
\documentstyle{article}

\begin{document}

\title{Semiclassical parabolic systems related to $M_{24}$}
\author{A. Fukshansky\footnotemark[1]
	 and G. Stroth\footnotemark[1] \\
	Martin-Luther-Universit\"at Halle-Wittenberg\\
	Fachbereich Mathematik und Informatik\\
	Institut f\"ur Algebra und Geometrie\\
 	06099 Halle, Germany}

\maketitle
\renewcommand{\thefootnote}{\fnsymbol{footnote}}
\footnotetext[1]{This work was supported by DFG, Str 203/10-1}
There is a long history of parabolic systems generalizing the buildings associatied to finite groups of Lie type. One of the main results until
now might be the classification of so called parabolic systems of classical type in \cite{Ni, Tim1, Tim2, Tim3, Str1, Str2, Str3, Str4, Str5, Hei1, Mei1}. See \cite{Mei2} for a survey. These parabolic systems mainly are related to groups of Lie type again. There is a second class of parabolic systems which has a large variety of sporadic examples. These are called parabolic systems of semiclassical type (see \cite{Str6}). In this paper we give one important step in the classification of semiclassical parabolic systems. Let us start with the definition. 

Let $G$ be a group generated by finite subgroups $P_1, ..., P_n$ such that
\begin{tabbing}

\quad\= (1) \=  $ P_i \notin \langle P_j | j \in \{ 1,...,n \} \backslash       \{ i \} \rangle $ \\

\>(2) \> For $ B = \bigcap^n_{i=1} P_i $ we have $P_i/B_{P_i}$ is a rank one group of Lie type \\
\>\> in characteristic two and  $ B = P_i \cap P_j$ for each pair $ P_i, P_j$.\\

\>(3)\>For each pair $i,j \in \{1,...,n\}$ let $ P_{ij} = \langle P_i,P_j \rangle
/ B_{\langle P_i,P_j \rangle}$ be one of\\
\>\> the following:

\end{tabbing}
                 \begin{tabbing}

                 \qquad \quad \= $(\alpha) \:$ \= $P_{ij} $ is a rank two Lie 
   		  group
                  in characteristic two,\\
                  \>$ (\beta)$\> $P_{ij} = P_iP_j$,\\             
                  \> $(\gamma)$\>$P_{ij} $ is the triple cover of $\Sigma_6$.\\
		 \end{tabbing}
\vspace{-1cm}

\begin{tabbing}
\quad \=(4) \= $B_G = 1$\\
\end{tabbing}
If in (3) the case ($\gamma$) never occurs the system is called of classical type. In this paper we assume that the case (3)($\gamma$) occurs for at least one choice $i, j$. Then the system is called semiclassical.

To each parabolic system we can associate a diagram $\Delta$ in the following way. The nodes of the diagram correspond to the groups $P_1,...,P_n$, denoted by $\{1,...,n\}$. For $i,j$ we connect the nodes $i$ and $j$ by the corresponding Dynkin diagram if we have (3)($\alpha$) and by 
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if we have (3)($\gamma$). If we will allow $\Sigma_6$ and $3\Sigma_6$ at the same place we also write
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$}. In the case (3)($\beta$) the nodes are not connected at all and by
 {$
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$} we denote any possible connection of the given kinds.

Since non-connected diagrams correspond to the product of the groups belonging to the connected parts we may assume
\begin{tabbing}
\quad \= (5) \= $\Delta$ is connected. \\
\end{tabbing}
According to \cite{Tim3} the semiclassical diagrams of rank three (i.e. $n=3$) are of the shape
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$},
{$
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$},
{$
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$}. 
The diagram 
{$
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$} is very interesting because it belongs to sporadic groups. According to \cite{Hei2} and \cite{Row} a semiclassical parabolic system with diagram {$
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$} has as group $G$ one of the following : $He, M_{24}$ or $3^7Sp_6(2)$. In the first two cases $|B|=2^{10}$ where $|B|=2^9$ in the third. Furthermore in all rank three cases above we have $P_i/B_{P_i} \cong \Sigma_3$.

So for this paper we make the following additional assumption
\begin{tabbing}
\quad \=(6) \= $P_i/B_{P_i} \cong \Sigma_3$\\
\end{tabbing}
Then an easy exercise shows $B_{P_i}=O_2(P_i) $ (see \cite{StrWo}). The aim of this paper is to finish the classificaion of this type of semiclassical systems of rank four. The general case will appear in a forthcoming paper.

So from now on we have $n=4$, i.e. $G=\langle P_1,P_2,P_3,P_4 \rangle$. For $ i \in \{1,2,3,4\}$ set $ G_i= \langle P_j | j \not= i \rangle$, and $K_i=B_{G_i}$. Then $G_i/K_i$ is a group which belongs to the corresponding rank three diagram. 
There is a natural subdivision concerning the rank three diagram of type {$
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$}. In \cite{Wie2} the case was treated that for any such subdiagram we have that $G_i/K_i \cong 3^7 Sp_6(2) $. Hence we will assume that $G_i/K_i \cong He$ or $M_{24}$ occurs at least once. So we will choose the numbering such that
\begin{tabbing}
\quad \=(7) \= $G_4/K_4 \cong He$ or $ M_{24}$.\\
\end{tabbing}
These parabolic systems we call parabolic systems of type $M_{24}$ of rank four.

There are two known diagrams with corresponding groups and parabolic systems of type $M_{24}$ : First \qquad {$
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$} \quad which represents the group $Co_1$ and second 
 {$\quad
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$} the corresponding group being one of $M(24), M(24)^\prime, 3M(24)$ or $3M(24)^\prime$ . In this paper we will prove


\newtheorem{theorem}{Theorem}

\begin{theorem}
	Let $ \{P_1,P_2,P_3,P_4\}$ be a parabolic system of rank four and of type $M_{24}$, and $\Delta$ be the corresponding diagram. Then one of the following holds
\begin{tabbing}

\quad \=(i) \quad \=$\Delta $ : \quad 

 {$
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$}
, \quad \quad \quad \= $G_4 , G_3 \cong He \mbox{ or } M_{24}$\\
\\
\> (ii) \> $\Delta$ : 
 {$
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$}, \> $G_1 \cong  G_4 \cong He$ \\
\>(iii) \>  $\Delta$  :  {$
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$}, \> $G \cong Co_1 $\\
\> (iv) \>$\Delta$ :  {$
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$}, \> The groups $G_i$ are isomorphic to the \\
\>\>\>corresponding groups in $M(24)$, \\
\>\>\> $3M(24), M(24)^\prime$ or $3M(24)^\prime$, and\\
\>\>\> there is a normal subgroup $K$ of $G$\\
\>\>\> such that $G/K \cong  M(24)^\prime $or $M(24)$.\\
\end{tabbing}
\end{theorem}

In fact in $(iv)$ we cannot say much more as the universal group $G$ is always infinite.\\

