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G. Stroth

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\begin{document}


\tit{2F -- modules with quadratic offender for the finite simple groups}

\indent
In this paper we are going to determine the irreducible faithful 
$2F$-modules for the finite quasisimple groups. Here we just concern about 
$2F$-modules over $GF(2)$. A module $V$ is called $2F$ -module for $G$, if 
there is a nontrivial elementary abelian subgroup $A$ in $G$ such that 
$|V/C_V(A)| \leq |A|^2$.  The group $A$ then will be called an offending 
subgroup or an offender. The offender is called quadratic if $[V,A,A] = 
1$. More precisely we prove\\\\
\noindent
{\bf Theorem} {\it Let $G$ be a quasisimple group and $V$ be an 
irreducible faithful $2F$-module in characteristic 
two for $G$ with a quadratic offender. Then one of the following holds
\begin{enumerate}
\item[(i)] $G/Z(G) \cong A_n$ and one of the following is true
\begin{enumerate}
\item[(a)] $V$ is the natural module
\item[(b)] $n \leq 8$ and $|V| = 16$
\item[(c)] $n = 6$, $|Z(G)| = 3$ and $|V| = 64$
\item[(d)] $n = 9$ and $|V| = 2^8$ is the spin module. 
\end{enumerate}
\item[(ii)] $G \cong 3M_{22}$ or $3U_4(3)$ and $V$ is the 12-dimensional 
$SU_6(q)$--module.
\item[(iii)] $G/Z(G) = G(q)$ is a group of Lie type , $q = 2^t$, and 
one of the following is true
\begin{enumerate}
\item[(a)] $G(q) \cong L_n(q)$, $Sp(2n,q)^\prime$, $\Omega^\pm(2n,q)$ or 
$U_n(q)$ and $V$ is the natural or dual module.
\item[(b)] $G(q) \cong L_n(q)$ and $V$ is the exterior square of the 
natural or dual module
\item[(c)]$G(q) \cong L_6(q)$, or $U_6(q)$ and $V$ is the exterior cube 
of the natural module.
\item[(d)] $G(q) \cong Sp(6,q)$, $Sp(8,q)$ or $Sp(10,q)$ and $V$ is 
the spin module
\item[(e)] $G(q) \cong \Omega^\pm(8,q)$, $\Omega^\pm(10,q)$ or 
$\Omega^+(12,q)$ and $V$ is the half spin module
\item[(f)] $G(q) \cong E_6(q)$ and $V = V(\lambda_1)$ or $V(\lambda_6)$.
\item[(g)] $G(q) \cong E_7(q)$ and $V = V(\lambda_7)$.
\item[(h)] $G(q) \cong F_4(q)$ and $V = V(\lambda_1)$ or $V(\lambda_4)$.
\item[(i)] $G(q) \cong G_2(q)^\prime$ and $V$ is the natural module.
\item[(j)] $G(q) \cong Sz(q)$ and $V$ is the natural module. 
\end{enumerate}
\end{enumerate}}


The proof of the theorem will depend on two main results. First of all we 
will use the classification of $F$--modules. This can be found in an 
unpublished paper \cite{FF} due to U. Meierfrankenfeld and the author. A 
preprint can be found on the homepage 
(http://coxeter.mathematik.uni-halle.de:8080/$\sim$stroth/rep\underline{ }html). But there is also
a classification in the literature. The $F$-modules for the sporadic groups, 
alternating groups and groups of Lie type in odd characteristic have been 
classified by M. Aschbacher \cite{Asch}. The $F$-modules for the groups of 
Lie type in characteristic two have been classified by B. Cooperstein 
\cite{Coop} and in an unpublished paper by B. Cooperstein and G. Mason 
\cite{CMFF}. 

We further will use the classification of quadratic modules in 
\cite{MeiStr}, \cite{MeiStr1} and \cite{Str} to end up with a very short 
list of modules and so it is easy to detect the $2F$-modules. 

In fact there 
are still some open question. The module $V(\lambda_1)$ for $E_6(q)$ is a 
$2F$--module. But we do not know whether it allows a quadratic offender. If 
not then the cases (iii)(f) - (h) will not show up. Further the paper just 
considers irreducible modules. So for general module one has to have an 
overview over the possible offenders, one has to study extensions of the 
modules above and/or of irreducible $2F$--modules by trivial modules. This
all has not been done so far but would be very useful for applications.

The main reason for the classification of these modules comes from the
application in the so called amalgam method. This method basically provides
us with $2F$--modules with quadratic offender for the groups involved, 
provided the parameter $b$ is not 1. So we will use the results in this 
paper in the revision of the classification of the finite simple groups of
characteristic two type. 
 
The notations will be standard. Concerning the representations of the 
groups of Lie type we follow \cite{Stei} and will use these results 
freely. In what follows $G$ will always be a quasisimple group. A Chevalley group or a group of Lie type will always mean a central factor of the corresponding universal Chevalley group.

\section{Preliminaries}


For convenience of the reader we first state the main resulut of \cite{FF}.


\begin{theorem}\label{ffmod} Let $E(G)= F^*(G)$ be a quasisimple group and 
$V$ be an irreducible faithful $F$-module in characteristic two for $G$. 
Then one of the following holds
\begin{enumerate}
\item[(i)] $E(G)/Z(E(G)) \cong A_n$ and one of the following is true
\begin{enumerate}
\item[($\alpha$)] $V$ is the natural module
\item[($\beta$)] $n \leq 8$ and $|V| = 16$
\item[($\gamma$)] $n = 6$, $|Z(E(G))| = 3$ and $|V| = 64$
\end{enumerate}
\item[(ii)] $E(G)/Z(E(G)) = G(q)$ is a group of Lie type , $q = 2^t$, and one of the following is true
\begin{enumerate}
\item[($\alpha$)] $G(q) \cong L_n(q)$, $Sp(2n,q)$, $\Omega^\pm(2n,q)$ or $U_n(q)$ and $V$ is the natural or dual module.
\item[($\beta$)] $G(q) \cong L_n(q)$ and $V$ is the exterior square of the natural or dual module
\item[($\gamma$)] $G(q) \cong Sp(6,q)$ and $V$ is the spin module
\item[($\delta$)] $G(q) \cong \Omega^+(8,q)$ or $\Omega^+(10,q)$ and $V$ is the half spin module
\item[($\epsilon$)] $G(q) \cong G_2(q)$ and $V$ is the natural module.
\end{enumerate}
\end{enumerate} 
\end{theorem}

\begin{lemma}\label{L2q}
Let $G$ be quasisimple with $G/Z(G) = L_2(q)$, $q = 2^n$, and $V$ be a faithful module in 
characteristic two. Then for any involution $a \in G$ we have $|V : 
C_V(a)| \geq q$. If $G/Z(G) = Sz(q)$ or $U_3(q)$, then for any involution 
$a$ we get $|V : C_V(a)| \geq q^2$.
\end{lemma}

\proof Let $p$ be a Zsigmondy prime which divides $q+1$ if $G/Z(G) = 
L_2(q)$ and $q^2+1$ else. Then we may assume that $a$ inverts some element 
$\omega$ of order $p$ in $G$. As $|[V,\omega]| \geq q^2$, we get the 
assertion. Suppose there is no such prime, then $q=8$ and $q+1= 9$. Then 
$a$ inverts some element $\omega$ of order nine, whose commutator has to 
be of order 64 at least.
\qed

\begin{lemma}\label{L2qff} Let $G$ be quasisimple with $G/Z(G) = Sz(q)$ or 
$U_3(q)$ and $V$ be an irreducible faithful $2F$--module for $G$ 
 with quadratic offender $A$. Then $V$ is the 
natural module.
\end{lemma}

\proof By \ref{L2q} we get $|A| = q$. Hence application of \cite[Corollary 
1]{Str} gives the assertion or $G \cong U_3(q)$ and $V$ is the eight -- 
dimensional  module. But in that case there are $g,h \in G$ with $G = 
\langle A, A^g,A^h \rangle$. So we get $|V| 
\leq q^6$,  and so $V$ has to be the natural module. \qed
\\
 
The next five lemmas are more or less well known results on the 
representations of groups of Lie type. We will use them freely  in the 
sequel.

\begin{proposition}\label{stein} Let $G = G(q)$ be a group of Lie type 
and $V$ be an irreducible module over $GF(q)$. Then 
$$V = V_1^{\sigma_1} \otimes \cdots \otimes V_l^{\sigma_l}$$
where the $V_i$ are basic irreducible $GF(q)G$--modules. Further distinct 
$l$--tuples $(V_1, \ldots V_l)$, $(V_1^\prime, \ldots , V_l^\prime)$ give 
nonisomorphic $GF(q)G$--modules.
\end{proposition}

\proof \cite[Th 41 and 43]{Stei} \qed

\begin{lemma}\label{split} Let $G = G(q)$ be a Chevalley group then 
$GF(q)$ is a splitting field for any irreducible module. 
\end{lemma}

\proof \cite[(7.5)]{Stei1} \qed

\begin{lemma}\label{split1} Let $G/Z(G) = G(q)$ be a Chevalley group and 
$V$ an absolutely irreducible $KG$--module for some $K \subseteq 
\widetilde{GF(q)}$. Let $\{\sigma_1, \ldots \sigma_r\} = Gal_{GF(p)}(K)$. 
Then $K = GF(p)(\chi)$ iff $V^{\sigma_1}$, \ldots , $V^{\sigma_r}$ are 
pairwise nonisomorphic $KG$--modules. \end{lemma}

\proof Choose notation such that $\sigma_1 = id$. Let $\{f_1, \ldots f_r\}$ 
be a $GF(p)$--basis for $K$ and $\{v_1,\ldots ,v_n\}$ be a $K$--basis for 
$V$. Then $\{f_iv_j \mid i = 1, \ldots , r, j = 1, \ldots n \}$ is a 
$GF(p)$--basis for $V$. Define
$$\phi_k : K \otimes V  \longrightarrow V^{\sigma_k} \mbox{ by } 
f \otimes f_iv_j \longrightarrow f_i^{\sigma_k}fv_j.$$
Then $\phi_k$ are epimorphisms. Let $n \in {\rm Ker} \phi_k$ for all $k$. 
Set
$$n = \sum_{i,j} f_{ij} \otimes f_iv_j$$
Then $$\sum_{i,j}f_{ij}f_i^{\sigma_k}v_j = 0 \mbox{ for all } k$$
This gives
$$\sum_{i}f_i^{\sigma_k}f_{ij} = 0 \mbox{ for all } k,j$$
By linear independence of automorphisms we have $f_{ij} = 0$ for all $i,j$ 
and so $\bigcap_{k} {\rm Ker }\phi_k = 0$. Set
$$D_i = \bigcap_{k \not= i} {\rm Ker }\phi_k$$
Then
$$K \otimes V = \bigoplus_{i=1}^{r} D_i$$
Now as $K \otimes V/{\rm Ker }\phi_i \cong D_i$, we have 
$D_i \cong V^{\sigma_i}$. 

Let now $K = GF(p)(\chi)$. Then 
$({\rm  Ker }\phi_1)^{\sigma_k} = {\rm Ker }\phi_k$, so $Gal_{GF(p)}(K)$ 
permutes the $D_i$ regularly. Hence by \cite[70.15]{CurRei} the $D_i$ are 
pairwise nonisomorphic.

Assume $K > GF(p)(\chi)$. Then each $V^{\sigma_k}$ appears 
$|K : GF(p)(\chi)|$ times among the $D_i$. Hence the $V^{\sigma_i}$ are not 
pairwise nonisomorphic.
\qed
\\

We like to consider representations of twisted Chevalley groups as well. 
Here we fix notation as follows. We have $^\sigma G(q) \leq G(q^\sigma)$, 
where $G(q^\sigma)$ is the corresponding untwisted group, $G \not= F_4(q)$ or 
$B_2(q)$. Now following \cite[chapter 9]{Stei1} we see that any basic module for 
$G(q^\sigma)$ reduced to $^\sigma G(q)$ remains irreducible. Moreover by 
\cite[9.3]{Stei1} all irreducible modules are given by the tensor product 
theorem. Further by \cite[7.5]{Stei1} $GF(q^\sigma)$ is a splitting field.

