q_3 \leq q_1$. We have $(r-1)^2 ~|~|L_1 \cap L_2|$. So let $Z_x \times Z_x \cong T = \langle \omega_1 \rangle \times \langle \omega \rangle \leq L_1 \cap L_2$. As we may assume $q_3 < r$, we have $[T,K_3] = 1$. Furthermore if $\bC_T(K_1) = 1$, we have $K_1 \cong (S)L_3(r)$. Now as $m_x(K_1T) \leq 3$, we see that $T \cap K_1 \not= 1$. Hence in any case we have some $\omega_2 \in T^\sharp$, with $N_G(\langle \omega_2 \rangle ) \leq M$. Application of (1.66) shows $E(L_2/Q_2) \cong (S)L_3(r)$ or $Sp_4(r)$. Let now $t \in Z_1$. then $|V_\alpha : \bC_{V_\alpha}(s)| \leq q_3r^2 < r^3$. But this contradicts (6.16).

So we have $[K_1,[K_2,Z_1]] \not= 1$. Now $[K_3,[K_2,Z_1]] = 1$. Furthermore $K_1$ is isomorphic to a subgroup of $SL_3(q_2)$ and so $K_1 \cong L_2(q_1)$ or $(S)L_3(q_1)$. In particular $3~|~|K_1|$. If $K_2 \cong Sz(q_2)$, then $|[t,Z_1]| \geq q_2^4$, a contradiction. So we have $3~|~|K_2|$ and then $K_3 \cong $ $L_2(q_3)$, $(S)L_3(q_3)$, $(S)U_3(q_3)$ or $Sz(q_3)$. So by symmetry we have $q_3 < r \leq q_1 < q_2$. Then we have $|V_\alpha : \bC_{V_\alpha}(Z_1)| \leq q_3^2r^2$. If $|V_\alpha : V_\alpha \cap L_1| \leq r$, we even  get $|V_\alpha : \bC_{V_\alpha}(Z_1)| \leq q_3^2r$ and on the other hand by (6.16) we may assume that there is $s \in Z_1$ with $|V_\alpha : \bC_{V_\alpha}(s)| \geq r^2$. This shows $K_3 \cong (S)L_3(q_3)$. If $|V_\alpha : V_\alpha \cap L_1| > r$ and $|Z_1 : Z_1 \cap Q_\alpha| > r$, we have with (6.16) that $|V_\alpha : \bC_{V_\alpha}(Z_1)| \geq r^4$, a contradiction. So we have $|Z_1 : Z_1 \cap Q_\alpha| \leq r$ and so $|Z_1 : \bC_{Z_1}(t)| \leq q_2^2r < q_2^3$. Now $K_2 \cong (S)L_3(q_2)$ and there is exactly one nontrivial irreducible $K_1$-module in $\bC_{Z_1}(K_2)$. But this would imply $[[K_1,\bC_{Z_1}(K_2)], K_3] = 1$ and $[[K_1,Z_1],K_3] = 1$, a contradiction as $[K_3, [K_2,Z_1]] = 1$ . So we have $K_3 \cong (S)L_3(q_3)$ and $|V_\alpha : V_\alpha \cap L_1| \leq r$. As $|Z_1 : \bC_{Z_1}(t)| \leq q_2^2r^2$, we get $r = q_1$ and $[\bC_{Z_1}(K_2), K_1]$ involves at most two nontrivial irreducible modules. Hence $[K_3,[\bC_{Z_1}(K_2),K_1]] = 1$, a contradiction again. This proves (1).

\begin{itemize}
\item[(2)]~~~~~~~~~We have $q_1 < r$ and $q_1^2 < r$ for $K_1 \cong (S)U_3(q_1)$.
\end{itemize}

Suppose false. Choose $\beta \in \Delta(\alpha)$ with $Z_1 \not\leq L_\beta$. Then we have $[Z_\beta, Z_1 \cap Q_\alpha] = 1$ by (1). Obviously $[Z_\beta, Z_1] \not= 1$. Let $t \in Z_\beta \cap L_1$ with $[t,Z_1] \not= 1$. Then $|Z_1 : \bC_{Z_1}(t)| \leq r^2 \leq q_1^2$ (or $q_1^4$ in case of $K_1 \cong (S)U_3(q_1)$). So there are exactly two nontrivial irreducible $K_1$--modules involved and $q_1 = r$ (or $q_1^2 = r$). But as $\bC_{K_2 \times K_3}([Z_1,K_1]) \not= 1$, we have a contradiction. So we have $Z_\beta \cap L_1 \leq Q_1$. Now $\langle Q_1, Z_\beta \rangle$ centralizes $Z_1 \cap Q_\alpha$. In particular $L_2 = S\bC_{L_2}(Z_1 \cap Q_\alpha)$. But $[V_\alpha, Z_3] = 1$ and so for $t \in V_\alpha \cap L_1$ we have $|[t,Z_1]| \leq r^2$. Again we get $V_\alpha \cap L_1 \leq Q_1$, contradicting (6.17). This proves (2).

We will show that $K_3 \cong $ $L_2(q_3)$, $(S)L_3(q_3)$, $(S)U_3(q_3)$ or $Sz(q_3)$. If not either $K_1 \cong Sz(q_1)$ or $K_2 \cong  Sz(q_2)$. Let $K_1 \cong Sz(q_1)$. Then $|V_\alpha : \bC_{V_\alpha}(Z_1)| \leq q_1r^i < r^{i+1}$, $i = 1,2$. By (6.16) we get $i = 2$ and so $Z_x \times Z_x \cong E \leq L_1 \cap L_2$. By (1.66) $N_G(\langle\tilde{\omega} \rangle) \not\leq M$ for all $\tilde{\omega} \in E$. So we have that $\bC_E(K_2) = 1$. This gives $K_2 \cong (S)L_3(q_2)$. But then either $m_x(K_2E) \geq 4$ or there is some $1 \not= \omega_2 \in E$ with $\omega_2 \in K_2$. In both cases $N_G(\langle \omega_2 \rangle) \leq M$, a contradiction. Assume now $K_2 \cong Sz(q_2)$. Then $|V_\alpha : \bC_{V_\alpha}(Z_1)| < r^4$ and so by (1.66) and (6.16) we have that $N_G(\langle \tilde{\omega} \rangle) \not\leq M$ for all $\tilde{\omega} \in L_1 \cap L_2$ with $o(\tilde{\omega}) = x$. Hence we have $(r-1)^2 \not|~|L_1 \cap L_2|$ as any $\tilde{\omega} \in L_1 \cap L_2$ with $o(\tilde{\omega}) = x$ has to centralize $K_1$ by (1.64) and (1.65). 

Now we get that $K_1 \cong (S)L_3(q_1)$. Let $\beta \in \Delta(\alpha)$ with $Z_1 \not\leq L_\beta$. Assume $[Z_1 \cap Q_\alpha, Z_\beta] = 1$. Then as before $Z_\beta \not\leq Q_2$ and so $\omega$ centralizes a subgroup of index $r$ in $Z_1$. Now we have $[[\omega, Z_1], K_1] = 1$ and so $[[K_1,Z_1],\omega] = 1$. But then $[K_2 \times K_3,[K_1, Z_1]] = 1$, a contradiction. 

Let $[Z_1 \cap Q_\alpha, K_{2\beta}] \not= 1$. Then we have $X \leq Z_\beta$, $|X| \geq q_2^4$ with $|Z_1 : \bC_{Z_1}(X)| \leq rq_2$, a contradiction. So we have $[Z_1 \cap Q_\alpha, K_{2\beta}] = 1$. Hence $[Z_1 \cap Q_\alpha, K_{1\beta}] \not= 1$, and there is $X \leq Z_\beta$ with $|X| = q_1^2$ and $|Z_1 : \bC_{Z_1}(X)| \leq q_1^2r \leq q_1^4$ as $r \leq q_1^2$. If $X \cap L_1 \not\leq Q_1$ then $X$ cannot act on $K_2$ as for $s \in K_2$ we have $|Z_1 : \bC_{Z_1}(s)| \geq q_2^2 > q_1^4$. Now we have that $X \cap L_1 \leq Q_1$. Hence as $q_1^2 \geq r$, we get $q_1^2 = r$ and there are exactly two nontrivial irreducible $K_1$--modules involved. Now $[[K_1,Z_1],K_2] = 1$. So we have $[K_3, [K_1,Z_1]] \not= 1$ and then $K_3 \cong L_2(q_3)$.

So in any case we may assume that $K_3 \cong $ $L_2(q_3)$, $(S)L_3(q_3)$, $(S)U_3(q_3)$ or $Sz(q_3)$. Let $q_i \leq q_j$, $\{i,j\} = \{2,3\}$. As before we get $q_i < r$ ($q_i^2 < r$ in case of $K_i \cong (S)U_3(q_i)$). Hence $[\omega,K_1K_i] = 1$ and so $N_G(\langle \omega \rangle) \leq M$, a contradiction.

\absa
{\bf (6.26)~Lemma.~}{\it Suppose there is exactly one component $K$ of $L_1/Q_1$ with $[V_\alpha \cap L_1 , K] \not= 1$. Then $K \not\!\!\unlhd L_1/Q_1$.}

\absa
Proof~. Suppose false. We have $m_p(K) = 1$. By (6.16) we have $|V_\alpha : V_\alpha \cap Q_1| \geq 16$. As $V_\alpha$ acts quadratically on $Z_1$, we have that $K$ is a group of Lie type in even characteristic or $U_4(3)$, $3\cdot M_{22}$ or $A_n$. As $m_3(K) \leq 3$, we get $K \cong 3\cdot M_{22}$ or $A_n$, $ 9 \leq n \leq 11$ in the latter. Furthermore just natural modules are involved. So assume first that $K$ is not a group of Lie type in even characteristic. As $[Z_1,K_1]$ is not centralized by $E \cong E_{p^2}$ we may assume with (4.4) that $m_p(L_1) \geq 4$. Furthermore there is $\beta \in \Delta (\alpha )$ with $Z_1 \not\subseteq L_\beta$. But there is some $t \in Z_1$ with $|Z_\beta : \bC_{Z_\beta}(t)| \leq 16$. As $p \geq 5$, we get that there is $X \leq Z_\beta$ with $|X| \geq 64$ which is centralized by some $E \cong E_{p^2}$ and $\Gamma_{E,1}(G) \leq M_\beta$, a contradiction.

So we have that $K$ is a group of Lie type in characteristic two. This now shows that $K$ is isomorphic to $L_n(q), n\leq 4$, $L_5(2)$, $L_6(2)$, $L_7(2)$, $U_n(q), n \leq 4$, $PSp_4(q)$, $PSp_6(q)$, $\Omega_8^-(q)$, $G_2(q)$, $^2F_4(q)$, $^3D_4(q)$, or $Sz(q)$.

If $K \cong L_n(2), n \geq 5$, then as before $K \cong L_7(2)$, $p = 5$ or $31$ and $|V_\alpha : V_\alpha \cap Q_1| \geq 2^8$. Furthermore there are at least 8 irreducible nontrivial $K$ - modules involved. By (6.16) $|Z_1 : Z_1 \cap Q_\alpha| \leq 2^6$. So there is $t \in Z_1 \cap Q_\alpha$ with $[t,Z_\beta] \not= 1$. If $[K_\beta, t ] = 1$, we see that $[[Z_\beta, K_\beta],t]$ contains the natural module. If $[K_\beta,t] \not= 1$, then as 8 modules are involved and $|[Z_\beta,t]| \leq 2^{12}$, we also see that there is a natural module involved. So now a 2-space is centralized by a $p$-element in $K$ and then there are at least 12 modules involved. But now for $t \in Z_1$ we have $|V_\alpha : \bC_{V_\alpha}(t)| \leq 2^6$, a contradiction.

So let $K \cong$ $^2F_4(q)$. Then $|V_\alpha : V_\alpha \cap Q_1| \leq q$ and so again $Z_1 \leq L_\beta$. Let $K \cong$ $^3D_4(q)$, then $|V_\alpha : V_\alpha \cap Q_1| \leq q^5$. We have $|V_\alpha : V_\alpha \cap Q_1| > q$ and so $Z_1$ just involves natural modules. The same applies for $K \cong G_2(q)$. In the former we get $[Z_1 \cap Q_\alpha , Z_\beta] = 1$. But this contradicts (6.16). So we have the latter and $|V_\alpha : V_\alpha \cap Q_1| \geq q^2$. But then $[Z_1,K]$ involves exactly one nontrivial irreducible module, a contradiction again.

Let $K \cong \Omega_8^-(q)$, $Sp_6(q)$, or $Sp_4(q)$. Let $[Z_1 \cap Q_\alpha , Z_\beta ] \not= 1$. If $[K_\beta,t] = 1$, then $K \cong Sp_6(q)$ and $|V_\alpha : V_\alpha \cap Q_1| = q^6$. Furthermore the natural module is involved. But now there is a direct sum of two such modules $Z_1^{(i)}, i = 1,2$ where $|Z_1^{(i)} : \bC_{Z_1^{(1)}}(V_\alpha)| = q^6$ and $[Z_1^{(i)} \cap Q_\alpha , Z_\beta] = 1$, contradicting (6.16). Hence there is $t$ with $[K_\beta,t] \not= 1$. As there is no field automorphism of order $p$, we get that $p^3~\Big|~|\bC(K_\beta)|$. Further $p \not\Big| q-1$. So we see that there are at least 6 nontrivial irreducible $K$-modules involved. Hence we have $K \cong \Omega_8^-(q)$ or $Sp_6(q)$, $|V_\alpha : V_\alpha \cap Q_1| = q^6$ and $t$ induces transvections on the natural module. As $V_\alpha$ has to act quadratically we get $K \cong \Omega_8^-(q)$. But now $|Z_1 \cap Q_\alpha : Z_1 \cap \bC(K_\beta)| = 2$. So $[Z_\beta, K_\beta]$ centralizes a subgroup of index $2\cdot |Z_1 : Z_1 \cap Q_\alpha|$ in $Z_1$, which is at least $q^{12}$. Hence by (6.16) and (6.4) we have that $|V_\alpha : V_\alpha \cap Q_1| \geq q^{11} $, a contradiction.

So we have that $[Z_1 \cap Q_\alpha, V_\alpha] = 1$. Now we see by (6.16) again that $|Z_1 : Z_1 \cap Q_\alpha| \leq q^3$ ($q^2$ in case of $K \cong Sp_4(q)$). In any case we see that there is at most one nontrivial irreducible $K$ - module involved, and so $L_2 \leq M$, a contradiction.

So we have $K \cong L_2(q)$, $L_3(q)$, $L_4(q)$, $U_3(q)$, $U_4(q)$, or $Sz(q)$. Suppose $[Z_1 \cap Q_\alpha , V_\alpha] \not= 1$ and $[K_\beta, t] = 1$. Then we get $K \cong L_4(q)$, $|V_\alpha : V_\alpha \cap Q_1 | = q^4$ and natural modules are involved. Now $Z_1$ is generated by elements $s$ with $|V_\alpha : \bC_{V_\alpha}(s)| \leq q^2$. As $\bC_{L_1/Q_1}(K)$ contains an elementary abelian group of order $p^3$, there is some $U \leq Z_\beta$, $|U| = q^4$ such that $U$ is centralized by $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \leq M_\beta$, a contradiction. So we have $[K_\beta,t] \not= 1$. Hence $K \not\cong U_3(q)$ or $Sz(q)$. In case of $K \cong L_4(q)$ or $U_4(q)$, we have that $p \not\!\Big| ~ |L_2(q)|$ and so there are at least 6 nontrivial irreducible $K$-modules involved, a contradiction. 

So we have $K \cong L_3(q)$ or $L_2(q)$. In case of $L_2(q)$ we just have one nontrivial irreducible involved, a contradiction. So we have $K \cong L_3(q)$ and exactly two nontrivial irreducible $K$-modules are involved. As $p \not\Big|~q-1$, we get $p \Big|~q+1$ and so $p^3 \Big|~|\bC_{L_1/Q_1}(K)|$. But then there is some $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \leq M$ which centralizes $[Z_1,K]$, a contradiction. So we have $[Z_1 \cap Q_\alpha, V_\alpha] = 1$. By (6.16) $|Z_1 : Z_1 \cap Q_\alpha| \leq q^2$. So there are at most two nontrivial irreducible $K$-modules involved and $V_\alpha$ induces transvections. But then $|V_\alpha : V_\alpha \cap Q_1 | = q^4$, a contradiction. 

\absa
{\bf (6.27)~Lemma.~}{\it There are at least two components $K_1$, $K_2$ of $L_1/Q_1$, where one of the $K_i$ might be a cyclic $p$-group normal in $L_1/Q_1$, with $[V_\alpha \cap L_1,K_{1}] \not= 1 \not= [V_\alpha \cap L_1, K_{2}]$. }

\absa
Proof.~ Suppose false. Then by (6.26) $K_1 \not\!\!\unlhd L_1/Q_1$. In particular $m_3(K_1) \leq 1$. So we have $K_1 \cong L_2(q)$, $Sz(q)$ or $J_1$. As $|V_\alpha : V_\alpha \cap Q_1| \geq 16$, as in the proof of (6.25), we get $K \cong L_2(q)$ or $Sz(q)$, $q$ even. So in $L_\beta/Q_\beta$ we have $K_{1\beta} \times K_{2\beta}$ and $K_{1\beta} \cong K_1 \cong K_{2\beta}$. Suppose there is $t \in Z_1 \cap Q_\alpha$ with $[t, K_{1\beta}] \not= 1$ or $[t, K_{2\beta}] \not= 1$. As $|Z_\beta : \bC_{Z_\beta}(t)| \leq q$, we see $K \cong L_2(q)$ and there is exactly one nontrivial irreducible module involved, which then is centralized by $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \leq M$ , a contradiction. So we have $[Z_1 \cap Q_3 , V_\alpha ] = 1$. Now we get $|Z_1 : Z_1 \cap Q_\alpha | < q$, a contradiction.

  
\absa
{\bf (6.28)~Lemma.~}{\it Assume (6.4) with $b=b_{1}>1$. Then $V_{\alpha} \leq L_1$.}

\absa
Proof~. Suppose false. By (6.16) - (6.27) we get that there are
exactly two components $K_{1}$, $K_{2}$ of $L_{1}/Q_{1}$ such that
$[V_{\alpha}\cap L_{1},K_{i}]\not= 1$, $i=1,2$. We have $m_p(K_1) = m_p(K_2) = 1$ for $p \in \sigma(M)$. Furthermore
$[V_{\alpha}\cap L_{1}, O^{p'}(\bC_{L_{1}/Q_{1}} (K_{1}\times K_{2}))]=1$. If $[[Z_1,K_1],K_2] = 1$ we have that $p~\not|~|\bC_{L_1/Q_1}(K_1 \times K_2)|$. Further by (4.4) we have $m_p(L_1) \geq 4$. Now there is $E \cong E_{p^2}$ with $E \cap K_1 \times K_2 = 1$ and $E$ induces field automorphism on $K_1$ and $K_2$ as well. This contradicts (4.12).
So we have
\begin{itemize}
\item[(1)] $[[Z_{1},K_{1}],K_{2}]\not= 1.$
\end{itemize}

As before we may assume
\begin{itemize}
\item[(2)] $|V_{\alpha} \cap L_{1}:\bC_{V_{\alpha}\cap L_{1}}
(K_{i})|\geq 8, i=1,2\, .$
\end{itemize}

Further by (4.4) we have 
\begin{itemize}
\item[(3)] $m_p(L_1) \geq 3$ and $p~|~|\bC_{L_1}(Z_1)|$ if $m_p(L_1) = 3$.
\end{itemize}

Next we choose notation such that $[K_{1},\omega ]\not= 1$. This is
possible by (1.63) as $[\omega ,\Omega_{1} (Z(S))]=1$. As
$m_{3}(K_{1})\leq 3$ we get with (1.5) and (1.12) that $K_{1}\cong
A_{n}$, $n\leq 11$, $M_{12}$, $M_{22}$, $M_{24}$, $3\cdot M_{22}$,
$J_{2}$, $L_{n}(q)$, $n\leq 4$, $L_{n}(2)$, $5\leq n\leq 7$,
$Sp_{2n}(q)$, $n\leq 3$, $U_{n}(q)$, $n\leq 4$, $U_{5}(4)$,
$\Omega^{\, -}_{\, 8} (q)$, $G_{2}(q)$, $^{3}D_{4}(q)$, $Sz(q)$,
$^{2}F_{4}(q)$, $q$ even. Set $q=2$ for $K$ sporadic or alternating.
As $V_\alpha \not\leq L_1$ we have $m_p(\bC_{L_1/Q_1}(K_1 \times K_2)) \leq 1$. If $m_p(L_1) \geq 4$, we get that there are outer automorphism of order $p$ in $L_1$. these have to be field automorphism. By (4.12) we get $K_1 \cong K_2 \cong L_2(q)$ or $Sz(q)$. If $m_p(L_1) = 3$ we get the same result with (3) as $m_p(\bC(Z_1)) \geq 1$. Now any involution in $K_1K_2$ is centralized by an element of order $p$ and so we get for $t \in V_\alpha \cap L_1 \setminus Q_1$ that $[t,Z_1]$ is normalized by $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \leq M$, a contradiction. So we have shown $V_\alpha \leq L_1$.


\absa
{\bf (6.29)~Lemma.~}{\it Let $b=b_{1}>1$. If $L_{2}$ is nonsolvable,
then $E(L_{2}/Q_{\, 2})\cong L_{2}(r)$, $L_{3}(r)$, $U_{3}(r)$,
$PSp_{4}(r)$, $L_{2}(r)\times L_{2}(r)$, $r$ odd, (in case of $L_2(r)$ or $L_2(r) \times L_2(r)$ we assume $r > 5$) $A_{6}$, $A_{9}$,
$3\cdot A_{6}$, $3A_{6} * 3A_{6}$ or $E(L_{2}/F(L_{2}/Q_{\,
2}))\cong L_{2}(r)$, $r$ odd.}

\absa
Proof~. Suppose false. Then (6.4) holds. By (6.28)
$V_{\alpha}\le L_{1}$. Furthermore by (6.27) there are two components $K_1$, $K_2$ of $L_1/Q_1$, where one of the $K_i$ might be a cylcic normal $p$-group, with $[V_\alpha , K_i] \not= 1$. We may assume that $m_3(K_1) \leq 1$ and so $K_1$ is solvable or $K_1 \cong L_2(q)$, $L_3(q)$, $U_3(q)$, $Sz(q)$ or $J_1$. Set $\tilde{Z_1} = [Z_1,K_1]$. Let $[\tilde{Z_1} \cap Q_\alpha , Z_\beta] \not= 1$ for some $\beta \in \Delta (\alpha)$ with $Z_1 \not\leq L_\beta$. Then $[\tilde{Z_1} \cap Q_\alpha, K_{1\beta} \times K_{2\beta}] \not= 1$ and $\bC_{\tilde{Z_1} \cap Q_\alpha}(K_{1\beta} \times K_{2\beta}) = 1$. Let $t \in \tilde{Z_1} \cap Q_\alpha$ with $[t, K_{1\beta}] \not= 1$. Then either $[K_1, Z_1]$ is irreducible or $K_1 \cong SL_3(q)$ and $[K_1, Z_1]$ involves exactly two natural modules. In the former, as by (4.4) we may assume that $m_p(L_1) \geq 4$, we get that $[K_1, Z_1]$ is centralized by $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \leq M$, a contradiction. So we have the latter. Now $\bC_{L_1/Q_1}(K_1)$ contains some $F \cong E_{p^3}$, and so as $p ~\not|~q-1$, we get some $E \leq F$, $|E| = p^2$ with $[E,[K_1,Z_1]] = 1$, a contradiction.
 
So we have $[\tilde{Z_1} \cap Q_\alpha , K_{1\beta}] = 1$. This implies $[\tilde{Z_1} \cap Q_\alpha, K_{2\beta}] \not= 1$. Now for $t \in \tilde{Z_1} \cap Q_\alpha$ we get $[[t,Z_\beta], K_{1\beta}] = 1$ and so $[[K_1,Z_1], K_2] = 1$. By (4.12) $m_p(\bC_{L_1/Q_1}(K_1 \times K_2)) \geq 1$. So there is some component $K_3$ of $L_1/Q_1$ or cyclic normal subgroup with $[K_1 \times K_2 , K_3] = 1$. We have $[t, K_{3\beta}] \not=1$. So $[K_2 \times K_3, [K_1, Z_1]] = 1$. But as $[t, K_{1\beta}^{L_\beta}] = 1$, we have $K_1 \unlhd L_1/Q_1$ and so $[K_1,Z_1] \cap Z_2 \not= 1$. So we get $L_2 \leq M$, a contradiction.

This shows $[\tilde{Z_1} \cap Q_\alpha , Z_\beta] = 1$ for $\beta \in \Delta (\alpha)$ with $Z_1 \not \leq L_\beta$. By (6.1) we have $[\tilde{Z_1} \cap Q_\alpha , V_\alpha] = 1$. Hence $|\tilde{Z_1} : \bC_{\tilde{Z_1}}(V_\alpha)| \leq r^2$. If $K_1$ is solvable or isomorphic to $J_1$ then $|V_\alpha : \bC_{V_\alpha}(\tilde{Z_1})| \leq 2$ contradicting (6.16). Let $K_1$ not isomorphic to $L_3(q)$. Then $|V_\alpha : \bC_{V_\alpha}(\tilde{Z_1})| \leq q$. By (6.16) $|V_\alpha : \bC_{V_\alpha}(\tilde{Z_1})| \geq r^2$, so $r^2 \leq q$ and then we get $r^2 = q$. But now by (6.16) again $| V_\alpha : \bC_{V_\alpha}(\tilde{Z_1})| \geq r^4$, a contradiction. So we have $K_1 \cong (S)L_3(q)$ and so $r = q$. But in this case $\tilde{Z_1}$ is irreducible. As $K_1 \unlhd L_1/Q_1$, we get $L_2 \leq M$, a contradiction.  


\absa
{\bf (6.30)~Hypothesis.~} Either $L_2$ is solvable or one of the groups in (6.29). If $L_2$ is solvable then $\sigma(L_2) = \emptyset$.

\absa 
 As $Z_{1}\not\le Q_{\alpha}$,
we see that $E(L_{2}/Q_{\, 2})\not\cong A_{9}$. Now we have $|V_{\alpha}
:V_{\alpha} \cap Q_{\, 2}|\leq 8$ or $E(L_{2}/Q_{\, 2})\cong
L_{2}(r)\times L_{2}(r)$, or $3\cdot A_{6} * 3\cdot A_{6}$ and $|V_{\alpha}
:V_{\alpha} \cap Q_{\, 2}|\leq 64$. But $V_{\alpha} $ has to act
quadratically which shows $|V_{\alpha} :V_{\alpha} \cap Q_{\, 2}|\leq
16$. So we have 
\\

\absa
{\bf (6.31)~Lemma.~}{\it Assume (6.30). Then  $|V_{\alpha} :V_{\alpha} \cap Q_{\, 2}|\leq
16$. If $E(L_{2}/Q_{\, 2})\not\cong 3\cdot A_{6} * 3\cdot A_{6}$ or
$L_{2}(r)\times L_{2}(r)$ then even  $|V_{\alpha} :V_{\alpha} \cap Q_{\, 2}|\leq 8$ 
 and $|V_{\alpha} :V_{\alpha} \cap Q_{\,
2}|\leq 4$ if $L_{2}$ is nonsolvable.}

\absa
{\bf (6.32)~Lemma.~}{\it Suppose (6.30) with $b = b_1 > 1$. Then $L_{1}$ is not as in (4.9) (ii) and also not the
special case.}

%b)~Let $K$ be a component of $L_{1}/Q_{1}$ with $[V_{\alpha} \cap
%L_{1},K]\not= 1$. If $m_{p}(K)\geq 2$ and $V_{\alpha} \not\le M$,
%then $K$ admits a quadratic fours group. If $p\not= 3$ then $|Z_{\alpha+1}
%\cap L_{1}:\bC_{Z_{\alpha+1} \cap L_{1}} (K)|>2$. Either $|Z_{\alpha +1}\cap
%L_{1}:Z_{\alpha +1}\cap Q_{1}|>2$ or $|Z_{1}\cap L_{\alpha
%+1}:Z_{1}\cap Q_{\alpha +1}|>2$.}

\absa
Proof~.~ Suppose false. We see that $|V_{\alpha} :\bC_{V_{\alpha} } (Z_{1})|\leq
2^{\, 2}\cdot |V_{\alpha} :V_{\alpha} \cap Q_{\, 2}|$. Hence with
(1.31) and (6.16), (6.28) we get that $|Z_{1}:Z_{1}\cap Q_{\alpha} |\leq 4$
and $|Z_{1}:Z_{1}\cap Q_{\alpha} |=2$ for $p=3$. 

Let first $L_{1}$ to be nonsolvable. Suppose there is some $t\in Z_{1} \cap L_{\beta}$
with $[Z_{\beta} ,t]\not= 1$ for $\beta\in\triangle (\alpha )$. Let
$[O_{r}(L_{\beta} ),Z_{\beta} ]\not= 1$. We have that $t$ inverts
some subgroup $R$ of order $r^{\, 2}$ in $O_{r}(L_{\beta} )$. As
$|[Z_{\beta} ,t]|\leq 2^{\, 6}$, this implies $|[R,Z_{\beta} ]|\leq
2^{\, 12}$ and $|[O_{r}(L_{\beta} ),Z_{\beta} ]|\leq 2^{\, 18}$. Now
Sylow $r$-subgroups of $L_{18}(2)$ have to be non abelian. This shows
$r=7$ and $|[Z_{\beta} ,t]|=2^{\, 6}$. But then $L_{\beta}$ does not
involve $A_{5}$ and so $|[Z_{\beta} ,t]|\leq 2|V_{\alpha}
:V_{\alpha}\cap Q_{\, 2}|\leq 2^{\, 5}$,  a contradiction.

This shows $[Z_{1},O_{r}(L_{1})]=1$. But then $[O_{r}(L_{1}),\Omega_{1} (Z(S))]=1$
and so by (4.10) $L_{2}\leq M$, a contradiction. 

This now proves $[Z_{1}\cap Q_{\alpha} ,V_{\alpha}]=1$. But then $V_{\alpha}$
centralizes a subgroup of index four in $Z_{1}$, a contradiction as
before as $[V_{\alpha}\cap L_{1},Z_{1}]\not= 1$ by (6.16).

So we have that $L_{1}$ is solvable. If $Z_{\alpha+1} \subseteq L_1$, then $Z_1 = \langle t ~|~|Z_{\alpha + 1}(t)| \leq 2 \rangle$. But then any such $t$ centralizes some $x \in Z_{\alpha + 1}$ such that $\bC_G(x) \leq M_{\alpha +1}$, a contradiction. So $Z_{\alpha + 1} \not\leq L_1$. Now we get $|Z_{\alpha + 1}\cap L_1 : Z_{\alpha + 1} \cap Q_1| \leq 4$, otherwise there is some $1 \not= x \in [Z_{\alpha +1}\cap L_1, Z_1]$ such that $\bC_G(x) \leq M$ and so $Z_{\alpha + 1} \leq L_1$, a contradiction. Hence we have $|Z_{\alpha + 1} : \bC_{Z_{\alpha + }}(Z_1) | \leq 16$. As $\bC_{Z_{\alpha +1}}(Z_1)$ does not contain some $x$ with $\bC_G(x) \leq M_{\alpha + 1}$ we get $m_{p}(L_{1})=3$ or $p=3$ and $m_{p}(L_{1})=4$. Suppose $m_p(L_1) = 4$. Then $|Z_{\alpha+1}:\bC_{Z_{\alpha +1}} (Z_{1})|\leq 8$, a contradiction again. 
So in
any case $m_{p}(L_{1})=3$. Now Sylow 2-subgroup of $L_{1}/Q_{1}$ are
cyclic. Let $t\in Z_{\alpha +1}\cap L_{1}$, $[t,Z_{1}]\not= 1$. Then
$t$ acts fixed point freely on $O_p(L_{1}/Q_{1})$. As $|[t,Z_{1}]|\leq
8$, this implies $p=3$ and $|[\omega ,Z_{1}]|=4$ for some $\omega\in
O_{3}(L_{1}/Q_{1})^{\,\sharp}$. But this  contradicts (1.52).

We now have that $[Z_{\alpha +1}\cap Q_{\, 2},Z_{1}]=1$ and $|Z_{\alpha
+1}:\bC_{Z_{\alpha +1}} (Z_{1})|\leq 4$. But then $p=3$ as before.
Now $|Z_{\alpha +1}:\bC_{Z_{\alpha +1}} (Z_{1})|\leq 2$ and then as
before $Z_{1}\le M_{\alpha +1}$, a contradiction.

%\absa
%b)~If any of the assertions is false we may assume that $|Z_1 \cap L_{\alpha+1} %: Z_1 \cap Q_{\alpha + 1}| = 2$ and so $|Z_{\alpha + 1} : \bC_{Z_{\alpha + %1}}(Z_1)| \leq 2^5$. This implies $p=3$, otherwise $Z_{1}$
%centralizes some $x\in Z_{\alpha +1}$ such that $\bC_{G} (x)\le
%M_{\alpha +1}$. Furthermore $|Z_{\alpha +1}:\bC_{Z_{\alpha +1}}
%(Z_{1})|\geq 2^{\, 4}$. Hence $|V_{\alpha} :V_{\alpha}\cap Q_{\,
%2}|\geq 2^{\, 3}$ and so $L_{2}/Q_{\, 2}$ is solvable. Now $|Z_{\alpha +1}
%:Z_{\alpha +1}\cap Q_{\, 2}|=8$ and so $Z_{\alpha +1}\not\le M$
%hence $|Z_{1}:Z_{1}\cap Q_{\,\alpha} |=8$ by symmetry. But now $V_{\alpha}$ is %an
%F-module, contradicting that $L_{2}$ is a $\{ 2,r\}$-group $r\geq 5$. 

\absa
{\bf (6.33)~Lemma.~} {\it Suppose (6.30). Then $[Z_1 \cap Q_\alpha , Z_{\alpha + 1}] \not= 1$.}

\absa
Proof~. Suppose false. Then $|Z_1 : \bC_{Z_1}(Z_{\alpha+1})| \leq 16$ and $|Z_1 : \bC_{Z_1}(Z_{\alpha+1})| \leq 8$ for $3 \in \sigma (M)$. Hence as $Z_1 \not\leq M_{\alpha + 1}$, we have $Z_{\alpha+1} \cap L_1 \not\leq Q_1$. So let $K$ be a component of $L_1/Q_1$ with $[K, Z_{\alpha + 1} \cap L_1] \not= 1$.\\

(1)~~~~~~~$[K, Z_{\alpha+1} \cap L_1] \leq K$\\


Let $t \in Z_{\alpha+1} \cap L_1$ with $K^t \not= K$. Let first $|Z_{\alpha + 1} \cap L_1 : \bC_{Z_{\alpha + 1} \cap L_1}(K)| \geq 4$. Then by (1.11) we have $K \cong L_2(q)$, $q$ even. As $|Z_1 : \bC_{Z_1}(t)| \leq 16$ we then get $q \leq 4$. Hence $K \cong A_5$ and $[K \times K^t, Z_1]$ is the $O_4^+(4)$-module. Finally $3 \not\in \sigma (M)$. Now for $s \in L_1 \cap Z_{\alpha + 1}$, with $[s,K] =1$, we see that $[s,[K \times K^t, Z_1]] = 1$. But as $\bC_{Z_1}(u) = \bC_{Z_1}(Z_{\alpha + 1} \cap L_1)$ for $u \in Z_{\alpha + 1} \cap K \times K^t$, we get that $[s,Z_1] = 1$. Hence $|Z_{\alpha +1 } \cap L_1 : Z_{\alpha + 1} \cap Q_1 | \leq 8$. As $5 \in \sigma (M)$ we now see that $Z_1 \leq M_{\alpha + 1}$, a contradiction. 

So we have $|Z_{\alpha + 1} \cap L_1 : \bC_{Z_{\alpha + 1} \cap L_1}(K)| = 2$. There is $U \leq [K \times K^t, Z_1]$ with $U \cap \bC(Z_{\alpha + 1}) = 1$ and $|U| \geq 4$. Now by (6.16) $|V_\alpha : \bC_{V_\alpha}(U)| \geq 2^6$. This shows that $|V_\alpha \cap L_1 : \bC_{V_\alpha \cap L_1}(K)| \geq 2^2$. In particular $K \cong A_5$ again. Further $|V_\alpha : V_\alpha \cap L_1| \geq 8$. If $3 \not\in \sigma (M)$ then $5 \in \sigma (M)$. As $[Z_1, K \times K^t]$ is generated by elements $x$ which centralize a subgroup of index $2^7$ in $V_\alpha$, we see that $[Z_1, K \times K^t] \leq \bigcap_{\beta \in \Delta (\alpha)} M_\beta$. So $[Z_1 , K \times K^t] \leq Q_\alpha$, a contradiction. Hence we have $3 \in \sigma (M)$. Now we get that $L_2$ is solvable with $m_r(L_2) = 3$. Furthermore $L_2$ possesses an $F_1$-module with offending foursgroup. This contradicts (1.36).\\

(2)~~~~~~~~~~$K/Z(K) \not\cong A_n, n \not= 8$\\


Suppose $K/Z(K) \cong A_n$. Assume furthermore $3 \in \sigma (M)$. Then $|[t, Z_1]| \leq 8$ for $t \in Z_{\alpha+1} \cap L_1$. By (1.30) $[K,Z_1]$ just involves natural modules or $n \leq 7$. So assume $n \geq 9$. Then any element in the natural module is centralized by $E \cong E_9$ with $\Gamma_{E,1}(G) \leq M$. Hence we get $L_2 \leq M$, a contradiction. 

So we have $n \leq 7$. This shows $|V_\alpha \cap L_1 : V_\alpha \cap \bC(K)| \leq 8$. Now as before there is no foursgroup $U$ with $\bC_U(Z_{\alpha + 1}) = 1$. So we get that $t$ induces a transvection and then $[K, Z_1]$ is the natural module. This again shows $K \cong A_6$ or $A_5$. Furthermore $K \not\!\!\unlhd L_1/Q_1$. But then again $L_2 \leq M$. 

So we have $3 \not\in \sigma (M)$. Hence $n \leq 11$. As $|V_\alpha \cap L_1 : \bC_{V_\alpha \cap L_1}(K)| \leq 2^5$, we see that $p = 5$ or 7. In the former $|V_\alpha \cap L_1 : \bC_{V_\alpha \cap L_1}(K)| \geq 2^4$ and so $n \geq 9$. As $3 \not\in \sigma (M)$, we see that $K \unlhd L_1/Q_1$ and so as $m_p(K) \leq 2$, we see that some $x \in \Omega_1(Z(S)) \cap [K, Z_1]$ is centralized by some $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \leq M$, contradicting $L_2 \not\subseteq M$. So we have $p = 7$ and $|V_\alpha \cap L_1 : \bC_{V_\alpha \cap L_1}(K)| \geq 2^2$. Now $n \geq 7$ and $K \unlhd L_1/Q_1$. But then as before we get $L_2 \leq M$, a contradiction.
\\

(3)~~~~~~~~~~~~~$K/Z(K)$ is not sporadic.\\

As there are no transvections we get $|V_\alpha \cap L_1 : \bC_{V_\alpha \cap L_1}(K)| \geq 4$. So $K$ is one of the groups in (1.12). By (1.30) $|Z_1 : \bC_{Z_1}(t)| \geq 8$ and so we even get $|V_\alpha \cap L_1 : \bC_{V_\alpha \cap L_1}(K)| \geq 16$. Hence by (1.12) we have $K \cong 3\cdot M_{22}$. Now by (1.30) we have $|Z_1 : \bC_{Z_1}(t)| = 16$ and so $3 \not\in \sigma(M)$. But then we even get $|V_\alpha : \bC_{V_\alpha}([Z_1,K])| \geq 2^{10}$, a contradiction.\\

(4)~~~~~~If $K/Z(K)$ is of Lie type, then the characteristic is two.\\


Suppose $K$ is of of odd characteristic. As there are no transvections we get
$|V_\alpha \cap L_1 : \bC_{V_\alpha \cap L_1}(K)| \geq 4$ and so by (1.12) $K/Z(K) \cong U_4(3)$. So $3 \in \sigma (M)$. Now $|V_\alpha : \bC_{V_\alpha}(K)| \leq 2^8$. As $[Z_1,K] \not\leq Q_\alpha$, we get that $m_3(L_1) \leq 6$ and so $K \unlhd L_1/Q_1$. By the construction of $L_1$ we see $\bC_{L_1/Q_1}(K) = Z(K)$. Now we have that $|Z_{\alpha+1} \cap L_1 : Z_{\alpha+1} \cap Q_1| = 2$. In particular $|Z_{\alpha + 1} : \bC_{Z_{\alpha + 1}}(Z_1)| \leq 16$. As $m_3(L_1) = 5$, we see that $Z_1 \leq M_{\alpha + 1}$, a contradiction.\\

(5)~~~~~~~~~~~~~~~$[E(L_1/Q_1), Z_{\alpha + 1} \cap L_1] = 1.$\\


Otherwise by (1) -  (4) there is $K \cong G(q), q = 2^m$, with $1 \not= [K, Z_{\alpha + 1} \cap L_1] \leq K$.  
 

Suppose $K \cong L_n(q)$. Then there are at most four nontrivial irreducible modules involved. Assume there are exactly four. Then $q = 2$ and these modules are natural ones. Further $3 \not\in \sigma (M)$ and so $n \leq 7$. If $n \geq 4$, we have $K \unlhd L_1/Q_1$ and so $E \cong E_{p^2}$ centralizes $[K, Z_1]$, a contradiction or $p = 7$ and $n \geq 6$. Then some $7$-element centralizes $[K,Z_1]$. But some $1 \not= x \in \Omega_1(Z(S)) \cap [Z_1, K]$ is centralized by a $7$-element in $K$, so we have the contradiction $L_2 \leq M$. So let $n = 3$, $K \cong L_3(2)$. Then $p = 7$. But now $[Z_1, K]$ centralizes in $V_\alpha$ a subgroup of index $2^6$, a contradiction.

Assume now there are exactly three modules involved. Again $q = 2$ and these modules are natural modules or $n = 4$ and one of them is the orthogonal module.
If $3 \not\in \sigma (M)$, we get a contradiction as before. So let $3 \in \sigma (M)$. If $n \geq 7$ any $x \in \Omega_1(Z(S))$ is centralized by $E \cong E_9$, a contradiction. So we have $n \leq 6$. As $|V_\alpha \cap L_1 : V_\alpha \cap \bC(K)| \geq 2^5$, we get $n \geq 5$. If $n = 5$, then $m_3(K) = 2$, and as any $x \in \Omega_1(Z(S)) \cap [Z_1,K]$ is centralized by some $3$-element in $K$, we get $L_2 \leq M$. So we have $n = 6$. But then the natural module and its dual is involved. Hence in one of these modules we get some $x \in \Omega_1(Z(S))$ which is centralized by some $E \cong E_{p^2}$ in $K$, a contradiction.

So let now be exactly two modules involved. Then $q \leq 4$. Suppose $q = 4$. then we have two natural modules or duals. Furthermore $3 \not\in \sigma (M)$. Let $n \geq 3$. Then we have $K \unlhd L_1/Q_1$. In particular we have $m_p(K) \geq 2$ if $p =5$ and $m_p(K) \geq 3$ else. As $3 \not\in \sigma (M)$ we have $K \cong L_4(4)$. This shows $p = 5$. But any $x \in \Omega_1(Z(S)) \cap [K, Z_1]$ is centralized by a $5$-element in $K$ and so $L_2 \leq M$, a contradiction. So $K \cong L_2(4)$. Now $x\in [Z_1,K]$ centralizes in $V_\alpha$ a subgroup of index $2^7$, which yields $x \in Q_\alpha$, as $p = 5$, a contradiction. 

So we have $q = 2$ again. By (1.30) we just have two nontrivial modules involved and we get a contradiction as before, or $n=4$, $q =2$ and we have two orthogonal modules. But this case we will handle lateron together with the orthogonal groups. 

So we are left with $[K,Z_1]$ involves exactly one irreducible nontrivial module and so $q \leq 16$. Let $[t,[K,Z_1]] \leq 8$, then $q \leq 8$ and by (1.30) the module is the natural one, or $n=2$, $q=4$ and we have the orthogonal module. Suppose $3 \in \sigma (M)$. Then we see that $n \leq 4$. Let $n = 4$, then $K \cong SL_4(8)$ or $L_4(2)$. Let $n \geq 3$, then $x \in \Omega_1(Z(S))$ is centralized by $E \cong E_9$, a contradiction. So we have $K \cong L_2(8)$ or $L_2(4)$. So for $x \in [Z_1,K]$ we have $|V_\alpha : \bC_{V_\alpha}(x)| \leq 2^6$. This shows $K \cong L_2(4)$. But then $|V_\alpha : \bC_{V_\alpha}(x)| \leq 2^5$, a contradiction. So we have $3 \not\in \sigma (M)$. Let $m_3(K) \geq 2$. Then $K \unlhd L_1/Q_1$. Hence we get $m_p(K) \geq 2$ for $p \Big|~q-1$ and $m_p(K) \geq 3$ for $p \not\Big|~q-1$. This shows $p = 7$ and $K \cong L_n(8)$, $4 \leq n \leq 7$. As no element in $\Omega_1(Z(S))^\sharp$ is centralized by $E \cong E_{7^2}$, we get a contradiction. So let $m_3(K) = 1$, i.e. $K\cong L_2(4)$, $L_2(8)$, $L_3(2)$ or $L_3(8)$. We have $K \not\cong L_2(4)$. Otherwise $p = 5$ and as any $x \in [Z_1,K]$ centralizes a subgroup of index $2^7$ in $V_\alpha$, we get a contradiction. So we have $p = 7$ and $K$ has at most two conjugates under $S$. As no $x \in \Omega_1(Z(S))^\sharp$ is centralized by $E \cong E_{7^2}$, we get $K \cong L_2(8)$ or $L_3(8)$. If $K \cong L_3(8)$, then there is some $\omega$ with $o(\omega) = 7$ and $[K,\omega] = 1$. Now as any $x \in \Omega_1(Z(S))$ is centralized by some 7-element in $K$ we see that there is $x \in \Omega_1(Z(S))^\sharp$ which is centralized by $E \cong E_{7^2}$, a contradiction. So we have $K \cong L_2(8)$ and $K$ is not normalized by $S$. As before there is $U \leq [K,Z_1]$, $|U| = 8$ with $U \cap \bC(Z_{\alpha + 1}) = 1$. By (6.16) $|V_\alpha : \bC_{V_\alpha}(U)| \geq 2^8$. But as $V_\alpha$ normalizes $K$, we get $|V_\alpha : \bC_{V_\alpha}(U)| \leq 2^7$, a contradiction.

So we have $|[t,[Z_1,K]]| = 16$. In particular $3 \not\in \sigma (M)$. Now $n \leq 7$. Suppose $K \unlhd L_1/Q_1$, further $q > 2$. If $n \geq 5$, then $p \Big|~q-1$. Now by (1.33) we have $E \cong E_{p^2}$ which centralizes $\Omega_1(Z(S))$, a contradiction. So we have $n \leq 4$. As $E(L_2/Q_2) \cong L_2(r) \times L_2(r)$, or $3\cdot A_6 * 3\cdot A_6$, we see with (6.16) that for some $U \leq [Z_1,K]$, $|U| = 16$, we have $|V_\alpha : \bC_{V_\alpha}(U)| \geq 2^{12}$. Hence $|V_\alpha \cap L_1 : \bC_{V_\alpha \cap L_1}(K)| \geq 2^8$. This shows $K \cong L_3(16)$, or $L_4(q)$. In the former $p = 5$ and $[K,Z_1]$ involves the natural module just once. But now we get some $x \in \Omega_1(Z(S))^\sharp$ which is centralized by $E \cong E_{25}$ in $L_1$, a contradiction. So we have $K \cong L_4(q)$. If $p \Big| q-1$, then $m_p(K) = 3$ and $E \cong E_{p^2}$ centralizes $\Omega_1(Z(S))$. So we have $p \not\!|~q-1$. Now $m_p(K) \leq 2$ and so some $E \cong E_{p^2}$ centralizing $K$ centralizes $[K,Z_1]$ too, a contradiction.

So we are left with $K \not\!\!\unlhd L_1/Q_1$ and then $m_3(K) = 1$, i.e. $K \cong L_2(q)$ or $L_3(q)$, or $3 \Big| |Z(K)|$ and so $K \cong SL_3(q)$. Let $K \cong L_3(q)$, $q > 2$ or $SL_3(q)$, $q > 4$. Then $p | q-1$ and so some $x \in \Omega_1(Z(S))^\sharp$ is centralized by $E \cong E_{p^2}$, a contradiction. So we have $K \cong L_2(4)$, $L_2(8)$, $L_2(16)$, $L_3(2)$ or $SL_3(4)$. As $|V_\alpha : V_\alpha \cap \bC(K)| \geq 2^{12}$, we get a contradiction.

Let next $K \cong Sp_{2n}(q)$, $n \geq 2$ or $\Omega_{2n}^\pm(q)$, $n \geq 3$. If $[K,Z_1]$ involves more than one nontrivial irreducible $K$-module then by (1.30) these are natural modules or $q=2$ , $n = 3$ and the spin module is involved. If four modules are involved we have $q = 2$ and $3 \not\in \sigma (M)$. So $n = 2,3$ or $K \cong \Omega_8^-(2)$. Now $m_p(\bC(K)) \geq 3$, contradicting $L_2 \not\leq M$. Let exactly three nontrivial irreducible modules be involved.  Again $q = 2$ and these modules are natural modules or one of them is the spin module for $Sp_6(2)$ or $\Omega_6^-(2)$. In the latter $3 \not\in \sigma (M)$ and $|V_\alpha \cap L_1 : \bC_{V_\alpha \cap L_1}(K)| \geq 2^8$, a contradiction. So we have three natural modules. If $3 \not\in \sigma (M)$ we get a contradiction as before. So assume that $3 \in \sigma (M)$. Then we get $K \cong Sp_4(2)$ or $\Omega_6^-(2)$. Furthermore we have $K \unlhd L_1/Q_1$. This shows $K \cong \Omega_6^-(2)$. But now $|V_\alpha : \bC_{V_\alpha}(K)| \geq 2^8$, a contradiction.

Let now exactly two nontrivial irreducible modules be involved. Again by (1.30) these are natural ones $q \leq 4$ or $K \cong Sp_6(2)$ or $\Omega_6^-(2)$ and there is a spin module. If $q = 4$ we have $3 \not\in \sigma (M)$. So we have $K \cong Sp_4(4)$, $Sp_6(4)$, $\Omega_6^\pm(4)$. But $K \cong Sp_4(4)$ is impossible. We have that $K \unlhd L_1/Q_1$ and so either $m_p(K) \geq 3$ or $p = 5$ and $m_5(K) = 2$. This in any case shows $p = 5$. As there is no $E_{p^2}$ centralizing some $x \in \Omega_1(Z(S))^\sharp$, we get $K \cong \Omega_6^-(4)$. But now we see $|Z_{\alpha + 1} : \bC_{Z_{\alpha+1}}(Z_1)| \leq 2^5$ and so $Z_1 \leq M_{\alpha +1}$, a contradiction.

So we have $q = 2$. Suppose $3 \not\in \sigma (M)$, then $K \cong Sp_6(2)$, $\Omega_6^-(2)$ or $\Omega_8^-(2)$ and $K \unlhd L_1/Q_1$. Further $m_p(K) =1 $ and so some $E \cong E_{p^2}$ centralizes some $x \in \Omega_1(Z(S))^\sharp$, a contradiction. 

So we have $3 \in \sigma (M)$. As no $x \in \Omega_1(Z(S))^\sharp$ is centralized by $E \cong E_9$, we see $K \cong \Omega_6^-(2)$. As $m_3(K) = 3$, we must have two natural modules and $|Z_\alpha \cap L_1 : \bC_{Z_\alpha \cap L_1}(K)| = 2$. This shows $|V_\alpha \cap L_1 : \bC_{V_\alpha \cap L_1}(K)| \leq 16$. Hence we have $U \leq Z_1$, $|U| = 4$ with $U \cap \bC(Z_{\alpha + 1}) = 1$. If $L_2$ is solvable then $|V_\alpha : \bC_{V_\alpha}(U)| \geq 2^8$ by (6.16), a contradiction. So $L_2$ is nonsolvable and $|[Z_1,t]| = 4$, In particular $|Z_{\alpha +1} : \bC_{Z_{\alpha+1}}(Z_1)| = 8$, contradicting $Z_1 \not\leq M_{\alpha + 1}$.

So we have that $[K,Z_1]$ contains exactly one nontrivial irreducible $K$-module. Let first $|[t, [Z_1,K]]| \leq 8$. Then we have the natural module or the spin module in case of $K \cong Sp_6(2)$ or $\Omega_6^-(2)$. Suppose $p~|~q-1$, then we get $K \cong Sp_4(q)$ or $\Omega_6^-(q)$ or $K \cong Sp_6(q)$ and we have the spin module. In the latter $q = 4$ but then $\Omega_1(Z(S))^\sharp$ is centralized by an elementary abelian group of order 9, a contradiction. So we have $K \cong Sp_4(q)$ or $\Omega_6^-(q)$ and the natural module. In both cases $m_p(K) = 2$. But then there is a $p$-element centralizing $[K,Z_1]$, a contradiction. So we have $p~\not|~q-1$. Then $K \cong Sp_4(q)$, $Sp_6(q)$, $\Omega_6^-(q)$, $\Omega_8^-(q)$, $\Omega_6^+(q)$. Further $m_p(K) \geq 3$. But this is impossible. 

So we have $|[t,[Z_1,K]]| = 16$ and $3 \not\in \sigma (M)$. This again shows $K \cong Sp_4(q)$, $Sp_6(q)$, $\Omega_6^-(q)$, $\Omega_8^-(q)$, $\Omega_6^+(q)$. Further $K \unlhd L_1/Q_1$. As before $p~|~q-1$. This shows $q = 8$ and $p = 7$ or $q =16$ and $p = 5$. Finally $m_p(K) \geq 3$. So $K \cong Sp_6(q)$, $\Omega_8^-(q)$ or $\Omega_6^+(q)$. By (1.22) in any case we have the natural module. But now some $x \in \Omega_1(Z(S))^\sharp$ is centralized by $E \cong E_{p^2}$ in $K$ with $\Gamma_{1,E}(G) \leq M$, a contradiction.

Let now $K \cong U_n(q)$ or $G_2(q)$. Then $q \leq 4$ and there are at most two nontrivial irreducible modules involved. Assume there are exactly two. Then $q = 2$ and these modules are natural ones by (1.30). Further $3 \not\in \sigma (M)$. So $K \cong U_4(2)$ or $G_2(2)^\prime$. But $U_4(2) \cong \Omega_6^-(2)$, a case just handled. So $K \cong G_2(2)^\prime$ and $K \unlhd L_1/Q_1$. Now there is $U \leq Z_1$, $|U| = 16$ with $U \cap \bC(Z_{\alpha + 1}) = 1$ and $|V_\alpha : \bC_{V_\alpha}(U)| \leq 16\cdot 8$, a contradiction. 

Hence $[K,Z_1]$ involves exactly one nontrivial irreducible $K$-module. If $K \cong U_n(q)$, $n \geq 5$, then $p~|~q+1$ and so some $x\in \Omega_1(Z(S))^\sharp$ is centralized by some $E \cong E_{p^2}$ with $\Gamma_{1,E}(G) \leq M$. So we have $K \cong U_4(2)$, $U_4(4)$, $U_3(4)$, $G_2(2)^\prime$ or $G_2(4)$. Suppose $q = 4$. Then we have $|V_\alpha \cap L_1 : \bC_{V_\alpha \cap L_1}(K)| \geq 2^8$ and just $K \cong U_4(4)$ is possible. Now we get $m_p(K) \geq 3$ and so $p = 5$. But now $|Z_{\alpha+1} \cap L_1 : Z_{\alpha + 1} \cap Q_1| = 4$ and we get $|Z_{\alpha + 1} : \bC_{Z_{\alpha +1}}(Z_1)| \leq 2^6$, contradicting $Z_1 \not\leq M_{\alpha + 1}$.

So we have $q = 2$. Now we see $|[[K,Z_1],t]| = 4$ and so just the natural module is involved in $[K, Z_1]$. If $3 \not\in \sigma (M)$, then $K \unlhd L_1/Q_1$ and we are done as $m_p(K) = 1$. So we have $3 \in \sigma (M)$. Again $K \unlhd L_1/Q_1$. This shows $K \cong U_4(2)$ and $m_3(\bC_{L_1/Q_1}(K)) =1 $. Now let $U$ be as before. We see that $|V_\alpha : \bC_{V_\alpha}(U)| \leq 2^7$, and so $L_2$ is nonsolvable and $|Z_1 : Z_1 \cap Q_\alpha| = 4$. Hence $|Z_{\alpha + 1} \cap L_1 : Z_{\alpha + 1} \cap Q_1| = 2$. So $Z_1$ centralizes a subgroup of index $8$ in $Z_{\alpha + 1}$, a contradiction. 

By (1.22) we are done. \\

(6)~~If $K \unlhd L_1/Q_1$, $K$ a $p$-group, $p \in \sigma (M)$, then $[K, Z_{\alpha + 1} \cap L_1 ] = 1$.\\

Suppose false. Choose $t \in Z_{\alpha + 1} \cap L_1$, $1 \not= \omega \in K$ with $\omega^t = \omega^{-1}$. Then as $|[Z_1,t]| \leq 16$, we get $o(\omega) = 3,5,7,17$. Suppose $p = 5$ or $17$. Then $|Z_{\alpha + 1} \cap L_1 : Z_{\alpha + 1} \cap Q_1| \geq 2^4$ and so there is $X \leq K((Z_{\alpha +1} \cap L_1)Q_1/Q_1)$ $X \cong D_{2p} \times D_{2p} \times D_{2p} \times D_{2p}$ with $(Z_{\alpha + 1} \cap L_1)Q_1/Q_1 \in Syl_2(X)$. But then there is 
$E \cong E_{p^2}$ centralizing $Z_1$ , a contradiction.

So we have $p = 7$ or $3$. If $p = 7$ the same argument as before shows that $|V_\alpha : V_\alpha \cap L_1| = 16$. Now $|V_\alpha \cap L_1 : V_\alpha \cap Q_1| \geq 2^8$. But then we see that there is some subgroup $U$ of $Z_1$ of order 8 centralizing a subgroup of index $2^5$ in $V_\alpha$, a contradiction. 

So $p = 3$ and $|[Z_1,t]| \leq 8$. If $|Z_{\alpha + 1} \cap L_1 : Z_{\alpha + 1} \cap Q_1| \geq 4$, then we find $U \leq Z_1$, $|U| = 4$ with $\bC_U(Z_{\alpha+1}) = 1$ and $|V_\alpha : \bC_{V_\alpha}(U)| \leq 2^5$, a contradiction. So we have $|Z_{\alpha + 1} \cap L_1 : Z_{\alpha + 1} \cap Q_1| = 2$. Now we get $m_3(L_1) = 4$. Further $|Z_{\alpha + 1} : Z_{\alpha + 1} \cap L_1| = 8$, and so $L_2$ is solvable. Now we see that $|[Z_1,t]| = 2$. Let $s \in Z_1$, $[s,t] \not= 1$ with $|V_\alpha \cap L_1 : \bC_{V_\alpha \cap L_1}(s)| = 2$. Then $|V_\alpha : \bC_{V_\alpha}(s)| \leq 16$ and so by (6.16) we have equality and $L_2$ is a $\{2,5\}$-group. Let $\omega$ be as before. Now $|[Z_1,\omega]| = 4$. Hence $\langle \omega \rangle \not\!\!\unlhd L_1/Q_1$. But then there is $E \cong E_9$ which centralizes a subgroup of index 16 in $Z_1$. As $Z_1 \not\leq M_{\alpha+1}$, we see that $|Z_1| \leq 2^8$. Furthermore $3~\not\!|~|\bC(Z_1)|$. So $K$ is isomorphic to a subgroup of $\Z_3 \wr \Z_3 \times \Z_3$. Now we have $|V_2 : \bC_{V_2}(Z_{\alpha+1})| \geq 2^{12}$, which shows that $|V_2 \cap L_{\alpha+1}/V_2 \cap Q_{\alpha +1}| \geq 2^9$. As $L_1$ is solvable this is impossible.

\absa
To prove the assertion it is enough to treat the case $|O_r(L_1/Q_1)| = r^3$ and $L_1/O_{2,r}(L_1) \cong L_2(r)$, $r$ odd. By (1.51) we have $[O_r(L_1) , Z_1] = 1$. But then by (4.10) $L_2 \leq M$, a contradiction.\\

\absa
{\bf (6.34)~Lemma.~}{\it If $b = b_1 > 1$ then $L_2$ is solvable with $\sigma (M) \not= \emptyset$.}

\absa
{Proof. } We choose notation as before.\\

(1)~~~~~~~~~~$[Z_{\alpha+1} \cap L_1 , Z_1] \not= 1$
\\


Otherwise $Z_{1}$ centralizes
a subgroup of index $2^{\, 4}$ in $Z_{\alpha +1}$. This shows $p=3$,
$|Z_{\alpha +1}:Z_{\alpha +1}\cap Q_{\, 2}|=16$ and Sylow 3-subgroup
of $L_{2}$ are cyclic, a contradiction to $m_{2}(L_{2}/Q_{\, 2}) = 1$.


Let $K$ be a component of $L_1/Q_1$ with $1 \not= [K, Z_{\alpha + 1} \cap L_1]$. Until further notice assume $m_p(K) \geq 2$.\\

(2)~~~~~~~~~~~~$[K, Z_{\alpha+1} \cap L_1] \leq K$\\

Otherwise $|Z_{\alpha+1} \cap L_1 : \bC_{Z_{\alpha+1} \cap L_1}(K)| = 2$. We have $|[[K \times K^t, Z_1],t]| \geq 4$, for some $t \in Z_{\alpha +1 } \cap L_1$. Suppose there is $U \leq [K \times K^t , Z_1]$ , $|U| = 4$ with $U \cap M_{\alpha + 1} = 1$. Then $|V_\alpha : \bC_{V_\alpha}(U)| \geq 64$, a contradiction.  So there is some $s \in [K \times K^t, Z_1]^\sharp$ with $s \in L_{\alpha + 1}$ and $[Z_{\alpha+ 1}, s] \not= 1$. Hence we have $|[Z_{\alpha + 1} , s]| \leq 2^5$. As $[Z_{\alpha +1}, K_{\alpha+1}]$ has to involve at least three nontrivial irreducible modules, we get that $s$ induces a transvection on one of them. So $K \cong A_n$, $L_n(2)$, $Sp_{2n}(2)$ or $\Omega_{2n}^\pm(2)$. Further $K$ is isomorphic to a subgroup of $L_5(2)$. This shows $K \cong L_4(2)$, $L_5(2)$ or $A_6$. In any case $p = 3$ and so even $|[Z_{\alpha + 1} , s]| \leq 2^4$. Hence $K \cong L_4(2)$ or $A_6$. But in both cases $[s,Z_1]$ is normalized by $E \cong E_9$, a contradiction.\\

(3)~~~~~~~~~~~~~~{ $K\not\cong A_{n}$}\\


Suppose $K\cong A_{n}$. As $m_{p}(K)\geq 2$, we have $n\geq 6$.
If $n\geq 12$, then $p=3$. Let $s \in Z_1 \cap L_{\alpha + 1}$. If $n\geq 13$, then $s$ is centralized by $E \cong E_9$ in $K$, a contradiction.

So we have $n\leq 12$. Suppose next that $|Z_{\alpha + 1} \cap L_1 : \bC_{Z_{\alpha+1} \cap L_1}(K)| = 2$ and $|[t, [Z_1,K]]| \geq 16$, for $t \in Z_{\alpha +1} \cap L_1$. Then there is $U \leq [Z_1,K]$, $|U| \geq 8$ with $|Z_{\alpha+1} : \bC_{Z_{\alpha+1}}(U)| \leq 32$. Further $U$ acts quadratically. So in any case we get with (1.30) or (1.12)(iii) that $[Z_1,K]$ just involves natural modules $T$, spin modules $T$ or $n \leq 8$ and the four dimensional module $T$ is involved. 

Suppose first $p = 3$ and natural modules are involved. If $n > 8$, then any element in the natural module is centralized by $E \cong E_9$, a contradiction. So $n \leq 8$. By (4.4)(ii) we may assume that $\bC_{L_{1}/Q_{1}} (K)$
contains an elementary abelian group of order $3^{\, 2}$. We have
$|[Z_{1},t]|\leq 2^{\, 7}$. As there are at least four conjugates of
$T$ under $\bC_{L_1/Q_1}(K)$ we see that $t$ has to induce a transvection on some of these
modules. So we get $U \cong E_{16}$ in $Z_1$ with $|Z_{\alpha +1} : \bC_{Z_{\alpha+1}}(U)| \leq 16.
$. This shows $|U : U \cap Q_\alpha| \geq 8$. Now $|V_\alpha : \bC_{V_\alpha}(U)| \geq 2^8$. Hence $V_\alpha \cap L_1 : \bC_{V_\alpha \cap L_1} (K)| \geq 2^5$, a contradiction.

Now we have $U$ as before $|U| = 16$ and so $U \cap Q_\alpha$ induces a quadratic fours group on $K_{\alpha + 1}$. So by (1.12) $s \in U$ corresponds to $(1,2)(3,4)$ or $(1,2)(3,4)(5,6)(7,8)$. As $\bC_{K_{\alpha+1}}(s)$ does not contain $E \cong E_9$, we get $n \leq 10$. By (1.12) we also see that $|[Z_{\alpha + 1} , s]| \leq 2^6$. So there is exactly one spin module involved. As $K$ is centralized by a 3-element in $L_1$, we see that the spin module is centralized by a 3-element in $L_1$. Now $[T,s]$ is normalized by $E \cong E_9$, a contradiction.

So assume $n = 7,8$ and we have the four dimensional module involved. The case $K \cong A_8 \cong L_4(2)$ will be treated among the groups of Lie type. So we have $n = 7$. Then $|Z_{\alpha+1} \cap L_1 : \bC_{Z_{\alpha +1} \cap L_1}(K)| \leq 4$ and so $|[s,Z_{\alpha + 1}]| \leq 2^5$. Now we get a contradiction as before. 

So we have $p=5$, $n=10,11$. Again by (4.4)(ii) we may assume that  $\bC_{L_1/Q_1}(K)$ contains $E \cong E_{25}$. Now we have at least 8 modules involved. Hence we get some $U \leq L_{\alpha + 1} \cap Z_1$, $U \cap Q_{\alpha + 1} = 1$, $|U| \geq 16$ and $|Z_{\alpha + 1} : \bC_{Z_{\alpha + 1}}(U)| \leq 2^9$, a contradiction.\\


(4)~~~~~~~~~~~~~~$K$ is not sporadic.\\


Suppose that $|V_\alpha \cap L_1 :\bC_{V_\alpha \cap L_1}(K)| = 2 = |Z_1 \cap L_{\alpha + 1} : Z_1 \cap Q_{\alpha + 1}|$. Then for $t \in Z_{\alpha + 1} \cap L_1$ we have $|Z_1 : \bC_{Z_1}(t)| \leq 2^5$. This shows with (1.25) that $K \cong M_{11}$, $M_{12}$, $M_{22}$, $M_{23}$, $M_{24}$, or $J_2$. Further $|Z_1 : \bC_{Z_1}(t)| \geq 8$. Hence we get $U \leq Z_1$, $|U| = 4$, $U \leq L_{\alpha + 1}$ with $U \cap Q_{\alpha + 1} = 1$, a contradiction.

So we have that $K$ admits a quadratic fours group. Hence by (1.12) $K\cong M_{12}$, $M_{22}$, $3\cdot M_{22}$,
$M_{24}$, $Co_1$, $Co_2$, $J_{2}$ or $3\cdot Suz$. Let $p=3$. Then for $M_{12}$,
$M_{22}$, $3\cdot M_{22}$ or $M_{24}$ we have by (4.4)(ii) that $3~|~|\bC_{L_1/Q_1}(K)|$ and so there are at least two
quadratic modules involved. So $|[Z_{1},t]|\geq 2^{\, 8}$, $2^{\,
6}$, $2^{\, 8}$, $2^{\, 8}$. But this contradicts the fact that
besides for $3\cdot M_{22}$ we always have  $|Z_{\alpha + 1} \cap L_1: Z_{\alpha + 1} \cap Q_1|\leq 4$ by (1.12). In case of $3\cdot M_{22}$ we get $|[Z_1, t]| \leq 2^7$, a contradiction again.  Let $K\cong
3Suz$, or $Co_1$. Then $|[Z_{1},t]|\geq 2^{\, 8}$. But $|Z_{\alpha + 1} \cap L_1: Z_{\alpha + 1} \cap Q_1|\leq 4$, a contradiction.

Also for $K\cong J_{2}$, or $Co_2$, we get $|[Z_{1},t]|\leq 2^{\, 6}$, $(2^{\, 5}$
if $p=3$). As $|[Z_1,t]| \geq 2^6$ and $p = 3$ for $K \cong Co_2$, we see $K \cong J_2$ and $Z_{1}$ involves exactly one irreducible
module. But $m_{p}(K)=2$, a contradiction as before.

\begin{itemize}
\item[(5)] $K$ is not a group of Lie type over a field of odd
characteristic, besides $K$ is a group of Lie type over a field of characteristic two, too.
\end{itemize}

Suppose $K \not\cong 3\cdot U_4(3)$. Then by (1.30) $|[[Z_1,K],t]| \geq 16$ and so we see that $K$ admits a quadratic fours group. So by (1.12) $K \cong 3\cdot U_4(3)$. Hence $p=3$. By
(1.55) $s \in Z_{\alpha + 1} \cap L_1$ is centralized by $E \cong E_9$ in $K_{\alpha + 1}$ and so $L_2 \leq M$, a contradiction.\\
 
(6) There is no component $K$ of $L_1/Q_1$ with $[K, Z_{\alpha+1} \cap L_1] \not= 1$, such that  $m_p(K) > 1$. \\


Assume that $K\cong G(q)$, $q$ even. We will establish (6) in a series
of steps.\\


(6.1) $K$ induces some $SC$-module or strong quadratic module
on $Z_{1}$.\\

We have $|[Z_{1},t]|\leq 16\cdot |Z_{1}\cap L_{\alpha +1}:Z_{1}\cap Q_{\alpha +1}|$. Then the assertion follows with (1.15) or $K\cong L_{3}(q)$,
$U_{3}(q)$, $L_{n}(2)$, $n\leq 7$, $Sp_{6}(2)$, $U_{4}(2)$,
$U_{5}(2)$, $G_{2}(2)^\prime$, $\Omega_8^{\,\pm} (2)$,
$^{2}F_{4}(2)^{\prime}$. Now in all these cases $[Z_{1},K]$ involves
exactly one nontrivial irreducible module. So if $m_{p}(K)=2$, we see 
that $[s,Z_{\alpha +1}]$ is normalized by $E\cong E_{p^{\, 2}}$ with $\Gamma_{E,1}
(G)\leq M_{\alpha+1}$, a contradiction. 