Combining our result with results due to G. Gillespie \cite{Gil} and C. Wiedorn \cite{Wie1}, \cite{Wie2} we have

\begin{theorem}
	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 diagram $\Delta$ and a subdiagram of type 
{$
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$}
. Then one of the following holds
\begin{tabbing}
\quad \=(i) \quad \=  $\{P_1,P_2,P_3,P_4 \} $ is of $M_{24}$-type. \= \\
\> (ii) \> $\Delta$ : 
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$} \quad \> $G_3/K_3 \cong A_7, K_1 = K_2 = 1, K_3 $ is \\ 
\>\>\>elementary abelian of order $2^6$ and\\
\>\>\>$G_1 \cong G_2 \cong 3^7Sp_6(2).$\\
\\ 
\>(iii) \>  $\Delta$  : 
 {$
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$}
\>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 lemma 1.7 (iv) {\sf ($\star$)}.\\


\> (iv) \>$\Delta$ :  {$
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$}, \> and there is a normal subgroup $K$ of \\
\>\>\>$G$ with $G/K \cong F_4(2).$
\\
\> (v) \>$\Delta$ :  {$
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$},\> $G \cong 3^{35}Sp_8(2)$.
\end{tabbing}
\end{theorem}

In fact in case $(iv)$ there is a nonsplit extension of an elementary abelian group of order $3^{833}$ by $ F_4(2)$ which belongs to this diagram \cite{Wie3}. It is an open question whether $G$ is isomorphic to this group. It is true if $K$ is abelian \cite{Wie4}.
\\

A few words on the proof might be helpful. The first step is to show that the diagrams are exactly those of the theorem. Furthermore, which will be done at the same time, we show that the $K_i$ and $G_i/K_i$ are as in the theorem. For this there is a well established method, the amalgam method. This relates the diagram with certain small modules (so called $F_1$--modules) for the groups $G_i/K_i$. The interplay of these modules then provides us with the results mentioned before. In our special case a great part of the work was already done by Chr. Parker \cite{Par1}, \cite{Par2}, who described the structure of $K_4$. After this we basically are left with the diagrams  {$
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$} and 
{$\quad
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$}. Here we are going to give the exact structure of $G_i$, $i = 1,2,3,4$, i.e. the multiplication table.  For this we start with the parabolic $G_4$ given by Parker's paper and set up generators and relations for this group. Those who are used to work with generators and relations will realize that our systems are made up by very special principles. Mostly one is interested in systems of generators and relations which use not too many of them. This is not our aim. In fact we use much to many generators. The point is that our set of generators and relations at the end should represent the geometry and one should be able to work in this geometry. So for the group $G_4$ for example we not just like to define this group. At the same time certain subsets of generators and relations should define the intersections of the other parabolics $G_i$ with $G_4$. So for the group $B$, a group of order $2^{21}$, we use 21 generators, where maybe 6 would suffice. But for our purpose it is much better to work with 21 generators. On the other hand to handle such a great set of generators with a computer becomes difficult. So we have to use short relations to overcome this problem. After we have $G_4$ in this way, we see $G_i \cap G_4$ for $i \not= 4$, and then we add one further generator $a_i$ for each $i$ to get $G_i = \langle G_i \cap G_4, a_i \rangle$ and the necessary relations. We show that this all can be done in a unique way, which then shows that the amalgam is uniquely determined. Using the Todd-Coxeter algorithm in case of {$
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$}, we see that our generators and relations define $Co_1$. In case of {$\quad
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\put(22,1){\makebox(0,0){$\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){$\scriptsize( \scriptstyle{\sim} \scriptsize) $}}
\end{picture}
$} this is not true, as we can show using some arguments due to A. Pasini.\\

Part of this work was done when the authors visited A. Pasini in 1996. We like to thank the university of Siena and CNR for their support and hospitality. We also like to thank A. Stein, who did many of the computer calculations for this paper.

\section{Preliminaries}

In this chapter we investigate the rank three case and prove some results on modules which will be of help later on. We use notations and assumptions as introduced before. 
For the proof of the main theorem we will distinguish two cases whether the following assumption is valid or not:
\\

(ASS) \quad $C_{G_4}(O_2(G_4)) \leq O_2(G_4)$.\\


If (ASS) is true, then we can use the results of Ch. Parker \cite{Par1}, \cite{Par2} and know that $G_4/K_4 \cong He$ is not possible, so $G_4/K_4 \cong M_{24}$ and we also know the structure of $K_4$, which is elementary abelian of order $2^{11}$ or $2^{12}$ and is the Golay code-- , the extended Golay code-- or the Todd--module.\\[0.6cm]
{\bf Definition:} Let $G$ be a group and $V$ a faithful $GF(2)G$-module. If there exists a nontrivial elementary abelian 2-group $A$ with $|V:C_V(A)| \leq |A| $ then $V$ is called an {\it$F$-module}. If $|V:C_V(A)| \leq 2|A| $ is true, $V$ is called an {\it $F_1$-module}. In both cases $A$ is called an {\it offending subgroup}. 

\newtheorem{lemma}{Lemma}[section]

\begin{lemma}
$M_{24}$ does not possess an F-module.
\end{lemma}

\begin{bf} Proof: \end{bf} See \cite{As}. $\Box$

\begin{lemma}
Assume (ASS). If $X_i \leq K_i, X_i \not= K_4, i=1,2 $ or $3$, $ X_i$ elementary abelian. Then $|X_i| < |K_4|$.
\end{lemma}

\begin{bf} Proof: \end{bf} Otherwise $ |X_i : X_i \cap K_4 | \geq |K_4 : X_i \cap K_4 | $ and so $| X_iK_4 / K_4| \geq | K_4 : C_{K_4}(X_i)|$. But this contradicts lemma 1.1. $\Box$

\begin{lemma}
Assume (ASS). Then $\{ P_i,P_j,P_k \}$ does not belong to the $A_7$ - geometry with diagram  {$
\begin{picture}(25,4)
\put(1,-2.5){i}
\put(11,-2.5){j}
\put(21,-2.5){k}
\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}
$}  or  {$
\begin{picture}(25,4)
\put(1,-2.5){i}
\put(11,-2.5){j}
\put(21,-2.5){k}

\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}
$} .
\end{lemma}

\begin{bf} Proof: \end{bf} We have $\langle P_i, P_k \rangle / O_2( \langle P_i,P_k \rangle) \cong (Z_3 \times Z_3) \cdot 2 $ and in the first system we have $\langle P_j, P_k \rangle /O_2( \langle P_j,P_k \rangle) \cong A_6 $. This shows that the possible diagrams are\\
 {$
\begin{picture}(25,12)
\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(12.8,10.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(1,2.5){1}
\put(10,2.5){2}
\put(21,-2.5){4}
\put(21,13){3}


\end{picture} 
$} \quad {$ \begin{picture}(5,20)
\put(2,5){,} 
\end{picture}
$} \quad
{$
\begin{picture}(15,20)
\put(-1,-0.5){1}
\put(13,-0.5){4}
\put(-1,11.5){2}
\put(13,11.5){3}

