
\documentstyle[12pt]{article}
\begin{document}
\noindent
\begin{center}
{\Large  The semiclassical parabolic systems}\\ \vskip1cm
{\bf A. Fukshansky, G. Stroth, C. Wiedorn\\
Martin-Luther-Universit\"at Halle-Wittenberg\\
Fachbereich Mathematik und Informatik\\
Institut f\"ur Algebra und Geometrie\\
06099 Halle, Germany}
\end{center}
\vskip1cm 

The classification of parabolic systems was launched for generalizing the
notion of a building related to groups of Lie type and intended for the
application to structure and classification of the finite simple groups.
One of the first major results in this area was the classification of the 
so called
parabolic systems of classical type due to S. Heiss [Hei1], Th. Meixner
[Mei1], R. Niles [Ni], G. Stroth [Str1, Str2, Str3, Str4, Str5] and
F. Timmesfeld [Tim1, Tim2, Tim3]. See [Mei2] for a survey.\\

For the convenience of the reader we give the definition of a classical
parabolic system. (Recall that for a group $G$ and a subgroup $X$ the
subgroup $\bigcap\limits_{g\in G}X^g$ is denoted ty $X_G$.)\\

Let $G$ be a group generated by finite subgroups $P_1,\ldots, P_n$ such
that
\begin{itemize}
\item [(1)] $P_i\not\in \langle P_j \, | \, j\in \{1,\ldots, n\} \setminus \{i\}\rangle,
\; i=1,\ldots, n$.
\item [(2)] If $B=\bigcap\limits_{i=1}^{n}P_i$, then $P_i/B_{P_i}$ is a
rank one group of Lie-type in characteristic $p$ for some prime $p$, 
$B /B_{P_i}$ is the Borel subgroup in  $P_i/B_{P_i}$, and
$B=P_i\cap P_j$ for any pair $P_i, P_j$. Further $B_G=1$.
\item [(3)] For each pair $i, j\in \{1,\ldots, n\}$ let $P_{i,j}=\langle P_i,
P_j\rangle$ be one of the following:
\begin{itemize}
\item [($\alpha$)] $P_{i,j}=P_iP_j$
\item[($\beta$)] $P_{i,j}/B_{P_{i,j}}$ is a rank two Lie type
group in characteristic $p$.
\end{itemize}
\end{itemize}

In fact all groups of Lie type have a classical parabolic system. It turns
out that among the sporadic simple groups we find $McL$, $Ly S$ and $Suz$
with such a parabolic system. A systematical search [RoStr] lead to a wider
class of parabolic systems including more sporadic groups. This is the
class of semiclassical systems which is the subject of this paper. For a
semiclassical system we just have to fix $p=2$ in the definition above and
add in (3) the further possibility
\begin{itemize}
\begin{itemize}
\item [($\gamma$)] $P_{i,j}/B_{P_{i,j}}\cong 3\Sigma_6$ or
$3A_6$.
\end{itemize}
\end{itemize}
Furthermore we assume that there is some pair $i,j$ realizing ($\gamma$).

The sporadic groups showing up in this context are $M_{24}, He, Co_1,
M(24)'$ and $F_1$.

Let now $G=\langle P_1,\ldots, P_n\rangle$ be a semiclassical parabolic
system. We call $n$ the rank of the system. To each system of rank $n$
we associate a diagram $\Delta$ in the following way. The nodes of the
diagram are $\{1,\ldots, n\}$ and $i,j$ is connected by the Dynkin diagram
corresponding to $P_{i,j}/B_{P_{i,j}}$, if this is a rank two Lie type group, by
\setlength{\unitlength}{1mm}
{$
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$}
 if $P_{i,j}$ is as in (3)$(\gamma)$ and not connected otherwise.

If we will allow 
{$
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$}
 and 
{$
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$} at the same place we also write 
{$
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$.} We will denote any possible connection between two nodes by
{$
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\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(2.5,0.8){........}
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$.}
Since non connected diagrams correspond to a product of groups we may assume
\begin{itemize}
\item [(4)] $\Delta$ is connected.
\end{itemize}

According to [Tim3] the semiclassical diagrams of rank three correspond
to diagrams 
{$
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$}, or 
{$
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\put(22,1){\makebox(0,0){$\circ$}}
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\put(12.5,1.7){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\end{picture}
$}. For the latter according to [Hei2] and
[Row] we have $G\cong He, M_{24}$ or $3^7Sp_6(2)$. The first step in a 
classification of semiclassical systems in general is the classification of
systems of rank four. Of course then one of the above diagrams is involved
in $\Delta$. The systems of rank four in which any subdiagram of shape
{$
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$}
 belongs to $3^7Sp_6(2)$ have been classified by G. Gillespie [Gil] and
C. Wiedorn [Wie1]. If some subdiagram of type 
{$
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\put(2,1){\makebox(0,0){$\circ$}}
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\put(12.5,1.7){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\end{picture}
$} belongs to
$M_{24}$ or $He$ the rank four systems are classified by [FuStr], provided
$P_i/B_{P_i}\cong \Sigma_3$ for all $1 \leq i \leq 4$. In (1.2) we will see that this always holds.
The rank four diagrams involving 
{$
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$} are handled in [Wie1].\\

The purpose of this paper is to finish the classification of semiclassical
parabolic systems, i.e. to give a classification of rank at least five.\\

To state the main result we need some notation and definitions. 
Let $G=\langle
P_i\mid i\in I\rangle$. For $J\subseteq I$ set $G_J=\langle P_i\mid i\in I
\backslash J
\rangle$, hence $G=G_\phi$. Set $K_J=B_{G_J}$.\\

For $\Delta$ and $J\subseteq I$ let $\Delta_J$ be the subdiagram obtained
from $\Delta$ by dropping all nodes in $J$ and all edges having one vertex in
$J$. Then $G_J/K_J$ is a parabolic system with diagram $\Delta_
J$.\\

Now we will give a list of some parabolic systems.
\begin{itemize}
\item[(i)] 
$\Delta$ is a complete bipartite graph all of whose edges are of type
{$
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$}. 
There exists $J\subseteq I$ such that $\Delta_J=$ 
{$
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$},
$K_J=1$ and $G_J$ is one of the groups in theorem (1.1)(a)(iv)($\star$)($\star \star$).
\item[(ii)]
The systems $He(n,m), n, m\ge 1:$
\\ \\
\begin{picture}(53,12)
\put(0,0.5){The diagram $\Delta$ is of type}
\end{picture} 
{$
\begin{picture}(35,12)
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
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\put(5,-6){\makebox(0,0){$\circ$}}
\put(5,8){\makebox(0,0){$\circ$}}

\put(12.5,1){\line(1,0){8.7}}

 \put(5.7,8.2){\line(1,-1){6.5}}
 \put(5,7.2){\line(1,-1){6.3}}

 \put(5,-5.1){\line(1,1){6.3}}
 \put(5.7,-6.2){\line(1,1){6.3}}


 \put(21.8,1.8){\line(1,1){6.4}}
 \put(22.6,1){\line(1,1){6.3}}

 \put(21.8,0.2){\line(1,-1){6.4}}
 \put(22.6,1){\line(1,-1){6.3}}

\put(23.5,5.5){\makebox(0,0){ $\scriptstyle{\sim}$}}
\put(23.5,-3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(10.2,5.5){\makebox(0,0){ $\scriptstyle{\sim}$}}
\put(10.2,-3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(11.6,-2.5){\scriptsize 1}
\put(20,-2.5){\scriptsize 2}
\put(31,-6.5){$b_m$}
\put(31,8){$b_1$}
\put(-0.5,-6.5){$a_n$}
\put(-0.5,8){$a_1$}

\put(4,0.5){.}
\put(4,3.5){.}
\put(4,-2.5){.}
\put(29,0.5){.}
\put(29,-2.5){.}
\put(29,3.5){.}
\put(35,0.5){.}
\end{picture}
$}\\ [1cm]
Let $J\subseteq I$ such that $\Delta_J$ is of type 
{$
\begin{picture}(25,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(12.5,1.7){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\end{picture}
$}, then 
$K_J=1$ and $G_J\cong He$.
\item[(iii)]
The systems $X(2,m), m\ge 0:$\\ \\
{$
\begin{picture}(55,12)
\put(0,1){The diagram $\Delta$ is of type}
\end{picture}
$}
{$
\begin{picture}(22,12)
\put(12,1){\makebox(0,0){$\circ$}}

\put(19,-6){\makebox(0,0){$\circ$}}
\put(19,8){\makebox(0,0){$\circ$}}
\put(5,-6){\makebox(0,0){$\circ$}}
\put(5,8){\makebox(0,0){$\circ$}}


 \put(5.4,7.6){\line(1,-1){6.0}}
 

 \put(5.4,-5.5){\line(1,1){6.0}}
 


 \put(11.8,1.9){\line(1,1){6.3}}
 \put(12.6,1){\line(1,1){6.3}}

 \put(11.8,0.2){\line(1,-1){6.3}}
 \put(12.6,1){\line(1,-1){6.3}}

\put(13.5,5.5){\makebox(0,0){ $\scriptstyle{\sim}$}}
\put(13,-3.5){\makebox(0,0){\tiny( $\! \! \scriptstyle{\sim} $\tiny)}}



\put(8,-0.2){\scriptsize 3}

\put(21,-6.5){\scriptsize 4}
\put(21,8){\scriptsize m+4}
\put(0,-6.5){\scriptsize 2}
\put(0,8){\scriptsize 1}


\put(19,0.5){.}
\put(19,-2.5){.}
\put(19,3.5){.}
\end{picture}
$}
{$
\begin{picture}(10,12)
\put(9,1){.}
\end{picture}
$}\\[1cm]
Let $J\subseteq I$ such that $\Delta_J$ is of type 
{$
\begin{picture}(25,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(12.5,1.7){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){\tiny( $\! \! \scriptstyle{\sim} $\tiny)}}
\end{picture}
$}, then $K_J=1$ and
$G_J\cong 3^7Sp_6(2)$ or $Sp_6(2)$. Set $J'=\{4,\ldots, m+4\}$, then $G_{J'}\cong 2^6
A_7$.
\item[(iv)]
The systems $Y(n,m), n,m\ge 1:$\\ \\
{$
\begin{picture}(60,12)
\put(0,1){The diagram $\Delta$ is of type}
\end{picture}
$}
{$
\begin{picture}(40,12)
\put(1,1){\makebox(0,0){$\circ$}}
\put(1.5,1){\line(1,0){8.5}}
\put(11,1){\makebox(0,0){$\circ$}}

\put(13,1.2){...}
\put(19,1){\makebox(0,0){$\circ$}}
\put(19.5,1){\line(1,0){8.5}}
\put(29,1){\makebox(0,0){$\circ$}}

\put(36,-6){\makebox(0,0){$\circ$}}
\put(36,8){\makebox(0,0){$\circ$}}




 \put(28.8,1.9){\line(1,1){6.3}}
 \put(29.6,1){\line(1,1){6.3}}

 \put(28.8,0.2){\line(1,-1){6.3}}
 \put(29.6,1){\line(1,-1){6.3}}

\put(30,5.5){\makebox(0,0){ $\scriptstyle{\sim}$}}
\put(30,-3.5){\makebox(0,0){\tiny( $\! \! \scriptstyle{\sim} $\tiny)}}

\put(24,-2.2){\scriptsize n+1}
\put(18,-2.2){\scriptsize n}

\put(31,-8.5){\scriptsize n+2}
\put(31,10){\scriptsize n+m+1}
\put(10,-2.2){\scriptsize 2}
\put(0,-2.2){\scriptsize 1}


\put(36,0.5){.}
\put(36,-2.5){.}
\put(36,3.5){.}
\end{picture}
$}
{$
\begin{picture}(3,12)
\put(0,1){.}
\end{picture}
$}
\\[0.7cm]
{$
\begin{picture}(75,12)
\put(0,1){Let $J\subseteq I$ such that $\Delta_J$ corresponds to}
\end{picture}
$}
{$
\begin{picture}(42,12)
\put(1,1){\makebox(0,0){$\circ$}}
\put(1.6,1){\line(1,0){8.5}}
\put(11,1){\makebox(0,0){$\circ$}}

\put(13,1.2){...}
\put(19,1){\makebox(0,0){$\circ$}}
\put(19.6,1){\line(1,0){8.5}}
\put(29,1){\makebox(0,0){$\circ$}}

\put(39,1){\makebox(0,0){$\circ$}}





 \put(29.5,1.7){\line(1,0){8.8}}
 \put(29.5,0.4){\line(1,0){8.8}}



\put(34,3.2){\makebox(0,0){\tiny( $\! \! \scriptstyle{\sim} $\tiny)}}


\put(26,-2.2){\scriptsize n+1}
\put(18,-2.2){\scriptsize n}


\put(10,-2.2){\scriptsize 2}
\put(0,-2.2){\scriptsize 1}



\end{picture}
$}
{$
\begin{picture}(13,12)
\put(2,1){,}
\end{picture}
$}
\\[0.7cm]
then $K_J=1$ or $G_J\cong 2^6\cdot A_7$. In the latter $n=1$ and there is 
at
most one $J\subseteq I$ such that $G_J\cong 2^6A_7$ and ($\alpha$)
below holds for all other $J$.\\
Further one of the following holds
\begin{enumerate}
\item[($\alpha$)]
For all $J$ of that type such that $K_J=1$ we have\\
$G_J\cong 3^{\frac{1}{3}(2^{n+2}-1)(2^{n+1}-1)}Sp_{2(n+2)}(2)$ or $Sp_{2(n+2)}(2)$.
\item[($\beta$)]
$n=3$ and for all $J$ we have $K_J=1$ and $G_J\cong F_1$
\item[($\gamma$)]
$n=2$ and for all $J$ we have $K_J=1$ and $G_J\cong Co_1$.
\item[($\delta$)]
$n=1$ and for all $J$ we have $K_J=1$ and exactly one of the 
following occurs: $G_J\cong M_{24}$ or $G_J\cong He$.
\end{enumerate}
This also includes $m=1$. Then $J=\phi$ and in $(\alpha)-(\delta)$ we have
$G_J=G$ is one of these groups.
\item[(v)]
The systems $M(n,m), n\ge 2, m\ge 1:$\\
We have a partition $I=\{1\} \cup I_n\cup I_m$, where $|I_n|=n$ and
$|I_m|=m$. The diagram induced on
$I_n\cup I_m$ is as in (i), i.e. a complete bipartite graph all of whose
edges are of type 
{$
\begin{picture}(15,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}

\put(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}
\put(7,3.5){\makebox(0,0){\tiny( $\! \! \scriptstyle{\sim} $\tiny)}}

\end{picture}
$}. There are no edges from 1 to any node in
$I_m$ and there is exactly one edge from 1 to $t\in I_n$. Let $u\in I_m$,
$v\in I_n \backslash \{t\}$, and set $J$=$I$ $\backslash \{1,u,v,t\}$. Then \\
\begin{center}
$\Delta_J=$
{$
\begin{picture}(35,5)
\put(1,-2.5){\scriptsize 1}
\put(11,-2.5){\scriptsize t}
\put(21,-2.5){\scriptsize u}
\put(31,-2.5){\scriptsize v}
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\put(12,1){\makebox(0,0){$\circ$}}
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\put(2.5,1){\line(1,0){8.7}}

\put(12.5,0.4){\line(1,0){8.9}}
\put(12.5,1.7){\line(1,0){8.9}}
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\put(22.5,0.4){\line(1,0){8.9}}
\put(22.5,1.7){\line(1,0){8.9}}
\put(27,3.5){\makebox(0,0){\tiny( $\! \! \scriptstyle{\sim} $\tiny)}}
\end{picture}
$}.
\end{center}
We have $K_J=1$ and $G_J$ is locally $M(24), M(24)',
3 M(24)$ or $3M(24)'$.\\
(In the sense of [Gold].)
\item[(vi)]
System of type $\tilde{F}_4:$\\
$\Delta$ is of shape 
{$
\begin{picture}(35,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(32,1){\makebox(0,0){$\circ$}}
\put(22.5,1){\line(1,0){8.7}}

\put(2.5,1){\line(1,0){8.7}}

\put(12.5,0.4){\line(1,0){8.9}}
\put(12.5,1.7){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\end{picture}
$.}\\
This system has been described in [Wie3].
\item[(vii)]
Square system:\\ \\
{$
\begin{picture}(25,15)
\put(1,6){$\Delta$ is of shape}
\end{picture}
$}
{$
\begin{picture}(20,15)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(2,11){\makebox(0,0){$\circ$}}
\put(12,11){\makebox(0,0){$\circ$}}

\put(-1.5,-0.2){\scriptsize 1}
\put(13,-0.2){\scriptsize 2}
\put(-1.5,11.8){\scriptsize 3}
\put(13,11.8){\scriptsize 4}

\put(2.5,1){\line(1,0){8.9}}

\put(1.6,1.6){\line(0,1){8.9}}
\put(2.4,1.6){\line(0,1){8.9}}

\put(11.6,1.6){\line(0,1){8.9}}
\put(12.4,1.6){\line(0,1){8.9}}

\put(2.5,11.4){\line(1,0){8.9}}
\put(2.5,10.6){\line(1,0){8.9}}

\put(7,13.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(20,6){,}
\end{picture}
$}
\begin{picture}(55,15)
\put(1,6){$G_3\cong G_4\cong A_7$ and $G_{1,2}\cong 3A_6$.}
\end{picture}
\\ 
\end{itemize}

In this paper we prove:\\[2ex]
{\bf Theorem.} {\it Let $G=\langle P_1,\ldots, P_n\rangle$ be a semiclassical
parabolic system of rank $n$ with connected diagram $\Delta$. Then it is
one of the systems from the list above.}\\

The remaining notations and definitions are standard. For convenience of the reader
we state the definitions around the coset graph used in this paper.\\ 

Let $\Gamma = \Gamma (G_1, G_2)$ be the coset graph for $G=\langle G_1,
G_2\rangle$, i.e. the vertices are the cosets of $G_1$ and $G_2$ and
$(G_1g,G_2h)$ is an edge iff $G_1g\cap G_2h\neq \phi$.  Then $G$ acts on
$\Gamma$ by right multiplication.
Let us identify the vertices $G_1, G_2$ with $1, 2 $ respectively. Then $G_i$
is the stabilizer in $G$ of the vertex $i$. For $\delta \in \Gamma$
denote by $G_\delta$ the stabilizer of $\delta$ in $G$.  Then $G_\delta$ is
 conjugate to $G_1$ or $G_2$.

Assume that $G_1
\cap G_2$ contains a Sylow 2-subgroup $S$ of $G_1$ and $G_2$. Then set for
$\delta \in \Gamma:$
$$Z_\delta=\langle \Omega_1(Z(T))\mid T\in Syl_2(G_\delta)\rangle \mbox{ and }$$
$$b_\delta= min\{d(\delta,\lambda)\mid \lambda \in \Gamma , Z_\delta \not\subseteq G_{\Delta 
(\lambda)}\},$$
where $G_{\Delta (\lambda)}$ is the pointwise stabilizer of the 
neighborhood of $\lambda$, and set
$$b=min \{b_\delta \mid \delta \in \Gamma\}.$$
As $G$ acts edge transitively on $\Gamma$, we get only two parameters
$b_1$ and $b_2$ and so $b=min \{b_1, b_2\}.$

By a {\it critical pair} we mean a pair $(\delta, \delta')$ of vertices of $\Gamma$ 
such that
\begin{itemize}
\item [(i)] $Z_\delta \not\subseteq G_{\Delta(\delta')},$
\item [(ii)] $d(\delta,\delta')=b.$
\end{itemize}
This coset graph will play an important role in the proof of the main theorem.\\

By [ShStr] a diagram of shape 
{$
\begin{picture}(42,5)
\put(1,1){\makebox(0,0){$\circ$}}
\put(1.6,1){\line(1,0){8.5}}
\put(11,1){\makebox(0,0){$\circ$}}

\put(13,1.2){...}
\put(19,1){\makebox(0,0){$\circ$}}
\put(19.6,1){\line(1,0){8.5}}
\put(29,1){\makebox(0,0){$\circ$}}

\put(39,1){\makebox(0,0){$\circ$}}





 \put(29.5,1.7){\line(1,0){8.8}}
 \put(29.5,0.4){\line(1,0){8.8}}



\put(34,3.2){\makebox(0,0){ $\scriptstyle{\sim} $}}


\put(27,-2){\tiny n-1}
\put(38.3,-2){\tiny n}


\put(10.3,-2){\tiny 2}
\put(0.3,-2){\tiny 1}



\end{picture}
$}
corresponds to a non-split extension of 
$Sp_{2n}(2)$ by an abelian group of order 
 $ 3^{\frac{1}{3}(2^{n}-1)(2^{n-1}-1)}$
if $G_{\{1, \ldots , n-3\}} \cong 3^7Sp_6(2)$.
If we do not care about the tilde, i.e. 
the diagram is of shape
{$
\begin{picture}(42,5)
\put(1,1){\makebox(0,0){$\circ$}}
\put(1.6,1){\line(1,0){8.5}}
\put(11,1){\makebox(0,0){$\circ$}}

\put(13,1.2){...}
\put(19,1){\makebox(0,0){$\circ$}}
\put(19.6,1){\line(1,0){8.5}}
\put(29,1){\makebox(0,0){$\circ$}}

\put(39,1){\makebox(0,0){$\circ$}}





 \put(29.5,1.7){\line(1,0){8.8}}
 \put(29.5,0.4){\line(1,0){8.8}}



\put(34,3.2){\makebox(0,0){\tiny( $\! \! \scriptstyle{\sim} $\tiny)}}


\put(27,-2){\tiny n-1}
\put(38.3,-2){\tiny n}


\put(10.3,-2){\tiny 2}
\put(0.3,-2){\tiny 1}



\end{picture}
$,}
then we will denote the corresponding group by
{$
\begin{picture}(46,6)
\put(0,0){ $\left( 3^{\frac{1}{3}(2^{n}-1)(2^{n-1}-1)}\right) Sp_{2n}(2)$.}
\end{picture}
$}
\vspace{1cm}
\noindent
\begin{center}
\boldmath \S\unboldmath {\bf 1 Preliminaries}
\end{center}
\vspace{1cm}

The proof of the main theorem will be
done by induction on the rank of the parabolic system. For
convenience of the reader we first state Timmesfeld's result
[Tim3] on rank three systems.\\[0.2cm]
{\bf (1.1) Theorem.}
{\it Let $G = \langle P_1,P_2,P_3 \rangle$ be a parabolic system of classical or semiclassical type with connected diagram $\Delta$.  Then one of the following
holds:\\
(a) \quad For all $i$ we have $P_i/O_2(P_i) \cong \Sigma_3$ and
\begin{tabbing}
\quad \= (i) \quad \=  $\Delta =$ {$
\begin{picture}(25,4)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}
\put(12.5,1){\line(1,0){8.7}}
\end{picture}
$}  and $G \cong A_7 $ or $A_8$.\\
\>(ii)\> $\Delta =$  {$
\begin{picture}(25,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(12.5,1.7){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\end{picture}
$} and $G \cong M_{24}, He $ or $3^7Sp_6(2)$ \\
\>(iii) \>  $\Delta =$ {$
\begin{picture}(25,4)

\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(12.5,1.7){\line(1,0){8.9}}
\end{picture}
$} and $G \cong A_7$ or $Sp_6(2)$.\\
\>(iv) \>  $\Delta = $ {$
\begin{picture}(25,5)
\put(1,-2){\scriptsize 1}
\put(11,-2){\scriptsize 2}
\put(21,-2){\scriptsize 3}\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}
\put(7,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(12.5,1.7){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}
\end{picture}
$},  $G_i/K_i \cong A_6, \Sigma_6, 3A_6$, or $ 3\Sigma_6, i \in \{1,3\} $, \\
\hspace*{3cm} \qquad \qquad $\;$ and there are two possibilities:\\
\\
\>\>{\sf ($\star$)} \quad \=$2^4 \leq |K_i| \leq 2^5, K_i $ is an indecomposable $G_i$-module and\\ \>\>\>  $C_{K_i}(G_i) = 1$, for $i= 1,3$. Moreover $\Omega_1(Z(S)) \lhd G_2$ and\\ \>\>\> $G_2/K_2 \cong \Sigma_3 \times \Sigma_3$.\\ \\
\>\>{\sf ($\star\star$)} \> $K_1 = K_3 = 1, \, G_2/K_2 \cong (Z\!\!\! Z_3 \times Z\!\!\! Z_3) Z\!\!\! Z_2.$\\
\>(v) \>  $\Delta =$ {$
\begin{picture}(25,4)

\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}
\put(2.4,1.6){\line(1,0){9}}
\put(2.4,0.4){\line(1,0){9}}
\put(12.5,1){\line(1,0){8.7}}
\put(12.4,1.6){\line(1,0){9}}
\put(12.4,0.4){\line(1,0){9}}
\put(1,-2){\scriptsize 1}
\put(11,-2){\scriptsize 2}
\put(21,-2){\scriptsize 3}

\end{picture}
$}
,
{$
\begin{picture}(25,4)

\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}

\put(12.5,1){\line(1,0){8.7}}
\put(12.4,1.6){\line(1,0){9}}
\put(12.4,0.4){\line(1,0){9}}
\put(1,-2){\scriptsize 1}
\put(11,-2){\scriptsize 2}
\put(21,-2){\scriptsize 3}

\end{picture}
$}, and $G_1 \cong G_2(2)$ or $G_2(2)^\prime$.\\
\end{tabbing}
(b) \quad There exists at least one $i$ with $P_i/O_2(P_i) \not\cong \Sigma_3$,
we have a classical parabolic system and one of the following holds: 
\begin{itemize}
\item[(i)] the diagram is of type $C_3$ and $G$ contains a normal
 subgroup isomorphic to the corresponding group of Lie type.
\item[(ii)] the diagram is a triangle and $B=1$.
\item[(iii)] $\Delta =$ 
{$
\begin{picture}(25,4)

\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}

\put(12.5,1){\line(1,0){8.7}}

\put(6.5,2.7){\small n}
\put(16.5,2.7){\small n}
\put(11,-2){\scriptsize 2}


\end{picture}
$}
and $P_2/O_2(P_2) \not\cong \Sigma_3$.
\item[(iv)] $\Delta =$ 
{$
\begin{picture}(29,4)

\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1.6){\line(1,0){8.9}}
\put(2.5,0.4){\line(1,0){8.9}}

\put(12.5,1){\line(1,0){8.7}}


\put(1,-2){\scriptsize 1}

\put(25,0){or}
\end{picture}
$}
{$
\begin{picture}(25,4)

\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}

\put(2.5,1.6){\line(1,0){8.9}}
\put(2.5,0.4){\line(1,0){8.9}}

\put(12.5,1.6){\line(1,0){8.9}}
\put(12.5,0.4){\line(1,0){8.9}}


\put(1,-2){\scriptsize 1}


\end{picture}
$} \qquad and $P_1/O_2(P_1) \cong L_2(4)$.
\end{itemize}
}
One of the facts which follows from (1.1) is that a diagram for a semiclassical system with $P_i/O_2(P_i) \cong \Sigma_3$ for all $i$ never contains a triangle. We will use this fact freely in what follows.\\

Next we will establish the classification of rank four semiclassical systems 
with connected diagram. For this first we prove that 
$P_i/O_2(P_i) \cong \Sigma_3$ for any such system.\\[2ex]
{\bf (1.2) Lemma.} {\it Let $\{P_1,P_2,P_3,P_4\}$ be a semiclassical parabolic 
system
of rank four with connected diagram. Then $P_i/O_2(P_i)\cong \Sigma_3$ for $i=1,2,3,4.$}\\[2ex]
{\bf Proof.} 
By (1.1) the parabolic subgroups $P_i$ in a semiclassical rank-three-system
satisfy $P_i/O_2(P_i)\cong \Sigma_3$ for all $ i$.
So we get that $\Delta$ is a string and we may choose notation such 
that $\Delta$ has the shape 
\begin{center}
{$
\begin{picture}(40,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(12.5,1){........}
\put(2.5,1){........}
\put(1,-2.5){{\scriptsize 1}}
\put(11,-2.5){{\scriptsize 2}}
\put(21,-2.5){{\scriptsize 3}}
\put(31,-2.5){{\scriptsize 4}}

\put(32,1){\makebox(0,0){$\circ$}}
\put(22.5,0.4){\line(1,0){8.9}}
\put(22.5,1.7){\line(1,0){8.9}}
\put(27,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(37,1){,}
\end{picture}
$}
\end{center}
where $P_1/O_2(P_1) \not\cong \Sigma_3$. As 
 $P_2/O_2(P_2) \cong \Sigma_3$  we get 
  $P_1/O_2(P_1) \cong L_2(4)$ by (1.1) and $\Delta$ is of
the form
\begin{center}
{$
\begin{picture}(45,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(12.5,1.7){\line(1,0){8.9}}
\put(-2,-3){ {\scriptsize $L_2(4)$}}
\put(32,1){\makebox(0,0){$\circ$}}
\put(22.5,0.4){\line(1,0){8.9}}
\put(22.5,1.7){\line(1,0){8.9}}
\put(27,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(37,0){or}
\end{picture}
$}
{$
\begin{picture}(40,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(12.5,1){\line(1,0){8.7}}
\put(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}
\put(-2,-3){{\scriptsize $L_2(4)$}}
\put(32,1){\makebox(0,0){$\circ$}}
\put(22.5,0.4){\line(1,0){8.9}}
\put(22.5,1.7){\line(1,0){8.9}}
\put(27,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(37,1){.}
\end{picture}
$}
\end{center}
The first diagram was shown not to exist by [Wie2; (4.13)].
So assume that we have 
{$
\begin{picture}(36,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(12.5,1){\line(1,0){8.7}}
\put(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}
\put(1,-2.5){{\scriptsize 1}}
\put(11,-2.5){{\scriptsize 2}}
\put(21,-2.5){{\scriptsize 3}}
\put(31,-2.5){{\scriptsize 4}}
\put(32,1){\makebox(0,0){$\circ$}}
\put(22.5,0.4){\line(1,0){8.9}}
\put(22.5,1.7){\line(1,0){8.9}}
\put(27,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(36,1){.}
\end{picture}
$}\\

Now we have $G_1/K_1\cong M_{24}, He$ or $3^7Sp_6(2)$. In the latter the
assertion follows with [Wie2; (4.12)] again. So assume that $G_1/K_1\cong
M_{24}$ or $He$. As $G_2/K_2\cong L_2(4)\times 3\Sigma_6$, we see that
$\mid K_1:O_2(K_1)\mid = 3$ 
and $K_2=O_2(K_2)$. Now we apply  [ParRow; Theorem  A] to $(P_1,G_1)$. This shows
${~I\!\!\!\!C}_{G_1}(O_2(K_1))\not\subseteq O_2(K_1)$.
As $P_1/O_2(P_1) \cong L_2(4)$, $B$ has no normal 
Sylow-3-subgroup, so ${~I\!\!\!\!C}_{K_1}(O_2(K_1))\subseteq O_2(K_1)$.
Thus $G_1=K_1{~I\!\!\!\!C}_{G_1}(O_2(K_1))$.
 As $Z(G_1\cap G_4/
K_4)=1$, we see $O_2(K_1)\subseteq K_4$. But for a Sylow 2-subgroup $S$ of
$G_1\cap G_4$ we have $\mid S:K_4\mid = 2^{12}$ while $\mid S:O_2(K_1)\mid =
2^{10}$ (notice that $G_4/K_4\cong \Omega_8^-(2)$). This contradiction proves the lemma.\hfill $\Box$
\\

Now (1.1) and (1.2) together with the results of [FuStr], [Gil], [Wie2], and [Wie3] yield\\[0.3cm]
{\bf (1.3) Theorem.}
{\it Let $ G=\langle P_1,P_2,P_3,P_4 \rangle ,\{ P_1,P_2,P_3,P_4 \}$ be a semiclassical para\-bolic system with connected diagram $\Delta$. Then one of the following holds
\begin{tabbing}


\quad \=(i) \quad \=$\Delta $ {\rm : }\quad 

 {$
\begin{picture}(25,12)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(19,-6){\makebox(0,0){$\circ$}}

\put(19,8){\makebox(0,0){$\circ$}}

\put(2.5,1){\line(1,0){8.7}}

 \put(11.8,1.7){\line(1,1){6.5}}
 \put(12.5,1){\line(1,1){6.5}}

 \put(11.8,0.4){\line(1,-1){6.5}}
 \put(12.5,1){\line(1,-1){6.5}}

\put(14,5.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(14,-3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(1,-2){\scriptsize 1}
\put(10,-2){\scriptsize 2}
\put(21,-6.5){\scriptsize 3}
\put(21,8){\scriptsize 4}


\end{picture}
$}
 \quad \quad \quad \= $G_4 , G_3 \cong He \mbox{ or } M_{24}$\\
\\
\> (ii) \> $\Delta$ {\rm :} 
 {$
\begin{picture}(35,10)
\put(1,-2){\scriptsize 1}
\put(11,-2){\scriptsize 2}
\put(21,-2){\scriptsize 3}
\put(31,-2){\scriptsize 4}

\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(32,1){\makebox(0,0){$\circ$}}
\put(12.5,1){\line(1,0){8.7}}

\put(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}
\put(7,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(22.5,0.4){\line(1,0){8.9}}
\put(22.5,1.7){\line(1,0){8.9}}
\put(27,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\end{picture}
$} \> $G_1 \cong  G_4 \cong He$ \\
\>(iii) \>  $\Delta$  {\rm :}   {$
\begin{picture}(35,10)
\put(1,-2){\scriptsize 4}
\put(11,-2){\scriptsize 3}
\put(21,-2){\scriptsize 2}
\put(31,-2){\scriptsize 1}
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(32,1){\makebox(0,0){$\circ$}}
\put(12.5,1){\line(1,0){8.7}}

\put(2.5,1){\line(1,0){8.7}}

\put(22.5,0.4){\line(1,0){8.9}}
\put(22.5,1.7){\line(1,0){8.9}}
\put(27,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\end{picture}
$} \> $G \cong Co_1 $\\
\> (iv) \>$\Delta$  {\rm :}   {$
\begin{picture}(35,10)
\put(1,-2){\scriptsize1}
\put(11,-2){\scriptsize2}
\put(21,-2){\scriptsize3}
\put(31,-2){\scriptsize4}
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(32,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}

\put(12.5,0.4){\line(1,0){8.9}}
\put(12.5,1.7){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(22.5,0.4){\line(1,0){8.9}}
\put(22.5,1.7){\line(1,0){8.9}}
\put(27,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}
\end{picture}
$} \> the groups $G_i$ are isomorphic to the \\
\>\>\>corresponding groups in $M(24)$, \\
\>\>\> $3M(24), M(24)^\prime$ or $3M(24)^\prime$.\\
\> (v) \> $\Delta$ {\rm :}  
 {$
\begin{picture}(25,10)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(19,-6){\makebox(0,0){$\circ$}}

\put(19,8){\makebox(0,0){$\circ$}}

\put(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}

 \put(12.2,1.4){\line(1,1){6.3}}


 \put(12.2,0.7){\line(1,-1){6.3}}
 

\put(7,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(1,-2){\scriptsize 3}
\put(10,-2){\scriptsize 4}
\put(21,-6.5){\scriptsize 2}
\put(21,8){\scriptsize 1}


\end{picture}
$} \quad \> $G_3/K_3 \cong A_7, K_1 = K_2 = 1, K_3 $ is \\ 
\>\>\>elementary abelian of order $2^6$ and\\
\>\>\>$G_1 \cong G_2 \cong 3^7Sp_6(2).$\\
\\ 
\>(vi) \>  $\Delta$ {\rm :} 
 {$
\begin{picture}(25,10)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(19,-6){\makebox(0,0){$\circ$}}

\put(19,8){\makebox(0,0){$\circ$}}

\put(2.5,1){\line(1,0){8.7}}

 \put(11.8,1.7){\line(1,1){6.5}}
 \put(12.5,1){\line(1,1){6.5}}

 \put(11.8,0.4){\line(1,-1){6.5}}
 \put(12.5,1){\line(1,-1){6.5}}

\put(12.8,5.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(14,-3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(1,-2){\scriptsize 3}
\put(10,-2){\scriptsize 4}
\put(21,-6.5){\scriptsize 2}
\put(21,8){\scriptsize 1}


\end{picture}
$}
\>if $G_2/K_2 \cong 3^7Sp_6(2)$ or $Sp_6(2)$ \\
\>\>\>then $K_1 = K_2 = 1$,  if $G_2/K_2 \cong A_7$ \\ 
\>\>\> then $K_1 = 1, G_2 = 2^6A_7$, and\\ 
\>\>\> $G_3/K_3$ as in (1.1)(a)(iv)  ($\star$).\\


\> (vii) \>$\Delta$ {\rm :} {$
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\put(12.5,0.4){\line(1,0){8.9}}
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\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(22.5,1){\line(1,0){8.7}}
\end{picture}
$} \>  there is a normal subgroup $K$ of \\
\>\>\>$G$ with $G/K \cong F_4(2).$
\\
\> (viii) \>$\Delta$ {\rm :}   {$
\begin{picture}(35,5)
\put(2,1){\makebox(0,0){$\circ$}}
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\put(27,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(12.5,1){\line(1,0){8.7}}
\end{picture}
$}\> $G \cong 3^{35}Sp_8(2)$.\\ \\
\>(ix) \>  $\Delta$ {\rm :}  \qquad
{$
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\put(2,-5){\makebox(0,0){$\circ$}}
\put(12,-5){\makebox(0,0){$\circ$}}
\put(2,5){\makebox(0,0){$\circ$}}
\put(12,5){\makebox(0,0){$\circ$}}

\put(-1.5,-6.2){\scriptsize 3}
\put(13.4,-6.2){\scriptsize 4}
\put(-1.5,5.8){\scriptsize 1}
\put(13.4,6){\scriptsize 2}

\put(2.5,-5){\line(1,0){8.7}}

\put(1.6,-4.4){\line(0,1){8.9}}
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\put(2.5,5.4){\line(1,0){8.9}}
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\put(7,7.5){\makebox(0,0){$\scriptstyle{\sim}$}}


\put(11.6,-4.4){\line(0,1){8.9}}
\put(12.4,-4.4){\line(0,1){8.9}}
\end{picture}
$}
\> $K_i=1$ for all $i$, $G_1\cong G_2 \cong A_7$ and\\
\>\>\> $G_3$ and $G_4$ are as in (1.1)(a)(iv)($\star \star$). \\ 
\\ 
\>(x) \>  $\Delta$ {\rm :} \qquad
{$
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\put(7,7.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(7,-7.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}
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\put(11.6,-4.4){\line(0,1){8.9}}
\put(12.4,-4.4){\line(0,1){8.9}}
\end{picture}
$}
\>  one of the following holds:\\
\>\>\> (1) $K_i=1$, $G_i$ is as in (1.1)(a)(iv)($\star \star$)\\ 
\>\>\> \quad $\:$ for all $i$,\\  
\>\>\> (2) we can choose numeration such\\
\>\>\> \quad $\:$ that $K_1 \!= \! K_3 \! = \! 1$, $G_1$ and $G_3$ are as\\
\>\>\> \quad $\:$ in (1.1)(a)(iv)($\star$) and $G_2/K_2$ and\\
\>\>\> \quad $\:$  $G_4/K_4$ are of type (1.1)(a)(iv)($\star \star$).\\
\>(xi) \>  $\Delta$ {\rm :}  
 {$
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 \put(11.8,1.7){\line(1,1){6.5}}
 \put(12.5,1){\line(1,1){6.5}}

 \put(11.8,0.4){\line(1,-1){6.5}}
 \put(12.5,1){\line(1,-1){6.5}}

\put(12.8,5.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(14,-3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(1,-2){\scriptsize 3}
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\put(21,-6.5){\scriptsize 4}
\put(21,8){\scriptsize 1}


\end{picture}
$}
\>  one of the following holds:\\
\>\>\> (1) $K_i=1$, $G_i$ is as in (1.1)(a)(iv)($\star\star$)\\ 
\>\>\> \quad $\:$ for $i=1,3,4$,\\  
\>\>\> (2) we can choose numeration such\\
\>\>\> \quad $\:$ that $K_1=K_3=1$, $G_1$ and $G_3$ are\\
\>\>\> \quad $\:$ as in (1.1)(a)(iv)($\star$) and $G_4/K_4$\\
\>\>\> \quad $\:$  is of type (1.1)(a)(iv) ($\star \star$).\\
\end{tabbing}}
\vspace{0.1cm}

In the remainder of this section we prove
some results on representations of the groups which will be 
involved in the proof of the main theorem.