Hence we have $K\cong L_{6}(2)$, $L_{7}(2)$, $Sp_{6}(2)$,
$U_{4}(2)$, $U_{5}(2)$ or $\Omega_8^{\,\pm} (2)$ and $p=3$. Now
$|[Z_{1},t]|\leq 8|Z_{1}\cap L_{\alpha +1}:Z_{1}\cap Q_{\alpha +1}|$. So by (1.15) we have that $K\cong U_{4}(2)$ or
$U_{5}(2)$ and $|Z_{1}\cap L_{\alpha +1}:Z_{1}\cap Q_{\alpha +1}|\geq
4$. By (1.14) we have equality and $|Z_{1}:Z_{1}\cap Q_{\, \alpha}
|=8$. But then  $|[Z_1,K] : [Z_1,K] \cap Q_\alpha| \geq 8$ and so  $|V_\alpha \cap L_1 :\bC_{V_{\alpha} \cap L_1} (K)|\geq 2^{\, 5}$, a contradiction.\\

(6.2) We have $K\cong L_{n}(q)$, $n\leq 7$, ($n\leq 4$ for
$q>2$), $Sp(2n,q)$, $n\leq 3$, $U_{n}(q)$, $n\leq 5$, $\Omega_8^{-}
(q)$, $\Omega_8^{+} (2)$, $G_{2}(q)$, $^{3}D_{4}(q)$.\\


By (6.1), (1.14) and (1.22) we know one irreducible module $T$ in
$[Z_{1},K]$. If $m_p(\bC_K(\bC_T(S \cap K))) \geq 2$, we see that there is $1 \not= x \in \Omega_1(Z(S)) \cap T^{L_1}$ with $m_p(\bC_K(x)) \geq 2$ or $T$ is the natural module for $L_{n}(q)$
or $\Omega^{+} (2n,q)$ and the graph automorphism is induced by
$L_{1}$. The former is not possible as $L_2 \not\leq M$. Now inspection gives (6.2).\\

We will check the groups in (6.2) to arrive at a contradiction in any case. Let $K\cong L_{n}(q)$. Let further first $q = 2$. If $m_p (K) = 2$ then $E \cong E_{p^2}$ centralizes $K$. As $Z_1 \cap L_{\alpha + 1}$ acts trivially on $\{T_{\alpha + 1}^{L_{\alpha + 1}}\}$.  Then $[\langle T_{\alpha + 1}^{L_{\alpha+1}}\rangle, Z_1 \cap L_{\alpha + 1}]$ is centralized by $E \cong E_{p^2}$, a contradiction. So we have $p = 3$ and $K \cong L_6(2)$ or $L_7(2)$. As $7^2 ~|~ |K|$, we have $K \unlhd L_1/Q_1$. But for $T$ we then have $E \leq \bC_K(\bC_T(S \cap K))$, $E \cong E_9$, a contradiction.

So we have $q > 2$ and $K \cong L_4(q)$ or $L_3(q)$. Now $T$ is the natural module or in case of $L_4(q)$ the orthogonal module. Let $|Z_1 \cap L_{\alpha+1} : Z_1 \cap Q_{\alpha + 1}| > q$, then we see that $m_p(K) = 2$, otherwise some $1\not= x \in [s,Z_{\alpha+1}]$ is centralized by $E \cong E_{p^2}$. But now by (1.16) $Z_1 \cap L_{\alpha +1}$ acts trivially on $\{T_{\alpha+1}^{L_{\alpha+1}}\}$ and so we get a contradiction as before . So we have $|Z_1 \cap L_{\alpha+1} : Z_1 \cap Q_{\alpha +1 }| \leq q$. In particular $|[Z_1,t]| \leq 16q \leq q^3$. Hence there are at most three nontrivial irreducible modules involved. Further $p \not= 3$. This shows $K \cong L_4(4)$, $p = 5$. Now $m_p(\bC(K)) \geq 2$. Then $[K,Z_1]$ is centralized by some 5-element. As there is a 5-element in $K$ which centralizes $[s,Z_{\alpha+1}]$, we have a contradiction.

Assume there are exactly two modules involved. If $p ~\not|~q-1$, there is a $p$-element centralizing $[K,Z_1]$. Furthermore both modules are normalized by $S$. But then some $1\not= x \in [S,T]^\sharp$ is centralized by $E \cong E_{p^2}$, a contradiction. So we have $p~|~q-1$. The same applies if $[Z_1,K]$ is irreducible. Now $K \unlhd L_1/Q_1$ and so we see $m_p(K) = 2$, otherwise some $x \in \Omega_1(Z(S))^\sharp$ is centrailzed by $E \cong E_{p^2}$. So we have $K \cong SL_3(q)$. If there is exactly one nontrivial irreducible module involved we have some $p$-element centralizing $[K,Z_1]$. But $x \in [s,Z_1]$ is centralized by a $p$-element in $K$ too, a contradiction. So we have exactly two nontrivial irreducible modules involved. Then $q \leq 8$, $p = 3$ or $7$. As $K \unlhd L_1/Q_1$ it is now easy to see that there is some $x \in \Omega_1(Z(S))^\sharp$ centralized by $E \cong E_{p^2}$, a contradiction.


Let $K\cong Sp_{2n}(q)$. In case $n=3$ we have that $T$ is the
spin module. We may assume $p \,\Big | \,|\bC_{L_{1}} (K)|$ by (4.4). So by (1.16) again $|Z_{\alpha+1}\cap L_{1}:Z_{\alpha + 1}\cap Q_{1}|\leq q$. Hence
$q^2 \leq |[Z_{1},t]|\leq 16q$. So $q \leq 16$. If there are at least two modules
involved we get $q=2$ and so $p = 3$. Now $|[Z_1,t]| = 16$ and so $|Z_1 : Z_1 \cap Q_\alpha| \geq 8$. This shows $|V_\alpha \cap L_1 : \bC_{V_\alpha \cap L_1}(K)| \geq 2^9$, a contradiction. So we just have one nontrivial irreducible module involved. Now there is $x \in [s, Z_{\alpha+1}]^\sharp$ centralized by $E \cong E_{p^2}$, as $p ~|~|\bC_{L_{\alpha+1}}([K,Z_1])|$. 

So let now $n=2$. We may assume that there is $E \cong E_{p^2}$ centralizing $K$. In particular there are at least two nontrivial modules and again $|Z_{\alpha+1} \cap L_1 : Z_{\alpha+1} \cap Q_1| \leq q$. Suppose $p~|~q+1$. Then either there is a $p$-element centralizing $[K,Z_1]$ or there are at least four nontrivial irreducible modules involved in $[K,Z_1]$ . In the former we get that $[s,Z_{\alpha+1}]$ is centralized by $E \cong E_{p^2}$, a contradiction. In the latter we have $q = 2$. But then $|V_\alpha \cap L_1 : V_\alpha \cap Q_1| \geq 2^5$, a contradiction. So we have that $p~|~q-1$ and then $q > 2$. Suppose there are three nontrivial irreducible modules involved. Then $q =4$ and $p = 3$. But now $|[Z_1,t]| \leq 8\cdot 4$, a contradiction. So there are exactly two nontrivial irreducible modules involved and $q \leq 16$. Further $K \unlhd L_1/Q_1$. Now $|\Omega_1(Z(S)) \cap [K, Z_1]| \leq q^2$ and there is $F \cong E_{p^4}$ acting on a group of order $q^2$ containing $\Omega_1(Z(S)) \cap [K,Z_1]$. In any case there is $E \cong E_{p^2}$ in $F$ centralizing some $x \in \Omega_1(Z(S)^\sharp$, a contradiction.

   



Let now $K\cong U_{n}(q)$. Let $n=5$. Then $T$ is not the natural module as otherwise some $1 \not= x \in [s,Z_{\alpha+1}]$ is centralized by some $E \cong E_{p^2}$, $p~|~q+1$ with $\Gamma_{E,1}(G) \leq M_{\alpha+1}$. Now from (1.34) (ii) we get
$|[Z_{1},t]|\geq q^{\, 6}$. Furthermore $Z_1 \cap L_{\alpha+1}$ does not contain transvections $s$ on the natural module, otherwise $[s,Z_{\alpha + 1}]$ is normalized by $E \cong E_{p^2}$. So we have $|Z_{1}\cap L_{\alpha
+1}:Z_{1}\cap Q_{\alpha +1}|\leq q^{\, 2}$. Now $q=2$ . Then $p = 3$ and
$|[Z_{1},t]| \leq 8q^{\, 2}$, a contradiction.  

Let now $n=4$. Then $p  \not\Big | \,q-1$, as otherwise for both modules, the
natural $U_{4}(q)$-module and the natural $\Omega^{-}_{\, 6}
(q)$-module, $[T,s]$ contains a subgroup $1\not= U$ normalized  by
$E\cong E_{p^2}$ with $\Gamma_{E,1} (G)\leq M$. So we must have  $p \, \Big | \,q+1$. Furthermore there are at least two nontrivial irreducible modules involved. Suppose both are natural modules. If there is $t \in Z_{\alpha + 1}$ with $|[[Z_1,K],t]| \geq q^8$, we get $q = 2$. But then $3 \in \sigma (M)$, a contradiction. So we have that any $t \in Z_{\alpha + 1}$ just induces transvections. In particular $|Z_{\alpha + 1} \cap L_1 : \bC_{Z_{\alpha+1} \cap L_1}(K)| \leq q$. Now we have $|U| = q^4$, $U \leq Z_1$, $[U, Z_{\alpha+1}] \not= 1$. If $q > 2$ we get $U_1 \leq U$, $|U_1| = q^2$, $U_1 \leq L_{\alpha+1}$. As $q^8 \leq |Z_{\alpha+1} : \bC_{Z_{\alpha+1}}(U_1)| \leq q\cdot 16$, we have a contradiction. So we get $q = 2$ and then $|U| = 16$. Now $|U \cap L_{\alpha+1}| = 2$. But then $|V_\alpha \cap L_1 : \bC_{V_\alpha \cap L_1}(K)| \geq 2^5$, a contradiction.

If $s \in N(K)$, then either $s$ centralizes a subgroup $E \cong E_{p^2}$ in $K$ or $s$ corresponds to an element of type $c_2$ (in Suzuki - notation) on the natural module. But in the latter $[s,Z_1]$ contains some $1\not= x$ with $\bC_K(x)$ involves $Sp_4(q)$. So in any case $V_2 \leq M_{\alpha+1}$ a contradiction.

Let now $n = 3$. Then as before $|[Z_1,t]| \ge q^5$. But $|[Z_1,t]| \le 16q$, a contradiction.

Let $K\cong \Omega_8^{-} (q)$. Then $T$ is the spin module and
$p \, \Big | \,q+1$. Furthermore $p~|~|\bC_{L_1/Q_1}(K)|$. Hence there are at least
two such modules involved. Now $|[Z_{1},t]|\geq q^{\, 8}$. But $|[Z_1,t]| \leq q^6\cdot 16$  and so $q\leq 4$. As elements of type $c_2$ or $a_2$ are centralized by $E \cong E_{p^2}$ in $K$, we get $|Z_{1}\cap L_{\alpha
+1}:Z_{1}\cap Q_{\alpha +1}|\leq q^{\, 4}$ and so we get $q=2$. But
in this case $p=3$ and so $|[Z_{1},t]|\leq 2^{\, 7}$, a
contradiction.

Let $K\cong \Omega_8^{+} (2)$. Then both natural modules are involved
and $p=3$. But in the natural module every element is centralized by
$E\cong E_{9}$ in $K$, a contradiction. 

Let $K\cong G_{2}(q)$. Then $|T|=q^{\, 6}$. Now $m_p(\bC_{L_1/Q_1}(K)) \geq 2$. So there are at least two nontrivial irreducible modules involved. Now $|[Z_1,t]| \leq 16q^3$. As $|[Z_1,t]| \geq q^4$, we have $q \leq 16$. If there are three such modules involved, we get $q=2$ and as usual $|V_\alpha \cap L_1 : \bC_{V_\alpha \cap L_1}(K)| > 2^5$, a contradiction. So we have exactly two natural modules involved. Let $|Z_1 \cap L_{\alpha +1 } : Z_1 \cap Q_{\alpha + 1}| > q$. Then with (1.16) we see that there is $E \cong E_{p^2}$ in $L_{\alpha +1}$ acting on $[Z_1 \cap L_{\alpha + 1} , Z_{\alpha + 1}]$, a contradiction to $Z_1 \not\leq M_{\alpha + 1}$. So $|Z_1 \cap L_{\alpha + 1} : Z_1 \cap Q_{\alpha + 1}| \leq q$. Now $|[Z_1,t]| \leq 16\cdot q$ and so $q = 2$ again. Hence $p = 3$. But then there is $\nu \in L_1$, $o(\nu) = 3$ and $[\nu,[Z_1,K]] = 1$. As $K \unlhd L_1/Q_1$, we get some $x \in \Omega_1(Z(S))^\sharp$ which is centralized by $E \cong E_9$, a contradiction. 


Let finally $K\cong$ $^{3}D_{4}(q)$. Then $T$ is the natural module.
Furthermore $|[t,T]|\geq q^{\, 6}$. In particular just one nontrivial irreducible module is involved and  $|Z_{\alpha + 1} \cap L_1 : \bC_{Z_{\alpha + 1} \cap L_1}(K) | \leq q^5$. Now $K \unlhd L_1/Q_1$ and so $T\cong
[Z_{1},K]$. This implies with (4.4) $m_{p}(K)\geq 3$, a contradiction.

So we have (6).\\


(7) If $K \unlhd L_1/Q_1$, $K$ a $p$--group with $[Z_{\alpha+1} \cap L_1, K] \not= 1$, then
$m_{p}(K)=1$. \\


Let $m_p(K) \geq 2$. Obviously
$|Z_1 \cap L_{\alpha+1}:Z_1\cap Q_{\alpha + 1}|\leq 4$. Let $m_p(K) \geq 3$. By \cite[(11.18)]{GoLyS} $s$ normalizes some $E \cong E_{p^2}$. Now $[Z_1,E] = \bigoplus_{|E : E_i| = p} \bC_{Z_1}(E_i)$. We have that $[\bC_{Z_1}(E_i), s] = 1$ for $[E_i, s ] \leq E_i$. As $[s,E] \not= 1$ there is $E_i$ with $[E_i,s] \not\leq E_i$. Now $\bC_{Z_1}(E_i) \cap \bC_{Z_1}(E_i^s) = 1$ and $\bC_{Z_1}(E_i) \oplus \bC_{Z_1}(E_i^s)$ is centralized by $E_i \cap E_i^s$. Let $E_i = \langle E_i \cap E_i^s, \rho \rangle$. Then $\rho \rho^s$ is centralized by $s$ and so $\rho \rho^s$ centralizes $[\bC_{Z_1}(E_i) \oplus \bC_{Z_1}(E_i^s), s]$, a contradiction.

So we have $m_p(K) = 2$. Now  $|V_2 :\bC_{V_2 }
(K_{\alpha+1})|\leq 2^{\, 6}$. Let $\beta \in \Delta (\alpha)$. Application of (6.16) shows $|[Z_\beta,K_\beta] : [Z_\beta,K_\beta] \cap Q_2| \leq 4$. Hence $[Z_1,K]$ is generated by elements $t$ with $|[Z_\beta,K_\beta] : \bC_{[Z_\beta,K_\beta]}(t)| \leq 8$. This shows $t \in M_\beta$. Hence $[Z_1,K] \leq \bigcap_{\beta \in \Delta (\alpha)} M_\beta \leq L_{\alpha + 1}$. Now there is some $\rho \in K$, $\rho^s = \rho^{-1}$ such that $\rho \in E \cong E_{p^3}$. Suppose $p > 3$. Then $|[\rho,Z_1]| \leq 2^6$. As $x \in [[\rho,Z_1],s]^\sharp$ is not centralized by a hyperplane in $E$, we get $p = 7$ and $|[\rho, Z_1]| = 2^6$. Now $|\bC_E([\rho, Z_1])| = 7$. Hence we get that the orbits of $E$ on $[\rho,Z_1]$ are 49, 7,7. We have that $\bC_{L_1/Q_1}(\rho)$ induces a subgroup of $GL_2(8)$ on $[\rho, Z_1]$, whose order is divisble by $7^2$. We see that there is a normal Sylow 7-subgroup and so $s$ acts on this group. As $s$ cannot invert a Sylow 7-subgroup of $GL_2(8)$ we see that there is some 7-element $\nu$ centralized by $s$. But then $[s,[\rho,Z_1]]$ is centralized by $\tilde{E} \cong E_{7^2}$ in $L_1$, a contradiction.

So let $p = 3$. Then we get $[V_2 : V_2 \cap \bC_{L_{\alpha + 1}/Q_{\alpha+1}}(K_{\alpha+1})| \leq 2^5$ and $|[Z_\beta,K_\beta] : [Z_\beta, K_\beta] \cap Q_2| \leq 2$ for $\beta \in \Delta (\alpha)$. Let $\rho$ be as before. We get $|[\rho, Z_1]| \leq 2^4$. But now we may argue as before.
\begin{itemize}
\item[(8)] If $K_{1}\times K_{2}\times K_{3} \leq L_1/Q_1$, $m_p(K_i) =1$ and the $K_i$ are components, or a cyclic $p$--group, $i = 1,2,3$, then $[Z_{\alpha +1}\cap L_{1},K_{i}]=1$ for suitable $i$.
\end{itemize}


Suppose false. Assume first $K_{1\alpha+1}^s \not= K_{1\alpha+1}$, where $K_{1\alpha+1}$ is the component in $L_{\alpha+1}$  corresponding to $K_1$. If $|V_2 \cap L_{\alpha+1} : \bC_{{V_2} \cap L_{\alpha+1}}(K_1)| = 2$, we get that $Z_{\alpha+1}$ centralizes a subgroup of index at most 32 in $V_2$. In particular $|Z_{\alpha+1} : Z_{\alpha+1} \cap Q_2| \leq 2$. Now there are at least two nontrivial irreducible $K_1$ - modules in $Z_1$, which tells us that for any $t \in Z_{\alpha+1} \cap L_1$ we have that $[K_1,Z_1]$ involves some nontrivial irreducible module $W$ with $|[W,t]| \leq 4$. This gives with 
(1.30) as all Sylow $r$-subgroups , $r$ odd, have to be cyclic and then that $K_1 \cong L_2(4)$ or $L_3(2)$. In both cases there is a subgroup $U$ of order 16 in $[Z_1,K_1]$ which centralizes a subgroup of index 64 in $V_\alpha$. Hence $|U \cap Q_\alpha| \geq 4$. But $U \cap \bC(Z_{\alpha+1}) = 1$, a contradiction.

So we have that $|V_2 \cap L_{\alpha+1} : \bC_{V_2 \cap L_{\alpha+1}}(K_{1\alpha +1})| > 2$ and then by (1.11) $K_1 \cong L_2(q)$ and $[Z_{\alpha+1}, \langle K_{1\alpha+1}, s \rangle]$ involves the orthogonal $O^+(4,q)$-module. Now $|V_2 \cap L_{\alpha+1} : V_2 \cap Q_{\alpha+1}| \leq 2q$ and $|Z_1 : \bC_{Z_1}(Z_{\alpha+1})| \leq 2^5q$. Furthermore at least two such irreducible modules are involved, which shows $|[Z_1,t]| \geq q^2$ and so $q \leq 32$ and $K_1^t \not= K_1$. Now for $s \in [\langle K_1,t \rangle, Z_1]$ we have $|Z_{\alpha+1} : \bC_{Z_{\alpha+1}}(t)| \leq 2^5$. If $q > 4$, we see that $[[\langle K_1 , t \rangle, Z_1] \cap Q_\alpha, Z_{\alpha+1}] = 1$. So $|[[\langle K_1,t \rangle, Z_1],t]| < 2^4$. So we have $q = 4$. As $|[[\langle K_1,t \rangle, Z_1],t]| < q^4$, we have at most three natural modules involved. This shows $p = 3$. We have $|Z_{\alpha+1} : Z_{\alpha+1} \cap Q_2| = 8$. So $|V_2 : V_2 \cap \bC(Z_{\alpha+1})| \geq 2^{12}$. This shows $|V_2 \cap L_{\alpha+1} : V_2 \cap Q_{\alpha+1}| \geq 2^9$, a contradiction.

So we have $[K_{i\alpha+1},s] \leq K_{i\alpha+1}$, for $i = 1,2,3$. In particular there are at least two $i$ with $[K_{i\alpha+1},s] \not= 1$.

Suppose $[[K_1,Z_1], K_2K_3] = 1$. Then $[V_2 \cap L_{\alpha+1}, K_{1\alpha+1}] = 1$ and $[[K_{1\alpha+1}, Z_{\alpha+1}], V_2 \cap L_{\alpha+1}] = 1$. Now $[K_{1\alpha+1} , Z_{\alpha+1}] \leq L_1$. Further $[K_{1\alpha+1}, Z_{\alpha+1}] \cap Q_1 = 1$. We may assume $m_3(K_2) \leq 1$. Assume furthermore $[[K_{1\alpha+1}, Z_{\alpha+1}], K_2] \leq K_2$. As $|[K_{1\alpha+1}, Z_{\alpha+1}]| > 2$, we get with (1.3) and (1.12) that $K_2 \cong L_2(q)$, $Sz(q)$, $U_3(q)$, or $L_3(q)$, $q$ even. As $[Z_1,K_2]$ involves at least two nontrivial irreducible modules we see that $K_2 \cong L_2(4)$, $L_3(4)$, or $L_3(2)$. As $Z_1 \not\leq M_{\alpha+1}$ we get $K_2 \cong SL_3(4)$ and $|[K_1,Z_1]| \leq 16$. Furthermore there are exactly two nontrivial irreducible $K_2$-modules involved which shows that $K_3$ is a subgroup of $L_2(4)$ and so $p = 5$. But then $[K_2,Z_1]$ is centralized by $E \cong E_{25}$, a contradiction.

So let $K_2^t \not= K_2$ for some $t \in [K_{1\alpha+1}, Z_{\alpha+1}]$. If $[K_{1\alpha+1}, Z_{\alpha+1}]$ induces a quadratic fours group on $K_2^tK_2$, we get $K_2 \cong L_2(q)$ and $q = 4$. Now $p = 5$ and we get a contradiction as before. So let $|[K_{1\alpha+1}, Z_{\alpha+1}] : \bC_{[K_{1\alpha+1}, Z_{\alpha+1}]}(K_2^tK_2)| = 2$. Then we get $|[K_{1\alpha+1}, Z_{\alpha+1}]| = 2$, a contradiction.

Now for any $i$  there is  some $j \not= i$ with $[[K_i,Z_1],K_j] \not= 1$. Let $[[K_1,Z_1],K_3] \not= 1$ and $[[K_1,Z_1],K_2] = 1$. Then we get $[[K_2, Z_1], K_1] = 1$ by the three subgroup lemma and so $[[K_2,Z_1], K_3] \not= 1$. Then also $[[K_3,Z_1],K_1] \not= 1 \not= [[K_3,Z_1], K_2]$. So we may choose notation such that $[[K_1,Z_1], K_2] \not= 1 \not= [[K_1,Z_1],K_3]$. 

Assume now $[K_i, Z_{\alpha+1} \cap L_1] \leq K_i$ for $i = 1,2,3$. Then $V_\alpha \cap L_1 \cap \bC(K_1) \leq \bC(K_1K_2K_3)$. Let $K_i$ be some component with $m_3(K_i) = 0$. We have that there are at least three nontrivial irreducible modules involved. Let now $1\not= t \in \bC_{Z_{\alpha+1} \cap L_1}(K_i) \setminus \bC(K_1K_2K_3)$. Then $[t, K_1] \not= 1$ and for $s \in Z_{\alpha+1} \cap L_1$ with $[K_i,s] \not= 1$, we have $[t,Z_1,s] = 1$, so $[[t,Z_1],K_i] = 1$, but then $[[K_1,Z_1],K_i] = 1$, a contradiction. 

So we have $\bC_{Z_{\alpha+1} \cap L_1}(K_i) \leq \bC(K_1K_2K_3)$. Let first $|V_\alpha \cap L_1 : \bC_{V_\alpha \cap L_1}(K)| = 2$. Then for $ s \in [K_1,Z_1]$ we have $|V_\alpha : \bC_{V_\alpha}(s)| \leq 32$. Suppose $p > 3$, then $[K_1,Z_1] \leq L_{\alpha+1}$. Let $s$ with $[K_{1\alpha+1},s] \not= 1$. Then $s$ induces a transvection on $[K_{1\alpha+1}, Z_{\alpha+1}]$. As now $K_2 \times K_3$ is a nonsolvable subgroup  of $L_5(2)$, we get $K_2 \cong \Z_3$ and $K_3 \cong A_5$ or $L_3(2)$, contradicting $p > 3$. So we have $[[K_1,Z_1],K_{1\alpha+1}] = 1$. Then $K_{1\alpha+1}$ acts nontrivially on $[s,Z_{\alpha+1}]$. In particular $K_1$ is a subgroup of $L_5(2)$. So $K_1 \cong L_2(4)$, $L_2(9)$, $L_2(7)$, $A_7$, or $A_8$, and $p = 5$ or $7$. But now we have $[[K_2,Z_1], K_3] \not= 1$ and then $[K_2,Z_1]$ involves at least four nontrivial irreducible modules. Now we may interchange roles. So we have $p = 3$ and $|V_\alpha : \bC_{V_\alpha}(s)| \leq 16$. Now there is $U \leq Z_1$, $|U| = 16$ with $U \cap \bC(Z_{\alpha+1}) = 1$. Hence there is $s \in U \cap L_{\alpha+1}$ with $[s,Z_{\alpha+1}] \not= 1$. As before we get that one of the irreducible modules admits transvections and so we get $K_1K_2K_3 \cong \Z_3 \times A_5 \times A_5$ or $\Z_3 \times A_5 \times L_3(2)$. Now as $[Z_1,K_1]$ centralizes a subgroup of index 16 in $V_\alpha$, we again see that $[Z_1,K_1] \leq L_{\alpha+1}$. But $|[Z_1,K_1] : [Z_1,K_1] \cap \bC(Z_{\alpha+1})| \geq 16$, which shows that $[K_1,Z_1,K_2] = 1$, a contradiction.

So we have $|V_\alpha \cap L_1 : V_\alpha \cap L_1 \cap \bC(K_1K_2K_3)| \geq 4$ and so $K_i \cong Sz(q)$. For $s \in [K_1K_2K_3, Z_1]$ we have $|Z_{\alpha+1} : \bC_{Z_{\alpha+1}}(s)| \leq 16q$. Furthermore there is some $s \in L_{\alpha+1} \setminus Q_{\alpha+1}$. This now shows $[s, K_{i\alpha+1}] = 1$ and even $[[s,Z_{\alpha+1}], K_{i\alpha+1}] = 1$. But then $[K_{1\alpha+1}, s] = 1$, which shows $K_i = K_1$. Now $[s, K_{2\alpha+1}] \not=1$ and so $[[K_{2\alpha+1},Z_{\alpha+1}],K_{1\alpha+1}] = 1$, a contradiction.

Now by (1.12) and (1.3) we have $K_1 \cong (S)U_3(q)$, $L_2(q)$ or $(S)L_3(q)$, $q$ even. We get $|[s,Z_{\alpha+1}]| \leq 16q$ (or $16q^2$ in case of $K_1 \cong (S)L_3(q)$), where $s \in Z_1$. Suppose $s \in L_{\alpha+1}$ If $[s, K_{1\alpha+1}] \not= 1$, we get $K_1 \cong SL_3(4)$ or $L_3(2)$. As $m_3(K_i) \geq 1$ for all $i$, we see $K_1 \leq L_3(2)$. Now $K_2K_3$ is a subgroup of $GL_6(2)$ and so $K_2 \cong K_3 \cong L_3(2)$ and there are at least six nontrivial irreducible $K_1$-modules involved. But now we get $U \leq Z_1$, $|U| = 2^6$ with $\bC_U(Z_{\alpha+1}) = 1$. Then $|U \cap L_{\alpha+1}| \geq 2^4$, a contradiction. So we have $[s,K_{1\alpha+1}] = 1$. Then $[K_{1\alpha+1}, [Z_{\alpha+1},s]] \not= 1$. This shows $K_1 \cong L_2(q)$, $q \leq 16$ or $K_1 \cong L_3(2)$. If $[K_2, Z_1, K_3] \not= 1$ we may argue as before. So we may assume $[K_2,Z_1,K_3] = 1$. Now we either have $[[[K_{2\alpha+1},Z_{\alpha+1}],s], K_{1\alpha+1}] = 1$ or $[[[K_{3\alpha+1}, Z_{\alpha+1}],s],K_{1\alpha+1}] = 1$, both yields a contradiction. 

This shows $[[Z_1,K_1] \cap L_{\alpha+1}, Z_{\alpha+1}] = 1$. Then $|[Z_1,K_1] : \bC_{[Z_1,K_1]}(Z_{\alpha+1})| \leq 16$. As there are at least four nontrivial irreducible modules involved we have $K_1 \cong L_3(2)$ or $A_5$ and there are exactly four nontrivial irreducible modules involved. Further $p > 3$. But then $K_2K_3$ is a subgroup of $L_4(2)$, a contradiction.

So we may assume $K_i^t \not= K_i$ for some $t \in Z_{\alpha+1} \cap L_1$. We may furthermore assume that $K_i^t = K_j$. Let $|V_\alpha \cap L_1 : \bC_{V_\alpha \cap L_1}(K_iK_j)| = 2$. Let further $s \in [\langle K_i, t \rangle, Z_1]$. Then $|V_\alpha : \bC_{V_\alpha}(s)| \leq 2^5$. Now by (6.16) we get $s \in L_{\alpha+1}$. If $[K_{1\alpha+1},s ] \not= 1$, we have $K_1 \cong L_3(2)$, $A_5$, or $K_1$ is a $3$-group. Now $K_2K_3$ is a subgroup of $L_5(2)$ and so we get $K_1K_2K_3 \cong A_5 \times A_5 \times \Z_3$ or $L_3(2) \times L_3(2) \times \Z_3$. In any case $p = 3$ and so we have $K_1K_2K_3 \cong A_5 \times A_5 \times \Z_3$. Now $s$ induces a transvection and so $s$ centralizes 3-elements in $K_{1\alpha+1}K_{2\alpha+1}K_{3\alpha+1}$. As there is a $3$-element centralizing $[K_1K_2K_3,Z_1]$ by (4.4), we get a contradiction. 

So we have $[K_{1\alpha+1},s] = 1$. Again $K_1$ is a subgroup of $L_5(2)$ and $m_3(K_1) = 1$. This shows $K_1 \cong L_3(2)$ or $A_5$ again. Now we may interchange the roles of 1 and 2 and get a contradiction as before. 

So we have $|V_\alpha \cap L_1 : \bC_{V_\alpha \cap L_1}(K_i \times K_j)| > 2$ and then by (1.11) $K_iK_j \cong L_2(q) \times L_2(q)$ and $O_4^+(q)$-modules are involved. Furthermore there are at least two of them. Now for $s \in Z_1 \cap L_{\alpha+1}$ we have that $|Z_{\alpha+1} : \bC_{Z_{\alpha+1}}(s)| \leq 32q$. On the other hand we have $|Z_{\alpha+1} : \bC_{Z_{\alpha+1}}(s)| \geq q^4$, which shows $q=2$, a contradiction. 

\begin{itemize}
\item[(9)] If $K_{1}\times K_{2} \leq L_1/K_1$ where $K_i$ are as in (8).
Then $[Z_{\alpha + 1} \cap L_1,K_{i}]=1$ for suitable $i \in \{1,2\}$.  
\end{itemize}


Suppose false. First of all $[Z_{\alpha +1}\cap L_{1},\bC_{L_{1}/Q_{1}}
(K_{1}\times K_{2})]=1$ by (8). Let first $|V_\alpha \cap L_1 : V_\alpha \cap Q_1| = 2$. Then $|Z_{\alpha+1} : \bC_{Z_{\alpha+1}}(s)| \leq 2^5$ for $s \in Z_1$. If $p > 3$, then $Z_1 \leq M_{\alpha+1}$ a contradiction. So $p = 3$ and we have $|Z_{\alpha+1} : \bC_{Z_{\alpha+1}}(s) | \leq 16$. Furthermore there is some $s$ with equality. We may assume $[K_{1\alpha+1},  s] \not= 1$. If $[K_{1\alpha+1}, s] \not\leq K_{1\alpha+1}$ we may define $K_{3\alpha+1} = K_{1\alpha+1}^s$. Now $\bC_{V_2}(K_{1\alpha+1}K_{3\alpha+1}) = 1$ and so either we have orthogonal $O_4^+(q)$ - modules involved or $|Z_1 \cap L_{\alpha+1} : Z_1 \cap Q_{\alpha+1}| = 2$ too. In the former there is some $s_1 \in Z_1 \cap L_{\alpha+1}$ with $|Z_{\alpha+1} : \bC_{Z_{\alpha + 1}}(s_1)| \geq q^2$ and so $q = 4$ and there is exactly one nontrivial irreducible $K_1K_3$ - module involved in $Z_1$. But this gives some $E \cong E_9$ in $M_{\alpha+1}$ centralizing $[Z_{\alpha+1}, s]$, a contradiction. So we have $|Z_1 \cap L_{\alpha+1} : Z_1 \cap Q_{\alpha+1}| = 2$. But we have $|Z_{\alpha+1} : Z_{\alpha+1} \cap L_1| = 2$ and so $|[Z_{\alpha+1}, s]| \leq 4$. This gives $K_1 \cong A_5$ and again there is just one nontrivial irreducible $K_1K_3$ - module involved in $Z_1$, a contradiction as before. So we have $[K_{1\alpha+1}, s] \leq K_{1\alpha+1}$. As we may assume that $m_p(L_1) \geq 4$, we have that $[K_1,Z_1]$ involves at least two nontrivial irreducible modules and so for some of them we have that $s$ centralizes a subspace of codimension two. As $p = 3$ we have $K_1 \cong L_3(2)$ or $L_2(4)$. Now in any case there is $x \in [Z_{\alpha+1}, s]^\sharp$ centralized by $E \cong E_9$ in $M_{\alpha+1}$, a contradiction. So we have $|V_\alpha \cap L_1 : V_\alpha \cap Q_1| \geq 4$.


Suppose $K_{1}\cong K_{2}\cong L_{2}(q)$ and
$\Omega_4^{+} (q)$-modules are involved in $Z_{1}$. Then $|[Z_{\alpha+1},s]|\leq 2^{\, 5}q$. Suppose $[K_1,s] \not= 1$. If $[K_{1\alpha+1},s] \leq K_{1\alpha+1}$, then we get $q \leq 2^5$ and there is just one $\Omega_4^+(q)$-module involved. Further $K_1K_2 \unlhd L_1/Q_1$ and so we get some $E \cong E_{p^2}$ centralizing some element in $\Omega_1(Z(S))^\sharp$, a contradiction. So we have $K_{1\alpha+1}^s \not= K_{1\alpha+1}$. Suppose $K_{1\alpha+1}^s = K_{2\alpha+1}$. Again we may assume that there are at least two nontrivial irreducible modules involved. Now $q \leq 32$ again. Furthermore there is no $p$-element centralizing $[Z_1, K_1K_2]$. This shows that there is some field automorphism of order $p$, i.e. $q=8$, $p=3$. But now for $t \in Z_{\alpha+1} \cap L_1$ we have $|[Z_1, t]| \leq 2^4$, a contradiction. So $K_{1\alpha+1}^s \not= K_{2\alpha+1}$. Then $[K_{2\alpha+1}] = 1$ and so $s$ centralizes $E \cong E_{p^2}$, a contradiction.

So we now have $[Z_{\alpha+1} \cap L_1, K_i] \leq K_i$, $i = 1,2$. Let $m_3(K_1) \leq 1$. Then $K_1 \cong Sz(q)$, $L_2(q)$, $L_3(q)$, $U_3(q)$, $J_1$ or $\Z_p$. 
If $q$ is odd or $K_1 \cong J_1$ or $\Z_p$ set $r = 2$ else $r = q$ or $r = q^2$ in case of $K_1 \cong L_3(q)$. Then we have that $|V_\alpha : \bC_{V_\alpha}([Z_1,K_1])| \leq 16r$. Let $s \in [Z_1,K_1]^\sharp \cap L_{\alpha+1}$ with $[Z_{\alpha+1}, s] \not= 1$. Assume further $[K_{1\alpha+1}, s] \not= 1$. Suppose $[K_{1\alpha+1}, s] \not\leq K_{1\alpha+1}$. If $|V_2 \cap L_{\alpha+1} : V_2 \cap Q_{\alpha+1}| = 2$, we have for $t \in Z_{\alpha+1}$ that $|[Z_1,t]| \leq 32$. As $[t,K_1] \not= 1$, we get $K_1 \cong L_2(2^f)$, $L_3(2^f)$ $f \leq 32$, $U_3(4)$, $L_2(25)$, $L_2(x)$, $x$ prime. As $|V_\alpha \cap L_1 : V_\alpha \cap Q_1| > 2$, and $[K_1,Z_1,K_2] = 1$ in the last two cases, we see that $K_{1\alpha+1}^s = K_{2\alpha+1}$. Now $K_1 \cong L_2(x)$ and $|[s,Z_{\alpha+1}] | \leq 2^5$. This shows $K_1$ is a subgroup of $L_5(2)$, so $K_1 \cong L_2(4)$ or $L_3(2)$. So we may assume that $K_1 \cong L_2(q)$, $q$ even, or $L_3(2)$. If $|V_2 \cap L_{\alpha+1} : V_2 \cap Q_{\alpha+1}| > 2$ we get $K_1 \cong L_2(q)$ too. Now $|[V_\alpha,s]| \leq 16q $ or $64$ in case of $L_3(2)$. As we may assume that there are at least two nontrivial irreducible modules involved, we get $|[V_\alpha,s]| \geq q^2$, $2^6$, respectively. Now in any case $K_1K_2 \unlhd L_1/Q_1$. Assume that we have a group of order $q^2$, in case of $L_3(2)$ of order $16$, centralized by $E \cong E_{p^2}$. In the latter we have $p = 3$ and so there is $E \cong E_9$ centralizing some element in $[\Omega_1(Z(S)), K_1K_2]^\sharp$. So we have $K_1 \cong L_2(q)$. If $|V_2 \cap L_{\alpha+1} : V_2 \cap Q_{\alpha+1}| > 2$ there are at least four nontrivial irreducible $K_1$- modules involved, so $q^4 \leq 32q$, a contradiction. Hence we have $|V_2 \cap L_{\alpha+1} : V_2 \cap Q_{\alpha+1}| = 2$, and then $q^2 \leq 32$, i.e. $q = 4$. We have $K_1 \cong L_2(4)$. Now we have $[K_1,Z_1,K_3] = 1$ and $[K_{1\alpha+1}, Z_{\alpha+1}]^s = [K_{3\alpha+1}, Z_{\alpha+1}]$. So $|[K_1,Z_1]| \geq q^4$, a contradiction.

So we have $[K_{1\alpha+1},s] \leq K_{1\alpha+1}$. Further we know that $[K_1,Z_1]$ involves at least two nontrivial irreducible modules. We see $|[s,Z_{\alpha+1}]| \leq 16r$. Now by (1.29) we have $K_1 \cong L_2(q)$, $L_3(q)$, $q \leq 16$, $\Z_3$, or $\Z_5$. Suppose there are exactly two nontrivial irreducible modules involved. If there is no field automorphism of order $p$ in $L_1$, we get $E \cong E_{p^2}$ centralizing $[K_1,Z_1]$, a contradiction. So there is a field automorphism and so $K_1 \cong L_2(8)$, $p = 3$. Now there is a 3-element centralizing $[Z_1,K_1]$. Furthermore $s$ centralizes some 3-element in $K_{1\alpha+1}\langle \nu \rangle$, where $\nu$ induces the field automorphism. But the $[Z_{\alpha+1},s]$ is normalized by $E \cong E_9$ in $L_{\alpha+1}$, a contradiction. So we have at least three modules involved. This shows $K_1 \cong L_2(4)$, $L_3(q)$, $q \leq 16$ or $\Z_3$. Let $K_1 \cong L_3(q)$, then $p~\not|~q-1$. So we see that in all this cases there are at least four modules involved and then $K_1 \cong L_3(4)$ or $L_3(2)$.

Let $K_1 \cong L_3(2)$. Then if $p = 7$, we get $E \cong E_{p^2}$ in $L_1$ centralizing $[Z_1,K_1]$. If $p = 3$, we see $\Z_3$ centralizing $[Z_1,K_1]$ and there is some $1\not= x \in [s,Z_{\alpha+1}]$ centralized by $E \cong E_9$ in $L_{\alpha+1}$, a contradiction. 

Let $K_1 \cong L_3(4)$. Then there are at most four nontrivial irreducible modules involved, and $p =5$ or $p = 7$. As there is no $E \cong E_{p^2}$ in $M$ centralizing $[Z_1,K]$, we get $p = 5$ and there are exactly four modules involved. Now $|Z_{\alpha+1} : Z_{\alpha+1} \cap Q_2| = 16$ and so $|V_2 \cap L_{\alpha+1} : V_2 \cap Q_{\alpha+1}| \geq 2^8$, which shows that there is a $5$-element centralizing $[K_{1\alpha+1}, Z_{\alpha+1}]$, and so there is $1\not= x \in [Z_{\alpha+1} ,s]$ centralized by $E \cong E_{p^2}$ in $M_{\alpha+1}$, a contradiction. 

Let $K_1 \cong L_2(4)$. Then we have three nontrivial irreducible modules and $p = 5$. But then $[K_1,Z_1]$ is centralized by $E \cong E_{p^2}$ in $M$ a contradcition.

So we have $K_1 \cong \Z_3$ and $|[K_1,Z_1]| \leq 2^8$. We have $[K_2, [K_1,Z_1]] = 1$. Let $U \leq [Z_{\alpha+1}, K_{1\alpha+1}]$, $|U| = 8$, $U \cap \bC(Z_1) = 1$. Then we see that $|U \cap \bC([K_1,Z_1])| \leq 2$. So $|U : U \cap Q_2| \geq 4$. Hence $|V_2 : \bC_{V_2}(U)| \geq 64$. This gives $|V_2 \cap L_{\alpha+1} : \bC_{V_2 \cap L_{\alpha+1}}(U)| \geq 8$, a contradiction.

So we finally have $[K_{1\alpha+1}, s] = 1$. Suppose $[K_{1\alpha+1}, [Z_{\alpha+1},s]] = 1$. We must have $[K_{2\alpha+1},s] \not=1$ and there is $K_3$ with $[K_{3\alpha+1},s] \not= 1$. Hence $[Z_1,K_1]$ is centralized by $E \cong E_{p^2}$ in $M$. So $\bC_{[Z_{\alpha+1},K_{1\alpha+1}]}(Z_1) = 1$. If $u \in [Z_{\alpha+1}, K_{1\alpha+1}]$ then $|V_2 : \bC_{V_2}(u)| \leq 16$. So we get $[Z_{\alpha+1}, K_{1\alpha+1}] \leq Q_2$. So $[[Z_{\alpha+1}, K_{1\alpha+1}],K_2] \not= 1$. Now we have that $[K_2,Z_1]$ involves at least two nontrivial irreducible modules. So there is $u \in [Z_{\alpha+1}, K_{1\alpha+1}]$ which centralizes codimension two in such a modules, and this has to be an F-modules. This shows that $K_2 \cong L_n(2)$, $ n \leq 7$, $L_n(4)$, $n \leq 3$, $Sp_{2n}(2)$, $n \leq 3$, $Sp_{2n}(4)$, $n \leq 3$, $\Omega_{2n}^+(2)$, $n \leq 4$, $G_2(2)^\prime$, $A_n$, $n \leq 11$. In any case there are at most four nontrivial irreducible modules involved . As there is no $E \cong E_{p^2}$ centralizing $[K_1,Z_1]$ in $M$, we get $p = 3$. But now there are at most three nontrivial irreducible modules involved, and we have $E \cong E_9$ centralizing $[K_2,Z_1]$, a contradiction. 

So we have $[K_{1\alpha+1}, [Z_{\alpha+1}, s]] \not= 1$. In particular $[K_1,Z_1,K_2] \not= 1$. So we see that $K_1 \cong \Z_p$, $L_2(q)$ or $L_3(q)$, $q$ even, where $q \leq 16$. Suppose $[K_{1\alpha+1}, [Z_{\alpha+1},s]]$ contains exactly one nontrivial irreducible module. Then either $K_1$ or $K_2$ has to admit a nontrivial field automorphism of order $p$. Suppose this is true for $K_1$ then $K_1 \cong L_2(8)$ and $p = 3$, $|[Z_{\alpha+1},s]| = 2^6$. But now there is $E \cong E_9$ centralizing $s$ modulo $Q_{\alpha+1}$, a contradiction. So we have that this is true for $K_2$ and then $K_2 \cong L_2(r^p)$. As there are at least four nontrivial irreducible $K_2$-modules involved we get $r^p \leq 8$ and so $p = 3$ and $K_2 \cong L_2(8)$. But now we must have $K_1 \cong L_3(16)$, a contradiction. 

So we have at least two nontrivial irreducible $K_1$-modules involved. This shows $K_1 \cong L_3(2)$ or $\Z_3$. Now we have that $K_2$ admits transvections. So $K_2 \cong L_n(2)$, $Sp_{2n}(2)$, $\Omega_{2n}^\pm(2)$, $A_n$. As $m_3(K_2) \leq 2$, we see that $K_2 \cong L_3(2)$, $L_4(2)$, $L_5(2)$, $A_5$, $A_6$, $A_7$. Let $K_1 \cong L_3(2)$. Then $|Z_{\alpha+1} : Z_{\alpha+1} \cap Q_2| = 16$, so $|V_2 \cap L_{\alpha+1} : V_2 \cap Q_{\alpha+1}| \geq 2^8$, a contradiction. So we have $K_1 \cong \Z_3$. Now $|Z_{\alpha+1} : Z_{\alpha+1} \cap Q_2| = 8$ and so $|V_2 \cap L_{\alpha+1} : V_2 \cap Q_{\alpha+1}| \geq 2^5$. As $m_3(K_2) = 1$, we have a contradiction again.

So it remains the case that $[[Z_1,K_1] \cap L_{\alpha+1}, Z_{\alpha+1}] = 1$. Then there is $t \in Z_{\alpha+1}^\sharp$ with $|[Z_1,K_1],t]| \leq 16$. We have $K_1 \cong L_2(q)$, $L_3(q)$, $q \leq 16$, $U_3(4)$, $L_2(r)$, $r$ odd, or $\Z_p$. If $|[[Z_1,K_1],t]| = 16$, we have $|V_\alpha \cap L_1 : \bC_{V_\alpha \cap L_1}([Z_1,K_1])| \geq 2^8$, a contradiction. So we have $|[[Z_1,K_1],t]| \leq 8$ and so $K_1 \cong L_2(q)$, $L_3(q)$ $q \leq 8$, or $\Z_p$. Now we get $|V_\alpha \cap L_1 : \bC_{V_\alpha \cap L_1}([Z_1,K_1])| \geq 2^5$, which gives $K_1 \cong L_3(8)$ and $[Z_1,K_1]$ involves exactly one nontrivial irreducible module. If $|[Z_1,K_1] : [Z_1,K_1] \cap Q_\alpha| \leq 4$, we get $K_1 \cong L_2(4)$ or $L_3(2)$. In any case there is just one nontrivial irreducible module involved and so $[K_1,Z_1]$ is centralized by $E \cong E_{p^2}$ in $M$. This shows $[V_2 \cap L_{\alpha+1} , K_{1\alpha+1}] = 1$. We have $[K_{1\alpha+1}, Z_{\alpha+1}] \leq L_1$ and $[K_{1\alpha+1}, Z_{\alpha+1}] \cap Z_1 = 1$. Furthermore $[[K_{1\alpha+1},Z_{\alpha+1}], [K_1,Z_1]] = 1$. Now we see that $|[[K_2,Z_1],t]| \leq 8$ for $t \in [K_{1\alpha+1},Z_{\alpha+1}]$. As $[K_{2\alpha+1},Z_{\alpha+1}]$ involves at least three nontrivial irreducible modules, we see that $t$ has to induce transvections. But then $K_2 \cong L_n(2)$, $Sp_{2n}(2)$, $\Omega_{2n}^\pm(2)$ or $A_n$ and we have exactly three modules involved. This implies $p = 3$ or $7$ and there is $E \cong E_{p^2}$ in $M$ centralizing $[K_2,Z_1]$, a contradiction.


\begin{itemize}
\item[(10)] We have $|O_r(L_1/Q_1)| = r^3$ for some $r \not\in \sigma (M)$.
\end{itemize}     


Suppose false. Then by (9) there is exactly one component $K$ with $[Z_1 \cap L_{\alpha+1}, K] \not= 1$. Further $m_p(K) = 1$. Let $K_{\alpha+1}$ the corresponding component, $U_1 = [Z_1, K]$ and suppose there is $s \in U_1$ with $1\not= [K_{\alpha+1},s] \leq K_{\alpha+1}$. 

Let first $m_3(K) \leq 1$. Then $K \cong L_2(q)$, $L_3(q)$, $U_3(q)$, $Sz(q)$, $J_1$ or $\Z_p$. Now $|[Z_{\alpha+1}, s]| \leq 16r$, where $r = q$ or $r =q^2$ if $q$ is even, and $r=2$ else. As $[[s,Z_{\alpha+1}]^\sharp$ contains no element which is centralized by $E \cong E_{p^2}$, we see that there are at least four nontrivial irreducible $K$-modules involved in $Z_1$. Hence we get $K \cong L_3(2)$ or $\Z_3$. In both cases $p = 3$ and so $|Z_{\alpha+1} : Z_{\alpha+1} \cap Q_2| \geq 4$. Now $|V_2 \cap L_{\alpha+1} : V_2 \cap Q_{\alpha+1}| \geq 8$. So there is some 3-element centralizing $[K,Z_1]$, which shows $K \cong \Z_3$. Now for $t \in Z_{\alpha+1} \cap L_1$, we have $|Z_1 : \bC_{Z_1}(t)| \geq 8$, which gives $|V_\alpha : \bC_{V_\alpha}(t)| \geq 64$ and $|V_\alpha \cap L_1 : V_\alpha \cap Q_1| \geq 8$, a contradiction.

So we have $m_3(K) \geq 2$ and so $p \not= 3$. Now there must be some component or normal $p$-subgroup $K_1$ or of $L_1/Q_1$, $K_1 \not= K$ with $[K_{1\alpha+1} , s] \not= 1$. Suppose $m_p(K_1) \geq 2$. Then $[K_1,Z_1,K] \not= 1$. Now let $r$ be as before and $t \in [K_{1\alpha+1}, Z_{\alpha+1}]$. Then $|[Z_1,t]| \leq 16r$. On the other hand $[t,K_1] = 1$. So $K_1$ acts faithfully on $[t,Z_1]$. We see that $K_1$ is solvable or $K_1 \cong L_3(q)$, $q\leq 16$. Suppose the latter. As $m_3(K_1) \leq 1$, we get $K_1 \cong L_3(q)$, $p = 7$. Now we see that $[KK_1,Z_1]$ involves exactly one nontrivial irreducible $KK_1$-module. As we may assume that $m_7(L_1) \geq 4$, we see that there is some $1\not= x \in \Omega_1(Z(S))$ centralized by $E \cong E_{49}$ in $M$ with $\Gamma_{E,1}(G) \leq M$, a contradiction. So $K_1$ is a $p$-group. Then for $t$ as before we get $|V_2 : \bC_{V_2}(t)| \leq 32$. As $p > 3$, we get $[K_{1\alpha+1}, Z_{\alpha+1}] \leq L_1$ and so $|[t,Z_1]| \leq 32$. On the other hand there has to be $E \cong E_{p^2}$ acting faithfully on $[Z_1,t]$, a contradiction. So we are left with $[K_{1\alpha+1},Z_{\alpha+1}] \cap Q_2 \leq Q_1$. Then $|[Z_{\alpha+1},s]| \leq 16$. This shows $K_1 \cong L_3(8)$, $U_3(4)$, $L_2(x)$, $x$ prime, $L_2(25)$. Besides of the first case we get $|V_2 : \bC_{V_2}(t)| \leq 64$ and so $[K_{1\alpha+1}, Z_{\alpha+1}] \leq L_1$, a contradiction. So we have $K_1 \cong L_3(8)$, $p = 7$. But we must have that there are at least two nontrivial irreducible modules in $Z_1$, a contradiction.

So we have $m_p(K_1) = 1$ and there is a third component $K_2$ such that $[K_{2\alpha+1}, s] \not= 1$. Also $m_p(K_2) = 1$ and $m_3(K_2) \leq 1$. We even may assume $m_3(K_2) = 0$ and so $K_2 \cong Sz(q)$ or $\Z_p$. Let $t \in [K_{2\alpha+1}, Z_{\alpha+1}] \cap L_1$ with $[t,Z_1] \not= 1$. Suppose further $[[t,Z_1],K_2] = 1$. In this case $[K_1, [K,Z_1]] \not=1$ as $[[K,Z_1], K_2] = 1$. Now choose $t_1 \in [K_{1\alpha+1}, Z_{\alpha+1}] \cap L_1$ with $[t,Z_1] \not= 1$. Then we get $[K_1,[t,Z_1]] \not= 1$. Further we have $K_1 \cong L_2(q)$, $L_3(q)$, or $K_1$ is a $p$-group. If $[Z_1,t]$ involves exactly one nontrivial irreducible $K_1$-module, we get some $x \in \Omega_1(Z(S))^\sharp$ which is centralized by $E \cong E_{p^2}$ or $K_1$ is a $p$-group. In the latter $|V_2 \cap L_{\alpha+1} : V_2 \cap Q_{\alpha+1}| = 2$ and so $Z_{\alpha+1} \leq L_1$. Now also $|[Z_1,t_1]| \leq 32$. As now $[K_1,Z_1]$ involves at least four nontrivial irreducible $K$-modules we get that it involves an F-module for $K$ in which $t_1$ centralizes a subgroup of index $4$. This now shows that $|\bC_{[Z_1,K_1]}(S \cap K)| \leq 32$. Now on this centralizer acts an elementary abelian group of order $p^3$ and so some $x \in \Omega_1(Z(S))^\sharp$ is centralized by some $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \leq M$, a contradiction.  

So we may assume $[K_{1\alpha+1}, Z_{\alpha+1}] \cap L_1 \leq Q_1$. Then $s$ centralizes a subgroup of index 16 in $[K_{1\alpha+1}, Z_{\alpha+1}]$. As $[K_{1\alpha+1}, Z_{\alpha+1}]$ involves at least four nontrivial irreducible modules we have that $s$ induces transvections. This shows $K_1 \cong L_3(2)$ or $L_2(4)$. As $[K_2,[K_1,Z_1]] \not=1 $ we see that $K_2 \cong \Z_p$. Now $|Z_{\alpha+1} : Z_{\alpha+1} \cap Q_2| = 16$ and then $|V_2 \cap L_{\alpha+1} : V_2 \cap Q_{\alpha+1}| \geq 2^8$, a contradiction. 

So we have $[[Z_1,t],K_2] \not= 1$. Now $K_2 \cong \Z_p$. Then $Z_{\alpha+1} \leq L_1$ and $|[t,Z_1]| \leq 32$. We have that $K_1$ is nonsolvable and there are more than one nontrivial irreducible $K_1$-modules involved. So again we see $|\bC_{[Z_1,K_1]}(S \cap K)| \leq 32$, which give the same contradiction as before.

So we are left with $[Z_{\alpha+1}, K_{2\alpha+1}] \cap L_1 \leq Q_1$. Now $|[[Z_{\alpha+1}, K_{2\alpha+1}],s]| \leq 16$ and so $K_2 \cong \Z_p$. Then again $Z_{\alpha+1} \leq L_1$, a contradiction.

We now have that $K_{\alpha+1}^s \not= K_{\alpha+1}$. Then $K \cong L_2(q)$, $L_3(2)$ or $J_1$. Let $|V_2 \cap L_{\alpha+1} : V_2 \cap Q_{\alpha+1}| = 2$, then $Z_{\alpha+1} \leq L_1$ and so we see that $K \cong L_2(q)$ and $|[t, Z_{\alpha+1}]| = q$. Then $[K \times K^s, Z_1]$ is irreducible and so there is some $x \in \Omega_1(Z(S))^\sharp$ centralized by $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \leq M$. So we have $|V_2 \cap L_{\alpha+1} : V_2 \cap Q_{\alpha+1}| > 2$ and then again $K \cong L_2(q)$ and natural $O_4^+(q)$ - modules are involved. Now for $t \in [K_{1\alpha+1}\times K_{1\alpha+1}^s, Z_{\alpha+1}]$ we have $|Z_1 : \bC_{Z_1}(t)| \leq 32q$ and as there are at least two orthogonal modules involved we see $|Z_1 : \bC_{Z_1}(t)| \geq q^4$ or $[K_{1\alpha+1} \times K_{1\alpha+1}, Z_{\alpha+1}] \cap L_1 \leq Q_1$. The former provides us with a contradiction. In the latter we have two such modules involved and then $q = 4$. Now $|Z_{\alpha+1} :Z_{\alpha+1} \cap Q_2| = 16$, a contradiction again.

So we have shown that $[[Z_1,K] \cap L_{\alpha+1}, K_{\alpha+1}] = 1$. Let $[K_{\alpha+1}, [Z_{\alpha+1},s]] = 1$. We have $K_1$ with $[K_{1\alpha+1}, s] \not= 1$ again. If $m_p(K_1) = 1$, we must have another $K_2$ with $[K_{2\alpha+1},s] \not= 1$ too. In any case $[K,Z_1]$ is centralized by $E\cong E_{p^2}$ with $\Gamma_{E,1}(G) \leq M$. Now we have $K \not\!\!\unlhd L_1/Q_1$ and so we get $[s, KK^x] = 1$, for some $x \not\in N_{L_1/Q_1}(K)$, a contradiction. So we have $[K_{\alpha+1}, [Z_{\alpha+1},s]] \not= 1$. Let $m_3(K) \leq 1$, then we see that $K \cong \Z_p$, $L_2(q)$, $L_3(q)$, $q$ even, $q \leq 16$. We have $[Z_{\alpha+1}, K_{\alpha+1}] \leq L_1$. In any case there is some $E \cong E_{p^3}$ centralizing $K_{\alpha+1}$. Suppose there is some nontrivial irreducible module in $[Z_{\alpha+1}, K_{\alpha+1}]$ centralized by $F \cong E_{p^2}$. As  this module intersects $Q_1$ nontrivially, we see $Z_1 \leq L_{\alpha+1}$, a contradiction. So we have that $K$ acts trivially on $\bC_{Z_1}(F)$ for any $F \leq E$, $|F| = p^2$. But $Z_{\alpha+1} = \langle \bC_{Z_{\alpha+1}}(F) ~\Big|~|F| = p^2 \rangle$, a contradiction.

So we have $m_3(K) \geq 2$ and $3 \not\in \sigma (M)$. As before there is an irreducible module in $Z_1$ whose dimension is less than the $2$-rank of $K$. With (1.5) we get that $K$ is a group of Lie type in characteristic two. This shows $K \cong L_4(q)$, $Sp_6(q)$, $L_5(2)$, $L_6(2)$ or $L_7(2)$. We have $p~\not|~q-1$. As there is $E \cong E_{p^3}$ in $\bC(K)$ there are at least two nontrivial modules in $Z_1$ which are centralized by some elementary abelian group of order $p^2$, or $[[E, Z_{\alpha+1}], K_{\alpha+1}] = 1$. Assume the former. Then we get $U \leq Z_{\alpha+1} \cap L_1$, $U \cap Q_1 = 1$ with $|U| \geq q^8$, $q^{12}$, $2^{10}$, $2^{12}$ and $2^{14}$ , respectively. This is impossible. So we have the latter. Now $[E, [Z_{\alpha+1}, K_{\alpha+1}]] = 1$ . Further 
$[Z_1,K]$ is the natural module. But then $[K_{1\alpha+1},[K_{\alpha+1}, Z_{\alpha+1}]] = 1$, a contradiction.

So it remains $[[K,Z_1] \cap L_{\alpha+1}, Z_{\alpha+1}] = 1$. Then for $t \in Z_{\alpha+1}$ we have $|[[K,Z_1],t]| \leq 16$ or $8$. Now we see that some $E \cong E_{p^2}$ centralizes $[K,Z_1]$. So $[V_2 \cap L_{\alpha+1}, K_{\alpha+1} ] = 1$. This gives that $[K_{\alpha+1}, Z_{\alpha+1}] \leq L_1$. Now we get the same groups for $K$ as before. Further $[K,Z_1]$ involves exactly one nontrivial irreducible module, which is the natural one. We have $[K_{\alpha+1}, Z_{\alpha+1}]Q_1/Q_1 \cap K \not= 1$. So $\bC_{Z_1}(K) \leq M_{\alpha+1}$. Now $Z_{\alpha+1}/\bC_{Z_{\alpha+1}}(K_{\alpha+1})$ has to be a nonsplit extension of the natural module by a trivial module. Application of (1.20) shows $K \cong Sp_6(q)$. Now $|Z_1 : Z_1 \cap M_{\alpha+1}| \leq q$. But $[K_{\alpha+1}, Z_{\alpha+1}]$ acts quadratically on $Z_{1}$, so some $x \in [K_{\alpha+1}, Z_{\alpha+1}]^\sharp$ centralizes some $y \in Z_1 \setminus [K,Z_1]\bC_{Z_1}(K)$, a contradiction.  So we have $Z_1 \leq M_{\alpha+1}$, a contradiction again. \\

(11)~~~~~~~~~~~~~~~~~~$Z_{\alpha+1} \cap L_1 \leq Q_1$.\\

Suppose false. By (10) we have $(L_1/Q_1)/O_r(L_1/Q_1) \cong L_2(r)$ and $|O_r(L_1/Q_1)| = r^3$. Now $|Z_{\alpha+1} \cap L_1/Z_{\alpha+1} \cap Q_1| \leq 4$. And so for $s \in Z_1$ we have $|Z_{\alpha+1} : \bC_{Z_{\alpha+1}} (s)| \leq 64$. By (6.16) we see that $|Z_{\alpha+1} : Z_{\alpha+1} \cap Q_2| \leq 2$. So $|Z_{\alpha+1} : \bC_{Z_{\alpha+1}}(s)| \leq 8$. Application of (1.51) shows $[O_r(L_1/Q_1),Z_1] = 1$. But this contradicts
(4.10). 

Now (11) contradicts (1) and we are done.

  



\absa
{\bf (6.35)~Proposition.~}{\it If $b=b_{1}$, then $b=1$.}

\absa
Proof~. By (6.34) it remains to treat the case $L_{2}/Q_{2}$ is
solvable with $\sigma (L_{2})\not= \emptyset$. By construction in
(4.11) we may assume that either $L_{1}$ is solvable too with
$\sigma (L_{1})\not= \emptyset$ or $E(L_{1}/Q_{1})\cong
K_{1}\times\ldots\times K_{t}$, $K_{i}\cong G(q)$, $q$ odd, $G(q)$ a
group of Lie type different from $U_{3}(3)$, $PSp_{4}(3)$, $^{2}G_{2}(3)$ 
and $3U_{4} (3)$. 
\begin{itemize}
\item[(1)] $[F^{*}(L_{1}/Q_{1}),Z_{1}]=Z_{1}$, $\bC_{Z_{1}} (F^{*}(L_{1}/Q_{1}))=1\, .$
\end{itemize}

Otherwise some $1\not= x\in\Omega_{1} (Z(S))$ is centralized by
$F^{*} (L_{1}/Q_{1})$ and so $L_{2}\le \bC_{G}(x)\le M$, a contradiction.

Application of \cite[(24.1)]{GoLyS}, \cite[(25.12)]{GoLyS} and (1.12) (ii) shows that $Z_{1}$ is
generated by elements $z$ such that 
\begin{itemize}
\item[(2)]\parbox{12cm}{$|V_{\alpha} :\bC_{V_{\alpha}} (z)|\leq 4|V_{\alpha}
:V_{\alpha}\cap L_{1}|$ and\\
$|V_{\alpha} :\bC_{V_{\alpha}} (z)|\leq 2|V_{\alpha} :V_{\alpha}\cap
L_{1}|$ for $K_{1}\not\cong L_{2}(7)$ or $L_{2}(9)$}
\end{itemize}

Suppose $V_\alpha \leq M$. Then $|V_\alpha : V_\alpha \cap L_1| \leq 2$ and so by (2) $Z_1 = \langle z \Big|~|Z_{\alpha+1} : \bC_{Z_{\alpha+1}}(z)| \leq 8 \rangle$ for $\alpha+1 \in \Delta (\alpha)$ with $Z_1 \not\leq M_{\alpha+1}$. But this is a contradiction. So we have $V_\alpha \not\leq M$ and then we have symmetry. So we may assume that $|V_{\alpha} :V_{\alpha} \cap Q_{\,
2}|=2^{\, s}\geq |V_{2}:V_{2}\cap Q_{\alpha} |$. Set 
$V_{\alpha} =(V_{\alpha} \cap Q_{\, 2})A,\; A=\langle z_{\beta_{1}}
,\ldots ,z_{\beta_{s}}\rangle ,z_{\beta_{i}} \in Z_{\beta_{i}}$,
$\beta_{i}\in\triangle (\alpha )$. As $Z_{\beta_1}$ is generated by elements $z$ with $|V_2 \cap L_{\beta_1} : \bC_{V_2 \cap L_{\beta_1}}(z)| \leq 4$, we may assume that $|V_2 \cap Q_\alpha : \bC_{V_2 \cap Q_\alpha}(z_{\beta_i})| \leq 4$. Hence $|V_2 : \bC_{V_2}(A)| \leq 2^{3s}$. On the other hand by (2.1) $L_2$ is a $\{2,r\}$-group with $r > 3$. Now $|T : \bC_T(A)| \geq 2^{2s}$ for any nontrivial irreducible $L_2/Q_2$-module $T$ with $O_2(\bC_{L_2/Q_2}(T)) = 1$. By (6.16) we get $|V_2 : \bC_{V_2}(A)| \geq 2^{4s}$, a contradiction.


\absa
{\bf (6.36)~Lemma.~}{\it Let $b=b_{1}=1$. Then there is
$3\in\triangle (2)$ such that $T=\langle Z_{3},Z_{1}\rangle /(Q_{\,
2}\cap\langle Z_{1},Z_{3}\rangle)\cong L_{2}(q)$, $Sz(q)$ or $D_{2t}$,
$t$ prime. Furthermore
$$Y=(Z_{1}\cap Q_{\, 2})\, (Z_{3}\cap Q_{\, 2})\lhd \langle
Z_{1},Z_{3}\rangle$$
and $Y/Z_{1}\cap Z_{3}$ is a direct sum of natural T-modules. Furthermore there is some $\omega \in T$ with $o(\omega) = x ~|~q-1$ and $N_G(\langle \omega \rangle) \not\leq M$.}

\absa
Proof~. We have $Z_{1}\not\le Q_{\, 2}$ and $Z_{1}Q_{\, 2}\lhd
S$. Furthermore $Z_{1}$ acts quadra\-tically on $Q_{\, 2}$. If $|Z_1 : Z_1 \cap Q_2| = 2$, we get some $g$ such that $\langle Z_1 , Z_1^g\rangle/(\langle Z_1, Z_1^g \rangle \cap Q_2) \cong D_{2t}$, $t$ odd. So assume $|Z_1 : Z_1 \cap Q_2| > 2$. Then by (1.12) $E(L_2/Q_2)$ is a product of groups of Lie type in characteristic two or $3\cdot A_6$, or there is some $g \in L_2$ such that $\langle Z_1 , Z_1^g\rangle/(\langle Z_1, Z_1^g \rangle \cap Q_2) \cong D_{2t}$, $t$ odd. Now with (1.1) it is easy to see that we get $g\in L_{2}$ such that $\langle Z_{3},Z_{1}\rangle /Q_{\,
2}\cap\langle Z_{1},Z_{3}\rangle\cong L_{2}(q)$, $Sz(q)$ or $D_{2t}$
or $E(L_{2}/Q_{\, 2})\cong Sp_{4}(q)$ or $3A_{6} * 3A_{6}$ and
then $|Z_{1}:Z_{1}\cap Q_{\, 2}|=q^{\, 2}$, $2^{\, 4}$ respectively.
If we have $L_2(q)$ or $Sz(q)$, then as $Z_1 \unlhd L_1 \cap L_2$, we see that $Z_1$ cannot contain a root group and so $\langle Z_1, Z_3\rangle $ contains some $\omega$, $o(\omega) = x ~|~q-1$, $x$ a Zsigmondy prime or $x = 9$ in case of $q-1 = 63$, such that $\omega \in L_1 \cap L_2$ and $N_G(\langle \omega \rangle) \not\leq M$.

If $E(L_2/Q_2) \cong Sp_4(q)$ or $3A_6*3A_6$ there is some $h\in L_{2}$ such that $\langle
Z_{1},Z_{1}^{\, h}\rangle Q_{\, 2}/Q_{\, 2}=E(L_{2}/Q_{\, 2})$. By
quadratic action we see that $Y/Z_{1}\cap Z_{3}$ just involves
natural modules (1.15) and (1.14).