\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){8.7}}

\put(12,1.8){\line(0,1){8.7}}

\put(2.5,11.7){\line(1,0){8.9}}
\put(2.5,10.4){\line(1,0){8.9}}
\put(7,13.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\put(2,1.8){\line(0,1){8.7}}
\end{picture}
$} \quad
{$ \begin{picture}(5,20)
\put(2,5){,} 
\end{picture}
$} \quad
{$
\begin{picture}(35,20)
\put(1,2.5){4}
\put(11,2.5){2}
\put(21,2.5){1}
\put(31,2.5){3}

\put(2,6){\makebox(0,0){$\circ$}}
\put(12,6){\makebox(0,0){$\circ$}}
\put(22,6){\makebox(0,0){$\circ$}}
\put(32,6){\makebox(0,0){$\circ$}}
\put(12.5,6){\line(1,0){8.7}}

\put(2.5,5.4){\line(1,0){8.9}}
\put(2.5,6.7){\line(1,0){8.9}}


\put(22.5,5.4){\line(1,0){8.9}}
\put(22.5,6.7){\line(1,0){8.9}}
\put(27,8.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\end{picture}
$} \quad \quad
{$ \begin{picture}(5,20)
\put(0.7,5){and} 
\end{picture}
$}
\\
{$
\begin{picture}(35,20)
\put(1,2.5){4}
\put(11,2.5){2}
\put(21,2.5){1}
\put(31,2.5){3}

\put(2,6){\makebox(0,0){$\circ$}}
\put(12,6){\makebox(0,0){$\circ$}}
\put(22,6){\makebox(0,0){$\circ$}}
\put(32,6){\makebox(0,0){$\circ$}}
\put(12.5,6){\line(1,0){8.7}}

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



\put(22.5,5.4){\line(1,0){8.9}}
\put(22.5,6.7){\line(1,0){8.9}}
\put(27,8.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\end{picture}
$} \quad {$ \begin{picture}(5,20)
\put(2,5){.} 
\end{picture}
$}
\\[0.4cm]
We have chosen notation such that $G_4/K_4 \cong M_{24}$ and  $G_3/K_3 \cong A_7$ . So we have $K_4 \leq K_3$. As $K_4 \not\!\!\lhd  G_3$ , we get some $x \in G_3 $ with $ K_4 \not= K^x_4 \leq K_3$ . But this contradicts lemma 1.2. 
$\Box$ \\
\begin {lemma}
Let $V$ be one of the following modules for $M_{24}$ : the Golay code of order $2^{11}$, the extended Golay code of order $2^{12}$ or the dual of the Golay code (the Todd code) of order $2^{11}$. Let $S \in Syl_2(M_{24})$. Then $|C_V(S)|=2$. If $V$ contains the Golay code then $C_{M_{24}}(C_V(S)) = 2^63\Sigma_6$ while $C_{M_{24}}(C_V(S)) = 2^4:A_8$ for the Todd--module $ V$. In any case if $t \in M_{24} $ is an involution then $|V:C_V(t)| \geq 2^4$.
\end{lemma}

{\bf Proof :} See \cite{Par1}, \cite{Par2}. $\Box$\\[0.3cm]
For the next lemma choose the numbering in the $M_{24}$-subdiagram as follows:
 \begin{center} {$
\begin{picture}(25,7)
\put(2,3){\makebox(0,0){$\circ$}}
\put(12,3){\makebox(0,0){$\circ$}}
\put(22,3){\makebox(0,0){$\circ$}}
\put(2.5,3){\line(1,0){8.7}}
\put(12.5,2.4){\line(1,0){8.9}}
\put(12.5,3.7){\line(1,0){8.9}}
\put(17,5.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(1,-0.5){1}
\put(11,-0.5){2}
\put(21,-0.5){3}

\end{picture}
$} \end{center}
\vspace{0.5cm}

\begin{lemma}
Assume (ASS). Then $G_3/K_3 \not\cong (3^7)Sp_6(2)$ and also $G_1/K_1 \not\cong (3^7)Sp_6(2)$.
\end{lemma}

{\bf Proof:}
Assume first  $G_1/K_1 \cong (3^7)Sp_6(2)$. By \cite{Par1}, \cite{Par2} we know $|K_4| \leq 2^{12}$. As $|S : K_4| = 2^{10}$ and $|S : K_1| = 2^9$, we see  $|K_1| \leq 2^{13}$. Let $\rho$ be an element of order three such that 
$N_{\langle P_2,P_3 \rangle}( \langle \rho \rangle) $ involves $\Sigma_6$.
Then $[\rho, K_1] \not= 1 $ and then $[O_3(G_1/K_1), K_1] \not= 1$ but that would force $|K_1| \geq 2^{14}$, a contradiction.\\ 
So we have $G_3/K_3 \cong (3^7)Sp_6(2)$. By lemma 1.2 we know that the Frattini subgroup $\Phi(K_3)$ is not trivial. Suppose first $\Phi(K_3) \subseteq K_4$. Then $[K_4,\Phi(K_3)] = 1$ and so $[\Phi(K_3), \langle K_4^{G_3} \rangle] = 1$. As $K_4 \not\leq K_3$ and $K_3K_4/K_4$ is abelian, we see $|K_3K_4/K_4| = 16$. Further $\langle P_1,P_2 \rangle = K_3C_{\langle P_1, P_2 \rangle}(\Phi (K_3))$. As we know that $C_{\langle P_1, P_2 \rangle}(\Phi (K_3))/K_4 \unlhd \langle P_1, P_2 \rangle /K_4$, we see $[\langle P_1, P_2 \rangle , \Phi (K_3)] = 1$. Hence we get $[K_3, \Phi (K_3)] = 1$ and so  $\Phi(K_3) \subseteq Z(G_3)$. As $|\Omega_1(Z(S))|=2$, we see $|\Phi(K_3)|=2$. Further $C_{G_4}(\Omega_1(Z(S)))$ involves $L_3(2)$ and so by lemma 1.4 $|K_4|=2^{11}$ and
$|K_3|=2^{12}$. As $|K_3K_4 : K_4| = 2^4$ we get $|K_3 \cap K_4|=2^8$ and $Z(K_3) \subseteq K_3 \cap K_4$. So $[G_3, Z(K_3)]=1$. Hence we have $Z(K_3)=K_3^\prime = \Phi(K_3)$ and then $K_3$ is extraspecial, a contradiction.
So we are left with $\Phi(K_3) \not\subseteq K_4$. As $K_3K_4/K_4$ is nonabelian we have $|\Phi (K_3)K_4/K_4| = 2$. Then also $|\Phi (K_3) : \Phi (K_3) \cap K_4| = 2$. As $[\Phi (K_3), K_4] \not= 1$, we see that $K_4$ induces transvections on $\Phi (K_3)$ to the hyperplane $\Phi (K_3) \cap K_4$. But then $|K_4: K_4 \cap K_3| = 2$, a contradiction. $\Box$

\begin{lemma}
Assume (ASS). For all $i \in \{1,...,4\}$ are equivalent:
\begin{tabbing}
\quad \= (i) \quad \= $\Omega_1(Z(S)) \not \!\! \lhd G_i$\\
\>(ii) \> $Z_i$ is an $F_1$-module
\end{tabbing}
\end{lemma}

{\bf Proof:} This is [Wie1, (2.1),(2.2)].$\Box$ \\[0.6cm]
The next lemma gives some information about rank three diagrams. The proofs are e.g. in \cite{Hei2}, \cite{Tim3}.