First of all we need some facts about representations of Chevalley
groups. We do not state results in full generality as we will just need them for untwisted groups
over $GF(2)$. The reader who is interested in the general theory should consult the
fundamental paper of R. Steinberg [Stein].\\[0.2cm]
{\bf (1.4) Lemma. [Sm]} {\it 
Let $G=G(q)$, $q=p^f$, be a Chevalley group
(twisted or not) and $V$ be a nontrivial irreducible
$GF(q)$--module for $G$. Let $P$ be
a parabolic subgroup of $G$ and $V_p={~I\!\!\!\!C}_V(O_p(P))$. Then
$V_p$ is an irreducible module for $P$.}\\

Let now $G=G(2)$ be untwisted of rank at least two, $V$ be an 
irreducible $GF(2)G$--module and
$\{P_1,\ldots , P_n\}$ be the set of minimal parabolic subgroups 
containing a given Sylow 2-subgroup $S$. Set $V_i={~I\!\!\!\!C}_V(O_2(P_i))$.
As $P_i/O_2(P_i) \cong \Sigma_3$ we have with (1.4) that $V_i$ is
either trivial or a two--dimensional $\Sigma_3$--module.
Let $\Lambda$ be the $n$--dimensional vectorspace over $GF(2)$ with
basis $\{\lambda_1,\ldots,\lambda_n\}$. Now set $a_i=0$ if
$V_i$ is trivial and $a_i=1$ else. By
\[ V \longrightarrow \lambda=\sum_{i=1}^n a_i\lambda_i \]
we get a mapping from the set of isomorphism types of irreducible
modules to the set of elements of $\Lambda$.
We denote $V$ by $V(\lambda)$. Then we have the following fundamental
theorem due to R. Steinberg.\\[0.2cm]
{\bf (1.5) Theorem.} {\it Let $G=G(2)$ be an untwisted Chevalley
group over $GF(2)$ of rank at least two and $V$ be an
irreducible $GF(2)$--module for $G$. Then
the mapping $V \longrightarrow V(\lambda)$ above is a bijective
mapping between the isomorphism classes of modules
 and the vectors of $\Lambda$.}\\

The module $V(\lambda)$, $\:\lambda= \sum_{i=1}^n \lambda_i$, is called the {\it Steinberg module} for $G$. Its dimension is the order of a Sylow 2-subgroup of $G$.

The theorem (1.5) will be used to identify modules. We
just have to check the action of the minimal parabolics on the centralizers of their unipotent radical.
Then the module is uniquely determined.\\[0.3cm]
{\bf (1.6) Definition.}
Let $G$ be some group acting faithfully on some $GF(p)$-module $V$.
\begin{itemize}
\item[(a)] The module $V$ is called an {\it $F$-module} if
$| V : {~I\!\!\!\!C}_V(A)| \leq | A|$
for some nontrivial elementary abelian p-subgroup $A$ of $G$.
\item[(b)] The module $V$ is called an {\it $F_1$-module} if 
$| V : {~I\!\!\!\!C}_V(A)| \leq p \cdot | A |$
for some  nontrivial elementary abelian p-subgroup $A$ of $G$.
\item[(c)] The group $A$ in (a) and (b) is called an
{\it offending subgroup}.
\end{itemize}
\vspace{0.3cm}
{\bf (1.7) Lemma.} 
{\it Let $G$ be one of $Sp_{2n}(q)$ or $\Omega^+_{2n}(q), n\geq 3,
q=2^f$ and $V$ be an $F_1$-module with offending subgroup $A$. Then one of the following holds:
\begin{itemize}
\item [(i)] $G\cong Sp_{2n}(q)$ and $V$ is the natural module,  $V$
is the 8-dimensional spin module and $n=3$, or 
$V$ is the 
16-dimensional
spin module, $n=4$, and $q=2$. In the latter $ V$ is not an $F$-module.
If $\: V$ is the natural module, $A\subseteq O_2(G_v), v \in V^\#$, then $|V
:{~I\!\!\!\!C}_V(A)|= |A|$ and $A$ contains a transvection.
\item[(ii)]$G\cong \Omega_{2n}^+(q),V$ is the natural module or $n=4$ or $5$ 
and $V$ is one of two 8-dimensional, resp. 16-dimensional halfspin modules.
\end{itemize}}
\noindent
{\bf Proof.} [Coop; (2.1)], [McCl; (6.40)].\hfill $\Box$\\[2ex]
{\bf(1.8) Lemma.} {\it Let $G=\Omega^+_{2n}(q),q=2^f,V$ an irreducible module 
over
$GF(q)$. Suppose $G=\langle P_1,\ldots, P_n\rangle$ where the $P_i$'s are
the minimal parabolic subgroups correspon\-ding to the  diagram
\begin{center}
{$
\begin{picture}(38,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(26.5,1){\makebox(0,0){$\circ$}}

\put(1,-2){{\scriptsize n}}
\put(11,-2){{\scriptsize n-1}}

\put(2.5,1){\line(1,0){8.7}}


\put(14,1){$\; . \; . \; .\;$}





\put(34.5,-6){\makebox(0,0){$\circ$}}

\put(34.5,8){\makebox(0,0){$\circ$}}



 \put(27.2,1.4){\line(1,1){6.3}}


 \put(27.2,0.7){\line(1,-1){6.3}}
 


\put(36,-6.5){\scriptsize2}
\put(36,8){\scriptsize1}


\end{picture}
$.}
\end{center}
Let $S\in Syl_2(\bigcap_{i=1}^nP_i).$
\begin{itemize}
\item[(i)] If $V$ in the natural module, then $N_G({~I\!\!\!\!C}_V(S))=
\langle P_1,\ldots, P_{n-1}\rangle$
\item[(ii)] If $V$ is a halfspin module, then $N_G({~I\!\!\!\!C}_V(S))=
\langle P_1,P_3,\ldots, P_n\rangle$ or \\ 
$\langle P_2, P_3,\ldots, P_n\rangle.$
\end{itemize}
If $q=2$ the converse of (i) and (ii) is also true}.\\[2ex]
{\bf Proof.}
 (i) and (ii) are well known. 
The last assertion follows from (1.5).\hfill
 $\Box$\\[2ex]
{\bf (1.9) Lemma. }{\it Let $\; G=Sp_{2n}(q)$, $\; q=2^f$, $\:V$ an irreducible $GF(q)$-module,
$G=\langle P_1,\ldots, P_n\rangle$ where the $P_i$'s are the minimal 
parabolic subgroups corresponding to the following diagram
\begin{center}
{$
\begin{picture}(48,5)
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\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}

\put(1,-2){{\scriptsize 1}}
\put(11,-2){{\scriptsize 2}}

\put(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}


\put(12.5,1){\line(1,0){8.7}}


\put(24,1){$\; . \; . \; .\;$}

\put(34,1){\makebox(0,0){$\circ$}}
\put(44,1){\makebox(0,0){$\circ$}}

\put(32.5,-2){{\scriptsize n-1}}
\put(43,-2){{\scriptsize n}}

\put(34.5,1){\line(1,0){8.7}}


\end{picture}
$}
\end{center}
Let $S\in Syl_2(\bigcap_{i=1}^nP_i)$.
\begin{itemize}
\item[(i)] If $V$ is the natural module, then $N_G({~I\!\!\!\!C}_V(S))=
\langle P_1,\ldots, P_{n-1}\rangle$
\item[(ii)] If $V$ in the spin module, then $N_G({~I\!\!\!\!C}_V(S))=
\langle P_2,\ldots, P_n\rangle$.
\end{itemize}
If $q=2$ the converse of (i) and (ii) is also true}.\\[2ex]
{\bf Proof.} Well known, the last assertion follows from (1.5).\hfill $\Box$\\[2ex]
{\bf (1.10) Lemma. } {\it 
Let $G\cong Sp_{2n}(2)$, $\,t \in G$ an involution
and $V$ a nontrivial irreducible $GF(2)G$--module. 
\begin{enumerate}
\item[(i)] 
If $n=3$ and $|V:{~I\!\!\!\!C}_V(t)| \leq 8$ then $V\cong V(\lambda_1)\,$ or
 $\,V(\lambda_3)$.
\item[(ii)] Let $n=3$ and $t$ a transvection in the natural representation  of $G$.
If $|V:{~I\!\!\!\!C}_V(t)| \leq 16$ then $V\cong V(\lambda_i),\:$ $ i=1,2,3.$
\item[(iii)] Let $n=4$ and $\,t$ an involution of type $a_2$ in the natural representation.
If $|V:{~I\!\!\!\!C}_V(t)| \leq 2^8$ then $V\cong V(\lambda_1)\,$ or
 $\,V(\lambda_4)$.
\item[(iv)] Let $n=4$ and
 $|V:{~I\!\!\!\!C}_V(t)| \leq 16$, then  $V\cong V(\lambda_1)\,$ or
 $\,V(\lambda_4)$.
\end{enumerate}
}
\noindent
{\bf Proof.} First of all we prove
\begin{itemize}
\item[($*$)] Let $G \cong Sp_4(2)$, $\,t\in G$ be an involution and
$V$ an irreducible $GF(2)G$--module with $|[V,t]|=2$.
Then $|V|=16$ and $V$ is one of the two natural modules.
\end{itemize}

We have $Sp_4(2) \cong \Sigma_6$. As $t$ cannot invert an element of order 5 in $A_6$, we get $t\not\in A_6$. So by suitable numbering we have 
$[t,P_1]=1$, $\,P_1$ being one of the two parabolic subgroups containing
a given Sylow 2-subgroup. Now the assertion follows with (1.9).

For the remainder of the proof choose numbering of the minimal parabolic
subgroups according to the  diagram
\begin{center}
{$
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\put(12,1){\makebox(0,0){$\circ$}}


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\put(11,-2){{\scriptsize 2}}

\put(1,-2){{\scriptsize 1}}


\end{picture}
$}
{$
\begin{picture}(40,5)
\put(16.5,1){\makebox(0,0){$\circ$}}
\put(26.5,1){\makebox(0,0){$\circ$}}
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\put(17,1){\line(1,0){8.7}}

\put(4,1){$\; . \; . \; .\;$}


\put(27,1.7){\line(1,0){8.9}}
\put(27,0.4){\line(1,0){8.9}}



\put(25.2,-2){\scriptsize n-1}
\put(35.8,-2){\scriptsize n}
 



\end{picture}
.$}
\end{center}
\vspace{0.3cm}

(i) Set $V_1={~I\!\!\!\!C}_V(O_2(G_1))$. By (1.9) we may assume
that $V_1$ is a nontrivial module.
Let $\langle i \rangle = Z(G_1)$.
 We may assume $t\in G_1\setminus O_2(G_1)$.
Hence $t$ is nontrivial on $[V,i]$ and $V/{~I\!\!\!\!C}_V(i)$.
This shows that $t$ induces a transvection on $V_1$. By ($*$) $V_1$ is one of the natural modules for $Sp_4(2)$. As $[V,i]= V_1$ by (1.4)
we now know that $t$ is not of type $c_1$ or $c_2$ in the natural representation. 

Let $t$ be of type $c_3$.
Then $t$ inverts some element $\nu$ of order 7 in $G$. This shows
$|[V,t]|=8$ and $|[V,\nu]|=64$. Set $V_3={~I\!\!\!\!C}_V(O_2(G_3))$.
By (1.9) we may assume that $V_3$ is nontrivial. If
$V_3$ is the eight--dimensional Steinberg module, then
$[V,\nu] \subseteq V_3$ and so $[V,G_3]=V_3$. 
But as $O_2(G_3) \not\cong V_3$ as $G_3/O_2(G_3)$--module, this is impossible.
So by (1.5) $V_3$ is the natural module or its dual. As
$O_2(G_3)$ is indecomposable, we get $|{~I\!\!\!\!C}_{V/V_3}(G_3)| \leq2$.
Hence $|[V,G_3]| \leq 2^7$ and so $|V| \leq 2^8$. 
In particular $V/V_1$ is an irreducible $G_1/O_2(G_1)$--module, which implies
$[V,O_2(G_1),O_2(G_1)]=1$. 
As $G=\langle {~I\!\!\!\!C}_G(x) \: | \: x\in O_2(G_1)^\# \rangle$,
we see that there is some $x\in O_2(G_1)^\#$ with ${~I\!\!\!\!C}_V(x) >V_1$.
So we have found $x\in O_2(G_1)$ of type $a_2$ with
$|V:{~I\!\!\!\!C}_V(x)|\leq 8$.
Hence we may assume that $t$ is of type $a_2$ right
from the beginning. So we may assume
that $t$ is in the center of $G_2$. Now
$G_2$ induces on $[V,t]$
a subgroup of $L_3(2)$. The largest factor
group of $G_2$ which can be embedded into $L_3(2)$ is $\Sigma_3$, so
we get $[V,t,O_2(G_2)]=1$. Application of [Str6; Theorem] gives the assertion.\\

(ii) We may choose $t$ such that $t\in Z(G_1)$. Hence 
$[V,t,O_2({~I\!\!\!\!C}_G(t))]=1$. The assertion follows with
 [Str6; Theorem].\\

(iii) We choose notation such that $G_2={~I\!\!\!\!C}_G(t)$. Set
$V_i={~I\!\!\!\!C}_V(O_2(G_i))$ for $i=1,2$. As $V_2 \leq [V,t]$ by (1.4)
 we get $|V_2| \leq2^8$. Let $|V_2|=2$, then by (1.5) we have $V=V(\lambda_2)$.
Now ${~I\!\!\!\!C}_V(Z(G_1))$ and $V/{~I\!\!\!\!C}_V(Z(G_1))$ are
nontrivial $G_1$--modules. As $t$ is conjugate to some involution
in $G_1\! \setminus \! O_2(G_1)$, we
see that $|[V,t]|\geq 4$. In particular $V_2\not=[V,t]$. As 
$W=O_2(G_2)/Z(O_2(G_2))$ is an irreducible $G_2/O_2(G_2)$--module
we get that $[V,t]/V_2$ involves a module isomorphic to $W$. Hence $|[V,t]|\geq 2^9$, a contradiction. So we
have $|V_2|>2$, Let $|V_2|=4$. By (1.5) we may assume $V=V(\lambda_1+\lambda_2)$.
In particular by (1.9) $V_1={~I\!\!\!\!C}_V(O_2(G_1))$ is the 
natural $Sp_6(2)$--module. Further we see that $[V,t]/V_2$ involves a nontrivial
$\Sigma_6$--module. In particular
${~I\!\!\!\!C}_{{~I\!\!\!\!C}_G(t)}([V,t]) \subseteq O_2({~I\!\!\!\!C}_G(t))$.

Let $H_1$ be a complement of $O_2(G_1)$ in $G_1$ with
$N_G(H_1) \cong H_1 \times \Sigma_3$ and $g\in G$ with $s=t^g \in H_1$ of type $a_2$ in the natural representation. Suppose that $V$ just
involves natural modules as non-trivial $H_1$--factors. Then there is some $\rho \in {~I\!\!\!\!C}_{H_1}(s)$, $\, o(\rho)=3$, such that $[\rho, [V,s]]=1$. But
this contradicts the fact that
${~I\!\!\!\!C}_{{~I\!\!\!\!C}_G(t)}([V,t]) \subseteq O_2({~I\!\!\!\!C}_G(t))$.
So $V$ involves at least one non-trivial $H_1$--module, which is not the natural
module.

Let $i$ be an involution in ${~I\!\!\!\!C}_G(H_1)$, then we have that $[V,i]$ and
$V/{~I\!\!\!\!C}_V(i)$ both involve the natural module. Let $U=\langle V_2^{G_1} \rangle$.
As $|V_2:V_2\cap V_1|=2$, we see with (1.9) and Gasch\"utz ([Hu; I.17.4]) that either
$|U:V_1|=2$ or
$U/V_1$ involves the natural module. In the former we would have $U=V_2V_1$ and so 
$[O_2(G_1),U]=[V_2,O_2(G_1)]\subseteq V_2$,
a contradiction. So we have at least three natural modules involved. As there has to be a  module involved that is not the natural one, by (i) this has to be the spin module and so 
$|[V,t]|=2^8$. 

Let $\nu \in {~I\!\!\!\!C}_G(H_1)$, $\,o(\nu)\!=\!3$. 
We have $|V/{~I\!\!\!\!C}_V(\nu)| \leq |V/{~I\!\!\!\!C}_V(i)|^2 \leq
2^{12}$. As $H_1$ acts on ${~I\!\!\!\!C}_V(\nu)$ and $|[V,\nu]|\leq 2^{12}$
we see that $[V,\nu]$ involves two natural modules and
${~I\!\!\!\!C}_V(\nu)$ involves one natural module and one spin module.
We know $[V,t]\cong [V,s] =W >W_1>W_2>1$,
$\, |W_2|=4=|W/W_1|$ and $|W_1/W_2|=16$, $\, W_1/W_2$ is a nontrivial 
$\Sigma_6$--module.
Since we have  $Z(O_2({~I\!\!\!\!C}_{H_1}(s)))=Z(O_2({~I\!\!\!\!C}_G(s)))$ we get
$[\nu, Z(O_2({~I\!\!\!\!C}_G(s)))]=1$ and so $[W_2,\nu]=1$ and 
$[W,\nu]\subseteq W_1$.
Hence $W_1=W_2 \oplus [W_1,\nu]$, since $|[[V,s],\nu]|=16$.
But now $$[[W_1,\nu],O_2({~I\!\!\!\!C}_{H_1}(s))] \subseteq W_2 \cap [W_1,\nu]=1.$$
This shows $[W_1,O_2({~I\!\!\!\!C}_G(s))]=1$, a contradiction.

Assume next $|V_2|=16$. Let $[{~I\!\!\!\!C}_{G_2}(Z(O_2(G_2))),[V,t]]=[V,t]$.
Then there is some $\rho\in G_2$, $\, o(\rho)=3$, with
$[[V,t],\rho]=1$ and $[O_2(G_2),\rho]=O_2(G_2)$.
Then $[[V,t],O_2(G_2)]=1$ and the assertion follows with [Str6; Theorem].

So we may assume  $V_2=[[V,t],{~I\!\!\!\!C}_{G_2}(Z(O_2(G_2)))]\not= [V,t]$.
In particular $[Z(O_2(G_2)),[V,t]]=1$. Application of [Str6; Corollary 2] implies $V=V(\lambda_i)$, $\, i=1,2,3,4.$ By (1.5)
obviously $V\not\cong V(\lambda_2)$. So we just have to consider $V\cong V(\lambda_3)$.
Then $V_1$ is not the natural or spin module. Let $H_1$ and $s$ be as before.  We see
$|[V_1,s]|\geq 16$ by (i) and so $|[V,t]|=2^8$. This then shows
$[{~I\!\!\!\!C}_V(Z(G_1)),G_1]=V_1$. Since
$O_2(G_1)\not\cong V_1$
as $G_1/O_2(G_1)$--module, we get $V_1={~I\!\!\!\!C}_V(Z(G_1))$ and so $[V,t]\subseteq V_1$ as $t\in O_2(G_1)$. In
particular $[[V,t],O_2(G_1)]=1$. 
Now $[[V,t],\langle O_2(G_1)^{G_2}\rangle]=1$.
But then we have $\langle O_2(G_1)^{G_2} \rangle = {~I\!\!\!\!C}_{G_2}(Z(O_2(G_2)))$, a contradiction.


We are left with $|V_2|=2^8$. Now $V_2=[V,t]$
and the assertion follows with [Str6; Theorem].\\

(iv) We may assume $t\in G_1\setminus O_2(G_1)$.
Suppose there is some $s\in O_2(G_1)$ of type $a_2$ in the natural
representation with $t\sim ts$. Then
$|[V,s]|\leq 2^8$ and the assertion follows with (iii).
So we may assume that there is no such $s$.
Then we see that $t$ has to induce a transvection on $O_2(G_1)$.
So $t$ is not of  type $c_2$. Moreover $t$ is not of type $a_2$, otherwise we are done by (iii). As we may assume that $V\not\cong V(\lambda_1)$,
we see $|{~I\!\!\!\!C}_V(O_2(G_1))| \geq 2^6$ and
so $|[V,Z(G_1)]| \geq 2^6$, i.e. $t$ is not of type $c_1$.
As $t$ has to induce a transvection on $O_2(G_1)$, we see
that $t$ is of type $c_3$. But there are always two commuting elements of type $c_3$ which just differ by an element of type $a_2$. This can easily be
seen in $O_6^+(2) \cong \Sigma_8$ where $c_3$--elements correspond to involutions of type $(12)(34)(56)$
and $a_2$--elements correspond to involutions of type $(13)(24)$.\hfill$\Box$\\[0.3cm]
{\bf (1.11) Lemma.} {\it Let $G\cong 3^7 Sp_6(2)$ and $V$ be a faithful 56-
dimensional $GF(2)G$-module. Let $t$ be an involution of $Sp_6(2)$ of
type $c_1,c_2$ or $a_2$ (on the natural $Sp_6(2)$-module). Then $|V: 
{~I\!\!\!\!C}_V(t)\mid \ge 2^{16}.$}\\[2ex]
{\bf Proof.} By [Sh] we have $V=V_1\bigoplus \cdots \bigoplus V_{28}$, where
$V_i={~I\!\!\!\!C}_V(E_i), E_i$ a hyperplane in $O_3(G)$ with $N_G(E_i)/
O_3(G)\cong O_6^-(2)$. Let $t\in N_G(E_i)$ be of type $c_1$ then $t\in N_G(E_i)
\backslash N_G(E_i)'$. In that case $\mid [V_i,t]\mid = 2$.
As $\mid {~I\!\!\!\!C}_{G/O_3(G)}(t):{~I\!\!\!\!C}_{N_G(E_i)/O_3(G)}(t)
\mid =16$, we have 16 $E_i$'s with $[E_i,t]\subseteq E_i$ and so $\mid [V,t]\mid
\ge 2^{16}$. Let $t\in N_G(E_i)$ be of type $c_2$ or $a_2$ and so $t\in N_G(E_i)'$. Hence $[V_i,t]=1$. As $\mid {~I\!\!\!\!C}_{G/O_3(G)}(t)
: {~I\!\!\!\!C}_{N_G(E_i)/O_3(G)}(t)\mid = 8$ or 4 respectively we get
that
there are at least 20 of the $E_i$'s which are not normalized by $t$.
Since $|V_i|=4$ this yields $\mid [V,t]\mid \ge 2^{20}$. \hfill $\Box$\\[0.3cm]
{\bf (1.12) Lemma.} 
 {\it 
(a)
Let $t \in Sp_{2n}(2)$, $\, n\geq 3,$ be an involution.
Then $Sp_{2n}(2)$ is generated by $2n+1$ conjugates of $t$.

(b) Let $G=\langle P_1,P_2,P_3,P_4\rangle$ be a semiclassical
parabolic system with diagram
{$
\begin{picture}(40,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}
\put(22.5,1){\line(1,0){8.7}}

\put(1,-2){{\scriptsize 1}}
\put(11,-2){{\scriptsize 2}}
\put(21,-2){{\scriptsize 3}}
\put(31,-2){{\scriptsize 4}}

\put(32,1){\makebox(0,0){$\circ$}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(12.5,1.7){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(37,0){.}
\end{picture}
$}
 Let $t$ be an involution in
$O_2(G_1)$. Then there are nine conjugates of $t$ which generate 
{$
\begin{picture}(4,5)

\put(0,0){$G$.}
\end{picture}
$}}\\[0.3cm]
{\bf Proof.} 
(a) Let $G_1$ be the stabilizer of a point in the natural representation. It
is enough to prove that $G_1$ is generated by $2n$ conjugates of $t$. This will be done by induction on $n$.

First we investigate $Sp_4(2) \cong \Sigma_6$. If $t$ is an involution
in $\Sigma_6\setminus A_6$ then $t$ is conjugate to $(12)$
or $(12)(34)(56)$ and we see that $\Sigma_6$ is generated by 5 conjugates.
If $t=(12)(34)$ then 
$\langle (12)(34), (13)(45), (13)(24), (13)(56) \rangle = A_6$.

We have that $O_2(G_1)$ contains three classes of involutions which are
of type $c_1, c_2,\, $ or $\, a_2$ in the natural representation.
As $n\geq3$, all these elements are conjugate into $G_1\setminus O_2(G_1)$,
so we may assume $t\in G_1\setminus O_2(G_1)$.
Now by induction and the remark about $\Sigma_6$, we
get that $G_1/O_2(G_1)$ is generated by $2n-1$ conjugates of $tO_2(G_1)$ or $n=3$ and these conjugates just
generate $A_6$. Now there is $s\in O_2(G_1)\setminus Z(G_1)$ with $t\sim ts$ in $G_1$.
Hence $G_1$ is generated by $2n$ conjugates or $n=3$ and these conjugates just generate $G_1^\prime$. So assume the latter. Choose some conjugate
$t^g\in G\setminus G_1$ and set $H=\langle G_1^\prime, t^g \rangle$. Let
$V$ be the natural module for $G$.
On $V^\#$ we have $G_1^\prime$--orbits of length $1$, $30$, and $32$.
Hence as neither $11$ nor $31$ divides $|Sp_6(2)|$ we
see that $H$ acts transitively on $V^\#$. So 
\[2^8 \cdot 3^4 \cdot 5 \cdot 7 = |G_1^\prime| \cdot 63 \: \Big| \: |H|.\]
As $Sp_6(2)$ is simple we get $H=Sp_6(2)$. So in any case $Sp_6(2)$
is generated by at most $7$ conjugates of $t$. The assertion now
follows by induction.\\

Let now $G$, $t$ be as in (b).
By (1.3)(vii) $G$ has a homomorphic
image $F_4(2)$. Hence there is some conjugate 
$t^g\in G_4\backslash O_2(G_4)$. By (a)
$G_4/O_2(G_4)$ is generated by seven conjugates of $t^g$. As $t^g$ does not
act trivially on $O_2(G_4)/Z(O_2(G_4))$ and $O_2(G_4)/Z(G_4)$ is an indecomposable $G_4$--module, we see that $G_4$ is generated by
eight conjugates of $t$. As any involution in $O_2(G_1)$ is conjugate to
some involution in $ O_2(G_1 \cap G_4)\setminus  O_2(G_1)$, we see that there is some
$t^h\in G_1\backslash O_2(G_1)$. Hence there is even some $t^x \in G_1
\backslash G_1\cap G_4$. But then $G=\langle G_4,t^x \rangle$ and
so $G$ is generated by nine conjugates of $t$.\hfill $\Box$
\\[2ex]
{\bf  (1.13) Lemma.} 
{\it Let $G=F_4(2)$ with diagram 
{$
\begin{picture}(35,3)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}
\put(22.5,1){\line(1,0){8.7}}

\put(1,-2){{\tiny 1}}
\put(11,-2){{\tiny 2}}
\put(21,-2){{\tiny 3}}
\put(31,-2){{\tiny 4}}

\put(32,1){\makebox(0,0){$\circ$}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(12.5,1.7){\line(1,0){8.9}}

\put(34,0){,}
\end{picture}
$}
 and let $G_1,G_4$ be the two maximal parabolics
sharing a Sylow 2-subgroup with $G_1/O_2(G_1)\cong Sp_6(2)\cong G_4/O_2
(G_4)$. Set $\langle i\rangle =Z(G_1)$ and $\langle j\rangle =Z(G_4)$. 
\begin{itemize}
\item[(a)]If $V$ is an irreducible $GF(2)G$-module with quadratically acting group
$A \leq O_2(G_1)$, $|A| \geq 2^5$, then $V$ is one of the two
26-dimensional modules.
\item[(b)] Let $V=V(\lambda_1)$ be one of the two 26-dimensional
irreducible $GF(2)$-modules. Then
\begin{itemize}
\item[(i)] $\mid [V,i]\mid \ge 2^9$ and $[V,i], V/{~I\!\!\!\!C}_V(i)$ both
involve the 8-dimensional spin module for $G_1/O_2(G_1)$ and ${~I\!\!\!\!C}_V
(i)/[V,i]$ involves the natural module. Further
$[V,j]$ and $V/{~I\!\!\!\!C}_V(j)$
are natural modules for $G_4/O_2(G_4)$, the module ${~I\!\!\!\!C}_V(j)/[V,j]$ is
isomorphic to $V(\lambda_2)$, the 14-dimensional irreducible
module for $Sp_6(2) \cong G_4/O_2
(G_4)$, and $[[V,j],O_2(G_4)]=1$. If $x$ is a transvection on $[V,j]$, then 
$|{~I\!\!\!\!C}_V(j) : {~I\!\!\!\!C}_{{~I\!\!\!\!C}_V(j)}(x)| \geq 2^5$.
\item [(ii)] Let $W\subseteq O_2(G_1)$ be a quadratic fours group on
$V$ with $i\in W$, then $W\subseteq Z(O_2(G_1)).$
\item[(iii)] If $W\subseteq O_2(G_1)$ is a quadratic fours group on $V$, then
we have $| W:W\cap Z(O_2(G_1))| \le 2.$
\item [(iv)] If $W\subseteq O_2(G_1)$ is a quadratic fours group with $j\in W$,
then $W\subseteq O_2(G_4)$.
\item [(v)] Let $A\subseteq Z(O_2(G_1))$ be a quadratic group on $V$ such 
that for all $a,b \in A^\sharp$ we have $ | V:{~I\!\!\!\!C}_V(a)| \le 2^6$ and
$| V:{~I\!\!\!\!C}_V(a)\cap {~I\!\!\!\!C}_V(b)| \le 2^7$, then $| A |\le 8.$
\noindent
\end{itemize}
\end{itemize}}
\noindent
{\bf Proof.}
We first prove that (a) is true if $O_2(G_1)$ contains a quadratically
acting fours group $\langle x,y\rangle$ with
 $\langle x,y\rangle \cap Z(O_2(G_1))=1$. There is some element $g \in O_2(G_1)$ 
with $x^g = x$, $y^g=yi$. Hence also $\langle x,i\rangle$ 
acts quadratically. And so $[V,i,\langle x^{{~I\!\!\!\!C}_G(i)}\rangle ]=1$. But
$\langle x^{{~I\!\!\!\!C}_G(i)} \rangle = O_2(G_1)$ and hence the
assertion follows with [Str6].
The same applies for $G_4$.

Set $B=A\cap Z(O_2(G_1))$.
So we may assume $|A:B|\leq 2$. Hence $|B\cap O_2(G_4)|\geq 8$.
Again we may assume $|B\cap O_2(G_4):B\cap Z(O_2(G_4))| \leq 2$.
Hence $4 \leq |B\cap Z(O_2(G_4))| \leq |Z(O_2(G_1)) \cap Z(O_2(G_4))|=4$.
So we get $\langle i,j\rangle = B\cap Z(O_2(G_4))$. Now either $A \not\subseteq
Z(O_2(G_1))$ or $A \not\subseteq
Z(O_2(G_4))$, and so either $O_2(G_1) \subseteq \langle A^{G_1}\rangle$ or 
$O_2(G_4) \subseteq \langle A^{G_4}\rangle$. So either 
$[V,i,O_2(G_1)]=1$ or $[V,j,O_2(G_4)]=1$. Now the assertion follows 
with [Str6].\\

(b) $\;$ (i) We have $[G_1,{~I\!\!\!\!C}_V(O_2(G_1))]=1$ while
$[G_4,{~I\!\!\!\!C}_V(O_2(G_4))] \neq 1$. Further ${~I\!\!\!\!C}_V
(O_2(G_4))$ is a six-dimensional $Sp_6(2)$-module by (1.4) and
the definition of $V(\lambda_1)$. As $G_4$ acts indecomposably
on $O_2(G_4)/\langle j\rangle$, we see that either $[V,j]={~I\!\!\!\!C}_V
(O_2(G_4))$ or $[V,j]$ also involves an eight-dimensional module. But then dim
[$V,j]\ge 14$, which is more than half of the dimension of $V$.

We have ${~I\!\!\!\!C}_V(O_2(G_1))$ is one-dimensional
and so by the same argument as above we see that $[V,i]$ involves an eight-dimensional module, which also shows $| [V,i]| \ge 2^9$ and that $[V,i]$ 
and $V/{~I\!\!\!\!C}_V(i)$ both involve the spin module.

Now $dim({~I\!\!\!\!C}_V(i)/[V,i]) \! \le \! 26 - 2 \cdot 9 = 8$.
So $[{~I\!\!\!\!C}_V(i) , O_2(G_1)] \subseteq [V,i]$.
Assume $[V,j][V,i]$ is $G_1$-invariant. Then also $[[V,j][V,i],O_2(G_1)]$ is
$G_1$-invariant. But the order of $[[V,j],O_2(G_1)]=[V,j]\cap [V,i]$ is $2^5$, a
contradiction. So $\langle [V,j]^{G_1}\rangle /[V,i]$ has to be a nontrivial module
which must involve the natural module by (1.5) because $[[V,j], G_1\cap G_4]\le
[V,i]$.