It remains the cases $G = F_4(q)$ or $B_2(q)$. Then there is a 
duality between the long and short roots. Just take all weights which 
vanish on all long roots. We call these modules restricted to $^\sigma 
G(q)$ the basic modules. Then again the tensor product theorem holds. If 
the rank is $\ell$ we now get $q^{\ell/2}$ modules. By \cite[12.2]{Stei1} 
$GF(q)$ is the field of definition for all these modules. \\

\begin{proposition}\label{deff} Let $G = A_{\ell}(q)$, $D_{\ell}(q)$, 
$E_6(q)$ or $D_4(q)$ and $V$ be a basic module with high weight $\lambda$ 
for ${^\sigma}G$.  Then the following holds for $\gamma$ the diagram 
automorphism
\begin{itemize}
\item[(1)] If $\lambda \not= \gamma (\lambda)$. Then $GF(q^\sigma)$ is the 
field of definition for $V$.
\item[(2)] If $\lambda = \gamma (\lambda)$ then $GF(q)$ is the field of 
definition for $V$.
\end{itemize}
\end{proposition}

\proof As $[\sigma \gamma,\, {^\sigma}G] = 1$, we have that $V$ and 
$V^{\sigma \gamma}$ are isomorphic. as $GF(q^\sigma)$-modules. So $V \cong 
V^\sigma$ iff $V \cong V^\gamma$. Now from \ref{stein} the fact that 
$GF(q^\sigma)$ is a splitting field and \ref{split1} we get that 
$GF(q^\sigma)$ is the field of definition if $V \not\cong V^\sigma$ and 
$GF(q)$ is the field of definition otherwise. So we just have to determine, 
when $V \cong V^\gamma$. \qed

\begin{lemma}\label{o2} Let $G = G(q), q = p^f,$ be a Chevalley group and 
$V$ be an irreducible module in characteristic $p$ over the splitting 
field. Let $P$ be a parabolic and $V_P = C_V(O_p(P))$. Then $V_P$ is an 
irreducible $P-module$.
\end{lemma}

\proof This is \cite{Sm}. \qed

\begin{Def}\label{pgv} Let $V$ be a faithful $GF(p)$-module for $G$. 
For $\epsilon \in R\!\!\!\!\!\!I$\,\, define ${\cal{P}}_\epsilon(G,V)$ 
to be the set of all non-identity $p$-subgroups $X$ of $G$ such that 
$|X|^\epsilon|C_V(X)| \geq |Y|^\epsilon|C_V(Y)|$ for all $Y \leq X$ 
(including $Y = 1$).
\end{Def}

If $V$ is a $2F$--module with quadratic offender there is always some 
quadratic offender in ${\cal{P}}_2(G,V)$.

\begin{lemma}\label{factor} Let $V$ be a $2F$-module or $F$--module for $G$ 
with quadratic offender and $V_1$ be an invariant subspace. Then $V_1$ is 
a trivial subspace or $V_1$ is a $2F$-module, $F$--module respectively, 
with quadratic offender too.
\end{lemma}

\proof Set $\mu = 2$ or $1$, according to $V$ be a $2F$--module or 
an $F$--module. Choose $A \in {\cal{P}}_\mu(G,V)$, $A$ quadratic on 
$V$. Set $B = C_A(V_1)$. Then $|V_1C_V(A)||B|^\mu \leq |C_V(A)||A|^\mu$. 
Hence $|V_1 : C_{V_1}(A)| \leq |A|^\mu/|B|^\mu$ and so $A/B$ offends on 
$V_1$. \qed




\begin{lemma}\label{closure} Let $V$ be a $GF(p)$--module for $G$.

(a) Let $A,B \in {\cal{P}}_\epsilon(G,V)$ be with $|C_V(A)||A|^\epsilon 
= |C_V(B)||B|^\epsilon$ be maximal, then $AB \in 
{\cal{P}}_\epsilon(G,V)$ and $|C_V(AB)||AB|^\epsilon = 
|C_V(A)||A|^\epsilon$.

(b) Let $A \in {\cal{P}}_\epsilon(G,V)$, $A \leq O_p(G)$. If 
$|C_V(A)||A|^\epsilon$ is maximal among all such $A$, then $\langle A^G 
\rangle$ is a (maybe nonabelian) $\epsilon$ --offender on $V$, i.e. $|V : 
C_V(\langle A^G \rangle)| \leq |\langle A^G \rangle|^\epsilon$.
\end{lemma}

\proof (a) We have $|C_V(AB)||AB|^\epsilon = 
|C_V(AB)||A|^\epsilon|B|^\epsilon/|A \cap B|^\epsilon$. Furthermore 
$$C_V(AB) = C_V(A) \cap C_V(B).$$ So we have \begin{eqnarray*} 
|C_V(AB)||AB|^\epsilon& = &|C_V(A) \cap C_V(B)||A|^\epsilon|B|^\epsilon/|A 
\cap B|^\epsilon\\ &= 
&|C_V(A)||C_V(B)||A|^\epsilon|B|^\epsilon/(|C_V(A)C_V(B)||A \cap 
B|^\epsilon).
\end{eqnarray*}
 As $C_V(A)C_V(B) \leq C_V(A \cap B)$, we get 
 $$|C_V(\langle A,B \rangle)||\langle A, B \rangle|^\epsilon \geq 
 |C_V(AB)||AB|^\epsilon \geq 
 |C_V(A)||A|^\epsilon|C_V(B)||B|^\epsilon/(|C_V(A \cap B)||A \cap 
 B|^\epsilon).$$ So $\langle A, B \rangle \in {\cal{P}}_\epsilon(G,V)$. 
 By maximality we get equality everywhere, the assertion.

(b) This follows by repeated application of (a).\qed
\\

\begin{lemma}\label{strong} Let $V$ be a $GF(p)$--module for $G$ with 
quadratically acting elementary abelian $p$--subgroup $A$. Let $g \in G$ 
and $a \in A$ with $a^g = az$. Then also $\langle z, C_A(g) \rangle$ acts 
quadratically. \end{lemma}

\proof We have $[V,C_A(g),a] = 1$. So also $[V,C_A(g),z] = 1$. As 
$[C_A(g), z] = 1$ the three subgroup lemma shows $[V,z,C_A(g)] = 1$, the 
assertion. \qed
\\ 

Let $W$ be a 
 Weylgroup with root system $\Phi$ and
fundamental roots  $\Pi$. We assume throughout this section
that the Dynkin diagram on $\Pi$ is connected. Let
$l=|\Pi|$, $I=\{1,2,\ldots l\}$ and $\Pi=\{\alpha_1,\alpha_2,\ldots
\alpha_l\}$ where we choose the labeling as follows:


$$ \anl$$

$$\cnl$$

$$\dnl$$

$$\esixl$$


$$\esevenl$$

$$\eeightl$$

$$\ffourl$$

$$\gtwol$$







\begin{lemma}\label{tensor1q} Let $G$ be quasisimple with $G/Z(G) = L_n(q)$ 
or $U_n(q)$ and $V = V(\lambda_1 + \lambda_{n-1})$. Assume $A \leq G$ with 
$[V,A,A] = 1$. Then $|A| \leq q$.
\end{lemma}

\proof We have that $V$ is the adjoint module. Let first $n = 3$ and 
$G/Z(G) = L_3(q)$. Then any elementary abelian subgroup intersects a root 
group. As $V$ is not strong quadratic \cite{Str}, we get the assertion. 
Let now $G/Z(G) = L_n(q)$. By induction as $V$ involves the adjoint module 
for $G_1/O_2(G_1)$ we have $|A/A \cap O_2(G_1)| \leq q$. So either we have 
the assertion or $A \cap O_2(G_1) \not= 1$. But then $V$ would be strong 
quadratic, a contradiction.

So let now $G/Z(G) \cong U_n(q)$. If $n = 3$ there is nothing to prove. 
Let $n = 4$. If $A \leq O_2(G_1)$, we get with \ref{strong} that $V$ is 
strong quadratic or $|A| \leq q$. So let $A \not\leq O_2(G_1)$. Then we 
have $|A /A \cap O_2(G_1)| \leq q$. Let $A \cap O_2(G_1) \not= 1$. As $V$ 
is not strong quadratic we get $|A| \leq q^2$ and $A \leq O_2(G_2)$. So 
let $x \in A^\sharp$. Then $[\langle A^{C_G(x)} \rangle, x] = 1$, and we 
get a quadratic fours group which intersects a root group and contains 
$x$, a contradiction. Let now $n \geq 5$. Then by induction $|A : A \cap 
O_2(G_1)| \leq q$. As $V$ is not strong quadratic, we see with 
\ref{strong} that $|A \cap O_2(G_1)| \leq q$ and $A$ induces transvections 
on $O_2(G_1)/Z(O_2(G_1))$ with $A \cap O_2(G_1) \leq [A, O_2(G_1)]$. But 
$\langle A, [A,O_2(G_1)] \rangle \leq U_4(q)$, and so as before we get 
strong quadratic action, a contradiction. \qed

\begin{lemma}\label{tensor2q} Let $X = V \otimes W$ be a faithful 
$GF(q)$--modules and $A$ be quadratic on $X$ . Then $|A| \leq q$. 
\end{lemma}

\proof Let $v \otimes w \in V \otimes W$, $a \in A$. Then $[v \otimes w, a] 
= [v,a] \otimes w + v \otimes [w,a] + [v,a] \otimes [w,a]$. Let $b \in A$ 
then $0 = [v \otimes w,a,b] = [v,a] \otimes [w,b] + [v,b] \otimes [w,a]$, 
hence $[v,a] = \lambda [v,b]$ and $[w,b] = \lambda^{-1} [w,a]$, for some 
$\lambda \in GF(q)^\sharp$. In particular $|A| \leq q$. \qed




 \begin{lemma}\label{tensorff} Let $X = V \otimes W$, be a faithful 
 $GF(q)$--modules. Suppose $X$ to be irreducible. If $X$ is a 
 $2F$--module with quadratic offender, then $G \cong L_2(q)$ and
 $X = V \otimes V^\sigma$, $V$ the natural module and $\sigma$ some field 
 automorphism of $GF(q)$. 
\end{lemma}

\proof Suppose by way of contradiction that $X$ is a $2F$--module with  
quadratic offender $A$. By \ref{tensor2q} we have $|A| 
\leq q$. Now we get that for $a \in A^\sharp$ the codimension of $C_X(a)$ 
is at least twice the codimension of the centralizers in the modules $V$ 
and $W$. This shows that the codimension is at least two. Hence we get $|X 
: C_X(a)| \geq q^2$, and so we have equality everywhere. Further we get
that $|V : C_V(a)| = q$. This gives $|V| = q^2$ and so $G \cong L_2(q)$
and the assertion follows with \ref{stein}.\qed


\begin{lemma}\label{transv} Let $G$ be quasisimple with $G/Z(G) = G(q)$ be 
of Lie type of rank at least two.\\
a) Let $A \leq G$, $|A| \leq q$ and $V$ be an irreducible module with $|V : 
C_V(A)| \leq |A|^2$ and $A$ quadratic, then $V$ is strong quadratic.

b) Let $R$ be a root group in $G$ with $|[V,R]| \leq q^2$ for some 
irreducible module $V$, then $G/Z(G) \cong L_n(q)$, $U_n(q)$, 
$\Omega^\pm(2n,q)$, $Sp(2n,q)$ and $V$ is the natural module, or $G \cong 
G_2(q)$ and $V$ is the $6$--dimensional module.
\end{lemma}

\proof a) Let $R$ be a root group with $[R,A] = 1$. Suppose first $R 
\geq A$, then obviously $[O_p(C_G(R)),[V,A]] = 1$ and so $V$ is strong 
quadratic. So let $R \not\geq A$. If $A \leq O_2(C_G(R))$ but $A \not\leq 
Z(O_2(C_G(R)))$, then there is some $a \in A$ and $r \in R$ such that $a 
\sim ar$. So $|[V,r]| \leq q^4$. As $C_G(r)$ acts on $[V,r]$ we see again 
that $V$ is strong quadratic or $C_G(R)$ possesses a factor embeddable 
into $GL(4,q)$. Further all involutions of $C_{C_G(R)}([V,r])$ are in $R$.
So we see that $G/Z(G) \cong L_4(q), U_4(q), G_2(q)$. Now we have that $a 
\in A^\sharp$ is not conjugate to $r$. This shows that 
$[[V,a],O_2(C_G(a))] = 1$. Hence $V$ is strong quadratic. 
 