\begin{lemma}
Let $G = \langle P_1,P_2,P_3 \rangle$ be a parabolic system of classical or semiclassical type with diagram $\Delta$, such that $P_i/O_2(P_i) \cong \Sigma_3$ for all $i$.  Then we have:
\begin{tabbing}
\quad \= (i) \quad \= If $\Delta =$ {$
\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}
$} , then $G \cong A_7 $ or $A_8$.\\
\>(ii)\>If $\Delta =$  {$
\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}
$}, then $G \cong M_{24}, He $ or $3^7Sp_6(2)$ \\
\>(iii) \> If $\Delta =$ {$
\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}
$}, then $G \cong A_7$ or $Sp_6(2)$.\\
\>(iv) \> If $\Delta = $ {$
\begin{picture}(25,5)
\put(1,-2.5){1}
\put(11,-2.5){2}
\put(21,-2.5){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){$\scriptsize( \! \scriptstyle{\sim} \! \scriptsize)$}}
\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){$\scriptsize( \! \scriptstyle{\sim} \! \scriptsize)$}}
\end{picture}
$}, then there are two possibilities:\\
\\
\>\>{\sf ($\star$)} \quad \= $G_i/K_i \cong A_6, \Sigma_6, 3A_6$ or $ 3\Sigma_6, (i \in \{1,3\} ), 2^4 \leq |K_i| \leq 2^5, K_i $\\ \>\>\> is an undecomposable $G_i$-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 ($\star\star$)} \> $K_1 = K_3 = 1, G_2/K_2 \cong (Z_3 \times Z_3) Z_2.$\\
\end{tabbing}
\end{lemma}
If we have the situation of lemma 1.7(iv)($\star$), we will call such a parabolic system a system of type $3U_4(3)$ or $U_4(3)$ according to $G_i/K_i$ involves $3A_6$ for at least one $i$ or not. \\

 
The following lemma is well known. For convenience of the reader we will give a proof.
\begin{lemma}
Let $G \cong \Omega^+(8,2)$ and $V$ be a 9-dimensional $GF(2)G$-module such that $[V,G]$ is the natural module. Then $[V,G]$ is a direct summand.
\end{lemma}

{\bf Proof:} Let $P \leq G$ be the parabolic fixing a singular vector $v_1 \in [V,G]$. Then $P$ is an extension of $E$, an elementary abelian group of order $2^6$,  by $\Omega^+(6,2)$. Now we have $V_1 \subseteq V$ such that $V_1/ \langle v_1 \rangle$ is the orthogonal module for $P/E$ and $[E, V_1] = \langle v_1 \rangle$. As $|V/V_1|=4$, we see that $[P,V] = V_1$. Let $S \in Syl_2(P)$. Then $|C_{V_1/ \langle v_1 \rangle}(S)| = 2$ and so $C_{V/ \langle v_1 \rangle} (E) > V_1/ \langle v_1 \rangle$. Let $V_2$ be the preimage of $C_{V/ \langle v_1 \rangle} (E)$. Then for $e \in E^\sharp$ we have $|V_2 : C_{V_2}(e)| = 2$ and so $|C_{V_2}(E)| \ge 4$.
As $C_{V_1}(E) = \langle v_1 \rangle$, we get  $|C_{V_2}(E)| = 4$. As $\Omega^+(6,2)$ is perfect, we see $[C_{V_2}(E),P]=1$. So we get that $S$ centralizes some vector $v \in V \setminus [V,G]$. Now by Gasch\"utz 
[Hu, (I.17.4)] $[V,G]$ is a direct summand. $\Box$

\section{Diagrams of rank four}

In this chapter we are going to prove the following

\newtheorem{Prop}{Proposition}[section]

\begin{Prop}
Let $\Gamma$ be a parabolic system of $M_{24}$-type of rank four. Then one of the following holds :
\begin{tabbing}
\quad \= (i) \quad \= $\Delta$ is a string,\\
\>(ii) \> $\Delta$ :  {$
\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,-2.5){1}
\put(10,-2.5){2}
\put(21,-6.5){3}
\put(21,8){4}


\end{picture}
$}, where $G_4 \cong M_{24}$ or $ He$ and $G_3 \cong M_{24} $ or $He$.\\
\\

\end{tabbing}
\end{Prop}



We establish proposition 2.1 in a series of lemmas. For the rest of this chaper we are working under the assumptions of proposition 2.1.

\begin{lemma}
Suppose the diagram $\Delta$ of $\Gamma$ is a star, then (ASS) does not hold.
\end{lemma}

{\bf Proof:} Suppose (ASS) holds. Then due to \cite{Tim3} we have two possible diagrams
\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){$\scriptsize( \!\scriptstyle{\sim} \! \scriptsize)$}}
\put(1,-2.5){1}
\put(10,-2.5){2}
\put(21,-6.5){4}
\put(21,8){3}


\end{picture}
$} \quad and   \quad{$
\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(12.2,0.7){\line(1,-1){6.3}}

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


\put(1,-2.5){1}
\put(10,-2.5){2}
\put(21,-6.5){4}
\put(21,8){3}


\end{picture}
$}.\end{center}
\vspace{1cm}

Now in the first case we see with lemma 1.7, \cite{Par1} and \cite{Par2} that $G_3/K_3 \cong M_{24}, (3^7)Sp_6(2) $ or $A_7$. In the first case $K_3$ is elementary abelian with $|K_3| = |K_4|$. But this contradicts lemma 1.2. The latter two cases contradict lemma 1.5 and lemma 1.3. In the second diagram we see with lemma 1.7, \cite{Par1} and \cite{Par2} that $G_1/K_1 \cong M_{24} $ or $3^7Sp_6(2)$ so we get a contradicion with lemma 1.2 and lemma 1.5 again.

\begin{lemma}
If $\Delta$ is a star then $\Delta$ is of shape  {$
\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){1}
\put(10,-2.5){2}
\put(21,-6.5){4}
\put(21,8){3}


\end{picture}
$} , where $G_4 \cong M_{24} $ or $He$ and $G_3 \cong M_{24}$ or $He$.
\end{lemma}

{\bf Proof:} By lemma 2.1 we have $G_4 = K_4C_{G_4}(K_4)$. There are again two possible diagrams:\\
\begin{tabbing}
\quad \= (1) \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(12.2,0.7){\line(1,-1){6.3}}

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

\put(1,-2.5){1}
\put(10,-2.5){2}
\put(21,-6.5){4}
\put(21,8){3}


\end{picture}
$} \qquad \qquad \qquad and \qquad \qquad\= (2) \quad \= {$
\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(12.8,5.5){\makebox(0,0){ $\scriptstyle{\sim}$}}

\put(14,-3.5){\makebox(0,0){$\scriptsize( \!\scriptstyle{\sim} \! \scriptsize)$}}
\put(1,-2.5){1}
\put(10,-2.5){2}
\put(21,-6.5){4}
\put(21,8){3}