We have $j$ is conjugate to a transvection $x$ in $G_4 / O_2(G_4)$. So
we see $|[{~I\!\!\!\!C}_V(j)/[V,j] , x]|\leq 2^4$. Now by (1.10)(ii)
${~I\!\!\!\!C}_V(j)/[V,j]$ involves just $V(\lambda_i),$ $\, i=1,2,3$. 
As for the spin module $W=V(\lambda_3)$ we would have 
$|[W,x]|=2^4$ and there are exactly
three $G_1\cap G_4 / O_2(G_1\cap G_4) $ - composition factors in 
${~I\!\!\!\!C}_V(j)/[V,j]$, we see that $V(\lambda_3)$ is not involved.
As $|{~I\!\!\!\!C}_V(j)/[V,j]|=2^{14}$ there cannot be involved three natural modules neither. So we must have ${~I\!\!\!\!C}_V(j)/[V,j] \cong V(\lambda_2)$. Now the assertion (i) concerning the action of $G_1$ and $G_4$ is proved.
The assertion about ${~I\!\!\!\!C}_{{~I\!\!\!\!C}_V(j)}(x)$ follows
from (1.10)(i).
\\

(ii) As just seen we have $[[V,i],O_2(G_1)]\neq 1.$ But as 
$[\langle W^{G_1}\rangle, [V,i]]=1$, we see $\langle W^{G_1}\rangle =Z(O_2
(G_1))$.\\

(iii) Let $W=\langle a,b\rangle$ with $W\cap Z(O_2(G_1))=1$. Then there
is $t\in O_2(G_1)$ with $a^t=a,b^t=bi$. Now also $\langle a,i\rangle$ is a
quadratic fours group, contradicting (ii).\\

(iv) If $W\not\subseteq O_2(G_4)$, then $\langle W^{G_4}\rangle =G_4$
and so $[[V,j],G_4]=1$, a contradiction.\\

(v) Suppose $\mid A\mid \ge 2^4$. As $A\subseteq Z(O_2(G_1))$, we have
for $B=A\cap O_2(G_4)$, $| B| \ge 2^3$. As $\mid Z(O_2(G_1))\cap Z(O_2
(G_4))| =4$, and $i\not\in A$ by (i), we see 
\[| B:B\cap Z(O_2(G_4))
| \ge 4.\]
 By conjugation in $G_1$ we may assume $\langle x \rangle  =
B\cap Z(O_2(G_4))\neq 1$. Let $j\not\in B$. Then we choose $b\in B\backslash
Z(O_2(G_4))$ and $a\in O_2(G_4)$ with $b^a =bj$.
 Hence $| V:{~I\!\!\!\!C}_V
(\langle b,x\rangle)| \le 2^7\ge | V:{~I\!\!\!\!C}_V(\langle 
bj,x\rangle)|$ and so $| V:{~I\!\!\!\!C}_V(\langle j,x\rangle)
| \le 2^8$. As $i\in \langle j,x\rangle$, this contradicts (i). So we 
have $j\in B$. This means that for $u\in Z(O_2(G_1))$ in any case $\langle i
\rangle u\cap A$ just contains one involution conjugate to $j$ or the
intersection is empty. Now by (iv) $A\subseteq \bigcap \limits_{a\in A^\sharp}
O_2({~I\!\!\!\!C}_G(a))$. But this implies that $A$ corresponds to a totally
isotropic subspace in the symplectic six-space $Z(O_2(G_1))/\langle i \rangle$
and so $| A| \le 8$.
\hfill $\Box$\\

Next we show that $G=F_4(2)$ does not possess an $F_1$-module. In fact P.
McClurg [McCl] has shown that there is just one candidate namely
the module $V(\lambda
_1)$. For proving that $V(\lambda_1)$ is not an $F_1$-module for
$F_4(2)$ he embeds $F_4(2)$ in $E_6(2)$ and quotes that $V(\lambda_1)$ for
$E_6(2)$ is not an $F_1$-module, unfortunately the latter is. So we give a proof using
the results of [McCl] concerning the codimensions of centralizers of involutions
which reduces to consider $V(\lambda_1)$.
An independent proof can be found in a forthcoming paper
due to U. Meierfrankenfeld and G. Stroth  where
$F_1$-modules for all Lie type groups are determined.\\[2ex]
{\bf (1.14) Lemma.} {\it $G=F_4(2)$ does not possess an $F_1$-module}.\\[2ex]
{\bf Proof.} As has been said before it is enough to treat the case $V=V(\lambda_1)$. Let
$A$ be an offending  subgroup in $G$, i.e. $1\neq A$ is elementary abelian 
and $| V:{~I\!\!\!\!C}_V(A)| \le 2| A|$. By [GoLySo; Theorem 3.3.3]
 we have $| A
| \le 2^{11}$ and so $| V:{~I\!\!\!\!C}_V(A)| \le 2^{12}$. We adopt
the notation from (1.13).

 Any involution in $O_2(G_1)$ is conjugate to some 
involution in $G_4\setminus O_2(G_4)$, hence we see 
with (1.13)(b)(i) that $|V:{~I\!\!\!\!C}_V(t)| \geq 2^6$
for any involution $t\in G$. This shows $|A|\geq 2^5$.\\

(1) $A \cap Z(O_2(G_4)) \not= 1$.\\

Suppose  that $A \cap Z(O_2(G_4)) = 1$. Then
 $|A \cap O_2(G_4) | \leq 2^4$ and so 
 $|A : A \cap O_2(G_4) | \geq 2$. As 
$[A , {~I\!\!\!\!C}_V( O_2(G_4)) ] \not= 1$ and so 
$|V:{~I\!\!\!\!C}_V(A)| \geq 2^7$, we see 
$|A: O_2(G_4) \cap A| \geq  4$. Now by (1.13)(b)(i) and
(1.7) we get 
\[2 |A| \geq |V : {~I\!\!\!\!C}_V(A)| \geq 4 \cdot 4 \cdot 4 \cdot 
|A: O_2(G_4) \cap A| .\]
Hence $|A \cap O_2(G_4)| \geq 2^5$, a contradiction.\\
 

(2) If $i\in A$, then $A\subseteq O_2(G_1)$.\\

Let $x \in A\backslash O_2(G_1)$. By (1.13)(b)(i) we
have that ${~I\!\!\!\!C}_V(i)$ involves the spin module $W_1$ and the
natural module $W_2$. If $|W_2:{~I\!\!\!\!C}_{W_2}(x)|= 2$, then
$|W_1:{~I\!\!\!\!C}_{W_1}(x)|=16.$
If $|W_2:{~I\!\!\!\!C}_{W_2}(x)|>2$, so
$|W_1:{~I\!\!\!\!C}_{W_1}(x)|\ge 4.$
Thus, if $i\in A$
then by (1.13)(b)(i) $$| V:{~I\!\!\!\!C}_V(A)| \ge | V:{~I\!\!\!\!C}_V(i)| \cdot
| {~I\!\!\!\!C}_V(i):{~I\!\!\!\!C}_V(\langle i,x\rangle )| \ge 2^9
\cdot 2^4=2^{13}$$ which is impossible.\\

(3)  If $j\in A$, then $A\le O_2(G_4)$.\\

Suppose $A \not\subseteq O_2(G_4)$. 
If $|A\cap Z(O_2(G_4))| \geq 2^5$ then there are elements $i_1,i_2 \in A\cap
 Z(O_2(G_4))$,
such that $i_1 \sim i_2 \sim i$ and $i_2 \not\in O_2({~I\!\!\!\!C}_G(i_1))$.
This contradicts (2). So we have $|A\cap O_2(G_4)| \leq 2^8$.
Now $$|V:{~I\!\!\!\!C}_V(A)|= |V:{~I\!\!\!\!C}_V(j)|
|{~I\!\!\!\!C}_V(j) : {~I\!\!\!\!C}_{{~I\!\!\!\!C}_V(j)}(A)|$$
and ${~I\!\!\!\!C}_V(j)$ involves a natural module and $V(\lambda_2)$.
If $A$ induces a transvection on the natural module then the codimension of 
the centralizer in $V(\lambda_2)$ is four. So in any case we get
$|{~I\!\!\!\!C}_V(j) : {~I\!\!\!\!C}_{{~I\!\!\!\!C}_V(j)}(A)| \geq 16 |A/A\cap O_2(G_4)|$ with (1.7). 
So $|A\cap O_2(G_4)| \geq 2^9 $, a contradiction.\\

Now suppose that we have $j\in A$. Then $A\le O_2(G_4)$ by (3) and $| A|
\ge 2^5 \: $. If $A\cap Z(O_2(G_4))=\langle j\rangle $, then $| A
|  = 2^5$ and ${~I\!\!\!\!C}_V(j)= {~I\!\!\!\!C}_V(A)$. But we have $\langle A^{G_4}
\rangle\! =\! O_2(G_4)$ and so $${~I\!\!\!\!C}_V(O_2(G_4))={~I\!\!\!\!C}_V
(\langle A^{G_4}\rangle)=\langle {~I\!\!\!\!C}_V(A)^{G_4}\rangle =\langle {~I\!\!\!\!C}_V
(j)^{G_4}\rangle
={~I\!\!\!\!C}_V(j)$$ in contradiction to ${~I\!\!\!\!C}_V(O_2(G_4))=
[V,j]$.

So $A\cap Z(O_2(G_4))\neq \langle j\rangle$ and by conjugation in $G_4$ we
may assume that $i\in A$. Now (1.13)(b)(i) implies that $| V:{~I\!\!\!\!C}_V
(\langle i,j\rangle )| \ge 2^{10}$, $| A| \ge 2^9$,  and $\,| A
\cap Z(O_2(G_2))| \ge 2^5$. 
Again there are $i_1,i_2 \in A$ with  $i_1\sim i\sim i_2$ but $i_1\not
\in O_2({~I\!\!\!\!C}_{G_2}(i_2))$, a contradiction to (2).

Hence $A$ does not contain any conjugate of $j$. Suppose $i\in A$. Then
by (2) $A\subseteq O_2(G_1)$ and by (1.13)(b) $| A| \ge 2^8$. So
$A\cap Z(O_2(G_1))\neq \langle i\rangle$ and $A$ contains a conjugate of
$j$, a contradiction.

Therefore, $A$ does not contain any conjugate of $i$ or $j$ and $A$ is not
an elementary abelian subgroup of $G$ of maximal rank, i.e. $| A|
\le 2^{10}$. Suppose $ij\in A$.
There is $H \leq G $, $H\cong Sp_8(2)$ such that $Z(O_2(G_1)) \subseteq H$
and $(H\cap G_1)O_2(G_1) = G_1$. Choose $x \in H$ an involution of type $c_2$ on
 the natural module of $H$. Then  $x$ is conjugate to some element of type $c_2$ in $G_1/O_1(G_1)$ and so $|W:{~I\!\!\!\!C}_W(x)| = 2^4$ for the
spin module $W$ of $G_1/O_2(G_1) \cong Sp_6(2)$. 
So by (1.13)(b)(i) we get $|V:{~I\!\!\!\!C}_V(x)| \geq 2^{10}$. We
are going to prove that $ij$ is of type $c_2$.
As $|[V,j]| = 2^6$ and as any involution in 
$Z(O_2(G_1))
 \backslash \langle i \rangle $ is either of type $c_2$ or $a_2$, we get
that  $j$ is of type $a_2$ and then $ij$ is of type $c_2$. So $|V:{~I\!\!\!\!C}_V(ij)| \geq 2^{10}$.
Hence, $| A| \ge 2^9$
and $|{~I\!\!\!\!C}_V(ij):{~I\!\!\!\!C}_V(A)| \! \le \! 2$. Since ${~I\!\!\!\!C}_V
(O_2(G_4)) \subseteq {~I\!\!\!\!C}_V(ij)$ and ${~I\!\!\!\!C}_V(O_2(G_4))$ is the
natural module for $G_4/O_2(G_4)$, we get $| A:A\cap O_2(G_4)| \le 2$.
But now $| A\cap O_2(G_4)| \ge 2^8 $ and $ | A\cap Z(O_2(G_4))| \ge 2^4$.
Then we may assume that $A\cap Z(O_2(G_4))$ contains some $x \in O_2(G_1)\backslash Z(O_2(G_1))$.
So $xij \sim xj$ in $G_1$. Hence $x \not\sim xij$ in $G_4$ and so
$x$ or $xij$ is conjugate to $i$ in $G_4$. 
This  proves
the lemma.\hfill $\Box$\\[2ex]
{\bf (1.15) Lemma}. 
{\it Let $G=\langle P_1,P_2,P_3,P_4\rangle$ be a semiclassical
parabolic sy\-stem with diagram
{$
\begin{picture}(40,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}
\put(22.5,1){\line(1,0){8.7}}

\put(1,-2){{\scriptsize 1}}
\put(11,-2){{\scriptsize 2}}
\put(21,-2){{\scriptsize 3}}
\put(31,-2){{\scriptsize 4}}

\put(32,1){\makebox(0,0){$\circ$}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(12.5,1.7){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(37,1){.}
\end{picture}
$} Let $V$ be a non-trivial $GF(2)G$--module.
\begin{itemize}
\item[(a)] If  for some involution $t \in
O_2(G_1)$ we have $|V  :  {~I\!\!\!\!C}_V(t)|  \leq 2^{12} $, then $[O_3(G_1),V]=1$.
\item[(b)] The module $V$ is not an $F_1$--module.
\end{itemize} }

\noindent
{\bf Proof.} 
We may assume $V$ to be irreducible.
Let $A$ be an offending subgroup for an $F_1$--module $V$. By
[Wie1] and [GoLySo; Theorem 3.3.3] we get that $|A| \leq 2^{11}$, hence
 $| V:{~I\!\!\!\!C}_V(a)| \le 2^{12}$
for any $a\in A^\sharp$. So $| V:{~I\!\!\!\!C}_V(a)| \le 2^{12}$
for some involution $a$ in both (a) and (b).
By (1.12)(b) we may generate
$G$ with 9 conjugates of $a$ and so $| V| \le 2^{108}$.
\begin{itemize}
\item[($*$)] If $[O_3(G_1),V]\not= 1 \not= [O_3(G_4),V]$ then
$|V:{~I\!\!\!\!C}_V(a)|\geq 2^{16}$ for any $a\in O_2(G_1)$.
\end{itemize}
Set $V_4=[V,O_3
(G_4)]$. Then $V_4=\bigoplus_{E \in {\cal H}} {~I\!\!\!\!C}_{V_4}(E)$,
 ${\cal H}$ the set of hyperplanes of
$O_3(G_4)$. Hence if ${~I\!\!\!\!C}_{V_4}(E)\neq 1$, we have at most 54
conjugates of $E$ and so $U=N(G_4(E)/O_{2,3}(G_4))$ must be
isomorphic to a subgroup of index at most 54 in $Sp_6(2)$.
This gives that $U\cong O_6^\pm (2)$,
so $| E^{Sp_6(2)}| =28$ or 36. As $2^3L_3(2)$ is not contained in
$GL_6(3)$ we get that $A_8\cong \Omega_6^+(2)$ cannot act on a hyperplane.
Hence $| E^{Sp_6(2)}| =28$ and so $| V_4|=2^{56}$. As $a$
is conjugate to some $b\in O_2(G_4\cap G_1)\backslash O_2(G_1)$ in $G$ we
see with (1.11) that $| [V_4,b]| \ge 2^{16}$ and so $| [V,a] |\ge
2^{16}$.\\

As $O_3(G_1) \cap O_3(G_4) \not= 1$ and $O_3(G_i)$ is an irreducible $G_i$--module,
either both $O_3(G_i)$, $i=1,4$, act trivially or non-trivially on $V$.
Hence ($*$) proves (a).\\

For (b) suppose first $[O_3(G_1),V]=1$. Then
$G/{~I\!\!\!\!C}_G(V)$ is an $F_4$--amalgam. By [Str3; (2.1)] 
$G/{~I\!\!\!\!C}_G(V) \cong F_4(2)$ But by (1.14) $F_4(2)$ does not possess an $F_1$--module.

Now $[O_3(G_1),V]\not=1$.
Then by ($*$) $A\cap O_2(G_1)=1$. This shows $|A| \leq 2^6$. Now choose
$b\in A^\#$. Then $[b,Z(O_2(G_1))]\not= 1$, so there is an $x\in Z(O_2(G_1))$, 
$1\not= x$, such that $xb \sim b$ in $G$. We get
$$|V:{~I\!\!\!\!C}_V(b)| = |V:{~I\!\!\!\!C}_V(bx)| \leq |V:{~I\!\!\!\!C}_V(A)|
\leq 2^7.$$ This shows $|V:{~I\!\!\!\!C}_V(x)|\leq 2^{14}$, a contradiction to ($*$).\hfill $\Box$\\[3ex]
{\bf (1.16) Lemma.} 
\begin{it}
 Let $G=\langle P_1,P_2,P_3,P_4\rangle$ be a semiclassical
parabolic system with diagram
{$
\begin{picture}(35,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}

\put(12.5,0.4){\line(1,0){8.9}}
\put(12.5,1.7){\line(1,0){8.9}}

\put(32,1){\makebox(0,0){$\circ$}}
\put(22.5,0.4){\line(1,0){8.9}}
\put(22.5,1.7){\line(1,0){8.9}}
\put(26.8,3.5){\makebox(0,0){$\tiny( \scriptstyle{\sim}\tiny) $}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}


\put(1,-2){{\scriptsize 1}}
\put(11,-2){{\scriptsize 2}}
\put(21,-2){{\scriptsize 3}}
\put(31,-2){{\scriptsize 4}}



\end{picture}
$}
and $V$ be a nontrivial $GF(2)G$-
module. Then we have
\begin{itemize}
\item [(i)] $W=O_2(G_4^{(\infty)})$ does not act quadratically
on $V$.
\item [(ii)] For $\langle i\rangle =Z(S)$, $S\in Syl_2(P_1)$ we have
\begin{itemize}
\item[($\alpha$)] There is no fours group $\langle i,x\rangle\subseteq S$ acting
quadratically on $V$.
\item[($\beta$)] $[[V,i],G_1^{(\infty)}]\neq 1.$
\item[($\gamma$)] $| [V,i] | > 2^7$.
\end{itemize}
\end{itemize}
\end{it}
\noindent
{\bf Proof.}  Let $U=S_{G_1}.$ By 
[FuStr] $U$ is an extraspecial group of order $2^{13}$ and $| W\cap
U| = 2^7$.

Suppose that $W$ acts quadratically on $V$. Then also $W\cap U$ acts
quadra\-tically. We have some element $g\in G_1$ with $(W\cap U)^g(W\cap U)=U$. Hence
$| (W\cap U)^g W / W | =2^6$, so $M_{24}$ possesses a quadratic subgroup
of order $2^6$ on some module. But this contradicts [MeiStr]. This is (i).

Suppose (ii)($\alpha$) is false.
Let first $x\not\in U$. Then $[x,U] \not\subseteq \langle i \rangle$.
Hence by [Wie4] $x$ does not induce a transvection
on $U/\langle i \rangle$. So there is a $g\in U$
with $x^g=xy\not=x$, $y^2=1$, and $y\not\in \langle i \rangle$.
In particular, $\langle i, xy \rangle$ acts quadratically and so
$\langle i,y\rangle$ acts quadratically. 
So without loss we may assume $\langle i,x\rangle \leq U$.
Hence $\langle x^{G_1} \rangle = U$ and so $[V,i,U]=1$.
This shows that any fours group in $U$ containing $i$ acts quadratically.
Set $R=\langle P_2,P_3 \rangle$. 
 Then $R$ induces on $(W\cap U)/\langle i\rangle$ the six-dimensional
$3\Sigma_6$-module and on $W/W\cap U$ the four-dimensional $\Sigma_6$-module.
If $y\in W\backslash W\cap U$, we have $| U/\langle i\rangle : 
{~I\!\!\!\!C}_{U/\langle i\rangle}(y)| =16$. Hence all $R$-orbits on
$W\backslash W\cap U$ have length divisible by 240. The $G_4$-orbits on $W^\sharp$ have length 1771 and 276. In particular there
are representatives for both orbits in $W\cap U\backslash \langle i\rangle$.
So choose $j\in W\cap U$ such that
 ${~I\!\!\!\!C}_{G_4}(j)$ involves $Aut(M_{22})$. We know
that $\langle i,j \rangle$ acts quadratically.
So $[V,j,\langle i^{{~I\!\!\!\!C}_{G_4}(j)} \rangle ]=1.$
Since $W/\langle j \rangle$ is an irreducible 
${~I\!\!\!\!C}_{G_4}(j)$-module we get 
$\langle j, i^{{~I\!\!\!\!C}_{G_4}(j)} \rangle =W.$
So $[V,j,W]=1$. But now $[V,\langle j^{G_4} \rangle ,W]=1$.
As $\langle j^{G_4} \rangle =W$ we get $[V,W,W]=1$ contradicting
(i). So ($\alpha$) is proved.

Now ($\beta$) and ($\gamma$) follow from the fact that $[V,i,U]\not= 1$
by ($\alpha$) and that
$G_1$ acts irreducibly on $U/\langle i \rangle$. This is (ii).\hfill $\Box$\\[2ex]
{\bf (1.17) Lemma.} {\it Assume (1.16). Then $V$ involves more than one 
nontrivial irreducible $G_4$-factor}.\\[2ex]
{\bf Proof.} Set $V_1= {~I\!\!\!\!C}_V(W)$ and $H_4=G_4^\prime$. If we have $[V,H_4]\le V_1$, then $[V,W,W]\le [V,H_4,W]\le [V_1,W]=1$,
contradicting (1.16)(i). Hence it must be $[V,H_4]\not\le V_1$. Assume now that $V$
involves exactly one nontrivial irreducible $H_4$-module. Then $[V_1,H_4]=1$,
and as ${H_4}'=H_4$, we see ${~I\!\!\!\!C}_{V/V_1} (H_4)=1$. So let $V_2/V_1$ be 
the nontrivial irreducible $H_4$-module in $V$. As $[W,V] \not\subseteq V_1$
by (1.16)(i), we see $V_2/V_1=[W,V]V_1/V_1$. In particular $V_2/V_1\cong
W$ as ${H_4}/W$-module. As $H^1(M_{24},W)$ is one-dimensional (see [Asch2;  22.8])
we see $| V:V_2
| \le 4$. Let $i\in W$ be as in (1.16)(ii). Then ${~I\!\!\!\!C}_{H_4}(i)/W
\cong 2^6\cdot 3\Sigma_6$. Let $\varphi \in {~I\!\!\!\!C}_{H_4}(i), o
(\varphi)=3$, with $| [\varphi,W]| = 2^6$. Then also $| [\varphi,
V]| = 2^6$ and $[\varphi, V]\cap V_1=1$. As $[V_2,W]\le V_1$, we have
$[[\varphi, V],i]\subseteq V_1\cap [\varphi, V]=1$. This shows $|
{~I\!\!\!\!C}_V(i):V_1| \ge 2^6$. As $| V:V_1| \le 2^{13}$, we see
$| V:{~I\!\!\!\!C}_V(i)| \le 2^7$, contradicting (1.16)(ii)($\gamma$).\hfill $\Box$
\\[2ex]
{\bf (1.18) Lemma.} {\it Assume (1.16).  Then there exists no subgroup $A\neq 1$
in $O_2(G_4)$ with $| V:{~I\!\!\!\!C}_V(A)|
\le 2| A|$}.\\[2ex]
{\bf Proof.} Suppose there is some $1\neq A\le O_2(G_4)$ with $| V:{~I\!\!\!\!C}_V(A)| \le 2| A|$. Suppose
further $i\in A$. By (1.16)(ii) $(\gamma)$ we have $| [V,i]| \ge 2^8$. Set
$H_1=G_1^{(\infty)}.$ We have $W\le H_1$. By (1.16)(i) we know $[[V,i],W]
\neq 1$ and so $[[V,i],H_1]\neq 1$. As $| A| \le 2^{12}$, we have
$| [V,i]| \le 2^{13}$. If $| [V,i]| = 2^{13}$, then $A=O_2
(G_4)$ and ${~I\!\!\!\!C}_V(A)={~I\!\!\!\!C}_V(i)$
is normal in $\langle P_1,P_2,P_3,P_4\rangle =G$, a contradiction. 

We have $| [V,i]| \le 2^{12}$ and so as $H_1$ acts irreducibly on
$U/\langle i\rangle, U=S_{H_1}$, we see $[[V,i],U]=1$,
contradicting (1.16)(ii)($\alpha$).

The $G_4$--orbits on $W^\#$ have lengths 1771 and 276 and now, as $A$ does
not contain any $G_4$--conjugate of $i$,  all
elements of $(A\cap W)^\sharp$ are in the orbit of length 276, in particular
$| A\cap W| \le 2^8$. If $| A\cap W| = 2^8$, choose $g\in
G_4$ with $(A\cap W)^g\neq A\cap W$. Then $| A
\cap W:A\cap A^g\cap W| \ge 2$ and so 
\[| (A\cap W)^g\cup (A\cap W)|
\ge 1+2\cdot 255-127=384>276+1,\]
 a contradiction. Hence we have $| A\cap
W| \le 2^7$ and so $| A| \le 2^8$. In particular $| [V,j]| \le
2^9$ for any $j\in A\cap W$. So we have: 
 If $A\cap W\neq 1$, then there is some $j\in A^{\sharp}$ such that
${~I\!\!\!\!C}_{G_4}(j)$ involves $M_{22}$ and
$| [V,j]| \le 2^9.$
Let now $\tilde {H}_{4}={~I\!\!\!\!C}_
{G_4}(j)$.We have $\tilde {H}_{4}/W\cong Aut(M_{22})$. 
As $W/ \langle j\rangle$ is an irreducible $\tilde {H}_{4}$-module
 and $| [V,j]| <  |  W/{\langle j\rangle} |$, we see
$[W,[V,j]]=1$. As $W=\langle j^g \, | \, g\in G_4 \rangle$,
we get $[V,W,W]=1$ contradicting (1.16)(i). 

So we are left with $A\cap W=1$, i.e. $| A| = 2$ and $| V:{~I\!\!\!\!C}_V(A)| \le 4$. Let $\langle a\rangle =A$. Then there is some conjugate
$ab \neq a$ with $b\in W$. Hence $| V:{~I\!\!\!\!C}_V(b)|\le 16$. By 
(1.16)(ii)($\gamma$) $b\not\sim i$ in $G_4$. So $b\sim j$ and we get
again $[V,W,W]=1$, a contradiction.\hfill $\Box$\\[2ex]
{\bf (1.19) Lemma.} {\it Assume (1.16). Set $U=S_{G_1}$. Then for $1\neq A
\subseteq U$, $A$ elementary abelian, we get $| V:{~I\!\!\!\!C}_V(A)|
>| A|$}. \\[2ex]
{\bf Proof.} Suppose there is some $A$ with $| V:{~I\!\!\!\!C}_V(A)|\le 
| A|$. We have $| A| \le 2^7$, as $A$ is elementary abelian.
In particular $| [V,t]| \le 2^7$ for any $t\in A^{\sharp}$. There is 
some element
$g\in G$ with $t^g\in G_4 \setminus O_2(G_4)$. By [Par; (3.7)] we have $| [\tilde{V},t^g]| \ge 2^4$ for
any nontrivial irreducible $G_4/O_2(G_4)$-module $\tilde{V}$ involved in $V$. This shows that $V$
involves exactly one nontrivial irreducible $G_4/
O_2(G_4)$-module, contradicting (1.17).\hfill $\Box$\\[2ex]
{\bf (1.20) Lemma.} {\it Let $G=L_5(2)$ with diagram
{$
\begin{picture}(35,4)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(32,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}
\put(12.5,1){\line(1,0){8.7}}
\put(22.5,1){\line(1,0){8.7}}
\put(1.3,-1.7){\tiny 1}
\put(11.3,-1.7){\tiny 2}
\put(21.3,-1.7){\tiny 3}
\put(31.3,-1.7){\tiny 4}
\end{picture}
$} 
and module $V=V(\lambda_2)$. Let $t\in G$ be an involution. If $t$ is a transvection in the natural representation, then
$|[V,t]|=8$. Else $|[V,t]|=16$. If $t_1 \not= t_2$ are transvections in the natural representations then 
${~I\!\!\!\!C}_V(t_1)\not= {~I\!\!\!\!C}_V(t_2)$.}\\[0.3cm]
{\bf Proof.} We have that $V(\lambda_2)$ is the exterior square of the natural
module (see [GoLySo; Example 1.14.9 (e)]). 
So $|V|=2^{10}$. Further by (1.4) we have that
$V_4={~I\!\!\!\!C}_V(O_2(G_4))$ is the orthogonal $G_4/O_2(G_4)$--module.
Hence $|V:V_4|=16$ and so $|[V,t]|\leq 16$ for $t\in O_2(G_4)$.
As $t$ is centralized by some $L_3(2)$ in $G_4$ which
itself has two nontrivial irreducible modules in $V_4$, we see
$|[V,t]|=8$. Let now $t\in G$ not be a transvection.
Then $t$ inverts some element $\omega$ of order $5$. 
As $|[V,\omega]|=2^8$, we
get $|[V,t]|\geq 16$. We may assume $\omega, t \in G_4\setminus O_2(G_4)$.
Then $|[V_4,t]|=4$ and $|[V/V_4,t]|=4$. Hence
$|{~I\!\!\!\!C}_V(t)|\geq64$. This shows $|[V,t]|=16$.

Suppose ${~I\!\!\!\!C}_V(t_1) = {~I\!\!\!\!C}_V(t_2)$.
We may assume $t_1 \in Z(G_1 \cap G_4)$.
Then $X=\langle {~I\!\!\!\!C}_V(t_1),{~I\!\!\!\!C}_V(t_2) \rangle \cong G_1, G_4$, or $G$
 as it has to be a parabolic subgroup.
So $X$ acts on ${~I\!\!\!\!C}_V(t_1)$ and 
$V/{~I\!\!\!\!C}_V(t_1)$. Since the latter is of order $8$ this is a contradiction. \hfill $\Box$\\

For the convenience of the reader we collect some well known basic results
on parabolic systems.\\[2ex]
{\bf (1.21) Lemma.} {\it Let $G=\langle G_1,G_2\rangle, K_i \unlhd G_i,i=1,2,$
and $K_1\subseteq G_2,K_2\subseteq G_1$. Suppose $(G_1\cap G_2)_G=1$. Let
$p$ be a prime number such that $p\not\;\!\!\;\!\!|$ $| K_1K_2/K_1|$ and $p
\not\;\!\!\!\!\; |$ $| 
K_1
K_2/K_2|$, then $p\not\;\!\!|$ $| K_1K_2|$}.\\[2ex]
{\bf Proof.} [Str3; (1.14)]\hfill $\Box$\\[2ex]
{\bf (1.22) Lemma.} {\it Let $\{P_1,\ldots,P_n\}$ be a parabolic system with
$B/B_{P_i}$ is a 2-group for all $i$. Then $B$ is a 2-group}.\\[2ex]
{\bf Proof.} [Str3; (1.15)]\hfill $\Box$\\

The next lemma is basically [StrWo; (2.1),(2.2)]. But as the assumptions
differ a little bit see [Wie1; 1.4] for a proof.\\[2ex]
{\bf (1.23) Lemma} {\it Let $G=
\langle G_1,G_2\rangle$. Set $B=G_1\cap G_2$ and $K_i=B_{G_i}$ for i=1,2.
Assume $B_{G_1}$ is a 2-group and the following holds.
\begin{itemize}
\item[(i)] There is $S\subseteq G_1\cap G_2, S\in Syl_2(G_2)$ and set $Z_i
=\langle \Omega_1(Z(S))^{G_i}\rangle,i=1,2$, then $Z_1\subseteq K_1$
and moreover any 2-element in $G_1 \cap G_2$ centralizing $Z_1$ is in $K_1$.
\item[(ii)] $| G_2:G_1\cap G_2| =3, G_2$ is finite and $Z_2\le O_2
(G_2)=S_{G_2}$.
\item[(iii)] $B_G=1$.
\end{itemize}
Then $Z_1$ is an $F_1$-module for $G_1$ with offending subgroup in $S_{G_2}K_1/ K_1,$
or for $\varphi \in \langle K_1^{G_2}\rangle, \varphi^3\in G_1\cap G_2$, we
have $| [\varphi, O_2(G_2)]| =4$ and $| [O_2(\langle S,
\varphi\rangle),\varphi]| =4$}.\\[3ex]
{\bf (1.24) Lemma ([Wie1; 1.5]).} {\it Let $\{P_1,\ldots ,P_n\}$ be a semiclassical parabolic
system. Assume (1.23) with $G_2=P_1$ and $G_1=\langle P_2,\ldots ,P_n\rangle$.
Let further $\langle P_1,P_2\rangle/B_{\langle P_1,P_2\rangle}$ be a rank two Lie 
group or
$3\Sigma_6,3A_6$ and $B_{\langle P_1,P_2\rangle}$ contain a Sylow 2-subgroup
of ${~I\!\!\!\!C}_
{\langle P_1,P_2\rangle}(B_{\langle P_1,P_2\rangle})$. Then $Z_1$ is an $F_1$--module with offending subgroup in $S_{G_2}.$}
\\[2ex]
\begin{center}
\boldmath \S\unboldmath {\bf 2 Extension of {$
\begin{picture}(40,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
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\put(22.5,1){\line(1,0){8.7}}


\put(32,1){\makebox(0,0){$\circ$}}
\put(12.5,0.4){\line(1,0){8.9}}
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\put(37,1){.}
\end{picture}
$}}
\end{center}
\vspace{1cm}

In this chapter we will assume that $\{P_1,\ldots, P_n\}$ is a semiclassical
parabolic system involving an $\tilde{F}_4$-subdiagram. 
The aim is to show
that $n=4$. Hence it is enough to show that there is no system $\{P_1,P_2,
P_3,P_4,P_5\}$ that involves an $\tilde{F}_4$-subdiagram. 
We choose notation such that $P_2,P_3,P_4$, and $P_5$ generate the
$\tilde{F}_4$-group. We recall that
$G_{1,2}/K_1 \cong G_{1,5}/K_1 \cong 2^{1+6+8}3^7Sp_6(2)$ and 
$G_{1,3}/K_{1,3} \cong G_{1,4}/K_{1,4} \cong L_3(2) \times \Sigma_3.$\\[2ex] 
{\bf (2.1) Lemma.} {\it There is no system with diagram $\Delta$ of shape
\begin{center}
{$
\begin{picture}(46,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}
\put(12.5,1){\line(1,0){8.7}}

\put(32.5,1){\line(1,0){8.7}}

\put(1,-2){{\scriptsize 1}}
\put(11,-2){{\scriptsize 2}}
\put(21,-2){{\scriptsize 3}}
\put(31,-2){{\scriptsize 4}}
\put(41,-2){{\scriptsize 5}}

\put(32,1){\makebox(0,0){$\circ$}}
\put(42,1){\makebox(0,0){$\circ$}}

\put(22.5,0.4){\line(1,0){8.9}}
\put(22.5,1.7){\line(1,0){8.9}}
\put(27,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(44,1){.}
\end{picture}
$}
\end{center}}
\noindent
{\bf Proof.} 
By [ShStr] we have $G_5/K_5 \cong 3^{35}Sp_8(2)$. 
By (1.22) and (1.2) we have that $B$ is a 2-group and so $K_i$ is a 2-group
for all $i$. Assume first ${~I\!\!\!\!C}_S(K_i)\subseteq K_i$ for $i=1$
and 5. By (1.15) $G_1/K_1$ does not possess an $F_1$-module. Application of (1.24) shows $\Omega_1(Z(S))=
Z_1$.

 We have $G_2/K_2
\cong \Sigma_3\times 3^7Sp_6(2)$ and $|G_2:G_2\cap G_1| = 3$. 
Now we investigate the coset graph $\Gamma(G_1,G_2)$. Let $(2,\alpha)
$ be a critical pair. Assume $\alpha \sim 2$. According to  [StrWo; (1.8)]
we may choose notation such that $Z_2$ is an $F$-module
 with offending subgroup $Z_\alpha/{~I\!\!\!\!C}_{Z_\alpha}
(Z_2)$. Since $Z(G_2)=1$ we get $|Z_2|=4$. Set $V_1=\langle {Z_2}^{G_1}
\rangle$. As $[Z_2,K_1]\subseteq Z_1$ we have $[V_1,K_1] \subseteq Z_1$.