Now choose $\tilde{Z}_{1}\le Z_{1}$ such that $|\tilde{Z}_{1}
:\tilde{Z}_{1}\cap Q_{\, 2}|=q$ and $\tilde{Z}_{1} Q_{\, 2}/Q_{\, 2}$
corresponds to a transvection group on the natural module. There are
two cases
\begin{itemize}
\item[(i)] $\langle\tilde{Z}_{1} ,\tilde{Z}_{1}^{\, h}\rangle
/(\langle \tilde{Z}_{1} ,\tilde{Z}_{1}^{\, h}\rangle\cap Q_{\, 2})\cong
L_{2}(q)$ or
\item[(ii)] $\langle\tilde{Z}_{1} ,\tilde{Z}_{1}^{\, h}\rangle\cong
E_{q^{\, 2}}$.
\end{itemize}

Suppose (i). Then $Y/Z_{1}\cap Z_{3}$ is a direct sum of natural
$L_{2}(q)$-modules by (1.27)(iv). Which gives that $|[\tilde{Z}_{1}
,Y/(Z_{1}\cap Z_{3})]|=|\bC_{Y/(Z_{1}\cap Z_{3})} (\tilde{Z}_{1} )|$. 
But this contradicts the fact that $Y/Z_{1}\cap Z_{3}$ involves some
natural module and $\tilde{Z}_{1}$ induces the transvection group on
this module.

So we have (ii). Then $\langle\tilde{Z}_{1} ,\tilde{Z}_{1}^{\,
h}\rangle Q_{\, 2}/Q_{\, 2}\le E$, $E\cong E_{q^{\, 3}}$ and
there is some $g\in N(E)$ with $\tilde{Z}_{1} Q_{\,
2}=\tilde{Z}_{1}^{\, hg} Q_{\, 2}$, $hg\in N(\tilde{Z}_{1} Q_{\,
2})$. As $Z_{1}\leq O_{2}(N_{L_{2}} (\tilde{Z}_{1} Q_{\, 2}))$, we
see that $Z_{1}^{\, hg}\le N(E)$ and so $Z_{1}^{\, h}\le
N(E)$. Now $\langle Z_{1},Z_{1}^{\, h}\rangle\le N(E)$, a
contradiction. This proves the first part of the assertion. 

As $\bC_{L_{2}} (Q_{\, 2})\le Q_{\, 2}$, we get $Y\not=
Z_{1}\cap Z_{3}$. Now the second part follows with (1.27)(iii).

\absa
{\bf (6.37)~Lemma.~}{\it Assume $b=b_{1}=1$. Let $K$ be some
component of $L_{1}/Q_{1}$, or normal $p$-group , $m_{p}(K)\geq 2$. Then $[K,Y]=1$. }

\absa
Proof~. Suppose $[K,Y]\not= 1$. Set $q = 2$ for $\langle Z_1, Z_3 \rangle/O_2(\langle Z_1, Z_3 \rangle) \cong D_{2t}$. 
\begin{itemize}
\item[(1)] $[K,Y]\le K$. Furthermore $[K,\omega ]\le K$ if $T \cong L_2(q)$ or $Sz(q)$.
\end{itemize}

Suppose first that (6.4) holds. Then 
$[Y,\omega]Q_{1}= YQ_{1}$. As $m_{p}(K)\geq 2$, we get by (1.63) that $[\omega,K]\le K$  or
$q=64$ and $\omega^{3}\in N_{L_{1}/Q_{1}} (K)$. If $[K, \omega] \leq K$ we see that $[Y,K] \leq K$. 

Assume $q = 64$ But then as $[Y,\omega^3] = Y$ we get $[Y,K] \leq K$ too. 
Now  $3\not\in \sigma (M)$. Hence $m_{3}(K)=1$,
$K^{\,\langle\omega\rangle} =K_{1}\times K_{2}\times K_{3}$. Let
$\nu\in L_{1}\cap L_{2}$, $o(\nu )=7$. Then $ 1 \not= [\nu ,K_{i}]\le
K_{i}$. As $m_{p}(K)\geq 2$, and there is some subgroup isomorphic to $E_{64}\Z_{21}$ normalizing $K$, we get with (1.5) that $K\cong L_{3}(r)$
or $U_{3}(r)$, $r$ even. Suppose that $\nu$ induces a field automorphism on $K_1 \times K_2 \times K_3$. Then $m_p(\bC_K(\nu)) \leq 1$ for any odd prime $p$. So $K \cong L_3(r)$ and then $E \cong E_{7^3}$ is contained in $\bC_{K_1 \times K_2 \times K_3}(\nu)$. This shows $m_7(M) \geq 4$ and so $7 \in \sigma (M)$ a contradiction as $N_G(\langle \nu \rangle) \not\leq M$. So we have that $\nu$ induces an inner automorphism. Then we see that $\nu$ is in a $2$-local of $K$, i.e. $o(\nu) ~|~r^2-1$. As $m_7(K) = 1$, we get that $K \cong L_3(r)$ and $7~|~r+1$, $p~|~r-1$ or $K \cong U_3(r)$, $7~|~r-1$ and $p~|~r+1$. In the last case there is $E \cong E_{p^3}$ which centralizes $\nu$ contradicting (1.63). So we have $K \cong L_3(r)$ and $7 ~|~r+1$, $p~|~r-1$. As $\nu$ is in a $2$-local it centralizes an element of order $r-1$ in $K_1$ and so it centralizes $E \cong E_{p^3}$ in $K_1 \times K_2 \times K_3$, a contradiction. 

So we have $\langle Z_{1},Z_{3}\rangle /\langle
Z_{1},Z_{3}\rangle\cap Q_{\, 2}\cong D_{2t}$. Hence $Z_{1}$ is an
$F_{1}$-module with offending subgroup $Y$. Now let $y\in Y$ with
$K^{\, y}\not= K$. Then $y$ acts on some $E\cong E_{p^{\, 4}}$. Hence
there is some $F\le E$, $|F|=p^{\, 2}$, $|\bC_{[y,Z_{1}]}
(F)|\geq 4$. But $|[y,Z_{1}]:[y,Z_{1}]\cap Z_{3}|=2$ and so some
$x\in (Z_{1}\cap Z_{3})^{\,\sharp}$ is centralized by $F$. As
$\Gamma_{F,1} (G)\leq M$, we get $\langle Z_{1},Z_{3}\rangle\le
M$, a contradiction.
\begin{itemize}
\item[(2)] $\bC_{Y} (K)=1$.
\end{itemize}



Let $y\in\bC_{Y} (K)$. Then $K$ acts on $[y,Z_{1}]$. So suppose first
$[K,[y,Z_{1}]]=1$. Then $[y,Z_{1}]\cap Z_{3}=1$. Hence
$|[y,Z_{1}]|=q$. As $\bC_{Z_{1}} (K^{\, S})=1$, we get some
$K_{1}\sim K$ in $L_{1}$ such that  $1\not= [K_{1},y]\le
K_{1}$. But then we see $[[K,Z_{1}],K_{1}]=1=[[K_{1},Z_{1}],K]$. So
there is some $y_{1}\in Y$ with $[y_{1},K_{1}]=1$ and
$[y_{1},K]\not= 1$. Hence also $|[y_{1},Z_{1}]|=q$. But then by (6.36)
$Z_1 \cap Y = \bC_{Z_{1}} (y)=\bC_{Z_{1}} (y_{1})=\bC_{Z_{1}} (y\, y_{1})$ which
is impossible, as $[y,[K_{1},Z_{1}]]=1$ and
$[y_{1},[K_{1},Z_{1}]]\not= 1$. 

So we have $[[y,Z_{1}],K]\not= 1$. Set $W=[y,Z_{1}]$. Then
$|W:\bC_{W} (Y)|\leq q$ by (6.36).  Furthermore $W\le Y$ and so
$[Y,W]\le Z_{1}\cap Z_{3}$ i.e. $Y$ acts quadratically on $W$.
As $|Y:\bC_{Y} (K)|\geq q$, by the action of $\omega$, we see that $W$
is an F-module for $K$. Hence by (1.18) we get $K\cong A_{n}$,
$L_{n}(r)$, $Sp_{2n}(r)$, $U_{n}(r)$, $\Omega_{2n}^{\,\pm} (r)$ or
$G_{2}(r)$, or $K$ is solvable. Furthermore we have that
$Z_{1}/\bC_{Z_{1}} (y)\cong W$ as K-module and $|[Y,Z_{1}/\bC_{Z_{1}}
(y)]|=|Z_{1}/\bC_{Z_{1}} (y)|/q$. 

Let $K$ be solvable. As some $A \leq Y$, $|A| = q$ with $A\langle \omega \rangle \leq N(K)$ and $A\langle \omega \rangle \cap \bC(K) = 1$ centralizes a subgroup of index $q$ in $W$, we see that $|Y:\bC_{Y}
(K)|=2$. This implies $q=2$ and $m_{3}(K)=2$. Finally $|W|=4$ as $Y$ induces the full transvection group on $W$.  Let
$y_{1}\in \bC_{Y} (K)^{\,\sharp}$, $y_{1}\not= y$. If
$|Z_{1}:\bC_{Z_{1}} (\langle y,y_{1}\rangle )|=8$ there is some $u\in
Z_{1}\setminus \bC_{Z_{1}} (\langle y,y_{1}\rangle )$,
$[u,E]\le \bC_{Z_{1}} (\langle y,y_{1}\rangle )$ with $E\cong
E_{9}$, $E\le K$. But $\bC_{Z_{1}} (E)\le\bC_{Z_{1}}
(y)$, as $[E,W]=W$. So $1\not= [\bC_{Z_{1}} (E),y_{1}]\le
Z_{1}\cap Z_{3}$, a contradiction.

This shows $|Z_{1}:\bC_{Z_{1}} (\bC_{Y} (K))|=4$. Furthermore
$|Z_{1}:Z_{1}\cap Z_{3}|>8$ and so $|\bC_{Y} (K)|\geq 4$. Hence $Z_{1}$
is an F-module for $\bC (K)$ with offending subgroup $\bC_{Y} (K)$.
But then the action of $K$ shows that we have two submodules
$T_{1},T_{2}$ of $Z_1$ such that $\bC_{Y} (K)$ induces the full group of
transvections on $T_{i}$ and $T_{1}\sim_{K} T_{2}$. This gives
$K_{1}\le \bC (K)$, $K_{1}\cong L_{n}(2)$, $|Y:Y\cap
Q_{1}|=2^{\, n}$. But now every $x\in T_{1}$ is centralized by $E\cong
E_{9}$, $E\le L_{1}$. Hence $[T_{1},\bC_{Y} (K)]\cap Z_{3}=1$.
The same is true for $T_{2}$. In particular $T_{1},T_{2}\le Y$ and so $|\langle T_{1},T_{2}\rangle :\bC_{\langle T_{1},T_{2}\rangle}
(y)|=2$ for $y\in \bC_{Y} (K)^{\,\sharp}$, a contradiction. So we
have that $K$ is nonsolvable.

Suppose first $q=2$, then $Y$ induces the full transvection group on
$W$ i.e. $K\cong L_{n}(2)$. Now $[Y,W]\le Z_{3}\cap Z_{1}$ and
is normalized by $L_{n-1}(2)$. This implies $m_{p}(L_{n-1}(2))\leq
1$. Hence $K\cong L_{4}(2)$, $p=3$ or $K\cong L_{6}(2)$, $p=7$. But
in both cases $[W,Y]$ contains a subgroup $U$ of dimension
$\frac{n}{2}$ normalized by $E\cong E_{p^{\, 2}}$, a contradiction.

So we have $q>2$. Now $Y\cap K\not= 1$. Let $K\cong U_{n}(r)$ or
$\Omega_{2n}^{\, \pm} (r)$, then $q\geq r^{\, 2}$ by (1.18). As $\omega \in L_1 \cap L_2$, we have that $\langle S, \omega \rangle$ is a $\{2,x\}$ - group, and so by (1.65) and (1.64) we have $q=r^{\, 2}$. But then $|[Y,W]|\not= |W|/q$. 

Let $K\cong L_{n}(r)$, $Sp_{2n}(r)$ or $G_{2}(r)$, then $q\geq r$. If
$q=r$, we get as before $K\cong L_{n}(r)$ and $W$ is the natural
module. Now as above ($K \cong L_4(2)$) we get a contradiction. So let $q>r$. By (1.65)
and (1.64) $q=4$, $r=2$ and $\omega\in K$. Now $m_{p}(K)\geq 3$,
$p\not= 3$ and $3 \not\in \sigma (M)$. But this is impossible. 

So we are left with $A_{n}$. Again by (1.9) $\omega\in K$ and so $m_{p}(K)\geq
3$, $p\not= 3$, $3 \not\in \sigma (M)$.  The same contradiction as before.
\begin{itemize}
\item[(3)] If $y\in Y$ then $|Z_{1}:\bC_{Z_{1}} (y)|\leq |Y|$.
Furthermore $K\cong G(r)$, a group of Lie type in characteristic two.
\end{itemize}

The first assertion is clear. Now we go over all possibilities for $K$. 

Let $K\cong A_{n}$. By (1.9) $q\leq 4$ or $q=64$. If $o(\omega )=3$,
or $9$ then $\omega\in K$ and so $m_{p}(K)\geq 3$, $p\not= 3$, $3 \not\in \sigma (M)$, a
contradiction.

Hence $q=2$ and so $Z_{1}$ is an F-module. By (1.42) $Z_{1}$ is the
natural module, or $n=7$ or $8$ and $Z_{1}$ involves a four-dimensional
module, or $n=9$ and $Z_{1}$ involves a spin module. Suppose that
$Z_{1}$ is natural. Let $3\in \sigma (M)$. Then some
$x\in\Omega_{1}(Z(S))^{\,\sharp}$ is centralized by an elementary
abelian group $E$ of order 9 and so $L_{2}\le M$, a
contradiction. So let $p=5$, $K\cong A_{10}$ or $A_{11}$. Then by (4.4) we may assume that 
$[K,Z_{1}]$ is centralized by $E_{25}$, a contradiction. Let next
$K\cong A_{9}$ and $Z_{1}$ involve a spin module. Now we have that
just one such module is involved. Hence as $m_{3}(K)=3$, we see that
$[Z_{1},K]$ is centralized by an element of order 3. As every element
in the spin module is centralized by a 3-element, we get some $E\cong
E_{9}$, $\Gamma_{E,1} (G)\leq M$, with $\bC_{\Omega_{1} (Z(S))}
(E)\not= 1$. Hence $L_{2}\le M$, a contradiction. 

So we have $K\cong A_{7} \,\,(K\cong A_{8}\cong GL_{4}(2)$ is a group of
Lie type), 
$p=3$. In that case by (4.4) any $z \in [K,Z_{1}]$ is centralized by an elementary
abelian group of order 9, a contradiction. 

Let $K$ be sporadic. By (1.9) $q\leq 4$ or $q=64$, or $q=8$ and
$K\cong J_{1}$. But $m_{p}(K)\geq 2$, so $K\cong J_{1}$ is not
possible. 

Let $q=4$ or $64$. Then $p\not= 3$. But $m_{p}(K)\geq 3$, a
contradiction to (1.4). So assume $q=2$. Then $Z_{1}$ is an F$_1$ - module
and so $K\cong M_{12}$, $M_{22}$, $M_{23}$, $M_{24}$ or $J_{2}$ by
(1.42). Now $p=3$ or $p=5$ in case $J_{2}$ and $m_{p}(K)=2$. Then
$\Omega_{1} (Z(S))^\sharp$ contains $x$, which is centralized by $E\cong
E_{9}$ in $M$, and so $L_{2}\le M$, a contradiction.

Let $K\cong G(r)$ be a group of  Lie type in odd characteristic. Suppose first
$q>2$. As $\omega$ acts fixed point freely on $Y$, we see that $Y$
corresponds to a subgroup of $K$. Now (1.41) yields a contradiction.
So we have $q=2$ and $Z_{1}$ is an F$_1$ - module. By (1.42) $K\cong
3U_{4}(3)$, $p=3$. But now $Y\cap K=1$, as every involution in $K$ is
centralized by $E\cong E_{27}$ by (1.55) and so some $1\not= z \in Z_1 \cap Z_3 \cap [W, Y \cap K]$ would be centralized by $F \cong E_9$. So we have $|Y|=2$ by (2).
Then $Y$ induces a transvection, a contradiction.

It remains $K$ to be solvable. Let $|Y:Y\cap Q_{1}|=2^{\, n}$. Now by \cite[(24.1)]{GoLyS}
there is $E\cong E_{p^{\, n}}$ such that $[E,y]=E$ for some $y\in Y$.
As $|Z_{1}:\bC_{Z_{1}} (y)|\leq |Y|$ there is some $F\cong E_{p^{\, 2}}$
in $E$ such that $|Z_{1}:\bC_{Z_{1}} (F)|\leq 16$. Now $\bC_{Z_{1}}
(F)\cap Z_{3}=1$ and so $|\bC_{Z_{1}} (Y):\bC_{\bC_{Z_{1}} (Y)}
(F)|\leq 4$. Hence we have 
$$ |Z_{1}\cap Z_{3}|\leq 4\, .$$

As $|[y,Z_{1}]|=|Y|$, we get $|Z_{1}|\leq |Z_{1}\cap Z_{3}|^{\,
2}\cdot q^{\, 2}$. Suppose first $|Y|\geq q^{\, 2}$. Then we see
$q\leq 4$ and so $|Z_{1}|\leq 2^{\, 8}$. If $q=4$, then $|Y|=16$ and
so $|Z_{1}|=2^{\, 8}$. But then the action of $K$ on $Z_{1}$ is
uniquely determined i.e. there is $y\in Y^{\,\sharp}$ with
$|[Z_{1},y]|=2$, a contradiction. This shows $q=2$ and so
$|Z_{1}|\leq 2^{\, 6}$. This now implies $|Z_{1}|=64$ and $|Y|=8$.
But now we have $\langle y_{1},y_{2}\rangle \leq Y$ with $|\langle y_{1},y_{2}\rangle
|=4$ and $|Z_{1}:\bC_{Z_{1}} (y)|=2=|Z_{1}:\bC_{Z_{1}} (y_{1})|$.
This would imply that $|Z_{1}:\bC_{Z_{1}} (\langle y_{1},y_{2}\rangle
)|=2$ by the action of $Y$ on $Z_{1}$. But this contradicts \cite[(26.11)]{GoLyS}.

So we have $|Y|=q$ and then $\bC_{Z_{1}} (Y)=\bC_{Z_{1}} (y)$ for every
$y\in Y\setminus Q_{1}$. Now 
$$|Z_{1}|=|Z_{1}\cap Z_{3}|q^{\, 2}$$
and $Y$ acts quadratically on $Z_{1}$. So by \cite[(24.1)]{GoLyS} and \cite[(25.12)]{GoLyS} there
is $t\in Z_{1}^{\,\sharp}$ with $|Y:\bC_{Y} (t)|=2$. This shows $q=2$
and so $|Z_{1}|=16$. But then there is $E\cong E_{9}$ with
$[Z_{1},E]=1$, whence $L_{2}\le M$, a contradiction.\\

Now we are able to prove the lemma. Let $T\le Z_{1}$ be an
irreducible Aut($K$)-submodule. Then $\bC_{K} (\bC_{T} (S\cap K))$ does
not contain an elementary abelian group of order $p^{\, 2}$. Now (3), (1.22) and (1.24) imply $K\cong L_{3}(r)$, $L_{4}(r)$, $L_{n}(2)$, $n\leq 6$, $U_{n}(r)$,
$n\leq 6$, $Sp_{4}(r)$, $Sp_{6}(r)$, $\Omega_8^{\, -} (r)$,
$G_{2}(r)$.

If $K\cong \Omega_8^{\, -} (r)$, then $T$ is the spin module and
$p\, \Big | \,r+1$. Now in the spin module the codimension of 
$\bC (v)$ is at
least four, $v$ an involution. Hence $T=[K,Z_{1}]\lhd L_{1}$ and so
$E\cong E_{p^{\, 2}}$ centralizes some $1\not= x\in\Omega_{1}
(Z(S))\cap T$ by (4.4), as $m_{p}(K)\leq 3$. 

Let $K\cong G_{2}(r)$, then also $T=[K,Z_{1}]\lhd L_{1}$ and also
some $1\not= x\in\Omega_{1} (Z(S))$ is centralized by $E\cong
E_{p^{\, 2}}$ again. 

Let $K\cong Sp_{6}(r)$, then $T$ is the spin module, $m_{p}(K)=3$. If
$T\not\!\!\unlhd L_{1}$, then $|[T,y]|=r^{\, 2}$ for every $y\in
Y\setminus Q_{1}$. Furthermore $|Y/Y\cap Q_{1}|\geq r^{\, 4}$.
Hence we see $|Y/Y\cap Q_{1}|=r^{\, 4}$ and $|\bC_{T} (Y)| > r^{\, 5}$.
But this is impossible. So we have $T\lhd L_{1}$ and the same
contradiction as before. 

Let $K\cong Sp_{4}(r)$, $T$ the natural module. As $|Y:Y\cap
Q_{1}|\leq r^{\, 3}$, we get that $T\lhd L_{1}$ and so we have the
same contradiction as before. 

Let $K\cong U_{6}(r)$. Then $T\cong V(\lambda_{1}
+\lambda_{l} )$ in the notation of (1.22). As $T$ is not an
$F_{1}$-module we see $q>2$. If $3~|~o(\omega)$ then $3 \not\in \sigma (M)$ and so $r > 2$. Hence by (1.65) and (1.64) we either have $q \leq r$ or $q = r^2$. But $p~|~r+1$ and so $p~|~q-1$, i.e. there is $\nu \in L_1 \cap L_2$, $o(\nu) = p$ and $N_G(\langle \nu \rangle) \not\leq M$. So we have $q \leq r$. Let $R$ be a transvection group in the natural representation with $[Y,R] \leq Q_1$. We have $|[T,R]| \geq r^6$, as $\bC_K(R)^\prime$ acts faithfully on $[T,R]$. If $R \cap Y \not= 1$, then $|Y : Y \cap Q_1| \geq r^6$ by (3). Now we have $|[T,R] : \bC_{[T,R]}(Y)| \leq q \leq r$. As $YQ_1/Q_1 \not\leq O_2(\bC_K(R))$, there is some module $V$ for $U_4(r)$ such that $|V : \bC_V(u)| \leq r$ for some 2-element $u \in U_4(r)$, contradicting (1.22) - (1.24). So we have $R \cap Y = 1$. As $[Y,Q_2] \leq Y$, we see that $Y \cap O_2(N_K(R))$ is abelian and $|O_2(N_K(R))/O_2(N_K(R)) \cap Q_2| \geq r^4 \geq q^4$. Hence $S/Q_2$ involves an elementary abelian subgroup of order $q^4$. By (1.7) we get either $q = r = 8$ and $E(L_2/Q_2) \cong SU_3(8)*SU_3(8)$ or $q=r=4$ and $E(L_2/Q_2) \cong SL_3(4)*SL_3(4)$. In the former $3 \in \sigma (M)$, contradicting (2.2). So we have the latter with $3 \not\in \sigma (M)$. Now $5 \in \sigma (M)$. We have $[R,Y] \leq Y \cap Q_1 \leq Z_1$, As $[R, \langle Z_1 , Z_3 \rangle] \leq Y$, we see $[R,Y] \leq Z_1 \cap Z_3$. So $|[Z_1, R] : Z_1 \cap Z_3 \cap [Z_1,R]| \leq q = 4$. Furthermore $[Z_1,R] \cap Z_3 \not= 1$. As $R$ is centralized by $F \cong E_{5^3}$, there is $E \cong E_{5^2}$ with $\Gamma_{E,1}(G) \leq M$ such that $\bC_{Z_1 \cap Z_3}(E) \not= 1$. But then $\langle Z_1, Z_3 \rangle \leq M$, a contradiction.

Let now $K\cong U_{n}(r)$, $n=4$ or $5$. Then $|Y:Y\cap Q_{1}|\leq
r^{\, 4}$. If $n=5$, then as before $R\not\le Y$ and $q = r = 4$. As $|[T,R] : \bC_{[T,R]}(Y)| \leq r$ we see that $Y \leq O_2(N_K(R))$ and so $|Y : Y \cap Q_1| \leq r^2$. But then we have a contradiction to (1.24).

So we have $n=4$. Let $T$ be the natural $U_{4}(r)$-module. Then we
get $|Y:Y\cap Q_{1}|\geq r^{\, 2}$ and then even $|Y:Y\cap
Q_{1}|=r^{\, 4}$, as any such subgroup of order $r^2$ in $U_4(r)$ contains an element $u$ with $|[T,u]| = r^4$.

Now $YQ_1/Q_1$ is uniquely determined and so the action of $Y$ on the natural module tells us $[Y,T,Y]=1$, i.e $Y$ acts quadratically and $T\lhd L_{1}$. But
$m_{p}(K)\leq 3$ and so by (4.4) some $1\not= x\in\Omega_{1} (Z(S))$ is
centralized by $E\cong E_{p^{\, 2}}$ with $\Gamma_{E,1} (G)\leq M$, a
contradiction. 

So we have $T$ is the natural $\Omega^{\, -} (6,r)$-module. As before
we get $T\not\!\!\unlhd L_{1}$. Then as $|[y,T]|=r^{\, 2}$ for every 
$y\in
(Y\cap K)^\sharp$, we see that $|Y|=r^{\, 4}$ and again $Y$ is uniquely determined. Now let $T^{\, g}\not= T$. Then
$|T^{\, g}\times T:\bC_{T^{\, g}\times T} (Y)|=r^{\, 10}$. On the
other hand $|Z_{1}:\bC_{Z_{1}} (Y)|\leq q\cdot r^{\, 4}$. This
implies $r^{\, 6}\leq q$, a contradiction.

Let $K\cong L_{3}(r)$. We may assume that $T\not\!\!\unlhd L_{1}$, as
$m_{p}(K)=2$. Then $T$ is the natural module and $|Y:Y\cap
Q_{1}|=r^{\, 2}$. Suppose $[T,Y,Y] \not= 1$. Then $|[T,Y,Y]| = q$ and $[T,Y,Y] \leq Z_1 \cap Z_3$. Then $Z_1 \cap Z_3$ contains a nontrivial subgroup which is normalized by $E \leq K$, $E \cong E_{p^2}$, a contradiction. But now $[Z_1,Y,Y] = 1$ and then $|Z_1 : \bC_{Z_1}(Y)| = q$. As $T \not\!\!\unlhd L_1$, we have $q \geq r^2$. But this contradicts (1.65) and (1.64).


So assume next $K\cong L_{4}(r)$. Again $T\not\!\!\unlhd L_{1}$. This shows
that $T$ is the natural module. Suppose there is some $y\in
Y^{\,\sharp}$ with $|[T,y]|=r^{\, 2}$. Then $|Y|=r^{\, 4}$ and
$[T,Y,Y]=1$. So $Y$ acts quadratically and $|Z_{1}:\bC_{Z_{1}}
(Y)|=q$. Now $q=r^{\, 4}$, a contradiction as before.
So we have that $|Y|\leq r^{\, 3}$ and $Y$ is a transvection group.
But every transvection group acts quadratically and so $|Z_1 : Z_1 \cap Y| = q \geq r^2$, contradicting (1.64) and (1.65).


Let finally $K\cong L_{n}(2)$, $5\leq n\leq 7$. Suppose first $p=3$. Then by
(1.65) we have $q = 2$ and so $Z_1$ is an $F_1$-module. By (1.22) - (1.24) we see that $K \cong L_5(2)$, otherwise some $x \in \Omega_1(Z(S))^\sharp$ is centralized by $E \cong E_9$, $E \leq K$, a contradiction to $L_2 \not\leq M$. So $K \cong L_5(2)$. Now by (1.22) - (1.24) natural modules, duals or $V(\lambda_1+\lambda_4)$ are involved. Let $T$ be one of these modules. As $m_3(K) = 2$, $T \not\!\!\unlhd L_1$. If $T \cong V(\lambda_1 + \lambda_4)$ we see that $|Y : Y \cap Q_1| = 2^6$. As $Y$ does not act quadratically, we see that $1\not= [Y,Z_1,Y] \leq Z_1 \cap Z_3$ is normalized by $N_K(Y) \cong YQ_1/Q_1(\Sigma_3 \times L_3(2))$ and so it is normalized by $E \leq K$, $E \cong E_9$, a contradiction. So $T$ is the natural module and as no $x \in \Omega_1(Z(S))^\sharp$ is centralized by $E \cong E_9$, $E \leq K$, we get that $T^*$, the dual module is also involved. Now there are at least two conjugates of each $T$ and $T^{\,\star}$.
Hence $Y$ is a transvection group. In particular $|Y|=16$ and
$|Z_{1}:\bC_{Z_{1}} (Y)|\geq 2^{\, 10}$. But $|Z_{1}:\bC_{Z_{1}}
(Y)|\leq 32$, a contradiction.

Let $K\cong L_{6}(2)$ or $L_{7}(2)$, $p=7$. As $m_{p}(K)=2$ we first
get that there are at least three conjugates of $T$. Furthermore $
T$ is the natural module. As $7 ~\not| \,|\bC_{K} (\bC_{T} (S\cap K))|$ we get with (4.4) that at least 6 nontrivial irreducible modules are involved. Now $K\cong L_{7}(2)$ and
$|Y|=2^{\, 6}$,  $Y$ is a transvection group. Then $|Z_{1}:\bC_{Z_{1}}
(Y)|\geq 2^{\, 12}$. But $|Z_{1}:\bC_{Z_{1}} (Y)|\leq 2\cdot 2^{\,
6}$, a contradiction.

\absa
{\bf (6.38)~Lemma.~}{\it Assume $b=b_{1}=1$. Let $K_{1}\times K_{2}$
be a direct product of two components of $L_{1}/Q_{1}$ of $p$-rank 1 or a component with a cyclic normal $p$-group. 
Then we may assume $[K_{2},Y]=1$. }

\absa
Proof~. Suppose $[K_{1},Y]\not= 1\not= [K_{2},Y]$.
\begin{itemize}
\item[(1)] $[K_{i},\omega ]\le K_{i}$, $i=1,2$,, or
$q=64$ and for $\nu =\omega^{\, 3}\nu_{1}$, $o(\nu_{1} )=7$,
$\langle\nu ,\omega\rangle\cong\bZ_{q-1}$ we have $[K_{i},\nu
]\le K_{i}$, or $q = 4$.
\end{itemize}

Suppose $[K_{1},\omega ]\not\le K_{1}$. Then we get
$m_{t}(K_{1})=1$ for every odd prime $t \,\Big | \,|K_{1}|$. Hence $K_{1}\cong
L_{2}(r)$, $Sz(r)$ or $J_{1}$ by (1.5). Let first $o(\omega )=x>3$. Then 
\begin{center}
$K_{1}^{\,\langle\omega\rangle} =K_{1}\times\ldots\times
K_{y},y=x\mbox{~or~} x=9,y=3\, .$
\end{center}
Let $\omega\in\langle\nu\rangle$, $o(\nu )=q-1$. Suppose first $y>3$.
If $1\not= \mu\in\langle\nu\rangle$ with $[K_{1},\mu ]\le
K_{1}$. Then $\mu$ induces an outer automorphism on $K_{1}$ with
$\bC_{K_{1}} (\mu )$ a 2-group. But $K_{1}$  does not possess such an
automorphism. This implies 
\begin{center}
$K_{1}^{\,\langle\nu\rangle} =K_{1}\times\ldots\times
K_{q-1}\, .$
\end{center}

Let $t\in Y$, $[K_{1},t]\not= 1$. Then $t$ acts on an elementary
abelian group $E\cong E_{p^{\, q-1}}$. We have
$|[Z_{1},t]:[Z_{1},t]\cap Z_{3}|\leq q$. If $[Z_{1},t]\cap Z_{3}\not=
1$. Then we see that there is $
F\cong E_{p^{\, 2}}\le E$, $\bC_{Z_{3}} (F)\not= 1$, a
contradiction. So we have $|[t,Z_{1}]|\leq q$. In particular $t\in
Z((Z_{1}\cap Q_{\, 2})(Z_{3}\cap Q_{\, 2}))$. Then $\bC_{E} (t)\not=
1$, otherwise $|[t,Z_1]| \geq \sqrt{|[E,Z_1]|} > q$.  But then we may assume that $[t,K_{i}]=1$  for some $i$. As $[Y,[t,Z_1]] \leq [Z_1,t] \cap Z_3 = 1$ and $[Y,K_i] \not= 1$, we see that $[[t,Z_1],K_i] = 1$.  But now $[[Z_1,K_i], t] =1$ and so $[Z_1,K_i] \leq Q_2$ and $[s,[Z_{1},K_{i}]]=1$, for
$s\in Y$, $[K_{2},s]\not= 1$ by the same argument as for $t$. But
this is not possible. 

So we are left with $q=64$, $\omega^{\, 3}\in N(K_{1})$, $\bC_{K_{1}}
(\omega )$ is not a 2-group. This implies $[K_{1},\nu_{1} ]\le
K_{1}$.
\begin{itemize}
\item[(2)] We may choose $K_{1},K_{2}$ such that $[K_{1}\times
K_{2}, Y]\le K_{1}\times K_{2}$.
\end{itemize}

If $q>4$ this follows from $[Y,\omega ]=Y$, $[Y,\nu ]=Y$,
respectively by (1). If $q=4$, $\omega$ induces at most one orbit of 
length 3 on $K_{1}^{Y\langle\omega\rangle}$. Thus either (2) holds or
$Y\langle\omega\rangle$ induces $A_{4}$ on
$K_{1}^{Y\langle\omega\rangle}$. Suppose the latter. Then as $3\not\in
\sigma (M)$, we have $K_{1}\cong Sz(r)$ and $\omega$ induces a field
automorphism on $K_{1}$. Now there is $\nu\in K_{1}$, $o(\nu )=p$,
$[\omega ,\nu ]=1$. In particular $\bC_{K_{1}^{Y\langle\omega\rangle}}
(\omega )\geq \langle\nu\rangle\times\tilde{K}$, $\tilde{K}\cong
K_{1}$. But then $N_{G}(\langle\omega\rangle )\le M$ by (1.63),
a contradiction.

So let $q=2$ and without loss $K_{2}^{\, y} =K_{3}\not= K_{2},K_{1}$
for some $y\in Y^{\,\sharp}$. Now $K_{2}\cong L_{2}(r)$, $Sz(r)$, or
$J_{1}$ again. Let $y_{1}\in Y^{\,\sharp}$ with $K_{2}^{\,
y_{1}}\not= K_{3},K_{2}$. Then $K_{2}^{\langle y,y_{1}\rangle}
=K_{2}\times K_{3}\times K_{4}\times K_{5}$ and so $y\, y_{1}$ acts
on $E\cong E_{p^{\, 4}}$, which shows that there is $1\not= x\in
Z_{1}\cap Z_{3}$ centralized by $F\cong E_{p^{\, 2}}$, $F\le
E$, a contradiction. So we have $[K_{2}\times K_{3},Y]\le
K_{2}\times K_{3}$.
\begin{itemize}
\item[(3)] We may choose $K_{1},K_{2}$ with (1), (2) and $\bC_{Y}
(K_{1}\times K_{2})=1$. 
\end{itemize}

Suppose false. Choose $1\not= y\in \bC_{Y} (K_{1}\times K_{2})$. Set
$W=[y,Z_{1}]$. Then $K_{1}\times K_{2}$ acts on $W$. Let first $[K_{1}\times
K_{2},W]=1$. Then $W\cap Z_{3}=1$ and so $|W|=q$. Hence $K_{1}\times
K_{2}\not\!\!\unlhd L_{1}/Q_{1}$. So we may assume that there is some $s\in
S$ with $K_{2}^{\, s}\not= K_{2}$ and $[y,K_{2}^{\, s}]\not= 1$. We
now have 
\begin{center}
$[K_{1}\times K_{2},Z_{1},K_{2}^{\, s}]=1\, .$
\end{center}

Furthermore $[[K_1 \times K_2,Z_1],Y] = 1$ and so $[K_1 \times K_2,Z_1] \leq Y$. So for $\tilde{y} \in Y$ we have $[\tilde{y},[K_i,Z_1]] \leq Z_1 \cap Z_3$.
Now
\begin{center}
$[K_{2},Z_{1},K_{1}^{s^{-1}}\times K_{2}^{\, s^{-1}}]=1\, .$
\end{center}

Let $\tilde{y}\in Y$ with $[K_{2},\tilde{y} ]\not= 1$. Then $[[K_2,Z_1], \tilde{y}]$ is centralized by a group of order $p^2$ in $K_1 \times K_2 \times K_1^{s^{-1}} \times K_2^{s^{-1}}$, a contradiction. 
 So we have 
$$[W,K_{1}\times K_{2}]\not= 1$$
and then 
$$1\not= |W:\bC_{W} (Y)|\leq q\, .$$

Suppose $[W,K_{i}]\not= 1$, $i=1,2\, .$ By (1) we have 
$|Y:\bC_{Y} (K_{i})|\geq q$, $i=1,2$, or $q=4$ and
$|Y:\bC_{Y} (K_{1})|=2$. In the former $W$ is an F-module for both
$K_{1}$ and $K_{2}$. In the latter $K_{1}\cong L_{2}(r)$, $Sz(r)$ or
$J_{1}$ and $|W:\bC_{W} (Y)|=4$. This implies $K_{1}\cong L_{2}(4)$
or $L_{3}(2)$ with (1.30). Hence also in this case just F-modules are involved in
$W$. 

If $W$ is an F-module, we also have $K_{1}\cong L_{2}(r)$ or
$L_{3}(r)$, or a 3-group,  if we choose notation such that $m_{3}(K_{1})\leq 1$. By
(1) we get $q=4$ or $r=q$, as $r\leq q$ anyway. Again for $q=4 $ we
may have $K_{1}\cong L_{3}(2)$. Hence in any case
$$K_{1}\cong L_{2}(q),L_{3}(q)\mbox{~or~} q=4\mbox{~and~} K_{1}\cong
L_{3}(2), \mbox{~or~} K_1 \mbox{~is a 3-group.}$$

Suppose first $[W,K_{1},K_{2}]=1$. There is $\tilde{y} \in
Y^{\,\sharp}$ with $[K_{1},\tilde{y} ]\not= 1\not= [K_{2},\tilde{y}
]$. As $|[W,\tilde{y} ]|\leq q$, we see that $K_{1}\cong L_{3}(2)$, $q = 4$, 
and $[W,K_{1}]$ is the natural module. Furthermore also $K_{2}\cong
L_{3}(2)$, $[W,K_{2}]$ is the natural module and $p=7$. Finally
$K_{1}^{\,\omega} =K_{2},K_{2}^{\,\omega} =K_{3}$. Now we have
$\bC_{Y} (K_{1}\times K_{2})=\bC_{Y} (K_{1}\times K_{2}\times
K_{3})$. But we have $\bC_{W} (Y)=\bC_{W} (\tilde{y} )$ which implies
$[W,K_{1}\times K_{2},K_{3}]\not= 1$, a contradiction. So we have 
$$[W,K_{1},K_{2}]\not= 1\, .$$

In particular there are at least two nontrivial $K_{1}$-modules
involved. This shows $q=4$ and $K_{1}\cong L_{3}(2)$. Now $K_{2}\cong
L_{3}(2)$. But then $[W,K_{1},K_{2}]=1$, as there are exactly two
$K_{1}$-modules involved. So we may assume 
$$[W,K_{1}]=1\mbox{~or~} [W,K_{2}]=1\, .$$

Choose notation such that $[K_{1},W]=1$. Hence
$[K_{1},Z_{1}]\le \bC_{Z_{1}} (Y)$. Let $K_{3}\le \bC
(K_{1}\times K_{2})$ with $[K_{3},Y]\not= 1$. Then 
$[[K_{1},Z_{1}],K_{3}]=1$. Hence we may assume
$[K_{3},Y]\le K_{3}$, otherwise there is $K_{3}\times
K_{4}\le \bC (K_{1})$ with $[K_{3}\times
K_{4},[K_{1},Z_{1}]]=1$, a contradiction. 

As $[K_{2},W]\not= 1$, we have $[K_{1},Y]\le K_{1}$ by (2) and
so $[K_{1}\times K_{3},Y]\le K_{1}\times K_{3}$. Now as
$[[K_{1},Z_{1}],K_{3}]=1$ we have $\bC_{Y} (K_{1}\times K_{3})=1$, as
shown before. Replace $\{ K_{1},K_{2}\}$ by $\{ K_{1},K_{3}\}$. 

We now choose notation such that $m_{3}(K_{1})\leq 1$; i.e.
$K_{1}\cong L_{2}(r)$, $L_{3}(r)$, $U_{3}(r)$ $Sz(r)$, $J_{1}$ or
solvable.
\begin{itemize}
\item[(4)] $\bC_{Y} (K_{1})=1\, .$
\end{itemize}

Suppose $Y_{1}=Y\cap \bC (K_{1})\not= 1$. Let $y\in\bC_{Y} (K_{1})$
and $W=[y,Z_{1}]$. Assume $[W,K_{1}]=1$. Then $[Z_{1},K_{1},K_{2}]=1$
by (3). Let $t\in [Z_{1},K_{1}]$ with $t\not\in Y$. Then
$[Y,t](Z_{1}\cap Z_{3})=Y\cap Z_{1}$. This shows
$[[Z_{1},K_{2}],Y]=1$, a contradiction. So we have
$[Z_{1},K_{1}]\le Y\geq [Z_{1},K_{2}]$. Let $K_{3}$ be a
component in $\bC_{L_{1}/Q_{1}} (K_{1}\times K_{2})$ or a normal $p$-group. Then $K_{3}$
acts on $[Z_{1},K_{1}]$ and $[Z_{1},K_{2}]$. If $[y, K_{3}]=1$,
we see that $K_{3}$ acts on $1\not=[y,[Z_{1},K_{2}]]\le
Z_{1}\cap Z_{3}$. Let $[y,K_{3}]\not= 1$. Then $K_{3}$ acts
trivially on $[Z_{1},K_{1}]$. Hence in both cases there is $E\cong
E_{p^{\, 2}}$ with $\Gamma_{E,1} (G)\leq M$, such that $E$ normalizes
$1\not= U\le Z_{1}\cap Z_{3}$, a contradiction to
$L_{2}\not\le M$. 

Hence $\bC_{L_{1}/Q_{1}} (K_{1}\times K_{2})=1$. By (4.4) we see $m_p(L_1) \geq 4$. But then we have
some $\mu\in L_{1}$, $o(\mu )=p$, such that $\mu$ induces a field
automorphism on $K_{1}$ and so $\mu$ acts on $\Omega_{1} (Z(S))\cap
[Z_{1},K_{1}]$ and then $\Omega_{1} (Z(S))\cap [Z_{1},K_{1}]\le
Z_{1}\cap Z_{3}$ is normalized by $E\cong E_{p^{\, 2}}$ with $\Gamma_{E,1}
(G)\leq M$, a contradiction. So we have
$$[W,K_{1}]\not= 1\, .$$

Now for $\tilde{y}\in Y$ we get $|W:\bC_{W} (\tilde{y} )|\leq q$. If
$\omega$ or $\nu$ act nontrivially on $K_{1}$, we get $r\geq q$ or
$K_{1}\cong U_{3}(r)$, $r^{\, 2}=q$. Hence we have $K_{1}\cong
L_{2}(q)$, $L_{3}(q)$ or $U_{3}(\sqrt{q} )$ and $W$ involves the
natural module just once. Suppose $q=4$ and $[K_{1},\omega
]\not\le K_{1}$. Now $|W:\bC_{W} (\tilde{y} )|\leq 4$. By
(1.30) we get $K_{1}\cong L_{2}(4)$ or $L_{3}(2)$, and there are at
most two nontrivial modules involved. Furthermore for $y_{1}\in\bC_{Y}
(K_{1})$ we get $[W,y_{1}]=1$. Now as $Z_{1}/\bC_{Z_{1}} (y)\cong W$
as $K_{1}$-module, we see that $Y$ has to induce the full group of
transvection. Hence $K_{1}\cong L_{2}(q)$, $SL_{3}(q)$ or
$L_{3}(2)$. In particular $|W|=q^{\, 2},q^{\, 3}$. In case
$K_{1}\cong L_{3}(2)$, we have $|Y:\bC_{Y} (K_{1})|\leq 4$. But $Y$
has to induce a group of order 16, a contradiction.

So we have $[\omega ,K_{1}]\le K_{1}$ and $|W|=q^{\, 2},q^{\,
3}$. Because of the action of $K_{1}$ we get that there are
$K_{2}$-submodules $T_{1},T_{2}$ for $K_{1}\cong L_{2}(q)$,
$T_{1},T_{2},T_{3}$ for $K_{1}\cong SL_{3}(q)$ with $|T_{i}:\bC_{T_{i}}
(\bC_{Y} (K_{1}))|\leq q$, $i=1,2,3$. Furthermore $[T_{i},\bC_{Y}
(K_{1})]=\bC_{T_{i}} (\bC_{Y} (K_{1}))$. 

Let first $q=2$. Then $K_{2}\cong L_{n}(2)$. As $m_{p}(K_{2})\leq 3$ for any odd prime $p$,
we get $n\leq 7$. As $3 \,\Big | \,|K_{2}|$, we get $3\not\in \sigma (M)$ or
$m_{3}(K_{2})=1$. Hence $n\leq 5$. But now there is $E\cong E_{p^{\,
2}}$ such that $[Z_{1},K_{1}\times K_{2}]$ is centralized by $E$ and
$\Gamma_{E,1} (G)\leq M$, a contradiction.

So we have $q\geq 4$. We have $\rho\in K_{1}$, $o(\rho )=q-1$, such
that $[\rho ,T_{1}]=T_{1}$. This shows $K_{2}\cong L_{n}(q)$ and so
$n\leq 5$ again. Let $n\geq 3$, then we get $K_{1}\cong L_{2}(q)$,
otherwise $m_{p}(K_{1})=2$. Now $p \,\Big | \,q^{\, 2}-1$ and 
so we have $n=3$.
As every $p$-element $\nu\in\bC (K_{1}\times K_{2})$ has to centralize
$T_{1}\bigoplus T_{2}$, we see $\bC (K_{1}\times K_{2})=1$. So by (4.4)
there is some $\mu ,o(\mu )=p$, $\mu$ induces a field
automorphism on $K_{1}$ or $K_{2}$. But this contradicts (4.12). 

So let $n=2$. By symmetry we may assume $K_{1}\cong L_{2}(q)$ too. If
$p \, \Big | \,q+1$, then we get a contradiction as before. 
Hence assume that
$p \, \Big | \,q-1$. Now if $[\omega ,K_{1}]\not= 1$, $[\omega ,K_{1}]\le
K_{1}$, then $\omega$ induces a diagonal automorphism and so $\omega$
centralizes $E\cong E_{p^{\, 2}}$ with $\Gamma_{E,1} (G)\leq M$,
contradicting $N_{G} (\langle\omega\rangle )\not\le M$ by
(1.63). Hence $q=4$ by (1). But now $K_{1}\cong L_{2}(4)$, $q-1=3$.
But $3\not\in \sigma (M)$ if $q=4$. This proves (4).

So we have $\bC_{Y} (K_{1})=1$. Then $[K_{1},Z_{1}]$ is an SC-module.
This implies $K_{1}\cong L_{2}(r)$, $L_{3}(r)$ or a $\{ 2,3\}$-group,
$r$ even by (1.25) - (1.27) and (1.29). 

Suppose that $[K_{1},Z_{1}]$ involves exactly one irreducible
nontrivial module. If $K_{1}\lhd L_{1}/Q_{1}$, then
$[K_{1},Z_{1}]\cap \Omega_{1} (Z(S))$ is normalized by $E\cong
E_{p^{\, 2}}$, a contradiction. So we have $K_{1}\not\!\!\unlhd
L_{1}/Q_{1}$. As $m_p(K_1) = 1$ this now gives $K_{1}\cong L_{2}(r)$. 
If $K_2$ is a $p$-group then (4) implies $|Y : Y \cap Q_1| = 2$. Hence $|Z_1 : \bC_{Z_1}(Y)| = 2$ and so $K_1 \cong L_2(2)$ is solvable too, a contradiction. So $K_2$ is nonsolvable. Now
$[[K_{1},Z_{1}],K_{2}]=1$. As $[Y,[K_{1},Z_{1}]]\not= 1$, and
$[K_{2},Y]\not= 1$, we see that $[K_{1},Z_{1}]\le Y\geq [K_{2},Z_{1}]$ again. By (4.4) we may assume that $m_p(L_1) \geq 4$. But now there is $E\cong E_{p^{\, 2}}$ with
$\Gamma_{E,1} (G)\leq M$, such that $\bC_{K_{1}} (E)$ contains some
$s$ which acts as some $t\in Y\setminus Q_{1}$ on
$[Z_{1},K_{1}]$. But then $1\not= [s,[Z_{1},K_{1}]]\le
Z_{1}\cap Z_{3}$, and so $E$ normalizes a nontrivial subgroup of
$Z_{1}\cap Z_{3}$, a contradiction.

So we have that $[K_{1},Z_{1}]$ involves two natural modules and
$K_{1}\cong SL_{3} (r)$. If $K_{1}\unlhd L_{1}/Q_{1}$, then as 
$p  \not\!\Big | \,r-1$ we see that there is
$1\not= U\le [K_{1},Z_{1}]\cap \Omega_{1}(Z(S))$ normalized by 
$E\cong E_{p^{\, 2}}$ with $\Gamma_{E,1} (G)\leq
M$, a contradiction. So we see $K_{1}\not\!\!\unlhd L_{1}/Q_{1}$ and then
$K_{1}\cong L_{3}(2)$. Now $[K_{2},[K_{1},Z_{1}]]=1$, as $K_{2}$ is
nonsolvable as above.  But now for $y\in Y\setminus Q_{1}$, we have
$|Z_{1}:\bC_{Z_{1}} (y)|\geq 8>|Y:Y\cap Q_{1}|=4$ or $2$, a
contradiction. 

\absa
{\bf (6.39)~Proposition.~}{\it $b\not= b_{1}$.}






\absa
Proof~. Suppose false. By (6.35) we know $b=b_{1}=1$. Choose $Y$ as
before. Let $L_1$ not be exceptional. Then by (6.38) either $[E(L_1/Q_1), Y] \not= 1$ and $[Y,K]\not= 1$ for exactly one component $K$
of $L_{1}/Q_{1}$ or $K = F(L_1/Q_1)$ and $[Y,K] \not=1 $. In any case $m_{p}(K)=1$.

Let $\tilde{K} =\bC_{L_{1}/Q_{1}} (K)$. Then $[Y,\tilde{K}
]=1$. Choose $y\in Y$. Set $W=[y,Z_{1}]$. Suppose $[W,\tilde{K} ]=1$.
Then we get $m_{p}(\bC_{M} (W))\geq 2$, as $m_{p}(\mbox{~Aut~}
(K))\leq 2$. But then $|W|=q$ and $W\cap Z_{3}=1$. We furthermore
have $[[Z_{1},K],\tilde{K} ]=1$. If $K\unlhd L_{1}/Q_{1}$, then we see
that $\Omega_{1} (Z(S))\cap [Z_{1},K]$ is centralized by $E\cong
E_{p^{\, 2}}$, $\Gamma_{E,1} (G)\leq M$, a contradiction. So
$K\not\!\!\unlhd L_{1}/Q_{1}$ and then $K\cong L_{2}(r)$, $Sz(r)$ or
$J_{1}$. As $\omega$ acts on $K$, we get $K\cong L_{2}(q)$ and
$[Z_{1},K]$ is the natural module, $|Z_{1}:Z_{1}\cap Z_{3}|=q^{\,
2}$. Hence $|Z_{1}:\bC_{Z_{1}} (K\times K^{\, s})|\leq q^{\, 4}$,
$s\in S$. As $|\bC_{Z_{1}} (K):\bC_{Z_{1}} (K)\cap Z_{3}|\leq q$, we
see that $|Z_1|\leq q^{\, 5}$ and so $K\times K^{\, s}\unlhd
L_{1}/Q_{1}$, for some $s\in S$. Now $\bC_{Z_{1}} (K\times K^{\,
s})=1$ and so $|Z_{1}|=q^{\, 4}$. Furthermore as $\omega$ induces a
diagonal automorphism we see that $p  \not\!\Big | \,q-1$. But now 
$p \, \Big | \,q+1$. Then
every $x\in [K,Z_{1}]$ is centralized by $E_{p^{\, 3}}$ and so
$Z_{3}\cap [K,Z_{1}]=1$, i.e. $\bC_{[K,Z_{1}]} (\omega )=1$. The same
is true for $[K^{\, s},Z_{1}]$ and so $\bC_{Z_{1}} (\omega )=1$,
contradicting $[Z_{1}\cap Z_{3},\omega ]=1$. 

We may assume $[W,\tilde{K} ]\not= 1$ for every $y\in Y^{\,\sharp}$.
But this implies $[W,Y]=1$ otherwise $[Y,W]\le Z_{1}\cap Z_{3}$
is normalized by $E\cong E_{p^{\, 2}}$, $\Gamma_{E,1}(G)\leq M$, a
contradiction. This now implies that $Y$ acts quadratically and so 
$|Z_{1}:\bC_{Z_{1}}(Y)|=q$. But now $[\tilde{K} ,W]=1$ as $|W|=q$ and 
$[W,\omega ]=W$. 

It remains the exceptional case. Here $|Y:Y\cap Q_{1}|\leq 4$ and so
$|Z_{1}:Z_{1}\cap Z_{3}|\leq 8$.  Inparticular $|\Omega_1(Z(M)) : \Omega_1(Z(M)) \cap Z_1 \cap Z_3 | \leq 8$. As $\Omega_1(Z(S)) \leq \Omega_1(Z(M))$, we have $\Omega_1(Z(M)) \cap Z_1 \cap Z_3 \not= 1$. But then there is $1\not= z\in
Z_{1}\cap Z_{3}$ centralized by $E\cong E_{p^{\, 2}}$ with $\Gamma_{E,1}
(G)\leq M$, a contradiction.











\begin{center}
\S~7 The case $[Z_2, Z_\alpha] \not= 1$
\end{center}

\vspace{1cm}

In this chapter we assume $b = b_2$. By (6.39) we know $\alpha \sim 2$. 
Let $(1,2,3,\ldots,\alpha)$ be a path of length $b+1$.

\absa
{\bf (7.1)~Lemma.~}{\it $Z_2$ is an F - module with offending subgroup 
$Z_{\alpha}Q_2/Q_2$. Furthermore one of the following holds :
\begin{enumerate}
\item[(1)] $E((L_2/Q_2)/F(L_2/Q_2)) \cong L_2(q)$, $q$ even, 
$Z_2/\bC_{Z_2}(E(L_2/Q_2))$ is the natural module
\item[(2)] $E(L_2/Q_2) \cong L_2(q) \times L_2(q)$, $q$ even, 
$Z_2/\bC_{Z_2}(E(L_2/Q_2))$ is a direct sum of two natural modules.
\item[(3)] $L_2$ is solvable and one of the following holds
\begin{enumerate}
\item[(a)] $|Z_2/\bC_{Z_2}(L_2)| = 4$, 
$L_2/\bC_{L_2}(Z_2) \cong \Sigma_3$
\item[(b)] $|Z_2/\bC_{Z_2}(L_2)| = 16$ and 
$L_2/\bC_{L_2}(Z_2) \cong \Sigma_3 \wr \bZ_2$
\end{enumerate}
\end{enumerate}
(Notice that (3) are special cases of (1) and (2).)}

\absa
Proof.~ As $[Z_2, Z_\alpha] \le Z_2 \cap Z_\alpha$, we see that $Z_2$ 
is an F - module with offending subgroup $Z_{\alpha}Q_2/Q_2$. Now 
application of (2.1) and 
(1.31) gives the structure of $L_2/Q_2$. Suppose $E(L_2/Q_2) \cong A_9$. Then $L_2 \cap M \le L_1$. So 
we see that $Q_3 \le Q_2$ which is 
not possible as $[Z_\alpha, Z_2] \not= 1$.

We have to show that orthogonal modules are not possible in case (1) or 
(2). In the case of the orthogonal module an offending group is not 
contained in $E((L_2/Q_2)/F(L_2/Q_2))$. But this contradicts 
$L_2 \cap M \le L_1$ and so $Q_3Q_2/Q_2 \le E((L_2/Q_2)/F(L_2/Q_2))$.

\absa
{\bf~(7.2) Lemma.~}{\it $Z_1 \le Z_2$, $L_1 = 
\bC_{L_1}(Z_1)(L_1 \cap L_2)$}

\absa
Proof.~Suppose false. As $b_1 > b$, we get $Z_1 \le L_\alpha$ and there 
is some $t \in Z_\alpha \setminus Q_2$ with $1 \not=[Z_2,t] = 
[Z_1Z_2,t] \le Z_2$ by (7.1). This shows $Z_1Z_2 \unlhd \langle L_1 \cap L_2, 
t \rangle = L_2$.

Suppose first $L_2/\bC_{L_2}(Z_2) \cong \Sigma_3$ or $L_2$ is 
nonsolvable. We may assume that $Q_1 = \bC_S(Z_1)$ and 
$Q_1 \cap Q_2 = \bC_S(Z_1Z_2)$. This shows $Q_1 \cap Q_2 \unlhd L_2$. 
Now $Q_1 \in $ Syl$_2(\langle Q_1^{L_2} \rangle)$. So we get some 
$\tilde{L_2} \le L_2$, $\tilde{L_2} \not\le M$, 
$\tilde{L_2}/O_2(\tilde{L_2}) \cong L_2(q)$ and 
$Q_1 \in$ Syl$_2(\tilde{L_2})$.

We are going to prove the same result for 
$L_2/\bC_{L_2}(Z_2) \cong \Sigma_3 \wr \bZ_2$. There is some $\omega \in L_2 \setminus M$, $o(\omega) = 3$, with 
$|Z_1 : \bC_{Z_1}(\omega)| \le 2$. If $m_p(L_1) \ge 3$, then $\omega$ centralizes some 
$x \in Z_1$, $\bC_G(x) \le M$, a contradiction.

According to (4.9) assume $m_p(L_1) = 1$. Let $P \in$ Syl$_3(L_2)$, then $[P,Z_1Z_2] = [P,Z_2] \cong E_{16}$. 
Let $x \in Z_1Z_2$ with $[Q_2,x] \leq Z_2$. As $[x,P] = 1$ and $Z_2 \not\leq Z_1$, we have $[x,Q_2] \leq \bC_{Z_2}(P)$, so $|Z_2| > 16$.  
Hence there is $1 \not= x \in \bC_{\Omega_1(Z(S))}(P)$. Set 
$\tilde{Z_1} = \langle x^{L_1} \rangle$. If $\tilde{Z_1}$ is an 
F - module for $L_1$, we see with \cite[(25.12)]{GoLyS} that 
$[O_r(L_1/Q_1), \tilde{Z_1}] = 1$. But this contradicts (4.10). 
Hence $J(S) = J(Q_1)$ and $|J(S) : J(S) \cap Q_2| = 4$. But now 
$[Q_1, [J(S), Z_2]] = 1$, and so $|Q_1 : Q_1 \cap Q_2| = 4$. In 
particular 
$\langle Q_1^{L_2} \rangle Q_2/Q_2 \cong \Sigma_3 \times \Sigma_3$, 
and then we get the same result as above.

Application of \cite[Theorem3]{Ste} provides us with a subgroup 
$1 \not= C \le Q_1$ which is normalized by automorphism of $Q_1$ of odd 
order and which is normal in $\tilde{L_2}$. Hence as $Q_1$ is normalized 
by some 
$E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \le M$, or by $O_r(L_1)$, we get the 
contradiction $\tilde{L_2} \le M$. 

This shows $Z_1 \le Z_2$ and furthermore $Q_1 < \bC_S(Z_1)$. But then by 
construction $L_1 = \bC_{L_1}(Z_1)(L_1 \cap L_2)$ and $Z_1 \le Z_2$.

\absa
{\bf~(7.3) Lemma.~}{\it $\bC_{Z_2}(L_2) = 1$ and (7.1)(2) does not 
occure.}

\absa
Proof.~By (7.2) $\bC_{Z_2}(L_2) \le Z_1$ and so $N_G(\bC_{Z_2}(L_2)) 
\geq \langle L_1, L_2 \rangle$. By (4.14) 
$O_2(\langle L_1, L_2 \rangle) = 1$ and so $\bC_{Z_2}(L_2) = 1$.

Let $L_2$ be as in (7.1)(2). Then as $L_2 \cap M \le L_1$, we see that 
$|Z_1| = q^2$. But now there is $1 \not= t \in Z_1$ with 
$\bC_{L_2}(t)/O_2(\bC_{L_2}(t)) \cong L_2(q)$. By (7.2) $\bC_G(t) \le M$ 
and then 
$L_2 = \langle S, \bC_{L_2}(t) \rangle \le M$, a contradiction. 

\absa
In the next series of lemmas we are going to prove $b = 2$. Let 
$O^2(L_2/\bC_{L_2}(Z_2)) \cong L_2(q)$. Then $Z_2$ is the natural module 
by (7.1) and $|Z_1| = q$. Set $V_1 = \langle Z_2^{L_1} \rangle$. 
We get $[Q_1, V_1] = Z_1$, $V_1^\prime = 1$ iff $b > 2$.

Let $L_2$ be as in (7.1)(3)(b). Then $|Z_1| = 2$. Let 
$\tilde{Z_2} = \bC_{Z_2/Z_1}(S)\, ({\mbox{mod}}\, Z_1)$ and 
$V_1 = \langle \tilde{Z_2}^{L_1} \rangle$. Also $V_1^\prime = 1$ 
if $b > 2$.

\absa
Suppose first that $V_1 = \langle Z_2^{L_1} \rangle$. Then we have 
$[Z_2, Z_\alpha] = Z_3 = Z_{\alpha - 1}$. If $b > 2$, then we have 
$[V_1,Z_2] = 1 = [V_1, Z_3] = [V_1, Z_{\alpha - 1}]$. As 
$V_1 \le L_{\alpha - 2}$, we get 
$V_1 \le L_{\alpha - 2} \cap M_{\alpha - 1} \le L_{\alpha - 1}$. Hence 
$[V_1, V_{\alpha - 1}] \le V_{\alpha - 1}$. Furthermore 
$V_{\alpha - 1} \le G_2$ but 
$Z_\alpha \not\le M$. Finally $\bC_{L_1}(V_1/Z_1) \le Q_1$. Otherwise 
some $1 \not= \omega \in L_1$, $\omega \in E \cong E_{p^3}$, centralizes 
$V_1$. Then $\omega$ 
centralizes $Z_3$ and so $\omega \in M_3$. By (4.2) $M = M_3$. The same 
applies for $m_p(L_1) = 1$ as then $[O_r(L_1), Z_3] = 1$, which 
contradicts (4.10). But 
$\langle L_1 \cap L_2, L_2 \cap L_3 \rangle = L_2$, a contradiction.

We are going to prove the same result for $L_2$ as in (7.1)(3)(b). We 
have $Z_{\alpha - 2} \le Q_1$. As $[V_1, Q_1] \le Z_1$, we get 
$[\tilde{Z_\beta}, Z_{\alpha - 2}] \le Z_1$, for every 
$\beta \in \Delta(1)$. Suppose 
$[\tilde{Z_\beta}, Z_{\alpha - 2}] \not= 1$ for some 
$\beta \in \Delta(1)$. Then we may assume that 
$[\tilde{Z_2}^g, Z_\alpha] = Z_3$, for $M^g = M_3$. But then 
$[Z_2, Z_\alpha ] =  \tilde{Z_2}^g$ and so 
$[\tilde{Z_2}^g, Z_\alpha] = 1$, a contradiction.

So we get $[Z_{\alpha - 2}, V_1] = 1$ and again 
$V_1 \le L_{\alpha - 1}$. This shows 
$[V_1, V_{\alpha - 1}] \le V_{\alpha - 1}$. Furthermore 
$Z_\alpha \not\le M$.

Suppose next $[V_1, V_{\alpha -1}] \le Z_{\alpha - 1}$. Then either 
$V_1 \le Q_{\alpha - 1}$ or we have (7.1)(3)(b) with $\bC_S(V_1) > Q_1$. 
By (4.10) in the latter $m_p(L_1) > 1$. Further 
$p \,\Big | \,|\bC_{L_1}(V_1)|$. As 
$L_2 = \langle \bC_{L_2}(\tilde{Z_2}), S \rangle \not\le M$, we get that 
$m_p(\bC_{L_1}(V_1)) = 1$. But now the construction in (4.13) gives a 
contradiction. So we may 
assume $V_1 \le Q_{\alpha -1}$. Then we see $V_1 \le L_\alpha$ and so 
for some $t \in Z_\alpha \setminus Q_2$ we get $[V_1,t] = [Z_2,t]$. Then 
$V_1Z_2 \unlhd \langle S, t \rangle = L_2$. This shows that we are in 
(7.1)(3)(b). Let $u \in V_1$, $u \not\in Z_2$, $[u,S] \le Z_2$. Then 
$[u,Q_2] \unlhd L_2$. As 
$[u,Q_2] \le V_1$, we see $Z_2 \le V_1$ and so 
$V_1 \unlhd \langle L_1, L_2 \rangle$, a contradiction.

\absa
{\bf~(7.4) Lemma.~}{\it Let $b > 2$. Then $V_1 \le L_{\alpha - 1}$ and 
$[V_1, V_{\alpha - 1}] \not\le Z_{\alpha - 1}$.}

\absa
{\bf~(7.5) Lemma.~}{\it Let $b > 2$. Then 
$[V_1, V_{\alpha - 1}, V_1] = 1$.}

\absa
Proof.~ Let $W_2 = \langle V_1^{L_2} \rangle$. Let $s \in L_2$. Then 
$\langle V_1, V_1^s \rangle \le Q_1 \cap Q_1^s$ as $b > 2$. Hence 
$[V_1, V_1^s] \le Z_1 \cap Z_1^s = 1$, as $M \not= M^s$. This shows that 
$W_2$ is abelian. As $[V_1, V_{\alpha - 1}] \le W_2$, we are done.

\absa
{\bf~(7.6) Lemma.~}{\it Let $L_2$ be as in (7.1)(3)(b). Then 
$3 \not\in \sigma(M)$, 
$[V_1, V_{\alpha - 1}] \le V_1 \cap V_{\alpha - 1}$ and 
$[V_{\alpha -1} \cap Q_1, V_1] = 1$.}

\absa
Proof.~We have $[Z_2,Z_\alpha \cap V_{\alpha - 1}] = 1$. This implies 
$[Z_2, V_{\alpha - 1}] = 1$. In particular $V_{\alpha -1} \le L_1$ and 
so
\setcounter{equation}{0}
\begin{equation}
[V_1,V_{\alpha - 1}] \le V_1 \cap V_{\alpha -1}
\end{equation}

Hence $V_1$ and $V_{\alpha - 1}$ act quadratically on each other. 
Furthermore as $m_3(L_2) \ge 2$ we have

\begin{equation}
3 \not\in \sigma(M)
\end{equation}

\begin{equation}
[V_{\alpha - 1} \cap Q_1, V_1] = 1
\end{equation}

We have $[V_{\alpha - 1} \cap Q_1, V_1] \le Z_1$. If 
$[V_{\alpha -1} \cap Q_1, V_1] \not= 1$, then $Z_\alpha$ centralizes 
$\langle Z_2 \cap V_3, z \rangle$ with $1\not= z \in [V_{\alpha-1} \cap Q_1, V_1]$. But 
$Q_2 \in $Syl$_2(\bC_{L_2}(\langle Z_2 \cap V_3, z \rangle ))$, 
and so $Z_\alpha \le L_1$, a contradiction.

\absa
Now we fix the following notation : $K$ is a component or normal $p$-subgroup of $L_1/Q_1$ with 
$1 \not= [V_{\alpha - 1}, K] \le K$ and $K_{\alpha - 1}$ is the corresponding 
component in $L_{\alpha - 1}/Q_{\alpha - 1}$. Furthermore we may assume that $Z_\alpha \not\le M$. We set $\tilde{V}_1 = [V_1, K]$. 

\absa
{\bf~(7.7) Lemma.~}{\it Let $\tilde{V}_1 \not\le Q_{\alpha-1} $ and $L_2/Q_2$ be 
as in (7.1)(3)(b).  Let $[K_{\alpha -1}, t] = 1$, for some 
$1 \not= t \in \tilde{V}_1$. Let $K_{\alpha -1}^{\!(1)}$ be a component 
of $L_{\alpha-1}/Q_{\alpha -1}$ with 
$[t,K_{\alpha -1}^{\!(1)}] \not= 1$. If $[[t,V_{\alpha - 1}], K_{\alpha -1 }] = 1$ and 
$K^{(1)}$ is the group corresponding $K_{\alpha-1}^{\!(1)}$ in 
$L_1/Q_1$, then 
$[\tilde{V}_1, K^{(1)}] = 1 = [[K^{(1)},V_1],K]$. Furthermore there is 
no $E \le L_1, E \cong E_{p^2}$, $\Gamma_{E,1}(G) \le M$ with 
$[\tilde{V}_1,E] = 1$ or 
$[[K^{(1)}, V_1], E] = 1$. In particular $m_p(K) = m_p(K^{(1)}) = 1$.}

\absa
Proof.~The first assertion is obvious by the three subgroup lemma. Assume $[E,\tilde{V}_1] = 1$. 
Then there is some $1 \not= x \in [\tilde{V}_1, V_{\alpha - 1}] \le 
V_{\alpha -1}$ with $E \le \bC_G(x)$ and so $\bC_G(x) \le M$. But 
$[V_{\alpha - 1}, Z_\alpha] = 1$ and so $Z_\alpha \le M$, a 
contradiction. Hence $m_p(K^{(1)}) = 1$.

Let now $[E,[\tilde{V}_1,K^{(1)}]] = 1$. Then by the same reason there 
is $1 \not= x \in [t, V_{\alpha - 1}]$ with 
$\bC_G(x) \le M_{\alpha - 1}$. But $[[\tilde{V}_1, V_{\alpha -1}], 
K^{(1)}] = 1$ and so $K^{(1)} \le M_{\alpha -1}$. 
But this shows $M = M_{\alpha -1}$ by (4.2), which contradicts 
$Z_\alpha \not\le M$.

\absa
{\bf~(7.8) Hypothesis.~}  $L_2/Q_2$ is as in (7 .1)(3)(b) and 
$\tilde{V}_1 \not\le Q_{\alpha -1}$, $K$ is a quasisimple component of 
$L_1/Q_1$ with $1 \not= [K, V_{\alpha - 1}] \le K$.