\end{picture}
$}\\
\\
\end{tabbing}

As $O_2( \langle P_2, P_3 \rangle)$ involves exactly one nontrivial $3\Sigma_6$-module which is 6-di- mensional, we see that $G_1/K_1 \cong M_{24}, He$ in case (1). As $Z( \langle P_2,P_3 \rangle / K_1) =1$, we get $K_4 \leq K_1$. Now $\Omega_1(Z(S)) \unlhd \langle P_1, P_2 \rangle$ and $\Omega_1(Z(S)) \unlhd \langle P_2, P_4 \rangle$, so $\Omega_1(Z(S)) \unlhd G_3$. By lemma 1.7 we have $G_3/K_3 \cong A_7$ or $A_8$. Now in the latter case we get $G_3 = C_{G_3}(K_3)K_3$, which contradicts the action of $\langle P_1, P_2 \rangle$ on $K_3$. So we have $G_3/K_3 \cong A_7$. Now $K_3/Z(K_3)$ is the natural $A_7$--module.  But $\langle P_1, P_2 \rangle$ induces orbits of length 7, 7, 21, 28 on $K_3/Z(K_3)$ and the orbits of length 7 generate a subgroup of order 8 in $K_3/Z(K_3)$, while in the action of $A_7$ on the natural module there is just one orbit of length 7 generating the whole module, a contradiction.\\
Suppose we have the second diagram.  Let $G_3/K_3 \cong A_7$. Then we obtain $\langle P_2, P_4 \rangle / O_2( \langle P_2, P_4 \rangle ) \cong A_6$ . By \cite{Tim3} we get that $G_1/K_1$ is parabolic of type $U_4(3)$. But then $O_2( \langle P_2, P_3 \rangle )$ involves a 4-dimensional $\Sigma_6$-module, a contradiction. \\
Hence $G_3/K_3 \not\cong A_7$. As $\langle P_1, P_2 \rangle$ induces a direct sum of two $L_3(2)$-modules on $O_2(\langle P_1, P_2 \rangle / C_{O_2 ( \langle P_1, P_2 \rangle )}(O^2( \langle P_1, P_2 \rangle))$, we see that $G_3/K_3 \not\cong (3^7)Sp_6(2)$. So $G_3/K_3 \cong M_{24} $ or $He$. Now $K_4K_3/K_3 \leq Z( \langle P_1, P_2 \rangle / K_3)$ and then $|K_3K_4/K_3| \leq 2$. As $K_4 \cap K_3 \lhd \langle G_4, G_3 \rangle = G$, we see $K_4 \cap K_3 = 1$. Hence $|K_4| = |K_3| \leq 2$. Suppose $|K_4| = |K_3| = 2$. From $G_4/K_4 \cong M_{24}$ we see that $|O_2(\langle P_1, P_3 \rangle)| = 2^9$ and $O_2(\langle P_1, P_3 \rangle)/K_4$ possesses no subgroup of index two normal in $\langle P_1, P_3 \rangle$. As $K_4 \leq O_2(\langle P_1, P_4 \rangle)$ too, we see that $K_4 \leq K_2$. Hence $|K_2| = 2^9$ and $|Z(K_2)| = 2^3$. Furthermore $G_2/K_2 = (Z_3 \times Z_3 \times Z_3)V_4$. As $3^2 \!\! \! \not \! |  ~|GL(3,2)|$ we see that $3^2 | \: |C_{G_2}(Z(K_2))|$. Hence $\langle P_1, P_3 \rangle$ acts in the same way on $Z(K_2)$ as $\langle P_1, P_4 \rangle$. But $C_{Z(K_2)}(\langle P_1, P_3 \rangle ) = K_4$ while $C_{Z(K_2)}(\langle P_1, P_4 \rangle ) = K_3$ and so $K_4 = K_3$, a contradiction. So we have $K_3 = K_4 = 1$ and the lemma is proved. $\Box$


\begin{lemma}
If $\Delta$ is a square, then (ASS) does not hold.
\end{lemma}


{\bf Proof:} Suppose false. Then we get with \cite{Tim3} that $\Delta$ has one of the following
shapes: \\
{$\begin{picture}(5,5)
\put(2,6){(1)}
\end{picture}$}
 \qquad {$
\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){4}
\put(13,-0.2){3}
\put(-1.5,11.8){1}
\put(13,11.8){2}

\put(2.5,1){.........}

\put(12,1.8){\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(2,1.8){\line(0,1){8.7}}
\end{picture}
$} \qquad \quad
{$\begin{picture}(5,5)
\put(2,6){(2)}
\end{picture}$}
\qquad
 {$
\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){4}
\put(13,-0.2){3}
\put(-1.5,11.8){1}
\put(13,11.8){2}

\put(2.3,1.5){\line(1,0){8.9}}
\put(2.3,0.6){\line(1,0){8.9}}


\put(12,1.8){\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,-2.5){\makebox(0,0){$\scriptsize( \!\scriptstyle{\sim} \! \scriptsize)$}}

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

\put(1.5,2){.}
\put(1.5,2.9){.}
\put(1.5,3.8){.}
\put(1.5,4.7){.}
\put(1.5,5.6){.}
\put(1.5,6.5){.}
\put(1.5,7.4){.}
\put(1.5,8.3){.}
\put(1.5,9.2){.}
\put(1.5,10.1){.}



\end{picture}
$} \qquad \quad {$\begin{picture}(5,5)
\put(2,6){(3)}
\end{picture}$}
 \qquad {$
\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){4}
\put(13,-0.2){3}
\put(-1.5,11.8){1}
\put(13,11.8){2}

\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(-1,6){\makebox(0,0){$\scriptsize( \!\scriptstyle{\sim} \! \scriptsize)$}}


\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(12,1.8){\line(0,1){8.7}}
\end{picture}
$}\quad.\\
\\
In case (1) we have $G_3/K_3 \cong M_{24}$ or $3^7Sp_6(2)$ by lemma 1.7, \cite{Par1} and \cite{Par2}. But this contradicts lemma 1.2 and lemma 1.5.\\
In case (2) we have $G_1/K_1 \cong M_{24}, (3^7)Sp_6(2)$ or $A_7$ by lemma 1.7 and \cite{Par1,Par2}.
Again the lemmas 1.2, 1.5 and 1.3 yield a contradiction.
\\
So we have that $\Delta$ is of type (3). Application of lemmas 1.2 and 1.3 yields $G_2/K_2 \cong (3^7)Sp_6(2)$. Furthermore by lemma 1.3 $G_1/K_1 \cong A_8$. By lemma 1.2 $K_4 \not\subseteq K_1$ and so $|K_4K_1/K_1| = 2^3$ and  then  $|K_4K_1/K_4| = 2^7$ . If $ t \in Z(K_1)$, then $|K_4 : C_{K_4}(t)| \leq 2^3$, as $|K_4 : K_4 \cap K_1| = 8$. By lemma 1.4 we get $[t, K_4] = 1$, i.e. $[Z(K_1), K_4] = 1$ and so $[Z(K_1), K_1\langle K_4^{G_1} \rangle] = 1$, i.e. $Z(K_1) \leq Z(G_1)$ and then $\Omega_1(Z(K_1)) = \Omega _1(Z(S))$. This shows $\Omega_1(Z(K_2)) \not= \Omega_1(Z(S))$ and then $|K_4 : K_4 \cap K_2 | \geq 2^4$ by lemma 1.4 again. But $O_2( \langle P_1, P_3 \rangle ) / K_2 \cong 2^{1+4} \times V_4 $, where $O_2( \langle P_1, P_3 \rangle / K_2) / Z(O_2( \langle P_1, P_3 \rangle ) / K_2) $ is an irreducible $\langle P_1, P_3 \rangle$ - module. As $K_4$ is abelian, this shows $|K_4 : K_4 \cap K_2| \leq 2^3$, a contradiction. $\Box$

\begin{lemma}
$\Delta$ is never a square.
\end{lemma}
{$
\begin{picture}(104,15)
\put(1,6){{\bf Proof:} By lemma 2.3 we may assume $G_4 = K_4C_{G_4}(K_4)$. Let $\Delta$ be }\quad \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){4}
\put(13,-0.2){3}
\put(-1.5,11.8){1}
\put(13,11.8){2}