 Assume first $V_1'=1$. Let $\beta\in \Delta(2)$ with $d(\beta,
\alpha)=d(2,\alpha)-1$. Then $V_\beta\subseteq G_\alpha$ but $V_\beta
\not\subseteq K_\alpha$. Hence $|V_\beta:V_\beta \cap G_{\alpha+1}|=2$. As
$Z_\beta = [Z_2,Z_\alpha]$ we have $[V_{\alpha+1},Z_\beta]=1$ and so
$V_{\alpha+1}\subseteq G_\beta.$ Now $[V_\beta \cap K_\alpha, V_{\alpha+1}]
\subseteq V_\beta$. As $[Z_2,Z_{\alpha+1}]\neq 1$ and $[V_\beta \cap
K_{\alpha+1},V_{\alpha+1}]\subseteq Z_{\alpha+1}$,
we get $[V_\beta \cap K_{\alpha+1},V_{\alpha+1}]=1$. Hence for $t\in V_
{\alpha+1}$, we get $|[V_\beta,t]|\le 2^{12}$. Suppose $V_{\alpha+1}
\subseteq K_\beta$. Then, as $Z_\beta \subseteq V_{\alpha+1}$ we get $[V_\beta,
V_{\alpha+1}]\subseteq V_{\alpha+1}$ and so $V_{\alpha+1}\lhd \langle V_\beta,
G_\alpha \cap G_{\alpha+1}\rangle =G_\alpha$, a contradiction.

So we have
$V_{\alpha+1}\not\subseteq K_\beta$. Now the application of (1.15)(a) shows that $G_\beta
/K_\beta{~I\!\!\!\!C}_{G_\beta}(V_\beta)$ is an $F_4(2)$-amalgam and so by
[Str3; (2.1)] is isomorphic to $F_4(2)$. By [Coop] we have $|[V_\beta,t]| \ge
2^6$ and so $|V_\beta \cap G_{\alpha+1}:V_\beta \cap K_{\alpha+1}|\ge 2^5$.
If there are two nontrivial irreducible modules involved, then
$|[V_\beta,t]|=2^{12}$ and ${~I\!\!\!\!C}_{V_\beta}(t)=
{~I\!\!\!\!C}_{V_\beta}(V_{\alpha+1})$ for all $t\in V_{\alpha+1}\backslash 
K_\beta$.
But this would imply that one of them is an $F_1$-module, which contradicts (1.14).
 So
we have that there is just one nontrivial irreducible module involved.
Now as $V_\beta \cap G_{\alpha + 1}$ acts quadratically on $V_{\alpha+1}$ we get
with (1.13)(a) that $V_{\alpha+1}$ and so also $V_\beta$ just involves 
a 26-dimensional and trivial $F_4(2)$-modules.
Let $V$ be the 26-dimensional  module. As $V_1=\langle 
Z_2^{G_1}\rangle$ and $|Z_2|=4$, we see that $$V_\beta/{~I\!\!\!\!C}_{V_\beta}
(G_\beta/K_\beta)\cong V \cong V_{\alpha + 1}/{~I\!\!\!\!C}_{V_{\alpha + 1}}
(G_{\alpha + 1}/K_{\alpha + 1}).$$ Now ${~I\!\!\!\!C}_V(G_{\alpha+1} \cap G_\alpha/K_
{\alpha+1})\neq 1$ and so this module is one-dimensional.
 Hence for any $i
\in Z(G_{\alpha+1} \cap G_\alpha/K_{\alpha+1})$ with $i\neq 1$ we have by (1.13)(b) $[[V,i],O_2(G_
{\alpha+1}\cap G_\alpha/K_{\alpha+1})]\neq 1$. As $V_\beta \cap G_{\alpha+1}$
acts quadratically we see with (1.13)(b)(ii),(iii) that $|V_\beta \cap G_{\alpha+1}:
V_\beta \cap K_{\alpha+1}|\le 2^7$. Suppose equality.
 Then by (1.13)(b)(iv) $(V_\beta \cap G_{\alpha+1})K_{\alpha+1}/K_{\alpha+1} \cap Z
(O_2(G_{\alpha+1} \cap G_\alpha/K_{\alpha+1}))$ does not cover $Z(O_2(G_{\alpha+1}
\cap G_\alpha/K_{\alpha+1}))$ modulo $i$ which is impossible by (1.13)(b)(ii),(iii).
 Hence $|V_\beta \cap G_{\alpha+1}:
V_\beta \cap K_{\alpha+1} | \le 2^6$. Now as $V$ is not an $F$-module, we
have $|V_{\alpha+1}:V_{\alpha+1} \cap K_\beta| \le 2^6$, too. Hence by (1.13)(b)(i) $i\not\in (V_\beta \cap G_{\alpha+1})K_{\alpha+1}/K_{\alpha+1}$. Set $$U=(V_\beta \cap G_{\alpha+1})K_{\alpha+1}/K_{\alpha+1}
\cap Z(O_2(G_{\alpha+1} \cap G_\alpha/K_{\alpha+1})).$$
Suppose that $|V_\beta \cap
G_{\alpha+1}:V_\beta \cap K_{\alpha+1}|=2^6$. Then for a suitable $s$ we get $(V_\beta \cap G_{\alpha+1})K_{\alpha+1}/K_{\alpha+1}=U\langle
s\rangle$. Now there is some $s_1\in O_2(G_{\alpha+1} \cap G_\alpha/K_
{\alpha+1})$ with $s^{s_1}=si$. So also $\langle i\rangle U$ acts quadratically
on $V$. As any coset of $\langle i\rangle$ in $\langle i\rangle U$ contains some conjugate
of the element $j$ in the notation of (1.13)(b), we see that there is some
conjugate $j^g$ such that $\langle i\rangle U\not\subseteq O_2({~I\!\!\!\!C}_
{G_{\alpha+1}/K_{\alpha+1}}(j^g))$, contradicting (1.13)(b)(iv).

So we have $|V_\beta \cap G_{\alpha+1} : V_\beta \cap K_{\alpha+1}|=2^5$.
Now $|V_{\alpha+1}:V_{\alpha+1} \cap K_\beta |\le 2^5$.
By (1.13)(b)(iii) we have $|U|\geq 16$ but by (1.13)(b)(v) we would get
 $|U|\leq 8$, a contradiction.

So let $V_1'\neq 1$. Then $b=2$ and $V_1=Z(V_1)X$ where X is extraspecial
of width $n$ for some $n$. Obviously $n\ge 13$. Let $g\in G_2$ with $Z_1\neq Z_1^g\le
Z_2\le V_1$. Then $|X^g:X^g\cap K_2|=2$. 
As $[K_1^{\prime}, V_1]=1 $ we get $Z_1^g\not\subseteq K_1'$.
In particular $X^g\cap K_1$ is
elementary abelian and so
$|X^g\cap G_1:X^g\cap K_1|\ge 2^{12}$. 
On the other hand $Z_1^g \subseteq Z_2 \subseteq K_1$ implies
 $(X^g\cap G_1/X^g\cap K_1)'=1$,
a contradiction as $O_2(G_1\cap G_2/K_1)$ does not contain elementary abelian
subgroups of order $2^{12}$.

So we are left with $\alpha \sim 1$. Then there is a fours group $\tilde{Z}_2
$ in $Z_2$ with $\tilde{Z}_2\lhd G_2$ and $\tilde{Z}_2\not\subseteq K_\alpha$.
Let first $b>1$. Set again $V_1=\langle \tilde{Z}_2^{G_1} \rangle$. Let
$\beta\in \Delta(2)$ with $d(\beta,\alpha)=b-1$. Then $[V_\beta,V_\alpha]
\subseteq V_\beta \cap V_\alpha$ and again $|V_\alpha:{~I\!\!\!\!C}_{V_\alpha}
(t)|\le 2^{12}$ for $t\in V_\beta$. As above $G_1/K_1{~I\!\!\!\!C}_{G_1}(V_1)
\cong F_4(2)$. Further $V_1$ involves just one nontrivial irreducible $F_4(2)$
- module, the 26-dimensional one. As $V_\beta$ acts quadratically on $V_\alpha
$, we again see with (1.13)(b) that $|V_\beta:V_\beta \cap K_\alpha | \le 2^6$ 
and a central involution $i$ in $G_{\alpha-1} \cap G_\alpha/K_\alpha$ is
not in $V_\beta$. By symmetry we  also have $|V_\alpha :V_\alpha \cap
K_\beta |\le 2^6$, and therefore, again by (1.13)(b)(v), we can conclude that $|V_\beta:V_\beta \cap
K_\alpha |=2^6=|V_\alpha:V_\alpha \cap K_\beta |$ and $V_\beta K_\alpha/
K_\alpha \not\subseteq Z(O_2(G_{\alpha-1} \cap G_\alpha/K_\alpha))$. But
then we may argue as above to get a quadratically acting subgroup in $Z(O_2
(G_{\alpha-1}\cap G_\alpha/K_\alpha))$ of order at least $2^6$, which again contradicts (1.13)(b)(iv).

Finally let $b=1$. Then $Z_2\not\subseteq K_1$ and $\tilde{Z}_2$
induces a transvection on $K_1$. By (1.12)(b) $G_1/K_1$ centralizes a subgroup of
index $2^9$ in $K_1$, contradicting ${~I\!\!\!\!C}_S(K_1)\subseteq K_1$.

So assume now ${~I\!\!\!\!C}_S(K_1)\not\subseteq K_1$, hence $G_1=K_1
{~I\!\!\!\!C}_{G_1}(K_1)$. As $K_1K_5/K_5$ is in the center of $G_1 \cap G_5/
K_5$, we see $|K_1K_5/K_5|\le 2$. Now from $G_5/K_5$ we see $G_1\cap G_5/K_1K_5\cong 3^72^7
Sp_6(2)$ or $3^72^6Sp_6(2)$, but there are no such factors of $G_1\cap G_5/K_1$.
So we have ${~I\!\!\!\!C}_S(K_1)\subseteq K_1$. As $G_1\cap G_5/K_1$
is 2-constrained we get from ${~I\!\!\!\!C}_S(K_5)\not\subseteq K_5$,
that $K_5 \subseteq K_1$. 
But $|S:K_5|=2^{16}$, while $|S:K_1|=2^{24}$. $\hfill\Box$ \\[2ex]
{\bf (2.2) Lemma.} {\it There is no semiclassical parabolic system with
diagram $\Delta$ of type 

\begin{center}
{$
\begin{picture}(25,15)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(2,11){\makebox(0,0){$\circ$}}
\put(12,11){\makebox(0,0){$\circ$}}
\put(17,6){\makebox(0,0){$\circ$}}

\put(-1.5,-0.2){\scriptsize 5}
\put(12.5,-0.2){ \scriptsize 1}
\put(-1.5,11.8){\scriptsize 4}
\put(13,11.8){\scriptsize 3}
\put(18.5,5.5){\scriptsize 2}

\put(2.5,1){\line(1,0){8.9}}

\put(2,1.6){\line(0,1){8.7}}




\put(2.5,11.4){\line(1,0){8.9}}
\put(2.5,10.6){\line(1,0){8.9}}

\put(7,13.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(12.5,1.5){\line(1,1){4}}
\put(12.5,10.5){\line(1,-1){4}}

\end{picture}
$.}
\end{center}}
\noindent       
{\bf Proof.} By (1.2) we see that $P_i/B_{P_i}\cong \Sigma_3$ for all $i$. So by (1.22) $B$
is a 2-group.

Assume first ${~I\!\!\!\!C}_S(K_i)\subseteq K_i$ for all $i$. 
Let $Z_1\le Z(G_1)$.
Then by (1.24) $Z_3$ and $Z_4$ both are $F_1$-modules. As $G_1
\cap G_i$ is the centralizer of $\Omega_1(Z(S))$ in $G_i$, for $i= 3,4$,
we see with (1.5) that 10-dimensional $L_5(2)$-modules
 $V(\lambda_2)$ or $V(\lambda_3)$ are involved. Since by (1.20) on this module $V$
we have $|[V,t]| = 2^3$ for transvections in $L_5(2)$
and $|[V,t]| = 2^4$ else, we see that just one such module is involved.
As $Z_3$ is generated by conjugates of $\Omega_1(Z(S))$, we see with a result
of Gasch\"utz [Hu; I 17.4] that $|Z_3|=2^{10}=|Z_4|$. Hence $|Z_3\cap
Z_4|=2^6$.

Let $\Gamma=\Gamma(G_3,G_4)$ be the coset graph and $b=b_\Gamma$. Let $(\alpha,
\beta)$ be a critical pair in $\Gamma$. Choose $\delta \in \Delta (\alpha)$
and $\gamma \in \Delta (\beta)$ with $d(\gamma,\alpha)=b-1$. By (1.20) we have
$|Z_\alpha K_\beta/K_\beta |=2^4=|Z_\beta K_\alpha/K_\alpha |$ and so $Z_
\alpha K_\beta/K_\beta=O_2(G_\gamma \cap G_\beta/K_\beta)$. In particular $
Z_\alpha \cap Z_\beta \supseteq
[Z_\alpha,Z_\beta]={~I\!\!\!\!C}_{Z_\beta}(O_2(G_\gamma \cap G_\beta))=Z_\beta
\cap Z_\gamma $.

 Assume $b>1$. Then
$[Z_\alpha,Z_\delta]=1$. Thus $[Z_\delta, Z_\beta \cap Z_\gamma]\subseteq
[Z_\delta,Z_\alpha]=1$. This
means $Z_\delta \subseteq {~I\!\!\!\!C}_{G_\gamma}(Z_\beta \cap Z_\gamma)
\subseteq G_\gamma \cap G_\beta \subseteq G_\beta$. Now $Z_\delta Z_\alpha
K_\beta=Z_\alpha K_\beta$ and so $[Z_\delta, Z_\beta]\subseteq [Z_\alpha, 
Z_\beta]\subseteq Z_\alpha$. Hence $Z_\alpha Z_\delta \lhd \langle G_\alpha
\cap G_\delta, Z_\beta \rangle$. Choose $\delta$ with $Z_\beta \not\subseteq
G_\delta$, then we have $Z_\alpha Z_\delta \lhd G_\alpha$. But then $Z_\alpha \cap Z_\delta=[Q_\alpha,
Z_\alpha Z_\delta] \lhd G_\alpha$, a contradiction.

So assume now $b=1$. Then $Z_3\not\le K_4$ and as $O_2(G_3 \cap G_4/K_4)$ is
elementary abelian, we have $Z_3\not\subseteq \Phi (K_3).$ As $Z_3$ is an
irreducible $G_3$-module, we have $Z_3\cap \Phi 
(K_3)=1$. Then $K_3$ is elementary
abelian. The same argument also shows that $K_4$ is 
 elementary
abelian. Now $[K_4,K_3]=[Z_4,Z_3]=Z_4\cap Z_3$. In particular $[G_3,K_3]=
Z_3$. Let $L$ be the $L_3(2)$ involved in $G_3\cap G_1$. Then from $G_3$, we
see that $L$ induces at most five 3-dimensional irreducible modules in $O_2
(G_1\cap G_3)$, while from $G_1/K_1$ in which $G_1\cap G_3/K_1$ corresponds
to a parabolic subgroup
in $F_4(2)$, we see that there are at least six such $L$-
modules involved, a contradiction.

So assume next $Z_1\not\le Z(G_1)$. Application of (1.24) shows that $Z_1$
is an $F_1$-module contradicting (1.15)(b).

So we are left with ${~I\!\!\!\!C}_S(K_i)\not\subseteq K_i$ for some $i$. If
$i=3$, then because of  $Z(G_3\cap G_4/K_4)=1$, we see $K_3=K_4=1$, contradicting
$|S|>2^{10}$. Hence ${~I\!\!\!\!C}_S(K_3)\subseteq K_3$ and $ {~I\!\!\!\!C}
_S(K_4)\subseteq K_4$. Let ${~I\!\!\!\!C}_S(K_2)\not\subseteq K_2$. Then
$|K_2K_1/K_1|\le 2$ and we get a contradiction to $|S:K_2|=2^{16}$ and
$|S:K_1|=2^{24}$. So also ${~I\!\!\!\!C}_S(K_2)\subseteq K_2$ and ${~I\!\!\!\!C}
_S(K_5)\subseteq K_5$. It remains ${~I\!\!\!\!C}_S(K_1)\not\subseteq K_1$.
Then again $|K_1K_2/K_2|\le 2$. But $G_1\cap G_2/K_1$ does not possess a
factor group isomorphic to $2^6Sp_6(2)$. $\hfill\Box$\\[2ex]
{\bf (2.3) Lemma}. {\it There is no semiclassical parabolic system with diagram
$\Delta$ of shape
\begin{center}
{$
\begin{picture}(35,12)
\put(19,1){\makebox(0,0){$\circ$}}


\put(9,1){\makebox(0,0){$\circ$}}
\put(1.5,-6){\makebox(0,0){$\circ$}}

\put(1.5,8){\makebox(0,0){$\circ$}}
\put(29,1){\makebox(0,0){$\circ$}}

\put(9.5,0.4){\line(1,0){8.9}}
\put(9.5,1.7){\line(1,0){8.9}}

 \put(2,8.2){\line(1,-1){6.3}}


 \put(2,-5.8){\line(1,1){6.3}}
 
\put(19.5,1){\line(1,0){8.7}}
\put(14,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(-1,-7.5){\scriptsize 2}
\put(-1,9){\scriptsize 1}

\put(8.5,-2){\scriptsize 3}
\put(18.5,-2){\scriptsize 4}
\put(28.5,-2){\scriptsize 5}
\put(32,1){.}


\end{picture}
$}
\end{center}}
\vspace{4ex}
\noindent
{\bf Proof.} First of all by (1.22) and (1.2) $B$ is a 2-group. Let ${~I\!\!\!\!C}_S
(K_i)\subseteq K_i$ for $i=1,2$. Then by (1.24) we may assume that $Z_1$ is
an $F_1$-module, a contradiction to (1.15)(b).

Hence we may assume ${~I\!\!\!\!C}_S(K_1)\not\subseteq K_1$. Now $|K_1K_2/
K_2|\!= \!|K_1K_2/
K_1|\le\! 2$  and we get $2^\prime$-elements
in $G_1\cap G_2$ centralizing $K_1$ and so also $K_2$. This implies that
${~I\!\!\!\!C}_S(K_2)\not\subseteq K_2$, too. In particular $K_1\cap K_2=1$
and $|K_1|=|K_2|\le 2$.  Let $X=G_{4,5}$. Then by (1.3)(v)  we have $K_5=O_2(G_1\cap G_5)$
and $X/K_5\cong 2^6A_7$. Further the action of $G_{1,4,5}/K_1$ on $K_5/K_1$ shows $2^5 \geq 
|Z(O_2(X))| \geq 2^4$ and $Z(O_2(X)) \subseteq Z(K_5)$. Now we have 
$ \langle Z(O_2(X))^{G_{1,5}} \rangle \subseteq Z(K_5)$.
Hence $|Z(K_5)|=2^3|Z(O_2(X))|$. But $X$
has to act nontrivially on $Z(K_5)/Z(O_2(X))$, a contradiction.
\hfill $\Box$
\\[2ex]
{\bf (2.4) Proposition.} {\it If $\{P_1,\ldots, P_n\}$ is a semiclassical parabolic
system with connected diagram $\Delta$ of rank at least five, then there
is no subdiagram of type
\begin{center}
{$
\begin{picture}(36,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}
\put(22.5,1){\line(1,0){8.7}}


\put(1,-2){{\scriptsize 2}}
\put(11,-2){{\scriptsize 3}}
\put(21,-2){{\scriptsize 4}}
\put(31,-2){{\scriptsize 5}}


\put(32,1){\makebox(0,0){$\circ$}}


\put(12.5,0.4){\line(1,0){8.9}}
\put(12.5,1.7){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(34,1){.}
\end{picture}
$}
\end{center} }
\noindent
{\bf Proof.} Suppose false. It is enough to prove the theorem for rank
$(\Delta)=5$. Let the node 1 be connected with 2. Then $\Delta_5$ is a
connected diagram having a residue belonging to $3^7Sp_6(2)$. By (1.3) we get that $\Delta_5$ is of shape 
\begin{center}
{$
\begin{picture}(36,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}
\put(12.5,1){\line(1,0){8.7}}


\put(1,-2){{\scriptsize 1}}
\put(11,-2){{\scriptsize 2}}
\put(21,-2){{\scriptsize 3}}
\put(31,-2){{\scriptsize 4}}


\put(32,1){\makebox(0,0){$\circ$}}


\put(22.5,0.4){\line(1,0){8.9}}
\put(22.5,1.7){\line(1,0){8.9}}
\put(27,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(34,1){.}
\end{picture}
$}
\end{center}
If 1 is not connected with 5, then $\Delta$ is as in (2.1), a contradiction.
So let 1 be connected with 5, then also $\Delta_2$ is of shape
\begin{center}
{$
\begin{picture}(36,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(22.5,1){\line(1,0){8.7}}
\put(12.5,1){\line(1,0){8.7}}


\put(1,-2){{\scriptsize 3}}
\put(11,-2){{\scriptsize 4}}
\put(21,-2){{\scriptsize 5}}
\put(31,-2){{\scriptsize 1}}


\put(32,1){\makebox(0,0){$\circ$}}


\put(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}
\put(7,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(34,1){.}
\end{picture}
$}
\end{center}
and so
$\Delta$ is as in (2.2), a contradiction.

So assume that 1 is connected neither to 2 nor to 5. Then we may assume that
1 is just connected to 3. Again by (1.3) $\Delta_2$ is of type
\begin{center}
{$
\begin{picture}(36,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}
\put(22.5,1){\line(1,0){8.7}}


\put(1,-2){{\scriptsize 1}}
\put(11,-2){{\scriptsize 3}}
\put(21,-2){{\scriptsize 4}}
\put(31,-2){{\scriptsize 5}}


\put(32,1){\makebox(0,0){$\circ$}}


\put(12.5,0.4){\line(1,0){8.9}}
\put(12.5,1.7){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(34,1){.}
\end{picture}
$}
\end{center}
and so $\Delta$ is as in (2.3), a contradiction. $\hfill\Box$\\[1cm]
\noindent

\begin{center}
\boldmath \S\unboldmath {\bf 3 Diagrams containing} 
{$
\begin{picture}(40,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}
\put(22.5,0.4){\line(1,0){8.9}}
\put(22.5,1.7){\line(1,0){8.9}}
\put(27,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(32,1){\makebox(0,0){$\circ$}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(12.5,1.7){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(37,1){.}
\end{picture}
$}
\end{center}
\vspace{1cm}

In this paragraph we will determine the possible semiclassical parabolic
systems $\{P_1,P_2,P_3,P_4,P_5\}$ of rank 5 whose diagram $\Delta$ involves a
subdiagram of the shape 
{$
\begin{picture}(40,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}
\put(22.5,0.4){\line(1,0){8.9}}
\put(22.5,1.7){\line(1,0){8.9}}
\put(27,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(32,1){\makebox(0,0){$\circ$}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(12.5,1.7){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(1,-2.5){\scriptsize1}
\put(11,-2.5){\scriptsize2}
\put(21,-2.5){\scriptsize3}
\put(31,-2.5){\scriptsize4}


\put(37,0){.}
\end{picture}
$}
 We have seen in (1.3) that the subdiagram
{$
\begin{picture}(26,6)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}




\put(12.5,0.4){\line(1,0){8.9}}
\put(12.5,1.7){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\end{picture}
$} 
belongs to $M_{24}$.
Further $G_{4,5}/K_5 \cong 2^{11}M_{24}$ or  $2^{12}M_{24}$, $K_{4,5}/K_5$ is elementary
abelian and is the Golay code module for $G_{4,5}/K_{4,5}$, $K_{1,5}/K_5$ is
an extraspecial 2-group of order $2^{13}$ and $G_{1,5}/K_{1,5}$ is a $3U_4(3)$-amalgam.
\\[2ex]
{\bf (3.1) Lemma.}{\it The diagram $\Delta$ is one of the following\\[0.2cm]
(1) {$
\begin{picture}(50,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}
\put(12.5,1){\line(1,0){8.7}}
\put(32.5,0.4){\line(1,0){8.9}}
\put(32.5,1.7){\line(1,0){8.9}}
\put(37,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}


\put(32,1){\makebox(0,0){$\circ$}}
\put(42,1){\makebox(0,0){$\circ$}}
\put(1,-2.5){\scriptsize5}
\put(11,-2.5){\scriptsize1}
\put(21,-2.5){\scriptsize2}
\put(31,-2.5){\scriptsize3}
\put(41,-2.5){\scriptsize4}



\put(22.5,0.4){\line(1,0){8.9}}
\put(22.5,1.7){\line(1,0){8.9}}
\put(27,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\end{picture}
$}\\[0.5cm]
(2) {$
\begin{picture}(50,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}
\put(32.5,1){\line(1,0){8.7}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(12.5,1.7){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}


\put(32,1){\makebox(0,0){$\circ$}}
\put(42,1){\makebox(0,0){$\circ$}}
\put(1,-2.5){\scriptsize1}
\put(11,-2.5){\scriptsize2}
\put(21,-2.5){\scriptsize3}
\put(31,-2.5){\scriptsize4}
\put(41,-2.5){\scriptsize5}



\put(22.5,0.4){\line(1,0){8.9}}
\put(22.5,1.7){\line(1,0){8.9}}
\put(27,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\end{picture}
$}\\[0.3cm]
(3) {$
\begin{picture}(35,12)
\put(2,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(29,-6){\makebox(0,0){$\circ$}}

\put(29,8){\makebox(0,0){$\circ$}}

\put(2.5,1){\line(1,0){8.7}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(12.5,1.7){\line(1,0){8.9}}

 \put(21.8,1.7){\line(1,1){6.5}}
 \put(22.5,1){\line(1,1){6.5}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}


 \put(21.8,0.4){\line(1,-1){6.5}}
 \put(22.5,1){\line(1,-1){6.5}}

\put(22.8,5.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(24,-3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(1,-2.5){\scriptsize1}
\put(10,-2.5){\scriptsize2}
\put(20,-2.5){\scriptsize3}
\put(31,-6.5){\scriptsize5}
\put(31,8){\scriptsize4}


\end{picture}
$}\\[0.5cm]
(4) {$
\begin{picture}(35,12)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(19,-6){\makebox(0,0){$\circ$}}

\put(19,8){\makebox(0,0){$\circ$}}
\put(26,1){\makebox(0,0){$\circ$}}

\put(2.5,1){\line(1,0){8.7}}

 \put(11.8,1.7){\line(1,1){6.5}}
 \put(12.5,1){\line(1,1){6.5}}

 \put(11.8,0.4){\line(1,-1){6.5}}
 \put(12.5,1){\line(1,-1){6.5}}

\put(12.8,5.5){\makebox(0,0){ $\scriptstyle{\sim}$}}

\put(14,-3.5){\makebox(0,0){$\scriptstyle{\sim} $}}

 \put(18.8,-5.3){\line(1,1){6.5}}
 \put(19.5,-6){\line(1,1){6.5}}

 \put(18.8,7.3){\line(1,-1){6.5}}
 \put(19.5,8){\line(1,-1){6.5}}

\put(24.4,5.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(24.4,-3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(1,-2.5){\scriptsize1}
\put(10,-2.5){\scriptsize2}
\put(20,-7.5){\scriptsize5}
\put(20,9){\scriptsize3}
\put(27,1){\scriptsize4}

\end{picture}
$}\\[0.8cm]
{$
\begin{picture}(5,15)
\put(0,5){(5)}
\end{picture}
$}
{$
\begin{picture}(25,15)

\put(2,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2,11){\makebox(0,0){$\circ$}}
\put(12,11){\makebox(0,0){$\circ$}}
\put(22,11){\makebox(0,0){$\circ$}}

\put(-1.5,-0.2){\scriptsize 1}
\put(23,-0.2){ \scriptsize 5}
\put(-1.5,11.8){\scriptsize 2}
\put(11,12.3){\scriptsize 3}
\put(23,11.8){\scriptsize 4}

\put(2.5,1){\line(1,0){18.5}}

\put(2,1.8){\line(0,1){8.7}}
\put(22,1.8){\line(0,1){8.7}}



\put(2.5,11.4){\line(1,0){8.9}}
\put(2.5,10.6){\line(1,0){8.9}}

\put(7,13.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(12.5,11.4){\line(1,0){8.9}}
\put(12.5,10.6){\line(1,0){8.9}}

\put(17,13.5){\makebox(0,0){$\scriptstyle{\sim}$}}


\end{picture}
$}}\\[2ex]
\noindent
{\bf Proof.} We choose the numbering as follows:\\
\begin{center}
 {$
\begin{picture}(40,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}

\put(2.5,1){\line(1,0){8.7}}
\put(22.5,0.4){\line(1,0){8.9}}
\put(22.5,1.7){\line(1,0){8.9}}
\put(27,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}


\put(32,1){\makebox(0,0){$\circ$}}


\put(1,-2.5){\scriptsize1}
\put(11,-2.5){\scriptsize2}
\put(21,-2.5){\scriptsize3}
\put(31,-2.5){\scriptsize4}



\put(12.5,0.4){\line(1,0){8.9}}
\put(12.5,1.7){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(37,1){,}
\end{picture}
$}
\end{center}
where 5 is the fifth vertex of the diagram.

First let us assume that the diagram is a string. If 1 is connected with 
5 then the residue of 4 has a diagram
of rank 4 which contains an $M_{24}$-subdiagram. So by (1.3) 
the residue of 4 has the
diagram 
\begin{center}
 {$
\begin{picture}(40,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}
\put(12.5,1){\line(1,0){8.7}}



\put(32,1){\makebox(0,0){$\circ$}}

\put(1,-2.5){\scriptsize5}
\put(11,-2.5){\scriptsize1}
\put(21,-2.5){\scriptsize2}
\put(31,-2.5){\scriptsize3}




\put(22.5,0.4){\line(1,0){8.9}}
\put(22.5,1.7){\line(1,0){8.9}}
\put(27,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(37,0){,}
\end{picture}
$}
\end{center} 
that is case (1). 

Let 5 be connected with 4.
Then the diagram $\Delta_1$ of the residue of 1 is a string
of rank 4, containing 
{$
\begin{picture}(24,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}

\put(12.5,0.4){\line(1,0){8.9}}
\put(12.5,1.7){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}
\put(7,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\end{picture}
$}.
By (1.2) we have $P_i/O_2(P_i) \cong \Sigma_3$ for all $i$. And so by (1.3) we get 
that $\Delta_1$ is of the shape 
\begin{center}
{$
\begin{picture}(40,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}

\put(22.5,1){\line(1,0){8.7}}
\put(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}
\put(7,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}


\put(32,1){\makebox(0,0){$\circ$}}




\put(12.5,0.4){\line(1,0){8.9}}
\put(12.5,1.7){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(37,0){,}
\end{picture}
$}
\end{center}
and this is case (2).

So we may assume that our diagram is not a string. Let 2 be connected with 5.
The residue of 4 contains an $M_{24}$-subdiagram
and by (1.3) it has to have the diagram 
\begin{center} 
{$
\begin{picture}(25,12)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(19,-6){\makebox(0,0){$\circ$}}

\put(19,8){\makebox(0,0){$\circ$}}

\put(2.5,1){\line(1,0){8.7}}

 \put(11.8,1.7){\line(1,1){6.5}}
 \put(12.5,1){\line(1,1){6.5}}

 \put(11.8,0.4){\line(1,-1){6.5}}
 \put(12.5,1){\line(1,-1){6.5}}

\put(12.8,5.5){\makebox(0,0){ $\scriptstyle{\sim}$}}

\put(14,-3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(1,-2.5){\scriptsize1}
\put(10,-2.5){\scriptsize2}
\put(21,-6.5){\scriptsize5}
\put(21,8){\scriptsize3}
\put(24,1){.}

\end{picture}
$}
\end{center}
\vspace{0.4cm}
\noindent
By (1.1) and (1.3) this forces the diagram of the residue of 1
to be a square and so of type 


\begin{center}
{$
\begin{picture}(26,15)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(2,11){\makebox(0,0){$\circ$}}
\put(12,11){\makebox(0,0){$\circ$}}


\put(2.5,1){\line(1,0){8.9}}

\put(1.6,1.6){\line(0,1){8.9}}
\put(2.4,1.6){\line(0,1){8.9}}
\put(-1,-2){\scriptsize5}
\put(13.5,-2){\scriptsize4}
\put(-1,13){\scriptsize2}
\put(13.5,13){\scriptsize3}



\put(2.5,11.4){\line(1,0){8.9}}
\put(2.5,10.6){\line(1,0){8.9}}

\put(7,13.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(-1,5.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(15,5.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(11.6,1.6){\line(0,1){8.9}}
\put(12.4,1.6){\line(0,1){8.9}}
\put(22,5.5){ or}
\end{picture}
$} 
\qquad
{$
\begin{picture}(17,15)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(2,11){\makebox(0,0){$\circ$}}
\put(12,11){\makebox(0,0){$\circ$}}


\put(2.5,1.4){\line(1,0){8.9}}
\put(2.5,0.6){\line(1,0){8.9}}
\put(7,-2.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}


\put(1.6,1.6){\line(0,1){8.9}}
\put(2.4,1.6){\line(0,1){8.9}}
\put(-1,5.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(2.5,11.4){\line(1,0){8.9}}
\put(2.5,10.6){\line(1,0){8.9}}

\put(7,13.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(-1,-2){\scriptsize5}
\put(13.5,-2){\scriptsize4}
\put(-1,13){\scriptsize2}
\put(13.5,13){\scriptsize3}


\put(11.6,1.6){\line(0,1){8.9}}
\put(12.4,1.6){\line(0,1){8.9}}
\put(15,5.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(18,5){.}
\end{picture}
$} 
\end{center} 
{$
\begin{picture}(91,15)
\put(0,5){Because of (1.3) there are no diagrams of type}
\end{picture}
\begin{picture}(26,15)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(2,11){\makebox(0,0){$\circ$}}
\put(12,11){\makebox(0,0){$\circ$}}


\put(2.5,1){\line(1,0){8.9}}

\put(1.6,1.6){\line(0,1){8.9}}
\put(2.4,1.6){\line(0,1){8.9}}



\put(2.5,11.4){\line(1,0){8.9}}
\put(2.5,10.6){\line(1,0){8.9}}

\put(7,13.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(-1,5.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(15,5.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(11.6,1.6){\line(0,1){8.9}}
\put(12.4,1.6){\line(0,1){8.9}}
\put(18,5.5){.}
\end{picture}
$}
{$
\begin{picture}(12,15)
\put(0,5){So we get}
\end{picture}
$}\\ 
{$
\begin{picture}(32,8)
\put(0,0){that $\Delta$ looks like}
\end{picture}
$}
{$
\begin{picture}(35,6)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(19,-6){\makebox(0,0){$\circ$}}

\put(19,8){\makebox(0,0){$\circ$}}
\put(26,1){\makebox(0,0){$\circ$}}

\put(2.5,1){\line(1,0){8.7}}

 \put(11.8,1.7){\line(1,1){6.5}}
 \put(12.5,1){\line(1,1){6.5}}

 \put(11.8,0.4){\line(1,-1){6.5}}
 \put(12.5,1){\line(1,-1){6.5}}

\put(12.8,5.5){\makebox(0,0){ $\scriptstyle{\sim}$}}

\put(14,-3.5){\makebox(0,0){$\scriptstyle{\sim} $}}

 \put(18.8,-5.3){\line(1,1){6.5}}
 \put(19.5,-6){\line(1,1){6.5}}

 \put(18.8,7.3){\line(1,-1){6.5}}
 \put(19.5,8){\line(1,-1){6.5}}

\put(24.4,5.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(24.4,-3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(1,-2.5){\scriptsize1}
\put(10,-2.5){\scriptsize2}
\put(20,-7.5){\scriptsize5}
\put(20,9){\scriptsize3}
\put(28,-1){\scriptsize4}

\end{picture}
$}
{$
\begin{picture}(25,8)
\put(0,0){and this is case (4).}
\end{picture}
$}\\ \\


Now let 3 be connected with 5. Then again by (1.3) the residue of 4 has to have the diagram 
{$
\begin{picture}(40,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}
\put(22.5,0.4){\line(1,0){8.9}}
\put(22.5,1.7){\line(1,0){8.9}}
\put(27,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(32,1){\makebox(0,0){$\circ$}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(12.5,1.7){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(37,1){,}
\end{picture}
$}
so we have case (3).