\absa
{\bf~(7.9) Lemma.~}{\it Assume (7.8). Let 
$|V_{\alpha - 1} : \bC_{V_{\alpha -1}}(K)| \le 4$. Then $K/Z(K) \cong A_7$, 
$A_9$, $A_{10}$, $A_{11}$, or $K/Z(K)$ is a group of Lie type in even 
characteristic.}

\absa
Proof.~By (7.6) $3 \not\in \sigma(M)$. Let $t \in \tilde{V}_1$. Then 
$|V_{\alpha - 1} : \bC_{V_{\alpha - 1}}(t)| \le 4$ by (7.6). If 
$[t, K_{\alpha - 1}] \not= 1$, we have the assertion with (1.30). 
So assume $[\tilde{V}_1, K_{\alpha -1 }] = 1$. Then 
$[K_{\alpha - 1}, [\tilde{V}_1, V_{\alpha -1}]] = 1$. Let $K_{\alpha-1}^{(1)}$ be a component or normal $p$-subgroup in $L_{\alpha-1}/Q_{\alpha-1}$ with $[t,K_{\alpha-1}] \not= 1$ and $K_1^{(1)}$ the corresponding group in $L_1/Q_1$.
Then 
$[K^{(1)}, V_1]$ involves at most two nontrivial irreducible 
$K^{(1)}$ - modules.  By (7.7) $m_p(K^{(1)}) = 1$. Hence by (4.4) $m_p(\bC_{L_1}(K^{(1)})) \geq 2$.  As $p > 3$ this shows that there is some $E \le M$, 
$E \cong E_{p^2}$, $\Gamma_{E,1}(G) \le M$, and $[[V_1,K^{(1)}], E] =1$. 
But this contradicts (7.7).

\absa
{\bf~(7.10) Lemma.~}{\it Assume (7.8). If $t \in V_{\alpha - 1}$ induces 
some $1 \not= s \in K$ on $\tilde{V}_1$, then 
$m_p(N_{L_1}(K)/\bC_{L_1}(K)) > 1$. If we have the assumptions of (7.7) then $m_p(N_{L_1}(K^{(1)})/\bC_{L_1}(K^{(1)})) > 1$.}

\absa
Proof.~Suppose $m_p(N_{L_1}(K)/\bC_{L_1}(K)) = 1$. Then there is 
some $E \cong E_{p^2}$, $E \leq \bC(K)$ with $\Gamma_{E,1}(G) \le M$ and some $\nu \in K $, 
$o(\nu) = r$, $r$ prime, $\nu^s = \nu^{-1}$. Hence $E$ acts on 
$[[\nu, \tilde{V}_1],t] \le V_{\alpha - 1}$, and so $Z_\alpha \le M$, a 
contradiction. The same argument works for $K^{(1)}$.

\absa
{\bf~(7.11) Lemma.~}{\it Assume (7.8). Then $K \cong G(r)$, a group of 
Lie type ine characteristic two.}

\absa
Proof. Suppose false. By (7.6), (7.9) and (1.12) $K \cong A_7$, $A_9$, 
$A_{10}$, $A_{11}$ or $3\cdot M_{22}$. Now 
$|V_{\alpha - 1} : \bC_{V_{\alpha -1}}(K)| \le 2^5$. Hence for 
$t \in \tilde{V}_{1}$ we have 
$|V_{\alpha -1} : \bC_{V_{\alpha -1}}(t)| \le 32$, where 
$|V_{\alpha-1} : \bC_{V_{\alpha-1}}(t)| \le 4$ for $K \cong A_7$ by 
(1.12). Suppose $[t,K_{\alpha-1}] = 1$. Then we may apply (7.7) and get 
$|V_{\alpha -1} : \bC_{V_{\alpha -1}}(t)| \ge 8$. Let $K \cong A_n$,
 then $K_{\alpha -1}^{\!(1)}$ is solvable or 
$K_{\alpha -1}^{\!(1)} \cong Sz(q)$. But the latter is impossible as 
$q \ge 8$ and (1.27) implies $|[V_{\alpha -1},t]| \ge q^2$. So 
$K_{\alpha-1}^{\!(1)}$ is solvable, contradicting (7.10). Hence as in 
case of $K \cong A_n$ we have $\bC_{\tilde{V}_1}(K_{\alpha-1}) =1$.

If $K \cong 3\cdot M_{22}$, then $K_{\alpha-1}^{\!(1)} \cong L_2(q)$, 
$Sz(q)$, $L_3(q)$, $U_3(q)$, or $J_1$. As $|[V_{\alpha-1},t]| \le 2^4$, 
we get with (1.25), (1.26) and (1.27) that 
$K^{(1)} \cong L_2(q), q \le 16$, $q$ even, or $L_3(q), q = 2, 8$. 
By (7.7) $m_p(K^{(1)}) = 1$. But this now contradicts (7.10) as $p > 3$.

So in any case $\tilde{V}_1$ acts faithfully on $K_{\alpha-1}$. Suppose 
first $m_p(K) = 1$. Then by (7.10) $\tilde{V}_1$ involves at least two 
nontrivial irreducible modules for $K$. By (1.25) 
$K \not\cong 3\cdot M_{22}$. 
Suppose there are at least three modules involved. As 
$|V_{\alpha-1} : \bC_{V_{\alpha -1}}(\tilde{V}_1)| \le 2^5$ we have 
$|\tilde{V}_1 : \tilde{V}_1 \cap Q_{\alpha-1}| = 2$. But then for 
$z \in V_{\alpha -1}$ we have $|\tilde{V}_1 : \bC_{\tilde{V}_1}(z)| \le 4$, 
a contradiction. Hence we have that there are at most two nontrivial 
irreducible modules involved, which are spin modules or natural modules. 
But now there is some $E \cong E_{p^2}$, 
$\Gamma_{E,1}(G) \le M$, $[E,\tilde{V}_1] = 1$, a contradiction.

So we have $p = 5$, $K \cong A_{10}$ or $A_{11}$. Furthermore for some 
$\omega \in M$, $o(\omega) = p$, we have $[\tilde{V}_1, \omega] =1$. 
As in the natural module every element is centralized by a 5 - element in 
$K$, we see that we have the spin module as a submodule. Now 
$|V_{\alpha -1} : \bC_{V_{\alpha-1}}(\tilde{V}_1)| \le 8$ by (1.12). 
But this contradicts (1.30).

\absa
As $3 \not\in \sigma(M)$ we now have :

\absa
{\bf~(7.12) Lemma.~}{\it Assume (7.8). If $K \cong G(r)$, $r$ even, then 
$K$ is one of the following
\begin{enumerate}
\item[(i)] $(S)L_n(r), n \le 7$
\item[(ii)] $(S)U_n(r), n \le 7$
\item[(iii)] $Sp(2n,r), n \le 3$
\item[(iv)] $\Omega^-(8,r)$
\item[(v)] $G_2(r)$, $^3D_4(r)$, $^2F_4(r)$, $Sz(r)$.
\end{enumerate}
}

\absa
{\bf~(7.13) Lemma.~}{\it Assume (7.8). Then $\tilde{V}_1$ acts 
faithfully on $K_{\alpha - 1}$}

\absa
Proof.~Suppose first $[K^{(1)}, \tilde{V}_1] = 1$. Then by (7.7) 
$m_p(K) = m_p(K^{(1)}) = 1$. Furthermore
by (7.10) $m_p(N_{L_1}(K^{(1)})/\bC_{L_1}(K^{(1)})) > 1$. Application of 
(1.67) shows that $K^{(1)}$ is one of the groups of (7.12).
Then we see that there is some $\omega \in L_{\alpha -1}$, 
$o(\omega) = p$, $\omega$ induces a field
automorphism on $K_{\alpha - 1}$. Hence we may assume 
$t \in \bC_{K_{\alpha -1}^{\!(1)}}(\omega)$, for some
$t \in \tilde{V}_1$. But now there is $E \cong E_{p^2}$ with 
$\Gamma_{E,1}(G) \le M_{\alpha - 1}$ and $E$ acts on 
$[t, V_{\alpha - 1}] \le \tilde{V}_1$.
So $K^{(1)} \le \bC_G([t, V_{\alpha -1}]) \le M_{\alpha - 1}$. But this 
contradicts (4.2).

So we have $[K_{\alpha - 1},[t,V_{\alpha-1}]] \not= 1$ for some 
$t \in \tilde{V}_1$, $[t,K_{\alpha-1}] = 1$. As 
$|V_{\alpha-1} : \bC_{V_{\alpha -1}}(t)| \le 2^{m_2(K)}$, we see that 
$K \cong L_n(r), 4 \le n \le 7$, or $Sp_6(r)$.
Furthermore as $[K_{\alpha-1}, [t, V_{\alpha -1}]] \not= 1$, we see that 
$[\tilde{V}_1, K_{\alpha - 1}] = 1$ and so
$[K_{\alpha-1}, [t, V_{\alpha-1}]] \not= 1$ for every 
$t \in \tilde{V}_1 \setminus Q_{\alpha -1}$. We have $m_3(K^{(1)}) \le 1$ 
and so
$K^{(1)} \cong L_2(q)$, $(S)L_3(q)$, $(S)U_3(q)$, $Sz(q)$, $J_1$, or solvable. 
If $q$ is odd, $K^{(1)}$ is solvable or $K^{(1)} \cong J_1$, we have
$t \in V_{\alpha -1}$, $[t,\tilde{V}_1] \not= 1$ and 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(t)| \le 4$ by (1.12), \cite[(24.1)]{GoLyS} and \cite[(25.12)]{GoLyS}. 
Hence $\tilde{V}_1$ involves at most two nontrivial irreducible 
modules. As $p \not= 3$, this would imply $[K^{(1)}, \tilde{V}_1] = 1$,
a contradiction. So we have $K^{(1)} \cong L_2(q)$, $(S)L_3(q)$, 
$(S)U_3(q)$, or $Sz(q)$, $q$ even. Now 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(t)| \le 2q$, $2q^2$ 
in case of $K^{(1)} \cong (S)L_3(q)$. As there are at least four 
irreducible
nontrivial $K^{(1)}$ - modules involved in $V_1$, we get that 
$[t, K^{(1)}] = 1$. 
Now $[[t,\tilde{V}_1], K^{(1)}] \not= 1$ and so we get 
$K^{(1)} \cong L_3(2)$, $|[\tilde{V}_1,t]| = 8$
and $\tilde{V}_1$ involves at least three irreducible nontrivial 
modules
on which $t$ induces transvections. As $m_3(K) \le 2$, we get 
$K \cong L_4(2)$ or $L_5(2), p = 7$. But now $m_p(K \times K^{(1)}) = 2$ 
and so there is some 
$\omega \in M$, $o(\omega) = 7$, $[\tilde{V}_1, \omega] = 1$. As every 
element in the natural
module is centralized by some $\nu \in K$, $o(\nu) = 7$, we see that some 
$1 \not= x \in [\tilde{V}_1, t]$ is 
centralized by $E \cong E_{49}$, $\Gamma_{E,1}(G) \le M$, a contradiction 
to $Z_\alpha \not\le M$. 

\absa
{\bf~(7.14) Lemma.~}{\it Let $L_2/Q_2$ be as in (7.1)(3)(b). There is no component $K$ of $L_1/Q_1$
with $1 \not= [K, V_{\alpha -1}] \le K$.}

\absa
Proof.

\setcounter{equation}{0}
\begin{equation}
 K \not\cong (S)L_n(r)
\end{equation}

\absa
By (7.13) we know that $\tilde{V}_1$ is an SC - module. Suppose 
furthermore every module
involved in $\tilde{V}_1$ is of type $V(\lambda)$, $\lambda$ some 
fundamental weight.
In what follows we will use (1.23) freely.

Let $n = 7$. Then by (1.23) we may assume that $V(\lambda_4)$ is not 
involved.
Let $V(\lambda_3)$ be involved. Then 
$|V_{\alpha - 1} : \bC_{V_{\alpha-1}}(K)| \ge r^{10}$.
Now $V_{\alpha-1}$ contains some $z$, $z$ a root element in $K$. 
But now $\langle V_{\alpha-1}^{\bC_K(z)} \rangle \ge O^2(\bC_K(z))$. 
This implies
$[\tilde{V}_1, z, O^2(\bC_K(z))] = 1$,
and so $V(\lambda_3)$ cannot be involved. Hence we have that only 
$V(\lambda_1)$, $V(\lambda_2)$ $V(\lambda_5)$, or $V(\lambda_6)$
may be involved. If both $V(\lambda_2)$ and $V(\lambda_5)$ are involved, 
then again $|V_{\alpha -1} : \bC_{V_{\alpha-1}}(K)| \ge r^{10}$ and
we get a contradiction as before. So we may assume that just 
$V(\lambda_1)$, $V(\lambda_6)$ and $V(\lambda_2)$ are involved.
Then $\bC_{\tilde{V}_1}(S)$ is centralized by $L_4(r)$ in $K$. As now 
$E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \le M$,
centralizes $\tilde{Z_2}$, we see $r = 2$. Furthermore there is a 
$p$ - element $\omega$
in $L_1$, $o(\omega) \ge 5$, such that $[\tilde{V}_1, \omega] \not= 1$ and 
$[\omega, K_1] = 1$. Let
$\hat{V_1} = [ \tilde{V}_1, \omega]$. Then $\hat{V_1}$ cannot involve 
$V(\lambda_2)$. So we have
natural submodules. As $7^2 \,\Big | \,|L_6(2)|$, we see that $m_p(K) = 1$. 
This now implies that we have at least 7 natural modules involved 
($p = 127$) or 
if $p \,\Big | \,|L_6(2)|$ there is a faithful $E_{p^2}$ and there are 
even 8 modules involved.

We have $|[V_{\alpha-1},t]| \le 2^{12}$ for $t \in \tilde{V}_1$. This 
shows that $\tilde{V}_1$ induces
transvections on the natural module and so 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(K_{\alpha-1})| \le 2^6$. Hence 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(z)| \le 2^7$ for 
every $z \in V_{\alpha-1}$. Then $p = 127$, and there are exactly 
7 natural modules involved. Now some
$\omega$, $o(\omega) = p$, centralizes $\tilde{V}_1$. But we have 
$Z_{\alpha-1} \le \tilde{V}_1$, and
so $\omega \in M_{\alpha-1}$. By (4.2) $M = M_{\alpha-1}$, a 
contradiction.

Let $n = 6$. If $|V_{\alpha-1} : \bC_{V_{\alpha-1}}(K)| > r^6$, we may 
argue as before to see that just 
natural or dual modules are involved. So let first 
$|V_{\alpha-1} : \bC_{V_{\alpha-1}}(K)| \le r^6$. Then by (1.23) 
$V(\lambda_3)$ is not involved. Furthermore either $V(\lambda_2)$ or 
$V(\lambda_4)$ is not involved. 
So in any case $\bC_{\tilde{V}_1}(S)$ is
centralized by $L_3(r)$. This implies $r = 2$, and there is $\omega$, 
$o(\omega) = p$,
$[\omega,K] = 1$ and $[\tilde{V}_1, \omega] \not= 1$. As 
$o(\omega) > 3$, we get again that
$\hat{V_1} = [\tilde{V}_1, \omega]$ just involves natural or dual 
modules.
As $p \,\Big | \,|L_5(2)|$, we get $p = 7$ and there are at least three 
natural modules involved, or
$p \not= 7$ and there are at least 8 natural modules involved. 
Now as in the case $n = 7$, we get $p = 7$. Then 
$|V_{\alpha-1} : \bC_{V_{\alpha-1}}(K)| \ge 8$. By (1.16) some 
$t \in V_{\alpha-1}$ induces some $s \in K$ on $\hat{V_1}$. 
Hence there is some $\nu \in K$, $o(\nu) > 2$ prime, and 
$\nu^s = \nu^{-1}$. Then we may assume $[\omega, \nu] = 1$, 
and so $\omega$ acts on $[[\hat{V}_1,\nu],s]$. As there is some
$\omega_1 \in K$, $o(\omega_1) = 7$, 
$\bC_{[[\hat{V_1},\nu],s]}(\omega_1) \not= 1$ and $[\omega_1, \omega] = 1$,
we get some $1 \not= U \le [V_1, V_{\alpha-1}]$ which is normalized by 
$E \cong E_{49}$ with $\Gamma_{E,1}(G) \le M$, a contradiction.

Let next $n = 5$. As before not both $V(\lambda_2)$ and $V(\lambda_3)$ 
are involved. So assume neither
$V(\lambda_2)$ nor $V(\lambda_3)$ are involved. Then 
$\bC_{\tilde{V}_1}(S)$ is centralized by $SL_3(r)$ and so $r = 2$. Hence 
$m_p(K) = 1$ and
so either $p \,\Big | \,|L_4(2)|$ and there are at least 6 natural modules 
involved or $p = 31$
and there are at least 5 natural modules involved. But
now as in the case $K \cong L_7(2)$ we get a contradiction.\\
\indent
So we have $V(\lambda_2)$ is involved and 
$|V_{\alpha-1} : \bC_{V_{\alpha-1}}(K)| \ge r^3$. As $V(\lambda_2)$ is 
involved just once, we
see that there has to be involved $V(\lambda_1)$ or $V(\lambda_4)$ too. 
This now gives
$|V_{\alpha-1} : \bC_{V_{\alpha-1}}(K)| \ge r^4$. As before we see 
$|V_{\alpha-1} : \bC_{V_{\alpha-1}}(K)| \le r^4$ and so we have 
$|V_{\alpha-1} : 
\bC_{V_{\alpha-1}}(K)| = r^4$. Now either $L_2(r) \times L_2(r)$ or 
$SL_3(r)$ centralizes $\bC_{\tilde{V}_1}(S)$. 
This implies $r = 2$. But then $m_p(K) = 1$ and $E \cong E_{p^2}$ with 
$\Gamma_{E,1}(G) \le M$ centralizes $\tilde{V}_1$, a contradiction.

Let now $n = 4$. Suppose $V(\lambda_2)$ is involved. Then 
$|V_{\alpha-1} : \bC_{V_{\alpha-1}}(K)| \le r^3$ and so there is at most 
one further nontrivial 
irreducible module involved which then has to be the natural module. Suppose 
$V(\lambda_2)$ is a submodule. Every $v \in V(\lambda_2)$
is centralized by $E \cong E_{p^2}$. If $p \,\Big | \,r^2 - 1$, we get 
$m_p(K) = 1$. But now $E \cong E_{p^2}$ centralizes $V(\lambda_2)$, a
contradiction.\\
\indent
So in any case by (1.20) we have a natural submodule and so 
$p \not\Big | \,r - 1$. Now $m_p(K) \le 2$
and so there is some $\omega \in \bC(K)$, $o(\omega) = p$ with 
$[\tilde{V}_1, \omega] \not= 1$. Hence there are at least two natural 
modules for $ r > 2$ and
at least three natural modules for $r = 2$ involved. Then 
$|V_{\alpha-1} : \bC_{V_{\alpha-1}}(K)| \ge r^2$ and
so there is some $t \in V_{\alpha-1}$ inducing some $1 \not= s \in K$ on 
$\tilde{V}_1$ by (1.16).
If $p  \,\Big | \,|SL_3(r)|$ we get as before some 
$1 \not= U \le [\tilde{V}_1, V_{\alpha-1}]$ 
normalized by some $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \le M$, a 
contradiction.

So we may assume that there are four natural modules involved. Now 
$|V_{\alpha-1} : \bC_{V_{\alpha-1}}(K)| = r^4$, and $|\tilde{V}_1 : 
\tilde{V}_1 \cap Q_\alpha| = r^3$. Hence 
$Z_{\alpha-1} \le [V_{\alpha-1}, \tilde{V}_1]$. But as $m_p(K) = 1$, we 
see that $Z_{\alpha-1}$
is centralized by $\omega$, $o(\omega) = p$, $\omega \in E$, 
$E \cong E_{p^3}$. This contradicts (4.2) and $M \not= M_{\alpha-1}$.

Let $n = 3$. Now we have at most two nontrivial modules involved. If 
$m_p(K) = 1$, we get some $\omega \in \bC(K)$,
$[\tilde{V}_1, \omega] \not= 1$, so there are exactly two nontrivial 
irreducible 
modules involved and then $p \,\Big | \,r + 1$. But now 
$|V_{\alpha-1} : \bC_{V_{\alpha-1}}(K)| = r^2$ and
some $t \in V_{\alpha-1}$ induces $s \in K$ on $\tilde{V}_1$ by (1.16). 
Hence again some 
$1 \not= U \le [V_1, V_{\alpha-1}]$ is normalized by $E \cong E_{p^2}$ with 
$\Gamma_{E,1}(G) \le M$, a contradiction.
This shows $m_p(K) = 2$ and so $p \,\Big | \,r - 1$, i.e. $r > 2$. By (1.20) 
the natural module
is a submodule of $\tilde{V}_1/Z_1$. Now in the natural module every 
element is centralized by $L_2(r)$ and so there is some $\omega$, 
$o(\omega) = p$, which does not normalize some natural module in 
$\tilde{V}_1$. But we may assume $[\omega,s] = 1$ for some 
$1 \not= s \in K$ where $s$ acts as $t \in V_{\alpha-1}$
on $\tilde{V}_1$. Now as before we get 
$1 \not= U \le [V_1, V_{\alpha-1}]$ normalized by $E \cong E_{p^2}$ with 
$\Gamma_{E,1}(G) \le M$.


Let finally $n = 2$. Now $|V_{\alpha - 1} : \bC_{V_{\alpha-1}}(K)| = r$ 
and $\tilde{V}_1/\bC_{\tilde{V}_1}(K)$
is the natural module. This implies that some 
$1 \not= x \in [V_{\alpha-1}, \tilde{V}_1]$ is centralized by 
$E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \le M$.


So we have seen that there is some module involved which is not a 
fundamental module. These modules are described in (1.22).

Let $T$ be such a module. If $\tilde{V}_1/\bC_{\tilde{V}_1}(K) \cong T$, 
then we see immediately that
$\bC_{\tilde{V}_1}(S)$ is centralized by $E \cong E_{p^2}$ with 
$\Gamma_{E,1}(G) \le M$, a contradiction.

Now application of (1.24) shows that $r$ is a square and either
$T \cong V(\lambda_1) \bigotimes V(\lambda_1)^\sigma$ or
$V(\lambda_n) \bigotimes V(\lambda_n)^\sigma$. As $3 \not\in \sigma(M)$ 
and (7.6), we get $3 \le n \le 4$. But now we see that 
$|V_{\alpha-1} : \bC_{V_{\alpha-1}}(K)| \ge r^{n-1}$, contradicting 
(1.15).

\absa
\begin{equation}
K \not\cong (S)U_n(r)
\end{equation}


By (1.22) every irreducible module involved in $\tilde{V}_1$ is a 
fundamental module or
$V(\lambda_1 + \lambda_l)$. But the latter contradicts (1.24) and 
(1.15). So we just
have fundamental modules involved.

Let $n = 6$ or $7$. Then $ p \,\Big | \,r+1$. If $V(\lambda_3)$ is
not involved then $\bC_{\tilde{V}_1}(S)$ is centralized by $SU_3(r)$, 
and
so by some $E \cong E_{p^2}$, $E \le K$ with $\Gamma_{E,1}(G) \le M$, a 
contradiction.

So we have $K \cong (S)U_6(r)$ and $V(\lambda_3)$ is involved. As 
$\bC_{V(\lambda_3)}(S)$ is centralized
by $(S)L_3(r^2)$, we see that there has to be some additional module 
involved. This has to be the
natural module. Now $|V_{\alpha-1} : \bC_{V_{\alpha-1}}(K)| \ge r^8$. 
But then $\tilde{V}_1$ just contains elements which induce transvections 
on the natural $K_{\alpha-1}$ - module. This implies 
$|\tilde{V}_1 : \tilde{V}_1 \cap Q_{\alpha-1}| \le r$.
Then $|\tilde{V}_1 : \bC_{\tilde{V}_1}(z)| \le 2r$ for every 
$z \in V_{\alpha-1}$, a contradiction.

If $n = 5$ then we just have natural modules involved, a contradiction as 
$p \,\Big | \,r+1$.

So we are left with $n = 4$. Let first 
$\tilde{V}_1/\bC_{\tilde{V}_1}(K) = T$ be irreducible.
If $T \cong V(\lambda_2)$, then $[t, \tilde{V}_1]$ contains elements 
centralized by $Sp_4(r)$. Hence $p \not\Big | \,r^2 - 1$.
This shows $m_p(K) = 1$. But then here is some 
$1 \not= x \in [\tilde{V}_1, V_{\alpha-1}]$ with $E \le \bC_G(x)$,  
$E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \le M$, a contradiction. So 
$T \cong V(\lambda_1)$, the natural module.

If there are two modules involved, then 
$|V_{\alpha-1} : \bC_{V_{\alpha-1}}(K)| = r^4$. If the natural 
$\Omega_6^-(r)$ - module is
involved we get $K \cong \Omega_6^-(2)$ by (1.39) and both modules are 
natural $\Omega_6^-(2)$ - modules. Hence either both modules
are natural $\Omega_6^-(2)$ - modules or $U_4(r)$ - modules. Now by 
(1.39) we get some $z \in \tilde{V}_{\alpha-1}$ with 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(z)| \ge r^4$, $2^6$ in case of
$\Omega_6^-(2)$ - modules. Hence 
$|\tilde{V}_1 : \tilde{V}_1 \cap Q_{\alpha-1}| \ge r^4/2$, $2^5$, 
respectively. The latter is not possible.
In the former we get that $T$ is the natural $U_4(r)$ - module. But then 
we get some $x \in \tilde{V}_1$ with 
$|[V_{\alpha-1},x]\bC_{\tilde{V}_1}(K)/\bC_{\tilde{V}_1}(K)| = r^4$, 
contradicting (1.50).

\absa
\begin{equation}
K \not\cong Sp_{2n}(r), n \le 3
\end{equation}

\absa
By (1.22) $\tilde{V}_1$ just involves $V(\lambda)$, $\lambda$ 
fundamental weight. Let first $n = 3$. If 
$|V_{\alpha-1} : \bC_{V_{\alpha-1}}(K)| > r^5$, 
then there is some root element $z \in V_{\alpha-1}$ with 
$\langle {V_{\alpha-1}}^{\bC_K(z)} \rangle \ge O^2(\bC_K(z))$.
This implies that $[[\tilde{V}_1,z],O^2(\bC_K(z))] = 1$ and so there are 
just natural modules involved.

Suppose $T$ is the natural module, $T$ a submodule of $\tilde{V}_1$. As 
every $x \in T$ is centralized by $Sp_4(r)$, we see that
$p \not\Big | \,r^2 - 1$. Hence $m_p(K) = 1$ and there is some $\omega$, 
$o(\omega) = p$, which acts nontrivially on the set of natural modules 
involved
in $\tilde{V}_1$. So there are at least three of them. Now $\tilde{V}_1$ 
centralizes in the natural module a subspace of codimension two. This
gives $|\tilde{V}_1 : \tilde{V}_1 \cap Q_\alpha| \le r^4$. Then 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(x)| \le 2r^4$ for $x \in V_{\alpha-1}$.
Hence $V_{\alpha-1}$ induces transvections on $T$, i.e. 
$|V_{\alpha-1} : \bC_{V_{\alpha-1}}(K)| \le r$. But then
$|V_{\alpha-1} : \bC_{V_{\alpha-1}}(\tilde{V}_1)| \le r$, contradicting 
the fact that $V_{\alpha-1}$ involves more than one natural module.

Let $T \cong V(\lambda_3)$ be a submodule of 
$\tilde{V}_1/\bC_{\tilde{V}_1}(K)$. Now 
$r^2 \le |V_{\alpha-1} : \bC_{V_{\alpha-1}}(K)| \le r^5$.
Hence there are at most two such modules involved. Then $m_p(K) \ge 2$, 
in particular $r \not= 2$. If there
are two such modules involved, then 
$|V_{\alpha-1} : \bC_{V_{\alpha-1}}(K)| \ge r^4$. But then for some 
$t \in T$ there is 
$|[t, V_{\alpha-1}]\bC_{\tilde{V}_1}(K)/\bC_{\tilde{V}_1}(K)| = r^4$, 
contradicting (1.50). Now let $T_{\alpha-1}$ be the corresponding module 
of $K_{\alpha-1}$. We have 
$[T \cap Q_{\alpha-1}, T_{\alpha-1}] = 1$, otherwise $Z_{\alpha-1} \le T$.
As $m_p(K) \ge 2$ we have $p \,\Big | \,r^2 - 1$ and so every $t \in T$ is centralized by some $\omega \in K$ with $o(\omega) = p$.  Now we have 
$[T, T_{\alpha-1}] \le T \cap T_{\alpha-1}$ and a contradiction with 
(1.50).

It remains the case that $V(\lambda_2)$ is a submodule. Then 
$|V_{\alpha-1} : \bC_{V_{\alpha-1}}(K)| \ge r^4$. If
$\tilde{V}_1/\bC_{\tilde{V}_1}(K) \cong V(\lambda_2)$, then 
$\bC_{\tilde{V}_1}(S)$ is
centralized by $L_2(r) \times L_2(r)$ and so $p \not\Big | \,r^2 - 1$, 
i.e. $m_p(K) = 1$. But now some 
$1 \not= U \le [\tilde{V}_1, V_{\alpha-1}]$ is 
normalized by $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \le M$, a 
contradiction.

Hence we have $|V_{\alpha-1} : \bC_{V_{\alpha-1}}(K)| = r^5$ and so some 
natural module is involved on which $\tilde{V}_1$ induces transvections. 
So $|\tilde{V}_1 : \tilde{V}_1 \cap Q_{\alpha-1}| \le r$ and then 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(z)| \le 2r$, for $z \in V_{\alpha-1}$, 
a contradiction.


Let now $n = 2$. There are just natural or dual modules involved. 
Suppose there are at least two such modules 
involved. As $|V_{\alpha-1} : \bC_{V_{\alpha-1}}(K)| \le r^3$, we get 
that
$\tilde{V}_1$ has to induce transvections on one of these modules. Hence 
$|\tilde{V}_1 : \tilde{V}_1 \cap Q_{\alpha-1}| \le r$ and so 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(z)| \le 2r$ for $z \in V_{\alpha-1}$. 
This shows $r = 2$.
But now $|V_{\alpha-1} : \bC_{V_{\alpha-1}}(K)| = 2$, a contradiction.

So we may assume that $\tilde{V}_1/\bC_{\tilde{V}_1}(K)$ is the natural
module. If $m_p(K) = 1$, we get a contradiction as before. So assume 
$m_p(K) = 2$,
i.e. $p \,\Big | \,r^2 - 1$. Now every $x \in \tilde{V}_1$ is centralized by 
some $\omega \in K$, $o(\omega) = p$.
We have $\bC_{\tilde{V}_1}(K) \cap [\tilde{V}_1, V_{\alpha-1}] = 1$. If 
$V_{\alpha-1}$ contains a group
$R$ with $|[\tilde{V}_1/\bC_{\tilde{V}_1}(K), R]| = r$, then 
$F \cong E_{p^3}$ acts on $[R, \tilde{V}_1]$ and so some 
$1 \not= x \in [\tilde{V}_1, V_{\alpha-1}]$ is centralized by
$E \le F$, $|E| = p^2$, a contradiction. Hence we get that 
$\langle V_{\alpha-1}, V_{\alpha-1}^g \rangle = O^2(P)$, for some 
$g \in K$ and some
parabolic $P$ of $K$. Now we see $|[O^2(P), \tilde{V}_1]Z_1/Z_1| = r^4$. 
Hence we have that $\tilde{V}_1/Z_1$ is the natural module. 
But then every $x \in \tilde{V}_1$ is centralized by $E \cong E_{p^2}$ with
$\Gamma_{E,1}(G) \le M$, a contradiction.   

\absa
\begin{equation}
K \not\cong \Omega_8^-(r)
\end{equation}

\absa
By (1.22) $\tilde{V}_1$ just involves $V(\lambda)$, $\lambda$ some 
fundamental weight.
Furthermore just $V(\lambda_1)$ or $V(\lambda_3)$ are involved by 
(1.15). Let
$V(\lambda_1)$ be a submodule. As every element in $V(\lambda_1)$ is 
centralized by $\Omega_6^-(r)$
or $Sp_6(r)$, we get $p \not\Big | \,r^2 - 1$. Hence $m_p(K) = 1$. If
there are two natural modules involved, then 
$|V_{\alpha-1} : \bC_{V_{\alpha-1}}(K)| \ge r^4$. So
for some module $\tilde{V}_1$ centralizes a subgroup of index $r^2$. By 
(1.45) we get $|\tilde{V}_1 : \tilde{V}_1 \cap Q_{\alpha-1}| \le r$.
Now $|\tilde{V}_1 : \bC_{\tilde{V}_1}(z)| \le 2r$ for 
$z \in V_{\alpha-1}$, a contradiction. Hence we have
that there is exactly one natural module involved. If $V(\lambda_3)$ is 
involved
then we get $|V_{\alpha-1} : \bC_{V_{\alpha-1}}(K)| = r^6$, but a group 
of that large order cannot act quadratically on the natural module.

So we have $\tilde{V}_1/\bC_{\tilde{V}_1}(K)$ is the natural module. But 
now some $\omega \in \bC(K)$, $o(\omega) = p$, centralizes $\tilde{V}_1$.
Hence there is some $\omega_1$, $o(\omega_1) = p$, which induces 
some field automorphism
on $K$. But this contradicts (4.12).

Let now $V(\lambda_3)$ be a submodule. Then 
$|V_{\alpha-1} : \bC_{V_{\alpha-1}}(K)| \ge r^4$. In particular we have 
$\tilde{V}_1/\bC_{\tilde{V}_1}(K) \cong V(\lambda_3)$. 
By (1.20) $\tilde{V}_1/Z_1 \cong V(\lambda_3)$. Furthermore as 
$V(\lambda_3)$ is not an F - module by (1.18), we see 
$Z_{\alpha-1} \le \tilde{V}_1$ and so $m_p(K) \ge 2$. 
Hence $p \,\Big | \,r^2 - 1$. This shows that $V_{\alpha-1}$ does not
contain elements of type $a_2$ or $c_2$, as those are centralized by 
$L_2(r^2) \times L_2(r)$,
$Sp_4(r)$, respectively, and so they are centralized by 
$F \cong E_{p^2}$. But now as 
$|V_{\alpha-1} : \bC_{V_{\alpha-1}}(K)| \ge r^4$, we see that there is 
some $t \in V_{\alpha-1}$ with 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(t)| \ge r^8$. This would imply 
$|\tilde{V}_1 : \tilde{V}_1 \cap Q_{\alpha-1}| \ge r^7$, a contradiction.

\absa
\begin{equation}
{\mbox {The lemma holds}}
\end{equation}

\absa
Suppose false. By (1.40), (1.27) and (1) - (4) we just have to treat the 
case $K \cong G_2(r)$.
Now $\tilde{V}_1/\bC_{\tilde{V}_1}(K)$ is the 6 - dimensional natural 
module. Then
every $x \in \tilde{V}_1$ is centralized by some $\omega \in M$ with 
$o(\omega) = p$. This
implies $[\tilde{V}_1 \cap Q_{\alpha-1}, V_{\alpha-1}] = 1$. Hence 
$V_{\alpha-1}$ offends as an F - module. This yields with (1.37) that 
$|V_{\alpha-1} : \bC_{V_{\alpha-1}}(K)| = r^3$. As 
$[\tilde{V}_1, V_{\alpha-1}] \cap \bC_{\tilde{V}_1}(K) = 1$,
we see that for 
$\langle V_{\alpha-1}, V_{\alpha-1}^g \rangle = O^2(P)$, $P$ some 
parabolic, we get $|[\tilde{V}_1, O^2(P)]Z_1/Z_1| = r^6$. Hence 
$\tilde{V}_1/Z_1$ is the natural module. But now every 
$x \in \tilde{V}_1$ is centralized by $E \cong E_{p^2}$ with 
$\Gamma_{E,1}(G) \le M$, a contradiction. 
  
\absa
{\bf~(7.15) Lemma.~}{\it Assume that $L_2/Q_2$ is as in (7.1)(3)(b) and 
$\tilde{V}_1 \not\le Q_{\alpha-1}$. If $K$ is some 
normal $p$-subgroup of $L_1/Q_1$, then 
$[K, V_{\alpha-1}] = 1$.}

\absa
Proof.~Suppose false. We have
$\tilde{V}_1 = \langle t \,\Big | \,|V_{\alpha-1} : \bC_{V_{\alpha-1}}(t)| 
\le 2 \rangle$. 
As $p > 3$, we see that $[\tilde{V}_1, K_{\alpha-1}] = 1$. But now by 
(7.7) $m_p(K) = m_p(K_1) = 1$. 
Furthermore with (7.10) we get a contradiction.
\\
\\

\absa
Let now $K$ be any component of $L_1/Q_1$ with $[V_{\alpha-1},K] \not= 1$. We choose $t \in V_{\alpha-1}$ with $K^t \not= K$. As 
$3 \not\in \sigma(M)$ by (7.6) we have $m_3(K) \le 1$, i.e. 
$K \cong L_2(r)$, $L_3(r)$, $U_3(r)$, 
$Sz(r)$ or $J_1$. If $m_p(K) \ge 2$, there is some 
$1 \not= u \in [V_1,t]$ centralized by $E \cong E_{p^2}$ with 
$\Gamma_{E,1}(G) \le M$. So we may assume 
$K \cong L_2(r)$, $Sz(r)$ or $J_1$.

Let $K_{\alpha-1} \times \tilde{K}_{\alpha-1}$ be the 
components corresponding to
$K \times K^t$ in $L_{\alpha-1}/Q_{\alpha-1}$. Suppose there is 
$z \in \tilde{V}_1 \setminus Q_{\alpha-1}$
with $[z, K_{\alpha-1} \times \tilde{K}_{\alpha-1}] = 1$. Then as
$[K_{\alpha-1} \times \tilde{K}_{\alpha-1}, [V_{\alpha-1}, z]] = 1$, we 
get $\bC_G([V_{\alpha-1}, z]) \le M_{\alpha-1}$. 
But this is impossible as there is some $p$ - element in $L_1$ 
centralizing some $1 \not= u \in [V_{\alpha-1},z]$.
So we have that $\tilde{V}_1$ acts faithfully on 
$K_{\alpha-1} \times \tilde{K}_{\alpha-1}$. This
implies $|V_{\alpha-1} : \bC_{V_{\alpha-1}}(K)| > 2$ and so by (1.11) 
$K \cong L_2(r)$, $r$ even. Now 
$|V_{\alpha-1} : \bC_{V_{\alpha-1}}(K)| = r^2$, but this is impossible 
as $|\bC_{K \times K^t}(t)|_2 = r > 2$. So we have

\absa
{\bf~(7.16) Lemma.~}{\it If $L_2/Q_2$ is as in (7.1)(3)(b), then 
$\tilde{V}_1 \le Q_{\alpha-1}$.}

\absa
{\bf~(7.17) Lemma.~}{\it $[V_{\alpha-1}, K] = 1$ for every component $K$ of $L_1/Q_1$, if $L_2/Q_2$ is as in (7.1)(3)(b)}

\absa
Proof.~By (7.16) we just have to handle the case 
$\tilde{V}_1 \le Q_{\alpha-1}$. Now $V_{\alpha-1}$ induces transvections to 
$Z_{\alpha-1}$ on $\tilde{V}_1$. Hence $K \cong L_n(2)$,
$Sp_{2n}(2)$, $\Omega_{2n}^\pm(2)$ or $A_n$. Furthermore 
$\tilde{V}_1/\bC_{\tilde{V}_1}(K)$ is irreducible. If $m_p(K) = 1$,
we get some $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \le M$, with 
$[\tilde{V}_1, E] = 1$, a contradiction to $Z_{\alpha-1} \le \tilde{V}_1$.

As $3 \not\in \sigma(M)$, we see that $K \cong L_6(2)$, 
$L_7(2), p = 7$ or $A_{10}, A_{11}, p = 5$. But in both cases 
($p = 7, p = 5$), there is $\omega \in M$, $o(\omega) = p$ with 
$[\omega, K] = 1$ and $[\omega, \tilde{V}_1] = 1$. Now every
$x \in \tilde{V}_1$ is centralized by $\omega_1 \in K$ with 
$o(\omega_1) = p$. This gives some 
$1 \not= x \in [\tilde{V}_1, V_{\alpha-1}]$ with
$\langle \omega, \omega_1 \rangle \le \bC_G(x)$, and so 
$Z_\alpha \le \bC_G(x) \le M$, a contradiction.

\absa
{\bf~(7.18) Proposition.~}{\it $L_2/Q_2$ is as in (7.1)(1) or 
(7.1)(3)(a).}

\absa
Proof.~ Let K be one of the exceptional components of $L_1/Q_1$ in (4.9),
$[K,V_{\alpha-1}] \not= 1$. Then $[K, V_{\alpha-1}] \le K$. This will
also handle the case of (4.9)(ii). Now
$|V_{\alpha-1} : \bC_{V_{\alpha-1}}(K)| \le 4$. Let 
$\tilde{V}_1 \le Q_{\alpha-1}$. Then $V_{\alpha-1}$
induces transvections on $\tilde{V}_1$ and so $[O_r(K), \tilde{V}_1] = 1$
and we have the exceptional case (4.9)(ii). But then (4.10) shows 
$Z_\alpha \le \bC_G([\tilde{V}_1, V_{\alpha-1}]) \le M$,
a contradiction. Thus we have $\tilde{V}_1 \not\le Q_{\alpha-1}$. If 
$t \in \tilde{V}_1$ with $[t, K_{\alpha-1}] = 1$,
then we see $[K_{\alpha-1},[t, V_{\alpha-1}]] = 1$. Now $m_p(K) = 1$. 
But then $m_p(N_{L_1/Q_1}(K)/\bC_{L_1/Q_1}(K)) = 1$, which contradicts 
(7.10).

Hence we have that $\tilde{V}_1$ acts faithfully on $K_{\alpha-1}$. Now 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(t)| \le 8$,
$t \in V_{\alpha-1}$. Application of (1.41) shows that we are in 
(4.9)(ii) and$K/O_r(K) \cong L_2(7)$ or $A_5$. As 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(t)| \le 8$, we see with (1.51)
that $[O_r(K), \tilde{V}_1] = 1$, the same contradiction as above.


\absa
From now on we have that $O^2(L_2/\bC_{L_2}(Z_2)) \cong L_2(q)$. 
Furthermore there is $C \cong \bZ_{q-1}$, $C \le L_1 \cap L_2$. We let 
$x$ be a Zsigmondy prime, $x \,\Big | \,q-1$ or $x = 9$ if $q = 64$, and 
$\omega \in C$ with $o(\omega) = x$.

We have $|Z_1| = q$ and $m_p(\bC_{L_1}(Z_1)) \ge 3$. Furthermore if 
$\omega\not= 1$, then $\omega \not\in \bC_{L_1}(Z_1)$.

\absa
If $b > 2$, then we know by (7.4) that $V_1 \le L_{\alpha-1}$ and 
$V_{\alpha-1} \le L_2$ but $V_{\alpha-1} \not\le M$. Hence
the main contradiction in that what follows will come from 
$V_{\alpha-1} \not\le M$.

\absa
We now fix some notation. By (7.4) $V_1 \not\le Q_{\alpha-1}$. Hence 
there is some component  or normal $p$-subgroup $K_{\alpha-1}$ of 
$L_{\alpha-1}/Q_{\alpha-1}$ with $[V_1, K_{\alpha-1}] \not= 1$.
Let $K$ be the corresponding group in $L_1/Q_1$. Set 
$\tilde{V}_{\alpha-1} = [V_{\alpha-1}, K_{\alpha-1}]$.
Until further notice assume $[V_1, K_{\alpha-1}] \le K_{\alpha-1}$.

\absa
{\bf~(7.19) Lemma.~}{\it Let $b > 2$. Then $K^\omega = K$.}

\absa
Proof.~Suppose false. Let $T = K_1 \times \ldots \times K_y$ be an orbit 
of $\langle \omega \rangle$, $y = x$, or $y = 3$ and $x = 9$. 
Then $\omega$ centralizes in $T$ a subgroup isomorphic to $K$, or 
$x = 9$ and $m_3(K) = 1$. Now (1.63) implies $m_p(K) = 1$, and so
$K \cong L_2(r)$, $Sz(r)$, $J_1$, or $x = 9$ and $y = 3$.

We have 
$\tilde{V}_{\alpha-1} = \langle t \,\Big | \,|V_1 : \bC_{V_1}(t)| 
\le qu \rangle$, where
$u = r$, or $u = 4$ for $K \cong L_2(7)$, $u = 2$ in the remaining cases. 
We show
\setcounter{equation}{0}
\begin{equation}
\tilde{V}_{\alpha-1} \le L_1
\end{equation}

\absa
On $V_1/Z_1$ there is some $E \cong E_{p^y}$. If 
$t \in \tilde{V}_{\alpha-1} \setminus L_1$,
then $[t, Z_1] = Z_3$ and so $|V_1 : \bC_{V_1}(t)Z_1| \le u$. Now we 
have that no $F \cong E_{p^2}$, $F \le E$, centralizes a group of order 
greater then $u$. This implies $u \not= 2$.

In $E$ there is some $F$ such that 
$|\bC_{V_1/Z_1}(F)| \ge 2^{(y-2)d(p)}$, $d(p)$ the minimal dimension of 
a cyclic $\bZ_p$ - module. 
Let $u = 4$, then we get $y = 3$ and $d(p) = 2$. But this implies 
$p = 3$, a contradiction to $o(\omega) = 3$, or $9$.

Let now $u = r$ and $y \ge 4$. Then we may choose $p$ with $d(p) \ge m$, 
$r = 2^m$. Now $|\bC_{V_1/Z_1}(F)| \ge 2^{2m} > u$,
a contradiction.

Hence $y = 3$, and so $d(p) \le m$. Let $z \in \tilde{V}_1$. Then we may 
assume $|\tilde{V}_{\alpha-1} : \bC_{\tilde{V}_{\alpha-1}}(z)| \le qr$. 
Let $[z, T_{\alpha-1}] \not= 1$. We have $q = 4$ or $q = 64$.

If $[K_1, V_1]$ involves at most two nontrivial irreducible modules, we 
see that 
$[K_3, [K_1,V_1]] = 1$ and so $|\bC_{V_1/Z_1}(F)| \ge r^2 = 2^{2m}$,
a contradiction.

So we have $|[\tilde{V}_{\alpha-1}, z]| \ge r^4$ and then $r^3 \le q$.
This implies $q = 64$ and $r = 4$, $K_1 \cong L_2(4)$, $p = 5$. But now
$|\bC_{V_1/Z_1}(F)| \ge 16 > 4 = u$. So we may assume 
$[\tilde{V}_1, T_{\alpha-1}] = 1$.
Then $[\tilde{V}_{\alpha-1}, \tilde{V}_1] \le Z_{\alpha-1}$ and
so obviously $\tilde{V}_{\alpha-1} \le L_1$.

\absa
By (1) we now have $\tilde{V}_{\alpha-1} \le L_1$. As 
$[V_1, \tilde{V}_{\alpha-1}] \not= 1$, we
get $\tilde{V}_{\alpha-1} \not\le Q_1$. Let 
$1 \not= t \in \tilde{V}_{\alpha-1}$. If $[T,t] = 1$,
then $[t,V_1]$ contains elements $z$ centralized by $F \cong E_{p^2}$, 
$F \le T$ and so $V_{\alpha-1} \le \bC_G(z) \le M$,
a contradiction. Hence 
$\tilde{V}_{\alpha-1}/\tilde{V}_{\alpha-1} \cap Q_1$ acts faithfully on 
$T$. We have that
$$ \tilde{V}_{\alpha-1} = \langle t \,\Big | \,|V_1/Z_1 : \bC_{V_1/Z_1}(t)| 
\le uq \rangle,$$ 
$u$ as above. Assume that $[[\tilde{V}_1, K_1], K_{y-1} \times K_y] = 1$.
Then we get again $z \in [[\tilde{V}_1, K_1],t]$ with 
$V_{\alpha-1} \le \bC_G(z) \le M$, a contradiction.

Hence as $y \ge 3$, we see that there are at least $2^{y-2}$ nontrivial 
irreducible $K_1$ - modules involved. Now if $K_1 \not\cong L_2(7)$,
then $uq \ge u^{2^{y-2}}$. Suppose $q \not= 64$. Then $q \le 2^{y-1}$ by 
Fermat's lemma. So
$$2^{y-1} \ge u^{2^{y-2} -1} \ge 2^{2^{y-1} - 2}.$$
\indent
Hence $y - 1\ge 2^{y-1} - 2$ and then $y = 3$, $u = 4 = q$. So in any 
case $y = 3$. But now
$$[K_3,[\tilde{V}_1, K_1]] = 1$$
\indent
As an element of order 3 acts transitively on $\{K_1, K_2, K_3\}$, we get
$$[[\tilde{V}_1, K_2], K_1] = 1$$
too, a contradiction. 

So assume $q = 64$. Then $y = 3$ and we have at least 4 nontrivial 
irreducible modules involved. Now
$$64u \ge u^4$$
and so $u = 4$, as $u > 2$. Hence 
$K_1 \times K_2 \times K_3 \cong L_2(4) \times L_2(4) \times L_2(4)$.
Choose $\nu \in L_1 \cap L_2$ with $o(\nu) = 7$. Then 
$[\nu, K_1 \times K_2 \times K_3] = 1$. By
(1.63) $N_G(\langle \nu \rangle) \le M$, a contradiction.

So it remains $K_1 \cong L_2(7)$. Then there are at least $3^{y-2}$ 
nontrivial irreducible modules involved. Hence 
$$4q \ge 2^{3^{y-2}}$$
\indent
This implies $y + 1 \ge 3^{y-2}$ or $q = 64$. But in both cases $y = 3$.
Now by the same argument as above, we have that there are at least 9 
nontrivial irreducible modules involved. So
$$2^9 \le 4q \le 2^8,$$
a contradiction.

\absa
{\bf~(7.20) Lemma.~}{\it Let $b > 2$. If $K_{\alpha-1}$ is sporadic or a 
group of Lie type $G(r)$, $r$ odd, then 
$m_p(K_{\alpha-1}) = 1$, $\tilde{V}_{\alpha-1} \le M$, 
$[\tilde{V}_{\alpha-1} \cap L_1, K] = 1$ and
$[\tilde{V}_1, K_{\alpha-1}] \le Q_{\alpha-1}$, or $K/Z(K) \cong L_2(5)$, 
$L_2(7)$, $L_2(9)$, $U_3(3)$, $PSp_4(3)$, or $^2G_2(3)^\prime$.}

\absa
Proof.~Let $|V_1 : \bC_{V_1}(K_{\alpha-1})| = 2^4$. Then
$$\tilde{V}_{\alpha-1} = \langle t \,\Big | \,|V_1 : \bC_{V_1}(t)| 
\le 2^uq \rangle$$

Furthermore $u \le 4$ by (1.12) or $K/Z(K) \cong U_4(3)$. Suppose $K$ is 
none of the groups listed in the lemma.

\absa
\setcounter{equation}{0}
\begin{equation}
\tilde{V}_{\alpha-1} \le M
\end{equation}

\absa
If $K \cong 3\cdot U_4(3)$, then $m_3(K) = 5$ and some $E \cong E_9$ 
centralizes a subgroup
of order $2^6$ in $\tilde{V}_1/\bC_{\tilde{V}_1}(K)$. So we have (1).

So assume (1) to be false. We have $u \ge 2$. Let $u = 2$. Then we get 
$p = 3$ and $m_3(L_1) = 3$. Application of
(1.12) shows $K \cong M_{12}$, $M_{22}$, $3\cdot M_{22}$, $M_{24}$, or 
$J_2$. In particular $m_3(K) = 2$. Now (7.19) shows that 
$\omega$ induces an automorphism on $K$ normalizing a Sylow
2 - subgroup. Hence we get $[\omega, K] = 1$, as $o(\omega) \not= 3$, 
by (1.9). But now (1.63)
implies $N_G(\langle \omega \rangle) \le M$, if $o(\omega) \not= 1$. 
Hence $\omega = 1$ and so $q = 2$. Let now
$z \in \tilde{V}_1$, $[z, K_{\alpha-1}] \not= 1$, which exists as 
$V_{\alpha-1} \not\le M$. Then 
$|V_{\alpha-1} : \bC_{V_{\alpha-1}}(z)| \le 8$ or $K \cong 3\cdot M_{22}$ 
and $|V_{\alpha-1} : \bC_{V_{\alpha-1}}(z)| \le 32$. Application of 
(1.25) shows $K \cong M_{22}$, or $3\cdot M_{22}$ and 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(K_{\alpha-1})| = 2$ in the former. But 
then $|\tilde{V}_1 : \bC_{\tilde{V}_1}(t)| = 2$, 
which implies $t \in M$, a contradiction.

So we are left with $K_{\alpha-1} \cong 3\cdot M_{22}$ and 
$\tilde{V}_{\alpha-1}$ involves just 12 - dimensional modules by (1.13) 
anyway. Furthermore
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(K_{\alpha-1})| \le 16$ and by (1.25) 
$|\tilde{V}_{\alpha-1} : \bC_{\tilde{V}_{\alpha-1}}(K)| \ge 8$. As 
$Z_1 \not\le V_{\alpha-1}$, we have $V_{\alpha-1} \cap Q_1 = 1$.
\
The 12 - dimensional $3\cdot M_{22}$ - module $W$ is not an 
F$_1$ - module. Otherwise an offending group $A$ is contained in 
$3\cdot M_{22}$, as $|W : \bC_W(x)| = 2^6$ for 
$x \in {\mbox {Aut}}(3\cdot M_{22}) \setminus 3\cdot M_{22}$. Hence as 
$W$ is not an F - module by (1.18), we see that $|A| = 8$ and 
$|W : \bC_W(A)| = 16 = |W : \bC_W(a)|$, $a \in A^\sharp$. Now 
$U = \langle \bC_{3\cdot M_{22}}(a) \,\Big | \,a \in A^\sharp \rangle 
\cong 2^4\cdot A_6$. But then we get that 
$|W : \bC_W(O_2(U))| = 16$, a contradiction to (1.18) again.

Hence $\tilde{V}_{\alpha-1}$ is not an F$_1$ - module for 
$3\cdot M_{22}$ and we get that 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(K_{\alpha-1})| \le 8$.
So we see that $|\tilde{V}_1 : \bC_{\tilde{V}_1}(K_{\alpha-1})| = 8$, 
$|\tilde{V}_{\alpha-1} : \bC_{\tilde{V}_{\alpha-1}}(K)| = 16$. 
But now $[\tilde{V}_1, Z(K_{\alpha-1})] = 1$ and then
$|\tilde{V}_{\alpha-1} : \bC_{\tilde{V}_{\alpha-1}}(\tilde{V}_1)| 
= 2^{2f}$, as $Z(K_{\alpha-1})$ acts
fixed point freely. But this contradicts 
$|\tilde{V}_{\alpha-1} : \bC_{\tilde{V}_{\alpha-1}}(\tilde{V}_1)| = 32$. 
This proves (1).

\absa
Now assume that for some $t \in V_{\alpha-1} \cap L_1$ we have $[K,t] \not= 1$.
If $K/Z(K) \cong U_4(3)$, and $Z(K) \not= 1$, we get with (1.55) that 
$t$ centralizes $F \cong E_{27}$ in $K$.
Now some $1 \not= u \in [\tilde{V}_1,t]$ is centralized by 
$E \cong E_9$, implying $Z_\alpha \le M$, a contradiction.

Suppose  $m_p(K) \ge 2$. If $K \cong G(r)$, then 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(K)| = 2$.
As $|\tilde{V}_1/Z_1 : \bC_{\tilde{V}_1/Z_1}(Z_\alpha)| > 2$, we see 
that
$\tilde{V}_1 \cap \bC_{L_{\alpha-1}/Q_{\alpha-1}}(K_{\alpha-1}) \not= 1$. 
Hence there is some component $K_{\alpha-1}^{\!(1)}$ with 
$[K_{\alpha-1}^{\!(1)}, K_{\alpha-1}] = 1$ and 
$[K_{\alpha-1}^{\!(1)}, \tilde{V}_{\alpha-1}] = 1$. The same holds for 
$\tilde{V}_1$. Furthermore as $M \not= M_{\alpha-1}$, we get with (4.2) 
that $Z_{\alpha-1} \cap \tilde{V}_1 = 1$
and so $|\tilde{V}_1 : \bC_{\tilde{V}_1}(t)| \le 2$, a contradiction.

So let now $K$ be sporadic. Suppose 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(t)| \le 2^u$. Then we get with (1.25)
and (1.11) that $u \ge 3$ and $K \cong 3\cdot M_{22}$, in which case 
$u = 4$. Hence $|\tilde{V}_1 : \bC_{\tilde{V}_1}(K_{\alpha-1})| \le 16$.
Now $\tilde{V}_1$ involves just one nontrivial irreducible module, the 
12 - dimesional module. If $m_p(K) = 1$, we get that 
$[\tilde{V}_1, E] = 1$ for some  
$E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \le M$, and so 
$Z_\alpha \le \bC([\tilde{V}_1, V_{\alpha-1}]) \le M$, a contradiction. 
This proves $p = 3$ and so $q = 2$. But now as before we get a 
contradiction.

So we have $|\tilde{V}_1 : \bC_{\tilde{V}_1}(t)| > 2^u$, in particular 
$\tilde{Z}_{\alpha-1} \cap \tilde{V}_1 \not= 1$. In this case by
(4.2) $|\tilde{V}_1 : \tilde{V}_1 \cap Q_{\alpha-1}| \le 2^u$, and so
$|\tilde{V}_1/Z_1 : \bC_{\tilde{V}_1/Z_1}(Z_{\alpha-1})| \le 2^u$. As 
$Z_{\alpha-1} \not\le M$, we see $u \ge 2$, $p = 3$ in case of $u = 2$. 
If $p = 3$, then as before we see $q = 2$. 
Furthermore we get $K \cong 3\cdot M_{22}$ in any case
and $|\tilde{V}_1 : \bC_{\tilde{V}_1}(K_{\alpha-1})| \le 16$ and 
$\tilde{V}_1$ just involves 12 - dimensional modules. We have 
$[\omega, K] = 1$ and 
$|\tilde{V}_1Z_2/Z_2 : \bC_{\tilde{V}_1Z_2/Z_2}(\omega)| \le 16$. 
Hence $|\tilde{V}_1 : \tilde{V}_1 \cap Z_2| \le 16$ for $\omega \not= 1$.
But $K \not\lesssim L_2(q)$, so we see $\omega = 1$. Now we may argue as 
before.

Hence we have $[\tilde{V}_{\alpha-1} \cap L_1, K] = 1$. If $m_p(K) \ge 2$, 
then $[\tilde{V}_{\alpha-1} \cap L_1, V_1]$ is centralized by 
$K$, so $V_1$ induces transvections on $\tilde{V}_{\alpha-1}$, 
a contradiction.

As $Z_\alpha \not\le M$, we have $|Z_\alpha : Z_\alpha \cap Q_2| = 2$ 
for $q = 2$ and so $\tilde{V}_{\alpha-1} \le Q_2$ in any case.
Now we have $[\tilde{V}_1, \tilde{V}_{\alpha-1}] = 1  $ and so 
$\tilde{V}_1 \le \bC(K_{\alpha-1})$.

\absa
{\bf~(7.21) Lemma.~}{\it Let $b > 2$. If $K \cong A_n$, $n = 7$ or 
$n \ge 9$, then $m_p(K_{\alpha-1}) = 1$, 
$\tilde{V}_{\alpha-1} \le M$ and $[\tilde{V}_{\alpha-1} \cap L_1, K] = 1$.
Furthermore $[\tilde{V}_1, K_{\alpha-1}] \le Q_{\alpha-1}$.}

\absa
Proof.~
\\
\setcounter{equation}{0}
\begin{equation}
q = 2
\end{equation}

\absa
Suppose $q > 2$. We have $[\omega, K] = 1$. Hence $m_p(K) = 1$ by (1.63) 
and so $p \not= 3$.
Now $[\tilde{V}_1, \omega] = \tilde{V}_1$ and as $n \le 11$, we get that
$|\tilde{V}_1/Z_1 : \bC_{\tilde{V}_1/Z_1}(\omega)| \le 2^5$, and so
$|\tilde{V}_1 : \tilde{V}_1 \cap Z_2| \le 2^5$. As 
$K \not\lesssim L_2(q)$, we get $q = 4$. But also 
$K \not\lesssim SL_3(4)$,
a contradiction.

\absa
Suppose next $p = 3$\\
\begin{equation}
|\tilde{V}_1 : \bC_{\tilde{V}_1}(K_{\alpha-1})| \le 2
\end{equation}

\absa
Suppose false. Then 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(K_{\alpha-1})| \ge 4$. Suppose 
furthermore that $\tilde{V}_1$ has a natural submodule $T$. In such a 
module every element is centralized by an elementary abelian group of 
order 9 in $K$, or $K \cong A_7$.

In the former $[\tilde{V}_{\alpha-1} \cap L_1, K] = 1$. As 
$[\tilde{V}_{\alpha-1} \cap Q_1, V_1] = 1$, we get that $\tilde{V}_1$ 
induces transvections on $\tilde{V}_{\alpha-1}$, a contradiction to 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(K_{\alpha-1})| \ge 4$.

Thus we have $K \cong A_7$. Again for some $t \in \tilde{V}_{\alpha-1}$ 
we have $[K,t] \not= 1$. Now $|[\tilde{V}_1, t]| \le 8$. This
implies that $\tilde{V}_1/\bC_{\tilde{V}_1}(K)$ is irreducible. But 
$m_3(K) = 2$. So there is some $\nu$, $o(\nu) = 3$, with 
$[\tilde{V}_1, \nu] = 1$. Now also every $x \in \tilde{V}_1$ is
centralized by $E \cong E_9$ in $L_1$, a contradiction.

So we have that $\tilde{V}_1/\bC_{\tilde{V}_1}(K)$ contains no natural 
submodule. By (1.12) this implies 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(K_{\alpha-1})| \le 8$. Let now 
$z \in \tilde{V}_1$. Then we see that 
$|\tilde{V}_{\alpha-1} : \bC_{\tilde{V}_{\alpha-1}}(z)| \le 16$. This 
shows that
$\tilde{V}_{\alpha-1}/\bC_{\tilde{V}_{\alpha-1}}(K_{\alpha-1})$ is a 
spin module and $n = 9$. Now we see that 
$Z_{\alpha-1} \le [t, \tilde{V}_1]$ for every 
$1 \not= t \in \tilde{V}_{\alpha-1} \cap L_1$. As every element in the spin
 module is centralized by some 3 - element in $A_9$, we get with (4.2) 
$M = M_{\alpha-1}$. But this contradicts $Z_\alpha \not\le M$.

\begin{equation}
p > 3
\end{equation}

\absa
If $p = 3$, then by (2) 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(K_{\alpha-1})| \le 2$. Now
$\tilde{V}_{\alpha-1} \le M$. As $p = 3$ , we have 
$L_2/Q_2 \cong \Sigma_3$ and so $\tilde{V}_{\alpha-1} \le L_1$. 
As $Z_\alpha \not\le M$, we get $[\tilde{V}_{\alpha-1}, K] \not= 1$.
But $\tilde{V}_{\alpha-1}$ has to induce transvections. This implies 
that $[\tilde{V}_{\alpha-1}, \tilde{V}_1]$ is centralized by 
$E \cong E_9$ in $L_1$, a contradiction.

\absa
Now by (3) we have $n \le 11$.

\begin{equation}
|\tilde{V}_1 : \bC_{\tilde{V}_1}(K_{\alpha-1})| \le 4
\end{equation}

\absa
Suppose false. Then we get 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(K_{\alpha-1})| \ge 8$. Hence
by (1.12)  $\tilde{V}_1$ involves just natural modules or 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(K_{\alpha-1})| = 8$ and
spin modules are involved.

As $[\tilde{V}_1, K_{\alpha-1}] \not= 1$, we see that 
$[\tilde{V}_{\alpha-1} \cap L_1, K] \not= 1$.
This shows that there are at most two natural modules
and at most one spin module involved. In both cases we see 
$[\bC_{L_1/Q_1}(K), \tilde{V}_1] = 1$. This
now shows $m_p(K) = 2$, i.e. $p = 5$, and some $\nu$ with $o(\nu) = 5$ 
centralizes $\tilde{V}_1$. Furthermore $n = 10$, or $n = 11$. 
But in the spin module for $A_n$ we get
$|[\tilde{V}_1, t]| > 2^4$ for 
$t \in \tilde{V}_{\alpha-1} \cap L_1$, $[t, K] \not= 1$,
a contradiction. Hence just natural modules are involved. In the 
natural module for $A_{10}$ or $A_{11}$ every element is 
centralized by some \ \ 5 - element in $K$. This implies that every 
$x \in \tilde{V}_1$ is centralized by $E \cong E_{25}$ with 
$\Gamma_{E,1}(G) \le M$, which contradicts $Z_\alpha \not\le M$. 
This proves (4).
  
\absa
Now as $p > 3$ we get from (4) that $\tilde{V}_{\alpha-1} \le M$. Suppose 
$[\tilde{V}_{\alpha-1} \cap L_1,K] \not= 1$. This is obviously true if 
$m_p(K) = 2$. We have that 
$|\tilde{V}_1/Z_1 : \bC_{\tilde{V}_1/Z_1}(Z_{\alpha-1})| > 4$ by (3). 
So by (4) there is some 
$1 \not= t \in \tilde{V}_1$, $[t, K_{\alpha-1}] = 1$. Hence
$p \,\Big | \,|\bC(\tilde{V}_1)|$. This implies that 
$Z_{\alpha-1} \cap \tilde{V}_1 = 1$ by (4.2).
Now by (4) $|[\tilde{V}_1, t]| \le 4$ for 
$t \in \tilde{V}_{\alpha-1} \cap L_1$, $[K,t] \not= 1$. So  
$\tilde{V}_1$ just involves spin modules or natural 
modules, by (1.30). Furthermore there are at most two natural 
modules involved. But if we have a natural submodule, 
then as above every $x \in \tilde{V}_1$ is centralized
by $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \le M$, which contradicts 
$Z_\alpha \not\le M$.

So we have the spin module as a submodule and now 
$\tilde{V}_1/\bC_{\tilde{V}_1}(K)$ is the spin module. 
This implies $m_p(K) = 2$, otherwise we may argue as before. 
Hence $p = 5$, $K \cong A_{10}$ or $A_{11}$. As $[\omega, K] = 1$ and
$[\omega, \tilde{V}_1] = \tilde{V}_1$ for $q > 2$, we see $q = 2$, and 
so for $z \in \tilde{V}_1$ we have
$|\tilde{V}_{\alpha-1} : \bC_{\tilde{V}_{\alpha-1}}(z)| \le 8$. But then 
as $|[\tilde{V}_{\alpha-1}/\bC_{\tilde{V}_{\alpha-1}}(K_{\alpha-1}), z]|^2 = 
|\tilde{V}_{\alpha-1}/\bC_{\tilde{V}_{\alpha-1}}(K_{\alpha-1})|$, 
we get $|\tilde{V}_{\alpha-1}/\bC_{\tilde{V}_{\alpha-1}}(K_{\alpha-1})| 
= 2^6$, a contradiction.

So we have $m_p(K) = 1$ and $[L_1 \cap \tilde{V}_{\alpha-1}, K] = 1$. 
Finally $[\tilde{V}_1, K_{\alpha-1}] \le Q_{\alpha-1}$ follows as in 
(7.20).

\absa
{\bf~(7.22) Lemma.~}{\it Let $b > 2$, $K \cong G(r)$, $r$ even. Then 
$m_p(K) = 1$, and $[\tilde{V}_{\alpha-1} \cap L_1, K] = 1$.}

\absa
Proof.~If $\nu \in K$, $o(\nu) = p$, then $\bC_{Z_2}(\nu) = Z_1$ 
by (4.2). We will assume $1 \not= [\tilde{V}_{\alpha-1} \cap K ] \le K$.

Set $\omega_1 = \omega$, or $\omega^3$ in case of $q = 64$. Let 
$[\omega_1,K] = 1$, $o(\omega_1) > 1$. Then by (1.63) $m_p(K) = 1$. 
Hence we have $K \cong L_2(r)$, $L_3(r)$, $U_3(r)$, or $Sz(r)$. 
Furthermore $[\omega_1, \tilde{V}_1] = \tilde{V}_1$. We get for 
$t \in \tilde{V}_{\alpha-1}$ that $|[\tilde{V}_1, t]| \le r$ or 
$r^2$ in case of $K \cong L_3(r)$. 
This now implies with (1.26) and (1.27) that $K \cong L_2(r)$ or 
$L_3(r)$ and by (1.26), (1.27) just natural or natural and 
dual modules are involved. There is $E \cong E_{p^2}$ with $E \le N(K)$ and
$E \cap K = 1$. Now $E$ acts on 
$\bC_{\tilde{V}_1/\bC_{\tilde{V}_1}(K)}(S \cap K)$ and 
so some $1 \not= x \in [\tilde{V}_1, t]$ is centralized by 
$F \cong E_{p^2}$ with $\Gamma_{F,1}(G) \le M$, and so $Z_\alpha \le M$, 
a contradiction (Recall that 
$\langle K, E \rangle \le \bC_{L_1/Q_1}(Z_1)$ by (7.2)). So we have

\setcounter{equation}{0}
\begin{equation}
{\mbox {If}}\,\, q > 2, \,\, {\mbox {then}} \,\, [\omega_1,K] \not= 1
\end{equation}

As $[\omega, Z_1] \not= 1$, we have $\omega \not\in K$. Hence (1.65)(i), 
(iii), or (v) holds. We have that (1.65)(vi) is not possible as in that 
case $q = 4$ but $3 \in \sigma(M)$. 
Now application of (1.64) shows $q \le r$, or $q = r^2$, 
$K \cong (S)U_n(r)$, $\Omega_{2n}^-(r)$, $^2E_6(r)$, or $^3D_4(r)$.