\put(2.5,1){.........}


\put(11.6,1.8){\line(0,1){8.9}}
\put(12.4,1.8){\line(0,1){8.9}}

\put(2.5,11.5){\line(1,0){8.7}}

\put(14.5,6){\makebox(0,0){$\scriptstyle{\sim}$}}

\put(1.5,2){.}
\put(1.5,2.9){.}
\put(1.5,3.8){.}
\put(1.5,4.7){.}
\put(1.5,5.6){.}
\put(1.5,6.5){.}
\put(1.5,7.4){.}
\put(1.5,8.3){.}
\put(1.5,9.2){.}
\put(1.5,10.1){.}



\end{picture}
$} \\
Let first \quad {$ 
\begin{picture}(48,5)
\put(4.5,1){.........}
\put(4,1){\makebox(0,0){$\circ$}}
\put(14,1){\makebox(0,0){$\circ$}}
\put(0.5,-0.2){4}
\put(15.5,-0.2){3}
\put(20.5,0.5){=}

\put(30,1){\makebox(0,0){$\circ$}}
\put(40,1){\makebox(0,0){$\circ$}}
\put(26.5,-0.2){4}
\put(42.5,-0.2){3}

\put(30.5,1){\line(1,0){8.7}}
\end{picture} $}.
Then by lemma 1.7 $G_1/K_1 \cong M_{24}$, $He $ or $3^7Sp_6(2)$.  As $O_2 (\langle P_2, P_3 \rangle ) $ involves exactly one nontrivial $3 \Sigma_6$-module, the 6-dimensional one, we get $G_1/K_1 \cong M_{24} $ or $He$. Hence we get $G_1 = K_1C_{G_1}(K_1) $ by lemma 2.3. As $Z( \langle P_2, P_3 \rangle ) / K_1) = 1 $, we see $ K_1 = K_4 = 1$. But now $O_2 ( \langle P_1, P_2 \rangle )$ is the only extraspecial subgroup of order $ 2^7$ in $S$, the same applies for $O_2( \langle P_3, P_4 \rangle )$. Hence $ O_2( \langle P_1, P_2 \rangle ) \lhd \langle P_1, P_2, P_3, P_4 \rangle = G$, a contradiction. \\
So we have \quad  {$ 
\begin{picture}(48,5)
\put(4.5,1){.........}
\put(4,1){\makebox(0,0){$\circ$}}
\put(14,1){\makebox(0,0){$\circ$}}
\put(0.5,-0.2){4}
\put(15.5,-0.2){3}
\put(20.5,0.5){=}

\put(30,1){\makebox(0,0){$\circ$}}
\put(40,1){\makebox(0,0){$\circ$}}
\put(26.5,-0.2){4}
\put(42.5,-0.2){3}

\put(30.5,1.4){\line(1,0){8.9}}
\put(30.5,0.7){\line(1,0){8.9}}
\put(35,3){\makebox(0,0){$\scriptsize( \!\scriptstyle{\sim} \! \scriptsize)$}}


\end{picture} $} 
by \cite{Tim3} applied to $G_1$. Let $G_3/K_3 \cong He $ or $M_{24}$. Then by lemma 2.3 we also get $G_3 = K_3C_{G_3}(K_3)$. As both $O_2( \langle P_2, P_3 \rangle )$ and $O_2( \langle P_1, P_4 \rangle )$ involve just one nontrivial $3 \Sigma_6$-module, the 6-dimen\-sional one, we see with lemma 1.7 that both $G_1/K_1$ and $G_2/K_2$ are not the $3U_4(3)$ or $U_4(3)$ amalgam. Hence we have $K_1 = O_2( \langle P_2, P_3 \rangle )$ $= O_2 ( \langle P_3, P_4 \rangle )$ $= O_2 ( \langle P_1, P_4 \rangle ) = K_2$. But then $K_1 = 1$, a contradiction as $|S| \geq 2^{10}$.

Let $G_3/K_3 \cong (3^7)Sp_6(2)$. Then $ O_2( \langle P_1, P_2 \rangle ) / K_3 $ is an indecomposable $L_2(7)$-module, while $O_2( \langle P_1, P_2 \rangle ) / K_4$ involves a direct sum of two $L_2(7)$--mod\-ules as $K_4$ just involves trivial $L_2(7)$--modules, a contradiction. 

Let $G_3/K_3 \cong A_7$. Then $O_2(\langle P_1,P_2 \rangle ) = K_4K_3$. Now $|K_4Z(K_3) : K_4| \leq 2$ and so $[\langle P_1,P_2 \rangle ^{(\infty)}, Z(K_3)] = 1$. Hence $[G_3, Z(K_3)] = 1$ as $G_3/K_3$ is simple. Now we see that $Z(K_3) \cap K_4$ is normalized by $\langle G_3, G_4 \rangle = G$, and so $Z(K_3) \cap K_4 = 1$. As $C_{\langle P_1, P_2 \rangle}(O_2(\langle P_1, P_2 \rangle)) \leq O_2(\langle P_1, P_2 \rangle)$, we have $C_{G_3}(K_3) \leq K_3$ and so $\Omega_1(Z(S)) \leq Z(K_3)$. As $Z(K_3) \cap K_4 = 1$, we get $K_4 = 1$. So $K_3 \cong 2^{1+3+3}$. But the action of $A_7$ on the extraspecial group does not fit with the action of $L_3(2)$ in $M_{24}$ or $He$. 