At last let 1 and 4 be connected with 5, that
is the diagram is a circle. But now 
by (1.3) the residues of 1 and of 2
have string diagrams and we get case (5).$\hfill\Box$\\[0.3cm]
Not all of these possible diagrams do exist, as the
next two lemmas will show.\\[2ex]
{\bf (3.2) Lemma.} {\it The cases (1) and (5) in (3.1) do not occur}.
\\[2ex]
{\bf Proof.} In both cases we have $G_4/K_4\cong Co_1$ by (1.3). First
let us assume ${~I\!\!\!\!C}_{G_4}(K_4)\subseteq K_4$.
Using [ParRow; Theorem B] we know that $K_4$ is extraspecial of order $2^{25}$. 
If we consider case (5), application of (1.3) to  $G_1$ and $G_2$
 shows 
$G_2/K_2\cong Co_1$ as well.
Now ${~I\!\!\!\!C}_{G_2}(K_2)\subseteq K_2$, too, and so again $K_2$ is
extraspecial of order $2^{25}$. But then 
\[Z(K_2) = \Omega_1(Z(S)) = Z(K_4) \lhd \langle G_2, G_4 \rangle = G,\]
a contradiction.

So let us consider case (1) and the diagram 
\begin{center}
{$
\begin{picture}(50,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}
\put(12.5,1){\line(1,0){8.7}}
\put(32.5,0.4){\line(1,0){8.9}}
\put(32.5,1.7){\line(1,0){8.9}}
\put(37,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}


\put(32,1){\makebox(0,0){$\circ$}}
\put(42,1){\makebox(0,0){$\circ$}}
\put(1,-2.5){\scriptsize5}
\put(11,-2.5){\scriptsize1}
\put(21,-2.5){\scriptsize2}
\put(31,-2.5){\scriptsize3}
\put(41,-2.5){\scriptsize4}



\put(22.5,0.4){\line(1,0){8.9}}
\put(22.5,1.7){\line(1,0){8.9}}
\put(27,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(47,1){.}
\end{picture}
$}
\end{center}
We have $G_3/K_3 \cong L_4(2) \times \Sigma_3$. Choose $g\in G_3\backslash G_4$
with $g^2 \in G_4$. Then we have $|K^g_4 : K^g_4 \cap K_3| =2$ and
$K^g_4 \cap K_4$ is elementary abelian. Therefore 
$|(K^g_4 \cap K_3)K_4 : K_4 | \geq 2^{11}$. By [Asch1]
$M_{24}$ does not possess an F-module, so 
$O_2(G_4 \cap G_5 / K_4) = J(S/K_4)$ and then this is
the only elementary abelian subgroup of order $2^{11}$ in $S/K_4$.
Hence
$X=(K^g_4 \cap K_3)K_4 = O_2(G_4 \cap G_5)$. 
As $N_{G_{3,4}}(X)K_{3,4} = G_{3,4}$ we get $X\lhd \langle G_{4,5},
G_{3,4} \rangle = G_4$,
a contradiction.

Only the case $G_4=K_4{~I\!\!\!\!C}_
{G_4}(K_4)$ is left to be considered. We know that  $Z(G_4\cap G_5/K_5)=1$ and so $K_4
\subseteq K_5.$ Since we have $|S:K_4|=2^{21}$ and $2^{21}\le |S:K_5|\le 2^{22}$
we see that $|K_4|\ge |K_5|$. So we get $K_4=K_5=1$. But now $O_2(G_4\cap G_5)$
is the Todd module in $G_4$ and the Golay code module in $G_5$
by [FuStr; 3.1], a contradiction. $\hfill\Box$
\\[2ex]
{\bf (3.3) Lemma.} {\it Case (2) in (3.1) does not occur}. \\[2ex]
{\bf Proof.}
Suppose there is case (2), so we have the diagram 
\begin{center}
{$
\begin{picture}(50,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}
\put(32.5,1){\line(1,0){8.7}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(12.5,1.7){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}


\put(32,1){\makebox(0,0){$\circ$}}
\put(42,1){\makebox(0,0){$\circ$}}
\put(1,-2.5){\scriptsize1}
\put(11,-2.5){\scriptsize2}
\put(21,-2.5){\scriptsize3}
\put(31,-2.5){\scriptsize4}
\put(41,-2.5){\scriptsize5}



\put(22.5,0.4){\line(1,0){8.9}}
\put(22.5,1.7){\line(1,0){8.9}}
\put(27,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(47,1){.}
\end{picture}
$}
\end{center}
Let first
${~I\!\!\!\!C}_S(K_5)\subseteq K_5$ and ${~I\!\!\!\!C}_S(K_1)\subseteq K_1$.
W.l.o.g. $Z_1\neq \Omega_1(Z(S))$. Let us consider the
graph $\Gamma=\Gamma(G_1,G_2)$. We see with  (1.24) that $Z_1$
is an $F_1$-module with offen\-ding subgroup contained in $O_2(G_1\cap
G_2/K_1)$, but this fact contradicts (1.18). So now w.l.o.g. we have ${~I\!\!\!\!C}_S(K_1)\not\subseteq K_1$. 
As $S_{G_1\cap G_5}/K_5$ is extraspecial we get
$|K_1K_5/K_5|\le 2$, and $(G_1\cap G_5 \cap G_4)^{(\infty)}$ lies in the centralizer
of the subgroup $K_1\cap K_5$ which has an index smaller or
equal to 4 in $K_5$.
But since $(G_1\cap G_5 \cap G_4)^{(\infty)}$ has no factorgroup
 isomorphic to a 
subgroup of $\Sigma_3$, it even
lies in the centralizer of $K_5$. So ${~I\!\!\!\!C}_S(K_5)\not\subseteq K_5,$
$|K_5K_1/K_1|\le 2$ and hence $K_5\cap K_1=1$.
In particular $|K_1|\le 2\ge |K_5|$. As $|S_{G_1\cap G_5}/K_1|
= |S_{ G_1\cap G_5}/K_5|$ 
we have $|K_1|= |K_5|
\le 2$.
So $O_2(G_4\cap G_5)$ is abelian and has the order $2^{11},2^{12}$, or
$2^{13}$ and $O_2(G_1\cap G_2)$ has the same order and is abelian as well.

There is only one abelian subgroup of order $|O_2(G_4\cap G_5/K_5)|$ in
$S/K_1$ and $S/K_5$ resp., and we get $O_2(G_1\cap G_2)=
O_2(G_4\cap G_5)\lhd \langle G_1\cap G_2,G_4\cap G_5\rangle =G$, but this
is a contradiction. $\hfill\Box$
\\

Now we will describe the groups that belong to the remaining possible 
diagrams. We change numbering a little bit.\\[2ex]
{\bf (3.4) Lemma}
{\it Let the diagram $\Delta$ be of type \begin{center}
{$
\begin{picture}(8,12)
\put(0,0){(1)}
\end{picture}
$}
{$
\begin{picture}(40,12)
\put(2,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(29,-6){\makebox(0,0){$\circ$}}

\put(29,8){\makebox(0,0){$\circ$}}

\put(2.5,1){\line(1,0){8.7}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(12.5,1.7){\line(1,0){8.9}}

 \put(21.8,1.7){\line(1,1){6.5}}
 \put(22.5,1){\line(1,1){6.5}}

 \put(21.8,0.4){\line(1,-1){6.5}}
 \put(22.5,1){\line(1,-1){6.5}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(22.8,5.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(24,-3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(1,-2.5){\scriptsize1}
\put(10,-2.5){\scriptsize2}
\put(20,-2.5){\scriptsize5}
\put(31,-6.5){\scriptsize3}
\put(31,8){\scriptsize4}

\put(35,0){or}
\end{picture}
$}
\qquad
{$
\begin{picture}(8,12)
\put(0,0){(2)}
\end{picture}
$}
{$
\begin{picture}(35,12)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(19,-6){\makebox(0,0){$\circ$}}

\put(19,8){\makebox(0,0){$\circ$}}
\put(26,1){\makebox(0,0){$\circ$}}

\put(2.5,1){\line(1,0){8.7}}

 \put(11.8,1.7){\line(1,1){6.5}}
 \put(12.5,1){\line(1,1){6.5}}

 \put(11.8,0.4){\line(1,-1){6.5}}
 \put(12.5,1){\line(1,-1){6.5}}

\put(12.8,5.5){\makebox(0,0){ $\scriptstyle{\sim}$}}

\put(14,-3.5){\makebox(0,0){$\scriptstyle{\sim} $}}

 \put(18.8,-5.3){\line(1,1){6.5}}
 \put(19.5,-6){\line(1,1){6.5}}

 \put(18.8,7.3){\line(1,-1){6.5}}
 \put(19.5,8){\line(1,-1){6.5}}

\put(24.4,5.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(24.4,-3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(1,-2.5){\scriptsize1}
\put(10,-2.5){\scriptsize2}
\put(20,-7.5){\scriptsize4}
\put(20,9){\scriptsize3}
\put(27,1){\scriptsize5}

\end{picture}
$}
\end{center}
\vspace{0.5cm}
Then $K_3=1=K_4$, and $G_3$ and $G_4$ are locally
$M(24)$, $M(24)^\prime$, $3M(24)$, or $3M(24)^\prime$}.\\[2ex]
{\bf Proof.}
Assume first ${~I\!\!\!\!C}_S(K_3)\subseteq K_3$ and ${~I\!\!\!\!C}_S(K_4)
\subseteq K_4$. W.l.o.g. assume $Z_3\neq \Omega_1(Z(S))$.
Consider (2).
By (1.3)(i) $K_5=O_2(G_4 \cap G_5) = O_2(G_5 \cap G_3)$.
In particular $K_3 \subseteq K_5$. But by (1.3)(x) if $K_3 \not\subseteq K_1$
then $K_5 \subseteq K_1$. So in any case $K_3 \subseteq K_1$.
 By (1.19) $Z_3$ is not an $F$-module with offending subgroup in $K_1/K_3$,
  so $J(K_1) \subseteq K_3$ and we
have $J(K_3)=J(K_1) \lhd \langle G_1, G_3 \rangle$, a contradiction.

Consider now (1).
By (1.3)(xi) either $K_3 \subseteq K_1$ or $K_4 \subseteq K_1$.
So if $\Omega_1(Z(S)) \not= Z_4$ then by symmetry we may assume $K_3 \subseteq K_1$
and get a contradiction as above.


So let $\Omega_1(Z(S))=Z_4$, then $\Omega_1(Z(S))\not= Z_1$.
Again by (1.19) we have $J(K_1K_3) \subseteq K_3$ and so $J(K_3)=J(K_1K_3)$.
But $G_{14}/K_1$ is a $(3)U_4(3)$-amalgam and by [Wie4] this amalgam does not have an $F$-module.
So $J(S) \subseteq K_1$. In particular $J(K_1K_3) \subseteq K_1$ and so
$J(K_1)= J(K_3)$. 
So in any case $J(K_1)=J(K_3) \lhd G$, a contradiction.

Now let ${~I\!\!\!\!C}_S(K_3)\not\subseteq K_3$. 
Then $K_3\subseteq K_4$ because
$Z(G_3\cap G_4/K_4)=1$. We know $|K_4:K_3|\le 2$, so ${~I\!\!\!\!C}_S(K_4)
\not\subseteq K_4$ and $K_4=K_3=1$. $\hfill\Box$
\\

We have proved the following\\[2ex]
{\bf (3.5) Proposition.} 
{\it Let $\{P_1,\ldots ,P_5\}$ be a semiclassical parabolic
system with connected diagram $\Delta$ which involves a subdiagram}
{$
\begin{picture}(40,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}
\put(22.5,0.4){\line(1,0){8.9}}
\put(22.5,1.7){\line(1,0){8.9}}
\put(27,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(32,1){\makebox(0,0){$\circ$}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(12.5,1.7){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(37,1){.}
\end{picture}
$}
{\it Then}
\begin{center}
{$
\begin{picture}(10,12)
\put(0,0){$\Delta =$}
\end{picture}
$}
{$
\begin{picture}(45,12)

\put(2,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(29,-6){\makebox(0,0){$\circ$}}

\put(29,8){\makebox(0,0){$\circ$}}

\put(2.5,1){\line(1,0){8.7}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(12.5,1.7){\line(1,0){8.9}}

 \put(21.8,1.7){\line(1,1){6.5}}
 \put(22.5,1){\line(1,1){6.5}}

 \put(21.8,0.4){\line(1,-1){6.5}}
 \put(22.5,1){\line(1,-1){6.5}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}


\put(22.8,5.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(24,-3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(1,-2.5){\scriptsize1}
\put(10,-2.5){\scriptsize2}
\put(20,-2.5){\scriptsize5}
\put(31,-6.5){\scriptsize3}
\put(31,8){\scriptsize4}

\put(35,0){\it or}
\end{picture}
$}
{$
\begin{picture}(10,12)
\put(0,0){$\Delta =$}
\end{picture}
$}
{$
\begin{picture}(35,12)

\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(19,-6){\makebox(0,0){$\circ$}}

\put(19,8){\makebox(0,0){$\circ$}}
\put(26,1){\makebox(0,0){$\circ$}}

\put(2.5,1){\line(1,0){8.7}}

 \put(11.8,1.7){\line(1,1){6.5}}
 \put(12.5,1){\line(1,1){6.5}}

 \put(11.8,0.4){\line(1,-1){6.5}}
 \put(12.5,1){\line(1,-1){6.5}}

\put(12.8,5.5){\makebox(0,0){ $\scriptstyle{\sim}$}}

\put(14,-3.5){\makebox(0,0){$\scriptstyle{\sim} $}}

 \put(18.8,-5.3){\line(1,1){6.5}}
 \put(19.5,-6){\line(1,1){6.5}}

 \put(18.8,7.3){\line(1,-1){6.5}}
 \put(19.5,8){\line(1,-1){6.5}}

\put(24.4,5.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(24.4,-3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(1,-2.5){\scriptsize1}
\put(10,-2.5){\scriptsize2}
\put(20,-7.5){\scriptsize4}
\put(20,9){\scriptsize3}
\put(27,1){\scriptsize5}

\end{picture}
$}
\end{center}
{\it and $K_3=K_4=1$}.
\\ \\
\noindent

\begin{center}
\boldmath \S\unboldmath {\bf 4 Diagrams containing}
{$
\begin{picture}(40,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}
\put(22.5,0.4){\line(1,0){8.9}}
\put(22.5,1.7){\line(1,0){8.9}}


\put(32,1){\makebox(0,0){$\circ$}}
\put(12.5,1){\line(1,0){8.7}}

\put(27,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(37,1){.}
\end{picture}
$}
\end{center}
\vspace{1cm}

The next goal is to determine all the possible semiclassical parabolic
systems $\{P_1,\ldots, P_k\}$ whose corresponding diagram $\Delta$ has a subdiagram of
type {$
\begin{picture}(40,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}
\put(22.5,0.4){\line(1,0){8.9}}
\put(22.5,1.7){\line(1,0){8.9}}


\put(32,1){\makebox(0,0){$\circ$}}
\put(12.5,1){\line(1,0){8.7}}

\put(27,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(37,1){.}
\end{picture}
$}
We are basically interested in $k=5$. But some of the results
of this paragraph also concern diagrams of rank greater than five.\\[2ex]
{\bf (4.1) Lemma.} 
{\it There is no semiclassical parabolic system $\{P_1,\ldots,
P_k\}$ with $k\ge 5$ and diagram 
\begin{center}
{$
\begin{picture}(20,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}
\put(7,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(12.5,1){\line(1,0){8.7}}
\put(1,-2){{\scriptsize k}}


\end{picture}
$}
{$
\begin{picture}(38,5)
\put(16.5,1){\makebox(0,0){$\circ$}}
\put(26.5,1){\makebox(0,0){$\circ$}}


\put(17,1){\line(1,0){8.7}}

\put(4,1){$\; . \; . \; .\;$}





\put(34.5,-6){\makebox(0,0){$\circ$}}

\put(34.5,8){\makebox(0,0){$\circ$}}



 \put(27.2,1.4){\line(1,1){6.3}}


 \put(27.2,0.7){\line(1,-1){6.3}}
 


\put(36,-6.5){\scriptsize2}
\put(36,8){\scriptsize1}
\put(25,-2){\scriptsize3}
\put(15,-2){\scriptsize4}
\end{picture}
$}
\end{center}
\vspace{0.4cm}
and $G_1/K_1\cong G_2/K_2\cong 3^{\frac{1}{3}(2^{k-1}-1)(2^{k-2}-1)}
Sp_{2k-2}(2)$}.\\[2ex]
{\bf Proof.} Suppose false.
Let at first ${~I\!\!\!\!C}_{G_1}(K_1)\subseteq K_1$ and ${~I\!\!\!\!C}_{G_2}
(K_2)\subseteq K_2$.\\

(1) We may assume $Z_1=\Omega_1(Z(S))$.\\

Suppose $Z_1\neq \Omega_1(Z(S))\neq Z_2$. We may assume that $Z_1$
is an $F$-module with an offending subgroup contained in $O_2(G_1\cap G_2)$.
Then by [Wie2; 2.8] $O_3(G_1/K_1)$ acts trivially on $Z_1$, so $Z_1$ is the natural module
by (1.7). In particular $\langle P_3,\ldots, P_k\rangle$ $\subseteq {~I\!\!\!\!C}
_G(\Omega_1(Z(S)))$ by (1.9). Further we conclude $P_1\not\subseteq {~I\!\!\!\!C}_G(\Omega_1
(Z(S)))$ $\not\supseteq P_2$.
So $Z_k\neq \Omega_1(Z(S))$ and $Z_k$ is an $F_1$--module 
for $G_k/K_k\cong \Omega^+_{2k-2}(2)$ by (1.24). Now by (1.7) and 
(1.8) we get ${~I\!\!\!\!C}_{G_k}(\Omega_1(Z(S)))=
\langle P_{k-2},\ldots ,P_2,  P_1 \rangle ,$ $ \langle P_{k-1},\ldots ,P_3,P_2\rangle$,
or $\langle P_{k-1},\ldots ,P_3,P_1
\rangle$,  a contradiction.

By symmetry we may choose notation such that $\Omega_1(Z(S))  \lhd 
G_1.$\\ 

(2)  $k=5$.\\

Assume $k\ge 6$.
By (1.24), (1.7), and (1.8) $Z_k$ contains a halfspin module $\tilde
{Z}_k$ of $\Omega^+_{2k-2}(2)$ and $k=6$. Consider the graph $\Gamma(G_6,G_5)$. Let
$(6,\alpha)$ be a critical pair. If $\alpha\sim 6, Z_6$ has to be an
$F$-module with offending subgroup
$Z_\alpha K_6 / K_6 \leq O_2(G_5\cap G_6 / K_6)$ and hence $Z_6$ is the halfspin module. But this contradicts
[Str3; (1.10)].
So $\alpha \sim 5$.

Let $b_\Gamma > 1$. Set $V_5=\langle \tilde{Z}_6^{G_5}\rangle. $
We have $V_5^\prime=1$ because $b_\Gamma \ge 3$.
Now we choose notation such that $d(5,\alpha)=d(6,\alpha)-1$. So $V_5
\subseteq G_\alpha$ and $V_\alpha \subseteq G_5$.
Let $\alpha+1\in \Delta (\alpha)$ be such that $Z_6\not\subseteq 
G_{\alpha +1}$. Then
$|Z_{\alpha+1}:{~I\!\!\!\!C}_{Z_{\alpha+1}}
(Z_6)|\le 2^9$, since $|Z_{\alpha+1}:
Z_{\alpha+1}\cap G_6|\le 2$ and 
\[|Z_{\alpha+1}\cap G_6:Z_{\alpha+1}
\cap K_6|\le |
K_5K_6/K_6|=|O_2(G_6\cap G_5/K_6)|=2^8.\]

Now ${~I\!\!\!\!C}_{\tilde{Z}_{\alpha+1}}(Z_6)\lhd \langle Z_6, 
K_{\alpha+1}\rangle$
and the fact that $\langle Z_6, K_{\alpha+1}\rangle$ acts transitively on
$\Delta(\alpha)$ implies ${~I\!\!\!\!C}_{\tilde{Z}_{\alpha+1}}(Z_6)\subseteq
\bigcap \limits_{g\in G_\alpha} \tilde{Z}_{\alpha+1}^{g} = 
\bigcap \limits_{g\in \langle Z_6, K_{\alpha+1}\rangle} \tilde{Z}_{\alpha+1}^{g}$.

We know that the dimension of $\tilde{Z}_{\alpha+1}$ is $2^4$, hence
$|\tilde{Z}_{\alpha+1}:{~I\!\!\!\!C}_{\tilde{Z}_{\alpha+1}} (Z_6)|\ge 2^8$.
This yields  $\tilde{Z}_6=Z_6$. Now
we have $|Z_{\alpha+1}:Z_{\alpha+1}\cap K_6|\ge 2^8$ and in the same way $|
Z_6:Z_6\cap K_{\alpha+1}|\ge 2^8$. Let $x\in Z_6\backslash K_\alpha$. Then
$Z_{\alpha-1}\cap Z_{\alpha+1}\lhd \langle K_{\alpha+1},x\rangle$ and so 
$Z_{\alpha-1}\cap Z_{\alpha+1}\lhd G_\alpha.$

We know that $Z_{\alpha+1}/Z_\alpha$ is an irreducible $(G_\alpha \cap G_{\alpha+1})$-
module, and this yields $Z_\alpha=Z_{\alpha-1}\cap Z_{\alpha+1}$ has order $2^8$.
Hence we get $|[Z_{\alpha+1},x]|=2^8$ and $Z_\alpha={~I\!\!\!\!C}_{
Z_{\alpha+1}}(Z_6)$.
Finally $[Z_{\alpha+1},x]\cap Z_\alpha=1$.

We have $|[Z_{\alpha + 1} \cap G_6,x]| \geq 2^7$ and $[Z_{\alpha + 1} \cap
G_6,x] \subseteq Z_6$. 
Furthermore $|Z_6\cap G_{\alpha+1}:Z_6\cap K_{\alpha+1}|\ge 2^7$. 
So there is some $y \in Z_6 \cap G_{\alpha + 1} $ such that
$[Z_{\alpha +1},y]={~I\!\!\!\!C}_{
Z_{\alpha+1}}(y)=Z_\alpha$. Then 
$|[Z_{\alpha + 1} \cap G_6 , y]| \geq 2^7$, 
thus $|Z_\alpha \cap Z_6| \geq 2^7$ and
$[Z_{\alpha + 1} \cap G_6 , y ] \cap [Z_{\alpha + 1} \cap G_6 , x] =1$.
Hence $|V_\alpha \cap Z_6| \geq 2^{14}$.
So $Z_{\alpha+1}$ lies in the centralizer
of a subgroup of index 4 in $Z_6$, but this contradicts the action on
the halfspin module.

So we are left to deal with $b_\Gamma=1$. Let $\tilde{Z}_6\subseteq K_5$.
Then for $\beta\in \Delta(5)$ we have $[\tilde{Z}_6,
\tilde{Z}_\beta]=1$
according to [Str3; (1.10)]. 
Hence $\tilde{Z}_6 \subseteq K_\beta$ and so $[\tilde{Z}_6,Z_\beta]=1$. 
Thus even $[\tilde{Z}_6,\langle Z_\beta^{G_5}
\rangle]=1$. Since $G_5=(G_6\cap G_5)\langle Z_\beta^{G_5}\rangle$,
we have $\tilde{Z}_6 \lhd \langle G_5,G_6\rangle= G$, but this is a
contradiction. 
Let now $\tilde{Z}_6\not\subseteq K_5$. Then $G_6\cap G_5$
normalizes a subgroup of index 2 in $\tilde{Z}_6$, and this is a contradiction
to the action on the halfspin module. We have proved (2).\\

 The following statement
is true:\\ 

(3) Let $\Gamma$ be the graph $\Gamma (G_1,G_2)$ and $(2,\alpha)$
a critical pair. Then $\alpha\sim 1$.\\

Suppose $\alpha\sim 2$. We may choose notation such that
$Z_\alpha \not\subseteq G_1$. Since $Z_2$ is an $F$-module, by (1.7) $Z_2$
has a submodule $\tilde{Z}_2$, which is a natural module. Furthermore, 
$\tilde{Z}_2$ 
has a subgroup $\tilde{\tilde{Z}}_2$ of index two, such that 
$[Z_\alpha,\tilde{\tilde{Z}}_2]\subseteq \Omega_1(Z(S))
\cap \tilde{Z}_2$. 
Since $|\tilde{Z}_2:  
\tilde{\tilde{Z}}_2|=2$ and there is no fours group of 
transvections in $Sp_8(2)$, we have $|\tilde{Z}_2:\tilde{Z}_2\cap 
K_\alpha| \le 4$.
So $|Z_\alpha : Z_\alpha \cap K_2| \leq 4$ and then also $
|Z_2 : Z_2 \cap K_\alpha| \leq 4$. If $Z_2$ involves two natural modules,
 $Z_\alpha$ has to induce transvections on both. In particular 
$|\tilde{Z_2} : \tilde{Z_2} \cap K_\alpha| =2$. But  now we argue as before
to get
$|Z_\alpha : Z_\alpha \cap K_2| =2= |Z_2 : Z_2 \cap K_\alpha|$,
a contradiction.
So $Z_2/\tilde{Z_2}$ is trivial. As $Z(G_2)=1$, we get with Gasch\"utz
([Hu; I 17.4])
$Z_2=\tilde{Z_2}$. Now $|Z_1|=2$ and  $|Z_3|=4$.
 Set $E_1=\langle Z_3^
{G_1}\rangle$.
As $[S,Z_3]=Z_1$ we have $[E_1,K_1] \subseteq Z_1$.

We want to show, that $[E_1,Z_2]=1$. Let us assume, that $[E_1,Z_2]\neq1$. As 
$E_1
K_2$ is a normal subgroup in $G_1\cap G_2$ this implies $|E_1:E_1\cap
K_2|=2$ or $2^7$.
Let $|E_1:E_1\cap K_2|=2$ at first. Then
 we have $E_1K_2/K_2 = Z(G_1\cap G_2/K_2)$. 
As $Z_3 \lhd S$, $Z_3 \subseteq Z_2$ and $E_1$ induces
transvections on $Z_2$, we see that $[E_1,Z_3]=1$ and
so $E_1^\prime = 1 $. In particular $Z_2 \not\subseteq E_1$.

Let $\beta\in \Delta(2)$ with $[Z_\beta, E_1]\neq 1$.
So
$[E_1\cap K_\beta,E_\beta\cap G_1]\subseteq E_1\cap Z_\beta=1$, since $E_1'=1$.
We have $[E_\beta \cap
K_1, E_1\cap G_\beta]\subseteq E_\beta \cap Z_1$.

Suppose $Z_1\subseteq E_\beta$. Then $E_\beta \subseteq G_1\cap G_2$. But
$[E_\beta,Z_2]\subseteq [E_\beta,K_\beta]=Z_\beta$. On the other hand $[Z_2,
G_1\cap G_2]\not\supseteq Z_\beta$, so $[E_\beta,Z_2]=1$, a contradiction
to $[E_1,Z_2]\not=1$. Therefore we have
$[E_\beta \cap K_1,E_1\cap G_\beta]=1$. Without loss we may assume that $|E_1\cap G_\beta
: E_1\cap K_\beta|\le |E_\beta\cap G_1 :E_\beta\cap K_1|$. Since we know that  $|E_1:E_1\cap
G_\beta|\le 2\ge |E_\beta :E_\beta\cap G_1|$, we get that $E_1$ is
an $F_1$-module for $G_1$ with offending subgroup $(E_\beta\cap G_1)K_1/K_1$
and by (1.7) it
must be the spin or the natural module. In any case by (1.9) $P_3$ 
centralizes $Z_3$ but this is not true.
So we have $|E_1:E_1\cap K_2|=2^7.$ Now
$[K_1,E_1]=Z_1$ implies $Z_2\not\subseteq
K_1$ and this contradicts $\alpha \sim 2$.

 We have proved that $[E_1,Z_2]=1$,
in parti\-cular $[E_1,Z_3]=1$ and $E_1'=1$. Furthermore we know that
$E_1$ is no $F_1$ - module. Let $E_1\subseteq K_{\alpha-1}\subseteq G_\alpha$.
If $[E_1,Z_\alpha]=[Z_2,Z_\alpha]$, then $E_1Z_2\lhd \langle G_1\cap G_2,
Z_\alpha\rangle = G_2$.
Hence $Z_2\cap [K_2,E_1Z_2]$ is a normal subgroup of $G_2$, so we have $Z_2
\subseteq E_1$ or $E_1\subseteq Z(K_2)$.
Since ${~I\!\!\!\!C}_S(E_1)\subseteq K_1, \: E_1\subseteq Z(K_2)$ would 
lead to the contradiction $K_1=K_2=1$. So we have $Z_2\subseteq E_1$, this
implies $E_1 \lhd \langle G_1,G_2\rangle =G$, a contradiction. Now $[E_1,
Z_\alpha]\neq [Z_2,Z_\alpha]$. If $|Z_2:Z_2\cap K_\alpha| \neq 2$, then there
exists an $x\in Z_\alpha$ such that $[Z_2,x]\neq 1$ and 
$[x,O_2(G_\alpha \cap G_{\alpha-1})] = Z_{\alpha-1}$.
Now we conclude $[E_1Z_2,x]=Z_{\alpha -1} = [x,Z_2] \subseteq [Z_\alpha,Z_2]
\subseteq Z_2$ and again $E_1Z_2\lhd \langle G_1\cap G_2,x\rangle$. 
But as $[x,Z_2]\neq Z_1$ and $x$ induces a transvection on $Z_2$ we have
$x\not\in G_1$. So $G_2=\langle G_1\cap G_2, x\rangle $ and we get a contradiction as before. 

Now we have proved $|Z_2:Z_2\cap K_\alpha|=2$. Then ${~I\!\!\!\!C}
_{Z_\alpha}(Z_2)$ induces transvections on $E_1.$ But $Z_1\not\subseteq 
Z_\alpha$, 
hence ${~I\!\!\!\!C}_{Z_\alpha}(Z_2)\not\subseteq K_1$ and so
$E_1$ is an $F$-module and we get a contradiction.

Thus $E_1\not\subseteq K_{\alpha-1}$. Choose $\beta \in \Delta(2)$ such that
$d(\beta,\alpha)=d(2,\alpha)-1$.
There is some $x \in Z_2$ which induces a transvection on 
$Z_\alpha$ as $Z_2$ offends on $Z_\alpha$.
 So $Z_{\alpha - 1} = \bigcap_{y \in Z_2\setminus K_\alpha} [y, Z_\alpha]$.
By symmetry we get  $Z_\beta = \bigcap_{y \in Z_\alpha\setminus K_2} [y, Z_2]$,
hence $Z_{\alpha - 1}= Z_\beta$. 
We have $[E_{\alpha-1}\cap K_1,E_1]\subseteq E_1\cap E_{\alpha-1}\cap [K_1,E_1]
\subseteq E_{\alpha-1} \cap Z_1=1$ since $[Z_1,Z_\alpha]\neq 1$ but $[E_
{\alpha-1},Z_\alpha]=1$.

If $Z_2
\subseteq E_1$, then $|K_1K_2:K_2|=2$ as $[K_1,E_1]\subseteq Z_1$. Also
$|K_1K_2:K_1|=2$. Since $Z_{\alpha-2}\subseteq E_{\alpha-1} \cap K_1$
we get $E_1 \subset K_{\alpha-2}$ and so $|E_1:E_1\cap K_{\alpha-1}|=2$.
Since we know that $|K_\beta K_2/K_2|=2$, we have  $|E_{\alpha-1}:E_{\alpha-1}\cap K_2|\le 2$
and $|E_{\alpha - 1}: E_{\alpha - 1} \cap K_1| \leq 4$. Hence for
$e\in E_1\backslash K_{\alpha-1}$ we get
$|E_{\alpha-1}:{~I\!\!\!\!C}_{E_{\alpha-1}}
(e)|\le 4$, and $E_1$ is an $F_1$-module, a contradiction.

Therefore $Z_2\not\subseteq E_1$ and then $Z_\beta\not\subseteq E_1$. 
Now we have $$[E_{\alpha-1}\cap G_1, E_1\cap K_{\alpha-1}]\subseteq Z_
{\alpha-1}\cap E_1=Z_\beta\cap E_1=1.$$ Thus $E_1$ is an $F_1$-module again,
because of $|E_{\alpha-1}:E_{\alpha-1}\cap G_1|\le 2$. This is a contradiction 
and we have proved (3).\\

By (1.24) $Z_2$ is an $F_1$--module. By (1) $Z(G_2)=1$ and so by (1.7)
and (1.9) a natural module  $\tilde{Z_2}$ is a submodule of $Z_2$.
Define $\tilde{Z}_3=Z_3\cap\tilde{Z}_2$ and $E_1=\langle \tilde{Z}_3^{G_1}
\rangle$.
As $ [\tilde{Z_3},K_1]\subseteq Z_1$ and ${~I\!\!\!\!C}_{G_1}(K_1) \subseteq K_1$
we get $E_1 \subseteq K_1$.\\ 

(4) $b_\Gamma>1$\\

Suppose $b_\Gamma=1$.
 First let $\tilde{Z}_2 \subseteq K_1$. Then $|E_1:E_1\cap K_2|
\le 2$. Since $[Z_2,E_1]\neq 1$ it follows that $|E_1:E_1\cap K_2|=2$. But
then $Z_2$ induces transvections on $E_1$ and $E_1$ would be  an 
$F$--module and  so $[P_3,\tilde{Z_3}]=1$, a contradiction.

So $\tilde{Z}_2\not\subseteq K_1$. Since $\Phi(K_2)\subseteq K_1$ and
because we could assume $\tilde{Z}_2\subseteq \Phi(K_2)$ if $\Phi(K_2)\neq 1$,
it follows that $\Phi(K_2)=1$. Clearly $[E_1\cap K_2,K_2]=1$.
Since $|K_1K_2/K_1|=|K_1K_2/K_2|\ge |E_1:E_1\cap K_2|,E_1$ is an $F$-module,
a contradiction.\\

Using (1) - (4) we want to derive a contradiction to our assumption 
${~I\!\!\!\!C}_{G_1}(K_1)\subseteq K_1$ and ${~I\!\!\!\!C}_{G_2}(K_2)
\subseteq K_2$ at the beginning of this proof.

Since  by (4) and (3) $b_\Gamma \geq 3$ we get $E_1'=1=[E_1,Z_2]$.
 Let $\beta\in \Delta(2)$ again,
with $d(\beta,\alpha)=d(2,\alpha)-1$. Then $[E_\beta, E_\alpha]\subseteq
E_\alpha\cap E_\beta$. Assume that $E_\alpha\subseteq K_\beta$. Thus $E_\alpha
\subseteq G_2$, and then $[E_\alpha,Z_2]\subseteq E_\alpha\cap Z_2$. As
$E_\alpha$ acts quadratically it follows that $|E_\alpha:E_\alpha\cap K_2|
\le 16$. Hence for any $t\in Z_2\backslash K_\alpha$ we have $|E_\alpha:
{~I\!\!\!\!C}_{E_\alpha}(t)|\le 16$. By [Wie2; 2.8] 
 $O_3(G_\alpha/K_\alpha)$ acts trivially on $E_\alpha$. Thus
by (1.10)(iv) $E_\alpha$ involves only natural and spin modules. In particular
by (1.9) $P_3\subseteq {~I\!\!\!\!C}_G(\tilde{Z}_3)$, a contradiction.

So we have to consider the case $E_\alpha \not\subseteq K_\beta$. Let $t\in
E_\beta$. Then we have $|E_\alpha \cap K_\beta:{~I\!\!\!\!C}_{E_\alpha\cap K_\beta}
(t)|\le 2$ and hence $|E_\alpha:{~I\!\!\!\!C}_{E_\alpha}(t)|\le 2|E_\alpha:
E_\alpha\cap K_\beta|\le 2^8$. Again by [Wie2; 2.8]
 $O_3(G_\alpha/K_\alpha)$ acts trivially.
Thus  $E_\alpha/Z_\alpha$ is an $Sp_8(2)$-module. 

If only the natural and
the spin module are involved, we get with (1.9) $[P_3,\tilde{Z}_3]=1$
again, and this would be a contradiction. 

Let $V_\alpha$ be an nontrivial module involved in 
$E_\alpha$ which is not a natural or a spin module.
By (1.10)(iii) we have that $E_\alpha$ cannot contain an 
element $t$ which is of type $a_2$ on $V_\beta$.
The same is true for $E_\beta$. So 
$|E_\alpha / E_\alpha \cap K_\beta| \leq 8$.
But now we get $|[E_\alpha, t]| \leq 16$, a contradiction by (1.10)(iv).