 So let now $A \leq Z(O_2(C_G(R)))$. Then $G \cong F_4(q)$ or $Sp(2n,q)$. 
Recall that for $L_3(q)$, $A$ intersects always a root group nontrivially. 
But then there is some $g \in G$ such that $A^g \leq C_G(R)$ but $A^g 
\not\leq O_2(C_G(R))$. So we may assume that $A \not\leq O_2(C_G(R))$
right from the beginning. 
Then there is even some $h \in C_G(R)$ and some $a \in A$ such that $a^h 
\in P \setminus O_2(P)$, where $P$ is some minimal parabolic of $C_G(R)$. 
Now $P$ acts on $[V,R]$ and $V/C_V(R)$. But by \ref{L2q} we have that 
$P/O_2(P)$ can just induce two
nontrivial modules in $V$. So we may assume that exactly one of those
is in $[V,R]$. This now shows that $|O_2(P)/R| \leq q^5$. In particular 
$G/Z(G) \cong L_4(q), Sp(4,q), U_4(q), G_2(q)$. But now we may argue as 
above. 
 Hence $V$ is strong quadratic. 

b) We may argue by induction. By a) we have that $V$ is strong quadratic. 
So by \cite{Str} $V = V(\lambda)$ for some weight $\lambda$. Further as 
$|[V,R]| \leq q^2$, we see that $[[V,R],C_G(R)] = 1$, or $C_G(R)$ has a 
factor $L_2(q)$, which acts nontrivially on $[V,R]$. Assume the latter. 
Then $G \cong Sp(2m,q)$, $
O^\pm(2m,q)$, or $G_2(q)$. In the first two cases we have the natural 
module, while for $G_2(q)$ we have the $6$--dimensional one.

So assume now that 
 $\lambda$ is the weight corresponding to $C_G(R)$, or $G/Z(G) \cong 
 L_n(q)$ and 
$\lambda = \lambda_1$ or $\lambda_{n-1}$. But the latter are the natural 
modules and we are done. So assume the former. 
Let $G$ be classical. Then we may assume that $G \cong Sp(2m,q)$, or 
$O^\pm(2m,q)$ and $V = V(\lambda_2)$. Now in all cases $C_V(O_2(G_1))$ is 
the natural module for $G_1/O_2(G_1)$. As there is a conjugate of $R$ in 
$G_1$ intersecting $O_2(G_1)$ trivially, and $[G_1,V] \not\leq 
C_V(O_2(G_1))$, we see that $G \cong Sp(2m,q)$. But now we see that
even $[O_2(C_G(R)),[V,R]] = 1$, and so $V$ cannot be $V(\lambda_2)$, by 
\cite{Str}.
Let now $G/Z(G) \cong E_6(q), E_7(q)$, or $F_4(q)$. Then we look at the 
endnode
of the diagram, which does not correspond to $\lambda$. Then induction 
gives us, that we must have $G \cong F_4(q)$ and $V = V(\lambda_1)$. But 
here
the parabolic $G_4$ induces more than two nontrivial modules, a 
contradiction. So we are left with $G \cong G_2(q)$ 
 or 
$^3D_4(q)$. If $G \cong G_2(q)$, then $V$ is the natural module. If 
$G \cong ~^3D_4(q)$, then $V$ is the natural $\Omega^+(8,q^3)$-module and 
so $|[V,R]| \geq q^3$. \qed

\begin{lemma}\label{l2offend} Let be quasisimple with $G/Z(G) \cong L_n(q)$, 
$n \geq 5$, let $V = V(\lambda_2)$ and $A$ be an offender as $F$-module. 
Then  $|A| = q^{n-1} = |V : C_V(A)|$.
\end{lemma}

\proof We prove the assertion by induction on $n$. Let first $n = 5$. Then 
$V$ is a $10$--dimensional module for $L_5(q)$. Let $G_4$ be the parabolic 
which has the following chief series in $V$
$$1  < V_1 < V,$$
where $V_1$ is the $6$--dimensional $L_4(q)$--module and $V/V_1$ the 
$4$--dimensional one. It means that $O_2(G_4)$ offends. Let $x \in 
O_2(G_4)^\sharp$ may assume that $C_G(x) = G_{14}$, so $|[V,x]| = q^3$, as 
all involutions of $O_2(G_4)$ are conjugate. Let now $A_1 \leq O_2(G_4)$, 
$|A_1| \geq q^2$. If $C_V(a) = C_V(A_1)$ for all $a \in A_1^\sharp$, then 
$C_V(A_1)$ is invariant under $\langle C_G(a) \mid a \in A_1^\sharp 
\rangle = O^\prime(G_4)$. But as $C_V(a) \not= V_1$, this is impossible. 
So we have shown that in fact $V_1 = C_V(A_1)$. Let now $A$ be any 
offender. As one can check easily $|V_1 : C_{V_1}(A)| \geq |A/A \cap 
O_2(G_4)|$. Hence we see $A \cap O_2(G_4) \not= 1$. So we get $|C_V(A \cap 
O_2(G_4)) : V_1| \leq q$. But then we see that $|A \cap O_2(G_4)| \geq 
q^3$. Now by the above we have $V_1 = C_V(A \cap O_2(G_4))$. This shows $A 
\geq O_2(G_4)$ and so $A = O_2(G_4)$. 

Let now $n > 5$. We first show $A \leq O_2(G_{n-1})$. Suppose false then 
we have
by induction and \ref{factor} that $A/A \cap O_2(G_{n-1})$ is uniqueley 
determined. So $O_2(G_{n-1})A = O_2(G_{n-1} \cap G_{n-2})$. This shows 
that also $A \cap O_2(G_{n-1})$ offends. So we may just prove that 
$O_2(G_{n-1})$ is a minimal offender. Let $x \in O_2(G_{n-1})^\sharp$, 
then we get that $|[V,x]| =q^{n-2}$. Hence for $A \leq O_2(G_{n-1})$ we 
have $|A| \geq q^{n-2}$. If $|A| = q^{n-2}$ then $C_V(A) = C_V(a)$ for all 
$a \in A^\sharp$. But then $G_{n-1} = \langle C_{G_1}(a) \,\mid \, a \in 
A^\sharp \rangle$ would act on $C_V(A)$,
a contradiction. This shows $A = O_2(G_{n-1})$ and we are done. \qed


\section{$2F$--modules for Lie type groups in even characteristic}

Throughout this chapter we will assume that $G$ is a quasisimple group 
with $G/Z(G)$ a group of Lie type over a field with $q = 2^n$ 
elements, including $G(2)^\prime$ and $A_6$. We additionally assume that 
$G^\prime \not\cong 3\cdot A_6$, as this will be handled in the last 
chapter together with the  alternating groups. As $G$ will act faithfully 
on a $GF(2)$--module, $G$ will always be a factor of the universal group. 

Further $V$ is an $2F$-module and 
 $A \in {\cal{P}}_2(G,V)$ is a quadratic offending subgroup. If $G_i$ is 
 a 
maximal parabolic in $G$ we set $Q_i = O_2(G_i)$.

\begin{lemma}\label{Vlambda2} Let $G/Z(G) = G(q)$ be classical or $F_4(q)$ 
and $V = V(\lambda_2)$ be a $2F$-module with quadratic offender $A$. Then 
$V$ is an $F$--module too.
\end{lemma}

\proof If $G \cong Sp(2n,q)$, $n > 2$, or $\Omega^{\pm}(2n,q)$, $n > 3$, 
then $V$ is the adjoint module. Let first $A \leq Q_2$. As $V$ is not 
strong quadratic by \cite{Str}, we see with \ref{strong} that $|A| \leq 
q$ or $A \leq Z(Q_2)$. By \ref{transv} we get $A \leq Z(Q_2)$ and $G 
\cong Sp(2n,q)$. Further $q^2 \geq |A| > q$. As $[[V,A],{Q_2}^\prime] 
\not=1 $, we see $G \cong Sp(6,q)$, $|A| = q^2$, and $|[V,A]| = q^4$. Let
$g \in G_2$ with $g \not\in N_{G_2}(A)$, which must exist. Then $|A \cap 
A^g| \geq q$ and $[V,A \cap A^g, \langle A, A^g \rangle] = 1$. As 
$\langle A, A^g \rangle$ contains root elements, we get that $V$ is 
strong quadratic, a contradiction. So we have $A \not\leq Q_2$. By 
\ref{strong}
we see that  $|A \cap Q_2 : A \cap Z(Q_2)| \leq q$. If $A \cap Q_2 
\not\leq Z(Q_2)$, we even get $A \cap Z(Q_2) = 1$. Now for $a \in A$ and 
$b \in A \cap Q_2$, we get with \ref{strong} $[C_{Q_2}(b),a] \leq A \cap 
Q_2$. This shows $|A : A \cap Q_2| \leq q$ and then $|A| \leq q^2$ and $G 
\cong Sp(2n,q)$. Now for the root group $R$ normal in $G_2$ we see that 
$[V,R]$ involves a module isomorphic to $Q_2/Z(Q_2)$. Further $C_V(R)$ 
involves some module $V(\lambda_2)$ for $Sp(2n-4,q)$. This shows $|V : 
C_V(A)| \geq q^5$, a contradiction. So we have $A \cap Q_2 \leq Z(Q_2)$.
Now with \ref{strong} we get $A \cap Q_2 = 1$. Let $a,b \in A^\sharp$, 
both not transvections on the natural module. Let $x \in C_{Q_2}(b)$, $x 
\not\in C_{Q_2}(a)$. Then by \ref{strong} $\langle b, [x,a] \rangle$ acts 
quadratically on $V$. Now $[C_{Q_2}([x,a]),b] \not\leq Z(C_{Q_2}([x,a]))$, 
as $|[Q_2,b]| \geq q^4$. Hence there is $\langle c,d \rangle \leq Q_2$, 
with $C_{Q_2}(c) \not= C_{Q_2}(d)$, which acts quadratically on $V$. But 
then by \ref{strong} $V$ is strong quadratic, a contradiction. So we see 
that $A$ contains a subgroup $B$ of index at most $q$, which does not 
contain any transvection such that $C_{Q_2}(b) = C_{Q_2}(B)$ for any $b \in 
B^\sharp$. Hence $|Q_2/Z(Q_2) : C_{Q_2/Z(Q_2)}(B)| \geq |B|^2$, as 
$Q_2/Z(Q_2)$ is a direct sum of two natural modules for $Sp(2n-4,q)$ or
$\Omega^\pm(2n-4,q)$. Again as $[V,R]$ involves a module isomorphic to 
$Q_2/Z(Q_2)$, we see that $|V : C_V(B)| \geq |B|^4$. This now shows $|B| 
\leq q$, as $|A| \leq |B|q$. Now by \ref{transv} we have that $A$ 
contains elements which are not transvections on the natural 
module. Then $|V : C_V(A)| \geq q^8$, which contradicts $|A| 
\leq q^2$.
 



Let now $G/Z(G) \cong U_n(q)$. Then $G$ is embedded in $SL_n(q^2)$ and 
the modules $V(\lambda_2)$ are the same for both groups. Let $n = 5$. If 
$A \leq Q_1$,
then we get $|A| = q^3$ and so $C_V(A) = C_V(R)$, $R$ a root group. But then
we get $C_V(R) = C_V(Q_1)$, a contradiction. So we have $A \not\leq Q_1$. 
Then $|C_V(Q_1) : C_{C_V(Q_1)}(A)| \geq |A/A \cap Q_1|^2$. This shows
that also $A \cap Q_1$ is an offender, a contradiction. 
 
Let next $n = 6$. If $A \leq Q_1$, then by \ref{closure} and \ref{transv}
we get $|V : C_V(Q_1)| \leq |Q_1|^2$, a contradiction. So we have $A 
\not\leq Q_1$. The same argument shows $A \not\leq Q_3$ and so $|A| \leq 
q^8$. If $x \in A \cap Q_1 \setminus R$, we have that $x$ is conjugated to
an element of type $c_2$ in $G_1/Q_1$. This shows $|V : C_V(x)| \geq q^{12}$.
If now $|A : A \cap Q_1| > q$, we see that $|V : C_V(A)| \geq |V : 
C_V(x)||C_V(Q_1) : C_{C_V(Q_1)}(A)| \geq q^{12}q^4$. Hence $|A| = q^8$ and 
we have equality everywhere. As now $|A \cap Q_1| \geq q^4$, we get $x,y \in 
A \cap Q_1 \setminus R$ with $G_1 = \langle C_{G_1}(x), 
C_{G_1}(y)\rangle$. This now shows $C_V(x) \not= C_V(y)$. But then $|V : 
C_V(A) | > q^{16}$. So we either have $A \cap Q_1 \leq R$ or $|A  : A \cap 
Q_1| \leq q$. Suppose $A \cap Q_1 \not\leq R$. Then we have $|V : C_V(x)| = 
q^{12} = |V : C_V(A)|$, as $|A| \leq q^6$. But now we argue as before.
So we have $A \cap Q_1 \leq R$ and so $|A| \leq q^5$. Now $A/A \cap Q_1$ 
cannot contain any element of type $c_2$ in $G_1/Q_1$. This shows $|A| \leq 
q^2$. But this contradicts $|V : C_V(a)| \geq q^6$ for any $a \in A^\sharp$.