So we have by \cite{Tim3} $G_3/K_3 \cong A_8$. We have $K_4 \subseteq K_3$ as $Z( \langle P_1, P_2 \rangle /K_3) = 1$. Hence $K_3/K_4 $ is elementary abelian. As $K_3$ involves exactly one nontrivial irreducible $ \langle P_1, P_2 \rangle /O_2( \langle P_1, P_2 \rangle )$-module, the natural one,  we see that $K_3$ involves exactly one natural $L_4(2)$-module $V$. As $G_1/K_1$ is not of type $U_4(3)$, we get with \cite{Tim3} that $ \langle P_3, P_4 \rangle / O_2 ( \langle P_3, P_4 \rangle ) \cong \Sigma_6$ or $ 3\Sigma_6$. Hence $G_2/K_2 \not\cong A_7$. Assume $G_2/K_2 \cong M_{24}$ or $He$. Then by lemma 2.3 $G_2 = K_2C_{G_2}(K_2)$ and $ |K_2| = |K_4|$. As $Z(\langle P_1, P_3 \rangle /K_4) = 1$, we see that $K_2 \leq K_4$ and so $K_2 = K_4 = 1$. As $S/K_2$ contains exactly one extraspecial subgroup of order $2^7$, we see that $O_2(  \langle P_1, P_2 \rangle ) = O_2(  \langle P_1, P_4 \rangle ) \lhd  \langle P_1, P_2, P_4 \rangle = G_3$, contradicting $|S : K_3| = 2^6$.

So we have $G_2/K_2 \cong (3^7)Sp_6(2)$. As $|K_2| \leq 4$, we have $K_2 \subseteq Z(G_2)$. As $|Z(S)| = |K_1| \cdot 2 = |K_2|$, we see that $K_2 = Z(S)$ and so $K_1 = 1, |K_2| = 2$. Hence $G_2 / O_3(G_2)$ is a covering group of $Sp_6(2)$. Now $K_3$ is the natural module for $L_4(2)$. In $Sp_6(2)$ we see that $S$ does not split over $K_3$. So by Gasch\"utz $G_3$ does not split over $K_3$. From $M_{24}$ or $He$, we see, that $S$ splits over $K_3$, and so by Gasch\"utz $G_3$ splits over $K_3$, a contradiction. $\Box$
\\


Now lemmas 2.1 to 2.4 prove proposition 2.1. $\Box$


\section{String diagrams}

In this chapter we will prove the following

\begin{Prop}
Let $\{P_1,P_2,P_3,P_4\}$ be a parabolic system of $M_{24}$--type. If the diagram $\Delta$ is a string then one of the following holds: 
\begin{tabbing}
\quad \= (1) \quad $\Delta :$  {$
\begin{picture}(35,10)
\put(1,-2.5){4}
\put(11,-2.5){3}
\put(21,-2.5){2}
\put(31,-2.5){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}
$}\\
\>(2) \quad $\Delta :$ {$
\begin{picture}(35,10)
\put(1,-2.5){1}
\put(11,-2.5){2}
\put(21,-2.5){3}
\put(31,-2.5){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(12.5,0.4){\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){$\scriptsize( \scriptstyle{\sim} \scriptsize) $}}
\end{picture}
$}\\
\> (3) \quad $\Delta :$  {$
\begin{picture}(35,10)
\put(1,-2.5){1}
\put(11,-2.5){2}
\put(21,-2.5){3}
\put(31,-2.5){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}
$} \quad, and $G_1 \cong G_4 \cong\ He$.\\
\\
\> In the cases (1), (2) (ASS) is true but not in case (3).
\end{tabbing}
\end{Prop}

We will prove proposition 3.1 in two lemmas 3.1 and 3.2.

\begin{lemma}
If $\Delta$ is a string diagram, then (ASS) holds or $G_1 \cong G_4 \cong\ He$ and $\Delta$ is of type {$
\begin{picture}(35,6)
\put(1,-2.5){1}
\put(11,-2.5){2}
\put(21,-2.5){3}
\put(31,-2.5){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}
$}
\end{lemma}

\vspace{0.4cm}
{\bf Proof:} Assume that (ASS) does not hold. By \cite{Tim3} $\Delta$ has one of the following shapes:
\begin{tabbing}
\quad \= (1) \quad \=  {$
\begin{picture}(35,6)
\put(1,-2.5){4}
\put(11,-2.5){3}
\put(21,-2.5){2}
\put(31,-2.5){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}
$} \\
\> (2) \>  {$
\begin{picture}(35,10)
\put(1,-2.5){1}
\put(11,-2.5){2}
\put(21,-2.5){3}
\put(31,-2.5){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(12.5,0.4){\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){$\scriptsize( \scriptstyle{\sim} \scriptsize) $}}
\end{picture}
$} \\
\> (3) \>  {$
\begin{picture}(35,10)
\put(1,-2.5){4}
\put(11,-2.5){3}
\put(21,-2.5){2}
\put(31,-2.5){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(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}
\put(7,3.5){\makebox(0,0){$ \scriptsize( \scriptstyle{\sim} \scriptsize) $}}

\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}
$} \\
\> (4) \>  {$
\begin{picture}(35,10)
\put(1,-2.5){4}
\put(11,-2.5){3}
\put(21,-2.5){2}
\put(31,-2.5){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(2.5,0.3){\line(1,0){8.9}}
\put(2.5,1.8){\line(1,0){8.9}}
\put(2.5,1){\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(27,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\end{picture}
$} \\
\> (5) \>  {$
\begin{picture}(35,10)
\put(1,-2.5){4}
\put(11,-2.5){3}
\put(21,-2.5){2}
\put(31,-2.5){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(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}
$}.\\
\end{tabbing} 
Suppose, we have (1), (3), or (4). Then $C_{K_3}(\langle P_1, P_2 \rangle^\prime ) \subseteq K_4$. As $\langle P_1, P_2 \rangle^\prime \unlhd G_3$, we have $C_{K_3}(\langle P_1,P_2 \rangle^\prime) \unlhd G_3$. As $G_4 = K_4C_{G_4}(K_4)$, we see that $C_{K_3}(\langle P_1, P_2 \rangle^\prime) = 1$.  As $G_3/K_3 \cong \Sigma_3 \times 3\Sigma_6$ or $(Z_3 \times 3A_6) \cdot2$, we get $|K_4| \leq 2$. Now in $G_1/K_1$ we see $C_{P_4}(O_2(P_4)) \subseteq O_2(P_4)$. This shows that $G_3/K_3 \cong (Z_3 \times 3A_6) \cdot 2$ and so $K_4 = 1$. In particular $|S|=2^{10}$.
Then choose $\rho \in \langle P_1, P_2 \rangle$ with $\rho O_2(\langle P_1, P_2 \rangle) \in O_3(\langle P_1, P_2 \rangle / O_2(\langle P_1, P_2 \rangle))$ and $o( \rho ) = 3$. Set $ P_5 = \langle \rho, S\rangle$. Inside of $M_{24}$ or $He$ one calculates that $\langle P_2, P_3, P_5 \rangle \cong 2^4A_8$ or $He$, respectively, where $\langle P_3, P_5 \rangle \cong 2^63\Sigma_6$ in the latter. Hence if we are in (1) then we see that $W := \langle P_5, P_2, P_3, P_4 \rangle$ is a parabolic system with the diagram  {$
\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(12.2,1.4){\line(1,1){6.3}}


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

\put(1,-2.5){4}
\put(10,-2.5){3}
\put(21,-6.5){2}
\put(21,8){5}


\end{picture}
$} \quad or \quad {$
\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(11.8,0.7){\line(1,-1){6.3}}