So our first assumption was wrong. Let now w.l.o.g.
${~I\!\!\!\!C}_{G_1}(K_1)$ $\not\subseteq K_1$.
It follows that $|K_1K_2/K_2|\le 2$.
We have $[(G_1\cap G_2)^{\infty}, K_1\cap K_2]=1$ and $|K_2:K_1\cap K_2|
\le 2$, therefore ${~I\!\!\!\!C}_{G_2}(K_2)\not\subseteq K_2$ and $K_1\cap
K_2=1$. So we get $|K_1|=|K_2|\le 2$. As $G_k/K_k\cong \Omega_{2k-2}^+(2)$
we get $|S|=2^{(k-1)^2}$ or $2^{(k-1)^2+1}$ and this yields $|K_k|=2^{k-1}$ or
$2^k$.
So ${~I\!\!\!\!C}_{G_k}(K_k)\not\subseteq K_k$. But 
as $Z(G_k \cap G_1/K_1)=1=Z(G_k \cap G_2/K_2)$ we have $K_k\subseteq K_1
\cap K_2=1$, again a contradiction and 
the proof of (4.1) is completed.\hfill $\Box$ \\[2ex]
{\bf (4.2) Lemma.} {\it Let $\{P_1,\ldots, P_k\}$ be a semiclassical parabolic
system with diagram
\begin{center}
{$
\begin{picture}(10,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}

\put(1,-2){{\scriptsize k}}
\end{picture}
$}
{$
\begin{picture}(38,5)
\put(16.5,1){\makebox(0,0){$\circ$}}
\put(26.5,1){\makebox(0,0){$\circ$}}
\put(36.5,1){\makebox(0,0){$\circ$}}

\put(17,1){\line(1,0){8.7}}

\put(4,1){$\; . \; . \; .\;$}








 \put(27,1.7){\line(1,0){8.9}}


 \put(27,0.4){\line(1,0){8.9}}
 \put(32,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}


\put(35.2,-2){\scriptsize1}
\put(25.2,-2){\scriptsize2}
\put(15.2,-2){\scriptsize3}
\end{picture}
$}
\end{center}
where $k\ge 5$. Then $G\cong 3^{\frac{1}{3}{(2^{k}-1)}
{(2^{k-1}-1})}
Sp_{2k}(2)$ or $k=5$ and  $G\cong F_1$.}\\[2ex]
{\bf Proof.}
Let $G_{\{4,5,\ldots, k\}}/K_{\{4,5,\ldots, k\}}\cong 3^7Sp_6(2)$. Then the claim 
follows with [ShStr]. 
So by (1.3) we may assume that $G_{\{5,6,\ldots, k\}}/K_{\{5,6,\ldots, k\}}\cong
Co_1$. First let $k=5$ and assume ${~I\!\!\!\!C}_{G_5}(K_5)\subseteq K_5$.
According to [ParRow;  Theorem B], $K_5$ is extraspecial of order $2^{25}$. 
Using [Iv] we get $G\cong F_1$.
Hence we may assume ${~I\!\!\!\!C}_{G_5}(K_5) \not\subseteq K_5$.
If $k=6$ [ParRow] implies that ${~I\!\!\!\!C}_{G_6}(K_6)\not\subseteq
K_6$. Let now $k=5$ or $6$ and ${~I\!\!\!\!C}_{G_k}(K_k) \not\subseteq K_k$.
Since $G_{k-1,k}/K_k$
is perfect and $G_{k-1}/K_{k-1} \cong \Sigma_3 \times M_{24}$  or
$\Sigma_3 \times Co_1$, we see $K_k\not\subseteq K_{k-1}$. But by general arguments $K_k \cap K_1 \subseteq K_{k-1}$.
As $Z(G_{1,k}/K_1)=1$, we have $K_k \subseteq K_1$.
And so $K_k \subseteq K_{k-1}$, a contradiction.\hfill $\Box$ \\[2ex]
{\bf (4.3) Lemma.} {\it There is no semiclassical parabolic system $\{P_1,\ldots ,
P_k\}$ with dia\-gram of type 
\begin{center}
{$
\begin{picture}(20,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}
\put(7,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(12.5,1){\line(1,0){8.7}}
\put(1,-2){{\scriptsize k}}


\end{picture}
$}
{$
\begin{picture}(38,5)
\put(16.5,1){\makebox(0,0){$\circ$}}
\put(26.5,1){\makebox(0,0){$\circ$}}


\put(17,1){\line(1,0){8.7}}

\put(4,1){$\; . \; . \; .\;$}





\put(34.5,-6){\makebox(0,0){$\circ$}}

\put(34.5,8){\makebox(0,0){$\circ$}}



 \put(27.2,1.4){\line(1,1){6.3}}


 \put(27.2,0.7){\line(1,-1){6.3}}
 


\put(36,-6.5){\scriptsize2}
\put(36,8){\scriptsize1}
\put(25,-2){\scriptsize3}
\put(15,-2){\scriptsize4}
\put(38,1){,}
\end{picture}
$}

\end{center}
\vspace{0.4cm}
where $G_1/K_1\cong Co_1$
or $F_1$}.\\[2ex]
{\bf Proof.} Assume false.
First let $k=5$. Then  by (1.3) $G_1/K_1\cong G_2/K_2\cong Co_1$. If ${~I\!\!\!\!C}_{G_1}
(K_1)\subseteq K_1$, then by [ParRow; Theorem B] we get $K_1$ is extraspecial of order $2^{25}$.
In the same
way we get $K_2$ is extraspecial of order $2^{25}$.
But then $\Omega_1(Z(S))\lhd \langle G_1,
G_2\rangle=G$ and this cannot hold.

Thus ${~I\!\!\!\!C}_{G_1}(K_1)\not\subseteq K_1.$ As $G_1\cap G_2/K_2$ is 
isomorphic to $2^{11}M_{24}$ we get $K_1=K_2=1$.
Hence $|S|=2^{21}$.

We have $G_3/K_3$
{$
\begin{picture}(4,3)
\put(1,0.7){$<$}
\put(1,-0.7){$\sim$}
\end{picture}
$}
$3\Sigma_6 \times \Sigma_3 \times \Sigma_3$. From
$G_1$ we know that $G_{1,2,3}$ induces in $K_{1,2,3}$
exactly one four-dimensional, two six-dimensional and one
trivial module for $3\Sigma_6$. So all the nontrival modules must be 
involved in $K_3$. This shows $|K_3| \geq 2^{16}$. As in
$G_1=Co_1$ we see $G_{1,3}/K_{1,3} \cong 3\Sigma_6 \times \Sigma_3$
we get that
$|K_3| = 2^{16}$ and so $G_3/K_3\cong (3A_6
\times Z\!\!\! Z_3\times Z\!\!\! Z_3)\cdot V_4$.
Further $|Z(K_3)|=16$ and $K_3/Z(K_3)$ is a direct sum of two six-dimensional
modules for $3A_6$.
Since also $G_{3,2}/K_{3,2}\cong 3\Sigma_6 \times \Sigma_3$
and $G_{5,4,3}/K_{5,4,3}\cong \Sigma_3\times \Sigma_3$ we conclude that
there are exactly 3 subgroups of order 3 in $O_3(G_3/K_3)$, which are normal
in $G_3/K_3$. None of them acts trivially on $K_3$. But since $K_3/Z(K_3)
\cong 2^6 \bigoplus 2^6$ there exists an element $\varrho \in G_3$ with
$o(\varrho)=3$ and $[\varrho, K_3]\subseteq Z(K_3)=\Phi(K_3)$. So $[\varrho,
K_3]=1$ and this is a contradiction.

We are left with the case $k=6$. Then 
\begin{center}
$\Delta = $
{$
\begin{picture}(50,5)
\put(11.5,1){\makebox(0,0){$\circ$}}
\put(21.5,1){\makebox(0,0){$\circ$}}
\put(31.5,1){\makebox(0,0){$\circ$}}
\put(1.5,1){\makebox(0,0){$\circ$}}

\put(22,1){\line(1,0){8.7}}
\put(12,1){\line(1,0){8.7}}
\put(2,1.7){\line(1,0){8.9}}
\put(2,0.4){\line(1,0){8.9}}
\put(7,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(39.5,-6){\makebox(0,0){$\circ$}}

\put(39.5,8){\makebox(0,0){$\circ$}}



 \put(32.2,1.4){\line(1,1){6.3}}


 \put(32.2,0.7){\line(1,-1){6.3}}
 


\put(44,1){,}
\end{picture}
$}

\end{center}
\vspace{0.4cm}
\noindent
but this amalgam contains
a subamalgam of type \\[0.3cm]
\begin{center}
{$
\begin{picture}(38,5)
\put(16.5,1){\makebox(0,0){$\circ$}}
\put(26.5,1){\makebox(0,0){$\circ$}}
\put(8.6,-6){\makebox(0,0){$\circ$}}
\put(8.6,8){\makebox(0,0){$\circ$}}
\put(17,1){\line(1,0){8.7}}


 \put(15.8,1.4){\line(-1,1){6.3}}


 \put(15.8,0.7){\line(-1,-1){6.3}}




\put(34.5,-6){\makebox(0,0){$\circ$}}

\put(34.5,8){\makebox(0,0){$\circ$}}



 \put(27.2,1.4){\line(1,1){6.3}}


 \put(27.2,0.7){\line(1,-1){6.3}}
 
\put(38,1){,}
\end{picture}
$}
\end{center}
\vspace{0.3cm}
([FuStr], Lemma 3.3) which does not exist by [Tim1].$\hfill\Box$
\\

The lemmas (4.1), (4.2), and (4.3) prove the following\\[2ex]
{\bf (4.4) Lemma.} {\it There is no semiclassical parabolic system $\{P_1,\ldots,
P_k\}, $ with $k>4$ and diagram $\Delta$ of type 
\begin{center}
{$
\begin{picture}(20,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}
\put(7,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(12.5,1){\line(1,0){8.7}}
\put(1,-2){{\scriptsize k}}

\put(12.5,1){\line(1,0){8.7}}

\end{picture}
$}
{$
\begin{picture}(38,5)
\put(16.5,1){\makebox(0,0){$\circ$}}
\put(26.5,1){\makebox(0,0){$\circ$}}


\put(17,1){\line(1,0){8.7}}

\put(4,1){$\; . \; . \; .\;$}





\put(34.5,-6){\makebox(0,0){$\circ$}}

\put(34.5,8){\makebox(0,0){$\circ$}}



 \put(27.2,1.4){\line(1,1){6.3}}


 \put(27.2,0.7){\line(1,-1){6.3}}
 


\put(36,-6.5){\scriptsize2}
\put(36,8){\scriptsize1}
\put(25,-2){\scriptsize3}
\put(15,-2){\scriptsize4}
\put(38,1){.}
\end{picture}
$}

\end{center}}
\vspace{0.4cm}
\hspace*{0.3cm} $\hfill\Box$\\[2ex]


\noindent
{\bf (4.5) Lemma.} 
{\it There is no semiclassical parabolic system $\{P_1,\ldots,
P_k\},$ with $ k>4$ and diagram $\Delta$ of type 
\begin{center}
{$
\begin{picture}(20,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}

\put(2.5,1.7){\line(1,0){8.9}}
\put(2.5,0.4){\line(1,0){8.9}}
\put(7,3.5){\makebox(0,0){$\scriptstyle{(\sim)}$}}

\put(12.5,1){\line(1,0){8.7}}
\put(1,-2){{\scriptsize 1}}
\put(11,-2){{\scriptsize 2}}
\end{picture}
$}
{$
\begin{picture}(38,5)
\put(16.5,1){\makebox(0,0){$\circ$}}
\put(26.5,1){\makebox(0,0){$\circ$}}
\put(36.5,1){\makebox(0,0){$\circ$}}

\put(17,1){\line(1,0){8.7}}

\put(4,1){$\; . \; . \; .\;$}

 \put(27,1.7){\line(1,0){8.9}}


 \put(27,0.4){\line(1,0){8.9}}
 \put(32,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}


\put(35.2,-2){\scriptsize k}
\put(25.2,-2){\scriptsize k-1}

\end{picture}
$.}
\end{center}}
 \vspace{2ex}
\noindent
{\bf Proof.} Suppose false.
If we have  $G_{\{1,\ldots, k-3\}}/K_{\{1,\ldots, k-3\}}\cong 3^7Sp_6(2)$ and $G_{\{4,\ldots, 
k\}}/K_{\{4,\ldots, k\}}$ is a group of Lie type or $3^7Sp_6(2)$ then the claim follows from [Wie2; (4.12)]. So $G_{\{1,\ldots,k-3\}}/
K_{\{1,\ldots,k-3\}}\cong M_{24}$ 
or
$He$. By (4.2) and (1.3) $k=5$ or $6$ and $G_1/K_1\cong Co_1$
or $F_1$. So by [FuStr; Lemma 3.3] we find a subamalgam of type
\begin{center}
{$
\begin{picture}(45,5)
\put(16.5,1){\makebox(0,0){$\circ$}}
\put(26.5,1){\makebox(0,0){$\circ$}}
\put(6.5,1){\makebox(0,0){$\circ$}}

\put(17,1){\line(1,0){8.7}}

\put(7,1.7){\line(1,0){8.9}}
\put(7,0.4){\line(1,0){8.9}}



\put(11,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(34.5,-6){\makebox(0,0){$\circ$}}

\put(34.5,8){\makebox(0,0){$\circ$}}



 \put(27.2,1.4){\line(1,1){6.3}}


 \put(27.2,0.7){\line(1,-1){6.3}}
 


\put(43,0){or}
\end{picture}
$}
\qquad \qquad
{$
\begin{picture}(48,5)
\put(16.5,1){\makebox(0,0){$\circ$}}
\put(26.5,1){\makebox(0,0){$\circ$}}
\put(6.5,1){\makebox(0,0){$\circ$}}
\put(-3.5,1){\makebox(0,0){$\circ$}}

\put(17,1){\line(1,0){8.7}}
\put(7,1){\line(1,0){8.7}}

\put(-3,1.7){\line(1,0){8.9}}
\put(-3,0.4){\line(1,0){8.9}}



\put(1,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(34.5,-6){\makebox(0,0){$\circ$}}

\put(34.5,8){\makebox(0,0){$\circ$}}



 \put(27.2,1.4){\line(1,1){6.3}}


 \put(27.2,0.7){\line(1,-1){6.3}}
 



\put(38,1){.}
\end{picture}
$}
\end{center}
\vspace{0.6cm}
But these diagrams do not exist according to (4.4) or [Str3]. $\hfill\Box$
\\[2ex]
{\bf (4.6) Lemma.} {\it Let $\{P_1,\ldots, P_k\},$ $k>4,$ be a semiclassical parabolic
system with diagram $\Delta$ of type
\begin{center}
{$
\begin{picture}(15,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}

\put(1,-2){{\scriptsize k}}
\end{picture}
$}
{$
\begin{picture}(14,12)
\put(1,1){$\; . \; . \; .\;$}
\end{picture}
$}
{$
\begin{picture}(25,12)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(19,-6){\makebox(0,0){$\circ$}}

\put(19,8){\makebox(0,0){$\circ$}}

\put(2.5,1){\line(1,0){8.7}}

 \put(11.8,1.7){\line(1,1){6.5}}
 \put(12.5,1){\line(1,1){6.5}}

 \put(11.8,0.4){\line(1,-1){6.5}}
 \put(12.5,1){\line(1,-1){6.5}}

\put(14,5.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(14,-3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(1,-2){\scriptsize4}
\put(11,-2){\scriptsize3}
\put(21,-6.5){\scriptsize1}
\put(21,8){\scriptsize2}


\end{picture}
$.}\end{center}
\vspace{0.4cm}
Then $K_1=K_2=1$, $G_1\cong G_2\cong Co_1$ or $F_1$ and $k=5$ or $6$, resp., or 
$G_1\cong 3^{\frac{1}{3}(2^{k-1}-1)(2^{k-2}-1)}Sp_{2k-2}(2)$ and 
$G_2\cong \left(3^{\frac{1}{3}(2^{k-1}-1)(2^{k-2}-1)}\right) Sp_{2k-2}(2)$}.\\[2ex]
{\bf Proof.}
We will prove $K_1=K_2=1$ by induction on $k$. 
By (1.3)(i)(vi) we have for a parabolic system with diagram
\begin{center}
{$
\begin{picture}(25,12)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(19,-6){\makebox(0,0){$\circ$}}

\put(19,8){\makebox(0,0){$\circ$}}

\put(2.5,1){\line(1,0){8.7}}

 \put(11.8,1.7){\line(1,1){6.5}}
 \put(12.5,1){\line(1,1){6.5}}

 \put(11.8,0.4){\line(1,-1){6.5}}
 \put(12.5,1){\line(1,-1){6.5}}

\put(14,5.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(14,-3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}


\put(21,-6.5){\scriptsize1}
\put(21,8){\scriptsize2}


\end{picture}
$}
\end{center}
\vspace{0.5cm}
that the kernels $K_1$ and $K_2$ are trivial or
$G_{2} \cong 2^6 \cdot A_7$.
The second case cannot occur as a residue of $\{5,\ldots , k\}$ since by [Str4]
there is no parabolic system with diagram
{$
\begin{picture}(35,3.5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}
\put(22.5,0.4){\line(1,0){8.9}}
\put(22.5,1.7){\line(1,0){8.9}}
\put(1,-1.3){\tiny 5}
\put(11,-1.3){\tiny 4}
\put(21,-1.3){\tiny 3}
\put(31,-1.3){\tiny 1}

\put(32,1){\makebox(0,0){$\circ$}}
\put(12.5,1){\line(1,0){8.7}}
\end{picture}
$}
where the 5-residue is the $A_7$--system.
So by induction we have $K_k=K_{1,k}=K_{2,k}$ and thus $K_1\subseteq K_k
\supseteq K_2$. We have $K_2K_1/K_1 \lhd G_1\cap G_2/K_1$.
 
Let $(G_1/K_1)/O_3(G_1/K_1)\cong Sp_{2k-2}(2)$.
By (1.3)(vi) we conclude that $G_{\{5,\ldots,k\}}/K_{\{5,\ldots,k\}} \cong (3^7)Sp_6(2)$.
Then, applying   [ShStr], we deduce that $( G_2/K_2 )/ O_3(G_2/K_2)  \cong Sp_{2k-2}(2)$.
We know that $K_k = K_{1,k}$, therefore $K_k/
K_1\not\subseteq O_2((G_1\cap G_2)/K_1)$.  We have that $O_2(G_1\cap G_2)/K_1$ involves exactly two irreducible
$G_1\cap G_2/O_2(G_1\cap G_2)$-modules, both of them being not trivial.
If $K_2 \not\subseteq K_1$ then $K_2$ involves at least one of these modules in particular $[K_2,K_k] \not\subseteq K_1$.
But
 since
$K_k/K_1$ is abelian we have  $[K_k,K_2]\subseteq K_1$, a contradiction.
Thus $K_2\subseteq K_1$. We have $|S/K_1|=|S/K_2|$
and so $K_1=K_2=1$.
 
By (4.2) and (1.3) we are left with $G_1/K_1\cong Co_1$ or
$F_1$. Applying (1.3) we see ${G_2}/K_2\cong Co_1$ or $F_1$ as well. If 
${~I\!\!\!\!C}
_{G_1}(K_1)\subseteq K_1$, then by [ParRow]  $G_1/K_1\cong
Co_1$ and $K_1\cong K_2$ is extraspecial of order $2^{25}.$ 
Thus we have  $Z(S)\lhd \langle G_1,G_2\rangle
=G$, a contradiction.

So ${~I\!\!\!\!C}_{G_1}(K_1)\not\subseteq K_1$ and 
${~I\!\!\!\!C}_{G_2}(K_2)\not\subseteq K_2$. In particular $K_1\cap K_2=1$.
Since $Z(G_1\cap G_2/K_1)=1$ if $G_1/K_1 \cong F_1$, in this case we have $K_1=K_2
=1$.

Now let $G_1/K_1\cong Co_1$ and $k=5$. Then $|Z(G_1\cap G_2/K_1)|=2$ and  $|K_1|=|K_2|\le 2$. We have that $G_{4,5}/K_{4,5}$
belongs to 
{$
\begin{picture}(25,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}
\put(7,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(12.5,1.7){\line(1,0){8.9}}


\put(12.5,0.4){\line(1,0){8.9}}

\end{picture}
$.}
By (1.3)(i) $K_{4,5,2}=K_{4,5,1}$. So (1.1)(a)(iv) implies
$G_{5,4,3}/K_{5,4,3} \cong (Z\!\!\! Z_3\times Z\!\!\! Z_3)\cdot Z\!\!\! Z_2$. 
Then we have $G_3/K_3 \cong L_3(2)\times [(Z\!\!\! Z_3\times Z\!\!\! Z_3)\cdot Z\!\!\! Z_2]$ and
$|Z(K_3/K_1)|=2^2$. If $K_1\neq 1$ then $K_1\not\!\!\! \lhd G_3$. Hence $|Z(K_3)|=2^3$.
Since $G_1\cap G_3$  and $G_2\cap G_3$ both induce $\Sigma_3$
on $Z(K_3), G_3$ acts on $Z(K_3)$ in the same way as $G_1\cap G_3$,
so $[K_1,G_3]=1$. This is a contradiction and we get again $K_1=1$.$\hfill\Box$ 
\\[2ex]
{\bf (4.7) Lemma.} 
{\it There is no semiclassical parabolic system $\{P_1,\ldots,
P_5\}$ with diagram $\Delta$ of type
\begin{center}
{$
\begin{picture}(35,12)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(29,-6){\makebox(0,0){$\circ$}}

\put(29,8){\makebox(0,0){$\circ$}}

\put(12.5,1){\line(1,0){8.7}}
\put(2.5,1){\line(1,0){8.7}}

 \put(22.3,1.3){\line(1,1){6.3}}
 

 \put(21.8,0.4){\line(1,-1){6.5}}
 \put(22.5,1){\line(1,-1){6.5}}


\put(24,-3.5){\makebox(0,0){$\scriptstyle{\sim}$}}


\put(31,-6.5){\scriptsize5}
\put(31,8){\scriptsize1}
\put(20.5,-2){\scriptsize4}
\put(11,-2){\scriptsize3}
\put(1,-2){\scriptsize2}
\put(34,0){or}

\end{picture}
$}
\qquad \qquad
{$
\begin{picture}(25,15)

\put(2,-4){\makebox(0,0){$\circ$}}
\put(12,-4){\makebox(0,0){$\circ$}}
\put(2,6){\makebox(0,0){$\circ$}}
\put(12,6){\makebox(0,0){$\circ$}}
\put(22,6){\makebox(0,0){$\circ$}}

\put(-1.5,-5.2){\scriptsize 2}
\put(12,-5.2){ \scriptsize 3}
\put(-1.5,6.8){\scriptsize 1}
\put(11,7.3){\scriptsize 4}
\put(23,6.8){\scriptsize 5}

\put(2.5,-4){\line(1,0){8.7}}

\put(2,-3.2){\line(0,1){8.7}}
\put(12,-3.2){\line(0,1){8.7}}



\put(2.5,6){\line(1,0){8.7}}



\put(12.5,6.4){\line(1,0){8.9}}
\put(12.5,5.6){\line(1,0){8.9}}

\put(17,8.5){\makebox(0,0){$\scriptstyle{\sim}$}}


\end{picture}
$} {.}
\end{center}}
\vspace{4ex}
\noindent
{\bf Proof.} Assume false.
Consider $G_2$. According to (1.3)(v) $G_{2,5}/K_{2,5}\cong A_7$ and
$G_{1,2}/K_{1,2} \cong G_{2,3}/K_{2,3} \cong 3^7 Sp_6(2)$. In the
first diagram $G_5/K_5\cong L_5(2)$, so $G_{2,5}/K_{2,5}\cong A_8$,
a contradiction. In the
second diagram using [Tim1] applied to $G_5$ we conclude that $G_{1,5}/K_{1,5}\cong A_7$ as
well. By (1.3)(viii) we have  $G_1/K_1\cong 3^{35}Sp_8(2)$ 
and so $G_{1,5}/K_{1,5}\cong A_8$, a contradiction again.
$\hfill\Box$\\[2ex]
\noindent
{\bf (4.8) Proposition.} 
{\it Let $\{P_1,\ldots , P_5\}$ be a semiclassical parabolic
system with connected diagram $\Delta$ containing a subdiagram of type
{$
\begin{picture}(40,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}
\put(22.5,0.4){\line(1,0){8.9}}
\put(22.5,1.7){\line(1,0){8.9}}


\put(32,1){\makebox(0,0){$\circ$}}
\put(12.5,1){\line(1,0){8.7}}

\put(27,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(37,1){.}
\end{picture}
$}
Then one of the following holds:
\begin{itemize}
\item [(i)] $\Delta = $
{$
\begin{picture}(50,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}
\put(32.5,0.4){\line(1,0){8.9}}
\put(32.5,1.7){\line(1,0){8.9}}


\put(32,1){\makebox(0,0){$\circ$}}
\put(42,1){\makebox(0,0){$\circ$}}

\put(12.5,1){\line(1,0){8.7}}
\put(22.5,1){\line(1,0){8.7}}

\put(37,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(47,1){,}
\end{picture}
$} $G\cong F_1$ or $3^{155}Sp_{10}(2)$
\item[(ii)] $\Delta =$
{$
\begin{picture}(35,12)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(29,-6){\makebox(0,0){$\circ$}}

\put(29,8){\makebox(0,0){$\circ$}}

\put(2.5,1){\line(1,0){8.7}}
\put(12.5,1){\line(1,0){8.7}}

 \put(21.8,1.7){\line(1,1){6.5}}
 \put(22.5,1){\line(1,1){6.5}}

 \put(21.8,0.4){\line(1,-1){6.5}}
 \put(22.5,1){\line(1,-1){6.5}}

\put(24,5.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(24,-3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}


\put(31,-6.5){\scriptsize1}
\put(31,8){\scriptsize2}


\end{picture}
,$}
$K_1=K_2=1$ and
$G_1\cong G_2\cong Co_1$ or
{$
\begin{picture}(45,12)
\put(0,1){ $G_1\cong$
$3^{35}Sp_8(2)$ 
and $G_2\cong 3^{35}Sp_8(2)$ or $Sp_8(2)$.}
\end{picture}
$}

{$
\begin{picture}(45,8)
\put(0,1){ }

\end{picture}
$}
\end{itemize}}
\noindent
{\bf Proof.}
We choose notation such that we have a subdiagram 
\begin{center}
{$
\begin{picture}(40,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}
\put(22.5,0.4){\line(1,0){8.9}}
\put(22.5,1.7){\line(1,0){8.9}}


\put(32,1){\makebox(0,0){$\circ$}}
\put(12.5,1){\line(1,0){8.7}}

\put(11,-2){\scriptsize 4}
\put(1,-2){\scriptsize 5}
\put(21,-2){\scriptsize 3}
\put(31,-2){\scriptsize 2}


\put(27,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(37,1){.}
\end{picture}
$}
\end{center}
Let us consider the residue of 5. It
contains the subdiagram 
{$
\begin{picture}(26,7)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}

\put(12.5,0.4){\line(1,0){8.9}}
\put(12.5,1.7){\line(1,0){8.9}}



\put(2.5,1){\line(1,0){8.7}}


\put(1,-2){\scriptsize 4}

\put(11,-2){\scriptsize 3}
\put(21,-2){\scriptsize 2}



\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\end{picture}
$}. 
 Suppose that the diagram of the  residue of 5 is 
connected. 

If this is a string then by (1.3)(iii)
it has one of the following shapes:
\begin{center}
{$
\begin{picture}(44,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(22.5,1){\line(1,0){8.7}}
\put(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}


\put(32,1){\makebox(0,0){$\circ$}}
\put(12.5,1){\line(1,0){8.7}}


\put(11,-2){\scriptsize 3}
\put(1,-2){\scriptsize 2}
\put(21,-2){\scriptsize 4}
\put(31,-2){\scriptsize 1}


\put(7,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(40,0){,}
\end{picture}
$}
\qquad
{$
\begin{picture}(44,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}
\put(22.5,0.4){\line(1,0){8.9}}
\put(22.5,1.7){\line(1,0){8.9}}

\put(32,1){\makebox(0,0){$\circ$}}
\put(12.5,1){\line(1,0){8.7}}


\put(11,-2){\scriptsize 3}
\put(1,-2){\scriptsize 2}
\put(21,-2){\scriptsize 4}
\put(31,-2){\scriptsize 1}

\put(7,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(27,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(40,0){,}

\end{picture}
$}
\end{center}
\begin{center}
{$
\begin{picture}(44,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(22.5,1){\line(1,0){8.7}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(12.5,1.7){\line(1,0){8.9}}


\put(32,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}


\put(11,-2){\scriptsize 3}
\put(1,-2){\scriptsize 4}
\put(21,-2){\scriptsize 2}
\put(31,-2){\scriptsize 1}


\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(40,0){, or}
\end{picture}
$}
\qquad
{$
\begin{picture}(40,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(12.5,1.7){\line(1,0){8.9}}
\put(22.5,0.4){\line(1,0){8.9}}
\put(22.5,1.7){\line(1,0){8.9}}

\put(32,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}


\put(11,-2){\scriptsize 3}
\put(1,-2){\scriptsize 4}
\put(21,-2){\scriptsize 2}
\put(31,-2){\scriptsize 1}


\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(26.7,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\end{picture}
$.}

\end{center}
The second diagram is not possible since  according to (1.3)(ii)
 $G_1/K_1\cong He$ in this case but $G_1/K_1$ has to be
isomorphic to $M_{24}$ or
$3^7Sp_6(2)$. The last two diagrams are not possible because of (2.4) and 
(3.5).
If the diagram is a star it is of shape
\begin{center}
{$
\begin{picture}(33,12)

\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(19,-6){\makebox(0,0){$\circ$}}

\put(19,8){\makebox(0,0){$\circ$}}


\put(2.5,1){\line(1,0){8.7}}

 \put(11.8,1.7){\line(1,1){6.5}}
 \put(12.5,1){\line(1,1){6.5}}

 \put(11.8,0.4){\line(1,-1){6.5}}
 \put(12.5,1){\line(1,-1){6.5}}

\put(14,5.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(14,-3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}


\put(21,-6.5){\scriptsize1}
\put(21,8){\scriptsize2}
\put(1,-2){\scriptsize4}
\put(10.5,-2){\scriptsize3}
\put(29,0){or}

\end{picture}
$}
\qquad
{$
\begin{picture}(25,12)

\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(19,-6){\makebox(0,0){$\circ$}}

\put(19,8){\makebox(0,0){$\circ$}}

\put(2.5,1){\line(1,0){8.7}}


 \put(12.3,1.3){\line(1,1){6.3}}
 

 \put(11.8,0.4){\line(1,-1){6.5}}
 \put(12.5,1){\line(1,-1){6.5}}


\put(14,-3.5){\makebox(0,0){$\scriptstyle{\sim}$}}


\put(21,-6.5){\scriptsize2}
\put(21,8){\scriptsize4}
\put(10.5,-2){\scriptsize3}
\put(1,-2){\scriptsize1}



\end{picture}
.$}

\end{center}
\vspace{0.5cm}
By (1.3) it is never a square.
Thus the possible diagrams are:
\begin{center}
{$
\begin{picture}(35,12)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(29,-6){\makebox(0,0){$\circ$}}

\put(29,8){\makebox(0,0){$\circ$}}

\put(2.5,1){\line(1,0){8.7}}
\put(12.5,1){\line(1,0){8.7}}

 \put(21.8,1.7){\line(1,1){6.5}}
 \put(22.5,1){\line(1,1){6.5}}

 \put(21.8,0.4){\line(1,-1){6.5}}
 \put(22.5,1){\line(1,-1){6.5}}

\put(24,5.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(24,-3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}


\put(31,-6.5){\scriptsize1}
\put(31,8){\scriptsize2}
\put(1,-2){\scriptsize5}
\put(11,-2){\scriptsize4}
\put(21,-2){\scriptsize3}
\put(34,0){,}
\end{picture}
$}
\qquad
{$
\begin{picture}(25,15)

\put(2,-4){\makebox(0,0){$\circ$}}
\put(12,-4){\makebox(0,0){$\circ$}}
\put(2,6){\makebox(0,0){$\circ$}}
\put(12,6){\makebox(0,0){$\circ$}}
\put(22,6){\makebox(0,0){$\circ$}}

\put(-1.5,-5.2){\scriptsize 5}
\put(12,-5.2){ \scriptsize 1}
\put(-1.5,6.8){\scriptsize 4}
\put(11,7.3){\scriptsize 3}
\put(23,6.8){\scriptsize 2}



\put(2,-3.2){\line(0,1){8.7}}
\put(11.5,-3.4){\line(0,1){8.9}}
\put(12.5,-3.4){\line(0,1){8.9}}
\put(2.5,-4.1){........}



\put(2.5,6){\line(1,0){8.7}}

\put(16,0){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(12.5,6.4){\line(1,0){8.9}}
\put(12.5,5.6){\line(1,0){8.9}}

\put(17,8.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(26,0){,}
\end{picture}
$}
\qquad 
{$
\begin{picture}(35,12)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(29,-6){\makebox(0,0){$\circ$}}

\put(29,8){\makebox(0,0){$\circ$}}

\put(2.5,1){\line(1,0){8.7}}
\put(12.5,1){\line(1,0){8.7}}

 \put(21.8,1.7){\line(1,1){6.5}}
 \put(22.5,1){\line(1,1){6.5}}

 
 \put(22.1,0.7){\line(1,-1){6.5}}

\put(24,5.5){\makebox(0,0){$\scriptstyle{\sim}$}}


\put(31,-6.5){\scriptsize1}
\put(31,8){\scriptsize2}
\put(1,-2){\scriptsize5}
\put(11,-2){\scriptsize4}
\put(21,-2){\scriptsize3}
\put(34,0){,}
\end{picture}
$}
\end{center}


\begin{center}
{$
\begin{picture}(25,15)

\put(2,-4){\makebox(0,0){$\circ$}}
\put(12,-4){\makebox(0,0){$\circ$}}
\put(2,6){\makebox(0,0){$\circ$}}
\put(12,6){\makebox(0,0){$\circ$}}
\put(22,6){\makebox(0,0){$\circ$}}

\put(-1.5,-5.2){\scriptsize 5}
\put(12,-5.2){ \scriptsize 1}
\put(-1.5,6.8){\scriptsize 4}
\put(11,7.3){\scriptsize 3}
\put(23,6.8){\scriptsize 2}



\put(2,-3.2){\line(0,1){8.7}}
\put(12,-3.2){\line(0,1){8.7}}

\put(2.5,-4.1){........}



\put(2.5,6){\line(1,0){8.7}}


\put(12.5,6.4){\line(1,0){8.9}}
\put(12.5,5.6){\line(1,0){8.9}}

\put(17,8.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(26,0){, \quad and}
\end{picture}
$}\qquad \qquad
{$
\begin{picture}(45,5)
\put(16.5,1){\makebox(0,0){$\circ$}}
\put(26.5,1){\makebox(0,0){$\circ$}}
\put(6.5,1){\makebox(0,0){$\circ$}}

\put(17,1){\line(1,0){8.7}}

\put(7,1.7){\line(1,0){8.9}}
\put(7,0.4){\line(1,0){8.9}}



\put(11,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(34.5,-6){\makebox(0,0){$\circ$}}

\put(34.5,8){\makebox(0,0){$\circ$}}



 \put(27.2,1.4){\line(1,1){6.3}}


 \put(27.2,0.7){\line(1,-1){6.3}}
 
\put(5,-2){\scriptsize 2}
\put(15,-2){ \scriptsize 3}
\put(24.5,-2){\scriptsize 4}
\put(36,-6.5){\scriptsize 5}
\put(36,8){\scriptsize 1}



\put(43,0){.}
\end{picture}
$}
\end{center}
\vspace{0.4cm}
In the first case we are done with (4.6). The third and the last case is not
possible by (4.7) and (4.4) respectively. Considering in the fourth diagram
the residue of 4 we conclude with (1.3)(ii) that it is of type
{$
\begin{picture}(34,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}

\put(2.5,1){\line(1,0){8.7}}


\put(22.5,1.7){\line(1,0){8.9}}
\put(22.5,0.4){\line(1,0){8.9}}


\put(32,1){\makebox(0,0){$\circ$}}
\put(12.5,1){\line(1,0){8.7}}


\put(11,-2){\scriptsize 1}
\put(1,-2){\scriptsize 5}
\put(21,-2){\scriptsize 2}
\put(31,-2){\scriptsize 3}

\put(27,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}


\end{picture}
$}
and so $\Delta$ is of the shape
\begin{center}
{$
\begin{picture}(25,10)



\put(2.5,-4){\line(1,0){8.7}}

\put(2,-3.2){\line(0,1){8.7}}
\put(12,-3.2){\line(0,1){8.7}}



\put(2.5,6){\line(1,0){8.7}}

\put(-1.5,-5.2){\scriptsize 5}
\put(12,-5.2){ \scriptsize 1}
\put(-1.5,6.8){\scriptsize 4}
\put(11,7.3){\scriptsize 2}
\put(23,6.8){\scriptsize 3}


\put(2,-4){\makebox(0,0){$\circ$}}
\put(12,-4){\makebox(0,0){$\circ$}}
\put(2,6){\makebox(0,0){$\circ$}}
\put(12,6){\makebox(0,0){$\circ$}}
\put(22,6){\makebox(0,0){$\circ$}}


\put(12.5,6.4){\line(1,0){8.9}}
\put(12.5,5.6){\line(1,0){8.9}}

\put(17,8.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(25,0){,}
\end{picture}
$}
\end{center}
 but this
contradicts (4.7).