So let now $n > 6$. Again $A \not\leq Q_1$. As $C_V(Z(Q_1))$ involves 
$V(\lambda_2)$ for $G_1/Q_1$, we get by induction that $|A \cap Q_1| 
\not= 1$. We have
$|C_V(Q_1) : C_{C_V(Q_1)}(A)| \geq q^2$. So we see that $|V : C_V(A)| 
>q^2|A/A \cap Q_1|^2q^{2(n-2)}$. This shows that $|A \cap Q_1| > q^{n-1}$,
a contradiction, as $Q_1$ does not contain abelian subgroups of that order.
  
So let finally $G \cong F_4(q)$. Then $C_V(Q_4) \cong V(\lambda_2)$ for 
$Sp_6(q)$. So we have $|C_V(Q_4)| = q^{14}$ and $A \leq Q_4$, as 
$V(\lambda_2)$ is not a $2F$--module with quadratic offender for $Sp_6(q)$. 
By \ref{transv} we have $|A| > q$. Let $A  \leq Z(Q_4)$. Then with 
\ref{closure} we get that also $Z(Q_4)$ is an offender in 
${\cal{P}}_2(G,V)$. Now $|Z(Q_4)Q_1/Q_1| = q$ is a long root group. But 
$C_V(Q_1)$ is the spin module for $G_1/Q_1$, and so $|C_V(Q_1) : 
C_{C_V(Q_1)}(Z(Q_4))| = q^4$, which contradicts \ref{factor}.
So we have $A \not\leq Z(Q_4)$. As $[V,Q_4,Q_4^\prime ] \not= 1$ by 
\cite{Str}, we get $|A : A \cap Z(Q_4)| \leq q$. Now the same applies for
$A \cap Q_1$, i.e. $|A \cap Q_1 : A \cap Z(Q_1)| \leq q$. Hence we see that
$|A| \leq q^4$. Recall that $Q_4^\prime \cap A = 1$ by \cite{Str} and the
same applies for $Q_1^\prime$ if $A  \cap Q_1 \not\leq Z(Q_1)$. Now as for 
$x \in Z(Q_1)$ we have $|V : C_V(x)| \geq  q^8$, as a conjugate of $x$ 
acts on two modules $V(\lambda_2)$ for $Sp_6(q)$, we get $|A| = q^4$. But 
as $A \not\leq Q_1$, we get that $C_{C_V(x)}(A) \not= C_V(x)$ for $x \in 
A \cap Z(Q_1)$,  a contradiction. \qed

\begin{lemma}\label{lambdai} Let $G/Z(G) = G(q)$ and $V$ be an irreducible 
faithful $2F$--module with quadratic offender. If $C_V(Q_i) = C_V(Z(S))$ 
for some $i$, then $V = V(\lambda_i)$ or $G/Z(G) \cong L_2(q)$ and $V = X 
\otimes X^\sigma$ for some field automorphism $\sigma$, $X$ the natural 
module.
\end{lemma}

\proof By \ref{stein} we have $V \cong V(\lambda_i)^{\sigma_1} \otimes 
V(\lambda_i)^{\sigma_2} \otimes \cdots \otimes V(\lambda_i)^{\sigma_k}$, for 
some field automorphisms $\sigma_1$, $\ldots , \sigma_k$. By \ref{tensorff} 
we get the assertion. \qed

\begin{lemma}\label{classicalff} Let $G/Z(G) = G(q) \not\cong L_n(q)$ be 
classical and $V$ be an irreducible faithful $2F$--module with quadratic 
offender $A \leq Q_1$. Then then pair $(G,V)$ is one of the theorem. 
\end{lemma}

\proof Suppose false. Application of \ref{transv} shows that we may assume 
that $|A| > q$. Now application of \ref{closure} shows that $Q_1$ offends on 
$V$. So $|V : C_V(Q_1)| \leq |Q_1|^2$. Furthermore by \ref{lambdai} we may 
assume that $C_V(Q_1)$ is a nontrivial $G_1/Q_1$--module. Let $G \cong 
\Omega^{\pm}(2n,q)$. Choose $a \in A^\sharp$. Set $B = \langle 
A^{C_{G_1}(a)} \rangle$. If $B$ does not contain root elements, we see 
$|B| \leq q^2$. But then $|A| \leq q$. So by \ref{transv} $V$ is 
strong quadratic. So in any case $V$ is strong quadratic and then by 
\cite{Str} $V \cong V(\lambda_n)$. But now also $|V/C_V(Q_1)| = 
|C_V(Q_1)|$ and
$C_V(Q_1)$ is the spin module for $G_1/Q_1$. So $|C_V(Q_1)| \leq |Q_1|^2$, 
which shows $G \cong \Omega^+(2n,q)$, $n =3,4,5,6$ or $\Omega^-(2n,q)$, 
$n = 3,4,5$. 

Let now $G/Z(G) \cong Sp(2n,q)$ or $U_n(q)$. Let first $G \not\cong 
Sp(4,q)$.
Let $R$ be a root group normal in $G_1$.
Then we see that $G$ acts strongly quadratic, as $|A| > q $ by \ref{transv}. 
Further we see that $V/C_V(Q_1)$ involves some module isomorphic to 
 $C_V(Q_1)$, as $C_V(Q_1) \leq [R,V]$ by \ref{o2}. As  $V 
 \not\cong V(\lambda_2)$ by \ref{Vlambda2}, we get with \cite{Str} 
 that $C_V(Q_1)$ is not the natural $G_1/Q_1$--module. Further by 
 \ref{closure} we see that $|V : C_V(Q_1)| \leq |Q_1|^2$. As $C_V(Q_1)$ 
 is some $V(\lambda_i)$, $i \not= 1$, for $G_1/Q_1$, we see that either 
 $[V,Q_1,Q_1] = 1$ or $G = Sp(2n,q)$, $n \leq 5$, and $V$ is the 
 spin module, or $G/Z(G) = U_n(q)$, $n \leq 6$, and $V = 
 V(\lambda_{[n/2]})$.
 Suppose first $[V,Q_1,Q_1] = 1$. Then $Q_1$ is abelian and so $G \cong 
 Sp(2n,q)$. Now we get that for $G_2$, we also have $[V,Q_2^\prime,Q_2] = 
 1$. So by \cite{Str}, again $V$ is the spinmodule and we are done. So we 
 are left with $U_5(q)$, $V = V(\lambda_2)$. Now $|[V,R]| = q^6$ and for 
 an involution $x \in Q_1 \setminus Q_1^\prime$, we have $|[V,x]| = q^8$.
 This shows $|A| = q^4$, but there is no such elementary abelian subgroup 
 in $Q_1$. 
  
 So we have to treat $G \cong Sp(4,q)$. Further we may assume that $V$ is 
 not strong quadratic. So we have that $A$ does not intersect any root 
 group. In  particular $|A| \leq q^2$. Using the closure under $G_1$, we 
 even see that $|A| \leq 2q$. If $q > 4$, we see with \ref{L2q} that $V$
 involves at most two nontrivial $G_1/Q_1$--modules and also two 
 $G_2/Q_2$--modules. This shows that $|[V,R]| \leq q^3$ and there is
 just once the natural module involved in $[V,Q]$. Now $|C_V(Q_1)| = q$, 
 or $q^2$. As $|V : C_V(Q_1)| 
 \leq q^6$, we get that $|V| \leq q^8$, a contradiction, as by 
 \ref{stein} then $V$ has to be the natural module. So we may assume  $q 
 = 4$ and $G_1$ induces three nontrivial modules. But then again $[V,R]$ 
 can just involve one such module, and so $|[V,R]| \leq q^3$. Again as 
 before, we get $|V| \leq q^8$.
 \qed

\begin{lemma}\label{rank2} Let $G/Z(G) = G(q)$ be of rank at most two and 
$V$ be a an irreducible faithful $2F$--module with quadratic offender $A$. 
Then $G/Z(G) \cong L_2(q)$, $L_3(q)$, $Sp(4,q)$, $U_4(q)$, 
$U_5(q)$, $\Omega^-(6,q)$ or $G_2(q)$ and $V$ is the natural module, or 
$G \cong L_2(q)$ and $V$ is a tensor product.
\end{lemma}

\proof If $G$ is of rank one the assertion follows with \ref{L2qff} and 
\ref{tensorff}. So let $G$ be of rank two. If $G/Z(G) \cong L_3(q)$, then 
by \ref{transv} $V$ is strong quadratic and the assertion follows with 
\cite{Str}. 

Let $G/Z(G) \cong Sp(4,q)$, $U_4(q)$, $U_5(q)$, or $\Omega^-(6,q)$. Then 
by \ref{classicalff} $A \not\leq Q_1$. If $G \cong Sp_4(q)$, we have 
$A \leq Q_2$ and we get the assertion 
with \ref{classicalff}. So let $G/Z(G) \cong U_4(q)$, $U_5(q)$, or 
$\Omega^-(6,q)$. By \ref{lambdai} $V_1 = C_V(Q_1)$ is a 
nontrivial $G_1/Q_1$--module. So by \ref{o2} it is the natural module or 
the tensor product in all cases. In particular $|A : A \cap Q_1| \leq 
q^2$ and $|A : A \cap Q_1| = q$ if $G/Z(G) \cong U_5(q)$. Now in all 
cases we have by \cite{Str} that $V$ is not strong quadratic. So we see 
that $|A| \leq q^2$. Now by \ref{lambdai} we have that $|C_V(Q_1) : 
C_{C_V(Q_1)}(A)| = q^2$,
$q$, respectively. This shows that in case of $U_5(q)$, we have $|V : C_V(A 
\cap Q_1)| =q^2$, and so for the root group $R$, we get $|V : C_V(R)| \leq 
q^4$, which contradicts the fact that $[V,R]$ contains a nontrivial 
$U_3(q)$--module. So we have $G \cong U_4(q)$. 
Further $A \not\leq Q_2$. 
As $V$ is not strong quadratic, we get that $A \cap Q_2 = 
1$. As $|A| > q$, there are $x,y \in A^\sharp$ such that $[C_{Q_2}(x),y]$ 
contains a root element. But then $V$ would be strong quadratic by 
\ref{strong}, a contradiction

Let now $G \cong G_2(q)$. If $V$ is strong quadratic then by \cite{Str} 
$V \cong V(\lambda_1)$, the assertion. So we may assume that $V$ is not 
strong quadratic. Now by \ref{transv} $|A| > q$. Further $A \not\leq 
Q_1$. So we get $|A| \leq q^2$. If $[A,Z(Q_2)] = 1$, we have $A \cap 
Z(Q_2) \not= 1$, and so $V$ would be strong quadratic. So we have that 
$|A : C_A(Z(Q_2))| \not= 1 \not= C_A(Z(Q_2))$ and $A \cap Z(Q_2) = 1$. 
Then there is some $x \in Z(Q_2)$ and $a \in A^\sharp$ with $[a,x] \not= 
1$. Now for $b \in C_A(Z(Q_2))^\sharp$ we get with \ref{strong} that 
$\langle [a,x], b \rangle$ is quadratic on $V$ and so $V$ is strong 
quadratic, a contradiction.

Let finally $G \cong \,^3D_4(q)$ or $^2F_4(q)$. Then in both cases 
$|A| \leq q^5$. If $V$ is strong quadratic, we get with \cite{Str} that 
$G \cong \,^3D_4(q)$ and $V$ is the natural $\Omega^+(8,q^3)$--module. 
If $A \not\leq Q_1$, then we see that $|V : C_V(A)| \geq q^{12}$, a 
contradiction. So we have $A \leq Q_1$. Now as $|V : C_V(Q_1)| = |Q_1|^2$, 
we get with \ref{closure} $|V : C_V(A)| = |A|^2$. As by \ref{deff} $V$ is 
defined over $GF(q^3)$,
we get that $|A| > q^4$ and so $A$ containes a root element. But then $A$ 
also
containes some $x$ which is conjugate to some element in $G_1 \setminus 
Q_1$.
Then $|V : C_V(x)| > q^{10}$, a contradiction.  