If $t \in \tilde{V}_{\alpha-1} \cap L_1$, $[t,K] \not= 1$, then 
$|[\tilde{V}_1/Z_1, t]| \le |\tilde{V}_1 : 
\bC_{\tilde{V}_1}(K_{\alpha-1})|u$, where $u = r$, or 
$u = r^2$ for $K \cong (S)U_n(r)$, $\Omega_{2n}^-(r)$, $^2E_6(r)$, or 
$u = r^3$ for $K \cong$ $^3D_4(r)$. Suppose $p \,\Big | \,r^2 - 1$.
\\
\\
\\
\\
\hspace{-0.5cm}(2)~~\parbox{12cm}{ $\tilde{V}_1$ involves some 
nontrivial irreducible module which is not a fundamental module.}
\\
\\

\absa
Suppose false. We have that $V(\lambda)$ for every fundamental weight 
$\lambda$ has to be involved. Application of (1.23) shows that 
$K \cong (S)L_n(r), n \le 4$, $(S)U_n(r), n \le 4$, $Sp_4(r)$, $Sp_6(r)$, 
or $Sz(r)$. 

If $K \cong Sp_6(r)$, then 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(K_{\alpha-1})| = r^6$ and then 
$\tilde{V}_1$ cannot act quadratically on the natural module.

Let $K \cong Sp_4(r)$. As no group of order $r^3$ act quadratically on 
both \,\, 4 - dimensional modules, we see 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(K_{\alpha-1})| = r^2$. 
But now $\tilde{V}_1$ contains some element $x$ whith 
$|\tilde{V}_{\alpha-1} : \bC_{\tilde{V}_{\alpha-1}}(x)| \ge r^4$, 
which implies $|\tilde{V}_{\alpha-1} \cap L_1 : 
\bC_{\tilde{V}_{\alpha-1} \cap L_1}(K)| = r^3$, a contradiction.

Let $K \cong (S)U_4(r)$. Then because of the natural 
$\Omega_6^-(r)$ - module we get 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(K_{\alpha-1})| = r^3$. Now 
$\tilde{V}_1$ contains $x$ which does not induce a 
transvection on the natural $SU_4(r)$ - module. Hence 
$|\tilde{V}_{\alpha-1} : \bC_{\tilde{V}_{\alpha-1}}(x)| \ge r^6$. But 
then $|\tilde{V}_{\alpha-1} \cap L_1 : 
\bC_{\tilde{V}_{\alpha-1} \cap L_1}(K)| = r^4$, a contradiction.

Let $K \cong (S)L_4(r)$. Because of the natural 
$\Omega_6^+(r)$ - module, we have 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(K_{\alpha-1})| = r^3 \ge 
|\tilde{V}_{\alpha-1} 
\cap L_1 : \bC_{\tilde{V}_{\alpha-1} \cap L_1}(K)|$. Hence 
$\tilde{V}_1$ consists of transvections. As the natural and the dual 
module are involved, we see
 $|\tilde{V}_{\alpha-1} : \bC_{\tilde{V}_{\alpha-1}}(\tilde{V}_1)| \ge r^7$, 
a contradiction.

So we have $K \cong L_2(r)$, $SL_3(r)$, $SU_3(r)$, or $Sz(r)$. 
Furthermore just natural or dual modules are involved.

Suppose first there is $E \cong E_{p^2}$, $E \le L_1$, $[K,E] = 1$. If 
$[\tilde{V}_1, \nu] =1 $ for some $\nu \in E^\sharp$, we get that 
$Z_{\alpha-1} \cap \tilde{V}_1 = 1$ and so 
$|[\tilde{V}_1, t]| \le |\tilde{V}_1 : \bC_{\tilde{V}_1}(K_{\alpha-1})|$. 
This implies $K \cong L_2(r)$ and $\tilde{V}_1/\bC_{\tilde{V}_1}(K)$ 
is the natural module. As $[E, \tilde{V}_1] \not= 1$, we get 
$p \,\Big | \,r -1 $. But now $t$ acts on $F \cong E_{p^3}$ and some 
$1 \not= U \le [\tilde{V}_1, t]$ is normalized by 
$\tilde{E} \le F$, $|\tilde{E}| = p^2$, a contradiction.

So we have that $E$ acts faithfully on 
$\tilde{V}_1/\bC_{\tilde{V}_1}(K)$. In particular 
$\tilde{V}_1/\bC_{\tilde{V}_1}(K)$ is not irreducible and so 
$K \cong L_2(r)$ or $SL_3(r)$. Furthermore $r = q$ and 
$p \,\Big | \,r - 1$. If $K \cong SL_3(r)$, we get that $\omega$ 
centralized $F \cong E_{p^2}$ with $\Gamma_{F,1}(G) \le M$, and so 
$N_G(\langle \omega \rangle) \le M$, a contradiction. So we have 
$K \cong L_2(r)$ and $\bC_{L_1/Q_1}(K) \lesssim GL_2(r)$. Hence 
$t \in \tilde{V}_{\alpha-1} \cap L_1$ normalizes 
$F \cong E_{p^3}$, a contradiction as before. 

Suppose now $m_p(K) = 1$. Then $m_p(\bC_{L_1/Q_1}(K)) = 1$ and there is 
some $\nu \in L_1$ with $o(\nu) = p$ which induces a field automorphism on 
$K$. By (4.12) we have $K \cong L_2(r)$ and 
$\bC_{L_1/Q_1}(K)$ involves $L_2(r)$ too. But then $t$ acts on 
$E \cong E_{p^3}$. This implies that there is some 
$1 \not= U \le [\tilde{V}_1,t]$ normalized by 
$F \cong E_{p^2}$ with $\Gamma_{F,1}(G) \le M$, a contradiction. 
We have that $\bC_{L_1/Q_1}(K)$ acts faithfully on 
$\tilde{V}_1/\bC_{\tilde{V}_1}(K)$.

We are left with $m_p(K) = 2$. Hence $K \cong SL_3(r)$, $ p \,\Big | \,r-1$, 
or $K \cong SU_3(r)$, $p \,\Big | \,r+1$. In the first case $\omega$ 
centralizes $E \cong E_{p^2}$ with $E \le K$, and so 
$N_G(\langle \omega \rangle) \le M$, a contradiction. Hence 
$K \cong SU_3(r)$ and $Z_{\alpha-1} \cap \tilde{V}_1 \not= 1$. But 
$t \in K$ acts on some $E_{p^2}$ and so $t$ acts on some $E_{p^3}$, 
a contradiction.
\absa
\setcounter{equation}{2}
\begin{equation}
p \not\Big | \,r^2 - 1
\end{equation}
\absa
Suppose false. By (2) we may assume that some $T$ not isomorphic to 
$V(\lambda)$ 
is involved in $\tilde{V}_1$. By (1.14) $\tilde{V}_1$ is not strongly 
quadratic. By (1.15) $\tilde{V}_1$ is an SC - module or 
$K \cong L_2(r)$, $U_3(r)$, or $Sz(r)$. Now application of (1.22) shows 
that for $K \cong G_l$, $V(\lambda)$, $\lambda \not= \lambda_1, 
\lambda_l$, has to be involved in $\tilde{V}_1$. 
Application of (1.38) and (1.26) show $K \cong L_2(r)$, or $SL_3(r)$. 
But then we may argue as above to get that 
$[\tilde{V}_1, \tilde{V}_{\alpha-1}]$ contains some 
$1 \not= x$ with $\bC_G(x) \le M$. Now $Z_\alpha \le M$, a contradiction.

\absa
\begin{equation}
{\mbox {If}} \,\,1 \not= [\tilde{V}_{\alpha-1} \cap L_1, K] \le K, 
\,\, {\mbox {then}} \,\, m_p(K) = 1
\end{equation}

\absa
Otherwise by (3) $p \not\Big | \,r^2 - 1$. Let $K \cong (S)L_n(r)$. Then 
we get for $r > 2$ that $n \le 4$ and $m_p(K) = 1$. If $r = 2$, then 
$p \not= 3$ and so $n \le 7$ and $m_p(K) = 1$ or $p = 7$ and $n = 6$ or 
$7$.

Suppose $m_p(K) = 2$. Then $\tilde{V}_1$ is either strongly quadratic or 
an SC - module by (1.15). Now $Z_2$ is not centralized by $\nu \in K$ with 
$o(\nu) = 7$. If $n = 7$ this implies that $V(\lambda_3)$ 
is involved. By (1.23) $V(\lambda_4)$ is not involved and so also 
$V(\lambda_5)$ is not involved, a contradiction.

Let $n = 6$. Then assume that $V(\lambda_3)$ is involved. Now neither 
$V(\lambda_2)$ nor $V(\lambda_4)$ are involved. Furthermore by (1.23) 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(K_{\alpha-1})| \ge 2^7$ 
and so by (1.15) and (1.14) just $V(\lambda)$, $\lambda$ a fundamental 
weight are involved in $\tilde{V}_1$. But we have that neither 
$V(\lambda_2)$ nor $V(\lambda_4)$ are involved.  Hence both 
$V(\lambda_1)$ and $V(\lambda_5)$ are involved. Now 
$\tilde{V}_{\alpha-1} \cap L_1/\bC_{\tilde{V}_{\alpha-1} \cap L_1}(K)$ 
is a transvection group in $K$. This implies 
$|\tilde{V}_{\alpha-1} \cap L_1 : \bC_{\tilde{V}_{\alpha-1} 
\cap L_1}(K)| \le 2^5$. As $[\omega, K] = 1$, 
we get $q = 2$. Hence for $z \in \tilde{V}_1$, we have 
$[\tilde{V}_{\alpha-1}, z]| \le 2^6$, contradicting (1.23).

So we have that $V(\lambda_3)$ is not involved. Then both $V(\lambda_2)$ 
and $V(\lambda_4)$ are involved. Now each $t \in \tilde{V}_{\alpha-1}$ 
with $[K,t] \not= 1$ induces an element on $V(\lambda_2)$ 
which centralizes a subspace of codimension 4. But these are exactly the 
transvections in $L_6(2)$ and we get the same contradiction as before.

Let $K \cong (S)U_n(r)$, we get $K \cong (S)U_4(r)$ or $U_5(4)$. Now 
$m_p(K) = 1$. If $K \cong Sp_{2n}(r)$, then $n \le 3$ and again 
$m_p(K) = 1$. If 
$K \cong \Omega_8^-(r)$ or $G_2(r)$, then as $p \,\not\Big | \,r^2 - 1$, we get 
$m_p(K) = 1$

\begin{equation}
{\mbox {If}} \,\, 1 \not= [\tilde{V}_{\alpha-1} \cap L_1, K], 
\,\, {\mbox {then}} \,\, [\tilde{V}_{\alpha-1} \cap L_1, K] \not\le K
\end{equation}

Suppose false. By (4) we have $m_p(K) = 1$. Furthermore 
$p \not\Big | \,r^2 - 1$. Let $E \cong E_{p^2}$ with $E \le \bC_{L_1}(Z_1)$, 
$[K,E] \le K$ and $E \cap K = 1$. Let $T_1$ be a submodule of 
$K$ in $\tilde{V}_1/\bC_{\tilde{V}_1}(K)$. Then $[E, T_1] \not= 1$. 
Suppose $[E, T_1] \le T_1$, then as $p \not\Big | \,r^2 - 1$ by (3) we 
have $\nu \in E$, $o(\nu) = p$ with $[\nu, T_1] = 1$ and $\nu_1 \in E$, 
$o(\nu_1) = p$, which induces a field automorphism on $K$. But now we 
may assume that $\nu_1$ acts on $[T_1,t]$ and so 
$\langle \nu, \nu_1 \rangle$ centralizes some element in 
$[\tilde{V}_1, \tilde{V}_{\alpha-1}]^\sharp$, a contradiction.

So we have $[E, T_1] \not\le T_1$. As $p \not\Big | \,r^2 - 1$ by (3), we 
see that there are at least three modules isomorphic to $T_1$ involved 
in $\tilde{V}_1$ or $K \cong (S)U_n(r)$, $n = 4$, or $5$, 
$p \,\Big | \,r^2+1$, and there are exactly two such modules.

Let first $K \cong \Omega_8^-(r)$. By (1.22) and (1.23) $T_1$ is the 
natural module. As $|\tilde{V}_1 : \bC_{\tilde{V}_1}(K_{\alpha-1})| > r$, 
we get with (1.45) that 
$|\tilde{V}_{\alpha-1} : \bC_{\tilde{V}_{\alpha-1}}(\tilde{V}_1)| \ge r^9$ 
and so $|\tilde{V}_{\alpha-1} \cap L_1 : \bC_{\tilde{V}_{\alpha-1} 
\cap L_1}(K)| \ge r^7$, a contradiction.

Let $K \cong Sp_6(r)$. We have 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(K_{\alpha-1})| \ge r^2$. As a 
transvection group is of order $r$, we see that $|\tilde{V}_{\alpha-1} 
\cap L_1 : \bC_{\tilde{V}_{\alpha-1} \cap L_1}(K)| \ge r^5$. Hence also 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(K_{\alpha-1})| \ge r^5$ and 
$\tilde{V}_1$ centralizes in $T_1$ a subgroup of index $r^2$. 
But this contradicts (1.46).

Let now $K \cong Sp_4(r)$. Then $p \,\Big | \,r^2 + 1$ and so there are four 
nontrivial irreducible modules involved. These have to be natural and 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(K_{\alpha-1})| = r^3$. 
But $\tilde{V}_{\alpha-1} \cap L_1$ has to induce transvections on $T_1$ 
and so $|\tilde{V}_{\alpha-1} \cap L_1 : \bC_{\tilde{V}_{\alpha-1} 
\cap L_1}(K)| \le r$. Hence for $z \in \tilde{V}_1$ we have 
$|\tilde{V}_{\alpha-1} : \bC_{\tilde{V}_{\alpha-1}}(z)| \le r^2$, a 
contradiction.

Let $K \cong (S)L_n(r)$. If $n = 6$ or $7$, then $p \ge 5$ and so there 
are at least four nontrivial irreducible modules involved. By (1.47) 
these are natural
 modules. Let $n \le 5$, then also by (1.47) these are natural.

Suppose $T_1$ to be the natural module. Assume $N_E(T_1) = 1$. Then we 
have at least six nontrivial irreducible modules involved and so 
$n \ge 5$. Let $n = 5$.
 Then $|\tilde{V}_{\alpha-1} \cap L_1 : \bC_{\tilde{V}_{\alpha-1} 
\cap L_1}(K)| \le 2^6$ and $q = 2$ by (1). This implies that 
$\tilde{V}_1$ induces transvections on 
$\tilde{V}_{\alpha-1}$. Hence 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(K_{\alpha-1})| \le 2^4$. But now 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(t)| \le 2^5$, a contradiction.

If $n > 5$, then $p \not= 7$, and so there are at least 8 modules 
involved. Again $\tilde{V}_1$ induces transvections. So 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(K_{\alpha-1})| \le 2^6$, 
a contradiction again.

Now we have $|N_E(T_1)| = p$. As all elements in $T_1^\sharp$ are 
conjugated under $K$, we get that $p \not\!\!\Big | \,\,r^m - 1$, $0 < m < n$. 
This implies that there are $n$ modules involved. Furthermore 
$\tilde{V}_{\alpha-1} \cap L_1$ induces transvections on $T_1$. This 
implies $|\tilde{V}_{\alpha-1} \cap L_1 : \bC_{\tilde{V}_{\alpha-1} 
\cap L_1}(K)| \le r^{n-1}$. This now yields 
$|\tilde{V}_{\alpha-1} \cap L_1 : \bC_{\tilde{V}_{\alpha-1} \cap L_1}(K)| 
= r^{n-1} = |\tilde{V}_1 : \bC_{\tilde{V}_1}(K_{\alpha-1})|$ and $q = r$.

We have $n > 2$ as $p \not\!\Big | \,r^2 - 1$. Then by (1.16) $t$ fixes any 
$T_1^\nu$. Hence $t$ acts as some $s \in K$ on 
$T_1 \oplus \ldots \oplus T_p$. But now 
$s$ inverts some $\mu \in K$ which is centralized by $F \cong E_{p^2}$. 
Hence $F$ normalizes some $1 \not= U \le [[\mu, \tilde{V}_1],t]$, a 
contradiction.

Let finally $K \cong (S)U_n(r)$, $n = 4$ or $5$. Now 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(K_{\alpha-1})| \le r^4$ and so there 
are at most three natural modules involved. Now 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(K_{\alpha-1})| \ge r^4$. 
But then $\tilde{V}_1$ contains some $x$ with 
$|T_1 : \bC_{T_1}(x)| = r^4$ and so $|\tilde{V}_{\alpha-1} : 
\bC_{\tilde{V}_{\alpha-1}}(x)| \ge r^8$, a contradiction as $q \le r^2$.

\begin{equation}
[\tilde{V}_{\alpha-1} \cap L_1, K] = 1
\end{equation}

Suppose $[\tilde{V}_{\alpha-1} \cap L_1, K] \not= 1$. Then by (5) there 
is some $t \in \tilde{V}_{\alpha-1} \cap L_1$ with $K^t \not= K$. Now 
$[\tilde{V}_{\alpha-1} \cap L_1, K \times K^t] \le K \times K^t$ 
by (1.11). Suppose first 
$|\tilde{V}_{\alpha-1} \cap L_1 : \bC_{\tilde{V}_{\alpha-1} 
\cap L_1}(K \times K^t)| > 2$. Then again by (1.11) 
$K \cong L_2(r)$, $r$ even, and $\tilde{V}_1$ involves the natural 
$\Omega_4^+(r)$ - module. Furthermore 
$|\tilde{V}_{\alpha-1} \cap L_1 : \bC_{\tilde{V}_{\alpha-1} 
\cap L_1}(K \times K^t)| \le 2r$.

Let $|\tilde{V}_{\alpha-1} \cap L_1 : 
\bC_{\tilde{V}_{\alpha-1} \cap L_1}(K \times K^t)| = 2$. If 
$m_p(K) \ge 2$, then $t$ acts on an elementary abelian group 
of order $p^4$ in $K \times K^t$, which implies $Z_\alpha \le M$, a 
contradiction. Hence $K \cong L_2(r)$, $Sz(r)$, or $L_3(2)$.

In both cases we have $[\omega_1, K \times K^t] \not= 1$, as 
$N_G(\langle \omega_1 \rangle) \not\le M$. Hence in the first case 
$q \le r$. This then implies $q = r$ and 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(K_{\alpha-1})| = r$.  We see that 
$Z_{\alpha-1} \le \tilde{V}_1$ and so $Z_2 \le \tilde{V}_1$. Hence 
$p \not\!\Big | \,|\bC_{L_1/Q_1}(K \times K^t)|$ as 
$\tilde{V}_1/\bC_{\tilde{V}_1}(K \times K^t)$ is an irreducible 
$\langle K, t \rangle$ - module.

In the second case, as 
$[\bC_{\tilde{V}_{\alpha-1} \cap L_1}(K \times K^t), \tilde{V}_1] = 1$, 
we get $|\tilde{V}_{\alpha-1} : \bC_{\tilde{V}_{\alpha-1}}(\tilde{V}_1)| 
\le 2q$. This implies $q = r$ , $K \cong L_2(r)$, or $q = 2 = r$, 
$K \cong L_3(2)$. Now also in this case as $|[t, \tilde{V}_1]| \ge r^2$, 
we see that $Z_{\alpha-1} \le \tilde{V}_1$ and we get that 
$\tilde{V}_1/\bC_{\tilde{V}_1}(K \times K^t)$ is a direct sum of two 
natural modules and so as $p \not\Big | \,r -1 $, 
we also get $p \not\Big | \,|\bC_{L_1/Q_1}(K \times K^t)|$.

We see that $t$ inverts $\nu \in K \times K^t$ with $o(\nu) = p$. If 
$E \le \bC(\nu)$ with $E \cong E_{p^2}$ and $[t,E] \le Q_1$, then $E$ acts on 
$[[\nu, \tilde{V}_1],t]$, and so $Z_\alpha \le M$, 
a contradiction. Hence there is $\tilde{\nu}$, $o(\tilde{\nu}) = p$, with 
$[\nu, \tilde{\nu}] = 1$ and $\tilde{\nu}^t = \tilde{\nu}^{-1}$. 
Furthermore there is $\nu_1$, $o(\nu_1) = p$, with 
$[\nu_1, \langle \nu, \tilde{\nu} \rangle ] = 1$ and $[\nu_1,t] \in Q_1$. 
Hence we may assume that $[\tilde{\nu},[[\nu, \tilde{V}_1],t]] = 1$ 
and so $\langle \tilde{\nu}, \nu_1 \rangle$ acts on 
$1 \not= U \le [ \tilde{V}_1, t]$, a contradiction.

\absa
\begin{equation}
m_p(K) = 1
\end{equation}

\absa
Suppose $m_p(K) \ge 2$. Then 
$[\tilde{V}_{\alpha-1} \cap L_1, \tilde{V}_1] = 1$. By (1.65) and (1.64) 
we have that $\tilde{V}_1$ induces transvections on $\tilde{V}_{\alpha-1}$
 and $K \cong L_n(r)$, $Sp_{2n}(r)$, $\Omega_{2n}^\pm(r)$, or $SU_n(r)$, or 
$|\tilde{V}_{\alpha-1} : \bC_{\tilde{V}_{\alpha-1}}(\tilde{V}_1)| = r^2$ 
and $K \cong \Omega_{2n}^-(r)$ and also $\tilde{V}_1$ is the natural module.

Suppose next $p \,\Big | \,r^2 - 1$. Then we see that 
$p \not\Big | \,|SL_{n-1}(r)|$, $|Sp_{2n-2}(r)|$, $|\Omega_{2n-2}^\pm(r)|$, or 
$|SU_{n-2}(r)|$, respectively. As $m_p(K) \ge 2$ 
this implies $K \cong SU_3(r)$, $p \,\Big | \,r+1$. Furthermore $q = r^2$. 
Now we have $\omega_2 \in L_2 \cap L_1$, $o(\omega_2) = p$. But then as 
$\omega_2 \not\in \bC_{L_1}(Z_1)$, we see that 
$\omega_2$ centralizes $E \cong E_{p^2}$ in $\bC_{L_1}(Z_1)$ with 
$\Gamma_{E,1}(G) \le M$. Hence $N_G(\langle \omega_2 \rangle) \not\le M$, 
a contradiction.

So we have $p \not\Big | \,r^2 - 1$. Hence $K \cong L_6(2)$, or $L_7(2)$, 
and $p = 7$. But then 
$7 \,\Big | \,|\bC_K(\bC_{\tilde{V}_1/\bC_{\tilde{V}_1}(K)}(S \cap K))|$
, a contradiction.
\\

\absa
{\bf~(7.23) Lemma.~}{\it Let $b > 2$, $K$ be an exceptional component, 
i.e. $O_r(K) \not= 1$. Then $\tilde{V}_{\alpha-1} \le M$ and 
$[\tilde{V}_{\alpha-1} \cap L_1, K] =1$. 
Furthermore $[\tilde{V}_1, K_{\alpha-1}] \le Q_{\alpha-1}$ and $L_1$ is 
not as in (4.9)(ii).}

\absa
Proof.~We have $|\tilde{V}_1 : \bC_{\tilde{V}_1}(K_{\alpha-1})| \le 4$. 
Suppose we do not have (4.9)(ii). Then 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(K_{\alpha-1})| \le 2$ 
by (1.12) and so $\tilde{V}_{\alpha-1} = \langle t \,\Big | 
\,|[t,\tilde{V}_1] : [t, \tilde{V}_1] \cap Z_{\alpha-1}| \le 2 \rangle$. 
This immediately shows $\tilde{V}_{\alpha-1} \le M$. Now let 
$t \in \tilde{V}_{\alpha-1} \cap L_1$, $[K,t] \not= 1$. By (4.10) we may 
assume that $\tilde{V}_{\alpha-1} \cap L_1/\tilde{V}_{\alpha-1} \cap Q_1$ 
acts faithfully on $V_1$. Hence $V_1$ centralizes a subgroup of index 
four in $\tilde{V}_{\alpha-1}$, a contradiction. Thus 
$[\tilde{V}_{\alpha-1} \cap L_1, K] = 1$.

\absa
Suppose now that we have (4.9)(ii). Set 
$\hat{G_2} = \langle Q_1, Z_\alpha \rangle$. Then 
$\hat{G_2}/\hat{G_2} \cap Q_1$ contains $L_2(q)$. Furthermore there is 
$\hat{V_1} \le V_1$, $|V_1 : \hat{V_1}| \le 4$ such that 
$\hat{V_1} \unlhd G_1$, $[\hat{G_2}, \hat{V_1}] \le Z_2$. Now 
$[Q_1 \cap Q_3, \hat{V_1}] = 1$ and $\langle 
Q_1 \cap Q_2, Q_2 \cap Q_3 \rangle/(Q_1 \cap Q_3)$ involves nontrivial 
$L_2(q)$ - modules. This shows $q \le 4$. Now let 
$x \in Q_3 \cap Q_2 \setminus Q_1$. Then $|[V_1,x]| \le 16$. 
We have $[O_r(L_1), V_1] \not= 1$ by (1.51). So $r = 5$, as 
$|[O_r(L_1),x]| = r^2$. Furthermore $|[O_5(L_1), V_1]Z_1/Z_1| \le 2^{12}$ 
as there is $\nu \in O_5(L_1)$, $o(\nu) = 5$, $|[\nu, V_1]| = 16$. But 
in $GL_{12}(2)$ there is no subgroup of type $5^3L_2(5)$.

This shows $Q_1 \cap Q_2 = Q_1 \cap Q_3$. Now we have a subgroup 
$\tilde{G_2} \le G_2$ with $Q_1 \in \mbox{Syl}_2(\tilde{G_2})$ and 
$\tilde{G_2}/O_2(\tilde{G_2}) \cong L_2(q)$. Further no nontrivial 
characteristic subgroup of $Q_1$ is normal in $\tilde{G_2}$. Hence by 
\cite[Theorem 1]{Ste} $O_2(G_2)$ involves exactly one nontrivial 
irreducible $\tilde{G_2}$ - module. Hence this is $Z_2$. Then $V_1 \le O_2(\tilde{G_2})$ is normal in 
$\langle L_1, \tilde{G_2} \rangle = G$, a contradiction.

\absa
{\bf~(7.24) Lemma.~}{\it Let $b > 2$, $K_{\alpha-1}$ be a component of 
$L_{\alpha-1}/Q_{\alpha-1}$ with $1 \not= [K_{\alpha-1}, V_1]$. Then 
$[K_{\alpha-1}, V_1] \le K_{\alpha-1}$.}

\absa
Proof.~Let $z \in V_1$ with $K_{\alpha-1}^z \not= K_{\alpha-1}$. Set 
$\tilde{V}_{\alpha-1} = [V_{\alpha-1}, K_{\alpha-1}^z \times 
K_{\alpha-1}]$.   Suppose first 
$|V_1 : \bC_{V_1}(K_{\alpha-1}^z \times K_{\alpha-1})| = 2$. Then 
$\tilde{V}_{\alpha-1} \le M$. Let $K^{(1)}$ be the component of 
$L_1/Q_1$ corresponding to
 $K_{\alpha-1}^z$. As $Z_\alpha \not\le M$ there is some component 
$K^{(2)}$ with $[K \times K^{(1)}, K^{(2)}] = 1$ and 
$[\tilde{V}_1, K^{(2)}] = 1$. Hence $Z_{\alpha-1} \cap \tilde{V}_1 = 1$. 
This implies that for $t \in \tilde{V}_{\alpha-1} \cap L_1$, we have 
$|\tilde{V}_1 : \bC_{\tilde{V}_1}(t)| \le 2$. Suppose $[t, K \times K^{(1)}] 
\not= 1$, then we may asume $[t,K] \not= 1$ and $[t, \tilde{V}_1]$ is 
centralized by $K^{(1)} \times K^{(2)}$, contradicting $Z_\alpha \le M$. 
Hence we have $[\tilde{V}_{\alpha-1} \cap L_1, K \times K^{(1)}] = 1$. 
This implies $[V_1, \tilde{V}_{\alpha-1} \cap L_1] = 1$. Now $V_1$ 
induces transvections on $\tilde{V}_{\alpha-1}$, a contradiction.

Now we have $K_{\alpha-1} \cong L_2(r)$, $r$ even, and 
$|V_1 : \bC_{V_1}(K_{\alpha-1}^z \times K_{\alpha-1})| \le 2r$ by 
(1.11). Furthermore $\tilde{V}_{\alpha-1}$ 
involves natural $\Omega_4^+(r)$ - modules. Let 
$\omega \in L_1 \cap L_2$, as before. Let $q > 2$. If 
$[\omega_1, K \times K^{(1)}] = 1$, $\omega_1 = \omega$, 
or $q = 64$ and $\omega_1 = \omega^3$, then by (1.63) $K \cong L_2(4)$
 and $p = 5$. Suppose $\bC_{V_1}(K_{\alpha-1}^z \times K_{\alpha-1}) 
\le Q_{\alpha-1}$. 
Then $|V_1/Z_2 : \bC_{V_1/Z_2}(Z_\alpha)| \le 8$, contradicting $p = 5$ 
and $Z_\alpha \not\le M$. Hence there is some component $K_2$, 
$[K \times K^{(1)}, K_2] = 1$ and $[\tilde{V}_1, K_2] = 1$. 
This implies $Z_{\alpha-1} \cap \tilde{V}_1 = 1$ and so 
$|\tilde{V}_1 : \bC_{\tilde{V_1}}(t)| \le 8$ for 
$t \in \tilde{V}_{\alpha-1} \cap L_1$, $[K,t] \not= 1$. 
Then $|\tilde{V}_{\alpha-1} \cap L_1 : \bC_{\tilde{V}_{\alpha-1} 
\cap L_1}(K \times K^{(1)})| \le 2$ and so $|\tilde{V}_{\alpha-1} : 
\bC_{\tilde{V}_{\alpha-1}}(u)| \le 2$ for $u \in \tilde{V}_1$, as 
$\tilde{V}_{\alpha-1} \le L_1$ by $q > 2$. This now shows 
$[\tilde{V}_1, K_{\alpha-1} \times K_{\alpha-1}^z] = 1$ 
and so $[\tilde{V}_1, \tilde{V}_{\alpha-1}] = 1$. As no element in 
$Z_{\alpha-1}$ is centralized by some element of order 5 
in $K \times K^{(1)}$, we get that $|[t, V_1]| \le 8$ for 
$t \in \tilde{V}_{\alpha-1}$. Hence $[K \times K^{(1)}, [t, V_1]] = 1$ 
and so as $Z_\alpha \not\le M$
, we have $[\tilde{V}_{\alpha-1}, V_1] = 1$, a contradiction.

This now implies $q = 2$ or $[\omega_1, K \times K^{(1)}] \not= 1$. 
Hence in all cases $q \le r$. As no $p$ - element in $K \times K^{(1)}$ 
centralizes some element in $Z_2 \setminus Z_1$, 
we get that $p \,\Big | \,r+1$. Let now $t \in \tilde{V}_{\alpha-1} \cap L_1$, 
$[t, K \times K^{(1)}] \not= 1$. Suppose there is some 
$\nu \in \bC(K \times K^{(1)})$, $o(\nu) = p$. If 
$[\nu, \tilde{V}_1] = 1$, then $Z_{\alpha-1} \cap \tilde{V}_1 = 1$ and 
so $|[t,\tilde{V}_1]| \le 2r$. This implies $K^t = K^{(1)}$. 
But then there is some $1 \not= U \le [\tilde{V}_1,t]$, $U$ is 
normalized by $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \le M$, a 
contradiction to $Z_\alpha \not\le M$. So we have 
$[\nu, \tilde{V}_1] \not= 1$. In particular there are at least two 
nontrivial irreducible $K \times K^{(1)}$ - modules involved. 
Again $K^t = K^{(1)}$. Now for $x \in \tilde{V}_1$ we have 
$|\tilde{V}_{\alpha-1} : \bC_{\tilde{V}_{\alpha-1}}(z)| \le 2r$ which 
implies $[\tilde{V}_1, K_{\alpha-1} \times K_{\alpha-1}^z] = 1$, 
as $\tilde{V}_{\alpha-1}$ involves two nontrivial irreducible modules. 
But then $\bC_G([\tilde{V}_1, \tilde{V}_{\alpha-1}]) 
\le M_{\alpha-1}$. Now $\bC_{K \times K^{(1)}}(t)$ centralizes 
$[t, \tilde{V}_1]$ and so $\bC_{K \times K^{(1)}}(t) \le M_{\alpha-1}$. 
But then with (4.2) we get $M = M_{\alpha-1}$, a contradiction.

We get that $p \not\Big | \,|\bC_{L_1/Q_1}(K \times K^{(1)})|$. Hence 
there is some $\nu \in L_1$, $o(\nu) = p$, $\nu$ induces a field 
automorphism of order $p$ on $K \times K^{(1)}$. 
Now if $t \in K \times K^{(1)}$, we get that $t$ acts on 
$E \cong E_{p^3}$ and so there is some $F \le E$, $|F| = p^2$ and 
$1 \not= U \le [\tilde{V}_1, t]$ with $F \le N(U)$. 
This implies $Z_\alpha \le M$, a contradiction. So we have 
$\nu^t = \nu^{-1}$. As all involutions in $(K \times K^{(1)})
\langle \nu \rangle t$ are conjugate, we again have that $t$ normalizes 
some $E \cong E_{p^3}$, the same contradiction as before.

So we have shown that $[\tilde{V}_{\alpha-1} \cap L_1 , K \times K^{(1)}] 
= 1$. We get that $[K, [t, V_1]] = 1$ for 
$t \in \tilde{V}_{\alpha-1} \cap L_1$. Hence
 $Z_{\alpha-1} \cap [t, V_1] = 1$. This implies $|[t,V_1]| \le 2r$ and 
so $[K \times K^{(1)}, [t, V_1]] = 1$. As $Z_\alpha \not\le M$, we see 
$[\tilde{V}_{\alpha-1} \cap L_1, V_1] = 1$. 
Now we have the contradiction $|V_1 : \bC_{V_1}(K_{\alpha-1} 
\times K_{\alpha-1}^z)| \le 2$, as 
$|\tilde{V}_{\alpha-1} : \bC_{\tilde{V}_{\alpha-1}}(V_1)| \le r$.
\\

\absa
{\bf~(7.25) Lemma.~}{\it Let $b > 2$ and $K$ be a normal $p$ - subgroup in $L_1/Q_1$. Then $[\tilde{V}_{\alpha-1} \cap L_1, K] = 1$ and 
$m_p(K) = 1$, $\tilde{V}_{\alpha-1} \le M$ and 
$[\tilde{V}_1, K_{\alpha-1}] =1$.}

\absa
Proof. ~Let 
$t \in \tilde{V}_{\alpha-1} \cap L_1$ with $[t, K] \not= 1$. If 
$m_p(K) \ge 3$, then by \cite[(11.18)]{GoLyS} $t$ acts on $E \cong E_{p^3}$ 
and so some $ 1 \not= U \le [\tilde{V}_1,t]$ is normalized by $F \le E$, 
$|F| = p^2$, contradicting $Z_\alpha \not\le M$. Hence $m_p(K) \le 2$. 
 Let 
$\tilde{V}_1 \cap \bC_{L_{\alpha-1}/Q_{\alpha-1}}(K_{\alpha-1}) 
\not= 1$. Let $u \in \tilde{V}_1$ and $K_{\alpha-1}^{(1)}$ be a component or normal $p$-subgroup of $L_{\alpha-1}/Q_{\alpha-1}$ with $[u, K_{\alpha-1}^{(1)}] = 1$ and $[K_{\alpha-1}^{(1)}, K_{\alpha-1}] = 1$. Then by (7.19) - (7.24) $[K^{(1)}, \tilde{V}_{\alpha-1} \cap L_1] = 1$ where $K^{(1)}$ is the group corresponding to $K_{\alpha-1}^{(1)}$ in $L_1/Q_1$. Hence if $m_p(K) = 2$ we again get $E \cong E_{p^3}$ normalized by $t$, a contradiction. 

So we may assume $m_p(K) = 1$. Now we have $[K_{\alpha-1}^{(1)}, \tilde{V}_{\alpha-1}] = 1$ and then $[K^{(1)}, \tilde{V}_1] = 1$.  Hence $Z_{\alpha-1} \cap  \tilde{V}_1 = 1$, and so 
$t$ induces a transvection on $\tilde{V}_1$, i.e. $p = 3$. So we have 
$|\tilde{V}_1/\bC_{\tilde{V}_1}(K)| = 4$. 
But then again some $E \cong E_9$ centralizes $\tilde{V}_1$, a 
contradicition.

So we have $|\tilde{V}_1 : \tilde{V}_1 \cap Q_{\alpha-1}| \le 4$. We get that 
$\tilde{V}_{\alpha-1} = \langle t \,\Big | \,|[\tilde{V}_1, t] : 
[\tilde{V}_1, t] \cap Z_{\alpha-1}| \le 2 \rangle$ by \cite[(24.1)]{GoLyS} and \cite[(25.12)]{GoLyS}. 
Hence $\tilde{V}_{\alpha-1} \le M$. As 
$Z_\alpha \not\le M$ this implies $p = 3$, 
$|\tilde{V}_1 : \tilde{V}_1 \cap Q_{\alpha-1}| = 4$ and 
$m_p(L_1) = 3$. As $p = 3$, we get $L_2/Q_2 \cong \Sigma_3$ if $q = 2$. 
Hence in any case $\tilde{V}_{\alpha-1} \le L_1$ and so 
$|\tilde{V}_{\alpha-1} : \bC_{\tilde{V}_{\alpha-1}}(\tilde{V}_1)| = 2$, 
contradicting $|\tilde{V}_1 : \tilde{V}_1 \cap Q_{\alpha-1}| = 4$ and 
\cite[(26.11)]{GoLyS}.

So we have $[\tilde{V}_{\alpha-1} \cap L_1, K] = 1$. If 
$m_p(K) \ge 2$, then $[\tilde{V}_{\alpha-1} \cap L_1, V_1] = 1$. 
Now $V_1$ induces transvections to the 
hyperplane $\tilde{V}_{\alpha-1} \cap L_1$. Hence 
$|V_1 : \bC_{V_1}(K)| = 2$ by \cite[(26.11)]{GoLyS} and $p = 3$. But now 
$\tilde{V}_{\alpha-1} \le L_1$ and 
so $[\tilde{V}_{\alpha-1}, V_1] = 1$, a contradiction.

That $[\tilde{V}_1, K_{\alpha-1}] = 1$ follows as in (7.20).
\\

\absa
{\bf~(7.26) Proposition.~}{\it $b = 2$}

\absa
Proof.~ Suppose $b > 2$. Let $K_1 \times \ldots \times K_l$ be the 
components or normal $p$-subgroups of $L_{\alpha-1}/Q_{\alpha-1}$ with $[V_1, K_i] \not= 1$, 
$i = 1,\ldots l$. Hence $[V_1, K_j] = 1$ for every component $K_j$, 
$j \not\in \{1,\ldots ,l\}$.

Suppose there is $x \in V_1 \setminus Z_2$, $[x, Z_\alpha] \le Z_\alpha$ 
and $\bC_G(x) \le M$. We have that $\langle Q_1, Z_\alpha \rangle$ 
normalizes $\langle x, Z_2 \rangle$. As $[Q_1, x] \le Z_1$, 
we get $x \in Z(\langle Q_1, Z_\alpha \rangle)$ or $[Q_1, x] = Z_1$ and 
so all elements in $xZ_2$ are conjugate under 
$\langle Q_1, Z_\alpha \rangle$. Hence in any case we get 
$\langle Q_1, Z_\alpha \rangle = O_2(\langle Q_1, Z_\alpha 
\rangle)\bC_{\langle Q_1, Z_\alpha \rangle}
(x)$. But $[Z_1, O_2(\langle Q_1, Z_\alpha \rangle)] = 1$. 
So we get $G_2 \le M$, a contradiction. We have

\absa
\setcounter{equation}{0}
\begin{equation}
{\mbox {There is no}} \quad x \in V_1 \setminus Z_2 \quad 
{\mbox {with}} \quad \bC_G(x) \le M \mbox{~and~} [Z_\alpha, x] \le Z_\alpha
\end{equation}

\absa
\begin{equation}
l > 1
\end{equation}

\absa
Suppose $l = 1$. Then by (7.19) - (7.25) we get $K \cong G(r)$, $r$ 
even, $m_p(K) = 1$. Let $\omega_1 = \omega$, or $q = 64$ and 
$\omega_1 = \omega^3$.

Assume $K \cong L_n(r)$. Then we get $n \le 4$, or $5 \le n \le 7$ and 
$r = 2$. Let $q > r$. Then by (1.65) and (1.64) $[\omega_1,K] = 1$, as 
$[\omega_1, Z_1] \not= 1$ and so (1.65)(ii) 
is not possible. Now $[\omega_1, \tilde{V}_1] = \tilde{V}_1$. 
We have $|V_1/Z_2 : \bC_{V_1/Z_2}(Z_\alpha)| \le 2^{m_2(K)}$. Let first 
$[\omega, K] = 1$. As 
$\bC_{\tilde{V}_1/Z_2}(\omega) = 1$, 
we get that $|\tilde{V}_1/Z_1| = q^u$ for some $u$.  If $n \le 4$, then $|\tilde{V}_1/Z_1| \le r^4q$, 
and so $|\tilde{V}_1/Z_1| \leq q^4$. As $K$ is not isomorphic to a subgroup of $SL_3(q)$, we see $|\tilde{V}_1/Z_1| = q^4$. Hence $K$ is isomorphic to a subgroup of $SL_4(q)$ and then $K \cong L_4(r)$, $q \ge r^2$, a 
contradiction. 

So let $5 \le n \le 7$. Then $|\tilde{V}_1/Z_2| \le 2^6$, $2^9$, 
$2^{12}$, respectively. Now $K \lesssim L_4(q)$, or $L_5(q)$. But both 
is not possible. Hence 
$[\omega, K] \not= 1$ and so $n \le 4$, $[\omega_1, K] = 1$, $q = 64$. 
Choose $\nu \in L_1 \cap L_2$, $o(\nu) = 7$, with $[\omega,\nu] =1$. If 
$[\nu,K] = 1$, we get
 $o(\omega_1\nu) = 21$, $[\omega_1\nu, K] = 1$, and we may argue as 
before. So assume $[\nu, K] \not= 1$. Then as $r < 64$, we get 
$7 \,\Big | \,r-1$ and so $r = 8$. Hence $\omega$ induces 
a field automorphism of order 3 on $K$. But $[\nu, \omega] = 1$. This is 
impossible as $\nu$ has to induce a diagonal automorphism of order 7. 

So we have $q \le r$. Let $t \in \tilde{V}_{\alpha-1} \cap L_1$ with 
$[\tilde{V}_1,t] \not= 1$. Then $|[t, \tilde{V}_1]| \ge r^n$ and so 
$|\tilde{V}_1 : \tilde{V}_1 \cap Q_{\alpha-1}| \ge r^{n-1}$. 
We have that there is a natural submodule $T_1 \le \tilde{V}_1$. Let 
$t_1 \in T_{\alpha-1}$. Then $[t_1, \tilde{V}_1] \not= 1$, 
and $|[t_1, \tilde{V}_1]| < r^n$. Hence $t_1 \not\in  L_1$ This shows 
that $\tilde{V}_1$ induces transvections on $T_{\alpha-1}$, i.e. 
$r = q$ and $|\tilde{V}_1 : \tilde{V}_1 \cap Q_{\alpha-1}| = 
r^{n-1}$. Now $Z_{\alpha-1} \le T_1$. This implies 
$T_1 = [\tilde{V}_1, \tilde{V}_{\alpha-1} \cap L_1]Z_1/Z_1$. This 
now yields that there are at most two natural modules involved. If 
$p \not\!\Big |\, r^2 - 1$, then there is some $\nu$, $o(\nu) = p$, with 
$[\nu,\tilde{V}_1] = 1$, a contradiction. 
So we have $p \,\Big | \,r^2 - 1$ and then $n \le 3$. As no $p$ - element in 
$L_1$ centralizes some $s \in Z_{\alpha-1}^\sharp$, we see $n = 2$ and
 $p \,\Big | \,r+1$. But now $L_2(r)\langle \nu \rangle$, $\nu$ a field 
automorphism of order $p$, acts on the natural module $T_1$. Hence every 
element is centralized by some $p$ - element, 
contradicting $Z_{\alpha-1} \le T_1$.

So we have shown $[\tilde{V}_{\alpha-1} \cap L_1, \tilde{V}_1] = 1$. Now 
$\tilde{V}_1$ induces transvections on $\tilde{V}_{\alpha-1}$, i.e. 
$|\tilde{V}_1 : \tilde{V}_1 \cap Q_{\alpha-1}| 
\le r^{n-1}$, and $\tilde{V}_{\alpha-1}/
\bC_{\tilde{V}_{\alpha-1}}(K_{\alpha-1})$ is the natural module. Assume 
$p \not\!\Big | \,r - 1$. Then for every 
$\nu \in \bC_{L_1/Q_1}(K)$, $o(\nu) = p$, we have 
$[\nu, \tilde{V}_1] = 1$. As $|\tilde{V}_1Z_2 \cap Q_{\alpha-1}| \ge r^3$, 
there is some $x \in \tilde{V}_1 \setminus Z_2$ 
with $[x,Z_\alpha] \le Z_\alpha$, $x$ is centralized by some
$E \cong E_{p^2}$ in $L_1$ with $\Gamma_{E,1}(G) \le M$. Hence 
$\bC_G(x) \le M$, a contradiction to (1).

So $p \,\Big | \,r-1$ and then $K \cong L_2(r)$, $r= q$. Now 
$p \,\Big | \,|\bC_{L_1/Q_1}(K)|$, furthermore we have some
$\mu \in L_1 \cap L_2$ with $o(\mu) = p$. As $[\mu, Z_1] = Z_1$, 
we get $p \,\Big | \,|\bC_K(\mu)|$ and $p \,\Big | \,|\bC_{\bC_{L_1/Q_1}(K)}(\mu)|$. 
Hence $\mu \in E \cong E_{p^3}$ with $E \le L_1$. This implies 
$N_G(\langle \mu \rangle) \le M$, a contradiction.

\absa
Suppose next $K \cong (S)U_n(r)$, $n \le 5$, or $\Omega_8^-(r)$. Let 
$q > r^2$. Assume first $[\omega,K] \not= 1$. Then by (1.65) and (1.64) 
$[\omega_1, K] = 1$, $q = 64$. Hence $r = 2$ or 
$4$ and $m_3(K) \le 2$. This implies $K \cong (S)U_n(4)$, $n \le 5$. 
Let $\nu \in L_1 \cap L_2$ with $o(\nu) = 7$ and $[\nu,\omega] = 1$. 
Then we see $[\nu,K] = 1$. Hence $[\langle \nu, \omega_1 \rangle, K] = 1$ 
and $[\langle \nu\omega_1 \rangle, \tilde{V}_1] = \tilde{V}_1$. So set 
$\omega_2 = \omega$ or $\nu\omega_1$. Then we get 
$[\omega_2, K] = 1$ and $[\omega_2, \tilde{V}_1] = \tilde{V}_1$. Now 
$|V_1/Z_2 : \bC_{V_1/Z_2}(\omega_2)| \le r$, $r^4$, $r^4$, 
$r^6$, respectively. Hence in any case we get $K \lesssim SL_3(q)$, a 
contradiction.

So we have $q \le r^2$. Suppose next $[t, \tilde{V}_1] \not= 1$ for some 
$t \in \tilde{V}_{\alpha-1} \cap L_1$. As $[t,\tilde{V}_1]$ is a 
$K$ - module, we see $K \cong \Omega_6^-(r)$ or 
$\Omega_8^-(r)$ and $|[t,\tilde{V}_1]| = r^6$, or $r^8$. Now $q = r^2$ 
and $|\tilde{V}_1 : \tilde{V}_1 \cap Q_{\alpha-1}| = r^4$, or 
$r^6$. In the last case $\omega$ centralizes a subgroup of order $p^2$ 
in $K$, a contradiction (recall 
$\langle \omega, K \rangle = \bZ_x \times K$ and so 
$m_x(\langle \omega, K \rangle ) = 4$). 
So we have $K \cong \Omega_6^-(r)$. But now some elementary abelian 
subgroup of order $r^4$ in $K$ acts quadratically on the 
natural module, a contradiction.

So we have $[\tilde{V}_1, \tilde{V}_{\alpha-1} \cap L_1] = 1$. Now we 
see that $|\tilde{V}_1 : \tilde{V}_1 \cap Q_{\alpha-1}| = r$ by (1.45) 
and $\tilde{V}_1/\bC_{\tilde{V}_1}(K)$ 
is the natural $SU_n(r)$ - module or $\Omega_{2n}^-(r)$ - module by (1.22), 
(1.23) and (1.26). If $p \not\Big | \,r^2 - 1$, we get some $\nu \in 
\bC(K)$ with $o(\nu) = p$ and $[\tilde{V}_1, \nu] = 1$. Now we may argue as 
before to get a contradiction with (1).

We are left with $K \cong SU_3(r)$, $p \,\Big | \,r-1$, $q = r^2$. Hence 
there is some $\mu \in L_1 \cap L_2$ with $o(\mu) = p$. Now we get a 
contradiction as before.

\absa
Let $K \cong Sp_{2n}(r)$, $n \le 3$. Let $q > r$. Then 
$[\omega_2, K] = 1$, $\omega_2 = \omega$, or $q = 64$ and 
$K \cong Sp_4(r)$, $r < 64$. In the latter let $\nu \in L_1 \cap L_2$ with 
$o(\nu) = 7$ and $[\nu,\omega] =1$. If $[\nu, K] \not= 1$, then 
$7 \,\Big | \,r-1$ and so $r = 8$. But then $\omega$ induces a field 
automorphism, contradicting $[\omega, \nu] = 1$. 
So $[\nu\omega_1, K] = 1$, and then $[\omega_2, K] = 1$, where 
$\omega_2 = \nu\omega_1$. In any case we have $|V_1/Z_2 : \bC_{V_1/Z_2}(\omega_2)| 
\le r^3$, or $r^6$. In the former we get the contradiction 
$Sp_4(r) \lesssim L_2(q)$. In the latter assume that $q \ge r^2$. 
Then we get 
$Sp_6(r) \lesssim L_3(q)$, a contradiction. So we have 
$K \cong Sp_4(r)$, $q \le r$, or $K \cong Sp_6(r)$, $q < r^2$. 

Suppose $[t,\tilde{V}_1] \not= 1$ for some $t \in \tilde{V}_{\alpha-1} 
\cap L_1$. Then $|[t, \tilde{V}_1]| = r^4$ in the first case while 
$r^6 \le |[t, \tilde{V}_1]| < r^8$ in the second case. 
In particular $[t, \tilde{V}_1]$ involves the natural module just once. 
Let $T_{\alpha-1}$ be the natural module. Then in the
 first case as $|\tilde{V}_1 : \tilde{V}_1 \cap Q_{\alpha-1}| = r^3$ 
there is some $t_1 \in T_{\alpha-1} \cap L_1$ with 
$1 \not= |[\tilde{V}_1, t_1]| \le r^3$,
 a contradiction. In the second case 
$|\tilde{V}_1 : \tilde{V}_1 \cap Q_{\alpha-1}| > r^4$. Now by (1.46) 
$|T_{\alpha-1} : \bC_{T_{\alpha-1}}(\tilde{V}_1)| \ge r^3$ and so there 
is some $t_1 \in T_{\alpha-1} \cap L_1$ with 
$1 \not= |[\tilde{V}_1,t_1]| \le r^3$, a contradiction again. So in any 
case $[\tilde{V}_1, \tilde{V}_{\alpha-1} \cap L_1] = 1$ and then 
$|\tilde{V}_1 : \tilde{V}_1 \cap Q_{\alpha-1}| = r$ and 
$\tilde{V}_1/\bC_{\tilde{V}_1}(K)$ is the natural $K$ - module. Now 
$p \not\Big | \,r-1$ as $m_p(K) = 1$ and so some $\nu \in \bC_{L_1/Q_1}(K)$, 
$o(\nu) = p$, centralizes $\tilde{V}_1$. But then as in the 
$L_n(r)$ - case we get a contradiction to (1).

Let finally $K \cong Sz(r)$, $^3D_4(r)$, $G_2(r)$, or $^2F_4(r)$. Assume 
$q > r$ or $q > r^5$ in case of $^3D_4(r)$. Then we get with (1.65) and 
(1.64) that $[\omega_1,K] = 1$. Let $q = 64$ and 
$[\omega,K] \not= 1$. Choose $\nu \in L_1 \cap L_2$ with $o(\nu) = 7$ and
$[\nu,\omega] = 1$. Let $[\nu, K] \not= 1$. Then we get 
$K \cong Sz(8)$, $G_2(8)$, $^2F_4(8)$ or $^3D_4(2)$. In any case 
$\omega$ induces an outer automorphism and so $[\omega, \nu] \not= 1$, a 
contradiction. Set $\omega_2 = \nu\omega_1$, $\omega_2 = \omega$ 
for $[\omega,K] = 1$, then $[\omega_2, K] =1$. But now 
$|V_1/Z_2 : \bC_{V_1/Z_2}(\omega_2)| \le r $, $r^5$, $r^3$, $r^5$ 
and so $Sz(r) \lesssim L_2(q)$, $^3D_4(r) \lesssim L_2(q)$, 
$G_2(r) \lesssim L_3(q)$, $^2F_4(r) \lesssim L_4(q)$, a contradiction. 
For $t\in \tilde{V}_{\alpha-1} \cap L_1$, 
we have $|[\tilde{V}_1, t]| \le r^2$, $r^{10}$, $r^4$, $r^6$, 
respectively. We get from (1.27), (1.40) and (1.22) that 
$K \cong G_2(r)$. But $G_2(r) \not\lesssim SL_4(r)$, a contradiction. 
Hence $[\tilde{V}_1, \tilde{V}_{\alpha-1} \cap L_1] = 1$. Then 
$|\tilde{V}_{\alpha-1} : \bC_{\tilde{V}_{\alpha-1}}(\tilde{V}_1)| \le r$ 
($r^5$ in case of $^3D_4(r)$). This  again contradicts (1.18), (1.40), 
(1.22) and (1.27). Hence we have (2).

\absa
Let now $l > 1$. $[V_1,K_1] \not= 1 \not= [V_1,K_2]$. We have 
$\tilde{V}_{\alpha-1} = [V_{\alpha-1}, K_1]$, 
${\tilde{V}_{\alpha-1}}^{(1)} = [\tilde{V}_{\alpha-1}, K_2]$.

\absa
\begin{equation}
{\tilde{V}_{\alpha-1}}^{(1)} = 1
\end{equation}

\absa
Suppose ${\tilde{V}_{\alpha-1}}^{(1)} \not= 1$. Let $K \times K^{(1)}$ be 
the corresponding components of $L_1/Q_1$. Then by (7.19) - (7.25)
 we have $[\tilde{V}_{\alpha-1} \cap L_1, K \times K^{(1)}] = 1$. 
Furthermore $m_p(K) = m_p(K^{(1)}) = 1$.
We may assume $m_3(K_1) \le 1$, hence $K_1 \cong Sz(r)$, $L_2(r)$, 
$U_3(r)$, $L_3(r)$, $J_1$, or solvable, or exceptional. 
Let $|V_1 : \bC_{V_1}(K_1 \times K_2)| = 2$. Then we get 
$\tilde{V}_{\alpha-1} \le L_1$. Let $t \in \tilde{V}_{\alpha-1}$,
$[t,V_1] \not= 1$. Then some $E \cong E_{p^2}$, $E \le K \times K^{(1)}$ 
acts on some $1 \not= U \le [V_1, t]$. Hence $Z_\alpha \le M$, a 
contradiction. From (1.12) we get that $K_1 \cong Sz(r)$, $L_2(r)$, 
$U_3(r)$, or $L_3(r)$, $r$ even. Suppose now 
$q > r$ or $q > r^2$ in case of $K_1 \cong U_3(r)$. Then by (1.64) and 
(1.65) $[\omega_2, K] = 1$, $\omega_2 = \omega$ if 
$[\omega, K] = 1$, $\omega_2 = \omega_1\nu$, $o(\nu) = 7$, for $q = 64$.
As before ${\tilde{V}_{\alpha-1}}^{(1)} \not\le M$. Hence some 
$t \in \tilde{V}_{\alpha-1} \setminus M$
centralizes in $V_1/Z_2$ a subgroup of index $r$ or $r^2$ in case of 
$K \cong L_3(r)$. Then $|V_1/Z_2 : \bC_{V_1/Z_2}(\omega_2)| \le r$ or $r^2$. 
This implies $Sz(r)$, $L_2(r)$, $U_3(r) \lesssim L_1(q)$ or 
$L_3(r) \lesssim L_2(q)$, respectively. But all these cases are impossible.

So $q \le r$, $q \le r^2$ in case of $K \cong U_3(r)$. Now 
$|\tilde{V}_{\alpha-1} : \bC_{\tilde{V}_{\alpha-1}}(V_1)| \le r$, or $r^2$. 
This implies  $K_1 \cong L_2(r)$, $U_3(r)$, or $L_3(r)$ and 
$\tilde{V}_{\alpha-1}/\bC_{\tilde{V}_{\alpha-1}}(V_1)$ is the natural 
$K_1$ - module. But this contradicts 
$[\tilde{V}_{\alpha-1}, K_2] = \tilde{V}_{\alpha-1}$.

\absa
By (3) we know $[[V_{\alpha-1}, K_1], K_2] = 1$. Set 
${\tilde{V}_{\alpha-1}}^{(i)} = [V_{\alpha-1}, K_i]$, $i = 1,2$. By 
(7.19) - (7.25) we know
$[{\tilde{V}_{\alpha-1}}^{(1)} \cap L_1,K] = 1$. Let 
$|V_1 : \bC_{V_1}(K_1)| = 2$. Then there is a fours group 
$W \le V_1$ such that $W$ is centralized by 
$E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \le M$, and 
$|W \cap Q_{\alpha-1}| \ge 2$, $W \cap Z_1 = 1$. 
But now $[W \cap Q_{\alpha-1}, Z_\alpha] \le Z_\alpha$, which 
contradicts (1). So we have 
that $K_1$, $K_2$ are sporadic, alternating or groups of Lie type in 
even characteristic. As $m_3(K) \ge 2$ for $K$ sporadic or 
alternating we may assume $K_2 \cong Sz(r)$, $L_2(r)$, $L_3(r)$, or 
$U_3(r)$, $r$ even.

Suppose for some $t \in {\tilde{V}_{\alpha-1}}^{(1)} \cap L_1$ we have 
$[K^{(1)}, t] \not = 1$. Let 
$[{\tilde{V}_{\alpha-1}}^{(1)} \cap L_1, \bC_{L_1/Q_1}(K^{(1)})] = 1$, then 
$tQ_1$ is centralized by $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \le M$. This 
shows that some $1 \not= U \le [V_1,t]$ is normalized by 
$E$, contradicting $Z_\alpha \not\le M$. So we have 
$[{\tilde{V}_{\alpha-1}}^{(1)} \cap L_1, \bC_{L_1/Q_1}(K^{(1)})] \not= 1$.
We know that $[{\tilde{V}_{\alpha-1}}^{(2)} \cap L_1, K^{(1)}] = 1$. 
Suppose there is $t_1 \in {\tilde{V}_{\alpha-1}}^{(2)} \cap L_1$
with $[K,t_1] \not= 1$. Then 
$\langle K^{(1)}, [\bC_{L_1/Q_1}(K^{(1)}), {\tilde{V}_{\alpha-1}}^{(1)} 
\cap L_1] \rangle$
centralizes $[\tilde{V}_1^{(1)}, t_1]$, a contradiction to 
$Z_\alpha \not\le M$.

Hence we get $[{\tilde{V}_{\alpha-1}}^{(2)} \cap L_1, K \times K^{(1)}] = 1$ 
and so $[{\tilde{V}_{\alpha-1}}^{(2)} \cap L_1, V_1] = 1$. 
We have ${\tilde{V}_{\alpha-1}}^{(2)} \not\le M$. Now as before
we see that 
${\tilde{V}_{\alpha-1}}^{(2)}/\bC_{{\tilde{V}_{\alpha-1}}^{(2)}}(K_2)$ 
is the natural module. But then $t$ in its action on
${\tilde{V}_{\alpha-1}}^{(2)}/\bC_{{\tilde{V}_{\alpha-1}}^{(2)}}(K^{(1)})$ 
is centralized
by $E \cong E_{p^2}$, which shows that $E$ normalizes some 
$1 \not= U \le [V_1,t]$, a contradiction to $Z_\alpha \not\le M$.

So we have $[{\tilde{V}_{\alpha-1}}^{(1)} \cap L_1, V_1] = 1$. Now 
$\tilde{V}_{\alpha-1} \not\le M$. Let $q > 2$. If $K_1$ is
sporadic or alternating, then $[\omega,K_1] = 1$ (recall 
$K_1 \not\cong J_1$ as $m_3(K_1) \ge 2$). Hence as
$|V_1/Z_2 : \bC_{V_1/Z_2}({\tilde{V}_{\alpha-1}}^{(1)})| \le 2^5$ and 
$Z_2 \cap \tilde{V}_1 = Z_1$,
we get $K_1 \lesssim GL_2(4)$, a contradiction. Hence $K \cong G(r_1)$, 
$r_1$ even. 

Suppose $q = 2$. Then $V_1$ induces transvections on 
${\tilde{V}_{\alpha-1}}^{(1)}$. Hence $K \cong G(2)$ or $A_n$. But in case 
of $K \cong A_n$ we get 
$|V_1 : \bC_{V_1}(K_1)| = 2$,  a contradiction as above.

Hence in any case $K_1 \cong G(r_1)$, $r_1$ even. Let first $K_1$ be 
untwisted and $q > r_1$. Then $[\omega_2, K_1] = 1$. 
Let $K_1 \cong L_n(r_1)$. Suppose 
$|V_1/Z_2 : \bC_{V_1/Z_2}(\omega_2)| \le r_1^4$.  
Then $|\tilde{V_1}/Z_1| \le q^3$, i.e. $K_1 \lesssim L_3(q)$. But 
this implies $n \le 3$ and so 
$|V_1/Z_2 :  \bC_{V_1/Z_2}(\omega_2)| \le r_1^2$ and then 
$K_1 \lesssim \Gamma_1(q)$,
a contradiction. Hence we have $r_1 = 2$ and $5 \le n \le 7$. 
If $n = 5$, then $|V_1/Z_2 : \bC_{V_1/Z_2}(\omega_2)| \le 2^6$
and as $q \ge 4$, we get $L_5(2) \lesssim L_3(4)$ or $L_2(8)$, a 
contradiction. So let $6 \le n \le 7$. Now $m_3(K) = 3$ and so $q \ge 8$. 
This now implies 
$L_6(2) \lesssim L_3(q)$ or $L_7(2) \lesssim L_3(q)$, or 
$L_7(2) \lesssim L_4(8)$, a contradiction again.

Let next $K_1 \cong Sp_{2n}(r_1)$, $n \le 3$. If $n = 2$, we get 
$Sp_4(r_1) \lesssim L_2(q)$, a contradiction. If $n = 3$, we get 
$Sp_6(r_1) \lesssim L_5(q)$. For $s \,\Big | \,r_1^2 - 1$, $s$ prime, 
we have an elementary abelian subgroup of order $s^3$ in $Sp_6(r_1)$. 
Hence $s \,\Big | \,q-1$. 
Now choose $s$ as a Zsigmondy prime. Then we get $r_1^2 \le q$. This now 
implies $Sp_6(r_1) \lesssim L_3(q)$, a contradiction.
Hence we have $r_1 = 8$. But now even $Sp_6(r_1) \lesssim L_4(q)$. As 
$Sp_6(r_1) \not\lesssim L_3(q)$, we get $q = 16$ and so 
$Sp_6(8) \lesssim L_4(16)$. But the order of $Sp_6(8)$ does 
not divide the order of $L_4(16)$, a contradiction.

Let $K_1 \cong G_2(r_1)$. Then we get $K_1 \lesssim L_2(q)$, a 
contradiction. For $K_1 \cong$ $^2F_4(r_1)$ we would get 
$K_1 \lesssim L_3(q)$ and for 
$K_1 \cong Sz(r_1)$ we would get $K_1 \lesssim \Gamma L_1(q)$, a 
contradiction.

Hence in all these cases we get that $q \le r_1$. Now 
$|\tilde{V}_{\alpha-1} : \bC_{\tilde{V}_{\alpha-1}}(V_1)| \le q \le r_1$. 
This implies that $V_1$ induces
transvections and so $K_1 \cong SL_n(r_1)$ or $Sp_{2n}(r_1)$, $q = r_1$. 
Furthermore
${\tilde{V}_{\alpha-1}}^{(1)}/\bC_{{\tilde{V}_{\alpha-1}}^{(1)}}(K_1)$ is 
irreducible.

So let now $K_1 \cong SU_n(r_1)$, or $\Omega_8^-(r_1)$, $q > r_1^2$. 
By (1.65) and (1.64) $[\omega_2, K_1] = 1$,
$\omega_2 = \omega$, or the element of order 21 in $L_1 \cap L_2$ for 
$q = 64$ and $[\omega, K ] \not= 1$.
Now $|V_1/Z_2 : \bC_{V_1/Z_2}(\omega_2)| \le r_1, r_1^4$, or $r_1^6$. 
This implies 
$K_1 \lesssim \Gamma L_1(q), L_2(q)$, or $L_3(q)$, a contradiction in 
all cases. Hence
$q \le r_1^2$ and so we have that 
${\tilde{V}_{\alpha-1}}^{(1)}/\bC_{{\tilde{V}_{\alpha-1}}^{(1)}}(K_1)$
is the natural $SU_n(r_1)$ - module or $\Omega_8^-(r_1)$ - module.
Furthermore in any case $q = r_1^2$. As 
$\langle \omega, K \rangle \cong K \times \bZ_x$, we
see that $(r_1 + 1)^3 \not\!\Big | \,\,|K_1|$. Hence $K_1 \cong SU_3(r_1)$ 
and 
${\tilde{V}_{\alpha-1}}^{(1)}/\bC_{{\tilde{V}_{\alpha-1}}^{(1)}}(K_1)$ is 
the natural module.

Let finally $K_1 \cong$ $^3D_4(r_1), q > r_1^5$. Then by (1.65) and 
(1.64) $[\omega_2, K_1] = 1$
Now $|V_1/Z_2 : \bC_{V_1/Z_2}(\omega_2)| \le r_1^5$. But then 
$K_1 \lesssim \Gamma L_1(q)$, a contradiction.

So we have $q \le r_1^5$, and 
$|{\tilde{V}_{\alpha-1}}^{(1)} : \bC_{{\tilde{V}_{\alpha-1}}^{(1)}}(V_1)| 
\le r_1^5$, a contradiction to (1.40). This shows

\begin{enumerate}
\item[\hspace{-0.3cm}(4)]  
$\tilde{V}_1/\bC_{\tilde{V}_1}(K)$ is irreducible. If $[s,K] \not= 1$, 
$s$ an involution, then there
is $1 \not= U \le [\tilde{V}_1, s]$ with $E \le N_G(U)$, 
$E \cong E_{p^2}$ and $\Gamma_{E,1}(G) \le M$.
\end{enumerate}

Let now $t_1 \in {\tilde{V}_{\alpha-1}}^{(2)} \cap L_1$. Then by (4) 
$[t_1, K] = 1$,
as $Z_\alpha \not\le M$. Now by (7.22) 
$[{\tilde{V}_{\alpha-1}}^{(2)} \cap L_1, K \times K^{(1)}] = 1$
and then $[{\tilde{V}_{\alpha-1}}^{(2)} \cap L_1, V_1] = 1$. In 
particular
${\tilde{V}_{\alpha-1}}^{(2)} \not\le M$. But then 
${\tilde{V}_{\alpha-1}}^{(2)}Q_2 \le {\tilde{V}_{\alpha-1}}^{(1)}Q_2$. 
Hence there is $t_2 = t_1t \in L_1$ with
$t \in {\tilde{V}_{\alpha-1}}^{(1)} \setminus L_1$ and 
$t_1 \in {\tilde{V}_{\alpha-1}}^{(2)}$. By (4) we get 
$[t_2, \tilde{V}_1] = 1$.
But now $[t, \tilde{V}_1] \le Z_{\alpha-1} \le Z_2$. Hence 
$\tilde{V}_1Z_2 \unlhd L_2$, a contradiction to (1).
\\

\absa
Now by (7.26) we have 
$\langle V_1, V_3 \rangle/O_2(\langle V_1, V_3 \rangle) \cong L_2(q)$ or 
$L_2/Q_2$ if $L_2$ is solvable.

Furthermore $V_1^\prime = Z_1$. Set 
$Y = (V_1 \cap O_2(\langle V_1, V_3 \rangle))(V_3 \cap 
O_2(\langle V_1, V_3 \rangle)) = (V_1 \cap Q_2)(V_3 \cap Q_2)$. 
Hence $Y \unlhd L_2$. Set $U = V_1 \cap V_3$. If $U = Y$ 
then $V_1 \cap Q_2 \unlhd L_2$ and so $[V_1 \cap Q_2, L_2] = Z_2$ and 
$[Q_2, O^2(L_2)] = Z_2$. This now
implies with (7.3) and (1.20) that $|Q_2 : Z_2| \le q$. Then 
$V_1 = Z_2Z_0$. Now $O^2(L_1)$
normalizes $Z_2$ and so $Z_2 \unlhd \langle L_1, L_2 \rangle$, 
contradicting (4.14).

So we have $U \not= Y$ and by (1.27)(iii) $Y/U$ is a direct sum of 
natural modules. Set $\hat{L_2} = \langle V_1, V_3 \rangle$.

\absa
{\bf~(7.27) Lemma.~}{\it If $x \in U$ with $\bC_G(x) \le M$, then 
$x \in Z_1$.}

\absa
Proof.~By way of contradiction we may assume $x \not\in Z_2$. Now 
$[x, \hat{L_2}] \le Z_2$. We have 
$\bC_{\hat{L_2}}(x) \le M \cap \hat{L_2}$. This
implies that $Z(\hat{L_2}) \cap \langle x, Z_2 \rangle \not= 1$. But 
$\hat{L_2} \unlhd L_2$ and so $L_2 = \hat{L_2}S$. Now $Z(L_2) \not= 1$, 
contradicting (7.3).

\absa
From now on we fix the following notation :
$X \le V_3$, $X \cap Q_1 = 1$, $|X| = q$, $XQ_1/Q_1 \unlhd S/Q_1$, 
$\omega \in L_1 \cap L_2$, $o(\omega) = x$, $[X, \omega] = X$.
Furthermore let $K$ be some component of $L_1/Q_1$ with $[X,K] \not= 1$.