There is the second diagram left. Considering the residue of 2 we get
 by (1.3)(ix)(x)
\begin{center}
{$
\begin{picture}(25,10)

\put(2,-4){\makebox(0,0){$\circ$}}
\put(12,-4){\makebox(0,0){$\circ$}}
\put(2,6){\makebox(0,0){$\circ$}}
\put(12,6){\makebox(0,0){$\circ$}}
\put(22,6){\makebox(0,0){$\circ$}}

\put(-1.5,-5.2){\scriptsize 5}
\put(12,-5.2){ \scriptsize 1}
\put(-1.5,6.8){\scriptsize 4}
\put(11,7.3){\scriptsize 3}
\put(23,6.8){\scriptsize 2}

\put(2.5,6){\line(1,0){8.7}}

\put(2,-3.2){\line(0,1){8.7}}
\put(11.4,-3.4){\line(0,1){8.9}}
\put(12.6,-3.4){\line(0,1){8.9}}



\put(2.5,-4){........}


\put(12.5,6.4){\line(1,0){8.9}}
\put(12.5,5.6){\line(1,0){8.9}}

\put(17,8.5){\makebox(0,0){$\scriptstyle{\sim}$}}


\end{picture}
$}{.}
\end{center}
\vspace{0.3cm}
Considering $G_4/K_4$ we get a contradiction to (1.3) again.

Let us now assume  that $\Delta_5$
is not connected. Then by [Str5] $\Delta$ is of type
\begin{center}
{$
\begin{picture}(50,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}

\put(2.5,1){\line(1,0){8.7}}
\put(32.5,0.4){\line(1,0){8.9}}
\put(32.5,1.7){\line(1,0){8.9}}

\put(1,-2){\scriptsize 1}
\put(11,-2){\scriptsize 5}
\put(21,-2){\scriptsize 4}
\put(31,-2){\scriptsize 3}
\put(41,-2){\scriptsize 2}

\put(32,1){\makebox(0,0){$\circ$}}
\put(42,1){\makebox(0,0){$\circ$}}

\put(12.5,1){\line(1,0){8.7}}
\put(22.5,1){\line(1,0){8.7}}

\put(37,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(50,0){or}
\end{picture}
$}
\qquad
{$
\begin{picture}(50,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}

\put(2.5,1.7){\line(1,0){8.9}}
\put(2.5,0.4){\line(1,0){8.9}}
\put(7,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(32.5,0.4){\line(1,0){8.9}}
\put(32.5,1.7){\line(1,0){8.9}}

\put(1,-2){\scriptsize 1}
\put(11,-2){\scriptsize 5}
\put(21,-2){\scriptsize 4}
\put(31,-2){\scriptsize 3}
\put(41,-2){\scriptsize 2}

\put(32,1){\makebox(0,0){$\circ$}}
\put(42,1){\makebox(0,0){$\circ$}}

\put(12.5,1){\line(1,0){8.7}}
\put(22.5,1){\line(1,0){8.7}}

\put(37,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(47,1){.}
\end{picture}
$}
\end{center}
The second diagram is out by (4.5), the first yields case (i) by
(4.2). Now the proposition is proved.$\hfill\Box$\\[2ex]
{\bf (4.9) Proposition.} 
{\it Let $\{P_1,\ldots , P_n\}$, $n\geq 5$, be a semiclassical parabolic
system with connected diagram $\Delta$ containing a subdiagram 
\begin{center}
{$
\begin{picture}(10,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}


\put(2.5,1){\line(1,0){8.7}}
\put(1,-2){{\scriptsize 1}}
\put(11,-2){{\scriptsize 2}}




\end{picture}
$}
{$
\begin{picture}(38,5)
\put(16.5,1){\makebox(0,0){$\circ$}}
\put(26.5,1){\makebox(0,0){$\circ$}}
\put(36.5,1){\makebox(0,0){$\circ$}}

\put(17,1){\line(1,0){8.7}}

\put(4,1){$\; . \; . \; .\;$}

\put(35,-2){\scriptsize n-1}
\put(25,-2){\scriptsize n-2}


\put(27,1.7){\line(1,0){8.9}}
\put(27,0.4){\line(1,0){8.9}}
\put(31,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(42,0){.}

\end{picture}
$}
\end{center}
Then $\Delta$ is one of the following
\begin{itemize}
\item [(i)] 
{$
\begin{picture}(20,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}

\put(2.5,1){\line(1,0){8.7}}
\put(11,-2){{\scriptsize 1}}
\put(21,-2){{\scriptsize 2}}
\put(1,-2){{\scriptsize n}}

\put(12.5,1){\line(1,0){8.7}}


\put(12.5,1){\line(1,0){8.7}}

\end{picture}
$}
{$
\begin{picture}(38,5)
\put(16.5,1){\makebox(0,0){$\circ$}}
\put(26.5,1){\makebox(0,0){$\circ$}}
\put(36.5,1){\makebox(0,0){$\circ$}}

\put(17,1){\line(1,0){8.7}}

\put(4,1){$\; . \; . \; .\;$}


\put(27,1.7){\line(1,0){8.9}}
\put(27,0.4){\line(1,0){8.9}}
\put(31,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}


\put(35,-2){\scriptsize n-1}
\put(25,-2){\scriptsize n-2}
 


\put(45,0){or}
\end{picture}
$}
\item[(ii)] 
{$
\begin{picture}(20,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}

\put(2.5,1){\line(1,0){8.7}}


\put(12.5,1){\line(1,0){8.7}}


\put(12.5,1){\line(1,0){8.7}}
\put(11,-2){{\scriptsize 2}}
\put(21,-2){{\scriptsize 3}}
\put(1,-2){{\scriptsize 1}}

\end{picture}
$}
{$
\begin{picture}(14,12)
\put(4,1){$\; . \; . \; .\;$}
\end{picture}
$}
{$
\begin{picture}(25,12)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(19,-6){\makebox(0,0){$\circ$}}

\put(19,8){\makebox(0,0){$\circ$}}

\put(2.5,1){\line(1,0){8.7}}

 \put(11.8,1.7){\line(1,1){6.5}}
 \put(12.5,1){\line(1,1){6.5}}

 \put(11.8,0.4){\line(1,-1){6.5}}
 \put(12.5,1){\line(1,-1){6.5}}

\put(14,5.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(14,-3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(22,-6.5){\scriptsize n}
\put(22,8){\scriptsize n-1}
\put(8,-2){\scriptsize n-2}



\end{picture}
$.}
\end{itemize}}
\vspace{0.5cm}
\noindent
{\bf Proof.} We prove the assertion by induction on $n$.
The case $n=5$ is Proposition (4.8). So we may assume that $n>5$.
We have to determine how the node $n$ is connected to the
given subdiagram.
By induction applied to $\Delta_1$ the node $n$ can just be
connected with the nodes $1$, $2$, or $n-2$.
Suppose $n$ is not connected with the nodes $1$ and $n-2$. Then
$n$ is just connected with $2$. By induction we get that
$\Delta_1$ has the shape
\begin{center}
{$
\begin{picture}(20,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}

\put(2.5,1){\line(1,0){8.7}}
\put(11,-2){{\scriptsize 2}}
\put(21,-2){{\scriptsize 3}}
\put(1,-2){{\scriptsize n}}

\put(12.5,1){\line(1,0){8.7}}


\put(12.5,1){\line(1,0){8.7}}

\end{picture}
$}
{$
\begin{picture}(40,5)
\put(16.5,1){\makebox(0,0){$\circ$}}
\put(26.5,1){\makebox(0,0){$\circ$}}
\put(36.5,1){\makebox(0,0){$\circ$}}

\put(17,1){\line(1,0){8.7}}

\put(4,1){$\; . \; . \; .\;$}


\put(27,1.7){\line(1,0){8.9}}
\put(27,0.4){\line(1,0){8.9}}
\put(31,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}


\put(25.2,-2){\scriptsize n-2}
\put(35.2,-2){\scriptsize n-1}
 



\end{picture}
,$}
\end{center}
but this is a contradiction to (4.4).

So $n$ is connected with $1$ or $n-2$.
Suppose $n$ is just connected with $1$.
By [Str5] applied to $\Delta_{\{4,\ldots, n-1\}}$ and (4.5) we get
that $\Delta$ must be of the shape
\begin{center}
{$
\begin{picture}(20,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}

\put(2.5,1){\line(1,0){8.7}}
\put(11,-2){{\scriptsize 1}}
\put(21,-2){{\scriptsize 2}}
\put(1,-2){{\scriptsize n}}

\put(12.5,1){\line(1,0){8.7}}


\put(12.5,1){\line(1,0){8.7}}

\end{picture}
$}
{$
\begin{picture}(40,5)
\put(16.5,1){\makebox(0,0){$\circ$}}
\put(26.5,1){\makebox(0,0){$\circ$}}
\put(36.5,1){\makebox(0,0){$\circ$}}

\put(17,1){\line(1,0){8.7}}

\put(4,1){$\; . \; . \; .\;$}


\put(27,1.7){\line(1,0){8.9}}
\put(27,0.4){\line(1,0){8.9}}
\put(31,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}


\put(25.2,-2){\scriptsize n-2}
\put(35.2,-2){\scriptsize n-1}
 



\end{picture}
,$}
\end{center}
this is case (i).

Let finally $n$ be connected with $n-2$. By induction we see that $\Delta_1$
is of the shape
\begin{center}
{$
\begin{picture}(20,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}

\put(2.5,1){\line(1,0){8.7}}


\put(12.5,1){\line(1,0){8.7}}


\put(12.5,1){\line(1,0){8.7}}
\put(1,-2){{\scriptsize 2}}
\put(11,-2){{\scriptsize 3}}


\end{picture}
$}
{$
\begin{picture}(14,12)
\put(4,1){$\; . \; . \; .\;$}
\end{picture}
$}
{$
\begin{picture}(25,12)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(19,-6){\makebox(0,0){$\circ$}}

\put(19,8){\makebox(0,0){$\circ$}}

\put(2.5,1){\line(1,0){8.7}}

 \put(11.8,1.7){\line(1,1){6.5}}
 \put(12.5,1){\line(1,1){6.5}}

 \put(11.8,0.4){\line(1,-1){6.5}}
 \put(12.5,1){\line(1,-1){6.5}}

\put(14,5.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(14,-3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(22,-6.5){\scriptsize n}
\put(22,8){\scriptsize n-1}
\put(8,-2){\scriptsize n-2}



\end{picture}
$.}
\end{center}
\vspace{0.5cm}
\noindent
If $n$ is not connected to $1$ this is case (ii). So suppose that
$n$ and $1$ are connected as well. Then
$\Delta_{\{2,\ldots, n-3\}}$ has the form 
{$
\begin{picture}(35,7)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(32,1){\makebox(0,0){$\circ$}}
\put(12.5,1.7){\line(1,0){8.9}}
\put(12.5,0.4){\line(1,0){8.9}}

\put(2.5,1){........}
\put(22.5,1.7){\line(1,0){8.9}}
\put(22.5,0.4){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}
\put(27,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(1,-2){{\scriptsize 1}}
\put(11,-2){{\scriptsize n}}

\put(20,-2){{\scriptsize n-2}}
\put(30,-2){{\scriptsize n-1}}

\end{picture}
$.}
By (1.3) we have that $\Delta_{\{2,\ldots, n-3\}}$ must
be
{$
\begin{picture}(35,7)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(32,1){\makebox(0,0){$\circ$}}
\put(12.5,1.7){\line(1,0){8.9}}
\put(12.5,0.4){\line(1,0){8.9}}

\put(2.5,1){\line(1,0){8.7}}
\put(22.5,1.7){\line(1,0){8.9}}
\put(22.5,0.4){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(27,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(1,-2){{\scriptsize 1}}
\put(11,-2){{\scriptsize n}}

\put(20,-2){{\scriptsize n-2}}
\put(30,-2){{\scriptsize n-1}}

\end{picture}
$.}
Now  $\Delta_{n-1}$ is a circle which has a subdiagram of the shape
\begin{center}
{$
\begin{picture}(20,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}

\put(2.5,1){\line(1,0){8.7}}
\put(11,-2){{\scriptsize 3}}

\put(1,-2){{\scriptsize 2}}

\put(12.5,1){\line(1,0){8.7}}


\put(12.5,1){\line(1,0){8.7}}

\end{picture}
$}
{$
\begin{picture}(40,5)
\put(16.5,1){\makebox(0,0){$\circ$}}
\put(26.5,1){\makebox(0,0){$\circ$}}
\put(36.5,1){\makebox(0,0){$\circ$}}

\put(17,1){\line(1,0){8.7}}

\put(4,1){$\; . \; . \; .\;$}


\put(27,1.7){\line(1,0){8.9}}
\put(27,0.4){\line(1,0){8.9}}
\put(31,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}


\put(36,-2){\scriptsize n}
\put(25.2,-2){\scriptsize n-2}
 



\end{picture}
.$}
\end{center}
So we get a contradiction by induction.\hfill
$\Box$\\[0.3cm]
\noindent
\begin{center}
\boldmath \S\unboldmath {\bf 5 Proof of the theorem}
\end{center}
\vspace{1cm}

In this chapter we are going to prove the main theorem. This will be done
by induction on the rank of the diagram.
In (2.4) we have shown that $\Delta$ has no residue of
$\tilde{F_4}$-type. Now we will show that $\Delta$ cannot contain a
residue of square type.\\[0.2cm]
\setlength{\unitlength}{1mm}
{\bf (5.1) Lemma. } {\it There is no parabolic system $\{P_1,P_2,P_3,P_4,P_5\}$
such that the diagram for $\{P_1,P_2,P_3,P_4\}$ is of type} 
\begin{center}
{$
\begin{picture}(15,15)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(2,11){\makebox(0,0){$\circ$}}
\put(12,11){\makebox(0,0){$\circ$}}

\put(-1.5,-0.2){\scriptsize 1}
\put(13,-0.2){ \scriptsize 2}
\put(-1.5,11.8){\scriptsize 3}
\put(13,11.8){\scriptsize 4}

\put(2.5,1){\line(1,0){8.9}}

\put(1.6,1.6){\line(0,1){8.9}}
\put(2.4,1.6){\line(0,1){8.9}}



\put(2.5,11.4){\line(1,0){8.9}}
\put(2.5,10.6){\line(1,0){8.9}}

\put(7,13.5){\makebox(0,0){$\scriptstyle{\sim}$}}


\put(11.6,1.6){\line(0,1){8.9}}
\put(12.4,1.6){\line(0,1){8.9}}
\end{picture}
$}
\end{center}
\noindent
{\bf Proof.} 
Suppose false.
We have that  
 $\{P_1,P_2,P_3,P_4\}$ is as in (1.3)(ix). Further 
$P_i/O_2(P_i)\cong \Sigma_3$ for  $1\leq i\leq 5$ by (1.2). 
 In particular we know that $G_{1,5}/K_{1,5}$ is of type (1.1)(a)(iv)($\star \star$) and
 $G_{3,5}/K_{3,5} \cong A_7$.
Let $\Delta$ be the diagram for 
$\{P_1,P_2,P_3,P_4,P_5\}$.
 As $\Delta$ does not contain triangles by (1.1)
we may assume that 5 is not connected with 1. 
In particular $\Delta_1$
is connected.

By (1.3) we get that $\Delta_1$ is one of the following three diagrams:
\begin{center}
{$
\begin{picture}(15,1)
\put(3,5){(1)}


\end{picture}
$}
{$
\begin{picture}(15,10)
\put(2,6){\makebox(0,0){$\circ$}}
\put(12,6){\makebox(0,0){$\circ$}}
\put(19,-1){\makebox(0,0){$\circ$}}

\put(19,13){\makebox(0,0){$\circ$}}

\put(2.5,6.6){\line(1,0){8.9}}
\put(2.5,5.4){\line(1,0){8.9}}
\put(7,8){\makebox(0,0){$\scriptstyle{\sim}$}}
 \put(11.8,6.7){\line(1,1){6.5}}
 \put(12.5,6){\line(1,1){6.5}}

 \put(11.8,5.4){\line(1,-1){6.5}}
 \put(12.5,6){\line(1,-1){6.5}}



\put(14,1.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(1,2.5){\scriptsize3}
\put(10,2.5){\scriptsize4}
\put(21,-1.5){\scriptsize5}
\put(21,13){\scriptsize2}
\put(18,5){,}

\end{picture}
$}
\quad 
{$
\begin{picture}(15,1)
\put(7,5){(2)}


\end{picture}
$}
{$
\begin{picture}(15,15)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(2,11){\makebox(0,0){$\circ$}}
\put(12,11){\makebox(0,0){$\circ$}}

\put(-1.5,-0.2){\scriptsize 5}
\put(13,-0.2){ \scriptsize 2}
\put(-1.5,11.8){\scriptsize 3}
\put(13,11.8){\scriptsize 4}

\put(2.5,1){\line(1,0){8.9}}

\put(1.6,1.6){\line(0,1){8.9}}
\put(2.4,1.6){\line(0,1){8.9}}



\put(2.5,11.4){\line(1,0){8.9}}
\put(2.5,10.6){\line(1,0){8.9}}

\put(7,13.5){\makebox(0,0){$\scriptstyle{\sim}$}}


\put(11.6,1.6){\line(0,1){8.9}}
\put(12.4,1.6){\line(0,1){8.9}}
\put(15.5,5){,}

\end{picture}
$}
\qquad 
{$
\begin{picture}(15,8)
\put(-2,5){or}
\put(5,5){(3)}


\end{picture}
$}
{$
\begin{picture}(15,15)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(2,11){\makebox(0,0){$\circ$}}
\put(12,11){\makebox(0,0){$\circ$}}

\put(-1.5,-0.2){\scriptsize 5}
\put(13,-0.2){ \scriptsize 2}
\put(-1.5,11.8){\scriptsize 3}
\put(13,11.8){\scriptsize 4}

\put(2.5,1.4){\line(1,0){8.9}}
\put(2.5,0.6){\line(1,0){8.9}}


\put(2.4,1.6){\line(0,1){8.9}}
\put(1.6,1.6){\line(0,1){8.9}}



\put(2.5,11.4){\line(1,0){8.9}}
\put(2.5,10.6){\line(1,0){8.9}}

\put(7,13.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(-3,5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}
\put(7,-2.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(11.6,1.6){\line(0,1){8.9}}
\put(12.4,1.6){\line(0,1){8.9}}
\end{picture}
$}
\end{center}
In case (1) $\Delta_3$ has the shape
\begin{center}
{$
\begin{picture}(35,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(32,1){\makebox(0,0){$\circ$}}

\put(2.5,1.7){\line(1,0){8.9}}
\put(2.5,0.4){\line(1,0){8.9}}
\put(7,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(12.5,1.7){\line(1,0){8.9}}
\put(12.5,0.4){\line(1,0){8.9}}

\put(22.5,1){\line(1,0){8.7}}

\put(1,-2){{\scriptsize 5}}
\put(11,-2){{\scriptsize 4}}
\put(21,-2){{\scriptsize 2}}
\put(31,-2){\scriptsize 1}
\put(33,0){,}
\end{picture}
$}
\end{center}
this cannot happen by (1.3) and [Str2].
In case (2) $\Delta_3$ has the shape 
\begin{center}
{$
\begin{picture}(25,12)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(19,-6){\makebox(0,0){$\circ$}}

\put(19,8){\makebox(0,0){$\circ$}}

\put(2.5,1.7){\line(1,0){8.9}}
\put(2.5,0.4){\line(1,0){8.9}}

 \put(12.4,1.6){\line(1,1){6.1}}
 

 \put(12.4,0.4){\line(1,-1){6.1}}

\put(1,-2){{\scriptsize 4}}
\put(11,-2){{\scriptsize 2}}
\put(21,-7){{\scriptsize 5}}
\put(21,9){\scriptsize 1}


\end{picture}
$.}
\end{center}
\vspace{0.5cm}
Since $G_{3,5}/K_{3,5} \cong A_7$ it follows with [Str2] that this cannot occur.
There is case (3) left. Now $\Delta_4$ has the shape
\begin{center}
{$
\begin{picture}(15,15)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(2,11){\makebox(0,0){$\circ$}}
\put(12,11){\makebox(0,0){$\circ$}}

\put(-1.5,-0.2){\scriptsize 5}
\put(13,-0.2){ \scriptsize 2}
\put(-1.5,11.8){\scriptsize 3}
\put(13,11.8){\scriptsize 1}

\put(2.5,1.4){\line(1,0){8.9}}
\put(2.5,0.6){\line(1,0){8.9}}

\put(2.4,1.6){\line(0,1){8.9}}
\put(1.6,1.6){\line(0,1){8.9}}



\put(2.5,11.4){\line(1,0){8.9}}
\put(2.5,10.6){\line(1,0){8.9}}


\put(-3.3,5.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}
\put(7,-2.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}


\put(12,1.8){\line(0,1){8.7}}
\end{picture}
$.}
\end{center}
Application of (1.3) resp. [Str2] shows that  $G_{3,4}/K_{3,4} \cong A_7$ .
But now $\Delta_3$ does not exist by [Str2].
\hfill $\Box$\\[0.3cm]
{\bf (5.2) Lemma.} {\it 
If $\Delta$ is one of the diagrams of the theorem then the corresponding
statements about the groups hold as well.}\\[0.3cm]
{\bf Proof.} 
We prove the claim by induction on the rank of $\Delta$. If $rank(\Delta) = 4$ then this is (1.3). So let $rank(\Delta)\geq 5$.

Let first $\Delta$ be a complete bipartite graph all of whose edges are
of type 
{$
\begin{picture}(15,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}

\put(2.5,1.4){\line(1,0){8.9}}
\put(2.5,0.6){\line(1,0){8.9}}
\put(7,2.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}
\end{picture}
$}
. Choose $x\in I$ such that $\Delta_x$ is connected, contains
{$
\begin{picture}(15,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}

\put(2.5,1.4){\line(1,0){8.9}}
\put(2.5,0.6){\line(1,0){8.9}}
\put(7,2.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}
\end{picture}
$}
, and $|K_x|$ is minimal. By induction there exists a subset $J$,
$x\in J\subseteq I$, such that $\Delta_J$ is of shape
{$
\begin{picture}(25,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(12.5,1.4){\line(1,0){8.9}}
\put(12.5,0.6){\line(1,0){8.9}}
\put(2.5,1.4){\line(1,0){8.9}}
\put(2.5,0.6){\line(1,0){8.9}}
\put(7,2.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}
\put(17,2.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}
\end{picture}
$}
and $K_J=K_x$. As $|I| \geq 5$ there is $y \not= x$, $y\in J$. Now $|K_y| \geq |K_x|$ and $K_y \subseteq K_J$, hence $K_y = K_x = 1$.

Let $\Delta$ be of type $He(n,m)$ and $J \subseteq I$ be any subset such that 
$\Delta_J$ is of shape
{$
\begin{picture}(25,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}

\put(12.5,1.4){\line(1,0){8.9}}
\put(12.5,0.6){\line(1,0){8.9}}
\put(17,2.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\end{picture}
$}.
Then again because of $|I|\geq 5$ there are $x,y \in J$, $x\not=y$. By induction applied to $\Delta_x$ resp. $\Delta_y$ we get
$K_x=K_J=K_y =1$.

Let $\Delta$ be of type $X(2,m)$ and $J\subseteq I$ such that
$\Delta_J$ is of shape 
{$
\begin{picture}(25,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}

\put(12.5,1.4){\line(1,0){8.9}}
\put(12.5,0.6){\line(1,0){8.9}}
\put(17,2.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}
\end{picture}
$}.
By [Str1; 4.3.1] applied to $\Delta_{J\setminus \{1,2\}}$ we have
$G_J/K_J \cong A_7$. Hence by (1.3) $  G_J/K_J \cong (3^7)Sp_6(2)$.
Let $J^\prime = \{1,4,\ldots , 4+m-1\}$. Then $K_{J^\prime} = K_4$ by
induction applied to $\Delta_4$. On the other hand
$\Delta_1$ is of type $Y(1,m+1)$ with
$G_{J^\prime}/K_{J^\prime} \cong 3^7Sp_6(2)$. So
$K_{J^\prime}=K_1$, hence $K_{J^\prime}=1$. Now
$|S|=2^9$ and $K_J=1$ for any $J$ with $G_J/K_J \cong (3^7)Sp_6(2)$.

Finally, if $J^\prime = \{4, \ldots , m+4 \}$ we see
$G_{J^\prime} \cong 2^6A_7$ by (1.3)(v) applied to $\Delta_{I\setminus 
\{1,2,3,m+4\}}$.

Let $\Delta$ be of type $Y(n,m)$. If $m=1$, we get the assertion
with (1.3) and (4.2), up to the fact that it was not
stated in (1.3) that for the diagram 
\vspace{0.5cm}
\begin{center}
{$
\begin{picture}(15,10)
\put(2,6){\makebox(0,0){$\circ$}}
\put(12,6){\makebox(0,0){$\circ$}}
\put(19,-1){\makebox(0,0){$\circ$}}

\put(19,13){\makebox(0,0){$\circ$}}

\put(2.5,6){\line(1,0){8.7}}

 \put(11.8,6.7){\line(1,1){6.5}}
 \put(12.5,6){\line(1,1){6.5}}

 \put(11.8,5.4){\line(1,-1){6.5}}
 \put(12.5,6){\line(1,-1){6.5}}



\put(14,1.5){\makebox(0,0){$\scriptstyle{\sim} $}}
\put(14,10.5){\makebox(0,0){$\scriptstyle{\sim} $}}

\put(1,2.5){\scriptsize1}
\put(10,2.5){\scriptsize2}
\put(21,-1.5){\scriptsize3}
\put(21,13){\scriptsize4}

\end{picture}
$}
\end{center}
with $G_3, G_4 \cong M_{24}, He$ always $G_3 \cong G_4$ holds. So suppose
$G_3 \cong M_{24}$ and $G_4 \cong He$.
Then by [FuStr; (3.3)] we would get a system with diagram
\begin{center}
{$
\begin{picture}(35,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(32,1){\makebox(0,0){$\circ$}}
\put(2.5,1){\line(1,0){8.7}}
\put(12.5,1){\line(1,0){8.7}}
\put(22.5,1.4){\line(1,0){8.9}}
\put(22.5,0.6){\line(1,0){8.9}}
\put(1,-2){{\scriptsize x}}
\put(11,-2){{\scriptsize 1}}
\put(21,-2){{\scriptsize 2}}
\put(31,-2){\scriptsize 3}
\put(27,2.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\end{picture}
$}
\quad
{$
\begin{picture}(5,5)
\put(1,0){or}

\end{picture}
$}
\quad
{$
\begin{picture}(15,7)
\put(2,-4){\makebox(0,0){$\circ$}}
\put(12,-4){\makebox(0,0){$\circ$}}
\put(2,6){\makebox(0,0){$\circ$}}
\put(12,6){\makebox(0,0){$\circ$}}

\put(-1.5,-5.2){\scriptsize x}
\put(13,-5.2){ \scriptsize 3}
\put(-1.5,6.8){\scriptsize 1}
\put(13,6.8){\scriptsize 2}


\put(2.5,-4){........}


\put(2,-3.1){\line(0,1){8.7}}



\put(2.5,6){\line(1,0){8.7}}

\put(14,0){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(11.6,-3.3){\line(0,1){8.9}}
\put(12.4,-3.3){\line(0,1){8.9}}
\end{picture}
$}
\end{center}
and $G_x/K_x \cong He$, a contradiction to (1.3).

 If $m=2$ and $n>1$, we may apply (4.6).
So let $m>2$. 
Let $J\subset I $ belong to the diagram $\:$
{$
\begin{picture}(42,5)
\put(1,1){\makebox(0,0){$\circ$}}
\put(1.6,1){\line(1,0){8.5}}
\put(11,1){\makebox(0,0){$\circ$}}

\put(13,1.2){...}
\put(19,1){\makebox(0,0){$\circ$}}
\put(19.6,1){\line(1,0){8.5}}
\put(29,1){\makebox(0,0){$\circ$}}

\put(39,1){\makebox(0,0){$\circ$}}





 \put(29.5,1.7){\line(1,0){8.8}}
 \put(29.5,0.4){\line(1,0){8.8}}



\put(34,3.2){\makebox(0,0){\tiny( $\! \! \scriptstyle{\sim} $\tiny)}}


\put(26,-2.2){\scriptsize n+1}
\put(18,-2.2){\scriptsize n}


\put(10,-2.2){\scriptsize 2}
\put(0,-2.2){\scriptsize 1}



\end{picture}
$.}
Then by induction we get that $G_J$ is as in 
{$
\begin{picture}(6,6)
\put(0,0){the}
\end{picture}
$}
definition of a $Y(n,m)$-structure. It just remains to show that there are
no $J_1,J_2,J_1\neq J_2$ such that $G_{J_1}\cong 2^6A_7=G_{J_2}$. But this
is equivalent to the following: 
There is no rank four classical system with diagram
\begin{center}
{$
\begin{picture}(25,15)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(19,-6){\makebox(0,0){$\circ$}}

\put(19,8){\makebox(0,0){$\circ$}}

\put(2.5,1){\line(1,0){8.7}}

 \put(11.8,1.7){\line(1,1){6.5}}
 \put(12.5,1){\line(1,1){6.5}}

 \put(11.8,0.4){\line(1,-1){6.5}}
 \put(12.5,1){\line(1,-1){6.5}}



\put(1,-2){{\scriptsize 1}}
\put(11,-2){{\scriptsize 2}}
\put(21,-7){{\scriptsize 3}}
\put(21,9){\scriptsize 4}



\end{picture}
$}
\end{center}
\vspace{0.4cm}
such that $G_4/K_4\cong G_3/K_3\cong A_7$. But this follows with [Str4].\\

Let finally $\Delta$ be of type $M(n,m)$. Let
$J=I\setminus \{1,t,u,v\}$ such that
$\Delta_J$ is of shape
{$
\begin{picture}(35,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(32,1){\makebox(0,0){$\circ$}}


\put(2.5,1){\line(1,0){8.7}}

\put(22.5,1.7){\line(1,0){8.9}}
\put(22.5,0.4){\line(1,0){8.9}}
\put(27,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(12.5,1.7){\line(1,0){8.9}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}






\put(1,-2){{\scriptsize 1}}
\put(11,-2){{\scriptsize t}}
\put(21,-2){{\scriptsize u}}
\put(31,-2){\scriptsize v}
\put(34,0){.}
\end{picture}
$}
Application of (1.3)(iv) shows that $G_J/K_J$ is locally isomorphic to 
$M(24),M(24)',3M(24)$ or $3M(24)'$. If $rank(\Delta)=5$ we get $K_J=1$ with (3.5). Otherwise there are 
$x,y \in J$, $x\not=y$, and we get $K_x=K_J=K_y=1$ by induction.

This proves the lemma.\hfill $\Box$\\[0.3cm]

Now we are going to show that a connected diagram $\Delta$ such that any subdiagram of $\Delta$ which contains 
{$
\begin{picture}(15,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}

\put(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}

\put(7,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\end{picture}
$}
is one of the diagrams of the theorem, is itself a diagram of the theorem.
For this we choose some $x\in I$ such that
$\Delta_x$ is connected. As there are at least two possible choices we 
may assume that $\Delta_x$ contains 
{$
\begin{picture}(15,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}

\put(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}

\put(7,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\end{picture}
$}.\\[3ex]
{\bf (5.3) Lemma.} 
{\it If $\Delta_x$ is a complete bipartite graph all of whose
edges are of type {$
\begin{picture}(15,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}

\put(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}

\put(7,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}
\put(14,0){,}
\end{picture}
$} then $\Delta$ is one of the diagrams of the
theorem.}\\[2ex]
{\bf Proof.} Let $I_1\cup I_2$ be the partion of the nodes of $\Delta_x$,
$n=|I_1|$, and $m=|I_2|$.
Assume first $I_1=\{1\}$, i.e. $\Delta_x$ is a star. Suppose $x$ is
connected to 1 in $\Delta$. As $\Delta$ does not contain triangles by 
(1.2) and (1.1) we get that $x$ is not connected with any $y\in I_2$. Hence the
diagram $\Delta$ has the shape of a star. If we have the subdiagram
 {$
\begin{picture}(15,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}

\put(1,-2){\tiny 1}
\put(11,-2){\tiny x}

\put(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}

\put(7,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}
\put(14,0){,}
\end{picture}
$} 
then $\Delta$ is 
as in the theorem (i). If we have the subdiagram  
{$
\begin{picture}(15,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}

\put(1,-2){\tiny 1}
\put(11,-2){\tiny x}

\put(2.5,1){\line(1,0){8.7}}


\put(14,0){,}
\end{picture}
$}
 then $\Delta$ is of the
shape $Y(1,m)$.
So assume now that $x$ is not connected with 1 in $\Delta$ but with some
$2\in I_2$. Assume there is some $3\in I_2$, such that 3 is not connected to
$x$ in $\Delta$. Then $\Delta_{I\setminus\{1,2,3,x\}}$ 
is of shape 
\begin{center}
{$
\begin{picture}(35,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(32,1){\makebox(0,0){$\circ$}}

\put(2.5,1.7){\line(1,0){8.9}}
\put(2.5,0.4){\line(1,0){8.9}}
\put(7,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(12.5,1.7){\line(1,0){8.9}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}
\put(22.5,1){........}

\put(1,-2){{\scriptsize 3}}
\put(11,-2){{\scriptsize 1}}
\put(21,-2){{\scriptsize 2}}
\put(31,-2){\scriptsize x}
\put(33,0){.}
\end{picture}
$}
\end{center} By (1.3) and [Str2] we see that
$\Delta_{I\setminus\{1,2,3,x\}}$ is of shape 
\begin{center}
{$
\begin{picture}(35,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(32,1){\makebox(0,0){$\circ$}}

\put(2.5,1.7){\line(1,0){8.9}}
\put(2.5,0.4){\line(1,0){8.9}}
\put(7,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(12.5,1.7){\line(1,0){8.9}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim} $}}
\put(22.5,1){\line(1,0){8.7}}

\put(1,-2){{\scriptsize 3}}
\put(11,-2){{\scriptsize 1}}
\put(21,-2){{\scriptsize 2}}
\put(31,-2){\scriptsize x}
\put(33,0){.}
\end{picture}
$}
\end{center}

 Let $4\in I_2, 4\neq 2$,
such that $x$ is connected with 4 in $\Delta$. Then we get that 
$\Delta_{I\setminus\{1,2,4,x\}}$ is of type 
\begin{center}
{$
\begin{picture}(15,10)

\put(2,-4){\makebox(0,0){$\circ$}}
\put(12,-4){\makebox(0,0){$\circ$}}
\put(2,6){\makebox(0,0){$\circ$}}
\put(12,6){\makebox(0,0){$\circ$}}

\put(-2,1){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}




\put(12,-3.2){\line(0,1){8.7}}

\put(1.4,-3.4){\line(0,1){8.9}}
\put(2.6,-3.4){\line(0,1){8.9}}



\put(2.5,-4){........}

\put(-1.5,-5.2){\scriptsize 4}
\put(12,-5.2){ \scriptsize x}
\put(-1.5,6.8){\scriptsize 1}
\put(12,6.8){\scriptsize 2}

\put(2.5,6.4){\line(1,0){8.9}}
\put(2.5,5.6){\line(1,0){8.9}}

\put(7,8.5){\makebox(0,0){$\scriptstyle{\sim}$}}


\end{picture}
$}
\end{center}
\vspace{0.4cm}
 contradicting (1.3). Hence $x$ is just connected with 2 and $\Delta$ is of shape
$M(m,1)$.