So 
we have that $V$ is not strong quadratic. In particular by \ref{transv} 
$|A| > q$ and then $A \not\leq Q_1$. Let $C_V(Q_1)$ be a trivial 
$G_1$--module and $R$ be a root group normal in $G_1$. Then we see that 
$[V,R]$ involves at least the eight-dimensional $L_2(q^3)$--module for $G 
\cong \, ^3D_4(q)$ or twice the natural $Sz(q)$--module for $G \cong 
\,^2F_4(q)$. Now $|[V,t]| \geq q^8$ for involutions $t \in G_1 \setminus 
Q_1$, which shows $|A| \geq q^4$. As $A \cap Z(Q_1) = 1$, we get $|A| = 
q^4$ and $C_V(A) = C_V(t)$ for  $t \in A \setminus Q_1$. As $|A \cap Q_1| 
\leq q$, we see that $G \cong \, ^3D_4(q)$ and $|A/A \cap Q_1| = q^3$. But
a Sylow 2-subgroup of $L_2(q^3)$ does not at quadratically on the 
eight--dimensional module, a contradiction. 

So we have with \ref{factor} that 
$C_V(Q_1)$ is a $2F$--module for $G_1/Q_1$, and so by \ref{L2qff} we get $G 
\cong \, ^3D_4(q)$ and $C_V(Q_1)$ is the natural $L_2(q^3)$--module, or 
$|C_V(Q_1) : C_{C_V(Q_1)}(A/A \cap Q_1)| = |A/A \cap Q_1|^2$. In the 
latter $A \cap Q_1$ is an offender too and so $A \cap Q_1 = 1$. This again 
shows $G \cong \, ^3D_4(q^3)$.  But then there is exactly one such module 
involved, i.e. $[V,G_1] \leq C_V(Q_1)$ and so $V$ would be strong 
quadratic, a contradiction. So we have the former. This then shows that
$[V,R]$ involves the eight dimensional module too, where $R$ as before is 
the root
group normal in $G_1$. Now for $t \in A \setminus Q_1$ we get $|V : 
C_V(t)| \geq q^{14}$, a contradiction. \qed


\begin{lemma}\label{Vlambdan} Let $G/Z(G) = G(q) \not\cong L_n(q)$  be 
classical 
 of rank at least three and $V = V(\lambda_n)$ be a $2F$--module with 
quadratic offender $A$. Then $(G,V)$ is one of the theorem. \end{lemma}

\proof By \ref{classicalff} we may assume $A \not\leq Q_1$. So by 
\ref{factor} and \ref{o2} it is enough to prove that $V(\lambda_n)$ is 
not a $2F$--module for $G/Z(G) \cong Sp(12,q)$, $\Omega^+(14,q)$, 
$\Omega^-(12,q)$, $U_5(q)$ and $U_8(q)$. The case $G/Z(G) \cong U_5(q)$ 
follows from \ref{rank2}. As the spin modules for $Sp(12,q)$ and 
$\Omega^-(12,q)$ are the same, it is enough to handle $G/Z(G) \cong 
Sp(12,q)$, $\Omega^+(14,q)$ and $U_8(q)$. As $V(\lambda_n)$
is not an $F$--module by \ref{ffmod}, we get that $A \cap Q_1 \not=1$. Let
$G/Z(G) \not\cong U_8(q)$. Then we see that $A/A \cap Q_1 \leq  
O_2(G_{12}/Q_1)$.
Now we see $|A  \cap Q_1| \leq q^3$ and $|C_V(Q_1) : C_{C_V(Q_1)}(A)| \geq 
|A/A \cap Q_1|^2/q^2$. Hence $|V : C_V(A \cap Q_1)| \leq q^8$, a 
contradiction, as $|V : C_V(x)| = q^{16}$, or $q^{32}$ for $x \in 
Q_1^\sharp$. 
 
So let now $G/Z(G) \cong U_8(q)$. Now $V$ involves twice $V(\lambda_3)$ 
and
once $V(\lambda_2)$ for $G_1/Q_1$. This shows that $|V : C_V(A/A \cap Q_1)| 
> |A/A \cap Q_1|^4$, as $V(\lambda_2)$ is not a $2F$--module with 
quadratic offender by \ref{Vlambda2} and $V(\lambda_3)$ is not an 
$F$--module by \ref{ffmod}. So $|A \cap Q_1|^2 > |A|$. On the other hand 
$|C_V(Q_1)| = q^{20}$ and so $|V : C_V(Q_1)| \geq |Q_1|^{3.8}$. Hence by 
\ref{closure} we get $|V : C_V(A\cap Q_1)| \geq |A \cap Q_1|^{3.8}$. So 
we get $|A|^2 \geq |V : C_V(A)| = |V : C_V(A \cap Q_1)||C_V(A \cap Q_1) 
: C_{C_V(A \cap Q_1)}(A)| \geq |A \cap Q_1|^{3.8}|A/A \cap Q_1|$. So $|A| 
\geq |A \cap Q_1|^{2.8} > |A|$, a contradiction. \qed

\begin{lemma}\label{Omega} Let $G \cong \Omega^-(2n,q)$, $n > 3$, and $V$ 
be a an irreducible faithful $2F$--module with quadratic offender $A$. Then 
$V \cong V(\lambda_1)$, or $n = 4,5$ and $V \cong V(\lambda_n)$. \end{lemma}

\proof By \cite{Str} and \ref{Vlambdan} it is enough to prove that $V$ is 
strong quadratic. So assume not. Then by \ref{classicalff} $A \not\leq 
Q_1$. If $A \cap Q_1 \not= 1$, there is some $t \in A \setminus Q_1$ such 
that $[t,Q_1]$ contains some root element and so $V$ is strong 
quadratic by \ref{strong}. Hence we may assume $A \cap Q_1 = 1$.
If there are $s,t \in A^\sharp$ with $C_{Q_1}(s) \not= C_{Q_1}(t)$, then 
there is some $g \in C_{Q_1}(s)$, say, with $t^g = t$ and $s^g = sa \not= 
s$. Now $\langle t,a \rangle$ acts quadratically by \ref{strong} and if 
$a$ is not in a root subgroup there is $h \in Q_1$ with $t^h = tb \not= 
t$, $b$ in a root subgroup. Now $\langle a, b \rangle $ acts 
quadratically by \ref{strong} and so $V$ is strong quadratic. So we may 
assume that $C_{Q_1}(A) = C_{Q_1}(t)$ for all $t \in A^\sharp$. Hence 
$|A| \leq q^{n-2}$. So $|V : C_V(t)| \leq q^{2n-4}$. For a root element 
$a$ we then get $|V : C_V(a)| \leq q^{4n-8}$. But as $V$ is not strong 
quadratic, we get $|[V,a]| > |Q_2/Q_2^\prime| = q^{4n-8}$, a 
contradiction. \qed


\begin{lemma}\label{sp2n} Let $G \cong Sp(2n,q)$, $n \leq 6$, and 
$V \cong V(\lambda_1 + \lambda_n)$. Then $V$ is not a $2F$--module with 
quadratic offender.
\end{lemma}

\proof By \cite{Str} we know that $V$ is not strong quadratic. Let $A$ be 
a quadratic offender. As $|V : C_V(Q_n)| > |Q_n|^2$ , we get with 
\ref{closure} that $A \not\leq Q_n$. But now there 
is some $t \in A \setminus Q_n$, with $[t,Q_n]$ contains a root element. As 
$A$ does not contain root elements, we see with \ref{strong} that $V$ is 
strong quadratic if
$A \cap Q_n \not= 1$. 
So we have $A \cap Q_n = 1$ and again with \ref{strong} $C_{Q_n}(A) = 
C_{Q_n}(a)$ for all $a \in A^\sharp$. This shows $|A| \leq q$ (recall that 
for $n = 3$ we have that $Q_n/U$ is dual to $U$, $U$ the natural 
submodule in $Q_n$, and so the stabilizer of a hyperplane in the natural 
module becomes the stabilizer of a point in the dual).  But as $V$ is not 
strong quadratic 
 this contradicts \ref{transv}. \qed

\begin{proposition}\label{fmodulesc} Let $G/Z(G) = G(q)$ be classical and 
$V$ be an irreducible faithful $2F$--module with quadratic offender $A$. 
Then one of the following holds
\begin{enumerate}
\item[(i)] $G(q) \cong L_n(q)$, $V \cong V(\lambda_1)$, $V(\lambda_2)$, 
$V(\lambda_{n-2})$, $V(\lambda_{n-1})$.
\item[(ii)] $G(q) \cong Sp(2n,q)$, $\Omega^{\pm}(2n,q)$, or $U_n(q)$ and 
$V \cong V(\lambda_1)$.
\item[(iii)] $G(q) \cong L_6(q)$, or $U_6(q)$ and $V \cong 
V(\lambda_3)$.
\item[(iv)] $G(q) \cong Sp(2n,q)$, $n = 3,4,5$, and $V$ is the spin 
module.
\item[(v)] $G(q) \cong \Omega^\pm(2n,q)$, $n = 4,5$, or $\Omega^+(12,q)$ 
and $V$ is the half spin module.
\item[(vi)] $G(q) \cong L_2(q^2)$ and $V \cong V_1^\sigma \otimes V_1$,
where $V_1$ is the natural module and $\sigma$ the field automorphism of 
order two.
\end{enumerate}
\end{proposition}

\proof By \ref{rank2} we may assume that the rank of $G$ is at least three. 
By \ref{Omega} we may assume $G(q) \not\cong \Omega^-(2n,q)$. By 
\ref{classicalff} and symmetry for $G(q) \cong L_n(q)$ we may assume that 
$A \not\leq Q_1$. By \ref{lambdai} we may assume that $C_V(Q_1)$ is a 
$2F$--module with quadratic offender. So by induction we first assume 
that $C_V(Q_1)$ is the natural module. Let $V$ be strong quadratic, we 
now get with \cite{Str} that
$V$ is the natural module in case of $L_n(q)$ or $V \cong V(\lambda_2)$
in which case the assertion follows with \ref{Vlambda2}.
So $V$ is not strong quadratic. In particular $|A \cap Q_1| \leq q^2$.
Let first $G(q) \cong Sp(2n,q)$ or $\Omega^+(2n,q)$. Then we get $A \cap 
Q_1 = 1$, or $G(q) \cong Sp(2n,q)$ and $A/A \cap Q_1$ induces a 
transvection on $Q_1$. So assume the latter. Then $|A| \leq q^3$. Choose 
$a \in (A \cap Q_1)^\sharp$. Then  $\langle (A \cap Q_1)^{C_{G_1}(a)} 
\rangle$ must not
contain a root element by \ref{strong}. So we have $|A \cap Q_1| \leq q$. 
Hence for
$a$ as above, we get $|V : C_V(a)| \leq q^4$. Then there is a root element
$r$ with $[r,G_2^\prime] = 1$ and $|V : C_V(r)| \leq q^8$. But as 
$|Q_2/Z(Q_2)| \geq q^8$ and $V$ is not strongly quadratic, we see that
$|[V,r]| > q^8$. So in both cases we have $|A \cap Q_1| = 1$.

Let now $G(q) \cong U_n(q)$. Then by \ref{strong} $|A \cap Q_1| \leq q$. 
Further $[A,C_{Q_1}(A \cap Q_1)] \leq A \cap Q_1$. This shows that $|A/A 
\cap Q_1| \leq q$ and then $|A| \leq q^2$. Hence for a root element $r$ 
we get $|[V,r]| \leq q^8$.
As $n \geq 6$, we get $[V,r] = C_V(Q_1)$, a contradiction. 

So in any case we 
have $A \cap Q_1 = 1$. Let $a,b \in A^\sharp$ with $C_{Q_1}(a) \not= 
C_{Q_1}(b)$.
Then by \ref{strong} we may assume that there is some $x \in Q_1^\sharp$ 
such that $\langle x,a \rangle$ acts quadratically on $V$. But then $V$ 
is strong quadratic or
$a$ induces a transvection on $Q_1$. Hence we have that $|A| \leq q^2$ and
$|Q_1 : C_{Q_1}(A)| = |A|$. If $C_{Q_1}(a) = C_{Q_1}(A)$ for all $a \in 
A^\sharp$, we also get $|Q_1 : C_{Q_1}(A)| = |A|$. So if $G(q) \cong 
Sp(2n,q)$
or $U_n(q)$, we now see that $C_V(R)$ involves exactly one nontrivial 
irreducible module, $C_V(Q_1)$, where $R$ is the root group normal in $G_1$.
But then we see that $|C_V(R) : C_V(Q_1)| \leq q, q^2$, respectively. In 
both cases we see that $C_V(Q_n \cap Q_1) \leq [V,R] = C_V(Q_1)$ and so 
$G_1 \cap G_n$ acts irreducibly on $C_V(Q_n)$, which is not possible, as 
$Q_n \not= Q_1 \cap Q_n$. 