\absa
{\bf~(7.28) Lemma.~}{\it $\bC_{V_1/Z_1}(Y) = Z_2/Z_1$.}

\absa
Proof.~Suppose false. Let $s \in \bC_U(Y)$. If $s \not\in Z_2$,
we may assume that $s \in Z(V_1)$. This now implies $s \in Z(\hat{L_2})$ 
and so $Z(L_2) \not= 1$, a contradiction. Thus 
$\bC_{U/Z_1}(Y) = Z_2/Z_1$. Let 
$U_1 \le V_1$, $[U_1,Y] \le Z_1$, $U_1 \cap U = 1$, $[\omega, U_1] = U_1$, 
$|U_1| = q$. Then we may assume that for some 
$t \in X$ we have $[t, V_1] \le Z_2U_1$. In particular 
$[X, [t,V_1]] \le Z_1$. So we
have that $X$ acts quadratically on $V_1/Z_1$ and for $t \in X$, 
$|[V_1/Z_1, t]| \le q^2$.

\absa
\setcounter{equation}{0}
\begin{equation}
Z(Y) = Z_2
\end{equation}

\absa
We have $Z(Y) \cap V_1 = (Z(Y) \cap Z(V_1))Z_2$,
$Z(Y) \cap V_3 = (Z(Y) \cap Z(V_3))Z_2$. As 
$Z(V_1) \cap Z(V_3) \le Z(\hat{L_2}) = 1$,
we see that
$$ Z(Y) = (Z(Y) \cap Z(V_1))(Z(Y) \cap Z(V_3)) $$
Now $Z(Y)$ is a direct sum of natural modules and so 
$[Z(Y), V_1] = Z(Y) \cap Z(V_1)$. In particular $Z(Y) \le Q_1$. 
This shows
$[Z(Y), V_1] = Z_1$. Hence $Z(Y) \cap Z(V_1) = Z_1$, and so 
$Z(Y) = Z_2$.

This now implies

\absa
\begin{equation}
[t,\tilde{V}_1/Z_1] \cap Z_2/Z_1 \not= 1, 
\,\, {\mbox {and}} \,\, q \le |[t, V_1/Z_1]| \le q^2.
\end{equation}

\absa
\begin{equation}
q > 2
\end{equation}

\absa
Suppose $q = 2$. We first show $[K,Y] \le K$. So assume there is some 
$y \in Y$
with $K^y \not= K$. Let first $\langle y \rangle = X$. Then 
$[\bC_{K \times K^y}(y), [V_1, y]] = 1$, as $|[V_1,y]| = 4$, a 
contradiction to $Z_2/Z_1 \le [V_1, y]$ by (2). So we have 
$[K,t] \le K$ for $\langle t \rangle = X$.
Now $\bC_{K \times K^y}(y) = K_1$ acts on 
$\hat{V_1} = [V_1,y] \le Q_2$. 
Furthermore $[t, \hat{V_1}] \le Z_2/Z_1$. So we may assume that 
$t$ induces a transvection on
some nontrivial irreducible $K_1$ - module $W$ in $V_1$. Now (1.30)
implies $K_1 \cong L_n(2)$, $Sp_{2n}(2)$, $\Omega_{2n}^\pm(2)$, or $A_n$, 
and $W$ is the natural module. In 
any case a 3 - ele\-ment in $K_1$ centralizes $Z_2/Z_1$.
This now implies that $3 \not\in \sigma(M)$. But  as $K_1 \cong K$, 
we see $m_3(K) = 1$,
whence $K \cong K_1 \cong L_3(2)$ or $A_5$, and $p = 7$, $p = 5$,
 respectively.
As $[t,K] \not= 1$, $[t,K^y] \not= 1$, we get $[[K,V_1],K^y] = 1$. Now 
there is $\nu \in L_1$, $o(\nu) = p$ with 
$[\nu, K \times K^y] = 1$. As $[K,V_1]/Z_1$ is irreducible, we get 
$[\nu,[K \times K^y, V_1]] = 1$. But $Z_2/Z_1 \le [K \times K^y, V_1]$,
a contradiction. So we have shown that $[K,Y] \le K$.

Now by (1.30) $K/Z(K) \cong A_n$, $L_n(s)$, $Sp_{2n}(s)$, 
$\Omega_{2n}^\pm(s)$, $ s\le 4$, $U_n(s)$, $G_2(s)$, $U_4(3)$, or
$K/O_r(K) \cong L_2(4)$,
$L_2(7)$, $[O_r(K), V_1] = 1$, or $K$ is solvable. The cases 
$K/O_r(K) \cong L_2(4)$, or $L_2(7)$ are not possible by (1.51).

Suppose first that $3 \in \sigma(M)$. Then 
$3 \not\!\Big | \,|\bC_K(\bC_{V_1/Z_1}(S \cap K))|$. Hence we get
$K \cong L_2(4)$ or $O_3(K) \not= 1$. If $K \cong L_2(4)$, then 
$|\tilde{V}_1| = 16$. 
If $O_3(K) \not= 1$, then $|[[t, O_3(K)], \tilde{V}_1]| \le 16$. In both 
cases $Z_2 \le \tilde{V}_1$ and some 3 - element in
$L_1$ centralizes $\tilde{V}_1$, a contradiction.

Hence we may assume $p > 3$. Now $K/Z(K) \cong A_n, n \le 11$, 
$L_n(4), n \le 4$, $L_n(2), n \le 7$, $Sp_{2n}(s), n \le 3, s \le 4$, 
$\Omega_8^-(s), s \le 4$, 
$U_4(2)$, $G_2(2)$, or $O_p(K) \not= 1$.

Suppose first $m_p(K) = 1$. Then there is $E \cong E_{p^2}$ with 
$\Gamma_{E,1}(G) \le M$ and $E \le \bC_{L_1}(K)$.
But as $\tilde{V}_1$ induces at most two nontrivial irreducible modules 
and $p \not\Big| \,4 - 1$, we get $[E,\tilde{V}_1] = 1$, a contradiction 
to $Z_2/Z_1 \le \tilde{V}_1$.

So we have $m_p(K) > 1$. This implies $K \cong A_{10}$, $A_{11}$, 
$L_4(4)$, $Sp_4(4)$, $Sp_6(4)$, $p = 5$ in this cases, or 
$K \cong L_6(2)$, $L_7(2)$, $p = 7$, or $O_p(K) \not= 1$.

If $m_p(K) = 2$, then there is some $\nu \in \bC_{L_1/Q_1}(K)$, 
$o(\nu) = p$. Again
$[\nu, \tilde{V}_1] = 1$, a contradiction. So we are left with 
$m_p(K) \ge 3$. As $p > 3$, we have $O_p(K) \not= 1$ and $t$ 
centralizes $E \cong E_{p^2}$ in $K$. Now $[[t, \tilde{V}_1], E] = 1$, 
but $Z_2/Z_1 \le [t, \tilde{V}_1]$, a contradiction. This proves (v).

\absa
As $q > 2$ by (3), we have a quadratic fours group on $V_1$. This implies 
that $K \cong G(r)$, $r$ even, $3\cdot U_4(3)$, sporadic, alternating or 
solvable, or 
$[O_r(K), V_1] = 1$ by (1.12), (1.51) and (1.11). 
The last case is impossible by (4.10).

We first again show $[\omega, K] \le K$. Suppose false. Let 
$K_1 \times \ldots \times K_y = K^{\langle \omega \rangle}$. Suppose 
$y > 3$. Then we have 
$K_1 \cong L_2(r)$, $r$ even, or $L_3(2)$. Furthermore let 
$\langle \nu \rangle \le L_1 \cap L_2$, $o(\nu) = q -1 $. Then 
${\mbox {gcd}}(o(\nu), |K|) = 1$, as 
$N_G(\langle \nu \rangle) \not\le M$. Let $ \mu \in \langle \nu \rangle$,
 with $[K_1, \mu] \le K_1$, then $\bC_{K_1}(\mu)$ is a
2 - group as for every odd prime $u$, $u \,\Big | \,|K_1|$, we get 
$u \in \sigma(M)$. But this is not possible for outer automorphisms of 
$K$. Hence we
have $K^{\langle \nu \rangle} = K_1 \times \ldots \times K_{q-1}$. 
Now $t \in X$ acts on $E \cong E_{p^{q-1}}$. But then some 
$F \cong E_{p^2}$ centralizes some $x \in Z_2 \setminus Z_1$, a 
contradiction.

So we have $y = 3$, and then $q = 4$ or $q = 64$. In both cases 
$3 \not\in \sigma(M)$. Let first $q = 64$. As $[\omega^3, Z_1] \not= 1$, 
we get $\omega^3 \not\in K_1 \times K_2 \times K_3$. 
But as $[\omega^3, K_1] \le K_1$, we get that $K_1$ is a 
$3^\prime$ - group. This shows $K_1 \cong Sz(r)$. As 
$[t,K_1]\not= 1$, we get with (1.64) $r \le q$. Hence $r = 8$ as 
$K_1$ possesses an automorphism of order 3.
Let now $\mu \in L_1 \cap L_2$ with $o(\mu) = 7$. Then $[\mu, K_1] \le K_1$ 
and $\langle \mu, K_1 \rangle \cong \bZ_7 \times K_1$, as 
$[\mu, Z_1] \not= 1$ and 
$[K_1,Z_1] = 1$. But then $m_7(\langle \mu, K_1, K_2, K_3 \rangle) = 4$, 
which shows $7 \in \sigma(M)$, and
$N_G(\langle \mu \rangle) \le M$, a contradiction.

So we are left with $q = 4$. Now $m_3(K) \le 1$ and so $K \cong L_2(r)$, 
$L_3(r)$, $U_3(r)$, or $Sz(r)$, $r$ even. We have $|[V_1,t]| \le 16$. 
Now (1.26) and (1.27) imply $r \le 16$ for 
$K \cong L_2(r)$ or $L_3(r)$, $r \le 4$ for $K \cong U_3(r)$ or 
$Sz(r)$. This shows $K \cong L_2(r), r \le 16$, 
$L_3(2)$, $L_3(8)$, or $U_3(4)$. But as $\bC_{K_1}(\omega) \cong K$, 
we get with (1.63)
that $K \cong L_2(r)$, $r \le 16$, or $L_3(2)$. As 
$[\omega, S]Q_1/Q_1 = Y$ is elementary abelian,
$K \cong L_3(2)$ is not possible.

If $[t,K_3] = 1$, then $K_3$ acts on $[V_1,t]$. As $|[V_1,t]| \le 16$ 
and $Z_2/Z_1 \cap [V_1,t] \not= 1$, we get $[K_3,[V_1,t]] \not= 1$. 
This shows $K \cong L_2(4)$. If $[K_1,t] \not= 1 \not= [K_2,t]$, 
then as $t \in Z(S)$ we have $t \in K_1$ and so by (1.27) again 
$K \cong L_2(4)$.

In both cases there are at most two nontrivial irreducible $K$ - modules 
involved.
Hence $[K_3, [V_1,K_1]] = 1$. But then as $\omega$ acts on 
$\{K_1, K_2, K_3\}$, we also get
$[K_2,[V_1,K_1]] = 1$. Now as $|[V_1,t]| \le 16$, we may assume 
$[t,K_3] = 1$ and so $K_3$ centralizes $[V_1,t]$, a contradiction.

\absa
So we have $[K, \omega] \le K$ and then also $[Y,K] \le K$, as 
$[Y,\omega] = Y$. Now as $[\omega,K] \not= 1$, we see that $K$ is 
solvable or $K \cong G(r)$ by (1.9).
Furthermore by (1.64) and (1.65) $q \le r$ or $K$ is twisted and 
$q = r^2$ or $K \cong$ $^3D_4(r)$. Recall that (1.65)(ii) and (iv) are 
not possible
as $[\omega, Z_1] \not= 1$ and so $\omega \not\in K$. Furthermore 
(1.65)(vi) is not possible as $3 \in \sigma(M)$ in this
case. Now application of (1.48) shows $K \cong SL_n(r)$, $Sp_{2n}(r)$, 
$SU(n,r)$, $\Omega_{2n}^\pm(r)$, $G_2(r)$, $Sz(r)$, $r^2, r = q$ or $q^2$, 
or $^3D_4(r)$.
If $K \cong$ $^3D_4(r)$ then by (1.40) $q^2 > r^5$. Furthermore we get 
from (1.65) and (1.64) $q = 64$ and $ r \le 4$,
or $q = 8$ and $ r = 2$. As $\omega$ acts on $K$ we see $r = 4$ in case 
of $q = 64$. Finally by (1.22) and (1.23) $\tilde{V_1}$ involves
just natural modules.

\begin{equation}
p \not\Big | \,r^2-1, \,\, {\mbox {if}} \,\, K \,\, {\mbox {is nonsolvable}}
\end{equation}

Suppose $p \,\Big | \,r^2 - 1$. Assume that $V_1$ is an SC - module for 
$K$. 
Let $V(\lambda_2)$ not be involved in $V_1$, then some $p$ - element 
centralizes $Z_2$ or $K \cong L_2(r)$ or $L_3(r)$. Let 
$V(\lambda_2)$ be involved. Then as $|[\tilde{V}_1, t]| \le r^2$,
we see $K \cong L_3(r)$, $L_4(r)$, $Sp_4(r)$ or $U_4(r)$. Hence in any 
case $K \cong L_2(r)$, $L_3(r)$, $L_4(r)$, $Sp_4(r)$, or 
$U_4(r)$. In case
of $U_4(r)$ both the natural $U_4(r)$ - module and the natural 
$\Omega_6^-(r)$ - module are involved. 
Hence $|[V_1,t]| \ge r^4$. This implies $q = r^2$ and so there is some
$\nu \in L_1 \cap L_2$ with $o(\nu) = p$.
But then $m_p(\bC_{\langle K, \nu \rangle}(\nu)) \ge 3$, contradicting 
$N_G(\langle \nu \rangle) \not\le M$.
If $K \cong Sp_4(r)$, we get that both natural modules are involved. 
Now $|[V_1,t]| \ge r^3$. But $r \ge q$, a contradiction.
Let $K \cong L_4(r)$, then also $[V_1,t]| \ge r^3$, as $V(\lambda_2)$ is 
involved and $V_1$ has to involve some further 
nontrivial irreducible module. But again $r \ge q$ gives a contradiction.

Let $K \cong L_3(r)$. Then $|[V_1,t]| = r^2$ and so $q = r$. Furthermore 
there are exactly
two (nonisomorphic) irreducible modules involved. As 
$p \not\Big | \,r - 1$, we get $p \,\Big | \,r+1$ and $m_p(K) = 1$.
Now there is some $\mu \in \bC(K)$, $o(\mu) = p$ with 
$[\tilde{V}_1, \mu] = 1$. But then $[\mu, Z_2] = 1$, a contradiction.

Let finally $K \cong L_2(r)$. Then $\tilde{V}_1$ involves exactly one 
nontrivial irreducible module, as
$\tilde{V}_1$ is an SC - module. Again some $\mu$ with $o(\mu) = p$ 
centralizes $\tilde{V}_1$. But $[X, V_1] \le [X,K] = \tilde{V}_1$,
and so $Z_2 \cap \tilde{V}_1 > Z_1$, a contradiction.

So assume now that $\tilde{V}_1$ is not an SC - module. Then we have that 
$K$ does not contain elementary abelian subgroups of order $q^2$. This
shows $K \cong L_2(r)$, $U_3(r)$ or $Sz(r)$. Let $1 \not= y \in \bC_Y(K)$. 
Suppose $[y, \tilde{V}_1] = 1$. Then some
$p$ - element $\mu \in \bC(K)$ centralizes $\tilde{V}_1$ and so $Z_2$, 
a contradiction. So we have $[y, \tilde{V}_1] \not= 1$. In particular
there are at least two nontrivial irreducible modules involved. Hence 
$K \cong L_2(q)$ or $U_3(\sqrt{q})$ and both modules
are natural modules. But as $\omega$ acts on $K$, we get that $|X| = q$ 
and so $K \cong L_2(q)$, $X \in {\mbox {Syl}}_2(K)$. As $\omega$ acts on 
$\bigcap\limits_{x \in X^\sharp}[V_1/Z_1,x] = T_1$, we have $T_1 = Z_2/Z_1$ 
or $T_1 = [V_1/Z_1,t]$. As $K$ admits some field automorphism
$\nu$ of order $p$ which normalizes $X$ and so also $T_1$, we get the 
latter. Now
$[V_1/Z_1, t] = [V_1/Z_1, X]$. Hence $\tilde{V}_1/Z_1$ is a direct sum 
of two natural modules.
But now again $\nu$ centralizes some $s \in Z_2 \setminus Z_1$, a 
contradiction to (4.2).

So we have $\bC_Y(K) = 1$. Hence $|Y : Y \cap Q_1| \le r$. As 
$[Y \cap Q_1, V_1] \le Z_1$, we get $V_3 \cap Q_1 = V_1 \cap V_3$. 
This implies
$|Y : V_1 \cap V_3| \le r^2$. As $r < q^2$ and $|Y : Y \cap Q_1| = q^u$ 
for some $u$, we get $|Y : Y \cap Q_1| = r = q$. In particular either 
$\tilde{V}_1/\bC_{\tilde{V}_1}(K)$ is irreducible or a direct sum of two 
natural $L_2(q)$ - modules. 
Furthermore $|Y/V_1 \cap V_3| = q^2$ and so by (1) $|V_1/Z_1| \le q^4$. 
Hence $K \cong Sz(q)$ or $L_2(q)$ and $|V_1/Z_1| = q^4$. But
now again some $\nu \in L_1$, $\nu$ inducing some field automorphism, 
$o(\nu) = p$, centralizes some $s \in Z_2 \setminus Z_1$, a contradiction.

\absa
\begin{equation}
K \,\, {\mbox {is a~}} p\mbox{--group}
\end{equation}

\absa
Suppose $K$ to be nonsolvable. By (4) $p \not\!\!\Big | \,r^2 - 1$. This now 
implies $K \cong L_n(r), n \le 4$, $L_n(2), 5 \le n \le 7$, 
$Sp_{2n}(r), n \le 3$, $SU_n(r), n \le 5$, $\Omega_8^-(r)$, $G_2(r)$ or 
$^3D_4(r)$.

Suppose $m_p(K) = 1$, $K \not\cong$ $^3D_4(r)$ or $SU_n(r)$. Then there 
is $\mu \in \bC_{L_1/Q_1}(K)$, $o(\mu) = p$. As $p \not\Big | \,r^2-1$
and there are at most two nontrivial irreducible modules involved, we 
see that $[\mu, \tilde{V}_1/Z_1] = 1$, contradicting 
$Z_2 \cap \tilde{V}_1 > Z_1$, by (2). 

So we have $m_p(K) > 1$ or $K \cong SU_n(r)$ or $^3D_4(r)$. As $q > 2$, 
we see that $L_n(2)$ is not possible. 
Hence in any case $K \cong SU_n(r)$ or $^3D_4(r)$. Let $K \cong SU_n(r)$. 
Then also $m_p(K) = 1$. Furthermore
$p \,\Big | \,r^2 + 1$, $\tilde{V}_1/\bC_{\tilde{V}_1}(K)$ is a direct sum of 
two natural modules
and so $r^2 = q$. Now $m_x(\langle \omega, K \rangle) \ge 4$ or 
$K \cong U_3(\sqrt{q})$. The
former is out as $N_G(\langle \omega \rangle) \not\le M$, the latter is 
out as $X \le K$ and $|X| = q$.

So we are left with $^3D_4(4)$ or $^3D_4(2)$. As $X \unlhd S/Q_1$ we see 
that $Z(S \cap K) \cap X \not= 1$. But $|Z(S \cap K)| \le 4$ and $\omega$ 
acts on $Z(S \cap K)$, so $[\omega^3, Z(S \cap K)] = 1$ and 
$[X, \omega^3] = X$, a contradiction. This proves (4).

\absa
Now we have that $K$ is a normal $p$--group. As $q > 2$, we get 
$m_p(K) \ge 3$. Let
$U = [K/\Phi(K), X]$. If $|\bC_U(t)| \ge p$, there is 
$t_1 \in X$ with $[\bC_U(t), t_1] \not= 1$. Now
$[t_1,[t,V_1]] = 1$ by quadratic action. Hence $[\bC_U(t), t_1]$ 
centralizes some
$z \in Z_2 \setminus Z_1$, a contradiction. This shows $[U,t] = U$, 
which contradicts $|X| \ge 4$.

\absa   
{\bf~(7.29) Lemma.~}{a) $Z_1 = Z(V_1).$\\
\\
\indent
b) {\it Let $q > 2$. Then $[K,\omega] \le K$.}}

\absa
Proof.~a) follows from (7.28) as $Z_2 \not\le Z(V_1)$.

b) Suppose $K_1 \times \ldots \times K_y = K^{\langle \omega \rangle}$, 
$y \ge 3$. Let
first $y > 3$. Then $m_p(K_1) = 1$ for every odd prime $p$. Now let 
$\langle \nu \rangle \le L_1 \cap L_2$, 
$\omega \in \langle \nu \rangle$, $o(\nu) = q-1$.
Let $\mu \not=1$, $\mu \in \langle \nu \rangle$, $[K_1, \mu] \le K_1$. 
Then $[K_i, \mu] \le K_i$, $i = 1,\ldots , y$.
As $N_G(\langle \mu \rangle) \not\le M$, we get $\bC_{K_1}(\mu)$ is a 
2 - group. But $K_1 \cong L_2(r)$, $Sz(r)$ or $J_1$ and so $\mu$ induces 
a field automorphism. Then $\bC_{K_1}(\mu)$ is not a 2 - group. 
So we have $K_1^{\langle \nu \rangle} = K_1 \times \ldots \times K_{q-1}$.
Suppose $y = 3$. Let $q = 64$ and $o(\omega) = 9$. As 
$[Z_1, \omega^3] \not= 1$, we have 
$\omega^3 \not\in K_1 \times \ldots \times K_y$.
Now $3 \not\in \sigma(M)$. Hence $3 \not\Big | \,|K_1|$. This implies 
$K_1 \cong Sz(r)$.
Now $\omega^3$ induces a field automorphism and so $r = r_1^3$. Hence 
$7 \,\Big | \,|K_1|$. Let now $\mu \in \langle \nu \rangle$, $o(\mu) = 7$. 
Suppose 
$[\mu, K_1] \le K_1$. Then $7 \,\Big | \,|\bC_{K_1}(\mu)|$ and so there is 
$E \cong E_{7^4}$, $\mu \in E$, $E \le L_1$, 
a contradiction to $N_G(\langle \mu \rangle) \not\le M$. Hence 
$K_1^{\langle \nu \rangle} = K_1 \times \ldots \times K_{21}$.
But as $\bC_{K_1}(\omega^3)$ is not a 2 - group this contradicts 
$N_G(\langle \omega^3 \rangle) \not\le M$. Hence in any case

\absa

\hspace{-0.6cm}(1)~~\parbox[t]{12cm} {$K^{\langle \nu \rangle} = 
K_1 \times \ldots \times K_{q-1}, 
\langle \nu \rangle \le L_1 \cap L_2, o(\nu) = q-1, K_1 
\cong L_2(r), Sz(r) \,\, {\mbox{or}} \,\,J_1$}

\absa
\setcounter{equation}{1}
\begin{equation}
q = 4
\end{equation}
Suppose $q > 4$. As 
$[\nu, S \cap K^{\langle \nu \rangle}] = Y \cap K^{\langle \nu \rangle}$
we see that there is $t \in Y \cap K^{\langle \nu \rangle}$ such that 
$\bC(t) \ge F \cong E_{p^{q-3}}$. Now
$F$ acts on $[t, V_1]$. We have $|[t, V_1] : [t,V_1] \cap V_3| = q$. 
Furthermore by (7.28) $[t, V_1] \cap V_3 \not= 1$.

There is $E \le F$, $E \cong E_{p^2}$ and 
$|\bC_{[t,V_1]}(E)| \ge 4^{q-5}$.
As $q \ge 8$, we have $4^{q-5} > q$ and so 
$\bC_{[t,V_1]}(E) \cap V_3 \not= 1$, which contradicts (7.3).

\begin{equation}
\bC_Y(K_1 \times K_2 \times K_3) = 1
\end{equation}

Let $y \in \bC_Y(K^{\langle \nu \rangle})$, $y \not= 1$. Then 
$K^{\langle \nu \rangle}$ acts on $[y,V_1]$ and 
$|[y,V_1] : \bC_{[y,V_1]}(t)| \le 16$, for 
$t \in Y \cap K^{\langle \nu \rangle}$. Now (1), (1.27) and (1.25) imply 
$K_1 \cong L_2(r)$, $r \le 16$.
But we have $[t,K_1] \not= 1 \not= [t,K_2]$ and so $t$ inverts an 
elementary abelian
group of order $17^2, 7^2$ for $r = 16$, $r = 8$, respectively. As 
$|[y,V_1] : \bC_{[y,V_1]}(t)| \le 16$, this is impossible.
Hence $K \cong L_2(4)$, $p = 5$. Now let $\rho \in K_1$, $o(\rho) = 5$. 
Then we see $|[[t,V_1], \rho]| \le 2^8$ and so $[V_1,t]$ involves at 
most two 
nontrivial irreducible $L_2(4)$ - modules. Hence we may assume 
$[[V_1,t],K_1, K_3] = 1$. As $\omega$ acts on $\{K_1,K_2,K_3\}$ 
the 3 - subgroup lemma implies $[[[V_1,t],K_1],K_2 \times K_3] = 1$. As 
$[t,[K_1,[V_1,y]]] \le V_1 \cap V_3$ this contradicts (7.3).

\absa
Now by (1), (2) and (3) $|Y : Y \cap Q_1| \le r^2$, ($64$ for 
$K \cong J_1$). As $[x,V_1] = Z_1$ for
every $x \in Z_2 \setminus Z_1$, (7.28) implies that $V_1/H$ is 
extraspecial for every hyperplane $H$ of $Z_1$.
Now $V_3 \cap Q_1 = V_3 \cap V_1$ is elementary abelian and so 
$|V_1/Z_1| \le q^2|Y : Y \cap Q_1|^2 \le 16r^4$ ($2^{16}$ for 
$K \cong J_1$).
Let $r = 2^m$. Then $(L_1/Q_1)^{(\infty)} \lesssim O^\pm(4m+4,2)$, 
($O^\pm(16,2)$ for $K \cong J_1$).
But if $L_2(r) \times L_2(r) \times L_2(r)$ or 
$Sz(r) \times Sz(r) \times Sz(r)$ acts on 
$V_1/Z_1$, then $|V_1/Z_1| \ge r^6$, and so $4m+4 \ge 6m$. This 
implies $m = 2$ and we have $L_2(4) \cong K_1$. As 
$7^3 \not\Big | \,|O_{16}^\pm(2)|$, we get $K_1 \not\cong J_1$.

So we have $|V_1/Z_1| = 2^{12}$ and $K_1 \cong L_2(4)$. Now 
$V_1/Z_1 = [V_1/Z_1,K_1] \oplus [V_1/Z_1,K_2] \oplus [V_1/Z_1,K_3]$.
Hence for some $t \in Y \setminus Q_1$, we have $[t,V_1/Z_1]$ is 
centralized by $\rho \in L_1$, $o(\rho) = 5 = p$. This implies
$[t,V_1/Z_1] \cap Z_2/Z_1 = 1$. But as 
$[t,V_1/Z_1] \cap \bC_{V_1/Z_1}(Y) \not= 1$, we get
a contradiction to (7.28).

\absa
{\bf~(7.30) Lemma.~}{\it $\bC_Y(K) = 1$.}

\absa
Proof.~
\\
\setcounter{equation}{0}
\begin{equation}
[K,Y] \le K
\end{equation}

\absa
This is clear by (7.29) if $q > 2$. So let $q = 2$, $y \in Y$ with 
$K^y \not= K$. Then
$[\bC_Y(K), K^y] = 1$. Let $\tilde{y} \in \bC_Y(K)^\sharp$, 
$W = [V_1, \tilde{y}]$. Then $K \times K^y$
acts nontrivially on $W$. As $W \le Q_2$, we see $|[W,y]| \le 4$. Now 
$[\bC_{K \times K^y}(y), [W,Y]] = 1$, contradicting (7.28).

\absa
We fix some notation. Let again $\tilde{y} \in \bC_Y(K)^\sharp$, 
$W = [V_1, \tilde{y}]$ and $\tilde{W} = [K, W]$. Then we have 
$|[t, \tilde{W}]| \le q^2$ and $|W : W \cap V_1 \cap V_3| \le q$.

\absa
Let $K$ be sporadic or alternating, $K \not\cong A_5$, $A_6$ or $A_8$. 
Then $q = 2$, otherwise $[\omega,K] \not= 1$ 
implies $[\omega, Z_1] = 1$.\\
\indent
Now $|[\tilde{W}, t]| \le 4$. This implies with (1.25) 
$K/Z(K) \cong A_n$. Suppose
$[\bC_Y(K), \tilde{W}] = 1$. Then we get that no $p$ - element 
centralizes $\bC_{\tilde{W}}(Y)$. This implies $p > 3$ and so
$n \le 11$. As $|\tilde{W} : \tilde{W} \cap V_1 \cap V_3| \le 2$, we get 
$|\tilde{W} : \bC_{\tilde{W}}(Y)| \le 2|Y : \bC_Y(K)|$.
As $|Y : \bC_Y(K)| \le 2^5$ this implies 
$|\tilde{W} : \bC_{\tilde{W}}(Y)| \le 2^6$. As
$\bC_{\tilde{W}}(Y) = Z_2/Z_1$, we see $|\tilde{W}| \le 2^7$. Hence we get 
$K \cong A_7$, $|\tilde{W}| = 16$ and $|Y : \bC_Y(K)| = 4$.
As $\bC_{V_1/Z_1}(K) = 1$, we get $W = \tilde{W}$ and so 
$|[V_1/Z_1, \tilde{y}]| = 16$. Furthermore
we have $Z_2 \le \tilde{W}$ and so if $\tilde{y_1} \in \bC_Y(K)$, 
$1 \not= \tilde{y_1} \not= \tilde{y}$, then
$[W, \tilde{y_1}] = [W, \tilde{y}]$. But this is impossible as for 
$w \in V_1 \setminus Q_2$ we have $[Y,w](V_1 \cap V_3) = V_1 \cap Q_2$. So we
get $|Y : \bC_Y(K)| = 2$ and then $|Y : Y \cap Q_1| \le 2^3$. This 
implies $|V_1/Z_1| \le 2^8$ and $L_1/Q_1 \lesssim O_8^\pm(2)$. As 
$m_p(L_1/Q_1) \ge 3$, $p \not= 3$, this impossible.

So we have $[\bC_Y(K), \tilde{W}] \not= 1$. Let 
$\tilde{y_1} \in \bC_Y(K)$ with $\tilde{W}^{y_1} \not= \tilde{W}$. Then 
as $|\tilde{W} : \bC_{\tilde{W}}(\tilde{y_1})| = 4$, we get 
$|\tilde{W}| = 4$. But then $[K,\tilde{W}] = 1$, a contradiction.

\absa
Let $K$ be a group of Lie type in odd charcteristic which is not a group 
of Lie type in even characteristic too. If $q = 2$, we get a 
contradiction with (1.30). So we have $q > 2$. Suppose first 
that $\omega$ induces a field automorphism on $K$. Then 
$E(\bC_K(\omega)) \cong G(p^h)$, where $K \cong G(p^f)$. If 
$p \in \sigma(M)$, we get $m_p(\bC_K(\omega)) = 1$. Hence 
$E(\bC_K(\omega)) \cong L_2(p)$ or $^2G_2(3)^\prime$ . Now 
$K \cong L_2(p^f)$ or $^2G_2(3^f)$. But
$[X, \omega] = X$ and so $o(\omega) = 3$ or $o(\omega) = 7$ in case of 
$K \cong$ $^2G_2(3^f)$. Now $|[\tilde{W}, t]| \le 16$, or 
$|[\tilde{W}, t]| \le 2^6$.
But this contradicts (1.29). So $p \not\in \sigma(M)$. In particular 
$m_p(K) \le 3$. Hence $f \le 3$ and so $o(\omega) = 3$ or $9$. 
Furthermore $K \cong L_2(p^3)$. But now again 
$|[\tilde{W},t]| \le 16$, a contradiction.

So we have that $\omega$ does not induce a field automorphism. Now
application of (1.10) shows that $o(\omega) = 3$, $9$ or 
$K \cong$ $^2G_2(r)$. In the latter we get $|[\tilde{W}, t]| \le 64$, a 
contradiction to (1.29). 
Hence $K \not\cong$ $^2G_2(r)$ and now $q = 4$ or $q = 64$. 

If $q = 64$, then there is $\nu \in L_1 \cap L_2$ with $o(\nu) = 7$. As 
$X \le K$, we see $[\nu,K] \le K$. Hence
$\nu$ induces a field automorphism on $K$ by (1.10). Now as before 
$K \cong L_2(p^7)$, a contradiction.

So we are left with $q = 4$, $o(\omega) = 3$. Now 
$|[\tilde{W}, t]| \le 16$. By (1.29) $K \cong L_3(3)$,
$L_2(25)$, $L_2(p)$ or $3\cdot U_4(3)$. As $3 \not\in \sigma(M)$, we 
have $K \not\cong 3\cdot U_4(3)$. As $L_3(3)$ and $L_2(25)$
do not possess an automorphism of order 3 normalizing a Sylow 
2 - subgroup, we get $K \cong L_2(p)$, $p \equiv 3,5 \,({\mbox{mod}}\, 8)$. We
have $|Y : Y \cap \bC(K)| = 4$. Suppose $[\tilde{W}, \bC_Y(K)] = 1$. Let 
$X = \langle t, t_1 \rangle$. We have
$\bC_{\tilde{W}}(Y) = \bC_{\tilde{W}}(X)$. Now 
$\bC_{\tilde{W}}(t) \not\le V_1 \cap V_3$ and
so $|\bC_{\tilde{W}}(t) : \bC_{\bC_{\tilde{W}}(t)}(t_1)| \le 4$. Hence 
$|\tilde{W} : \bC_{\tilde{W}}(Y)| \le 2^6$.
As $\bC_{\tilde{W}}(Y) \not\le Z_2/Z_1$ by (7.28), we get 
$|\tilde{W}| \le 2^8$. But $L_2(p) \not\lesssim GL_8(2)$,
as $p > 7$. So we have $[\tilde{W}, \tilde{y_1}] \not= 1$, for some 
$\tilde{y_1} \in \bC_Y(K)$. This implies $|\tilde{W}| = 4$, a 
contradiction. So we have shown
\\
\\
\hspace{-0.3cm}(2)~~\parbox[t]{12cm} { $K \cong G(r)$, $r$ even, or 
a normal $p$-subgroup, or $O_r(K) \not= 1$ for some $r$, $r$ odd, and 
$K/O_r(K) \cong L_3(r)$, or $L_2(r)$.}
\\

\absa
If $O_r(K) \not=  1$, then we have $q \le 4$. So 
$|[\tilde{W},t]| \le 16$. By (1.51) $[O_r(K), \tilde{W}] = 1$. 
Hence $[[\tilde{y}, V_1], O_r(K)] = 1$. But then $[Z_2, O_r(K)] = 1$, so 
either we have (4.9)(ii)
or we may consider $L_1$ as a component $L_3(r)$, $r$ odd. The latter 
contradicts (2). The former is a contradiction to (4.10). So

\setcounter{equation}{2}
\begin{equation}
K \cong G(r), r \,\, {\mbox{even, or a normal}} \,\, p\mbox{--subgroup}
\end{equation}

Let first $K \cong G(r)$. We have $|[\tilde{W}, t]| \le q^2$. Let  
$K$ be untwisted. Then application of (1.65) and (1.64) 
shows $q \le r$. Now we get with (1.48) that $K \cong L_n(r)$, 
$Sp_{2n}(r)$, $\Omega_{2n}^+(r)$, or $G_2(r)$. Suppose next 
$[\tilde{W}, \bC_Y(K)] = 1$. Then 
$\bC_{\tilde{W}}(S \cap K) \le Z_2/Z_1$. Hence 
$p \not\Big | \,|\bC_K(\bC_{\tilde{W}}(S \cap K))|$.

Let $V(\lambda_2)$ be involved in $\tilde{W}$. Then by (1.23) 
$K \cong L_3(r)$ or $L_4(r)$.
If $K \cong L_4(r)$, then as $|[\tilde{W}, t]| \le r^2$, we get that 
$\tilde{W} \cong V(\lambda_2)$.
Now $\bC_{\tilde{W}}(S \cap K)$ is centralized by $L_2(r)$ in all cases but 
$K \cong L_2(r)$ or $L_3(r)$. This shows $p \not\Big | \,r^2-1$.

Assume now $p \not\Big | \,r^2-1$. Then $K \cong L_n(r), n \le 4$, $L_n(2), 
n \le 7$, $Sp_4(r)$, $Sp_6(r)$, $G_2(r)$.

If for some $\tilde{y} \in \bC_Y(K)$, $[\tilde{W},y] \not= \tilde{W}$, 
then $|[\tilde{W},y]| \le q^2 \le r^2$ and so $K \cong L_2(q)$. So we have
\\
\\
\\
\hspace{-0.6cm}(4)~~\parbox[t]{12cm}{ If $K \cong G(r)$, untwisted, 
then $K \cong L_n(r), n \le 4$, $L_n(2), n \le 7$, $Sp_4(r)$, $Sp_6(r)$ 
or $G_2(r)$. If $[\tilde{W}, \bC_Y(K)] \not= 1$, then $K \cong L_2(q)$.}

\absa
Suppose next that $K$ is twisted, $K \not\cong$ $^3D_4(r)$. Let 
$q \le r$. Then by (1.48) $K \cong SU_n(r)$, $\Omega_{2n}^-(r)$ or
$Sz(r)$. Furthermore $q = r$.

Let $K \not\cong Sz(r)$ or $SU_3(r)$. If $p \,\Big | \,r^2-1$, then as 
$\tilde{W}$ involves the natural module, we see that 
$p \,\Big | \,|\bC_K(Z_2)|$, a contradiction. So $p \not\Big | \,r^2-1$. 
This shows $K \cong SU_4(r)$, $SU_5(4)$, or $\Omega_8^-(r)$.

Let now $q > r$. Then by (1.65) and (1.64) $q = r^2$ and 
$K \cong SU_n(r)$ or $\Omega_{2n}^-(r)$. Now let $K \not\cong SU_3(r)$, 
$SU_4(r)$, and $SU_5(r)$.
Then by (1.48) $\tilde{W}$ involves just natural modules and so 
$\bC_{\tilde{W}}(S \cap K)$ is
centralized by $L_2(r)$. As $p \,\Big | \,r^2 - 1$ or 
$K \cong \Omega_8^-(r)$, we 
get $|[\tilde{W}, \bC_Y(K)]| \not= 1$. But for 
$\tilde{y} \in \bC_Y(K)$, we have $|[\tilde{W}, \tilde{y}]| \le r^4$, a 
contradiction. We have
\\
\\
\\
\hspace{-0.6cm}(5)~~\parbox[t]{12cm}{ If $K \cong G(r)$, then 
$K \cong L_n(r), n \le 4$, $L_n(2), n \le 7$, $Sp_4(r)$, $Sp_6(r)$, 
$G_2(r)$, $q \le r$, 
$SU_n(q)$, $SU_n(\sqrt{q}), n \le 5$, $Sz(q)$, $\Omega_8^-(q)$, 
$\Omega_8^-(\sqrt{q})$, or $^3D_4(r)$. If 
$[\tilde{W}, \bC_Y(\tilde{W}] \not= 1$, then $K \cong L_2(q)$. }
\\

\absa
As $V_1/H$ is extraspecial for every hyperplane $H$ of $Z_1$ by (7.28), 
we get that $|W| \le q^2|Y : \bC_Y(K)|$, if $[\bC_Y(K), W] = 1$.

\absa
\setcounter{equation}{5}
\begin{equation}
K \cong L_2(r), \,\, {\mbox{or}} \,\, K \,\, {\mbox{is a normal}}\,\,p\mbox{--subgroup}
\end{equation}

\absa
Suppose false. Then $[\bC_Y(K), W] = 1$. Let first 
$K \cong SU_n(q), n \le 5$, $\Omega_8^-(q)$ or $Sz(q)$. Then 
$|W| \le q^2q, q^2q^4, q^2q^4, q^2q^6, q^2q$, respectively. But then by 
(1.48) we have $K \cong SU_4(q) \cong \Omega_6^-(q)$, 
$|W| = q^6$, or $K \cong \Omega_8^-(q)$ and $|W| = q^8$. In both cases 
$W$ is the natural module for $K$. Now choose 
$t \in X^\sharp$. As $t \in Z(S \cap K)$ we get that $|[W,t]| = q^2$.
Hence $W = [V_1,\tilde{y}]$ for every $\tilde{y} \in \bC_Y(K)^\sharp$. 
So $|\bC_Y(K)| \le q$. Now $|V_1/Z_1| \le q^{12}$ or $q^{16}$. As 
$|[V_1,y]| = q^6, q^8$, we get $|V_1/Z_1| = q^{12}$ or $q^{16}$ and 
$V_1/\tilde{W}$ is the natural module too. Hence $|\bC_Y(K)| = q$. Now 
$\bC_{L_1/Q_1}(K)$ acts on $\bC_{V_1/Z_1}(S)$ and so 
$\bC_{L_1/Q_1}(K) \cong L_2(q)$, as it contains $p$ - elements and a 
Sylow 2 - subgroup
of $L_2(q)$. Now we get that $p \,\Big | \,q^2 - 1$. But as $\tilde{W}$ is the 
natural module $Z_2/Z_1$ is centralized by $L_2(q)$, contradicting (4.2).

As $\omega$ acts fixed point freely on $X \le Z(S \cap K)$, we see that 
$K \cong SU_n(\sqrt{q})$, $\Omega_8^-(\sqrt{q})$ are not possible, since 
$|Z(S \cap K)| = \sqrt{q}$.

Let now $K \cong L_n(r)$, $n \le 4$, $Sp_4(r)$, $Sp_6(r)$ or $G_2(r)$. 
Then $|W| \le q^2r^2$, $q^2r^4$, $q^2r^3$, $q^2r^6$, $q^2r^3$. As 
$q \le r$ we have $K \not\cong G_2(r)$.
Furthermore $W$ involves exactly one nontrivial irreducible module which 
is the natural module or
$K \cong L_4(r)$, $r = q$ and $\tilde{W}$ is the orthogonal module, or 
$K \cong Sp_6(r)$, $r = q$,
and $\tilde{W}$ is the spin module. As $Z_2 \le \tilde{W}$ we get 
$\tilde{W} \cap [V_1,\tilde{y_1}] \not= 1$ for all 
$\tilde{y_1} \in \bC_Y(K)$. Hence $\tilde{W} = [[V_1, \tilde{y_1}], K]$. 
Set
$$W_1 = \langle [V_1, \tilde{y}_1] \,\Big | \,\tilde{y}_1 \in 
\bC_Y(K) \rangle $$

Now $[W_1,K] = \tilde{W}$. As $\bC_{V_1/Z_1}(K) = 1$ we see by (1.20) 
that $|W_1 : \tilde{W}| \le r$. But
$|W_1(V_1 \cap V_3) : V_1 \cap V_3| = | \bC_Y(K)|$. As 
$|\tilde{W} : \tilde{W} \cap V_1 \cap V_3| \le q$, we have
$|\bC_Y(K)| \le qr$. This now implies 
$|Y : Y \cap Q_1| \le qr^3, qr^5, qr^4, qr^7$, respectively. As 
$p \not\Big | \,r^2-1$, we get that 
$|\bigoplus\limits_{i = 0}^{p-1} \tilde{W}^{\mu^i}| \ge |\tilde{W}|^3$ 
for $\mu \in \bC_{L_1/Q_1}(K)$, $o(\mu) = p$.

So we have $|V_1/Z_1| \ge r^9, r^{12}, r^{12}, r^{18}$, respectively. As 
$|V_1/Z_1| \le q^2|Y : Y \cap Q_1|^2$, we get $|V_1/Z_1| \le q^4r^6$, 
$q^4r^{10}$, $q^4r^8$, $q^4r^{14}$.

Let $K \cong (S)L_3(r)$. If $|W| = r^3$, then 
$|Y : Y \cap Q_1| \le qr^2$ and so $|V_1/Z_1| \le q^4r^4 \le r^8$, 
a contradiction. So we have
$W > \tilde{W}$. By (1.20) $K \cong L_3(2)$. Now 
$|V_1/Z_1| \le 2^{10}$. But
$m_7(O_{10}^\pm(2)) = 1$, contradicting $p = 7$.

Let $K \cong (S)L_4(r)$. We have $|\tilde{W}| = r^4$. By (1.20) 
$\tilde{W} = W$. Now $|\bC_Y(K)| \le q$ and then $|V_1/Z_1| \le q^4r^8$. 
This shows 
$q = r$ and $|V_1/Z_1| = r^{12}$. If $q > 2$ then 
$\langle \omega, K \rangle \cong \bZ_x \times K$ and so
$m_x(\langle \omega, K \rangle) = 4$, whence $x \in \sigma(M)$, a 
contradiction to 
$N_G(\langle \omega \rangle) \not\le M$. This shows $q = 2$ and $p = 7$. But 
$m_7(O_{12}^\pm(2)) \le 2$, a contradiction.

Let $K \cong Sp_4(r)$. Then $|V_1/Z_1| = r^{12} = q^{12}$ and 
$|\bC_Y(K)| = q^2$, $|W| = q^5$. But
in this case $W_1 = W$. Hence $[V_1, \tilde{y_1}] = W$ for all 
$\tilde{y_1} \in \bC_Y(K)^\sharp$. This implies $|\bC_Y(K)| \le q$, 
a contradiction.

Let finally $K \cong Sp_6(r)$. Then $r = q$ and $|W : \tilde{W}| = q$. 
But then again
$|\bC_Y(K)| \le q$, which implies $|Y : Y \cap Q_1| \le q^7$, a 
contradiction as $|V_1/Z_1| \le q^{16}$.

Let now $K \cong L_n(2), 5 \le n \le 7$, $p \not= 3$ . We have 
$q = 2$. Furthermore $|W| \le 2^8, 2^{11}, 2^{14}$, respectively. 
Hence in any case just natural modules are involved. By
(1.20) we get $|W| = 2^n$ or $n = 7$ and $|W| = 2^{14} = |\tilde{W}|$. 
Suppose the former. Then $Z_2 \le W$ and so $W = [V_1, \tilde{y}_1]$ for every 
$\tilde{y_1} \in \bC_Y(K)^\sharp$. Now
$|\bC_Y(K)| \le 2$ and so $|Y : Y \cap Q_1| \le 2^7, 2^{10}, 2^{13}$, 
respectively. Furthermore
$|V_1/Z_1| \le 2^{16}, 2^{22}, 2^{28}$. As $p \not\!\Big | \,|\bC_K(Z_1)|$, 
we get $n \not= 6$, $p = 31$ for $n = 5$ and $p = 127$ for $n = 7$,
But $|O_{16}^\pm(2)|_{31} = 31$, $|O_{28}^\pm(2)|_{127} = 127^2$, a 
contradiction.

Hence $W = \tilde{W}$, $|W| = 2^{14}$. But $|[W,u]| = 4$ for every 
$u \in Y$ with $[K,u] \not= 1$. This implies that $u$ induces a 
transvection on the natural module. Now $|Y : \bC_Y(K)| \le 2^6$. 
Hence $|W| \le 2^8$, a contradiction.

Let finally $K \cong$ $^3D_4(r)$. By (1.40) $q > r^2$. By (1.65) and 
(1.64) $q = r^3$ or $ q = 64$ and $r = 2$.
As $[\omega,X] = X$, and $X \le Z(S \cap K)$, we get a contradiction 
since $|Z(S \cap K)| = r$.
\\
\\
\\
\hspace{0.6cm}(7)~~\parbox[t]{13cm}{ If $K$ is a normal $p$--subgroup, then $q = 2$, 
and $K \cong \bZ_5$ or $p = 3$ and $m_3(K) \le 2$.}
\\

\absa
Let $K$ be a normal $p$-subgroup, $Y_1 \le Y$, $\bC_{Y_1}(K) = 1$. Let 
$|Y_1| \ge 16$. Then there is $K_1 \le K$, 
$K_1 \cong D_{2p} \times \ldots \times D_{2p}$, $|K_1| = p^n|Y_1|$, 
where $|Y_1| = 2^n$ and $Y_1 \in {\mbox {Syl}}_2(K_1)$ by \cite[(24.1)]{GoLyS}.
For $t \in Y_1$, we have $[W,t] \le V_1 \cap V_3$. Let $t \in Y_1$ with 
$p^3 \,\Big | \,|\bC_{K_1}(t)|$. Then by (7.3) we get $[W,t] = 1$, 
a contradiction. So we have $|Y_1| \le 8$, i.e. $q \le 8$. If $q = 8$, 
we can choose $t$ such that $K_2 \cong D_{2p} \times D_{2p}$ 
acts on $[W,t]$. Now for $t_1 \in Y \cap K_2$,
we get $1 \not= [[W,t],t_1] \le Z_2$. We may choose $t_1$ with 
$\bC_{K_2}(t_1) \ge D_{2p}$. As $[Y \cap \bC_{K_2}(t_1), Z_2] = 1$, we 
get some $x \in Z_2 \setminus Z_1$, with $p \,\Big | \,|\bC_{K_1}(x)|$, 
a contradiction to (4.2).

So we have $q \le 4$. Let $q = 4$. Then $p \ge 5$. We have 
$\langle X, \omega \rangle \cong A_4$. As $|W : W \cap V_3| = 4$, we see 
that $m_p(K) = 2$. Now for a critical subgroup 
$C$ of $K$ we get $\Omega_1(C) \cong \bZ_p \times \bZ_p$ or 
$E_{p^{1+2}}$. But no such group admits $A_4$, a contradiction.

Now $q = 2$, $|W : W \cap V_3| = 2$. Again $m_p(K) \le 2$. As 
$|[W,t]| \le 4$, we see $p \in \{3,5\}$. If $p = 5$, we get 
$\bC_K(t) = 1$ and so $K \cong \bZ_5$. This proves (7).

\absa
Now by (6) and (7) we have $K_1 \times \ldots \times K_l \le L_1/Q_1$, 
$K_i \cong L_2(r_i)$, $q \le r_i$, or $K_i$ is a normal $p$--subgroup, 
$i = 1, \ldots , l$.
Furthermore $\bC_Y(K_1 \times \ldots \times K_l) = 1$. Finally at most 
one of the $K_i$ is a $p$-subgroup.

\setcounter{equation}{7}
\begin{equation}
{\mbox {If}} \quad q = 2, \,\, {\mbox {then}} \,\, K \not\cong L_2(r)
\end{equation}


Suppose false. Then as $|[W,t]| \le 4$, we see $K \cong L_2(4)$. But as 
$q \not= r$, we get that $|W| \le rq^2 = 16$. So $W = \tilde{W}$ and then 
$[\tilde{W}, \bC_Y(K)] = 1$. In particular 
$\tilde{W} = [V_1, \tilde{y_1}]$ for every 
$\tilde{y_1} \in \bC_Y(K)^\sharp$. This now implies $|\bC_Y(K)| = 2$, 
and then $|Y : Y \cap Q_1| \le 8$. Now $|V_1/Z_1| \le 2^8$. 
Hence $L_1/Q_1 \lesssim O_8^\pm(2)$. As $p^3 \,\Big | \,|L_1/Q_1|$,
we see $p = 3$. Now $V_1/W \cong W$ and so an element of order 5 acts 
fixed point freely on $V_1/Z_1$. As $3 \not\!\Big | \,|\bC_{L_1}(Z_2)|$, we 
see that $|Z_2^{L_1}| = 135$ and so all involutions in 
$V_1$ are conjugate. But there is some $\rho \in L_1$, $o(\rho) = 3$,
with $|\bC_{V_1}(\rho)| = 2^5$. Now $\Omega_1(\bC_{V_1}(\rho)) \ge Z_1$, 
and so there are involutions $i$ centralized by $\rho$.
But this contradicts $i \sim j$, $j \in Z_2 \setminus Z_1$, and 
$3 \not\Big | \,|\bC_{L_1}(j)|$.

\absa
Now by (7) and (8) we get that $K_i \cong L_2(r_i)$, $r_i \ge q > 2$, 
for all $i$.

\begin{equation}
l = 2
\end{equation}

Assume $l > 2$. Let $t_i \in K_i$. Then 
$|[V_1,t_1] : [V_1,t_1] \cap V_3| \le q$. Hence 
$[[V_1,t_1],t_2] \le V_1 \cap V_3$ and
$|[[V_1,t_1], t_2]| \le q^2$. This implies $K_3 \cong L_2(q)$. Now 
interchange $K_1, K_2, K_3$, which implies $K_i \cong L_2(q)$ for all $i$.
But then there is some $\mu \in L_1 \cap L_2$ with $o(\mu) = p$. 
This gives $N_G(\langle \mu \rangle) \le M$, as $[\mu, K_i] \le K_i$. 
But this contradicts $L_2 \not\le M$.

\absa
By (9) we get $Y \le K_1 \times K_2$, $K_i \cong L_2(r_i)$, $i = 1,2$. 
Hence there is some $\rho \in L_1$, $\rho \not\in K_1 \times K_2$ with
$o(\rho) = p$. Let $t_1 \in K_1$, $t_2 \in K_2$. We may assume 
$[\rho, t_i] = 1$. Set $W = [V_1,t_1]$.
Assume $[Y \cap K_1, W] = 1$. Then we have $|W| \le q^2r_2 \le r_2^3$. 
As $K_2$ acts on $W$ we get with (1.32) $\tilde{W} = [W,K]$ 
is of order $r_2^2$ or $r_2 = u^3$ and
$|\tilde{W}| = u^8$. We have that $\rho$ acts on 
$\bC_{\tilde{W}}(S \cap K) \le Z_2/Z_1$. Now there is $\rho_1 \in K_2$
with $\langle \rho_1 \rangle$ acts transitively on 
$\bC_{\tilde{W}}(S \cap K)^\sharp$. This shows that some $p$ - element in 
$\langle K_1 \times K_2, \rho \rangle$ centralizes some 
$x \in Z_2 \setminus Z_1$, contradicting (4.2).

So we have $[W, Y \cap K_1] \not= 1$. By (4) $K_2 \cong L_2(q)$, 
$|W| = q^4$. Now we may interchange $K_1$ and $K_2$ and so 
$K_1 \cong K_2 \cong L_2(q)$.
As $V_1/\bC_{V_1}(t_1) \cong W$, we see $|V_1/Z_1| \ge q^8$. 
But $|Y : Y \cap Q_1| \le q^2$ and so 
$|V_1/Z_1| \le q^2|Y : Y \cap Q_1|^2 \le q^6$, a contradiction.

\absa
{\bf~(7.31) Lemma.~}{\it We have $q = 2$ and $[Y,K] \not\le K$.}

\absa 
Proof.~Let $q > 2$. As $\bC_Y(K) = 1$ by (7.30) and $[Y,\omega] = Y$, 
$[Q_2, \omega] \le Y$, we see that $Y \le K$. Now $\bC_{L_1/Q_1}(K)$ 
acts on $\bC_{V_1/Z_1}(Y) = Z_2/Z_1$. If there is $\nu \in \bC_{L_1/Q_1}(K)$ with 
$o(\nu) = p$ and $[Z_1, \nu] = 1$, then $p \,\Big | \,q-1$. But now there is some
$\mu \in L_1 \cap L_2$ with $o(\mu) = p$. Then $[\mu,Z_1] \not= 1$. Hence 
$p \,\Big | \,|\bC_K(\mu)|$, $p \,\Big | \,|\bC_{\bC_{L_1/Q_1}(K)}(\mu)|$, and 
so $m_p(\bC_{L_1}(\mu)) \ge 3$, which contradicts 
$N_G(\langle \mu \rangle) \not\le M$. Hence we have shown that 
$\bC_{L_1}(Z_1)/Q_1 \lesssim {\mbox {Aut}}(K)$.

We are going to prove the same result for $q = 2$. 
Let $K_1 \le \bC_{L_1/Q_1}(K)$. By (7.30) $[Y,K_1] \not= 1$. 
Hence we may assume that $K$ is nonsolvable. Let $V_1^{(1)} \le V_1/Z_1$ 
be an irreducible $K$ - submodule, such that $N_Y(V_1^{(1)})$ is maximal. 
Suppose $Y \not= N_Y(V_1^{(1)})$. Then there is 
$y_1 \in Y \setminus N_Y(V_1^{(1)})$ with 
${V_1^{(1)}}^{y_1} = V_1^{(2)} \not= V_1^{(1)}$. Set 
$[V_1^{(1)} \oplus V_1^{(2)}, y_1] = \tilde{V}$. Then 
$|\tilde{V} : \tilde{V} \cap V_3| \le 2$. Let 
$y_2 \in Y \setminus \langle N_Y(V_1^{(1)}), y_1 \rangle$. We have 
$|\tilde{V} : \bC_{\tilde{V}}(y_2)| \le 4$. As $|V_1^{(1)}| > 4$ 
($K$ is nonsolvable), we get 
$\bC_{V_1^{(1)} \oplus V_1^{(2)}}(y_2) \not= 1$. Hence 
$W_1 = (V_1^{(1)} \oplus V_1^{(2)})^{y_2} \cap (V_1^{(1)} 
\oplus V_1^{(2)}) \not= 1$. Now $W_1$ is a $K$ - module. If 
$W_1 \not= V_1^{(1)} \oplus V_1^{(2)}$, then $W_1 \cong V_1^{(1)}$ and 
$y_2 \in N_Y(W_1)$. But $N_Y(V_1^{(1)}) \le N_Y(V_1^{(1)} \oplus V_1^{(2)})$ 
and so $N_Y(V_1^{(1)}) \le N_Y(W_1)$. By the maximality we now get 
$y_2 \in N(V_1^{(1)} \oplus V_1^{(2)})$. So we have shown 
$|Y : N_Y(V_1^{(1)})| \le 2$. By (7.28) $N_{L_1/Q_1}(K)$ 
acts on $V_1^{(1)} \oplus V_1^{(2)}$, as 
$Z_2/Z_1 \le V_1^{(1)} \oplus V_1^{(2)}$. In particular 
$\bC_{K_1}(V_1^{(1)} \oplus V_1^{(2)}) = 1$. Let $U = S \cap K_1$. 
Then by (7.30) $[U,Y] = 1$. As $|\bC_{V_1^{(i)}}(Y)| = 2$, we get 
$|U : \bC_U(V_1^{(1)} \oplus V_1^{(2)})| = |Y : N_Y(V_1^{(1)})| \le 2$. 
Hence $K_1$ is solvable. Let $E \le K_1$ with $E \cong E_{p^3}$. 
Then there is $F \le E$, $|F| = p$ with 
$[F, V_1^{(1)} \oplus V_1^{(2)}] = 1$. 
But this contradicts $Z_2/Z_1 \le V_1^{(1)} \oplus V_1^{(2)}$ and (4.2). 
We have $m_p(K_1) \le 2$ and so $|Y :\bC_Y(K_1)| \le 4$. 
By (7.30) we get $|Y : Y \cap Q_1| \le 4$ and so $|V_1/Z_1| \le 2^8$, if 
$|Y : N_Y(V_1^{(1)})| = 2$. If $Y = N_Y(V_1^{(1)})$, then 
$m_p(K_1) = 1$, and so $|Y : Y \cap Q_1| = 2$, which shows 
$|V_1/Z_1| \le 2^6$.

Suppose first $|V_1/Z_1| \le 2^6$. As $m_p(L_1/Q_1) \ge 3$, 
we get $L_1/Q_1 \lesssim O_6^-(2)$. Now as $K$ is nonsolvable 
$5 \,\Big | \,|K|$. But in $O_6^-(2)$ no 5 - element is centralized by a 
3 - element, a contradcition. So we have $L_1/Q_1 \lesssim O_8^\pm(2)$ 
and $m_3(K_1) = 2$. But in $O_8^\pm(2)$ no 5 - or 7 - element 
$\tau$ with $|[V_1/Z_1, \tau]| \ge 2^6$ is centralized by $E \cong E_9$, 
a contradiction again. So we have

\setcounter{equation}{0}
\begin{equation}
\bC_{L_1}(Z_1)/Q_1 \lesssim {\mbox {Aut}}(K)
\end{equation}

In particular $m_p(K) \ge 2$.

\begin{equation}
{\mbox {If}} \,\, q > 2, \,\, {\mbox {then}} \,\, K \cong G(r)
\end{equation}

Suppose first that $K$ is sporadic or alternating. Then as 
$[\omega, K] \not= 1$ and $K$ has no outer automorphism of odd order, 
we see that $K \cong A_5$ or $J_1$ by (1.9), contradicting 
$m_p(K) \ge 2$. Suppose that $K$ is a normal $p$-subgroup. Now $m_p(K) \ge 3$. 
Let $t \in X^\sharp$. Then $|[V_1,t] : [V_1,t] \cap V_3| = q$. As 
$\bC_{V_1/Z_1}(Y) = Z_2/Z_1$, we see $Z_2 \cap [V_1,t] \not= 1$. Let 
$|Y : Y \cap Q_1| \ge 2^5$. Then as $[t, V_1,s] \le V_1 \cap V_3$ for 
$s \in Y$, we see that there is some 
$1 \not= u \in V_1 \cap V_3 \setminus Z_1$, $u$ is centralized by 
$E \cong E_{p^2}$ by \cite[(25.12)]{GoLyS}. But this contradicts (7.3). So we have 
$|Y| \le 2^4$. Now $|V_1/Z(V_1)| \le 2^{16}$. If $q = 4$ or $16$, 
we have $p \not= 3$. Let $|Y| = 16$. Then we see that 
$m_p(O_{16}^\pm(2)) \ge 5$, a contradiction. Hence $q \le 8$. 
Let $q = 4$, then $|Y| = 4$ and so $|V_1/Z(V_1)| \le 2^8$. 
But now $m_p(O_8^\pm(2)) \le 2$, a contradiction. So we have $q = 8$, 
$|Y| = 8$ and $|V_1/Z(V_1)| \le 2^{12}$. Now we get $p = 3$. By (7.28)
$Z(V_1) = Z_1$ and so $|V_1/Z_1| \le 2^{12}$.  As $m_3(O_3(K)) = 6$, 
we get some $\rho \in O_3(K)$, $|\bC_{V_1}(\rho)/Z_1| = 2^{10}$. Hence 
$\bC_{V_1}(\rho) \cap Z_2 \not\ge Z_1$, a contradiction.


\begin{equation}
{\mbox{If}} \,\, q = 2, \,\, {\mbox{then}} \,\, K \cong G(r), r \,\, {\mbox{even}}
\end{equation}

   
Choose $t \in X$. Then $|[t,V_1] : [t, V_1] \cap V_3| = 2$. 
Suppose there is $E \cong E_{p^3}$, $E \le \bC(t)$, then there is 
$u \in V_3 \cap V_1$, $u \not\in Z_1$, $u$ centralized by 
$F \le E$, $|F| = p^2$. But this contradicts (7.3). 
Furthermore $|V_1 : \bC_{V_1}(t)| \le 2|Y|$. 

Let first $K$ be sporadic. Then by (1.25) 
$K \cong M_{11}$, $M_{12}$, $M_{22}$, $M_{23}$, $M_{24}$, $J_2$, $HiS$, 
$Co_2$ or $Co_1$. Now in any case $m_p(K) \ge 3$. This shows 
with (1.4) $p = 3$ and $K \cong Co_1$ or $Co_2$. But then $\bC(t)$ 
involves $Sp_6(2)$ or $U_4(3)$. In both cases $m_3(\bC(t)) \ge 3$, 
a contradiction.

Let $K \cong A_n$. Then also $m_p(K) \ge 3$, and then $p = 3$ by (1.4). 
As $m_3(K) \ge 3$, we get $n \ge 9$. Furthermore we get $n \le 18$. 
We have that $|Y : Y \cap Q_1| \le 2^{\lfloor\frac{n}{2}\rfloor}$. 
Hence $|V_1/Z_1| \le 2^{n+2}$. As $3 \not\!\Big | \,|\bC_K(Z_2)|$, 
we get with (1.9) that $|K|_{2^\prime} \le 2^{n+2} - 1$, 
a contradiction.

Let now $K$ be a group of Lie type in odd characteristic which is not 
a group of Lie type in even characteristic too. By (1.41) we get 
$K \cong 3\cdot U_4(3)$, $L_3(3)$, $U_4(3)$ or $L_4(3)$. 
As $m_3(K) \ge 3$, we get $K \not\cong L_3(3)$. By (1.10) 
$\bC_K(Z_2) \in {\mbox {Syl}}_2(K)$, as $3 \not\Big | \,|\bC_K(Z_2)|$. 
But $|Y : Y \cap Q_1| \le 2^6$ and so $|V_1/Z_1| \le 2^{14}$ and 
$|K|_{2^\prime} > 2^{14} - 1$, a contradiction.

Let now $K$ be a normal $p$-subgroup. Let $|Y : Y \cap Q_1| \ge 16$. Then 
by \cite[(24.1)]{GoLyS} there is $y \in Y \setminus Q_1$, $m_p(\bC_{K}(y)) \ge 3$. 
We have $|[y,V_1] : [y,V_1] \cap V_3| = 2$. 
By (7.3) we see $[y,V_1] \cap V_3 = 1$. But then $[Y,[y,V_1]] = 1$ 
and so $Z_2 = [y,V_1]$ by (7.28), a contradiction. So we have 
$|Y : Y \cap Q_1| \le 8$ and then $|V_1/Z_1| \le 2^8$. 
This now implies $p = 3$.

Let first $|V_1| = 2^7$. Then $L_1/Q_1 \lesssim O_6^-(2)$. 
As $V_1 \cap V_3$ is elementary abelian we get $|V_1 \cap V_3| \le 8$ 
and so $|Y : Y \cap Q_1| = 8$. This now implies 
$\Sigma_3 \times \Sigma_3 \times \Sigma_3 \lesssim L_1/Q_1$. For 
$y \in Y \setminus Q_1$, we have $|[V_1,y]Z_1/Z_1| \le 8$. 
But then some 3 - element in $L_1$ centralizes $Z_2$, 
a contradiction to (4.2).

Hence we have $|V_1| = 2^9$ and $L_1/Q_1 \lesssim O^+(8,2)$. 
Let $\hat{Q}_1 = \bC_{Q_1}(V_1)$. Then $\hat{Q}_1 \cap V_1 = Z_1$, 
$\hat{Q}_1 \le Q_2$ and $\bC_G(x) \le M$ for every 
$x \in \hat{Q}_1^\sharp$. We have 
$[V_3 \cap L_1, \hat{Q}_1] \le \hat{Q}_1 \cap V_3 \le Z_1$. Let 
$u \in \hat{Q}_1$, $u \not\in Q_3$. Then $u$ acts on an 
elementary abelian group of order $3^3$. Suppose 
$[V_3 \cap L_1,u] = Z_1$. We have $|[V_3,u]| \le 4$. 
But then there is some 3 - element $\nu$ in $L_3$ with 
$[\nu,Z_1] = 1$, a contradiction to (4.2). So we have 
$[V_3 \cap L_1, \hat{Q}_1] = 1$. As $[\hat{Q}_1, V_1 \cap L_3] = 1$, 
we see that $V_1 \cap L_3$ acts on $[V_3, \hat{Q}_1]$. 
By (7.28) we get $[V_3, \hat{Q}_1] \le Z_3$. Now as 
$[V_3 \cap L_1, \hat{Q}_1] = 1$, we get 
$|\hat{Q}_1 : \hat{Q}_1 \cap \hat{Q}_3| \le 2$. As $M \not= M_3$, 
we have $|\hat{Q}_1| = 2$ and so $\hat{Q}_1 = Z_1$. This 
implies $Q_1 = V_1$. Now choose $x \in V_3 \cap L_1$, such that 
$x$ centralizes $E \cong E_9$ in $O_3(L_1/Q_1)$. 
We have $|[x,V_1]Z_1/Z_1| = 16$ and $E$ acts faithfully on 
$[x,V_1]$. Hence $[x,V_1]Z_1$ is elementary abelian of order 32 
and if we choose $x$ with $x^2 = 1$, then 
$\langle x, [x,V_1], Z_1 \rangle = F$ is elementary abelian of order 64. 
Furthermore $F$ is the only elementary abelian group of order 
64 in $\langle V_1,x \rangle$. This shows that $E \le N_G(F)$ and so 
$N_G(F) \le M$.

Now let $\rho \in L_2$ with $o(\rho) = 3$. Set 
$F_1 = \langle V_1 \cap V_3, x, x^\rho \rangle$. Then $|F_1| = 2^7$. 
Hence $F \le F_1$. As $|V_1 \cap V_3 : \bC_{V_1 \cap V_3}(x)| = 2$, 
we see that $|Z(F_1)| = 2^4$. Hence there is exactly one elementary 
abelian group of order 64 in $F_1$. This implies 
$\rho \in N_G(F) \le M$, a contradiction.

\absa
\begin{equation}
K \cong G(r), r \,\, {\mbox{even}} 
\end{equation}

\absa
By (2) and (3) it is enough to treat the case that $K \cong G(r)$, 
$r$ odd, and $K$ is not a group of Lie type in characteristic two. 
Furthermore $q > 2$. As $[\omega,Z_1] = Z_1$, 
we get that $\omega$ induces an automorphism on $K$ which normalizes a 
Sylow 2 - subgroup of $K$. In case of $q = 64$ let 
$\nu \in L_1 \cap L_2$ with $o(\nu) = 7$. Then also $\nu$ 
induces an automorphism on $K$ which normalizes a Sylow 2 - subgroup.

Suppose that $\nu$ or $\omega$ induce field automorphisms. Let 
$K_1 = \bC_K(\omega)$. Then $m_p(K) \le 1$. Let $r = p^f$, 
$p \in \sigma(M)$. Then we get $K_1 \cong L_2(p)$ and 
$K \cong L_2(p^f)$. Now $m_2(L_1/Q_1) = 2$, and so 
$|Y : Y \cap Q_1| = 4$. Hence $|V_1/Z_1| \le 2^8$ and so 
$K \lesssim O_8^\pm(2)$. This implies $p = 3$. But then also 
$f = 3$ and so $o(\omega) = 3$ , contradicting $3 \in \sigma(M)$.

So we have $p \not\in \sigma(M)$. Now $m_p(K) \le 3$. But as 
$f \ge 3$, we get $K \cong L_2(r) = L_2(p^3)$ again. This contradicts 
$m_u(K) \ge 3$ for $u \in \sigma(M)$.