Assume now that $x$ is connected with all nodes in $I_2$.
Let $a\in I_2$. As $|I_2| \geq 3$ we may choose $b\in I_2\setminus\{a\}$
such that $\Delta_{I\setminus\{1,a,b,x\}}$
contains 
 {$
\begin{picture}(15,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}

\put(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}

\put(1,-1.4){{\tiny 1}}
\put(11,-1.4){{\tiny c}}

\put(7,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\end{picture}
$} 
 for some $c\in \{a,b\}.$
As $\Delta_{I\setminus\{1,a,b,x\}}$ has the shape of a square we get with
(1.3) and (5.1) that we have the subdiagram
 {$
\begin{picture}(15,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}

\put(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}

\put(1,-1.4){{\tiny1}}
\put(11,-1.4){{\tiny x}}

\put(7,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\end{picture}
$}
and so $\Delta$ is a complete bipartite
graph all of whose edges are of type 
{$
\begin{picture}(15,8)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}



\put(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}

\put(7,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}
\put(14,0){.}
\end{picture}
$} 

Assume now $|I_1|\ge 2\le |I_2|$. Then we may assume that $x$ is connected
with 1 in $I_1$. As we have no triangles in $\Delta,$ $\,x$ is not connected
to any element in $I_2$. Let $2\in I_1$, such that 2 is not connected to $x$.
Then for $a\in I_2$, we have by [Str2] and (1.3) that 
$\Delta_{I\setminus\{1,2,a,x\}}$ is of type 
\begin{center}
{$
\begin{picture}(35,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(32,1){\makebox(0,0){$\circ$}}

\put(22.5,1.7){\line(1,0){8.9}}
\put(22.5,0.4){\line(1,0){8.9}}
\put(27,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(12.5,1.7){\line(1,0){8.9}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(2.5,1){\line(1,0){8.7}}

\put(1,-2){{\scriptsize x}}
\put(11,-2){{\scriptsize 1}}
\put(21,-2){{\scriptsize a}}
\put(31,-2){\scriptsize 2}
\put(33,0){.}
\end{picture}
$}
\end{center}  As before we see that $x$
is not connected to any element in $I_1\setminus \{1\}$ and so $\Delta$ is of type
$M(n,m)$.

If $x$ is connected to all elements in $I_1$, then as above we see that for
$b\in I_1$, we have a subdiagram
{$
\begin{picture}(15,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}

\put(1,-2){\scriptsize x}
\put(11,-2){\scriptsize b}

\put(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}

\put(7,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}
\put(14,0){,}
\end{picture}
$} and so $\Delta$ is a complete bipartite graph
all of whose edges are of type 
{$
\begin{picture}(15,8)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}


\put(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}

\put(7,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}
\put(14,0){.}
\end{picture}
$}\hfill $\Box$\\[2ex]
{\bf (5.4) Lemma.} {\it If $\Delta_x$ is of type $He(n,m)$ then $\Delta$ is as
in the theorem}.\\[2ex]
{\bf Proof.} Let $a$ be some end node of $\Delta_x$ connected with $x$. Then
$\Delta_{I\setminus\{1,2,a,x\}}$ is of shape 
\begin{center}
{$
\begin{picture}(45,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(32,1){\makebox(0,0){$\circ$}}

\put(22.5,1){\line(1,0){8.7}}



\put(12.5,1.7){\line(1,0){8.9}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(2.5,1){........}

\put(1,-2){{\scriptsize x}}
\put(11,-2){{\scriptsize a}}
\put(21,-2){{\scriptsize 1}}
\put(31,-2){\scriptsize 2}
\put(40,0){or}
\end{picture}
$}
\quad
{$
\begin{picture}(15,10)

\put(2,-4){\makebox(0,0){$\circ$}}
\put(12,-4){\makebox(0,0){$\circ$}}
\put(2,6){\makebox(0,0){$\circ$}}
\put(12,6){\makebox(0,0){$\circ$}}

\put(-0.8,-5){\scriptsize x}
\put(-0.8,7){\scriptsize a}
\put(14,-5){\scriptsize 2}
\put(14,7){\scriptsize 1}
\put(1.5,-3){.}
\put(1.5,-2.1){.}
\put(1.5,-1.2){.}
\put(1.5,-0.3){.}
\put(1.5,0.6){.}
\put(1.5,1.5){.}
\put(1.5,2.4){.}
\put(1.5,3.3){.}
\put(1.5,4.2){.}
\put(1.5,5.1){.}


\put(12,-3.2){\line(0,1){8.7}}




\put(2.5,-4){........}


\put(2.5,6.4){\line(1,0){8.9}}
\put(2.5,5.6){\line(1,0){8.9}}

\put(7,8.5){\makebox(0,0){$\scriptstyle{\sim}$}}


\end{picture}
$}
\end{center}
\vspace{0.4cm}
but this contradicts
(1.3) as $G_{I\setminus \{a,1,2\}} \cong He$. So we have that $x$ is not connected to any end node. We may
assume that $x$ is connected to 1 but not to 2. Let $b$ be an
end node not connected with 1. 
By induction and the shape of $\Delta_{bx}$ we get $\Delta_{b}$ is of type $He(n+1,m-1)$ or
$Y(1,n+1)$. So in any case we have a subdiagram
{$
\begin{picture}(15,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}

\put(1,-2){\scriptsize x}
\put(11,-2){\scriptsize 1}

\put(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}

\put(7,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\end{picture}
$}
 and then $\Delta$ is of type
$He(n+1,m)$.\hfill $\Box$\\[2ex]
{\bf (5.5) Lemma.} {\it If $\Delta_x$ is of type $X(2,m)$, then $\Delta$ is one
of the diagrams of the theorem.}\\[2ex]
{\bf Proof.} Let $x$ be connected with $m+4$. Now 
$\Delta_{I\backslash \{2,3,m+4,x\}}$ is a rank four diagram with a
$3^7Sp_6(2)$-residue. By (1.3) we get that the diagram
is of shape 
\begin{center}
{$
\begin{picture}(35,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(32,1){\makebox(0,0){$\circ$}}

\put(2.5,1){\line(1,0){8.7}}

\put(12.5,1.7){\line(1,0){8.9}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim} $}}

\put(22.5,1){\line(1,0){8.7}}



\put(1,-2){{\scriptsize 2}}
\put(11,-2){{\scriptsize 3}}
\put(19,-2){{\scriptsize m+4}}
\put(31,-2){\scriptsize x}
\put(34,0){,}
\end{picture}
$}
\end{center} contradicting (2.4).

Let $x$ be connected with $y$, $4\le y<m+4$. 
By the same arguments as before we may assume that the subdiagram 
{$
\begin{picture}(15,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}

\put(1,-2){\scriptsize 3}
\put(11,-2){\scriptsize y}

\put(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}



\end{picture}
$}
has no tilde. Now $\Delta_{I\backslash \{3,y,m+4,x\}}$ is of shape 
\begin{center}
{$
\begin{picture}(35,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(32,1){\makebox(0,0){$\circ$}}

\put(2.5,1.7){\line(1,0){8.9}}
\put(2.5,0.4){\line(1,0){8.9}}
\put(7,3.5){\makebox(0,0){$\scriptstyle{\sim} $}}

\put(12.5,1.7){\line(1,0){8.9}}
\put(12.5,0.4){\line(1,0){8.9}}


\put(22.5,1){........}



\put(-0.6,-2){{\scriptsize m+4}}
\put(11,-2){{\scriptsize 3}}
\put(21,-2){{\scriptsize y}}
\put(31,-2){\scriptsize x}
\put(33,0){,}
\end{picture}
$} 
\end{center}
contradicting (1.3).
Hence $x$ is not connected to any $y$ with $4\le y\le m+4$. 

Let first $m\neq
0$. Then $\Delta_4$ is connected and has a subdiagram of shape 
\begin{center}
{$
\begin{picture}(25,12)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(19,-6){\makebox(0,0){$\circ$}}

\put(19,8){\makebox(0,0){$\circ$}}

\put(2.5,1.7){\line(1,0){8.9}}
\put(2.5,0.4){\line(1,0){8.9}}

 \put(12.4,1.6){\line(1,1){6.1}}
 

 \put(12.4,0.4){\line(1,-1){6.1}}

\put(7,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(0,-2){{\scriptsize m+4}}
\put(11,-2){{\scriptsize 3}}
\put(21,-7){{\scriptsize 2}}
\put(21,9){\scriptsize 1}


\end{picture}
$}.\end{center}
\vspace{0.8cm}
Hence by induction
we have that $\Delta_4$ is of type $X(2,m)$ and so $\Delta$
is of type $X(2,m+1)$.

Let now $m=0$. We may assume that $\Delta_1$ is connected. As $\Delta_1$
contains a $3^7Sp_6(2)$-residue, we get with (1.3) that
$\Delta_1$ has one of the following shapes.
\begin{itemize}
\item[(1)] 
{$
\begin{picture}(35,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(32,1){\makebox(0,0){$\circ$}}

\put(2.5,1){\line(1,0){8.7}}

\put(22.5,1.7){\line(1,0){8.9}}
\put(22.5,0.4){\line(1,0){8.9}}
\put(27,3.5){\makebox(0,0){$\scriptstyle{\sim} $}}

\put(12.5,1){\line(1,0){8.7}}



\put(1,-2){{\scriptsize x}}
\put(11,-2){{\scriptsize 2}}
\put(21,-2){{\scriptsize 3}}
\put(31,-2){\scriptsize 4}
\put(34,0){,}
\end{picture}
$}
\item[(2)] 
{$
\begin{picture}(25,12)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(19,-6){\makebox(0,0){$\circ$}}

\put(19,8){\makebox(0,0){$\circ$}}

\put(2.5,1.7){\line(1,0){8.9}}
\put(2.5,0.4){\line(1,0){8.9}}

 \put(12.4,1.6){\line(1,1){6.1}}
 

 \put(12.4,0.4){\line(1,-1){6.1}}

\put(7,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(1,-2){{\scriptsize 4}}
\put(11,-2){{\scriptsize 3}}
\put(21,-7){{\scriptsize x}}
\put(21,9){\scriptsize 2}



\end{picture}
$}
\item[(3)] 
 {$
\begin{picture}(25,15)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(19,-6){\makebox(0,0){$\circ$}}

\put(19,8){\makebox(0,0){$\circ$}}

\put(2.5,1){\line(1,0){8.7}}

 \put(11.8,1.7){\line(1,1){6.5}}
 \put(12.5,1){\line(1,1){6.5}}

 \put(11.8,0.4){\line(1,-1){6.5}}
 \put(12.5,1){\line(1,-1){6.5}}

\put(14,5.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(14,-3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}


\put(1,-2){{\scriptsize 2}}
\put(11,-2){{\scriptsize 3}}
\put(21,-7){{\scriptsize x}}
\put(21,9){\scriptsize 4}



\end{picture}
$}
\end{itemize}
\vspace{0.4cm}

The first case contradicts (4.8). In (2) and (3) 
$x$ cannot be connected to $1$ since there are no triangles in $\Delta$ 
by (1.1).
Hence for $\Delta$ we get the following two possibilities.\\[0.4cm]
\begin{itemize}
\item[(1)] 
{$
\begin{picture}(38,5)
\put(10.5,1){\makebox(0,0){$\circ$}}
\put(2.6,1){\makebox(0,0){$\circ$}}
\put(20.5,1){\makebox(0,0){$\circ$}}
\put(2.6,-6){\makebox(0,0){$\circ$}}
\put(2.6,8){\makebox(0,0){$\circ$}}

\put(11,1.7){\line(1,0){8.9}}
\put(11,0.4){\line(1,0){8.9}}

\put(3.1,1){\line(1,0){6.3}}

 \put(9.8,1.4){\line(-1,1){6.3}}


 \put(9.8,0.7){\line(-1,-1){6.3}}

\put(15.5,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}



\put(-0.3,0){{\scriptsize 1}}
\put(-0.3,-6){{\scriptsize x}}
\put(-0.3,8){{\scriptsize 2}}
\put(19.5,-2){\scriptsize 4}
\put(9.5,-2){{\scriptsize 3}}




 



 

\end{picture}
$}
\item[(2)] 
{$
\begin{picture}(38,16)
\put(10.5,1){\makebox(0,0){$\circ$}}


\put(2.6,-6){\makebox(0,0){$\circ$}}
\put(2.6,8){\makebox(0,0){$\circ$}}


 \put(9.8,1.4){\line(-1,1){6.3}}


 \put(9.8,0.7){\line(-1,-1){6.3}}




\put(17.5,-6){\makebox(0,0){$\circ$}}

\put(17.5,8){\makebox(0,0){$\circ$}}


 \put(10.3,1.7){\line(1,1){6.5}}
 \put(11,1){\line(1,1){6.5}}

 \put(10.3,0.4){\line(1,-1){6.5}}
 \put(11,1){\line(1,-1){6.5}}

\put(12.5,5.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(12.5,-3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(19,-6){{\scriptsize x}}
\put(19,8){{\scriptsize 4}}

\put(-0.3,-6){{\scriptsize 1}}
\put(-0.3,8){{\scriptsize 2}}

\put(9.5,-2){{\scriptsize 3}}

\end{picture}
$}
\end{itemize}
\vspace{0.4cm}

The second diagram is of type $X(2,2)$. So assume, that we have the first. Then $\Delta_4$
is a $D_4$-diagram and so by [Tim1] $G_4/K_4\cong \Omega^+_8(2)$. Then
$G_{x,4}/K_{x,4}\cong A_8$, but in the residue of $x$ we conclude with
(1.3)(v) that $G_{x,4}/K_{x,4}\cong A_7$,
 a contradiction.\hfill $\Box$\\[2ex]
{\bf (5.6) Lemma.} 
{\it If $\Delta_x$ is of type $Y(n,m)$, then $\Delta$ is one of
the diagrams in the theorem}.\\[2ex]
{\bf Proof.} By (4.9) we may assume $m>1$. Let $x$ be connected with
$m+n+1$. Application of (4.8) to $\Delta_{\{1,\ldots, n-2, n+2,\ldots, n+m\} }$ shows
$n=1$ and (2.4) shows that $\Delta_{\{3,\ldots, m+1\}}$ is of the shape 
\begin{center}
{$
\begin{picture}(35,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(32,1){\makebox(0,0){$\circ$}}

\put(22.5,1.7){\line(1,0){8.9}}
\put(22.5,0.4){\line(1,0){8.9}}
\put(27,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(12.5,1.7){\line(1,0){8.9}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(2.5,1){\line(1,0){8.7}}

\put(1,-2){{\scriptsize 1}}
\put(11,-2){{\scriptsize 2}}
\put(20,-2){{\scriptsize m+2}}
\put(31,-2){\scriptsize x}
\put(33,0){.}
\end{picture}
$} 
\end{center}Let
$3\le y<m+2$ be not connected with $x$. Then 
$\Delta_{I\setminus\{2,x,y,m+2\}}$ is of the shape 
\begin{center}
{$
\begin{picture}(35,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(32,1){\makebox(0,0){$\circ$}}

\put(2.5,1.7){\line(1,0){8.9}}
\put(2.5,0.4){\line(1,0){8.9}}
\put(7,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(12.5,1.7){\line(1,0){8.9}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(22.5,1.7){\line(1,0){8.9}}
\put(22.5,0.4){\line(1,0){8.9}}


\put(1,-2){{\scriptsize y}}
\put(11,-2){{\scriptsize 2}}
\put(17.8,-2){{\scriptsize m+2}}
\put(31,-2){\scriptsize x}
\put(33,0){,}
\end{picture}
$} 
\end{center} contradicting (1.3). So we have that $x$ is connected with all $y$, $\, 3\le y\le m+2$.
Further all these edges are of type 
{$
\begin{picture}(15,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}


\put(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}

\put(7,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim}\:\!\tiny)$}}

\end{picture}
$}
. As we know from 
$\Delta_{\{3,\ldots,m+1\}}$ that $\Delta_{\{3,\ldots,m+1,x\}}$ is a 
$M_{24}$-residue, we have that $\Delta_{\{3,\ldots,m+2,x\}\backslash \{y\}}$ 
is a $M_{24}$-residue for all $y$ with $3\le y\le m+2$ by (1.3)(i). By (1.3)(iv)
$\Delta_{\{3,\ldots,m+2\}\backslash \{y\}}$ is locally $M(24), M(24)',
3M(24)'$ or $3M(24)$ and so $\Delta$ is of type $M(2,m).$

So assume now that the node $x$ is not connected with $n+m+1$. Let then $n+2\le y<n+m+1$
be connected with $x$. Then we may assume that the subdiagram
{$
\begin{picture}(15,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}

\put(-0.3,-2){\scriptsize n+1}
\put(11,-2){\scriptsize y}

\put(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}



\end{picture}
$}
has no tilde.
Hence the shape of 
$\Delta_{I\backslash \{n+m+1,n+1,y,x\}}$ is 
\hspace*{0.1cm}
$\; \;$
{$
\begin{picture}(35,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(32,1){\makebox(0,0){$\circ$}}

\put(2.5,1.7){\line(1,0){8.9}}
\put(2.5,0.4){\line(1,0){8.9}}
\put(7,3.5){\makebox(0,0){$\scriptstyle{\sim} $}}

\put(12.5,1.7){\line(1,0){8.9}}
\put(12.5,0.4){\line(1,0){8.9}}


\put(22.5,1){........}



\put(-2.6,-2){{\scriptsize n+m+1}}
\put(9.5,-2){{\scriptsize n+1}}
\put(21,-2){{\scriptsize y}}
\put(31,-2){\scriptsize x}
\put(33,0){.}
\end{picture}
$}
Application of 
(1.3) gives a contradiction.\\ 

Hence $x$ is not connected to
any $y$ with $n+2\le y\le n+m+1$. Now $\Delta_{n+2}$ is connected
and contains the subdiagram 
{$
\begin{picture}(25,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}


\put(2.5,1){\line(1,0){8.7}}

\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim} $}}

\put(12.5,1.7){\line(1,0){8.9}}
\put(12.5,0.4){\line(1,0){8.9}}

\put(23,0){.}
\end{picture}
$} By induction we get that 
$\Delta_{n+2}$ is of type $Y(n,m),$ $X(2,m\! - \! 1),$ $He(1,m\!- \! 1 \!)$ or $Y(n\!+\!1,m\! - \! 1)$.
Hence we get that $\Delta$ is of type $Y(n,m+1), X(2,m),He(1,m)$ or
$Y(n+1,m)$, respectively.\hfill $\Box$\\[2ex]
\noindent
{\bf (5.7) Lemma.} {\it If $\Delta_x$ is of type $M(n,m)$, then $\Delta$ is one
of the diagrams of the theorem}.\\[2ex]
{\bf Proof.} Let $y\in I_m$ connected with $x$. As $\Delta$ does not contain
triangles we have that $x$ is not connected with any element in $I_n$. Now
$\Delta_{I\backslash \{1,t,y,x\}}$ is of the shape 
{$
\begin{picture}(37,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(32,1){\makebox(0,0){$\circ$}}

\put(22.5,1.7){\line(1,0){8.9}}
\put(22.5,0.4){\line(1,0){8.9}}
\put(27,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(12.5,1.7){\line(1,0){8.9}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(2.5,1){\line(1,0){8.7}}

\put(1,-2){{\tiny 1}}
\put(11,-2){{\tiny t}}
\put(21,-2){{\tiny y}}
\put(31,-2){\tiny x}
\put(35,0){}
\end{picture}
$} by (1.3) where $t$ is the unique element connected to $1$. Let $u \in I_m$ not connected with $x$, then 
$\Delta_{I\backslash \{t,y,u,x\}}$ is of the shape 
{$
\begin{picture}(37,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(32,1){\makebox(0,0){$\circ$}}

\put(22.5,1.7){\line(1,0){8.9}}
\put(22.5,0.4){\line(1,0){8.9}}
\put(27,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(12.5,1.7){\line(1,0){8.9}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(2.5,1.7){\line(1,0){8.9}}
\put(2.5,0.4){\line(1,0){8.9}}
\put(7,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(1,-2){{\tiny u}}
\put(11,-2){{\tiny t}}
\put(21,-2){{\tiny y}}
\put(31,-2){\tiny x}


\put(35,0){,}
\end{picture}
$} contradicting
(1.3). So we have that $x$ is
 connected with all elements in $I_m$. Now
$\Delta$ is of type $M(n+1,m)$.

Now let $x$ be not connected with any element in $I_m$. Let $x$ be connected
with $t$. Choose $v\in I_n\backslash \{t\}$. Then we have for $u\in I_m$ that
$\Delta_{I\backslash\{t,v,u,x\}}$ is of shape 
\begin{center}
{$
\begin{picture}(40,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(32,1){\makebox(0,0){$\circ$}}

\put(22.5,1.7){\line(1,0){8.9}}
\put(22.5,0.4){\line(1,0){8.9}}
\put(27,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(12.5,1.7){\line(1,0){8.9}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(2.5,1){........}

\put(1,-2){{\scriptsize x}}
\put(11,-2){{\scriptsize t}}
\put(21,-2){{\scriptsize u}}
\put(31,-2){\scriptsize v}
\put(40,0){or}
\end{picture}
$} 
\qquad\quad
{$
\begin{picture}(15,10)

\put(2,-4){\makebox(0,0){$\circ$}}
\put(12,-4){\makebox(0,0){$\circ$}}
\put(2,6){\makebox(0,0){$\circ$}}
\put(12,6){\makebox(0,0){$\circ$}}

\put(-0.8,-5){\scriptsize x}
\put(-0.8,7){\scriptsize t}
\put(14,-5){\scriptsize v}
\put(14,7){\scriptsize u}
\put(1.5,-3){.}
\put(1.5,-2.1){.}
\put(1.5,-1.2){.}
\put(1.5,-0.3){.}
\put(1.5,0.6){.}
\put(1.5,1.5){.}
\put(1.5,2.4){.}
\put(1.5,3.3){.}
\put(1.5,4.2){.}
\put(1.5,5.1){.}


\put(12.6,-3.4){\line(0,1){8.9}}
\put(11.4,-3.4){\line(0,1){8.9}}
\put(15.2,1){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}


\put(2.5,-4){........}


\put(2.5,6.4){\line(1,0){8.9}}
\put(2.5,5.6){\line(1,0){8.9}}

\put(7,8.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(20,0.5){.}
\end{picture}
$}
\end{center}
\vspace{0.4cm}

Let $x$ not be connected with $v$. Then by (1.3) we
have that $\Delta_{I\backslash\{t,v,u,x\}}$ is of type 
{$
\begin{picture}(37,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(32,1){\makebox(0,0){$\circ$}}

\put(22.5,1.7){\line(1,0){8.9}}
\put(22.5,0.4){\line(1,0){8.9}}
\put(27,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(12.5,1.7){\line(1,0){8.9}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(2.5,1){\line(1,0){8.7}}

\put(1,-2){{\scriptsize x}}
\put(11,-2){{\scriptsize t}}
\put(21,-2){{\scriptsize u}}
\put(31,-2){\scriptsize v}
\put(35,0){.}
\end{picture}
$}
Now $\Delta_{I\backslash\{1,t,u,x\}}$ if of type
\begin{center}
{$
\begin{picture}(25,12)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(19,-6){\makebox(0,0){$\circ$}}

\put(19,8){\makebox(0,0){$\circ$}}

\put(2.5,1.7){\line(1,0){8.9}}
\put(2.5,0.4){\line(1,0){8.9}}

 \put(12.4,1.6){\line(1,1){6.1}}
 

 \put(12.4,0.4){\line(1,-1){6.1}}

\put(7,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(1,-2){{\scriptsize u}}
\put(11,-2){{\scriptsize t}}
\put(21,-7){{\scriptsize x}}
\put(21,9){\scriptsize 1}


\put(24,0){,}
\end{picture}
$}
\end{center}
\vspace{0.4cm}
with
$M_{24}$-residues, which contradicts (1.3).

So we have that $x$ is connected with all $v\in I_n$. By (1.3) we see that $\Delta_{I\backslash\{t,v,u,x\}}$ has the shape
\begin{center}
{$
\begin{picture}(15,10)

\put(2,-4){\makebox(0,0){$\circ$}}
\put(12,-4){\makebox(0,0){$\circ$}}
\put(2,6){\makebox(0,0){$\circ$}}
\put(12,6){\makebox(0,0){$\circ$}}

\put(-0.8,-5){\scriptsize x}
\put(-0.8,7){\scriptsize t}
\put(14,-5){\scriptsize v}
\put(14,7){\scriptsize u}

\put(12.6,-3.4){\line(0,1){8.9}}
\put(11.4,-3.4){\line(0,1){8.9}}
\put(15.2,1){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(2.6,-3.4){\line(0,1){8.9}}
\put(1.4,-3.4){\line(0,1){8.9}}
\put(-2,1){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}



\put(2.5,6.4){\line(1,0){8.9}}
\put(2.5,5.6){\line(1,0){8.9}}
\put(7,8.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(2.5,-4.6){\line(1,0){8.9}}
\put(2.5,-3.4){\line(1,0){8.9}}
\put(7,-7){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(20,0.5){.}
\end{picture}
$}
\end{center}
\vspace{0.8cm}
Furthermore by (1.3) we have that $\Delta_{I\backslash\{1,t,x,v\}}$
is of shape 
{$
\begin{picture}(35,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(32,1){\makebox(0,0){$\circ$}}

\put(22.5,1.7){\line(1,0){8.9}}
\put(22.5,0.4){\line(1,0){8.9}}
\put(27,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}


\put(1,-2){{\scriptsize 1}}
\put(11,-2){{\scriptsize t}}
\put(21,-2){{\scriptsize x}}
\put(31,-2){\scriptsize v}

\put(12.5,1.7){\line(1,0){8.9}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(2.5,1){\line(1,0){8.7}}


\end{picture}
$} and that $\Delta$ is of type
$M(n,m+1)$. 

Now suppose that $x$ is not connected with $t$ but with some 
$y\in I_n \setminus \{ t \}$.
Choose $u\in I_m$. Then $\Delta_{I\backslash\{t,y,u,x\}}$ is of shape
\begin{center}
{$
\begin{picture}(45,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(32,1){\makebox(0,0){$\circ$}}

\put(22.5,1.7){\line(1,0){8.9}}
\put(22.5,0.4){\line(1,0){8.9}}
\put(27,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(12.5,1.7){\line(1,0){8.9}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}
\put(2.5,1){........}

\put(1,-2){{\scriptsize x}}
\put(11,-2){{\scriptsize y}}
\put(21,-2){{\scriptsize u}}
\put(31,-2){\scriptsize t}
\end{picture}
$} 
\end{center}
\vspace{0.4cm}
This is a contradiction to (1.3) and (3.5).

We are left with the case that just 1 is connected with $x$. Then
$\Delta_{I\backslash\{1,t,x,u\}},$ $u\in I_m$ has the shape 
{$
\begin{picture}(35,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(32,1){\makebox(0,0){$\circ$}}

\put(2.5,1){........}

\put(22.5,1.7){\line(1,0){8.9}}
\put(22.5,0.4){\line(1,0){8.9}}
\put(27,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}


\put(12.5,1){\line(1,0){8.7}}

\put(1,-2){{\scriptsize x}}
\put(11,-2){{\scriptsize 1}}
\put(21,-2){{\scriptsize t}}
\put(31,-2){\scriptsize u}
\end{picture}
$}
with an $M_{24}$-residue 
{$
\begin{picture}(25,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}


\put(12.5,1.7){\line(1,0){8.9}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}


\put(2.5,1){\line(1,0){8.7}}


\put(1,-2){{\scriptsize 1}}
\put(11,-2){{\scriptsize t}}
\put(21,-2){\scriptsize u}
\put(24,0){.}
\end{picture}
$}
Application of (1.3) shows that
$\Delta_{I\backslash\{1,t,x,u\}}$ is of type 
\begin{center}
{$
\begin{picture}(35,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(32,1){\makebox(0,0){$\circ$}}


\put(2.5,1){\line(1,0){8.7}}

\put(22.5,1.7){\line(1,0){8.9}}
\put(22.5,0.4){\line(1,0){8.9}}
\put(27,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}


\put(12.5,1){\line(1,0){8.7}}





\put(1,-2){{\scriptsize x}}
\put(11,-2){{\scriptsize 1}}
\put(21,-2){{\scriptsize t}}
\put(31,-2){\scriptsize u}
\put(34,0){.}
\end{picture}
$}
\end{center}
Now choose $v\in I_n$. Then $\Delta_{I\backslash\{1,x,t,u,v\}}$ is of type
\begin{center}
{$
\begin{picture}(45,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(32,1){\makebox(0,0){$\circ$}}
\put(42,1){\makebox(0,0){$\circ$}}

\put(2.5,1){\line(1,0){8.7}}

\put(22.5,1.7){\line(1,0){8.9}}
\put(22.5,0.4){\line(1,0){8.9}}
\put(27,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}


\put(12.5,1){\line(1,0){8.7}}

\put(32.5,1.7){\line(1,0){8.9}}
\put(32.5,0.4){\line(1,0){8.9}}
\put(37,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}



\put(1,-2){{\scriptsize x}}
\put(11,-2){{\scriptsize 1}}
\put(21,-2){{\scriptsize t}}
\put(31,-2){\scriptsize u}
\put(41,-2){\scriptsize v}

\end{picture}
$}
\end{center}
contradicting (4.8).\hfill $\Box$\\

\noindent
{\bf Proof of the theorem. }\\
Let $\{P_1,\ldots, P_n\}$ be a minimal counterexample to the theorem with
connected diagram $\Delta$. Then by induction any subdiagram containing 
{$
\begin{picture}(15,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}

\put(2.5,1.7){\line(1,0){8.9}}
\put(2.5,0.4){\line(1,0){8.9}}
\put(7,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}


\end{picture}
$}
fulfills the theorem. So we may apply (5.3) -- (5.7) to show that
$\Delta$ has one of the shapes of the theorem.
But now by (5.2) the corresponding statements about the groups also
hold, a contradiction.

This proves the theorem. \hfill $\Box$\\[2ex]

\begin{thebibliography}{xxxxxxx}
\bibitem[Asch1]{} M. Aschbacher, $GF(2)$-representations of finite groups,
Amer. J. of Mathematics 104, 1982, 683-771.
\bibitem[Asch2]{} M. Aschbacher, 3-Transposition Groups, Cambridge University Press 1997,
\bibitem[Coop]{} B. Cooperstein, An enemies list for factorization theorems,
Comm. Algebra 6, 1978, 1239-1288.
\bibitem[FuStr]{} A. Fukshansky, G. Stroth, Semiclassical parabolic systems
related to $M_{24}$, Geom. Dedicata, to appear.
\bibitem[Gil]{} G. Gillespie, Die Nichtexistenz einer bestimmten Klasse
paraboli- scher Systeme mit quadratischem Diagramm, Diplomarbeit FU Berlin 1993.
\bibitem[Gold]{} D. M. Goldschmidt, Automorphisms of trivalent graphs,
Ann. of Math., 111(1980), 377-404.
\bibitem[GoLySo]{} D. Gorenstein, R. Lyons, R. Solomon, The classification of the finite simple groups,
Number III, Mathematical Service and Monographs 40.3. AMS 1998.
\bibitem[Hei1]{} S. Heiss, Extensions for chamber systems of type $A_3$ for
$A_7$, J. Algebra 123, 1989, 120-150.
\bibitem[Hei2]{} S. Heiss, On a minimal parabolic system of type $M_{24}$,
J. Algebra 142, 1991, 188-200.
\bibitem[Hu]{} B. Huppert, Endliche Gruppen I, Springer 1967.
\bibitem[Iv]{} A. Ivanov, A geometric characterization of the monster,
In: Groups, Combinatorics and Geometry, London Math. Soc. LN 165 (eds.
M. Liebeck, J. Saxl), 1992, 46-62.
\bibitem[McCl]{} P. McClurg, The classification of $F_1$-pairs, PhD.-
thesis, Univ. of Ca\-lifornia at Santa Cruz, 1980.
\bibitem[Mei1]{} Th. Meixner, Parabolic systems: The $GF(3)$-case, Trans.
Amer. Math. Soc. 338, 1993, 615-637
\bibitem[Mei2]{} Th. Meixner, Groups acting transitively on locally
finite classical Tits Chamber Systems, In: Finite geometries, buildings
and related topics (Pingree Park 1988), 45-65, Oxford Univ. Press 1990.
\bibitem[MeiStr]{} U. Meierfrankfenfeld, G. Stroth, Quadratic $GF(2)$-
modules for sporadic groups and alternating groups, Comm. Algebra 18,
1990, 2099-2140.
\bibitem[Ni]{} R. Niles, BN-pairs and finite groups with parabolic type 
subgroups, J. Algebra 75, 1982, 484-492
\bibitem[Par]{} Chr. Parker, Groups containing a subdiagram 
{$
\begin{picture}(25,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}


\put(12.5,1.7){\line(1,0){8.9}}
\put(12.5,0.4){\line(1,0){8.9}}
\put(17,3.5){\makebox(0,0){$\tiny( \!\scriptstyle{\sim} \:\! \!\tiny)$}}

\put(2.5,1){\line(1,0){8.7}}
\put(24,0){,}
\end{picture}
$}
Proc. London Math. Soc. 65, 1992, 85-120.
\bibitem[ParRow]{} Chr. Parker, P. Rowley, Amalgams which involve sporadic
simple groups I, II, J. Alg. 193, 1997, 441-528.
\bibitem[RoStr]{} M. Ronan, G. Stroth, Minimal parabolic systems for the
sporadic groups, Europ. J. Comb. 5, 1984, 59-91.
\bibitem[Row]{} P. Rowley, On the minimal parabolic system related to $
M_{24}$, J. London Math. Soc. 40, 1989, 40-56.
\bibitem[Sh]{} S. Shpectorov, Natural representations of some tilde and
Petersen-type geometries, Geom. Dedicata 54, 1995, 87-102.
\bibitem[ShStr]{} S. Shpectorov, G. Stroth, Classification of certain types
of tilde geometries, Geom. Dedicata 49, 1994, 155-172. 
\bibitem[Sm]{} S. Smith, Irreducible modules and parabolic subgroups, 
 J. Algebra 75,
1982, 286-289.
\bibitem[Stein]{} R. Steinberg, Lectures on Chevalley groups, Lecture Notes, Yale University, 1967.
\bibitem[Str1]{} G. Stroth, One node extensions of buildings, Geom. Dedicata
25, 1988, 71-120
\bibitem[Str2]{} G. Stroth, Chamber systems, geometries and parabolic systems
whose diagram contains only bonds of strength 1 and 2, Invent.Math. 102,
1990, 209-234
\bibitem[Str3]{} G. Stroth, The nonexistence of certain Tits geometries with
affine diagram, Geom. Dedicata 28, 1988, 277-319
\bibitem[Str4]{} G. Stroth, Geometries of type $M$ related to $A_7$, Geom.
Dedicata 20, 1986, 265-293.
\bibitem[Str5]{} G. Stroth, A local classification of finite classical
Tits geometries of characteristic $\neq 3$, Geom. Dedicata 28, 1988,
93-105.
\bibitem[Str6]{} G. Stroth, Strong quadratic modules, Israel J. Math. 79,
1992, 257-279.
\bibitem[StrWo]{} G. Stroth, S.K. Wong, Minimal parabolic systems and small
representations, Comm. Algebra 15, 1987, 751-790.
\bibitem[Tim1]{} F. Timmesfeld, Tits geometries and parabolic systems in
finitely generated groups, I, II, Math. Z. 184, 1983, 377-396, 449-487.
\bibitem[Tim2]{} F. Timmesfeld, Locally finite classical Tits chamber systems
of large order, Invent. Math. 87, 1987, 603-641.
\bibitem[Tim3]{} F. Timmesfeld, Classical locally finite Tits chamber systems
of rank 3, J. Algebra 124, 1989, 9-59.
\bibitem[Wie1]{} C. Wiedorn, A Tilde Geometry related to $F_4$, J. Alg. to
appear.
\bibitem[Wie2]{} C. Wiedorn, Some semiclassical parabolic systems of rank 4, J. Alg. to appear.
\bibitem[Wie3]{} C. Wiedorn, A Tilde Geometry for $F_4(2)$, Beitr\"age
zur Alg. und Geom., 38, 1997, 337-342.
\bibitem[Wie4]{} C. Wiedorn, An $F_1$--Module for the $3U_4(3)$--Amalgam,
J. Alg. 195, 1997, 337-356.
preprint.
\end{thebibliography}




\end{document}