Let $G(q) = \Omega^+(2n,q)$, then $|A| \leq q^{n-1}$. The same argument 
as before also shows that even $A \cap Q_2 = 1$ and further $C_{Q_2}(a) = 
C_{Q_2}(A)$ for $a \in A^\sharp$. This now shows $|A| \leq q^{n-2}$. So 
for $a \in A^\sharp$ we have $|V : C_V(a)| \leq q^{2n-4}$. This shows 
that for a root
element $r$ we have $|V : C_V(r)| \leq q^{4n-8}$. We may assume $r \in 
Z(Q_2)$. We have $|Q_2| = q^{1+(4n-8)}$. As $V \not\cong V(\lambda_2)$ by
\ref{Vlambda2}, we get that $C_V(Q_2)$ is a nontrivial module on which
$O^2(G_1^\prime \cap G_2)$ acts trivially. Hence $|C_V(Q_2)| \geq 
q^2$ by \ref{o2}. Further by \ref{o2} we also have $C_V(Q_2) \leq 
[V,r]$. Now we see that $[V,r]$ has to
involve at least two natural modules for $\Omega^+(2n-4,q)$. Hence 
$|[V,r]| \geq q^2q^{2(2n-4)} = q^{4n-6}$, a contradiction.

So we are left with $G(q) \cong L_n(q)$. Again $V$ is not strong 
quadratic. So
$A \cap Q_{n-1} = 1$. Hence by induction, we have $C_V(Q_{n-1})$ is the 
trivial, the natural module or then $V(\lambda_{n-3})$--module. This shows 
that we have $V \cong V(\lambda_{n-1})$, $V(\lambda_2)$, or $V(\lambda_1 + 
\lambda_{n-1})$. But the latter is out by \ref{tensor1q} and \ref{transv}.
 
 
So we may assume that $C_V(Q_1)$ is not natural. Let first this module 
correspond to $\lambda_n$. Then $G(q) \cong Sp(2n,q)$, $\Omega^+(2n,q)$ , 
or $U_6(q)$. If $V$ is strong quadratic the assertion follows with 
\ref{Vlambdan}. So we may assume that $V$ is not strong quadratic. As before
we see that $A \cap Q_1 = 1$. If $A \leq Q_n$, we see that $A$ cannot be 
an $F$--module offender on $C_V(Q_1)$. If there is a root group $R$ 
normal in $G_1$, then by \ref{o2} we have $C_V(Q_1) \leq [V,R]$ and so 
also $V/C_V(R)$ involves a module isomorphic to $C_V(Q_1)$. Hence there 
is no root group normal in $G_1$, which shows that $G(q) \cong 
\Omega^+(2n,q)$. Now as above, we get that $|A| \leq q^{n-2}$ and 
$|[V,r]| \leq q^{4n-8}$. As $C_V(Q_2)$ involves the half spin module for 
$\Omega^+(2n-4,q)$ and the parabolic in $G_2$ not in $G_1$ acts 
nontrivially on $C_V(Q_2)$, otherwise $V \cong V(\lambda_n)$, we see that 
two half spin modules are involved. Now $C_V(Q_2) \not\cong Q_2/Z(Q_2)$ 
and so $[V,Z(Q_2)]$ has to involve another nontrivial 
$\Omega^+(2n-4,q)$--module. This shows $|[V,r]| \geq q^{2^{n-2}+2n-4}$. 
Hence we get $n = 4$. But in this case we have symmetry and might assume 
that $A \not\leq Q_n$ too. 

So we have $A \not\leq Q_n$. If $C_V(Q_n)$ is the trivial module, we get 
$V \cong V(\lambda_n)$ with \ref{lambdai} again. So by \ref{Vlambdan} we 
have the assertion.

So let $C_V(Q_n)$ be a $2F$--module too. Then we get 
$V \cong V(\lambda_1 + \lambda_n)$. Now \ref{tensor1q} and \ref{transv} 
yield $G(q) \cong Sp(2n,q)$ or $\Omega^+(2n,q)$. If $G(q) \cong Sp(2n,q)$ 
then by induction we have $n \leq 6$ and we get a contradiction with 
\ref{sp2n}. So let $G(q) \cong \Omega^+(2n,q)$. Now for $G_{n-1}$ we get 
that $C_V(Q_{n-1})$ is $V(\lambda_1 + \lambda_n)$ for $G_{n-1}/Q_{n-1}$. 
As this is not a $2F$--module with quadratic offender,  we get $A 
\leq Q_{n-1}$. Now again we see that $|A| \leq q^{n-2}$ and get the same 
contradiction as before.


So by induction the only case left is $G(q) \cong L_n(q)$ and $C_V(Q_1)$ 
corresponds to $\lambda_{n-2}$ or $\lambda_3$, or $G(q) \cong L_7(q)$ or 
$U_8(q)$ and $C_V(Q_1)$ corresponds to $\lambda_4$. Assume that $G(q) 
\cong L_n(q)$ and $C_V(Q_1)$ corresponds to $\lambda_{n-2}$.  If $V$ is 
strong quadratic we get with \cite{Str} that $V \cong 
V(\lambda_{n-2})$. So let $V$ not be strong quadratic. Then 
$C_V(Q_{n-1})$ is not the natural module. In particular it is not a 
$2F$--module with quadratic offender. Hence by \ref{factor} $A \leq 
Q_{n-1}$ which contradicts \ref{transv} and the fact that $V$ is not 
strong quadratic.

So let $C_V(Q_1)$ correspond to $\lambda_3$, or $n = 7$ and $C_V(Q_1)$ 
corresponds to $\lambda_4$.  Then either $C_V(Q_{n-1})$ is a $2F$--module 
or $V$ is strong quadratic. In both cases we get $V \cong V(\lambda_3)$, 
or $n = 7$ and $V \cong V(\lambda_4)$, which is just the dual of 
$V(\lambda_3)$. Hence we may assume $V = V(\lambda_3)$ in all cases. 
 Further we may assume that $n > 5$, otherwise $\lambda_3 = 
 \lambda_{n-2}$ 
or $\lambda_{n-1}$. 

Let first $n = 7$. Now $|V| = q^{35}$. Again $A 
\not\leq Q_1$ and  $A \not\leq Q_6$. Now for $G_1$ we get a series of 
submodules $1 < V_1 <  V$, where $V_1$ is the $V(\lambda_3)$--module 
and $V/V_1$ is the $V(\lambda_2)$--module for $G_1/Q_1$. By 
\ref{l2offend} we see that $|V/V_1 : C_{V/V_1}(A/A \cap Q_1)| \geq |A/A 
\cap Q_1|$. As $V(\lambda_3)$ is not an $F$--module for $L_6(q)$ by 
\ref{ffmod}, we see that $A \cap Q_1 \not= 1$. So we may assume
that $A \cap Z(Q_{1,6}) \not= 1$. As $|V : C_V(Q_{1,6})| = q^{25} > q^{22} = 
|Q_{1,6}|^2$, we see that $A \cap Q_{1,6}$ must not be an offender. But 
by quadratic action we have $[V,Z(Q_{1,6}),A] = 1$, and so $A \leq Q_{1,6}$,
a contradiction.

Let now $n > 7$, then $C_V(Q_{n-1})$
by induction is not a $2F$--module with quadratic offender. So $A \leq Q_{n-1}$. But $|V : 
C_V(Q_{n-1}) | = q^{{n-1}\choose{2}} > q^{2(n-1)}$, which contradicts 
\ref{closure}.



So we are left with $G(q) \cong U_8(q)$ and $C_V(Q_1)$ is the 
$V(\lambda_3)$--module for $G_1/Q_1$. As this module is involved at least 
twice in $V$ and $V(\lambda_3)$ is not an $F$--module for $U_6(q)$ by 
\ref{ffmod}, we see that $A \cap Q_1 \not= 1$. Let $R$ be the root 
group normal in $G_1$. Then for $r \in R^\sharp$ we get $|V : C_V(r)| 
\leq |A|^8$. As $[V,r] \geq C_V(Q_1)$, we see $|[V,r]| \geq q^{20}$. 
Hence $|A| > q^2$. So we see that either $|A \cap Q_1| > q$ or $A$ 
contains some $a$ with $|Q_1/Z(Q_1) : C_{Q_1/Z(Q_1)}(a)| \geq q^4$. So 
by \ref{strong}  we now see that
$V$ is strong quadratic  and then $V \cong V(\lambda_4)$ by \cite{Str}.
Now we get a contradiction with \ref{Vlambdan}. \qed

\begin{proposition}\label{En} Let $G/Z(G) = E_n(q)$, $n = 6,7,8$, and $V$ 
be an irreducible faithful $2F$--module with quadratic offender $A$. Then 
$n = 6$ and $V \cong V(\lambda_1)$ or $V(\lambda_6)$, or
$n = 7$ and $V \cong V(\lambda_7)$. 
 \end{proposition}

\proof Suppose false. Let first $G/Z(G) = E_6(q)$. If $A \leq Q_4$, then 
$V$ is strong quadratic and by 
\cite{Str} we have $V \cong V(\lambda_1)$ or $V(\lambda_6)$. 
So let $A \not\leq 
Q_4$. So we may assume that  $V$ is not strong quadratic. Hence we 
have with \ref{strong} that $A \cap Q_4 = 1$ and $C_{Q_4}(a) = 
C_{Q_4}(A)$ for all $a \in A^\sharp$. 
 If $C_V(Q_4)$ is nontrivial, then it is a $2F$--module by \ref{closure}. If 
$A \leq Q_1$ or $A \leq Q_6$, then we now get $|A| \leq q^5$. Now elements 
in different root groups in $A$ have different centralizer in $Q_4$, and 
so $|A| \leq q$,  which contradicts \ref{transv}. Hence we have $A 
\not\leq Q_1$ and $A \not\leq Q_6$. By symmetry we may assume that 
$C_V(Q_1)$ is nontrivial. So it is the natural module or the half spin 
module. If we have
the half spin module it corresponds to $\lambda_4$ or $\lambda_2$. So we get 
that $C_V(Q_6)$ is the $V(\lambda_2)$ or $V(\lambda_1 + 
\lambda_4)$--module,
which both are not $2F$--modules with quadratic offender, contradicting 
\ref{factor}. So $C_V(Q_1)$ is the natural module. This then tells us 
that $V \cong V(\lambda_6)$.

So we may assume that $C_V(Q_4)$ is a trivial module. Now $[V,Z(Q_4)]$
involves the 20-dimensional $L_6(q)$--module, and so $V$ involves this 
module twice. But it is not an $F$--module by \ref{ffmod}, contradicting 
$A \cap Q_4 = 1$. 


Let $G/Z(G) \cong E_7(q)$. If $V$ is strong quadratic then by \cite{Str} 
$V \cong V(\lambda_7)$. So we may assume that $V$ is not strong 
quadratic. Now
if $A \leq Q_1$, then $|A| \leq q$, contradicting \ref{transv}.
So we have $A \not\leq Q_1$ and $|A \cap Q_1| \leq q$. As there are no
elements in $\Omega^+(12,q)$ which centralize codimension two in the half 
spin modules, we even get $A \cap Q_1 = 1$. As the half spin module is not 
an  $F$--module for $\Omega^+(12,q)$, we see that this module is not 
involved in $[V, Z(Q_1)]$. So we have $C_V(Q_1)$ is an $F$--module and then
it has to be the natural one. Now as $A$ has to be an $F$--offender on
$[V,Z(Q_1)]$, we see that $A \not\leq Q_7$. As $V$ is not strong 
quadratic we get with \ref{factor} that $V \cong V(\lambda_1 + 
\lambda_7)$.
This shows with \ref{factor} $A \leq Q_4$. As $Z(Q_4)$ is the natural 
module and there is no
quadratic group containing a root subgroup properly, we see that $|A| \leq 
q^7$, otherwise there is $b \in Q_4$ with $[b,A] \not= 1$ and $C_A(b) 
\not= 1$. So there is some root element $r \in [b,A]$ and then by 
\ref{strong} $\langle r, C_A(b) \rangle$ acts quadratically on $V$.  Now 
as $[V,Z(Q_1)]$ has to contain another nontrivial module besides 
$C_V(Q_1)$, we see that $|C_V(Q_1) : C_{C_V(Q_1)}(A)| \leq q^3$.
This now shows $|A| \leq q^3$, but then we have $|V : C_V(A)| \leq q^6$,
which contradicts the fact the $V$ involves at least four nontrivial 
$G_1/Q_1$--modules and $G_1$ cannot induce transvections on some module.
 