So we have shown that neither $\omega$ nor $\nu$ induces a field 
automorphism. Application of (1.10) shows that $o(\omega) = 3$ or 
$K \cong$ $^2G_2(r)$.

Suppose $K \cong$ $^2G_2(r)$. Then $|Y : Y \cap Q_1| \le 8$ and so 
$|V_1/Z_1| \le 2^{12}$, $K \lesssim O_{12}^\pm(2)$. But then by order 
consideration we get $r = 3$, a contradiction as 
$L_2(8)\cdot 3 \cong $ $^2G_2(3)$.

So we have $q = 4$, $o(\omega) = 3$. This now implies that 
$3 \not\in \sigma(M)$. Let $m_2(L_1/Q_1) \le 6$. Then we get 
$K \lesssim O_{16}^\pm(2)$ and so either $K \cong L_2(r)$ 
or $K$ contains a nonabelian subgroup of order $u^3$ for some odd prime 
$u$. The latter now implies $r = 3^f$. As $m_3(K) \le 3$, we get 
with (1.5) $K \cong L_3(3)$, $U_3(3)$, $PSp_4(3)$, or $G_2(3)$. 
But as $3 \not\in \sigma(M)$ and $m_p(K) = 1$ for $p \not= 3$, $p$ odd, 
this is a contradiction.

Let $K \cong L_2(r)$. Then $|Y| = 4$ and so $|V_1/Z_1| \le 2^8$. 
As $r = p^f$, $f \ge 3$, we get a contradiction. 

So we have $m_2(L_1/Q_1) > 6$. Application of (1.3) shows that 
$m_3(K) = 3$, $K \cong Sp_6(r)$, $\Omega_8^-(r)$, $L_6(r)$, $U_6(r)$, 
$L_7(r)$, or $U_7(r)$. Now $3 \not\Big |  \,\,r$ and $3 \not\Big | \,r-1$, 
$3 \not\Big | \,r+1$ in case of $K \cong U_6(r)$ or $U_7(r)$. Hence 
never $K$ possesses a diagonal automorphism of order 3. 
As $\omega$ does not induce a field automorphism, we see 
$\langle \omega, K \rangle \cong \bZ_3 \times K$, a contradiction to 
$m_3(\bZ_3 \times K) = 4$ and $3 \not\in \sigma(M)$.
\\
\begin{equation}
p \not\Big | \,r^2 - 1
\end{equation}

\absa
Let $K \not\cong L_2(r)$, $Sz(r)$, or $U_3(r)$. Then by (1.53) there 
is some minimal parabolic $P$ of $K$ such that $Y \le O_2(P)$ and 
$P/O_2(P) \cong L_2(r)$ in case of $^2F_4(r)$. Hence 
$P/O_2(P) \cong L_2(r)$ or $U_3(r)$ and so by (7.28) 
$\bC_{V_1/Z_1}(O_2(P)) \le Z_2/Z_1$. If $[\omega,P] \le P$, then 
there is some $\nu \in P$, $o(\nu) = p$ with 
$[Z_2,\nu] = 1$, a contradiction to (4.2) and $M \not= M_3$. 
Hence we may assume $[\omega,P] \not\le P$. Then we have 
$K \cong \Omega_8^+(r)$ and $\omega$ induces a graph 
automorphism of order $3$. In particular $3 \not\in \sigma(M)$. 
But this contradicts $m_3(\Omega_8^+(r)) \ge 4$.

So assume $K \cong L_2(r)$, $Sz(r)$ or $U_3(r)$. As $m_p(K) \ge 2$, 
we see $K \cong U_3(r)$, $p \,\Big | \,r+1$. Now 
$|Y : Y \cap Q_1| \le r$ and so $|V_1/Z_1| \le r^4$. But then 
$|[V_1,t]Z_1/Z_1| \le r^2$, and so by (1.26) $|V_1/Z_1| \ge r^6$, 
a contradiction.

\absa
By (5) $K \cong L_n(r), n \le 4$, $L_n(2), 5 \le n \le 7$, 
$U_n(r), n \le 5$, $Sp_{2n}(r), n \le 3$, $G_2(r)$, $^3D_4(r)$, 
$^2F_4(r)$, $\Omega_8^-(r)$. But we have $m_p(K) \ge 2$. 
This implies $K \cong L_6(2)$ or $L_7(2)$, $p = 7$, $^2F_4(r)$, 
$p \,\Big | \,r^2+1$, or $^3D_4(r)$, $p \,\Big | \,r^6-1$.

If $K \cong$ $^2F_4(r)$, or $^3D_4(r)$, then by (1.53) there is 
a parabolic $P$ with $Sz(r) \lesssim P/O_2(P)$, or 
$L_2(r^3) \lesssim P/O_2(P)$ and $Y \le O_2(P)$. 
But then we get the same contradiction as in (5).

So we are left with $K \cong L_6(2)$ or $L_7(2)$ and $p = 7$. 
Then $q = 2$. But as $m_7(L_1) \ge 3$, this contradicts (1).

\absa
{\bf~(7.32) Proposition.~}{\it If $b = b_2$, then $[Z_2, Z_{\alpha}] =1$. }

\absa
Proof.~Suppose $[Z_2,Z_\alpha] \not= 1$. Then by (7.31) we have 
$q = 2$ and $[K,Y] \not\le K$. Let $y \in Y$, $K^y \not= K$. As 
$Y \unlhd S$, we see that $K$ possesses abelian Sylow 2 - subgroups. 
Hence $K \cong L_2(r)$, or $J_1$. Now $\bC_{K^y \times K}(y) \cong K$ 
acts on $[y,V_1]$. If $1 \not= t \in \bC_{K^y \times K}(y) \cap Y$, then 
$|[[y,V_1],t]| \le 4$. By (7.30) $Z_2/Z_1 \le [y,V_1]$ and so 
$\bC_{K^y \times K}(y)$ acts faithfully on $[y,V_1]$. Application 
of (1.27) shows $K \cong L_2(4)$. By (7.30) we see 
$|Y : Y \cap Q_1| \le 8$ and so $|V_1/Z_1| \le 2^8$. Now 
$L_1/Q_1 \lesssim O_8^\pm(2)$. Furthermore $p = 3$. But then 
$3 \,\Big | \,|\bC_{L_1/Q_1}(K^y \times K)|$. As 
$5^2 \,\Big | \,|L_1/Q_1|$, we have $L_1/Q_1 \lesssim O_8^+(2)$. 
Now we have $|Z_2^{L_1}| = 135$ as no 3 - element centralizes 
$Z_2$. But as $E \cong E_{27}$ is contained in $L_1$, we get some 
3 - element $\rho$ with $|\bC_{V_1}(\rho)| \ge 2^5$. Now $\rho$ 
centralizes some involution $i \not\in Z_1$. As all involutions 
in $V_1 \setminus Z_1$ are conjugate under $L_1$ 
(there are exactly 270 such involutions), we get 
$i \sim j \in Z_2 \setminus Z_1$, a contradiction.

















\begin{center}
\S~8 The case $[Z_2,Z_\alpha] = 1$
\end{center}

\vspace{1cm}

In this final chapter we consider the case $b = b_2$. By (7.32) we know $[Z_2, Z_\alpha] = 1, \alpha \sim 1$. 
We are going to work under

\absa
{\bf (8.1)~Hypothesis.~} $L_1$ is not as in (4.9)(ii).

\absa
The construction of $L_1$ in (4.4) now implies $L_1 = \bC_{L_1}(Z_1)(L_1 \cap L_2)$, $m_p(\bC_M(Z_1)) \ge 3$. 
In particular $Z_1 \le Z_2$, or we are in the situation of (4.13).

\vspace{5mm}

Suppose first $b > 1$. Then $b \ge 3$ and we may apply (5.6). Remember 
that in the case of (4.13) we have the situation of (5.6) anyway.

Set
\begin{center}
$W_1 = \langle {Z_2}^{L_1} \rangle$ and 
$ V_1 = \langle (\bC_{Z_2/Z_1}(Q_1)$ \, mod \, $Z_1)^{L_1} \rangle$.
\end{center}

We have
$$[V_1, Q_1] \le Z_1.$$

Now $W_1 = V_1$ in case of $E(L_2/Q_2) \cong L_2(q)$ and $Z_2$ is a 
direct sum of at most two natural modules.

If $E(L_2/Q_2) \cong A_9$, $E(L_2/Q_2) \cong L_2(q)$ and $Z_2$ is the 
orthogonal module, or $E(L_2/Q_2) \cong L_2(q) \times L_2(q)$ and 
$Z_2$ is the orthogonal
 module, then $W_1$ involves at least three nontrivial irreducible 
$L_1$ - modules as has been seen in \S~5. 

If $E(L_2/Q_2) \cong L_2(4) \times L_2(4)$ and $Z_2$ is a sum of two 
orthogonal $L_2(4)$ - modules, then $J(S) \not\le Q_1$. Then $m_p(L_1) \geq 3$. Now by (5.4) 
for every $x \in \bC_{Z_2}(Q_1)^\sharp$ we have $\bC_G(x) \le M$, a contradiction, as 
$\langle \bC_{L_2}(x) \Big | 1 \not= x \in \bC_{Z_2}(Q_1) \rangle = 
L_2$, or we have (5.4)(i). In this case $m_p(\langle J(S)^{L_{1}} \rangle ) \leq 1$. Set $X = \langle J(S)^{L_1} \rangle Q_1/Q_1$. As $m_p(N(J(S))) \geq 2$ we get with (1.63) that $J(S) \not\leq Q_2$. Let $U \in $Syl$_3(\langle J(S)^{L_1} \rangle )$ with $|U \cap G_2| = 9$. Then $m_p(N_{L_1}(U)) \geq 2$. As $3 \not\in \sigma(M)$, we have $m_3(U) \leq 3$. If $m_p(\bC(U)) \geq 2$, then we get $G_2 = N_{G_2}(U \cap G_2)(G_2 \cap M) \leq M$ by (1.63). So we have $m_p(\bC(U)) \leq 1$ and then $p = 5$ or $13$ and $m_p(\bC(U)) = 1$. Further $m_p(X) = 1$ and $m_p(L_1) = 3$. Let $C$ be a critical subgroup of $U$. Then $C \cong E_{27}$ or $E_{3^{1+4}}$. Let $K_1$ be a component of $L_1/Q_1$. Then we have $C \cap K_1 = 1$ or $C \leq K_1$. If $K_1 \leq X$, we get $C \leq K_1$ and so $K_1$ possesses an outer automorphism of order $p$. Now with (1.67) and $m_3(K_1) = 3$, we see that $K_1$ is a group of Lie type in characteristic two. Now as $m_p(K_1) = 1$, this automorphism has to be a field automorphism. But this contradicts (4.12). So we have $F^*(L_1/Q_1) \cap X \leq F(L_1/Q_1)$ This shows $\bC_X(F(X)) \leq F(X)$ and $m_r(F(X)) \leq 3$ for any odd prime $r$. So $C$ cannot act nontrivially on any Sylow $r$-subgroup of $F(X)$ for $r \not= 3$, which shows $C \leq F(X)$. But now some elementary abelian group $F$ of order $p^3$ normalizes $C$ and so some $F_1 \leq F$, $|F_1| = p^2$ centralizes $U \cap L_2$, a contradiction. \\
\\
\indent
In all cases $[W_3, W_\alpha] \le W_3 \cap W_\alpha$ and we 
have quadratic action.

\absa
{\bf (8.2)~Hypothesis.~} There is a component $K_\alpha$ of 
$L_\alpha/Q_\alpha$ with $1 \not= [Z_2, K_\alpha] \le K_\alpha, 
m_p(K_\alpha) \ge 2$.

\absa
Set $\tilde{V_\alpha} = [V_\alpha, K_\alpha]$. Let $K_3$ be the 
corresponding component to $K_\alpha$ in $L_3/Q_3$. 

\absa
{\bf (8.3)~Lemma.~}{\it Assume (8.1) and (8.2). Then if $K_\alpha/Z(K_\alpha)$ is a group of 
Lie type in odd characteristic it is a group of Lie type in even 
characteristic too.}

\absa
Proof.~ Suppose false.\\
\setcounter{equation}{0}%
\begin{equation}
 K_\alpha/Z(K_\alpha) \cong U_4(3) 
\end{equation}


If $|W_3 : \bC_{W_3}(K_\alpha)| \ge 4$, then (1) follows from (1.12). 
So assume $|W_3 : \bC_{W_3}(K_\alpha)| \le 2$. The same applies for 
$|W_\alpha : \bC_{W_\alpha}(K_3)|$. Let $x \in \tilde{V_\alpha}$. Then 
$|[W_3,x] : [W_3,x] \cap Z_\alpha| \le 2$. Suppose 
$[x,K_3] \le Q_3, x \not\in Q_3$. As $m_p(K_3) \ge 2$ and 
$p \not\!\!\Big|~ |\bC_{K_3}(u)|$, for $ u \in Z_\alpha^\sharp$ by (4.2) and (5.7), we 
get $|[W_3,x]| = 2$. As $p \Big | |\bC_{L_3}(\tilde{V_3})|$ we have that $p \Big | |\bC_{L_\alpha}(\tilde{V_\alpha})|$. 
But $[W_3,x]$ is 
centralized by $E \cong E_{p^2}, \Gamma_{E,1}(G) \le M_3$, and so (4.2) 
and (5.7) yield a contradiction. We have $x \in Q_3$.

So we get $|\tilde{V_\alpha} : \tilde{V_\alpha} \cap L_2| \le 2$. As 
$Z_2$ cannot induce transvections on $\tilde{V_\alpha}$, we get 
$[\tilde{V_\alpha} \cap L_2, Z_2] \not= 1$. Let $x \in \tilde{V_\alpha} \cap L_2$. Then again 
$|[Z_2,x] : [Z_2,x] \cap Z_\alpha| \le 2$. By (5.7) we have 
$Z_\alpha \cap Z_\beta = 1$ for
 all $\beta \in \Delta(2)$. This now implies $E(L_2/Q_2) \cong L_2(4)$ 
or $L_2(4) \times L_2(4)$, or $L_2$ is solvable. Furthermore 
$|Z_\alpha| \le 4$. Hence $
|[W_3,x]| \le 8$ for $x \in \tilde{V_\alpha}$. Application of (1.41) 
shows $[\tilde{V_\alpha}, K_3] \le Q_3$ and so 
$[\tilde{V_\alpha}, Z_2] \le Z_2 \cap \tilde{V_\alpha}$. Now we see that there is $y \in Z_2, [y,K_\alpha] \not= 1, 
|\tilde{V_\alpha} : \bC_{\tilde{V_\alpha}}(y)| \le 4$, and so again 
$K_\alpha/Z(K_\alpha) \cong U_4(3)$ by (1.41).
\\
\begin{equation}
 L_2/Q_2 \cong F_{10}, F_{20}, or \Sigma_3.
\end{equation}


By (1) $3 \in \sigma(M)$ and so $m_3(L_2) \le 1$. Let 
$\omega \in L_3 \cap L_2, o(\omega) > 3$. We have 
$[\omega, K_3] \le K_3$ by (1.63) and then $[\omega, K_3] = 1$. But this contradicts $N_G(\langle \omega \rangle) \not\le M$.

Now by (2) $|Z_\alpha| \le 4$. For $x \in \tilde{V_\alpha}$ we have 
$|[W_3,x]| \le 2^6$. Suppose $|W_3 : \bC_{W_3}(K_\alpha)| \ge 4$. 
Then by (1.12) just 12 - 
dimensional modules are involved. Now let $x \in \tilde{V_\alpha}, 
1 \not= [x,K_3] \le K_3$. Then $W_3$ involves exactly one irreducible 
nontrivial module, and
 by (5.5) $L_2/Q_2 \cong \Sigma_3$. (Remember that (5.5) may be applied. 
If $m_p(L_1) = 2$, then by the construction in (4.13) $L_1 < \hat{L_1}, 
\hat{L_1}$ satisfies (3.2). Furthermore $K_1$ is also a component of $\hat{L_1}$.) But 
then $3 \Big | |\bC_{K_3}(Z_2)|$, a contradiction to (4.2) and (5.7).

Let ${K_3}^x \not= K_3$. Then $x$ acts fixed point freely on a group of 
order $3^4$. Hence there is $\rho \in K_3 \times {K_3}^x, o(\rho) = 3, 
|[W_3, \rho]| =4$. But then again there is exactly one nontrivial irreducible $K_3$ - module involved, a contradiction to (4.2) and (5.7).

Suppose $|W_3 : \bC_{W_3}(K_\alpha)| = 2$. Then $|[W_3,x]| \le 2^3$, and 
so ${K_3}^x \not= K_3$, but $x$ inverts $E \cong E_{3^4}$, a contradiction.

So we have $[\tilde{V}_\alpha, K_3] \le Q_3$. Let $\tilde{V}_\alpha \not\leq Q_3$. Then $[K_3,[V_3, \tilde{V}_\alpha]] \leq Z_3$ and so by the 3 - subgroup lemma we get $[\tilde{V}_3, \tilde{V}_\alpha] \leq Z_3$. Hence a $p$-element centralizes $\tilde{V}_3$. If $[\tilde{V_3}, \tilde{V_\alpha}] \not= 1$, then we have a contrdiction to (4.2) and (5.7). So we have $[\tilde{V}_3, \tilde{V}_\alpha] = 1$. Now as above we arrive at a contradiction. So we see $\tilde{V_\alpha} \le Q_3$ and $[\tilde{V_\alpha}, Z_2] \le Z_2 \cap \tilde{V_\alpha}$. By (2) $Z_2$ induces transvections on $\tilde{V_\alpha}$, a contradiction to (1.18).

\absa
{\bf (8.4) Lemma.~}{\it Assume (8.1) and (8.2). Then $K_\alpha/Z(K_\alpha)$ is not sporadic}

\absa
Proof.~ Suppose $K_\alpha/Z(K_\alpha)$ to be sporadic. Let $\omega \in L_2 \cap L_3, 
o(\omega) = r > 1$, $N_G(\langle \omega \rangle) \not\le M$. Then as 
$m_p(K_3) \ge 2$,
 we get $[\omega, K_3] \le K_3$ with (1.63) and $[K_3, \omega] \not= 1$. 
Now (1.9) implies $o(\omega) = 3$. This implies $|Z_\alpha| \le 4$. 

Let $|W_3 : \bC_{W_3}(K_\alpha)| = 2$. Then for $x \in \tilde{V_\alpha}$ 
we have $|[W_3,x]| \le 8$. Suppose $ 1 \not= [K_3,x] \le K_3$. Then 
by (1.25) $W_3$ in
volves exactly one nontrivial irreducible module, $K_3/Z(K_3) \cong 
M_{22}$ and $L_2/Q_2 \cong \Sigma_3$ by (5.5). Furthermore by (1.30) 
$|\tilde{V_3}/\bC_{\tilde{V_3}}(K_3)| = 2^{10}$. But now $3 \Big | |\bC_{K_3}(Z_2)|$, a 
contradiction as $3 \in \sigma(M)$.

If ${K_3}^x \not= K_3$, there is some 5- or 7-element $t$ in ${K_3}^x 
\times K_3$ with $|[W_3,t]| = 16$ or $2^6$, and again there is just 
one nontrivial irreducible $K_3$ - module in $W_3$. Hence we have $[\tilde{V_\alpha}, K_3] = 1$. 

Let $|W_3 : \bC_{W_3}(K_\alpha)| \ge 4$. Then by (1.12) 
$K_\alpha/Z(K_\alpha) \cong M_{12}$, $M_{22}$, $M_{24}$, $J_2$, $Co_1$, 
$Co_2$ or $Sz$. Now $K_\alpha \cong 3 \cdot M_{22}$ or $|W_3 : \bC_{W_3}(x)| \le 16$ for 
$x \in \tilde{V_\alpha}$. In the latter as before 
$[\tilde{V_\alpha}, K_3] = 1$. Let $K_\alpha \cong 3 \cdot M_{22}$ and $|W_3 : \bC_{W_3}(x)| \le 64$. 
Then as $11 \Big | |K_\alpha|$, we see $[K_3, x] \le K_3$. But then 
there is also just one nontrivial module involved by (1.25). This is the 12 - dimensional 
module by (1.13). Furthermore by (5.5) $L_2/Q_2 \cong \Sigma_3$. But 
then $3 \in \sigma(M)$ and $3 \Big | |\bC_{K_3}(Z_2)|$, a contradiction. So in any case
$[\tilde{V_\alpha}, K_3] = 1.$

Let $x \in \tilde{V_\alpha} \setminus Q_3$. Then $[K_3, [V_3,x]] = 1$. 
This shows with (4.2) and (5.7) that $[V_3, \tilde{V_\alpha}] = 1$. 
Hence $\tilde{V_\alpha} \le Q_3$. We now have that $Z_2$ induces transvections on 
$\tilde{V_\alpha}$, contradicting (1.18).

\absa
{\bf (8.5) Lemma.~}{\it Assume (8.1) and (8.2). Then $O_r(K_\alpha/Z(K_\alpha)) \not= 1$ for 
some $r$,  or $K_\alpha/Z(K_\alpha)$ is a group of Lie type in even characteristic.}

\absa
Proof.~ Suppose false. Then $K_\alpha/Z(K_\alpha)$ is an alternating 
group $A_n, n \ge 9$, or $n = 7$. \\
\setcounter{equation}{0}
\begin{equation}
 |Z_\alpha| \le 4  
\end{equation}


This follows from (1.63) as in (8.4).
\begin{enumerate}
 \item[\hspace{-0.2cm} (2) \hspace{0.2cm}] There is no submodule 
${\tilde{V_\alpha}}\!^{(1)}$ of $\tilde{V_\alpha}$, such that $[K_\alpha, {\tilde{V_\alpha}}\!^{(1)}] \not= 1$ 
and every $v \in {\tilde{V_\alpha}}\!^{(1)}$ is centralized by $ E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \le 
M_\alpha$.
\end{enumerate}

\absa
Suppose false. As $M_3 \not= M_\alpha$ by (5.7), we have 
$[{\tilde{V_\alpha}}\!^{(1)}, {\tilde{V_3}}\!^{(1)}] =1$. Let $x \in 
{\tilde{V_\alpha}}\!^{(1)} \setminus Q_3$. Then $[V_3,x]$ is normalized by $K_3$ and so there is some 
$t \in [V_3,x]$, with $ p \Big | |\bC_{K_3}(t)|$. But $t \in \tilde{V_\alpha}$, 
and so by (4.2), $M_3 = M_\alpha$, a contradiction.

We have ${\tilde{V_\alpha}}\!^{(1)} \le L_2$. Furthermore $Z_3 \cap 
[{\tilde{V_\alpha}}\!^{(1)}, Z_2] = 1$. This shows that $L_2/Q_2$ is 
nonsolvable, in particular $3 
\not\in \sigma(M)$. Hence $E(L_2/Q_2) \cong L_2(4)$ and $Z_2$ is the 
orthogonal module, or $E(L_2/Q_2) \cong L_2(4) \times L_2(4)$ and 
$Z_2$ is the orthogonal 
$O^-(4,4)$ - module. But ${\tilde{V_\alpha}}\!^{(1)} \le Q_3$ and so 
${\tilde{V_\alpha}}\!^{(1)}Q_2/Q_2 \le E(L_2/Q_2)$. This then yields 
$[{\tilde{V_\alpha}}\!^{(1)}, Z_2] \cap Z_3 \not= 1$, a contradiction.\\
\setcounter{equation}{2}
\begin{equation}
K_\alpha \cong A_7, A_{10} \quad {\mbox {or}} \quad A_{11}, 
\quad {\mbox {where}} \quad p = 5 \quad {\mbox {in the last two cases.}}
\end{equation}

\absa
Otherwise $p = 3$. Let ${\tilde{V_\alpha}}\!^{(1)}$ be an irreducible 
submodule of $V_\alpha/\bC_{V_\alpha}(K_\alpha)$. By (2) 
${\tilde{V_\alpha}}\!^{(1)}$ is not the 
permutation module. Hence by (1.12)(iii) $|W_3 : \bC_{W_3}(K_\alpha)| 
\le 8$ and so $|W_3 : \bC_{W_3}(x)| \le 32$ for every $x \in V_\alpha$. 
Let $[x, K_3] \not= 1$. Then by (1.30) there is exactly one nontrivial submodule involved 
in $\tilde{V_\alpha}$. Now by (5.5) $Z_2$ is an $F$ - module and so 
$|Z_\alpha| = 2$. Hence $|W_3 : \bC_{W_3}(x)| \le 16$. By (2) and (1.30) we have 
$|W_3 : \bC_{W_3}(x)| = 16$ and so by (1.12)(iii) $K_\alpha \cong A_9$ 
and the spinmodule is involved. Now $Z_\alpha \le \tilde{V_3}$. But then $Z_\alpha$ is centralized 
by a 3 - element in $K_3$. Hence we have $M_3 = M_\alpha$ by (4.2), 
contradicting (5.7). So we have $[\tilde{V_\alpha}, K_3] = 1$. Then $\tilde{V_\alpha} 
\le L_2$ and there is $y \in Z_2$ with $[y, K_\alpha] \not= 1$ and 
$|\tilde{V_\alpha} : \bC_{\tilde{V_\alpha}}(y)| \le 4$. But then again (2) and (1.30) yield a 
contradiction.\\
\begin{equation}
 K_\alpha \cong A_7 
\end{equation}


Let $K_\alpha \cong A_{10}$ or $A_{11}$. For $x \in \tilde{V_\alpha}$ we 
have $|W_3 : \bC_{W_3}(x)| \le 2^7$. Suppose 
$|W_3 : \bC_{W_3}(K_\alpha)| \ge 16$. Then by (1.12) just permutation modules are involved. If $x \in 
\tilde{V_\alpha}$ induces some $s \in K_3$ on $\tilde{V_3}$, we see that 
at most three permutation
 module are involved. But $5 \Big | |\bC_{L_1}(K_1)|$ and so every 
$v \in {\tilde{V_\alpha}}\!^{(1)}$ is centralized by $E \cong E_{25}$. 
This contradicts (2).

So we have $|\tilde{V_\alpha} : \bC_{\tilde{V_\alpha}}(K_3)| \le 2$ for 
$|W_3 : \bC_{W_3}(K_\alpha)| \ge 16$. Now for $z \in \tilde{V_3}$, we 
have $|\tilde{V_\alpha} : \bC_{\tilde{V_\alpha}}(z)| \le 8$ and if $[\tilde{V_3}, K_\alpha] \not= 1$, we see that there are at most three permutation 
modules involved, the same
 contradiction as before. Hence $[\tilde{V_3}, K_\alpha] = 1$ and so 
$[\tilde{V_3}, \tilde{V_\alpha}] = 1$.

If $|W_3 : \bC_{W_3}(K_\alpha)| \le 8$, we have $|W_3 : \bC_{W_3}(x)| 
\le 32$ for $x \in \tilde{V_\alpha}$. As before $|\tilde{V_\alpha} : 
\bC_{\tilde{V_\alpha}}(K_3)| \le 2$, the same contradiction as above. Hence we have 
$[\tilde{V_\alpha}, \tilde{V_3}] =1$. Now $[[\tilde{V_\alpha}, V_3], K_3] = 1$. But then $[\tilde{V_\alpha}, V_3] = 1$, i.e. $\tilde{V_\alpha} \le L_2$. This shows there is 
$z \in Z_2$ with $|\tilde{V_\alpha} : \tilde{V_\alpha} \cap \bC(z)| \le 4$. 
Now $\tilde{V_\alpha}$ involves at most two permutation modules, a contradiction.


Finally we have to treat the case $K_\alpha \cong A_7$. Let $x \in 
\tilde{V_\alpha}$ with $[x, K_3] = 1$. Suppose $[[x,V_3],K_3] \not= 1$. 
Then $|[V_3,x]| \ge 16$ and so $[V_3,x] \cap Z_\alpha \not= 1$. But every $s \in [x,V_3]$ is 
centralized by a 3 - element in $K_3$. This contradicts (4.2). Hence we 
have $[K_3,[x,V_3]] = 1$. This shows $[x,V_3] = 1$, i.e. $x \in Q_3$. 

Let 
${\tilde{V_\alpha}}\!^{(1)}/\bC_{{\tilde{V_\alpha}}\!^{(1)}}(K_\alpha)$ be the 
permutation module. If $x \in {\tilde{V_\alpha}}\!^{(1)}$ with $[K_3,x] \not= 1$, then $x$ corresponds to a 
transvection, otherwise some $1 \not= v \in [x, \tilde{V_3}]$ is 
centralized by $E \cong E_9$ in $K_3$ 
and by an element of order three  in $K_\alpha$, contradicting (4.2). Furthermore 
$Z_3 \cap {\tilde{V_\alpha}}\!^{(1)} = 1$, i.e. $|{\tilde{V_\alpha}}\!^{(1)} : 
\bC_{{\tilde{V_\alpha}}\!^{(1)}}(\tilde{V_3})| = 2$. By (2) there is some 3 - element $\rho \in \bC_{L_3/Q_3}(K_3)$ with ${({\tilde{V_3}}\!^{(1)})}^\rho \not= 
{\tilde{V_3}}\!^{(1)}$. This implies 
that $Z_\alpha \cap [{\tilde{V_\alpha}}\!^{(1)}, 
{({\tilde{V_3}}\!^{(1)})}^{\langle \rho \rangle}] \not= 1$. 
But $[\bC_{K_3}({\tilde{V_\alpha}}\!^{(1)}), [{({\tilde{V_3}}\!^{(1)})}^{\langle \rho \rangle}, {\tilde{V_\alpha}}\!^{(1)}]] = 1$, 
contradicting (4.2) and (5.7). Hence we have shown 
${\tilde{V_\alpha}}\!^{(1)} \le L_2$. But then $[{\tilde{V_\alpha}}\!^{(1)}, Z_2] \cap Z_3 \not= 1$, a contradiction.


Let now ${\tilde{V_\alpha}}\!^{(1)}/\bC_{{\tilde{V_\alpha}}\!^{(1)}}(K_\alpha)$ be the 4 - dimensional module. Then again ${\tilde{V_\alpha}}\!^{(1)} \not\le L_2$ and so there is $x \in {\tilde{V_\alpha}}\!^{(1)}, [x, K_3] \not= 1$. We have that 
two such modules are involved by (2). Hence $|[x, \tilde{V_3}]| \ge 16$. 
Now let $\rho$ 
be as before. Then every element in $[{\tilde{V_3}}\!^{(1)}, \rho]$ is 
centralized by a 3 - element in $L_3$. As $Z_\alpha \cap 
[[{\tilde{V_3}}\!^{(1)}, \rho],x] \not= 1$, this contradicts (4.2) and (5.7).

Now we treat the general case. Let $x \in \tilde{V_\alpha}$. We may 
assume that neither the 6 - dimensional nor the 4 - dimensional module 
is a submodule of $\tilde{V_\alpha}/\bC_{\tilde{V_\alpha}}(K_\alpha)$. Hence by (1.12) 
$|W_3 : \bC_{W_3}(K_\alpha)| \le 2$. This implies $|\tilde{V_3} : 
\bC_{\tilde{V_3}}(x)| \le 8$. By (1.30) $[x, K_3] = 1$. Now $\tilde{V_\alpha} \le L_2$ and so $Z_2$ 
induces transvections on $\tilde{V_\alpha}$, contradicting (1.18) again. 

\absa
{\bf (8.6) Lemma.~}{\it Assume (8.1) and (8.2). Then $O_r(K_\alpha/Z(K_\alpha)) \not= 1$ for 
some $r$.}

\absa
Proof.~By (8.5) we may assume $K_\alpha/Z(K_\alpha) \cong G(r)$, a group of Lie type 
over a field $GF(r)$ of
characteristic two.

Let $x \in \tilde{V_\alpha}$ with $[x,K_3] \not= 1$. Then 
$|W_3 : \bC_{W_3}(x)| \le |Z_\alpha||W_3 : \bC_{W_3}(K_\alpha)|$.
\begin{enumerate}
\item[(1)] If $W_3$ is an SC - module, then it involves some irreducible module which is not 
$V(\lambda)$ for some basic weight $\lambda$
\end{enumerate}

\absa
Suppose false. Let first $p \Big | r^2 - 1$. If $\omega$ is a $p$ - 
element,
with $N_G(\langle \omega \rangle ) \le M$ in some minimal parabolic of 
$K_\alpha$,
such that $\omega$ centralizes $\bC_{U_\alpha}(S)$ for any composition 
factor $U_\alpha$ of
$W_\alpha$, then we have that $\omega$ centralizes $Z_2$, a 
contradiction to (4.2).

Hence $V(\lambda)$ for any basic weight $\lambda$ is involved. 
Application of (1.23) shows $K_\alpha/Z(K_\alpha) \cong (S)L_n(r), n \le 4$,
$(S)U_n(r), n \le 4$, $Sp_4(r)$, or $Sp_6(r)$.

If $K_1 \cong Sp_6(r)$, then by (1.64) $L_2$ is solvable or $q \le r$. 
We see that $|W_3 : \bC_{W_3}(K_\alpha)| \ge r^5$, but such a group does 
not act 
quadratically on the natural module.

Let $K_1 \cong Sp_4(r)$. As no group of order greater than $r^2$ acts 
quadratically on both natural modules, we have $|W_3 : \bC_{W_3}(K_\alpha)| \le r^2$. 
Let ${\tilde{V_\alpha}}\!^{(1)}$ be a natural submodule of 
$\tilde{V_\alpha}/\bC_{\tilde{V_\alpha}}(K_\alpha)$.
Then every $v \in {\tilde{V_\alpha}}\!^{(1)}$ is centralized by some $p$ - 
element in $L_\alpha$. Hence 
$[{\tilde{V_\alpha}}\!^{(1)}, {\tilde{V_3}}\!^{(1)}]$ does not 
contain any element centralized by $E \cong E_{p^2}$ in $L_1$, and vice 
versa.

Let $[{\tilde{V_\alpha}}\!^{(1)} , K_3] = 1$. We have $|[v,V_3]| \le r^4$ 
for $v \in {\tilde{V_\alpha}}\!^{(1)}$.
Now if $[K_3, [v, V_3]] \not= 1$, then $|[v, V_3]| = r^4$ and $Z_\alpha 
\cap [v, V_3] \not= 1$. But every $x \in [v, V_3]$ is centralized
by some $p$ - element in $K_3$, contradicting (4.2) and (5.7). So we 
have $[V_3, {\tilde{V_\alpha}}\!^{(1)}] = 1$,
and so ${\tilde{V_\alpha}}\!^{(1)} \le L_2$. Now we see 
$[{\tilde{V_\alpha}}\!^{(1)} , Z_2] \cap Z_3 \not= 1$,
a contradiction. Hence we have $[{\tilde{V_\alpha}}\!^{(1)} , K_3] \not=1$ 
and we get that $[{\tilde{V_\alpha}}\!^{(1)}, K_\alpha]Z_\alpha/Z_\alpha$ is 
the
natural module. Furthermore there is some $p$ - element $\omega \in 
L_\alpha$, with ${({\tilde{V_\alpha}}\!^{(1)})}^\omega \not= 
{\tilde{V_\alpha}}\!^{(1)}$. But
$[{\tilde{V_\alpha}}\!^{(1)}, \tilde{V_3}] \le {\tilde{V_\alpha}}\!^{(1)} \cap 
\tilde{V_3}$ and so $[{\tilde{V_\alpha}}\!^{(1)} \cap Q_3, \tilde{V_3}] 
= 1$. This 
implies $|\tilde{V_3} : \bC_{\tilde{V_3}}(K_\alpha)| = r$ and so 
$|{\tilde{V_\alpha}}\!^{(1)} : \bC_{{\tilde{V_\alpha}}\!^{(1)}}(K_3)| = r^2$.
But by symmetry $|\tilde{V_\alpha} : \bC_{\tilde{V_\alpha}}(K_3)| = r$, 
a contradiction.

Suppose next $K_\alpha \cong (S)U_4(r)$. Then the natural $SU_4(r)$ - 
module and the
natural $SO_6^-(r)$ - module are involved. Now for $x \in 
\tilde{V_\alpha}$, we have
$|W_3 : \bC_{W_3}(x)| \ge r^4$ (if $x$ is a unitary transvection) or 
$|W_3 : \bC_{W_3}(x)| \ge r^6$.
As a group of order $r^4$ does not act quadratically on the 
$SO_6^-(r)$ - module and the $U_4(r)$ - module by (1.39), we get 
$|W_3 : \bC_{W_3}(x)| < r^4|Z_\alpha|$. Suppose $|\tilde{V_\alpha} : 
\bC_{\tilde{V_\alpha}}(K_3)| > r$. Then we get $r^2 < |Z_\alpha|$
and then by (1.65) and (1.64) $r^2 = q$, $|Z_\alpha| = q^2$, and so we 
get some $\omega \in L_2 \cap L_3$, 
$o(\omega) = q - 1$, $[Z_3, \omega] = Z_3$, $[\omega,K_3] \le K_3$. Now 
as $p \Big | q-1$, and
$\langle \omega, K_3 \rangle$ possesses abelian Sylow $p$ - subgroups of 
order $p^3$, we get $N_G(\langle \omega \rangle) \le M$, a contradiction.

Hence we have $|\tilde{V_\alpha} : \bC_{\tilde{V_\alpha}}(K_3)| \le r$. 
Let
${\tilde{V_\alpha}}\!^{(1)}/\bC_{{\tilde{V_\alpha}}\!^{(1)}}(K_\alpha)$ be an 
irreducible $K_\alpha$ - module
(i.e. the natural $SU_4(r)$ - module or the natural $SO_6^-(r)$  - 
module). In both cases every $v \in {\tilde{V_\alpha}}\!^{(1)}$ is 
centralized
by some $p$ - element in $K_\alpha$. Hence 
$[\bC_{{\tilde{V_\alpha}}\!^{(1)}}(K_3), {\tilde{V_3}}\!^{(1)}] = 1$. As 
$|{\tilde{V_\alpha}}\!^{(1)} :
 \bC_{{\tilde{V_\alpha}}\!^{(1)}}(K_3)| \le r$, we see that 
${\tilde{V_3}}\!^{(1)}$ induces a group of transvections
on ${\tilde{V_\alpha}}\!^{(1)}$. This shows that $r = 2$ and 
${\tilde{V_\alpha}}\!^{(1)}/\bC_{{\tilde{V_\alpha}}\!^{(1)}}(K_\alpha)$ is
the natural $SO_6^-(2)$ - module. But then $[{\tilde{V_\alpha}}\!^{(1)}, 
{\tilde{V_3}}\!^{(1)}]$ is centralized by $A_6$ in $K_\alpha$ and $K_3$ as 
well.
As $3 \in \sigma(M)$, a contradiction.

So we have $K_\alpha \cong (S)L_3(r)$ or $(S)L_4(r)$. If $K_\alpha \cong 
SL_4(r)$, then the natural module, the dual module and the natural 
$SO_6^+(r)$ -
module are involved. Now by (1.39) $|W_3 : \bC_{W_3}(K_\alpha)| 
\le r^3$. This shows that for
$x \in V_\alpha$, $|W_3 : \bC_{W_3}(x)| \le r^3|Z_\alpha|$. As 
$p \Big | r^2 - 1$, we get $p = 3$ for $r = 2$. Now application of 
(1.65) shows that 
(1.65)(i) holds. Then by (1.64) $q \le r$ and we get 
$|W_3 : \bC_{W_3}(x)| \le r^5$. Hence $x$ induces a transvection
on the natural module. Furthermore 
$|W_3 : \bC_{W_3}(K_\alpha)| \ge r^2$. By (1.16)
 $W_3$ normalizes ${\tilde{V_\alpha}}\!^{(1)}$ where 
${\tilde{V_\alpha}}\!^{(1)}/\bC_{{\tilde{V_\alpha}}\!^{(1)}}(K_\alpha)$ is the 
natural module for $SL_4(r)$ or 
$SO_6^+(r)$. In the latter every $v \in {\tilde{V_\alpha}}\!^{(1)}$ is 
centralized by $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \leq M_\alpha$ in $K_\alpha$,
a contradiction. So we have the natural $SL_4(r)$ - module. Now 
$|[W_3 , {\tilde{V_\alpha}}\!^{(1)}]| \le r^5$.
This implies $|{\tilde{V_\alpha}}\!^{(1)} : \bC_{{\tilde{V_\alpha}}\!^{(1)}}(K_3)| 
\le r$. As before
$[{\tilde{V_\alpha}}\!^{(1)}, K_3] = 1$. As above there is some $\rho \in 
L_\alpha$, $o(\rho) = p$ with ${({\tilde{V_\alpha}}\!^{(1)})}^\rho \not= 
{\tilde{V_\alpha}}\!^{(1)}$.
 As$|[W_3, {\tilde{V_\alpha}}\!^{(1)}]| \le r^5$ we see $U_\alpha = 
({\tilde{V_\alpha}}\!^{(1)} \oplus {({\tilde{V_\alpha}}\!^{(1)})}^\rho)Z_\alpha/
Z_\alpha \unlhd L_\alpha/Z_\alpha$. Furthermore every $x \in U_\alpha$ is centralized by 
some $p$ - element in $L_\alpha$. Hence
$[U_\alpha \cap \bC(K_3), U_3] = 1$. Now $|[U_\alpha, U_3]Z_3/Z_3| = 
r^4$ and some group $E \cong E_{p^3}$ with $E \le \bC(Z_3)$ acts on $[U_\alpha, U_3]Z_3/Z_3$. 
Hence there is some $v \in [U_\alpha, U_3]^\sharp$, $v$ centralized by 
$F \le E$, $|F| = p^2$, a contradiction to (4.2) and (5.7).

Let finally $K_\alpha \cong (S)L_3(r)$. Then the natural module and the 
dual module are involved. Furthermore
$p \Big | r - 1$. Hence in both modules every element is centralized by 
a $p$ - element in $K_\alpha$. Now let
${\tilde{V_\alpha}}\!^{(1)}$ be as above. Then there is $\rho \in L_\alpha$ 
with $o(\rho) = p$ and ${({\tilde{V_\alpha}}\!^{(1)})}^\rho \not= 
{\tilde{V_\alpha}}\!^{(1)}$. If $U_\alpha = {\tilde{V_\alpha}}\!^{(1)} \oplus 
{({\tilde{V_\alpha}}\!^{(1)})}^\rho$
is normal in $L_\alpha$ we argue as before, as again every $v \in 
U_\alpha$ is centralized by some $p$ - element.
Hence we have three natural modules involved and at least one dual 
module. This shows that for $x \in \tilde{V_\alpha}$ we have  
$|W_3 : \bC_{W_3}(x)| \ge r^4$. Hence as $|W_3 : \bC_{W_3}(K_\alpha)| 
\le r^2$, we get $|Z_\alpha| = r^2$. As $q \le r$ by (1.65) and (1.64) 
we see
that $q = r$ and $|Z_\alpha| = q^2$. But then for $\omega \in L_3 \cap 
L_2$, $o(\omega) = p$, we have
$m_p(\langle \omega, K_3 \rangle) = 3$, and so $N_G(\langle \omega 
\rangle) \le M_3$, a contradiction.

So we have seen that $p \not\Big | r^2 - 1$. This implies that 
$K_\alpha \cong L_6(2)$, or $L_7(2)$, $p = 7$.
Let $K_\alpha \cong L_7(2)$. Then we may assume that $V(\lambda_3)$ is 
involved.
By (1.23) all other modules involved are natural modules or dual modules 
and so $V(\lambda_4)$ and $V(\lambda_5)$
are not involved. Then $7 \Big | |\bC_{K_\alpha}(Z_{\alpha - 1})|$ , a 
contradiction.

Let $K_\alpha \cong L_6(2)$. If $V(\lambda_3)$ is involved we have that 
$V(\lambda_1)$ and $V(\lambda_5)$
are involved too. As $|Z_\alpha| \le 4$, we get $|W_3 : 
\bC_{W_3}(K_\alpha)| \ge 2^6$. Let $x \in 
\tilde{V_\alpha} \setminus \bC_{\tilde{V_\alpha}}(K_3)$. Suppose $x$ 
does not induce a transvection on the natural module
. Then we get that $|W_3 : \bC_{W_3}(x)| \ge 2^{10}$ and so $|W_3 : 
\bC_{W_3}(K_\alpha)| \ge 2^8$.
Hence we may asume that $W_3/\bC_{W_3}(K_\alpha)$ contains transvections.

If $V(\lambda_3)$ is not involved, then $V(\lambda_2)$ and 
$V(\lambda_4)$ are involved and so again 
$|W_3 : \bC_{W_3}(K_\alpha)| \ge 2^6$. Let $x \in K_\alpha$ be an 
involution which does not 
induce a transvection on the natural module. Then $x$ inverts some 
element $\nu$ of order $5$ in $K_\alpha$.
Let $P$ be a parabolic with $P/O_2(P) \cong L_5(2)$ such that 
$\bC_{V(\lambda_2)}(O_2(P)) = W$
is the module $V(\lambda_2)$ for $L_5(2)$. Now let $P_1$ be a parabolic 
in $L_5(2)$ such that $P_1/O_2(P_1) \cong A_8$ and $\bC_W(O_2(P_1))$
is the orthogonal $A_8$ - module while $W/\bC_W(O_2(P_1))$ is the 
natural $L_4(2)$ - module.
This shows $|[V(\lambda_2), \nu]| = 2^{12}$ and so $|[V(\lambda_2), x]| 
\ge 2^6$. Hence also transvections are involved in 
$W_3/\bC_{W_3}(K_\alpha)$.

Let $w$ be a transvection in $W_3/\bC_{W_3}(K_\alpha)$. Then $\langle 
W_3^{\bC_{K_\alpha}(w)} \rangle = \bC_{K_\alpha}(w)$. But
then $[[w,W_\alpha], \bC_{K_\alpha}(w)] = 1$ and so just natural modules 
are involved, a contradiction. This proves (1).\\ 
\setcounter{equation}{1}
\begin{equation}
W_3 {\mbox {~ is not an SC - module }}
\end{equation}

\absa
Suppose false. Then by (1) and (1.38) we have that $K_\alpha \cong 
(S)L_3(r)$, and
by (1.22) $V(\lambda_1) \bigotimes V(\lambda_1)^\sigma$ is involved, 
$\sigma$ an automorphism of
order $2$ of $GF(r)$. Let $x \in \tilde{V_\alpha}$ with $[x, K_3] \not= 1$.
Then we see that $|W_3 : \bC_{W_3}(x)| \le r^2|Z_\alpha|$. As $q \le r$, 
we get
$|W_3 : \bC_{W_3}(x)| \le r^4$. By (1.14) we have 
$|W_3 : \bC_{W_3}(K_\alpha)| \le r$ and so $|W_3 : \bC_{W_3}(x)| \le r^3$.
This shows $|W_3 : \bC_{W_3}(x)| = r^3$ and $|Z_\alpha| = r^2$. Hence 
as $r \ge 4$, we have $q = r$ and so for $\omega \in L_2 \cap L_3, 
o(\omega) = p$
, we see that $m_p(\langle \omega, K_3 \rangle) = 3$. But this 
contradicts $N_G(\langle \omega \rangle) \not\le M_3$.\\

\begin{equation}
W_3 {\mbox { ~is a strong quadratic module }}
\end{equation}

\absa
Suppose false. Let first $q \le r$. Then $|Z_\alpha| \le r^2$ and so $|W_3 : 
\bC_{W_3}(x)| \le |W_3 : \bC_{W_3}(K_\alpha)|\cdot r^2$.
Let $u = 2 ^{m_2(K_3)}$. Then we get $|W_3 : \bC_{W_3}(K_\alpha)| > u/r^2$ 
by (2).

Application of (1.15) shows $K_\alpha/Z(K_\alpha) \cong SL_3(r)$, 
$SU_3(r)$, or $Sp_4(2)^\prime$. If 
$K_\alpha/Z(K_\alpha) \not\cong Sp_4(2)^\prime$, then 
$|W_3 : \bC_{W_3}(K_\alpha)| \le r$ and if 
$K_\alpha/Z(K_\alpha) \cong Sp_4(2)^\prime$, then 
$|W_3 : \bC_{W_3}(K_\alpha)| = 4$.

Suppose first $K_\alpha/Z(K_\alpha) \not\cong Sp_4(2)^\prime$. Now 
$|W_3 : \bC_{W_3}(x)| \le r$. Let $x \in \tilde{V_\alpha}$ with $[x, K_3] 
\not= 1$, then 
$|W_3 : \bC_{W_3}(x)| \le r^3$. Then we see with (1.26) that either 
$W_3$ involves
exactly one nontrivial irreducible $K_3$ - module, or every irreducible 
$K_3$ - module
involved is an SC - module. In the latter we argue as above.

Hence by (5.5) we have that $Z_2$ is an F - module and then $|Z_\alpha| 
\le r$. Remember (5.5) is applicable as $L_1$ is contained in
$\tilde{L_1}$ satisfying (5.2) and $K_3 \not\cong L_3(2)$ or $A_{2^m}$.
Now by (2) $K_\alpha \cong SU_3(r)$ and by (1.26) just the natural module 
is involved. By (1.20) we have
$\tilde{V_\alpha}Z_\alpha /Z_\alpha$ is the natural module. But now every 
$v \in \tilde{V_3}$ is
centralized by a $p$ - element in $L_3$. As $Z_\alpha \cap [x, \tilde{V_3}] 
\not= 1$,
this is a contradiction to (4.2) and (5.7).         

We are left with $K_\alpha/Z(K_\alpha) \cong A_6$. As $|W_3 : \bC_{W_3}(K_\alpha)| 
= 4$, we see $|W_3 : \bC_{W_3}(x)| \le 16$
for $x \in \tilde{V_\alpha}$. Let 
${\tilde{V_\alpha}}\!^{(1)}/\bC_{{\tilde{V_\alpha}}\!^{(1)}}(K_\alpha)$ be an 
irreducible
submodule of $\tilde{V_\alpha}$. Then 
$|{\tilde{V_\alpha}}\!^{(1)}/\bC_{{\tilde{V_\alpha}}\!^{(1)}}(K_\alpha)| = 16$, 
or $64$.
Every $v \in {\tilde{V_\alpha}}\!^{(1)}$ is centralized by a 3 - element in 
$K_\alpha$. Hence we have
$[{\tilde{V_\alpha}}\!^{(1)} , K_3] \not= 1$ and $|{\tilde{V_\alpha}}\!^{(1)} : 
\bC_{{\tilde{V_\alpha}}\!^{(1)}}(K_3)| = 4$. The same
applies for ${\tilde{V_3}}\!^{(1)}$ and so there are exactly two such 
modules involved and for $\rho \in \bC_{L_\alpha /Q_\alpha}(K_\alpha), 
o(\rho) = 3$, we have 
$U_\alpha = {({\tilde{V_\alpha}}\!^{(1)})}^\rho \oplus {\tilde{V_\alpha}}\!^{(1)} 
\not= {\tilde{V_\alpha}}\!^{(1)}$. If ${\tilde{V_\alpha}}\!^{(1)}$ is the 
natural
module, every $v \in U_\alpha$ is centralized by a 3 - element, a 
contradiction as $Z_3 \cap U_\alpha \not= 1$. Hence 
${\tilde{V_\alpha}}\!^{(1)}$ is the $3\cdot 
A_6$ - module. But now $[U_\alpha, U_3] \ge \langle Z_3, Z_\alpha 
\rangle$ and $[U_3, U_\alpha]$ is
normalized by $E_9$ in $L_\alpha$ and $L_3$ as well, a contradiction to 
$M_3 \not= M_\alpha$
by (5.7).

We have $q > r$. Suppose $r = 2$ and (1.65)(ii) holds. Now $3 \not\in 
\sigma(M)$. This shows $K \cong L_6(2)$
or $L_7(2)$, $p = 7$ and $q = 4$. Further $\omega \in K$. Hence $[\omega, Z_1] = 1$. This shows with (5.6) that $|Z_2| = 16$
and $|Z_1| = 2$. By (2) and (1.15) we now have a contradiction. So (1.65)
(ii) does not hold.
Then by (1.64) and (1.65) $K_\alpha/Z(K_\alpha) \cong (S)U_n(r)$, $^2E_6(r)$, 
$\Omega_{2n}^-(r), q = r^2$, $^3D_4(r)$, or $^2F_4(r)$. Let $u$ be as 
before. Then $|W_3 : 
\bC_{W_3}(K_\alpha)| > u/r^4$. Now application of (1.15) shows $K_\alpha/Z(K_\alpha) 
\cong \Omega_8^-(r)$, $SU_4(r)$, $SU_5(r)$, or $^3D_4(r)$.
Hence $|W_3 : \bC_{W_3}(K_\alpha)| \le r^3, r^2, r^2, r$, respectively.

Suppose first that any module involved in $W_3$ is an SC - module. Then 
by (1.23) $V(\lambda_1)$, $V(\lambda_2)$ and $V(\lambda_3)$ are involved 
in case of
$K_\alpha \cong \Omega_8^-(r)$. But now $|W_3 : \bC_{W_3}(x)| \ge r^{16}$, 
a contradiction.
In case of $K_\alpha \cong SU_4(r)$, we have $V(\lambda_1)$ and 
$V(\lambda_2)$ are involved.
But now we may argue as in (1).

Hence we have that $W_3$ involves exactly one nontrivial irreducible 
module. By (5.5) $Z_2$ is an F - module and as $q > 2$, we get
$|Z_\alpha| \le q$. Hence $|W_3 : \bC_{W_3}(K_\alpha)| > u/q$. Now the
assertion follows with (1.15).

\begin{equation}
 p \not\Big | r^2 - 1 
\end{equation}

\absa
Suppose false. By (3) and (1.14) just $V(\lambda)$, $\lambda$ some 
fundamental
weight is involved in $W_3$. Hence $V(\lambda)$ for every weight 
$\lambda$ is involved. This
shows $K_\alpha \cong (S)L_n(r)$, $(S)U_n(r)$, $Sp_{2n}(r)$ or $F_4(r)$.

Let $q \le r$. Then if every $V(\lambda)$ which is involved in $W_3$ is 
an SC - module,
we see $K_\alpha \cong (S)L_n(r), n \le 7$, $(S)U_n(r), n \le 6$, 
$Sp_{2n}(r), n \le 3$. We may assume that $K_\alpha \cong (S)L_n(r), 
n \ge 5$ or $(S)U_n(r), 
n \ge 5$. All other cases can be handled as in (1).

Now in $SL_5(r), SL_6(r), SL_7(r)$, we have $|W_3 : \bC_{W_3}(x)| 
\ge r^8, r^{16}, r^{32}$, respectively, by (1.23).
But we have in any case $|W_3 : \bC_{W_3}(x)| \le r^8, r^{11}, r^{14}$. 
Hence we have $K_\alpha \cong SL_5(r)$
and $|W_3 : \bC_{W_3}(K_\alpha)| = r^6$. But then just natural modules 
can be involved,
as $W_3$ acts quadratically. If $K_\alpha \cong SU_5(r)$ or $SU_6(r)$ we 
get
$|W_3 : \bC_{W_3}(x)| \ge r^8$, or $r^{16}$. But $|W_3 : \bC_{W_3}(x)| 
\le r^6$, or $r^{11}$.

Hence we have that there is some $V(\lambda)$ which is not an SC - 
module. This shows that $W_3$ involves exactly one nontrivial 
irreducible module. Hence there is just one weight. This implies 
$K_\alpha \cong SU_3(r)$, and we argue as above.

Let now $q > r$. As $p \Big | r^2 - 1$ we have $r \not= 2$ by (1.65). 
Hence (1.64) and (1.65) show $q = r^2$
and $K_\alpha \cong (S)U_n(r)$. By (1.23) $n \le 6$, and we just have 
to
handle $n = 5$ or $6$. Now $|W_3 : \bC_{W_3}(x)| \le r^8, r^{13}$, 
respectively.  This gives $K_\alpha \cong (S)U_5(r)$ and
$|W_3 : \bC_{W_3}(K_\alpha)| = r^4$. Now let $t \in W_3$ be an
unitary transvection. Then $\langle {W_3}^{\bC_{K_\alpha}(t)} \rangle = 
O^{2^\prime}(\bC_{K_\alpha}(t))$.
Hence $[[t, W_3], \bC_{K_\alpha}(t)] = 1$ and so just natural modules 
are involved in $W_3$, a contradiction.
\begin{equation}
[\tilde{V_\alpha}, K_3] = 1 
\end{equation}

Otherise by (4) we have $K_\alpha \cong L_6(2)$ or $L_7(2), p = 7$. 
Now $|W_3 : \bC_{W_3}(x)| \le 2^{11}$ or $2^{14}$, respectively. But 
then
we get a contradiction as before.

Let now $x \in \tilde{V_\alpha}$, $[[x, V_3], K_3] \not= 1$. We have 
that $[x, V_3]$ is an SC  - module. Let first $q \le r$. Then 
$|[x, V_3]| \le ur$,
$u$ as before. This now implies with (1.23) $K_\alpha \cong (S)L_n(r)$, 
$(S)U_n(r)$, $Sp_{2n}(r)$, $\Omega_{2n}^{\pm}(r)$ and just natural modules 
are 
involved. Hence there is ${\tilde{V_\alpha}}\!^{(1)}$ such that 
${\tilde{V_\alpha}}\!^{(1)}/\bC_{{\tilde{V_\alpha}}\!^{(1)}}(K_\alpha)$ 
is the natural module. Choose $x \in {\tilde{V_\alpha}}\!^{(1)}$. Suppose 
some $w \in [x, V_3]$ is centralized by $E \cong E_{p^2}$
in $K_\alpha$. Then we get a contradiction to (4.2) and (5.7). This 
implies that
$K_\alpha \cong (S)L_n(r), n \le 4$, $\Omega_6^-(r)$, $Sp_4(r)$, or 
$L_6(2)$.
In any case $[x, V_3] \cap Z_\alpha = 1$. Hence we have $K_\alpha 
\cong SL_4(r)$ or $L_6(2)$. Now
$|V_3 : \bC_{V_3}(K_\alpha)| = r^4$ or $2^6$. But as $x \in 
{\tilde{V_\alpha}}\!^{(1)}$ we get $[V_3 , x] \cap 
\bC_{{\tilde{V_\alpha}}\!^{(1)}}(K_\alpha) \not= 1$,
a contradiction.

So we have $[{\tilde{V_\alpha}}\!^{(1)}, V_3] = 1$ and so 
${\tilde{V_\alpha}}\!^{(1)} \le L_2$. But now in any case 
${\tilde{V_\alpha}}\!^{(1)} \cap Z_3 \not= 1$,
a contradiction as in the natural module every element is centralized by 
some
element of order $p$. 

So we have $q > r$. Suppose first that $r = 2$ and we have (1.65)(ii), 
(iii) or (vi). Then $3 \not\in \sigma (M)$. This shows $K_\alpha 
\cong L_6(2)$ or 
$L_7(2), p = 7$. Furthermore $|Z_1| = 2$. Then we get the same 
contradiction as before. Now (1.64) and (1.65) yield
$K_\alpha \cong SU_n(r)$, $\Omega_{2n}^-(r)$, $^2E_6(r)$, $^2F_4(r)$, or 
$^3D_4(r)$. Again $[x,V_3]$ is an SC - module
and so $^3D_4(r)$ and $^2F_4(r)$ are not possible by (1.22) and (1.23).
If $K_\alpha \cong$ $^2E_6(r)$, then by (1.22) and (1.23) $V(\lambda_1)$ 
is involved. But dim$V(\lambda_1) = 54$, while $|W_3 : \bC_{W_3}(K_\alpha)| \le q^{20}$, a contradiction.

Hence we have that $K_\alpha \cong SU_n(r)$, or $\Omega_{2n}^-(r)$ and 
just natural modules are involved. But now we argue as before, or 
$K_\alpha \cong SU_4(r),
 [x, V_3]$ is the natural module. Then $[x, V_3] \ge Z_\alpha$. But 
every $v \in [x, V_3]$ is centralized by a $p$ - element in $K_3$,
a contradiction to (4.2) and (5.7). Hence also in this case we get 
${\tilde{V_\alpha}}\!^{(1)} \le L_2$ and then we have the same contradiction 
as before.

\absa
{\bf (8.7) Lemma.~}{\it Assume (8.1). Let $K_\alpha$ be a component of 
$L_\alpha /Q_\alpha$ or $K_\alpha$ normal in $L_\alpha/Q_\alpha$ with $O_r(K_\alpha) 
\not= 1$ for some odd $r$. If
$ 1 \not= [Z_2,K_\alpha]$ and $m_p(K_\alpha) \ge 2$, then $[Z_2, K_\alpha] \not\le K_\alpha$.}

\absa
Proof.~ By (8.6) we just have to treat the case $O_r(K_\alpha) \not= 1$. 
Then $\tilde{V_\alpha} = \langle t \Big | |[W_3,t] : [W_3,t] \cap Z_\alpha| \le 2 
\rangle $, by \cite[(24.1)]{GoLyS}, \cite[(25.12)]{GoLyS} and (1.12).

Suppose first $|\tilde{V_\alpha} : \bC_{\tilde{V_\alpha}}(K_3)| \ge 4$. Then $r = p \in \sigma(M)$. By \cite[(24.1)]{GoLyS} there is $D_{2p} \times D_{2p} \cong U_3 \le K_3
\tilde{V_\alpha}$, 
$\tilde{V_\alpha} \cap U_3 \in$ Syl$_2(U_3)$. If $|[V_3, t]| > 2$, then 
there is $\rho \in U_3$, $o(\rho) = p$, $\bC_{Z_\alpha}(\rho) \not= 1$.
But this contradicts (4.2) and (5.7). Hence $|[W_3,t]| = 2$ and so 
$p = 3$ and $|[\rho, V_3]| = 4$. Now
$[\rho, V_3]$ is centralized by $E \cong E_9$ in $L_3$. Choose
$x \in [\rho, V_3] \setminus \bC(t)$. Then $|[x, V_\alpha]| = 2$ and so 
$3 \Big | |\bC_{L_\alpha}([x, V_\alpha])|$, contradicting (4.2) and (5.7).
 
So we have $|\tilde{V_\alpha} : \bC_{\tilde{V_\alpha}}(K_3)| \le 2$. The 
same applies for $\tilde{V_3}$. Hence for $x \in \tilde{V_\alpha}$ with
$[K_3,x] = 1$, we see that $[V_3,x] = 1$. This shows $|\tilde{V_\alpha} : 
\tilde{V_\alpha} \cap L_2| \le 2$. Let $[\tilde{V_\alpha} \cap L_2, Z_2] \not= 1$.
Then for $x \in \tilde{V_\alpha} \cap L_2$, $[x,Z_2] \cap Z_3 \not= 1$. As 
$|[Z_2,x] : [Z_2,x] \cap Z_\alpha| \le 2$
we see that $|[Z_2,x] \cap Z_3| \le 2$. This shows with (1.31) and (5.6) 
that $|Z_3| = 2$ and $E(L_2/Q_2) \cong L_2(4)$
or $L_2$ is solvable. Let $t \in \tilde{V}_\alpha$ with $|[W_3,t] : [W_3,t] \cap Z_\alpha| \leq 2$ and $t \not\in L_2$. Now $|[\tilde{V_3}, t]| \le 4$. 
So we get $3 \in \sigma (M)$ and there is $\rho \in K_3$, $o(\rho) = 3$ with 
$|[V_3, \rho]|
\le 16$. Hence $[V_3, \rho]$ is centralized by some 3 - element in 
$L_3$, a contradiction as above.
We have $\tilde{V_\alpha} \le L_2$ or $[\tilde{V}_\alpha \cap L_2, Z_2 ] = 1$ and so $Z_2$ is generated by elements 
inducing transvections on $\tilde{V_\alpha}$. Hence $3 \in \sigma (M)$ 
and then 
$L_2/Q_2 \cong \Sigma_3$. Now let $\rho_1 \in K_\alpha$, $z \in Z_2$ with 
$\rho_1^z = \rho_1^{-1}$. Then $|[\tilde{V_\alpha}, \rho_1]| = 4$.
Further there is  $E \le L_\alpha$, $E \cong E_9$ with $[E,[\tilde{V_\alpha}, \rho_1]] 
= 1$. But
$[\tilde{V_\alpha}, Z_2] = Z_3$, and so $[Z_3, E] = 1$. This shows 
$E \le M_3$, contradicting $M_\alpha \not= M_3$
by (5.7).  

\absa
{\bf (8.8) Lemma.~}{\it Assume (8.1). Let $K_\alpha$ be a component of 
$L_\alpha /Q_\alpha$ with $m_p(K_\alpha) \ge 2$. Then 
$[K_\alpha , Z_2] = 1$}

\absa
Proof.~ Suppose $[K_\alpha, Z_2] \not= 1$. As $m_p(K_\alpha) \ge 2$, we 
get
from (1.11) and (8.7) that $|W_3 : \bC_{W_3}(K_\alpha)| = 2$.
As we may interchange the roles of $3$ and $\alpha$, we get 
$|\tilde{V_\alpha} : \bC_{\tilde{V_\alpha}}(K_3)| \le 2$ too. As for 
$v \in \tilde{V_\alpha}$, we 
have $|[\tilde{V_3} , v] : [\tilde{V_3}, v] \cap Z_\alpha| \le 2$ and 
$v$ acts on $E \cong E_{p^4}$, for $[v, K_3] \not= 1$,
we see $[\tilde{V_\alpha}, K_3] = 1$, and then $[\tilde{V_\alpha}, V_3] 
= 1$. This implies
$\tilde{V_\alpha} \le L_2$. Now we get a contradiction as in (8.7).

\absa
{\bf (8.9) Lemma.~}{\it Assume (8.1). Let $K^{(\alpha)} = K_1^{(\alpha)} 
\times K_2^{(\alpha)} \times K_3^{(\alpha)} \le L_\alpha /Q_\alpha$ where $K_i^{(\alpha)}$ is either a component or a normal $p$-subgroup and  
$m_p(K_i^{(\alpha)}) = 1$.
Then we may assume that $[K_3^{(\alpha)}, Z_2] = 1$.}

\absa
Proof.~ Let $[K_i^{(\alpha)}, Z_2] \not= 1$ for $i = 1,2,3$. Set $\tilde{V_\alpha} = [ V_\alpha, K^{(\alpha)}]$. Assume that 
$\tilde{V_\alpha}/\bC_{\tilde{V_\alpha}}(K^{(\alpha)}) = 
{\tilde{V_\alpha}}\!^{(1)} \oplus 
\tilde{V_\alpha}^{(2)}$ with $[K_1^{(\alpha)} \times K_2^{(\alpha)}, 
\tilde{V_\alpha}^{(2)}] = 1$ and
$[{\tilde{V_\alpha}}\!^{(1)}, K_3^{(\alpha)}] = 1$. Let $x \in 
\tilde{V_\alpha}^{(2)}$. Suppose $[K^{(3)}, x] \not= 1$. Suppose that $[\tilde{V}_3,x]$ is centralized by $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \leq M_\alpha$. But $[\tilde{V_3}, x]$ contains an element
which is centralized by 
$\rho \in K^{(3)}$ with $o(\rho) = p$. This contradicts (4.2) and (5.7).
So we may assume that $[\tilde{V}_3,x]$ is not centralized by such an elementary abelian group. This shows that $[V_3,K_3^{(\alpha)}] \not\leq K_3^{(\alpha)}$. Suppose first that $|V_3 : \bC_{V_3}(K_3^{(\alpha)})| > 2$. Then by (1.11) $K_3^{(\alpha)} \cong L_2(r)$ and $[\langle {K_3^{(\alpha)}}^{V_3} \rangle, V_\alpha] = [K_3^{(\alpha)}, V_\alpha]$ is centralized by $K_1^{(\alpha)} \times K_2^{(\alpha)}$. Then we see that $[\tilde{V_3}, x]$ is centralized by $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \leq M_\alpha$ , a contradiction. Suppose that $|V_3 : \bC_{V_3}(K_3^{(\alpha)})| = 2$. If $|V_3 : \bC_{V_3}(K^{(\alpha)})| > 2$, then we may assume that $[[K_1^{(\alpha)},V_{\alpha}], {K_3^{(\alpha)}}^{V_3}] = 1$ and some $y \in [K_1^{(\alpha)},V_\alpha]$ acts nontrivially on $\tilde{V_3}$ and then also on $K^{(3)}$. So we just can interchange roles. 

This means that we may assume that $|\tilde{V}_3 : \bC_{\tilde{V}_3}(K^{(\alpha)})| = 2$. But then also by symmetry $|\tilde{V_\alpha} : \bC_{\tilde{V_\alpha}}(K^{(3)})| = 2$. This shows $|[\tilde{V}_3, \tilde{V}_\alpha] : [\tilde{V}_3, \tilde{V}_\alpha] \cap Z_\alpha| \leq 2 \geq |[\tilde{V}_3, \tilde{V}_\alpha] : [\tilde{V}_3, \tilde{V}_\alpha] \cap Z_3|$. Hence we see that $|[\tilde{V}_3, \tilde{V}_\alpha]| \leq 4$. But this contradicts $K_3^{(\alpha)}$ is not normalized by $\tilde{V_3}$.

Hence $[\tilde{V_\alpha}^{(2)}, K^{(3)}] = 1$. This implies by the same 
reason as before $[\tilde{V_\alpha}^{(2)}, V_3] = 1$,
and so $\tilde{V_\alpha}^{(2)} \le L_2$. But then $Z_3 \cap 
[\tilde{V_\alpha}^{(2)}, Z_2] = 1$ and so $[\tilde{V_\alpha}^{(2)}, Z_2] = 1$, which implies $[K_3^{(\alpha)}, Z_2] = 1$,  a contradiction.

Hence we have that $[[K_i^{(\alpha)}, \tilde{V_\alpha}], K_j^{(\alpha)}] 
\not= 1$ for every pair
$\{i,j\}$. As $m_p(K_i^{(\alpha)}) = 1$, we get that $m_3(K_3^{(\alpha)}) 
\le 1$. Hence 
$K_3^{(\alpha)} \cong L_2(r)$, $U_3(r)$, $L_3(r)$, $Sz(r)$, $J_1$, or 
a $p$-group. The same applies for $K_2^{(\alpha)}$.
We have $|W_3 : \bC_{W_3}(K^{(\alpha)})| = 2$, if $r$ is odd. In this 
case we have that $|\tilde{V_\alpha} : \bC_{\tilde{V_\alpha}}(K^{(3)})| 
\le 2$. Furthermore let $t \in \tilde{V_\alpha}$ with $[t, K^{(3)}] = 1$ and $[t, V_3] \not= 1$. 
Then $|[t, V_3]| = 2$. Hence
there is $U \le \bC(K^{(3)})$ with $p \Big | |U|$ and $[U, \tilde{V_3}] = 1$. 
As $[t, V_3] \le \tilde{V_\alpha}$, we get that $[t, V_3]$ is 
centralized by a 
$p$ - element in $L_\alpha$ and $E \cong E_{p^3}$ in $L_3$, a 
contradiction to (4.2) and (5.7).
This implies $|\tilde{V_\alpha} : \tilde{V_\alpha} \cap L_2| = 2$. Let 
$t \in \tilde{V_3} \cap L_2$ with $[Z_2, t] \not= 1$.
Then $|[Z_2, t] : [Z_2, t] \cap Z_\alpha| \le 2$. As $[Z_2, t] \cap Z_3 
\not= 1$, we now see $|Z_3| = 2$, and so
$|[Z_2, t]| \le 4$. This shows that $Z_2$ contains elements $z$ with 
$|[\tilde{V_\alpha}, z]| \le 8$,
a contradiction as $\tilde{V_\alpha}$ involves at least four nontrivial 
irreducible $K_3^{(\alpha)}$ - modules. So we have $|W_3 :\bC_{W_3}(K^{(\alpha)})| > 2$, and then $r$ is even.

Now we choose notation such that $K_3^{(\alpha)} \cong G(r_3)$, $K_2^{(\alpha)} \cong G(r_2)$ with 
$r_3 \ge r_2$. We see
that for $v \in \tilde{V_\alpha}$, $[v, K^{(3)}] \not= 1$, we get 
$|[v, \tilde{V_3}] \ge r_3^4$.
Hence we get $|Z_\alpha| \ge r_3^3$, $r_3^2$ (in case of 
$ K_3^{(3)}/Z(K_3^{(3)}) \cong L_3(r_3)$). Suppose there is some $\omega \in L_{\alpha - 1} \cap L_\alpha$, 
$o(\omega) = q - 1$, i.e. $L_2$ is nonsolvable, or $SU_3(2) \unlhd 
L_2/Q_2$. Now
$|Z_\alpha| \le q^2$, and so $r_3^3 \le q^2$, or $r_3^2 \le q^2$. 
Let $[\omega, K_3^{(\alpha)}] \le K_3^{(\alpha)}$ and $[\omega, 
K_3^{(\alpha)}] \not= 1$.
Then application of (1.64) and (1.65) shows that $K_3^{(\alpha)} \cong 
SL_3(r_3) = SL_3(q)$,
$SU_3(r_3) = SU_3(\sqrt{q})$, or $L_3(2)$. If $K_3^{(\alpha)} \cong L_3(2)$, then $q = 4$ and $\omega \in K_3^{(\alpha)}$. Hence 
$[\omega, Z_\alpha] = 1$.
Now we see $|Z_\alpha| = 2 < 4 = r_3^2$, a contradiction. 

In case of $K_3^{(\alpha)} \cong SU_3(\sqrt{q})$, we get four natural 
modules involved and so 
$|[v, \tilde{V_3}]| \ge q^4$, i.e. $|Z_\alpha| > q^2$, a contradiction.

Hence we have $K_3^{(\alpha)} \cong SL_3(q)$ and there are exactly four 
natural modules involved. Now $K_1^{(\alpha)} \times K_2^{(\alpha)} 
\lesssim L_2(q) 
\times L_2(q)$. But $|W_3 : \bC_{W_3}(K^{(\alpha)})| = q^2$, 
a contradiction.

Hence let now $[\omega, K_3^{(\alpha)}] \not\le K_3^{(\alpha)}$. Then 
$K_3^{(\alpha)} \cong Sz(r)$,
$L_2(r)$, or $L_3(2)$, as $m_p(K_3^{(\alpha)}) = 1$. Let 
$\langle (K_3^{(\alpha)})^{\langle \omega \rangle} \rangle = X_1 \times \cdots \times X_t$, $X_1 \cong
 K_3^{(\alpha)}$. As no $X_i, X_j$, $ i \not= j$ can centralize 
$[X_1, V_\alpha]$, otherwise we argue as before, we see that 
$|Z_\alpha| \ge r^{2^{t-1}-2}$.
But we may assume that $2^{t-1} - 1 \ge q - 1$ or $2^{2t - 2} \ge r^{2^{t-1} - 2} \ge 2^{2^{t-1}-2}$. So we have $t = 3$. Now $|Z_\alpha| \ge 4^2$, $q = 4$ for $K_3^{(\alpha)} \cong L_3(2)$ or $ q= 64$ and $K_3^{(\alpha)} \cong L_2(4)$,
$L_2(8)$ or $L_2(16)$. If $q = 64$ there is $\nu \in L_3 \cap L_2$, 
$o(\nu) = 7$. This shows $[K_3^{(\alpha)}, \nu] \not\le K_3^{(\alpha)}$,
as $[\omega, \nu] = 1$. But then we have 63 conjugates of 
$K_3^{(\alpha)}$, a contradiction. So we are left with 
$K_3^{(\alpha)} \cong L_3(2), q = 4$.
By the choice of $r_3$, we may assume $K^{(\alpha)} = X_1 \times X_2 
\times X_3$. But now there are 9 natural modules involved and so we get
$|Z_\alpha| \ge 2^7  > q^2$, a contradiction.

We are left with $[K_3^{(\alpha)}, \omega] = 1$. But as $r_2 \le r_3$, 
we also get $[K_2^{(\alpha)}, \omega] = 1$, or $[K_2^{(\alpha)}, \omega] 
\not\le 
K_2^{(\alpha)}$ and $r_2 < r_3$. The latter implies $q = 4$ and so 
$K_2^{(\alpha)} \times K_3^{(\alpha)} \lesssim \bC_{K^{(\alpha)}}(\omega)$,
a contradiction to (1.63). Hence $[\omega, K_2^{(\alpha)}] = 1$. By 
(1.63) $K_2^{(\alpha)} \cong K_3^{(\alpha)} \cong A_5$, $ p = 5$. But 
now $K_1^{(\alpha)}$ 
is of the same shape as $K_2^{(\alpha)}$ and so again $[K_1^{(\alpha)}, 
\omega] = 1$, a contradiction to (1.63).

We are left with $L_2$ solvable. Now $|Z_\alpha| \le 4$. Hence 
$K_3^{(\alpha)} \cong L_3(2) \cong K_2^{(\alpha)}$. This
shows that also $K_1^{(\alpha)}$ has cyclic Sylow 3 - subgroups and so 
we also get
$K_1^{(\alpha)} \cong L_3(2)$. Now we have 6 modules involved. This 
shows $|[x,\tilde{V_3}]| \ge 2^6 > 4|Z_\alpha| = 16$, a contradiction.

\absa
{\bf (8.10) Lemma.~}{\it Assume (8.1) and let $K_1^{(\alpha)} \times K_2^{(\alpha)} 
\le L_\alpha /Q_\alpha$, $K_i^{(\alpha)}$ components or normal $p$-subgroups with $m_p(K_i^{(\alpha)}) = 1$, $i = 1,2$.  Then we 
may assume $[K_2^{(\alpha)}, Z_2] = 1$.}

\absa
Proof.~ Assume ${K_1^{(\alpha)}}^{W_3} \not\leq K_1^{(\alpha)} \times K_2^{(\alpha)}$. Then we have that ${K_1^{(\alpha)}}^{W_3} \times K_2^{(\alpha)}$ fulfills (8.9). So either $[K_1^{(\alpha)}, Z_2] = 1$ or $[K_2^{(\alpha)},Z_2] = 1$. Hence we may assume that $[K_1^{(\alpha)} \times K_2^{(\alpha)}, W_3] \leq K_1^{(\alpha)} \times K_2^{(\alpha)}$. Suppose first $[[K_1^{(\alpha)}, V_\alpha], K_2^{(\alpha)}] = 1$. 
Let $v \in [K_1^{(\alpha)} , V_\alpha] = {\tilde{V_\alpha}}\!^{(1)}$ with 
$K_1^{(3)} \geq [v,K_1^{(3)}] \not = 1$. First of all $[v, {\tilde{V_3}}\!^{(1)}] \cap Z_\alpha = 1$.
 Furthermore, suppose that there is $\nu \in K_1^{(3)}, o(\nu)$ odd, 
$\nu^v = \nu^{-1}$, and $\rho \in L_3$, $[\rho, \langle \nu, v \rangle] = 1$, 
$o(\rho) = p, \rho \not\in K_2^{(3)}$. Then $\rho$ acts on 
$[[\nu, {\tilde{V_3}}\!^{(1)}],v]$ and so some $U \subseteq [v, {\tilde{V_3}}\!^{(1)}]$ is normalized by $E \cong E_{p^2}$ with 
$\Gamma_{E,1}(G) \le M_3$, and centralized by some $p$ - element in 
$L_\alpha$. This shows with (4.2) $M_3 = M_\alpha$, contradicting (5.7).

As $m_p(K_1^{(3)}) = 1$, there is some $p$ - element 
$\rho \not\in K_2^{(3)}$, $\rho$ induces an outer automorphism on 
$K_1^{(3)}$. This implies with (1.67) that $K_1^{(3)} \cong G(r)$, a group of Lie type. As $m_p(K_1^{(3)}) = 1$, 
we see that $\rho$ has to induce a field automorphism. Now by (4.12) 
$K_1^{(3)} \cong 
L_2(r)$ or $Sz(r)$, $r$ even. Hence we may assume 
$v \in \bC_{K_1^{(3)}}(\rho)$, a contradiction.

So we have shown that $[v, K_1^{(3)}] = 1$. Assume now 
$[v, K_2^{(3)}] \not= 1$. We may choose notation such that 
$m_3(K_2^{(3)}) \le 1$. Hence by (1.5) $K_2^{
(3)} \cong L_2(r)$, $U_3(r)$, $Sz(r)$, $L_3(r)$, $J_1$ or a $p$-group. Now 
as before there has to be some outer automorphism $\rho$ of $K_2^{(3)}$ with 
$o(\rho) = p$. This again has to be a field automorphism and so it centralizes $v$, a 
contradiction as before. We have
\begin{enumerate}
\item[\hspace{-0.2cm} (1) \hspace{0.2cm}] If 
$[[K_1^{(\alpha)}, V_\alpha], K_2^{(\alpha)}] = 1$, 
then for ${\tilde{V_\alpha}}\!^{(1)} = [V_\alpha, K_1^{(\alpha)}]$, we have 
$[{\tilde{V_\alpha}}\!^{(1)}, K_1^{(3)} \times K_2^{(3)}] = 1$, or ${(K_1^{(3)})}^v = K_2^{(3)}$ 
for some $v \in {\tilde{V_\alpha}}\!^{(1)}$, 
$|{\tilde{V_\alpha}}\!^{(1)} : \bC_{{\tilde{V_\alpha}}\!^{(1)}}(K_1^{(3)})| = 2. $
\end{enumerate}

\absa
We now show\\
\setcounter{equation}{1}
\begin{equation}
 [[K_1^{(\alpha)}, V_\alpha], K_2^{(\alpha)}] \not= 1  
\end{equation}

\absa
By (1) we may assume $|{\tilde{V_\alpha}}\!^{(1)} : 
\bC_{{\tilde{V_\alpha}}\!^{(1)}}(K_1^{(3)} \times K_2^{(3)})| \le 2$. 
Choose $t \in {\tilde{V_\alpha}}\!^{(1)}$ with $ [t, K_1^{(3)} \times K_2^{(3)}] = 1$. Then $K_1^{(3)} \times K_2^{(3)}$ acts on 
$[t, V_\alpha]$. Suppose $t \not\in L_2$. We may also use (1) with $3$ 
and $\alpha$ interchanged. This together with (8.3) - (8.9) shows that 
$|{\tilde{V_3}}\!^{(1)} : {\tilde{V_3}}\!^{(1)} \cap Q_\alpha| \le 2$. Hence 
$|[{\tilde{V_3}}\!^{(1)}, t] : [{\tilde{V_3}}\!^{(1)}, t] \cap Z_\alpha| \le 2$. By (4.2) and (5.7) this shows 
$|[{\tilde{V_3}}\!^{(1)}, t]| \le 2$. Hence $[t, {\tilde{V_3}}\!^{(1)}] = 1$ and 
so $[K_1^{(3)}, [V_3, t]] = 1$. If $[V_3,t] \not= 1$ we have $\bC_{L_\alpha}({\tilde{V_\alpha}}\!^{(1)})$ 
contains $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \le M_\alpha$. So we get 
$[V_3, t] = 1$. This shows ${\tilde{V_\alpha}}\!^{(1)} \le L_2$.

By (4.2) and (5.7) we have $[{\tilde{V_\alpha}}\!^{(1)}, Z_2] \cap Z_3 = 1$. 
But this would imply $[{\tilde{V_\alpha}}\!^{(1)}, Z_2] = 1$, a contradiction.\\
\begin{equation}
[{\tilde{V_\alpha}}\!^{(1)}, K_1^{(3)} \times K_2^{(3)}] = 1  
\end{equation}

\absa
Suppose false. We choose notation such that $m_3(K_2^{(3)}) \le 1$. In particular 
$K_2^{(3)} \cong L_2(r)$, $L_3(r)$, $U_3(r)$, $Sz(r)$, $J_1$, or 
a $p$-group. Suppose first that $|W_3 : \bC_{W_3}(K_2^{(\alpha)})| = 2$. Choose $x \in 
\tilde{V_\alpha}$, $[x, K_1^{(3)} \times K_2^{(3)}] = 1$. Now 
$|[x,V_3] : [x,V_3] \cap Z_\alpha| \le 2
$. As   $K_1^{(3)} \times K_2^{(3)}$ acts on $[x,V_3]$, we see 
$|[x,V_3]| \le 2$ and $[K_1^{(3)} \times K_2^{(3)}, [x, V_3]] = 1$. 
Then there is some $p$ -
 element $\rho \in L_\alpha$ with $[\tilde{V_\alpha}, \rho] = 1$. Let 
$x \not\in L_2$. So $[x, V_3]$ is centralized  by $\rho \in 
L_\alpha$ and by $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \le M_3$. Application of (4.2) and (5.7) 
provides us with a contradiction.

 Now for 
$x \in \tilde{V_\alpha} \cap L_2$ we have 
$|[Z_2,x] : [Z_2,x] \cap Z_\alpha| \le 2$. Furthermore if $[Z_2,x] \not= 1$, then $[Z_2, x] \cap Z_3 \not= 1$. 
This implies $|Z_3| = 2$ and so $|[Z_2,x]| \le 4$. Application of (1.43) 
shows that in any case for $z \in Z_2$ with $[z, K_1^{(\alpha)} \times K_2^{(\alpha)}] \not= 1$ 
we get $|[\tilde{V_\alpha}, z]| \le 4|\tilde{V}_\alpha : \tilde{V}_\alpha \cap L_2|$. By (1.12) $|\tilde{V}_\alpha : \tilde{V}_\alpha \cap L_2| \leq 2$ or $K_2^{(\alpha)} \cong L_2(r), U_3(r), Sz(r)$ or $L_3(r)$, $r$ even, and $|\tilde{V}_\alpha : \tilde{V}_\alpha \cap L_2| \leq r$ or $r^2$ in case of $K_2^{(\alpha)} \cong L_3(r)$. Set $r = 2$ in the first case. As there are at least two 
nontrivial $K_2^{(\alpha)}$ - modules involved, we see that $z$ has to induce a transvection 
on one of these modules. Now we get $K_2^{(\alpha)} \cong L_2(4)$, 
$L_3(r)$ , $r$ even, or a 3-group. In any case, as $p \not \Big | r-1$ in case of $K_2^{(\alpha)} \cong L_3(r)$, some $\rho \in L_\alpha$ with $o(\rho) = p$ centralizes 
$\tilde{V_\alpha}$. Hence $[Z_2,x] \cap Z_3 = 1$ and then $[Z_2, x] = 1$.
 This implies that $|\tilde{V_\alpha} : \bC_{\tilde{V_\alpha}}(Z_2)| = r^2$ and $K_2^{(\alpha)} \cong L_3(r)$ and exactly two natural modules are involved. Then $K_1^{(\alpha)}$ is a subgroup of $L_2(r)$. But now by (2) $|\tilde{V}_\alpha : \bC_{\tilde{V}_\alpha}(K_1^{(\alpha)})| \leq r$, a 
contradiction. So we have
\begin{enumerate}
\item[(i)] $\hspace{3.5cm} |W_3 : 
\bC_{W_3}(K_2^{(\alpha)})| \ge 4 $
\end{enumerate}

\absa
Application of (1.12) now again implies $K_2^{(\alpha)} \cong L_2(r)$, 
$U_3(r)$, $Sz(r)$ or $L_3(r)$, $r$ even. Next we show
\begin{enumerate}
\item[(ii)] $ K_1^{(\alpha)} \times K_2^{(\alpha)} \cong L_2(r) \times 
L_2(r) $ and $\tilde{V_\alpha}/\bC_{\tilde{V_\alpha}}(K_1^{(\alpha)} \times K_2^{(\alpha)})$ is the $O^+(4,r)$ - module, $L_2$ is nonsolvable and $E(L_2/Q_2) \cong G(q)$ with $q = r$.
\end{enumerate}

\absa
Let $\alpha + 1 \in \Delta (\alpha)$ with $[Z_{\alpha + 1}, K_1^{(3)} 
\times K_2^{(3)}] \not= 1$. Suppose first $[\tilde{V_3} \cap 
L_{\alpha + 1}, Z_{\alpha + 1}] = 1$. Then $|\tilde{V_3} : \bC_{\tilde{V_3}}(Z_{\alpha + 1})| \le r$ 
or $r^2$ in case of $L_3(r)$. As $[Z_{\alpha + 1}, \tilde{V_3}] \not= 1$,
 this implies
 $|\tilde{V_3} : \bC_{\tilde{V_3}}(Z_{\alpha + 1})| = r^2$ and there are 
exactly two $L_3(r)$ - modules involved. Now $K_1^{(\alpha)} \lesssim 
L_2(r)$, which contradicts $|\tilde{V_3} : \bC_{\tilde{V_3}}(K_1^{(\alpha)})| = r^2$.

So we have $[\tilde{V_3} \cap L_{\alpha + 1}, Z_{\alpha + 1}] \not= 1$. 
Then even $[\tilde{V_3} \cap L_{\alpha + 1}, Z_{\alpha + 1}] \cap 
Z_{\alpha} \not= 1$. 
Furthermore for $x \in Z_{\alpha + 1}$ with $[x, \tilde{V_3}] \not= 1$, we 
see that $|\tilde{V_3} \cap L_{\alpha + 1} : \bC_{\tilde{V_3} \cap 
L_{\alpha + 1}}(x)| \ge r$. If $L_2$ is solvable this implies $r = 2$ and then 
$K_1^{(\alpha)} \cong K_2^{(\alpha)} \cong L_3(2)$. But now there is 
some $p$ - element $\rho \in L_3
$ with $[\rho, \tilde{V_3}] = 1$. As $[x, \tilde{V_3}] \cap Z_\alpha 
\not= 1$, this contradicts (4.2) and (5.7).

We have $L_2$ is nonsolvable and so $q \ge r$. Suppose $q > r$ ($q > 
r^2$ for $K_2^{(\alpha)} \cong U_3(r)$). Let $\omega \in L_2 \cap L_3$, 
$o(\omega) = x$, $x$ a Zsigmondy prime or $x = 9$ for $q = 64$. Suppose $[\omega, K_2^{(3)}] \not\leq K_2^{(3)}$. If $[K_2^{(3)}, V_3]$ is centralized by $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \leq M_3$, then we see $M_3 = M_\alpha$, a contradiction. Let $K_2^{(3)} \times \ldots \times K_{2i}^{(3)} = {K_2^{(3)}}^{\langle \omega \rangle}$ with $o(\omega) = y$, $y$ a Zsigmondy prime in $q-1$. Then $i = y$. Now for $x \in V_{\alpha}^{(1)}$ with $[K_2^{(3)}, x] \not= 1$, we see that $|[V_3^{(1)}, x]| \geq r^{2^{y-2}}$. But as $y | 2^{y-1} - 1$, we see that $|[V_3^{(1)}, x]| \leq r^22^{y-1}$. So $2^{y-1} \geq r^{2^{y-2}-2}$. This shows $y = 3$, $q = 4$ and so $r =2$. Then $K_2^{(3)} \cong L_3(2)$. But now we see that there are at least 6 nontrivial irreducible modules involved, a contradiction.  So we have $q = 64$. We see that there are at least 8 nontrivial modules involved and so $r^8 \leq qr^2$. Now again $r=2$ and $K_3^{(2)} \cong L_3(2)$. But now we have at least 9 nontrivial irreducible modules involved, a contradiction. So we have $[\omega, K_2^{(3)}] \leq K_2^{(3)}$ and  then 
$[\omega, K_2^{(3)}] = 1$. Hence by (1.63) either $K_1^{(3)} \times K_2^{(3)} 
\cong A_5 \times A_5$ or $[\omega, K_1^{(3)}] \not= 1$. Assume the latter. By (i) and (1.12) we have that 
$K_1^{(3)}$ is either a group of Lie type over a field of characteristic 
two or a sporadic or
 alternating group. Suppose $K_1^{(3)}$ is not a group of Lie type over 
a field of characteristic two. Then by (1.9) $q = 4$. This shows 
$r = 2$ and so $K_2^{(3)} \cong L_3(2)$. Hence $m_3(K_1^{(3)}) = 2$. Now 
$3 \not\in \sigma (M)$ and so $\omega \in K_1^{(3)}$. But then 
$\omega$ is centralized by $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \le M$, if $m_p(L_1) \ge 3$. If $m_p(L_1) = 2$, then 
by the construction in (4.13) we have $L_1 \le \tilde{L_1}$ with 
$m_p(\tilde{L_1}) = 3$ and $K_1^{(3)}$ is a component in $\tilde{L_1}$ too. Hence $\omega$ is 
centralized by $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \le M$ and $E \le 
\tilde{L_1}$. As $\omega \in 
K_1^{(3)}$ we have $[\omega, Z_1] = 1$. Hence $O_2(\bC_M(\omega)) 
\not= 1$. As $p \not= 3$, we may apply (1.63). This shows $N_G(\langle 
\omega \rangle) \le M$
, a contradiction to (1.66). 

Now $K_1^{(3)} \cong G(r_1)$, $r_1$ even. Assume $q \le r_1$. Let $t \in 
Z_{\alpha + 1}$ be as before.  Suppose $K_1^{(3)} \not\cong L_3(r)$. Then there are at least two nontrivial irreducible modules involved and $r_1^2 \leq |[\tilde{V_3}, x]| \le qr < q^2 
\le r_1^2$. Suppose $K_1^{(3)} \cong L_3(r)$, then there are at least three nontrivial irreducible modules involved and so $r_1^3 \leq |[\tilde{V_3}, x]| \le qr^2  < q^3 \le r_1^3$. In both cases we have a contradiction.

Assume $q > r_1$. By (1.66) we see that $\omega \not\in K_1^{(3)}$. 
Otherwise as above we get $N_G(\langle \omega \rangle) \le M$. 
Now (1.64) and (1.65) imply $q 
= r_1^2$ and $K_1^{(3)} \cong U_n(r_1)$, $\Omega_{2n}^-(r_1)$, 
$^2F_4(r_1)$ or $^3D_4(r_1)$. The last two cases are not possible as 
neither $^3D_4(r_1)$ nor $^2F_4(r_1)$ do possess SC - modules by (1.22). In the other two cases we 
get $r_1^2 = q = r$, a contradiction.

So we are left with $r = 4$ and $K_1^{(3)} \times K_2^{(3)} \cong A_5 
\times A_5$ and $5 \in \sigma (M)$. Now some $t \in \tilde{V_3}$ acts on 
$E \cong E_{25}$ and $
\bC_E(t) = 1$, and centralizes some $\rho \in L_\alpha$ with $o(\rho) = 5$ and 
$[E, \rho] = 1$. This shows that there is 
$U \le [t,\tilde{V_\alpha}]$, $|U| = 16$, $U$ is centralized by a 5 - element in $L_\alpha$. But 
$|[t, \tilde{V_\alpha}] : [t, \tilde{V_\alpha}] \cap Z_3| \le 4$, and so 
we get with (4.2) $M_3 = M_\alpha$, a contradiction to (5.7).

We have shown $q = r$ or $q \le r^2$ for $K_2^{(3)} \cong U_3(r)$. In 
the latter $|\tilde{V_3} : \bC_{\tilde{V_3}}(t)| \le r^2$ for every 
$t \in Z_{\alpha + 1}
$, a contradiction as at least two nontrivial irreducible 
$K_2^{(3)}$ - modules are involved in $\tilde{V_3}$ and (1.26). The same 
applies for $K_2^{(3)} \cong Sz(q)$. Hence we have $K_2^{(3)} \cong L_2(q)$ or $L_3(q)$. Suppose 
we have $K_2^{(3)} \cong L_3(q)$. Then there are at most three 
nontrivial irreducible
 modules involved. This shows $K_1^{(3)} \lesssim L_3(q)$. As 
$m_p(K_1^{(3)}) = 1$, we get $K_1^{(3)} \cong L_2(q)$. By interchanging 
the roles of $K_1^{(3)}$ 
and $K_2^{(3)}$ we may assume $K_2^{(3)} \cong L_2(q)$. Let 
$t \in Z_{\alpha + 1}$. Then $|[\tilde{V_3}, t]| \le q^2$. If 
$[t, K_2^{(3)}] \not= 1$ we get the assertion. 

So assume $[t, K_2^{(3)}] = 1$. Then $|[t, \tilde{V_3}]| = q^2$. As 
$|Z_\alpha| \le q^2$ we get  $|[u, \tilde{V_\alpha}]| \le q^3$ for $u \in \tilde{V_3}$ with 
$[u, K_2^{(\alpha)}] \not= 1$,. In particular there are at most three 
nontrivial irreducible $L_2(q)$ - modules involved. This implies 
$K_1^{(3)} \cong SL_3(q)$ or $L_2(q)$. As $p \Big | q^2 - 1$, we see in the former that 
$[t , \tilde{V_3}]$ is centralized by some $p$ - element in $L_3$. But 
$|Z_\alpha \cap [t, \tilde{V_3}]| \not= 1$, a contradiction.

So we have $K_1^{(3)} \cong L_2(q) \cong K_2^{(3)}$. Now we get 
$\tilde{V_3}/\bC_{\tilde{V_3}}(K_1^{(3)} \times K_2^{(3)}) 
\cong V_1^{(3)} \oplus V_2^{(3)}$, $V_1^{(3)}$ is the $O_4^+(q)$ - module and $[V_2^{(3)}, K_2^{(3)}] = 1$, 
$V_2 ^{(3)}$ is the natural $K_1^{(3)}$ - module. Let 
$u \in \tilde{V_3}$. Then we get 
$|[u, \tilde{V_\alpha}]| = q^3$ and so $Z_3 \le \tilde{V_\alpha}$. But 
in $V_1^{(\alpha)} \oplus V_2^{(\alpha)}$ any subgroup of order $q^2$ 
contains elements 
centralized by a $p$ - element, as this is true for any subgroup of 
order $q$ in $\bC_{V_1^{(\alpha)}}(W_3)$. This is a contradiction to 
(4.2) and (5.7). Hence
 (ii) holds.

\absa
We now can prove (3).  Suppose we are in the situation of (4.13). Then there is a $p$-element in $M_3$ centralizing $\tilde{V}_3$. But as $Z_\alpha \cap [\tilde{V}_3,t] \not= 1$ we have a contradiction to (4.2) and (5.7). So we have $m_p(L_1) \geq 3$. Now as $p \not \Big | |L_2 \cap L_3|$, we have $p | q+1$. Suppose there is some $\nu$ with $o(\nu) = p$ such that $\nu$ induces a field automorphism on $K_3^{(1)}$. Then by (4.12) we have a subgroup isomorphic to $D_{2p}$ in Out$(K_3^{(1)} \times K_3^{(2)})$. So we have some $\rho$ with $o(\rho) = p$ such that $[\rho,\tilde{V}_3] \leq \bC_{\tilde{V}_3}(K_3^{(1)} \times K_3^{(2)})$. Hence $[[\tilde{V}_3, K_3^{(1)} \times K_3^{(2)}], \rho] = 1$ as $[\rho, Z_3] = 1$. So we see $[\rho, \tilde{V}_3] = 1$, the same contradiction as before.  This contradiction proves (3).

\absa
Let now $t \in \tilde{V_\alpha}$. By (3) 
$[t, K_1^{(3)} \times K_2^{(3)}] = 1$. We argue as before. Let 
$[[t, \tilde{V_3}], K_1^{(3)} \times K_2^{(3)}] \not= 1$. Then we may assume that $t \in Z_{\alpha + 1}$ and so 
$|[t, \tilde{V_3}]| \le rq$ or $r^2q$. In particular $q > r$, $q > r^2$ 
for $K_2^{(3)} \cong U_3(r)$. Hence $L_2$ is nonsolvable and for $\omega \in L_2 \cap L_3$, 
$o(\omega) = x$, $x$ as above, we see that $[\omega, K_2^{(3)}] = 1$ as before. 
Now $K_1^{(3)} \cong G(r_1)$ and also $[K_1^{(3)}, [V_3, t]] \not= 1$, so $q > r_1$, 
$q > r_1^2$ for $K_1^{(3)} \cong U_n(r_1)$, $\Omega_{2n}^-(r_1)$, or 
$^2E_6(r_1)$, and $q > r_1^3$ 
for $K_1^{(3)} \cong$ $^3D_4(r_1)$. Hence we see 
$[\omega, K_1^{(3)}] = 1$, too. Application of (1.56) shows 
$K_1^{(3)} \cong K_2^{(3)} \cong A_5$ and $p = 5$. Now
 $|[V_3, t] : [V_3, t] \cap Z_\alpha| \le 4$. But there is 
$\rho \in K_1^{(3)} \times K_2^{(3)}$, $o(\rho) = 5$, with 
$|\bC_{[V_3, t]}(\rho)| \ge 16$, a contradiction to (4.2) and (5.7).

Hence we have $\tilde{V_\alpha} \le L_2$. Now again $q \ge r^2$ 
($q \ge r^4$ for $K_2^{(\alpha)} \cong U_3(r)$). As there is 
$t \in Z_2$, with $|[\tilde{V_\alpha}, t]| \le q$, we get as before $K_1^{(3)} \cong K_2^{(3)} \cong A_5$, 
$p = 5$. Now for $u \in \tilde{V_\alpha}$, we have 
$|[Z_2,u] : [Z_2,u] \cap Z_\alpha| \le 4$. Hence $|Z_\alpha| \le 4$, as $Z_3 \le [Z_2, u]$. This shows 
$q \le 4$. But this contradicts $q \ge r^2 = 16$.

\absa
{\bf (8.11) Proposition.~}{\it Assume (8.1). Then $b = 1$.}

\absa
Proof.~ Let $K_\alpha$ be a component or normal $p$-subgroup in $L_\alpha/Q_\alpha$ with $[K_\alpha, Z_2] \not= 1$. By (8.8) we have $m_p(K_\alpha) = 1$. Assume first 
$[\tilde{V_\alpha}, K_3] \not= 1$ too. By (8.10) we have $[K_\alpha,Z_2] \leq K_\alpha$ and $[\tilde{V}_\alpha, K_3] \leq K_3$.\\
\setcounter{equation}{0}
\begin{equation}
|\tilde{V_\alpha} : \bC_{\tilde{V_\alpha}}(K_\alpha)| \ge 4
\end{equation}

\absa
Otherwise for $v \in \tilde{V_\alpha}$, we have 
$|[\tilde{V_3}, v] : [\tilde{V_3}, v] \cap Z_\alpha| \le 2$. This 
implies $m_p(\bC_{L_3}(K_3)) = 1$. Hence $K_3
$ admits an outer automorphism $\rho$ of order $p$. This yields that 
$K_3$ is a group of Lie type. As $m_p(K_3) = 1$, we get that $\rho$ 
induces a field automorphism. Hence we get $K_3 \cong L_2(r)$, or $Sz(r)$ by (4.12). In any 
case we may assume that $v$ is centralized by $E \cong E_{p^2}$ in 
$L_3$. As $p \not\Big | |\bC_{L_3}(u)|$ for $u \in {Z_\alpha}^\sharp$, we get 
$|[\tilde{V_3}, v]| = 2$. This contradicts (1.27).
\\
\begin{equation}
K_3 \cong G(r) , \quad {\mbox {a group of Lie type in even characteristic}} 
\end{equation}

By (1) we see that $K_3$ admits a quadratic fours group. Now application 
of (1.12) shows $K_3/Z(K_3) \cong U_4(3)$, $M_{12}$, $M_{22}$, $M_{24}$, 
$J_2$, $Co_1$
, $Co_2$, $Sz$, or $A_n, n \le 11$. In any case there is 
$E \cong E_{p^2}$, $E \le \bC_{L_3/Q_3}(K_3)$. As $m_3(K_3) \ge 2$, we 
have $p > 3$. Suppose $|W_3 : \bC_{W_3}(K_\alpha)| = 4$, then we see that $|[v, \tilde{V_3}]| = 4$ as before. Now 
(1.30) yields $K_3/Z(K_3) \cong A_n$, as $K_3/Z(K_3) \not\cong U_4(3)$ 
by $m_p(K_3) = 1$. Suppose $|W_3 : \bC_{W_3}(K_\alpha)| > 4$. Then 
$K_3/Z(K_3) \cong A_n$ or $M_{22}$. In any case, $K_3/Z(K_3) \cong A_n$ 
or $M_{22}$, there is $E \cong E_{p^2}$, $E \le \bC_{L_3/Q_3}(K_3)$. Now $[E, \tilde{V_3}] \not= 1$. Hence 
there are at least three nontrivial irreducible modules $T_1, T_2, T_3$ 
involved, which 
are centralized by a $p$ - element. This shows that 
$[v, \langle T_1, T_2, T_3 \rangle] \cap Z_\alpha = 1$. Now 
$|[v, T_1]| = 2$ and so we have that $|\tilde{V_3} : \tilde{V_3} \cap Q_\alpha| = 2$. But then 
$|[v, \langle T_1, T_2, T_3 \rangle ]| = 2$, a contradiction.
\\
\begin{equation}
{\mbox {Let}}\quad [\tilde{V_\alpha}, K_3] \not= 1. \quad 
{\mbox {Then}}\quad m_p(\bC_{L_3/Q_3}(K_3)) \ge m_p(L_3) -1.
\end{equation}

Suppose false. Then $K_3$ admits a field automorphism. By (4.12) $K_3 
\cong L_2(r)$, or $Sz(r)$ and there is $\tilde{K_3} \le 
\bC_{L_3/Q_3}(K_3)$, $\tilde{K_3}
 \cong K_3$. If $[\tilde{K_3}, \tilde{V_3}] =1$, we have $\tilde{V_3} 
\cap Z_\alpha =1 $. Hence $K_3 \cong L_2(r)$ and there is just one 
natural module involved (1.27). As we have some $\rho \in L_3$, $\rho$ induces a field 
automorphism on $K_3$, we see that there is some 
$v \in [\tilde{V_3}, \tilde{V_\alpha}]$, which is centralized by $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \le M$, a 
contradiction.

So we have $[\tilde{K_3}, \tilde{V_3}] \not= 1$. As $r \ge 8$, we get 
$|Z_\alpha| \ge 8$. Hence $L_2$ is nonsolvable. Let 
$\omega \in L_1 \cap L_2$, $o(\omega)
$ odd. Then $[\omega, K_3] \not=1 $ or $[\omega, \tilde{K_3}] \not= 1$ 
by (1.63). In particular $q \le r$. Now for 
$z \in Z_2 \setminus Q_\alpha$, we have $|\tilde{V_\alpha} : \bC_{\tilde{V_\alpha}}(z)| \le qr$. This shows that 
$q = r$ and $|[\tilde{V_\alpha}, z]| = r^2$. Now $\tilde{K}_\alpha$ acts on 
$[z, \tilde{V_\alpha}]
$ and induces the natural module. Furthermore some $\rho_\alpha$ of 
order $p$ with $[\rho_\alpha, Z_\alpha] = 1$ induces a field 
automorphism on $K_\alpha$ and acts on $[\tilde{V_\alpha}, z]Z_\alpha/Z_\alpha$. Now every 
$v \in [\tilde{V_\alpha}, z]$ is centralized by some $p$ - element 
$\tilde{\rho_\alpha} \in L_\alpha$ with $N_G(\langle \tilde{\rho_\alpha} \rangle) \le M_\alpha$. 
But $[z, \tilde{V_\alpha}] \cap Z_3 \not= 1$, as can be seen from the 
action of $\tilde{V_\alpha} \cap L_2$ on $Z_2$ by (5.6). This shows $M_3 = M_\alpha$ with 
(4.2), contradicting (5.7).
\\
\begin{equation}
 [\tilde{V_\alpha}, K_3] = 1 
\end{equation}

Let first $p \Big | r^2 - 1$. Then $K_3 \cong Sz(r)$, $L_2(r)$, 
$L_3(r)$, or $U_3(r)$. By (3) $m_p(\bC_{L_3/Q_3}(K_3)) \ge m_p(L_3) - 1$. Suppose 
there is $\rho \in M_3$ with
 $[\rho, \tilde{V_3}] = 1$ and $N_G(\langle \rho \rangle) \leq M_3$. Then $\tilde{V_3} \cap Z_\alpha = 1$. Hence 
$K_3 \cong L_2(r)$ and $\tilde{V_3}/\bC_{\tilde{V_3}}(K_3)$ is the 
natural module, or $K_3 \cong L_3(r)$ and there are at most two nontrivial modules involved, 
which have to be the natural module and its dual. In both cases we get 
$p \Big | r -1$. Hence as $m_p(K_3) = 1$, we see $K_3 \cong L_2(r)$, and there is 
$v \in [\tilde{V_3}, \tilde{V_\alpha}]$ centralized by $E \cong E_{p^2}$ 
in $M_3$, a contradiction.

So we have $m_p(L_3) \geq 3$ and $[\rho, \tilde{V_3}] \not= 1$ for all $\rho \in L_3$ with $o(\rho) = p$. Choose $\rho$ with $[\rho,K_3] = 1$.  Now $[\bC_{\tilde{V_3}}(\rho), K_3] \not= 1$. This shows $Z_\alpha \cap [\bC_{\tilde{V_3}}(\rho), K_3] = 1$ and so again $K_3 \cong L_2(r)$ or $L_3(r)$. Let ${\tilde{V_3}}\!^{(1)}$ be a 
nontrivial module in $[\bC_{\tilde{V_3}}(\rho), K_3]$. Then 
${\tilde{V_3}}\!^{(1)}$ is the natural
 module. We have $|[{\tilde{V_3}}\!^{(1)}, {\tilde{V_\alpha}}\!^{(1)}]| = r$ in 
case of $K_3 \cong L_2(r)$ and so we see that any element in 
${\tilde{V_3}}\!^{(1)}$ is centralized by $E \cong E_{p^2}$. As the same holds for 
${\tilde{V_\alpha}}\!^{(1)}$, we get a contradiction to (4.2) and (5.7).

So we have $p \not\Big | r^2 - 1$. Choose $\rho$ with $o(\rho) = p$ and $[K_3, \rho] = 1$ or in case of $m_p(L_3) = 2$ with $[\tilde{V}_3, \rho] = 1$. Now 
$\bC_{\tilde{V_3}}(\rho)$ involves at least three nontrivial irreducible 
$K_3$ - modules $T_i$,
 $i = 1,2,3$. As $T_i \cap Z_\alpha = 1$, we see $|[T_i, t]| \le r^2$, 
$i = 1,2,3$, for $t \in \tilde{V}_\alpha$, and so $K_3 \cong L_3(r)$, $L_4(r)$, $L_5(2)$, $L_6(2)$, 
$L_7(2)$, $Sp_4(r)
$, $Sp_6(r)$ or $\Omega_8^-(r)$. In the last three cases by (1.46) and 
(1.45) $|\tilde{V_3} : \bC_{\tilde{V_3}}(K_\alpha)| \le r$ and then 
$|[\tilde{V_\alpha},T_1 \oplus T_2 \oplus T_3]| \le r$, a contradiction.

In case of $K_3 \cong L_3(r)$, we have 
$|\tilde{V_3} : \bC_{\tilde{V_3}}(K_\alpha)| \le r^2$, a contradiction 
too. Hence we have $K_3 \cong L_4(r)$, $L_5(2)$, 
$L_6(2)$, or $L_7(2)$.

Suppose $K_3 \cong L_4(r)$. Then 
$|\tilde{V_3} : \bC_{\tilde{V_3}}(K_\alpha)| = r^3$. Furthermore 
$p \Big | |L_3(r)|$. Now every element in $T_1$ is centralized by $E \cong E_{p^2}$. But then $[\tilde{V_3}, \tilde{V_\alpha}]$ 
contains some $v \not= 1$, centralized by $E \cong E_{p^2}$ in $L_3$ with $\Gamma_{E,1}(G)\leq M_3$ and some 
$p$ - element in $L_\alpha$, contradicting (4.2) and (5.7).

Let $K_3 \cong L_5(2), L_6(2)$, or $L_7(2)$. Now $|[T,t]| \le 2^2, 2^3, 
2^4$, respectively. By (1.47) we see that $T$ is the natural module, or 
$K_3 \cong L_7(2)$ and $|[T,v]| = 2^4$. Now there are exactly three nontrivial 
irreducible modules involved and so $p = 7$. But $m_7(L_7(2)) = 2$. 
Hence in any case $T_i$ are natural modules. Furthermore $p \not|~ |L_{n-1}(2)|, n = 5,6,7$. 
This shows $p = 31$ for $n = 5$, or $p = 127$ for $n = 7$. In particular 
there are $n$ modules $T_1,\dots, T_n$. Now $\tilde{V_3}$ induces transvections and 
so $|\tilde{V_3} : \tilde{V_3} \cap Q_\alpha| \le 2^{n-1}$, a 
contradiction.
\\
\\
\indent
We will assume $\tilde{V_\alpha} \not\le L_2$. By symmetry also 
$\tilde{V_3} \not\le L_{\alpha + 1}$. Hence there is a component 
$K_3^{(1)}$ with $[K_3^{(1)}, \tilde{V_\alpha}] \not= 1 \not= [K_\alpha, \tilde{V_3}]$. 
Let $K_\alpha^{(1)}$ be the corresponding component of 
$L_\alpha /Q_\alpha$. Hence $[K_3, \tilde{V_\alpha}] = 1 = [K_\alpha^{(1)}, \tilde{V_3}]$.
\begin{enumerate}
\item[\hspace{0.2cm} (5) \hspace{0.2cm}] One of $K_3$ or $K_3^{(1)}$ is a 
group of Lie type over a 
field of characteristic two. If $K_3$ or $K_3^{(1)}$ is not a group of 
Lie type then 
$K_3/Z(K_3)$ or $K_3^{(1)}/Z(K_3^{(1)})$ is isomorphic to 
$A_n, 7 \le n \le 11, n \not= 8$, $M_{12}$, $M_{22}$, $M_{24}$, or 
$J_2$. In particular $m_3(K_3/Z(K_3)) \ge 2$ or $m_3(K_3^{(1)}/Z(K_3^{(1)})) \ge 2. $
\end{enumerate}

Assume false. Let $|W_3 : W_3 \cap Q_\alpha| = 2$. Then for 
$v \in \tilde{V_\alpha}$, we have 
$|[v,V_3] : [v, V_3] \cap Z_\alpha| \le 2$. As $m_p(K_3^{(1)}) = 1$ we have $|[v,V_3]| \geq 4$. We may assume $v \in Z_{\alpha+1}$. Now by (8.3) - (8.10) we have $F^\star(L_3/Q_3) = K_3^{(1)}\bC_{F^\star(L_3/Q_3)}(v)$. Hence $F^\star(L_3/Q_3) = K_3K_3^{(1)}$ and $m_p(L_3) \geq 3$. Furthermore there is a fieldautomorphism induced. Now by (4.12) we get that $v$ centralizes some $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \leq M_3$ which acts on $[v, V_3]$. So we have a contradiction. 
Hence $|W_3 : W_3 \cap Q_\alpha| > 2$. By the same argument we have that $|W_\alpha : W_\alpha \cap Q_3| > 2$. Now if $K_3$ is not a group of Lie type we have  
$K_3/Z(K_3) \cong A_n, n \le 11
$, $M_{12}$, $M_{22}$, $M_{24}$, or $J_2$. If the same applies for $K_3^{(1)}$ we have $m_3(K_3^{(1)} \times K_3) \ge 4$, a contradiction. So one of both is a group of Lie type.
\\
\setcounter{equation}{5}
\begin{equation}
L_2 \quad {\mbox {is nonsolvable}} 
\end{equation}

\absa
Choose notation such that $m_3(K_3) \le 1$. By (5) $K_3$ is a group of 
Lie type and so $K_3 \cong L_2(r)$, $L_3(r)$, $U_3(r)$ or $Sz(r)$. Now for $v \in \tilde{V}_\alpha$ we have 
$|[v, V_3]| \le 2r$, or $2r^2$ in case $K_3 \cong SL_3(r)$. But 
$[K_3, [v, V_3]] \not= 1$. This implies $K_3 \cong L_3(2), p = 7$ and 
$|[v, V_3]| = 8$. Now $p \not\Big|~ |\bC_{L_3/Q_3}(K_3 \times K_3^{(1)})|$. 
Application of (4.12) gives a contradiction.

\absa
We are now going to prove that $\tilde{V}_\alpha \leq L_2$. So assume false. Choose notation such that $G(r_1) \cong K_3$ is a group of Lie type. Furthermore if $K_3^{(1)}$ is a group of Lie type too we assume 
$K_3^{(1)} \cong G(r_2), r_1 \ge r_2$.  By (6) $q > 2$.

Suppose $q \le r_1$. Then we get 
$|[v, V_3]| \le |W_3 : \bC_{W_3}(K_\alpha)|r_1$. This shows 
$K_3 \cong L_2(r_1)$, $|[v, V_3]| = r_1^2$; $K_3 \cong L_3(r_1)$, 
$|[v,V_3]| = r_1^3$;
$K_3 \cong L_4(r_1)$ or $K_3 \cong Sp_4(r_1)$, $|[v,V_3]| = r_1^4$; 
$K_3 \cong Sp_6(r_1)$. Suppose $[v,V_3] \cap Z_\alpha \not= 1$. There is some
$\rho \in \bC_{M_3}(K_3 \times K_3^{(1)})$ with $o(\rho) = p$. As $[\rho,[V_3,v]] \not= 1$, we get with (4.13) that we may assume that $\rho \in L_3$. Now $\rho$ acts on $[v, V_3]$ 
and we see $o(\rho) \Big | r_1 - 1$. As $m_p(K_3) = 1$, this implies 
$K_3 \cong L_2(r_1
)$. But again every $x \in [v,V_3]^\sharp$ is centralized by some 
$p$ - element in $K_3 \times \langle \rho \rangle$, contradicting (4.2) 
and (5.7).

We are left with $K_3 \cong L_4(r_1), |[v,V_3]| = r_1^4$, or 
$K_3 \cong Sp_6(r_1), |[v, V_3]| = r_1^6$. Now 
$[K_\alpha, \tilde{V_\alpha}]$ involves the natural
 module. We have $m_3(K_3^{(1)}) \le 1$. This implies with (5) 
$K_3^{(1)} \cong L_2(r_2)$, $L_3(r_2)$, $U_3(r_2)$, or $Sz(r_2)$. Let 
$\rho \in \bC_{L_3/Q_3}(K_3) \times K_3^{(1)})$ with $o(\rho) = p$. Then $[\rho, [V_3, v]] = 1$, as 
$o(\rho) \not\!\Big | ~r_1 - 1$. This shows that 
$[[K_3^{(1)}, V_3], \rho] = 1$. Hence $[K_3^{(1)}, V_3] \cap Z_\alpha = 1$ in any case. Now let $u \in V_3$. Then 
$|\tilde{V_\alpha} : \bC_{\tilde{V_\alpha}}(u)| \le r_2q$. As 
$K_\alpha$ acts on $[K_\alpha^{(1)}, V_\alpha]$, we get that 
$|[K_\alpha^{(1)}, V_\alpha] : \bC_{[K_\alpha^{(1)}, V_\alpha]}(u)| \le r_2$. But $K_\alpha^{(1)}$ acts on $[[K_\alpha^{(1)}, V_\alpha],u]$, a contradiction. 

We have $q > r_1$. Suppose there is $\omega \in L_2 \cap L_3$, 
$o(\omega) = x$, $x$ a Zsigmondy prime, or $x = 9$ for $q = 64$ with 
$[\omega, K_3] \not= 1$. If $[K_3, \omega] \not\leq K_3$, we may assume $[V_\alpha, K_3^{\langle \omega \rangle}] = 1$. Now $V_3$ is generated by elements $x$ with $m_p(\bC_{M_3}(x)) \geq 2$ and so $\bC_G(x) \leq M_3$. Hence $[x, \tilde{V}_\alpha]$ is centralized by $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \leq M_3$. But there is $x \in \tilde{V}_3$ with $m_p(\bC_{L_\alpha}(x)) \geq 2$ and so some $1 \not= y \in [x,\tilde{V}_\alpha]$ is centralized by some $p$-element in $L_\alpha$, contradicting (4.2) and (5.7). So we have $[K_3, \omega] \leq K_3$. As $N_G(\langle \omega \rangle) \not\le M_3$, we see that (1.65)(ii) is not 
possible. In these cases $\omega$ would be centralized by $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \le M_3$. As $[Z_1, \omega] = 1$ and $p \not= 3$, we get a 
contradiction with (1.63).

Now with (1.64) and (1.65) we get $q = r_1^2$, 
$K_3 \cong U_n(r_1), n \le 5$, $\Omega_8^-(r_1)$, $^2F_4(r_1)$, or 
$^3D_4(r_1)$. As $[v, V_3]$ is an SC - module
 for $K_3$, the cases $K_3 \cong$ $^2F_4(r_1)$ and $K_3 \cong$ 
$^3D_4(r_1)$ are not possible by (1.22). Hence we have 
$K_3 \cong \Omega_6^-(r_1), |[V_3,v]| = r_1^6$; $K_3 \cong \Omega^-(8,r_1), |[V_3,v]| = r_1^8$. In both cases 
$[V_3,v] \cap Z_\alpha \not= 1$. Let 
$\rho \in \bC_{L_3/Q_3}(K_3 \times K_3^{(1)})$ with $o(\rho) = p$. Then $[\rho,[V_3,v]] \not= 1$. This implies 
$p \Big | r_1 - 1$, contradicting $m_p(K_3) = 1$. Hence we have 
$[\omega, K_3] = 1$.

Let now $[\omega, K_3^{(1)}] \not= 1$. Again $[\omega, K_3^{(1)}] \leq K_3^{(1)}$.  Suppose $K_3^{(1)}$ is not a group 
of Lie type. Then by (5) and (1.9) $o(\omega) = 3$. Now $K_3 \cong L_3(2)$ and so $\omega \in K_3^{(1)}$. But then $\bC_M(\omega) \ge E$, $E \cong E_{p^2}$ with 
$\Gamma_{E,1}(G) \le M_3$. As $\bC_M(\omega) \ge K_3$, we get 
$N_G(\langle \omega \rangle) \le M_3$ by (1.63), a contradiction. Now with (1.64) and (1.65) we 
see $r_2^2 = q$ and $K_3^{(1)} \cong U_n(r_2)$, $\Omega_8^-(r_2)$, 
$^2F_4(r_2)$ or $^3D_4(r_2)$. Let $v \in [K_3^{(1)}, V_3]$. Then we see that 
$|[v, V_\alpha]| \le r_2^6, r_2^6, r_2^{10}, r_2^6$, respectively. As 
$[v, V_\alpha]$ is an SC - module
 for $K_3^{(1)}$, we see with (1.22) that 
$K_3^{(1)} \cong \Omega_6^-(r_2), |[v, V_\alpha]| = r_2^6$, or 
$K_3^{(1)} \cong \Omega_8^-(r_2), |[v,V_\alpha]| = r_2^8$. But now we get the same contradiction as before. This implies 
$[\omega, K_3 \times K_3^{(1)}] = 1$. By (1.63) we get 
$K_3 \times K_3^{(1)} \cong L_2(4) \times L_2(4)$ and $p = 5$. Now $|[v, V_3] : [v,V_3] \cap Z_\alpha| \le 4$. 
As $E_{25}$ acts on $[v, V_3]$, we get some $\rho \in L_3$ with 
$o(\rho) = 5$ and $N_G(\langle \rho \rangle) \le M_3$, and $\bC_{Z_\alpha}(\rho) \not= 1$, 
contradicting (4.2) and (5.7). So we have shown
\\
\begin{equation}
 \tilde{V_\alpha} \le Q_3 \le L_2 
\end{equation}

\absa
We now investigate $[\tilde{V_\alpha}, Z_2]$. Suppose $L_2$ to be 
solvable. Then we have transvections on $\tilde{V_\alpha}$. This implies 
$K_\alpha \cong A_n$, $L_n(2)$, $Sp_{2n}(2)$ or $\Omega_{2n}^{\pm}(2)$. Furthermore as 
$K_\alpha$ does not possess an outer automorphism of order $p$, we get 
$[z, \tilde{V_\alpha}]$ is centralized by $E \cong E_{p^2}$ with $\Gamma_{E,1}(G) \le M_\alpha$, 
where $[z, \tilde{V_\alpha}] \cap Z_3 \not= 1$. But this contradicts 
(4.2) and (5.7). So we
have $L_2$ is nonsolvable. Now there is $z \in Z_2$ with 
$1 \not= [z, \tilde{V_\alpha}] \le Z_3$. As $E \cong E_{p^2}$ acts on 
$[v, V_3]$ we get some $1 \not= u \in Z_3$ centralized by some $p$ - element in $L_\alpha$, 
contradicting (4.2) and (5.7). This final contradiction proves the proposition.

\absa
{\bf (8.12) Proposition.~}{$b = 1$}

\absa
Proof.~ By (8.11) we just have to show that (8.1) holds. Hence assume 
$L_1/O_2(L_1)$ is an extension of an $r$ - group $R$ by $L_2(r)$ times 
$L_1 \cap L_2$. Now
 we have 
$|Z_2 : Z_2 \cap Q_\alpha| \le 4 \ge |Z_{\alpha + 1} : Z_{\alpha + 1} 
\cap Q_3|$. Choose $\alpha + 1 \in \Delta (\alpha)$ with 
$[Z_2, Z_{\alpha + 1}] 
\not= 1$.\\
\\
\setcounter{equation}{0}
\begin{equation}
{\mbox {We may assume}}\quad [Z_{\alpha + 1} \cap Q_3 , Z_2] \not= 1 
\end{equation}

\absa
Otherwise we have $[Z_{\alpha + 1} \cap Q_3 , W_3] = 1$. Hence 
$W_3 \cap Q_\alpha$ centralizes a subgroup of index four in 
$Z_{\alpha + 1}$. If $[W_3 \cap Q_\alpha, Z_{\alpha + 1}] = 1$, then $Z_{\alpha + 1}$ centralizes a subgroup 
of index four in $W_3$. Now by (1.51) either $Z_{\alpha + 1} \le Q_3$ or 
$O_r(L_3/Q_3)
$ centralizes $W_3$. By (4.10) the latter is not possible as this would 
imply $L_2 \le M_3$. So we have $[Z_{\alpha + 1}, W_3] = 1$, a 
contradiction. This implies $[W_3 \cap Q_\alpha, Z_{\alpha + 1}] \not= 1$. In particular 
$W_3 \cap Q_\alpha$ centralizes a subgroup of index four in 
$Z_{\alpha + 1}$. By (1.43) we get $E((L_2/Q_2)/F(L_2/Q_2)) \cong L_2(4)$, $L_3(2)$, 
$L_2(4) \times L_2(4)$, $L_3(2) \times L_3(2)$, or $L_2/Q_2$ is 
solvable. In any case $Z_{\alpha + 1}$ is generated by elements $z$ such that $|W_3 : \bC_{W_3}(z)| \le 16$. 
Application of (1.51) gives $[O_r(L_3/Q_3), W_3] = 1$, a contradiction 
to (4.10).
\\
\begin{equation}
 [Z_{\alpha + 1} \cap Q_3, Z_2 \cap Q_\alpha] \not= 1 
\end{equation}

\absa
Otherwise $Z_{\alpha + 1} \cap Q_3$ centralizes a subgroup of index four 
in $Z_2 \cap Q_\alpha$. Now any $z \in Z_2$ centralizes a subgroup of 
index 16 in $W_\alpha$. Application of (1.51) and (4.10) yields a contradiction as 
$Z_2 \not\le Q_\alpha$.
\\
\begin{equation}
L_2 \quad {\mbox {is nonsolvable and}} \quad |W_\alpha \cap Q_3 : 
W_\alpha \cap Q_2| \ge 16 
\end{equation}

\absa
Otherwise we have that 
$Z_2 = \langle z \Big | |W_\alpha : \bC_{W_\alpha}(z)| \le 32 \rangle$. 
By (1.51) $[O_r(L_\alpha /Q_\alpha), W_\alpha] = 1$, contradicting (4.10). Now application of (1.7), (2.1) together with (1.12) shows 
that $E(L_2/Q_2)$ is as in (1.7)(i) or (iii).
\\
\begin{equation}
|Z_{\alpha + 1} \cap Q_3 : Z_{\alpha + 1} \cap Q_2| \ge 4 
\end{equation}

\absa
Otherwise $z \in Z_2 \cap Q_\alpha$ centralizes a subgroup of index 
eight in $Z_{\alpha + 1}$. Now we apply (1.26) and (1.27). This shows 
that $E(L_2/Q_2) \cong L_2(8)$, $L_2(8) \times L_2(8)$, $L_2(4) \times L_2(4)$, $A_6$, 
$3\cdot A_6$, $3\cdot A_6 * 3\cdot A_6$. As 
$|W_\alpha \cap Q_3 : W_\alpha \cap Q_2| \ge 16$, we get $E(L_2/Q_2) \cong L_2(8) \times L_2(8)$, 
$L_2(4) \times L_2(4)$, or $3\cdot A_6 * 3\cdot A_6$. Now in any case by 
quadratic action we have that
$E(L_2/Q_2) = X_1X_2$, $Z_2/\bC_{Z_2}(X_1X_2) = ([X_1,Z_2] \oplus 
[X_2,Z_2])\bC_{Z_2}(X_1X_2)/\bC_{Z_2}(X_1X_2)$. But now again 
$Z_2 = \langle z \Big | |W_\alpha : \bC_{W_\alpha}(z)| \le 32 \rangle$, contradicting (1.51) and 
(4.10). 

\absa
Now we know :
\begin{center}  
$|Z_2 : \bC_{Z_2}(Z_{\alpha + 1} \cap Q_3)| \le 4|Z_{\alpha + 1} 
\cap Q_3 : Z_{\alpha + 1} \cap Q_2| \le |Z_{\alpha + 1} \cap Q_3 : 
Z_{\alpha + 1} 
\cap Q_2|^2$.
\end{center}

\absa
Hence we may apply (1.35). This now yields the following cases :
\begin{enumerate}
\item[(i)] $E(L_2/Q_2) \cong L_2(q), q \ge 16$, $Z_2$ is the natural 
module
\item[(ii)] $E(L_2/Q_2) \cong L_2(16)$, $Z_2$ is the orthogonal module
\item[(iii)] $E(L_2/Q_2) \cong L_2(q) \times L_2(q), q \ge 4$, $Z_2$ is 
a direct sum of two natural modules, or $ q = 4$ and $Z_2$ is a direct 
sum of two orthogonal modules.
\item[(iv)] $E(L_2/Q_2) \cong L_2(16) \times L_2(16)$, $Z_2$ is a direct 
sum of two orthogonal modules.
\item[(v)] $E(L_2/Q_2) \cong L_2(4) \times L_2(4)$, 
$Z_2 = V_1 \oplus V_2$, $V_i, i = 1,2$ is a direct sum of two natural 
modules.
\item[(vi)] $E(L_2/Q_2) \cong L_3(2) \times L_3(2)$, 
$Z_2 = V_1 \oplus V_2$, $V_i, i = 1,2$ is a direct sum of the natural 
module and its dual.
\item[(vii)] $E(L_2/Q_2) \cong 3\cdot A_6 * 3\cdot A_6$, 
$Z_2 = V_1 \oplus V_2$, $V_i$ is the direct
sum of the natural $A_6$ - module and its dual.
\item[(viii)] $E(L_2/Q_2) \cong SL_3(4)$ and $Z_2$ is a sum of the natural module and its dual.
\item[(ix)] $E(L_2/Q_2) \cong A_9$.
\end{enumerate}

\absa
If $E(L_2/Q_2)$ possesses two components, then the projection of 
$W_\alpha \cap Q_3/W_\alpha \cap Q_2$ onto each of these components is 
of order at least 16 by
 (1.51) and (4.10). Hence we just have (i), (iii), (viii) or (ix), as for $L_2(16)$ on 
the orthogonal module no Sylow 2 - subgroup acts quadratically by 
(1.27)(iv). As $W_\alpha \cap Q_3 \leq L_2 \cap L_3$ and $W_\alpha \cap Q_3 \not\leq Q_2$ we see that $E(L_2/Q_2) \not\cong A_9$. 

So we have $Z_2$ is either the natural module or a direct sum of two 
natural modules. In the latter $Z_1$ is centralized by $O_r(L_1/Q_1)$ 
and $\bC_{L_2}(x)
 \not\le M$, for some $x \in Z_1^\sharp$. This contradicts (4.10). So we 
have
\\
\begin{equation}
E(L_2/Q_2) \cong L_2(q), q \ge 16, Z_2 {\mbox {~is the natural 
module.}}
\end{equation}

\absa
But then 
$1 \not= [Z_{\alpha + 1} \cap Q_3, Z_2 \cap Q_\alpha] \le Z_\alpha 
\cap Z_3$. Hence $M_\alpha = M_3$. By (1.51) and (4.12) 
$\bC_G(x) \le M_3$ for every $x \in Z_1^\sharp$. Hence application of (5.7) gives a contradiction. 
This proves the proposition. 

\absa
{\bf (8.13) Proposition.~}{$b > 1$}

\absa
Proof.~ By (5.7) we just have to show that $Z_2$ is one of the modules 
in (5.6) or the exceptional case of (5.7) holds.
Assume $b = 1$. Then $Z_2 \not\le Q_1$. Application of (1.62) shows\\
\begin{enumerate}
\item[($\star$)] There is some $g \in L_1$ such that for 
$X = \langle Z_2, Z_2^g \rangle$ we have $X/O_2(X) \cong D_{2r}$, $r$ 
odd, $L_2(r)$,
$Sp_4(r)$, $Sz(r)$, $r$ even, and for 
$Y = (Z_2 \cap O_2(X))(Z_2^g \cap O_2(X))$ we have 
$|Z_2 : \bC_{Z_2}(Y)| \le |Y : Y \cap Q_2|^2$, or $Z_2$ is an 
F - module.
\end{enumerate}

\absa
For what follows we will assume that $Z_2$ is neither an F - module nor 
one of the modules from (5.6). This in particular implies 
$[Z_2, Y, Y] \not= 1$.
Suppose first that $L_2$ is nonsolvable. As $[Z_2,Y,Y] \not= 1$, we get
(1.35)(5), (6), (8), (9), (11), (12), (13), or (14). Now (1.35)(5) is 
(5.6)(ii), (1.35)(9) is (5.6)(iii). If we have (1.35)(14), then 
$Q_1 \le Q_2$ and so 
$[Z_2, Q_1] = 1$, i.e. $Z_2 \le Q_1$. Hence $b > 1$.

It remains (1.35)(6), (8), (11), (12), or (13). As 
$m_p(\bC_{L_1}(Z_1)) \ge 3$,
we get $ q = 4$ in (8) and $q = 2$ in (11).

Suppose we have (1.35)(11), (12), or (13). As 
$|Z_2 : \bC_{Z_2}(Y)| \le |Y : Y \cap Q_2|^2$,
we see that $|Y : Y \cap Q_2| = 8$ or we have (1.35)(13). Let first 
$|Y : Y \cap Q_2| = 8$. As $X$ cannot act transitively on 
$(Y/Z_2 \cap Z_2^g)^\sharp$, 
otherwise $Y$ would be abelian, we now see that $X/O_2(X)$ is dihedral. 
Hence for every $t \in Y$ we have $|Z_2 : \bC_{Z_2}(t)| \le 8$. But for 
$t \in Y \cap 
L_1^\prime$, we have $|Z_2 : \bC_{Z_2}(t)| \ge 16$, a contradiction.

Hence we have (1.35)(13) and $|Y : Y \cap Q_2| > 8$. We have furthermore 
$|Y : Y \cap Q_2| \le 64$.
In any case there is $t \in Y$ with $|Z_2 : \bC_{Z_2}(t)| \ge 2^7$. In 
particular
$X/O_2(X)$ is not dihedral. Suppose $X/O_2(X) \cong L_2(r)$. As $X$ 
cannot act transitively on $(Y/Z_2 \cap Z_2^g)^\sharp$, we get $r \le 8$. Now again $|Z_2 
: \bC_{Z_2}(t)| \le 2^6$, a contradiction. If $X/O_2(X) \cong Sz(8)$, 
then for $t \in Y \setminus Q_2$, we
have $[S \cap X, t](Z_2 \cap Z_2^g) = Z_2 \cap O_2(X)$. Hence
$[t, Z_2 \cap O_2(X)] = 1$, and so $|[t, Z_2]| \le 8$, a contradiction. 
Let finally $X/O_2(X) \cong Sp(4,r)$. As not all elements in 
$(Y/Z_2 \cap Z_2^g)^\sharp
$ are conjugate, we see that $X/O_2(X) \cong Sp_4(2)$. Now 
$|[Z_2 , t]| \le 2^7$ for every $t \in Y$, as 
$|Z_2 : Z_2 \cap O_2(X)| = 4$. This implies $|Y : Y \cap Q_2| = 16$, otherwise $Y$ contains $t \in E(L_2/Q_2)$ which does not 
belong
to either of the components and then $|[t, Z_2]| \ge 2^8$. But $|Y : Y \cap Q_2| > 8$ yields  
$|[t, Z_2]| \le 2^5$ for every $t \in Y$, a contradiction.

Suppose now (1.35)(6) or (8), $q = 4$. If $|Y : Y \cap Q_2| \le 4$, then 
we get $X/O_2(X)$ is dihedral and so for  $t \in Y$,
$|Z_2 : \bC_{Z_2}(t)| \le 4$. This proves that we have (8). Now 
$|Y : Y \cap Q_2| \le 16$. We
have that $E(L_2/Q_2) = X_1 \times X_2$, $X_i \cong L_2(4)$ and 
$Z_2 = Z_2^{(1)} \oplus Z_2^{(2)}$,
$[Z_2^{(1)}, X_2] = [Z_2^{(2)}, X_1] = 1$, and $Z_2^{(i)}$ is a direct 
sum of two orthogonal $X_i$ - modules, $i = 1,2$. This implies for 
$t \in Y \cap 
E(L_2/Q_2)$, that $|[t, Z_2]| \ge 16$, and $|[t, Z_2]| \ge 2^8$, for 
$t \not\in X_1 \cup X_2$. As $|Y/Y \cap Q_2| \le 16$, we have 
$|Y/Z_2 \cap Z_2^g| \le 2^8$. This shows that $X/O_2(X) \cong Sp_4(2)$, $L_4(2)$, or 
dihedral. Hence
for every $t \in Y$, we have $|[Z_2, t]| \le 2^5$. But then we get 
$|Y : Y \cap Q_2| \le 8$. Now
we have that $X/O_2(X)$ is dihedral and then $|[Z_2, t]| \le 8$ for 
$t \in Y$. Hence $|Y \cap E(L_2/Q_2)| = 1$. Now $|Y : Y \cap Q_2| = 4$, 
and $|[Z_2, t]| \le 4$ for $t \in Y$. But $Y$ contains some $t$ with 
$[X_1, t] \not= 1 \not= [X_2,t]$ and so $|[Z_2,t]| \ge 16$, a 
contradiction.

We are left with (1.35)(15) and (16). We have $|Y : Y \cap Q_2| \le 8$. 
Hence either $Y$ acts quadratically on $Z_2$ and
$|Z_2 : \bC_{Z_2}(Y)| \le 4$, i.e. $Z_2$ is an F - module, or $X/O_2(X)$ 
is dihedral and $|Z_2 : \bC_{Z_2}(t)| \le 8$
for every $t \in Y$. Furthermore $|Y : Y \cap Q_2| \ge 4$. Suppose the 
latter.
Then $L_2$ is a $\{2,3\}$ - group. Furthermore we may assume that 
(1.35)(16)(iii), (v), or (vi) holds. In any case $Z_2$ is not an F - 
module.
Hence $|[Z_2,t]| \ge 4$ for $t \in Y \setminus Q_2$. If we have 
(1.35)(16)(iii), we have $|Y : Y \cap Q_2| = 4$ and 
$|Z_2 : \bC_{Z_2}(Y)| = 8$.
Now $|Z_2 : \bC_{Z_2}(O_3(L_2/Q_2))| \le 2^6$, contradicting 
$|Z_2| = 2^8$. Let $L_2$
be as in (1.35)(16)(v). Then $S/Q_2 \lesssim Sp_4(3)\cdot 2$. But $Y$ 
does not contain
some $t$ with $[t, O_3(L_2/Q_2)] = O_3(L_2/Q_2)$. Hence again 
$|Y : Y \cap Q_2| = 4$. Let
$Y = \langle y_1, y_2 \rangle (Y \cap Q_2)$. Then 
$W = [O_3(L_2/Q_2), y_1] \cong 3^{1+2}$. Now $|[W, Z_2]| = 16$. 
As $[Z_2,Y,Y] \not= 1$, we see $[Y,[Z_2,W]] \not= 1$. Hence 
$[\bC_{Z_2}(W), Y] = 1$. But then
$[Y, O_3(L_2/Q_2)] = W$, a contradiction.

So we are left with (1.35)(16)(vi). By (1.8)(iv) $|Y : Y \cap Q_2| = 4$. 
But now $|Z_2 : \bC_{Z_2}(Y)| < |Y : Y \cap Q_2|^2$. This contradicts 
(1.36). This 
final contradiction proves the lemma.

\absa
Now the main theorem follows from (8.12), (8.13), (7.32) and (6.39) .

 










 










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