Let finally $G(q) \cong E_8(q)$. As there is no strong quadratic module, 
we see that $A \cap Q_8 = 1$. As there are at least two nontrivial 
$E_7(q)$--modules involved in $V$ and $E_7(q)$ does not possess an
$F$--module, we get a contradiction. \qed

\begin{lemma}\label{F4} If $V$ is an irreducible faithful $2F$--module with 
quadratic offender $A$ for
the group $G \cong F_4(q)$, then $V \cong V(\lambda_1)$ or $V(\lambda_4)$.
\end{lemma}

\proof By \cite{Str} and \ref{Vlambda2} we may assume that $V$ is not 
strong quadratic. Let $A \leq Q_1$, the we get $A \leq Z(Q_1)$. Now if $A 
\leq Q_4$ too, then we get $A \leq Z(Q_1) \cap Z(Q_4)$, so $|A| \leq 
q^2$. As $A$ cannot contain any root element, we even see $|A| \leq 
q$, which contradicts \ref{transv}. So we may assume $A \not\leq 
Q_1$. Let $A \cap Z(Q_1) \not = 1$. Then there is $x \in Q_1$ with $[x,A] 
\not\leq Z(Q_1)$. So by \ref{strong} there is
a quadratic group $B$ in $Q_1$ with $B \not\leq Z(Q_1)$ but $B \cap Z(Q_1) 
\not =1$. Hence again by \ref{strong} we get a subgroup acting strongly 
quadratic. So we have $A \cap Z(Q_1) = 1$. This first with 
\ref{strong} shows $|A \cap Q_1| \leq q$. If $A \cap Q_1 \not= 1$, then 
we choose  $x \in Z(Q_1)$ and proceed as before. So we have $A \cap Q_1 
= 1$. Now further we see that $C_{Z(Q_1)}(a) = C_{Z(Q_1)}(A)$
for all $a \in A^\sharp$. Hence $|A| \leq q^3$. As $|A| > q$ by 
\ref{transv}, we see that $A \not\leq Q_4$. But now we also get $A \cap Q_4 
= 1$. So we get that both
$[V,Q_i^\prime]$, $i = 1,4$, are $F$--modules. This shows that we may 
assume that $C_V(Q_1)$ is the natural module and $|A| = q^3$. Now 
$[V,Q_1^\prime]/C_V(Q_1)$ is a trivial module and so 
$C_{Q_1}([V,Q_1^\prime])$ covers $Q_1/Z(Q_1)$ which shows that $V$ is 
strong quadratic. \qed

\begin{lemma}\label{twistff} The group $G/Z(G) \cong \,^2E_6(q)$ 
does not possess a $2F$--module with quadratic offender. 
\end{lemma}

\proof Suppose false. Let $A$ be a quadratic offender. Assume 
first that $V$ is not strong quadratic. Let $A \leq Q_1$. By 
\ref{transv}, we have $|A| > q$. Now we see with \ref{strong} that $V$ is 
strong quadratic, a contradiction. So we have $A \not\leq Q_1$. Further 
$|A \cap Q_1| \leq q$. Let $|A/A \cap Q_1| \leq q$. Then $A \cap Q_1 
\not= 1$. Let $1 \not= a \in A \cap Q_1$, then $|V : C_V(a)| \leq q^4$, 
so we get $|V : C_V(Z(Q_1))| \leq q^8$, and so $V$ is strong quadratic, a 
contradiction. So we have $|A/A \cap Q_1| > q$. As $|[Q_1/Z(Q_1),a]| > 
q^2$ for all $a \in A \setminus A \cap Q_1$, we get that $A \cap Q_1 = 1$.

Now we get that there are at least two nontrivial 
$G_1/Q_1$--modules involved, which are isomorphic to $Q_1/Z(Q_1)$, or 
$C_V(Q_1)$ is a nontrivial module not isomorphic to $Q_1/Z(Q_1)$. Then by 
\ref{factor} and \ref{o2} we get that $C_V(Q_1)$ is the natural 
$U_6(q)$--module. As $V$ is not strong quadratic we  have that 
$[V,Z(Q_1)]$ involves a further
nontrivial module. As $Q_1/Z(Q_1)$ is 
not an $F$--module for $U_6(q)$, we get a 
contradiction in the former. In the latter by \ref{ffmod} we may assume 
that $V$ involves four natural modules. So we get for the natural 
module $V_1$ that $|V_1 : C_{V_1}(A)| \leq q^4$. But then we see that 
$|A| \leq q^4$ and so $|V : C_V(A)| \leq q^8$. Now even $|V_1 : 
C_{V_1}(A)| \leq q^2$ and so $|A| \leq q$. But then $|V : C_V(A)| \leq 
q^2$, a contradiction.



So we have that $V$ is strong quadratic and then by \cite{Str} we have $V \cong V(\lambda_6)$. In particular  $|V| = q^{54}$ and by 
\ref{deff} $V$ is defined over $GF(q^2)$. As a $G_1$--module we have the 
following series $1 < V_1 < V_2 < V$, where $V_1$ and $V/V_2$ are natural 
modules and $V_2/V_1$ is the $V(\lambda_2)$--module for $U_6(q)$. 

If $A \leq Q_1$, then we will just consider offender  $B 
\in {\cal{P}}_2(Q_1,V)$ with $|C_V(B)||B|$ maximal. If $Z(Q_1) \cap B = 
1$. Then choose $x \in Q_1 \setminus C_{Q_1}(B)$. By \ref{closure} 
$\langle B, B^x \rangle$ is also an offender. So we may assume that 
$Z(Q_1) \cap B \not=1$. As $|B| \leq q^{11}$, we get $|C_V(Z(Q_1)) : 
C_{C_V(Z(Q_1))}(B/B \cap Z(Q_1))| \leq |B/B \cap Z(Q_1)|$. So by 
\ref{closure} we get $q^{30} = |C_V(Z(Q_1)) : C_V(Q_1)| 
\leq |Q_1/Z(Q_1)| = q^{20}$, a contradiction. 

So we have $A \not\leq Q_1$. In particular by quadratic action $A \cap 
Z(Q_1) = 1$, and so $|A| \leq q^{12}$. Then  $|V : C_V(A)| \leq q^{24}$. 
Suppose first that $|C_V(Q_1) : C_{C_V(Q_1)}(A)| \geq q^6$. As by 
\ref{Vlambda2} $V_2/V_1$ is not a $2F$--module with quadratic 
offender, we get that
$|A|^2 > q^{12}|A/A \cap Q_1|^2$. This shows $|A \cap Q_1| > q^6$ and further
$|V : C_V(A \cap Q_1)| \leq q^{18}$. If $|C_V(Q_1) : C_{C_V(Q_1)}(A)| = 
q^4$, we get $|A/A \cap Q_1| \leq q^4$. Hence $|A \cap Q_1| > q^4$. Let 
finally $|C_V(Q_1) : C_{C_V(Q_1)}(A)| = q^2$, then $|A/A \cap Q_1| \leq q$.
Then we see that for $a \in A \setminus A \cap Q_1$ we have $|V : C_V(a)| 
\geq q^{12}$, as in the $V(\lambda_2)$--module for $U_6(q)$ we get  an 
index $q^8$. Hence again $|A \cap Q_1| > q^4$. So in both cases there is 
some $x$
with $|A \cap Q_1| = q^{4+x}$, and $|V : C_V(A \cap 
Q_1)| \leq q^{12+2x}$, $q^{8+2x}$, respectively. Now set $A_1 = A \cap Q_1$.
Then we have $|V : C_V(A_1)| \leq q^{a}$ and $|A_1| \geq q^{b}$, where $b = 
6,4+x,4+x$ and $a = 18, 12 + 2x, 8 + 2x$. Now choose $y \in Q_1$ with 
$[A_1,y] \not= 1$. Set $A_2 = A_1 \cap A_1^y$ and $B_1 = \langle A_1, A_1^y 
\rangle$. Now we get
$$|V : C_V(B_1)| \leq q^{2a}/|V : C_V(A_2)|$$
By \ref{closure} we have that $|A_1|^{1.5} \leq |C_V(Z(Q_1)) : 
C_{C_V(Z(Q_1))}(A_1)| \leq |V : C_V(B_1)|/q^{12}$. This shows
$$q^{2a-(1.5b + 12)} \geq |V : C_V(A_2)| = q^t$$
Here we get $t \leq 15, 8 + 2.5x, 2.5x - 2$. Now we play the same game with 
$A_2$. Choose $z \in Q_1$ with $[A_2,z] \not=1$. Set $B_2 = \langle A_2, 
A_2^z \rangle$ and $A_3 = A_2 \cap A_2^z$. Then we get
$$|V : C_V(B_2)| \leq q^{2t}/|V : C_V(A_3)|$$
and
$$ |A_2|^{1.5} \leq |C_V(Z(Q_1)) : 
C_{C_V(Z(Q_1))}(A_2)| \leq |V : C_V(B_2)|/q^{12} $$
As $|A_1 : A_2| \leq q$, we have $|A_2| \geq q^{b-1}$. Now
$$q^{2t - (1.5b + 10.5)} \geq |V : C_V(A_3)| = q^s$$
where $s = 10.5, 3.5x - 0.5, 3.5x - 20.5$. As $A_3 \not= 1$, we see that $s 
\geq 12$. So we have the second case with $x > 3$. Now we may apply the 
procedure once more and get some $A_4 \not= 1$ with $|V : C_V(A_4)| = 
q^r$, where $r \leq 2s -(1.5b + 9) = 5.5x - 16$. Once more we get $A_5$ 
with
$|V : C_V(A_5)| \leq q^{2r -(1.5b + 7.5)} \leq q^{9.5x - 45.5}$. As $|A_5| 
\geq q^4$, we can proceed twice and get $A_7$ with $|V : C_V(A_7)| \leq 
q^{33.5x -217.5}$. But then $x > 6$, a contradiction, as $q^{10} \geq |A 
\cap Q_1| = q^{4+x}$. \qed
\\

So we have shown

\begin{proposition}\label{ffmodules1} Let $G/Z(G) = G(q)$ be of Lie type, 
$V$ be an irreducible faithful $2F$--module with quadratic offending group. 
Then one of the following holds
\begin{enumerate}
\item[(i)] $G(q) \cong L_n(q)$, $V \cong V(\lambda_1)$, $V(\lambda_2)$, 
$V(\lambda_{n-2})$, $V(\lambda_{n-1})$.
\item[(ii)] $G(q) \cong Sp(2n,q)$, $\Omega^{\pm}(2n,q)$ or $U_n(q)$ and 
$V \cong V(\lambda_1)$.
\item[(iii)] $G(q) \cong L_6(q)$ or $U_6(q)$ and $V \cong 
V(\lambda_3)$.
\item[(iv)] $G(q) \cong Sp(2n,q)$, $n = 3,4,5$, and $V$ is the spin 
module.
\item[(v)] $G(q) \cong \Omega^\pm(2n,q)$, $n = 4,5$, or $\Omega^+(12,q)$ 
and $V$ is the half spin module.
\item[(vi)] $G(q) \cong L_2(q^2)$ and $V \cong V_1^\sigma \otimes V_1$,
where $V_1$ is the natural module and $\sigma$ the field automorphism of 
order two.
\item[(vii)] $G(q) \cong E_6(q)$ and $V \cong V(\lambda_1)$ or 
$V(\lambda_6)$.
\item[(viii)] $G(q) \cong E_7(q)$ and $V \cong V(\lambda_7)$.
\item[(ix)] $G(q) \cong F_4(q)$ and $V \cong V(\lambda_1)$ or 
$V(\lambda_4)$.
\end{enumerate}
\end{proposition}

  


\section{$2F$-modules for alternating, sporadic and Lie type groups in odd 
characteristic}

Throughout this chapter we will assume that $G$ is a perfect central 
extension of an alternating group, a sporadic group or a group of Lie type 
over a field of odd characteristic, which is not a group of Lie type over a 
field of characteristic 2 too. 
 Further $V$ is a $2F$-module and $A \in {\cal{P}}_2(G,V)$ an offending 
 subgroup which acts quadratically. 

\begin{lemma}\label{alternating} Let $G = A_n$ or $G = 3A_m$, $m = 6,7$ , 
and $V$ be an irreducible faithful $2F$--module with quadratic offender 
$A$,  then either $V$ is the permutation module or
\begin{enumerate}
\item[(i)] $G \cong A_8 \cong L_4(2)$ and $V$ is the natural $L_4(2)$-module.
\item[(ii)] $G \cong A_7$ and $V$ is as in (i).
\item[(iii)] $G \cong 3A_6$ and $|V| = 2^6$.
\item[(iv)] $G \cong A_5 \cong L_2(4)$ and $V$ is the natural 
$L_2(4)$-module.
\item[(v)] $G \cong A_9$ and $V$ is the eight dimensional spin module.
\end{enumerate}
\end{lemma}

\proof Let first $n \geq 9$. Suppose $|A| = 2$. Then $|V : C_V(A)| 
\leq 4$. Now  we see that there is some involution $b \not= a$ 
with $[[V,a],b] = 1$. So we have a quadratic fours group. The same applies 
for $|A| \geq 4$. 
So we assume that we have a quadratic fours group. By 
\cite{MeiStr} we have that $V$ is either the natural module or the spin 
module. So assume $V$ to be the spin module. Let $a\in A^{\sharp}$, where we 
set $A = \langle a, b \rangle$ in the first case. Let $k$ be the number of
fixed points of $a$. Then there is 
$K\leq C_{G}(a), K\cong \Sigma_{k}$. Furthermore 
$C_{C_G(a)}(K^\prime)$ is an extension 
of a 2--group of order $2^m$ by $A_{m}$, $m = (n-k)/2$. Now choose 
$a \in A$ with $m > 2$ if possible. Suppose first 
$[A, C_{C_G(a)}(K^\prime)] \not= 1$. 
If $m \ge 5$, then $\Sigma_m$ is nonsolvable and so 
$C_{C_G(a)}([V,a])$ contains an elementary abelian subgroup of $O_2(C_G(a))$ 
of order $2^{m-1}$. But then this group contains a conjugate of (12)(34) 
which contradicts \cite[(4.3)]{MeiStr}.\\
\indent
Let $m = 4$. Then $a \sim (12)(34)(56)(78)$. Furthermore as we may 
assume that there is no $x \sim (12)(34)$ which is contained in 
$\langle A^{C_G(a)} \rangle$ we see that $A$ is conjugate to a 
subgroup of 
$\langle (12)(34)(56)(78), (13)(24)(57)(68), (15)(26)(37)(48) \rangle$. 
We will  handle this case later. \\
\indent
So let $[A, O^{2^\prime}(C_{C_G(a)}(K^\prime))] = 1$. 
If $[A, K^\prime] \not= 1$, then $[K^\prime, [V,a]] = 1$. 
If $k \ge 4$, then $K^\prime$ contains some $
x \sim (12)(34)$. This again contradicts \cite[(4.3)]{MeiStr}. 
Let $k \le 3$. As $[A, O^{2^\prime}(C_{C_G(a)}(K^\prime))] = 1$ 
and $m > 2$ this is impossible. So we are 
left with $[A, K^\prime] = 1 = [A, C_{C_G(a)}(K^\prime)]$. But this is 
impossible with $m > 2$.\\
\indent
So we have $m = 2$ for all $a \in A^\sharp$. We may 
assume $a = (12)(34) \in A$. Now $A \ge \langle a, b \rangle$, $b = 
(13)(24)$, or $(12)(56)$. Let $[b,K^\prime] \not= 1$. Then $b = 
(12)(56)$ and so $K^\prime$ contains no involutions by \cite[(4.3)]{MeiStr}. 
This shows $k \le 3$ and so $A \le A_7$, a contradiction.\\
\indent
So we have seen that $|A| \leq 8$ and so $|V : C_V(a)| \leq 2^6$ for all 
$a \in A^\sharp$. Furthermore as $n > 8$, we see that there is some $a \in 
A^\sharp$ which inverts $(1,2,3) = \rho$. But $\rho$ is fixed point freely 
on $V$. This shows $|V| \leq 2^{12}$. Let $n > 9$, then $U = \langle 
\omega_1, \omega_2 \rangle$, $\omega_1 = (1,2,3,4,5)$, $\omega_2 = 
(6,7,8,9,10)$ act on $V$. As $\langle \omega_1 \rangle$, $\langle \omega_2 
\rangle$ act fixed point freely and all other subgroups of order five in $U$ 
are conjugate in $G$, we get $|V| \geq 2^{16}$, a contradiction. So we 
have $n = 9$. 

The cases of $n = 5,6,8$ were done before, as $G$ is a group of Lie type in 
characteristic two. So let $n = 7$. If $|A| = 4$, then by \cite{MeiStr} 
$|V| = 16$, which is (ii). So assume $|A| = 2$ and there is no quadratic 
fours group. Now we may embedd $A$ into a subgroup $U \cong \Sigma_5$. 
Further $V$ involves exactly one nontrivial $U$--module. So there is a 
submodule $V_1$
which is the permutation module or an extension of a trivial module by 
the $L_2(4)$--module, with $V_1 \geq [V,A]$. But now we see that there
is a fours group centralizing $[V,A]$.\\ 
\indent
Let now $G \cong 3A_6$. As $G$ is generated by four involutions or three
fours groups, we get $|V| \leq 2^{12}$. Now the assertion follows with
\cite[page 4]{Brauer}. 
\\
\indent
Let $G \cong 3A_7$. Then we have a subgroup $U \cong 3A_6$
and so again $|V| \leq 2^{12}$. This time we get with 
\cite[page 13]{Brauer} $|V| = 2^{12}$. Further we get $|A| = 4$. And so 
$|V : C_V(A)| = 16$. This shows that the two $3A_6$--modules involved in 
$V$ have to be isomorphic. Let $V_1$ be a $3A_6$--submodule. Then  
$|C_V(A)| = 16$. There is another conjugay class of fours groups in 
$A_7$. If $C$ is in this class then we get $|C_{V_1}(C)| = 4$. Let $B$ be 
the fours group in $A_7$ which centralizes an element of order three in 
$A_7$.
Then we have that $U \cong 3^{1+2}$ acts on $C_V(B)$. This shows that 
$|C_V(B)| = 2^6$. In particular
$|C_{V_1}(B)| = 16$.   So $ B \sim A$, a contradiction.
\qed

\begin{lemma}\label{sporadic} Let $G/Z(G)$ be sporadic and $V$ be an 
irreducible faithful  $2F$--module with quadratic offender $A$. Then $G 
\cong 3M_{22}$ and $V$ is the 12-dimensional module comming from the 
embedding into $SU_6(2)$.
\end{lemma}

\proof Let  first $|A| = 2$. As any sporadic group but $M_{11}$ contains an 
elementary  abelian subgroup of order eight, we get a quadratic fours group. 
So we may assume that $G \cong M_{11}$ or there is a quadratic fours group. 
So assume the latter. Then by \cite{MeiStr} we just 
have to handle the cases $E(G) \cong M_{12}$, $M_{24}$, $3Suz$, 
$3M_{22}$, $J_2$, $Co_1$, $Co_2$. Furthermore $V$ is known. In all cases 
besides $G \cong 3M_{22}$ we have $|A| \leq 4$.

Let $G \cong M_{12}$. There is a subgroup $U$ in $G$ , $U \cong M_{10}$. 
This shows $|[V,a]| \geq 16$ for all $a \in A^\sharp$. So we have $|A| = 
4$ and $C_V(a) = C_V(A)$ for all $a \in A^\sharp$. But the centralizer of 
$a$ is a maximal subgroup of $G$, a contradiction.
\\
\indent
Let $G \cong M_{24}$. If $t$ is a non 2-central involution, then it
inverts an element of order 11 in $G$, so $|[V,t]| \geq 32$. Hence $A$ just 
consists of 2--central involutions and so we have that $|A| = 4$ and $C_V(a) 
= C_V(A)$ for all $a \in A^\sharp$. Now we get that $U \cong E_{16}A_8$ 
acts on $C_V(A)$. Hence $[C_V(A), O_2(U)] = 1$ and so $V$ would be an 
$F$--module, contradicting \ref{ffmod}.
\\
\indent
Let $G \cong 3Suz$. If $A$ contains a 2-central involution $a$, we see 
that $[[a,V], C_G(a)^\prime] = 1$. Now any fours group containing $a$ is 
quadratic, which contradicts \cite{MeiStr}. So we have $A$ just consists of
non 2--central involutions. But then $U \cong SL_3(4)$ would centralize 
$[V,a]$, which  contradicts the fact that an element of order three in 
$Z(U)$ acts fixed point freely on $V$.\\
\indent
Let $G \cong Co_1$ or $Co_2$.  As $3Suz$ embedds into $Co_1$ and the 
modules are the same and up to conjugation the quadratic subgroups are 
the same, we get a contradiction in that case as before. So assume that 
$G \cong Co_2$. Then $A$ consists of 2--central involutions and
for $a \in A^\sharp$, we have $[[V,a],C_G(a)] = 1$. But now any fours group
containig $a$ would be quadratic, contradicting \cite{MeiStr}.\\
\indent
Let $G \cong J_2$. Then $V$ is the $G_2(4)$--module. Hence for the 2-central
involutions $z$ we have $|[V,z]| = 16$, while for the non 2--central ones $r$
we have $|[V,r]| = 64$. So $|A| = 16$ and $C_V(A) = C_V(a)$. But $C_G(a)$ 
is a maximal subgroup in $G$, a contradiction.\\
\indent
Let finally $G \cong 3M_{22}$. Then the assertion follows with 
\cite{MeiStr}.
\\
\indent
So we are left with $G \cong M_{11}$ and $|A| = 2$. Let $A \leq T$, $T$
a Sylow 2--subgroup of $G$. Then $T$ acts transitively on an elementary 
abelian subgroup $U$ of order 9. This shows $|[V,U]| \geq 2^8$. As $A$ 
inverts $U$, we get $|[V,A]| \geq 2^4$, a contradiction. \qed

\begin{lemma}\label{oddchar} Let $V$ be an irreducible faithful 
$2F$--module with quadratic offender $A$. If $G$ is some covering group of 
a group of Lie type in odd characteristic then it is a group of Lie type in 
even characteristic too,
or $G \cong 3\cdot U_4(3)$ and $V$ is the 12--dimensional module, comming 
from the embedding into $SU_6(2)$.
\end{lemma}

\proof Suppose false. Assume first that we have a quadratic fours group. So 
by \cite{MeiStr1} we get $G \cong 3\cdot U_4(3)$ and $|V| = 2^{12}$. 
\\
\indent
So we may assume $|A| = 2$ and there are no quadratic fours groups. 
In particular for $A \leq T$, $T$ a Sylow 2--subgroup, we have that 
$C_T(A)$ has a cyclic subgroup $U$ of index at most two. Now $C_T(U) = 
U$, and so by \cite[5.4.8]{Go} we have that $T$ is dihedral or 
semidihedral. Hence we have that $G/Z(G) \cong L_2(q)$, $L_3(q)$, or 
$U_3(q)$, $q$  odd. \\
\indent
Let $G \cong U_3(q)$ or $L_3(q)$. Then $A$ acts on a special group $U$ of 
order
$q^3$, centralizing the center and inverting $U/Z(U)$. As $[[V,Z(U)],A] 
\not=1$, we see that $q = 3$. So we have $G \cong L_3(3)$, as $U_3(3) \cong 
G_2(2)^\prime$. But now there is also an involution $t \in G$ which
inverts $Z(U)$. As all involutions are conjugate, we get that $|[V,Z(U)]| = 
16$ and so $U \leq GL(4,2)$, a contradiction.
\\
\indent
So it remains $G \cong L_2(q)$. Then $G$ is generated by three conjugates
of $A$. Hence we have $|V| \leq 2^6$, and so $G \leq GL(6,2)$. Now it is 
easy to see that $G \cong L_2(5)$, $L_2(7)$, or $L_2(9)$, which all are 
also Lie groups in characteristic two. \qed 
 



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Fachbereich Mathematik und Informatik\\
Institut f\"ur Algebra und Geometrie\\
Martin-Luther-Universit\"at Halle - Wittenberg\\
06099 Halle\\
Germany\\
stroth@coxeter.mathematik.uni-halle.de




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