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

\put(1,-2.5){4}
\put(10,-2.5){3}
\put(21,-6.5){2}
\put(21,8){5}


\end{picture}
$} \quad in case of $G_4 \cong He$. By 
proposition 2.1 we have a\\
\\
\\
contradiction in the latter. Hence $W$ has a factor $\Omega^+(8,2)$ by \cite{Tim2}. But then $|S| \geq 2^{12}$, a contradiction. If we are in (3) and $G_4\cong M_{24}$, we get a diagram of type 
 {$
\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(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}
\put(6,3.5){\makebox(0,0){ $ \scriptsize ( \scriptstyle{\sim} \scriptsize ) $}}


 \put(12.3,1.3){\line(1,1){6.2}}

 \put(12.3,0.7){\line(1,-1){6.2}}



\put(1,-2.5){4}
\put(10,-2.5){3}
\put(21,-6.5){2}
\put(21,8){5}


\end{picture}
$}, where $W_4/O_2(W_4) \cong A_8$.
\\
\\

But then by proposition 2.1, \cite{Str3} or  \cite{Wie1} we have $P_5 
\subseteq \langle P_4, P_3, P_2 \rangle$. So we see that $G_1$ contains 
a subgroup $2^4 : A_8$. Hence we have $G_1 \cong M_{24} $ as $  He$ does not contain $2^4:A_8$.
Now $W_5 \cong W_2 \cong M_{24}$. But in $W_2 \cong M_{24}$ we have $W_{4,2} \cong 2^{1+6}L_3(2) \cong W_{5,4}$ in $W_5 \cong M_{24}$. So $W_{5,4} = W_{4,2}$. Then $W_4 = \langle W_{5,4},W_{4,2} \rangle \cong 2^{1+6}L_3(2)$, a contradiction. 


If we are in (3) and $G_4 \cong He$ we get a diagram of type ~~
{$
\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(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}
\put(6,3.5){\makebox(0,0){ $ \scriptstyle{\sim}$}}


 \put(12.3,1.3){\line(1,1){6.2}}


 \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){$\scriptsize( \!\scriptstyle{\sim} \! \scriptsize)$}}
\put(-1,0){5}
\put(10,3){3}
\put(21,-7){4}
\put(21,8){2} 

\end{picture}
$}.
\\
\\
\\
If $P_5 \leq \langle P_2, P_3, P_4 \rangle$, we have $G_1 \cong He$. If $P_5 \not\leq \langle P_2, P_3, P_4 \rangle$, then by proposition 2.1 we see that $G_1 \cong M_{24}$ or $He$. If $G_1 \cong M_{24}$, we may interchange 1 and 4 and argue as before to achieve a contradiction. So we have $G_1 \cong He \cong G_4$.


Let $\Delta$ be of type (4). As $|S| = 2^{10}$ and $\langle P_3, P_4 \rangle $ involves $U_3(3)$, we see that $2^4 \leq |K_1| \leq 2^5$. Now have that $G_1 = 
K_1C_{G_1}(K_1)$. In $G_1/K_1$ we see that $\langle P_2, P_3 \rangle /K_1$ is an extension of an elementary abelian group of order eight by $L_3(2)$. Hence $O_2(\langle P_2, P_3 \rangle)$ involves exactly one nontrivial $L_3(2)$--module. But from $G_4$, we see that there are exactly two of them, a contradiction.

Assume next that $\Delta$ is of type (5). Then $G_1/K_1 \cong He$ or $M_{24}$ too, as $3^7Sp_6(2)$ does not involve $2^63\Sigma_6$. As $Z( \langle P_2, P_3 \rangle / K_4 ) = 1$, we see that $K_1 = K_2 = 1$. Now $S$ contains exactly one extraspecial subgroup $W$ of order $2^7$ which yields $W= O_2( \langle P_1, P_2 \rangle ) =  O_2( \langle P_3, P_4 \rangle )$, a contradiction. 

Let finally $\Delta$ be of type (2). We have $G_3/K_3 \cong L_3(2) \times \Sigma_3$. Again by lemma 1.7(iv) we get $K_4 \subseteq K_1$  and so $ \langle K^{P_4}_4 \rangle \subseteq K_1$. As $G_3 = \langle P_1, P_2, P_4 \rangle$, we see $K_4 \subseteq K_3$. But then $G_3 \cap G_4/K_3 \cong L_3(2) \times Z_2$ but $ \langle P_1, P_2 \rangle / K_4 $ possesses no subgroup of index two, a contradiction. $\Box$


\begin{lemma}
Let $\Delta$ be a string diagram and assume (ASS). Then $\Delta$ is of type 
  {$
\begin{picture}(35,10)
\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}
$} or  {$
\begin{picture}(35,10)
\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){$\scriptsize( \scriptstyle{\sim} \scriptsize)$}}
\end{picture}
$}.


\end{lemma}

{\bf Proof:} The possible other diagrams are
\begin{tabbing}
\quad \=(1) \quad \=  {$
\begin{picture}(35,10)
\put(1,-2.5){4}
\put(11,-2.5){3}
\put(21,-2.5){2}
\put(31,-2.5){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(2.5,0.4){\line(1,0){8.9}}
\put(2.5,1.7){\line(1,0){8.9}}
\put(7,3.5){\makebox(0,0){$ \scriptsize( \scriptstyle{\sim} \scriptsize) $}}

\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}
$} \\
\> (2) \>  {$
\begin{picture}(35,10)
\put(1,-2.5){4}
\put(11,-2.5){3}
\put(21,-2.5){2}
\put(31,-2.5){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(2.5,0.3){\line(1,0){8.9}}
\put(2.5,1.8){\line(1,0){8.9}}
\put(2.5,1){\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(27,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}
\end{picture}
$} \\
\> (3) \>  {$
\begin{picture}(35,10)
\put(1,-2.5){4}
\put(11,-2.5){3}
\put(21,-2.5){2}
\put(31,-2.5){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(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}
$}.\\
\end{tabbing}
By lemmas 1.3 and 1.5 and \cite{Par1}, \cite{Par2} we have in case (1) and (3) that $G_1/K_1 \cong M_{24}$ too. As $|K_4| = |K_1|$ this contradicts lemma 1.2. So we are left with (2). 

Let $\rho \in \langle P_1, P_2 \rangle$ with $o(\rho) = 3$ and $\rho O_2(\langle P_1, P_2 \rangle)  \in Z( \langle P_1, P_2 \rangle^\prime /O_2( \langle P_1, P_2 \rangle ))$. Set $P_5 = \langle \rho, S\rangle$. Then  $W := \langle P_5, P_2, P_3, P_4 \rangle$ is a parabolic system with the diagram   {$
\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(2.5,1.8){\line(1,0){8.9}}
\put(2.5,0.3){\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(1,-2.5){4}
\put(10,-2.5){3}
\put(21,-6.5){2}
\put(21,8){5}


\end{picture}
$} 
where $W_4/O_2(W_4) \cong A_8$. As a Sylow 2-subgroup of\\[0.6cm]
$\langle P_3, P_2, P_4 \rangle $ and   $\langle P_3, P_4, P_5 \rangle$ is as in $G_2(2)$ which is not isomorphic to a Sylow 2-subgroup of $A_8$, we get $P_4 \not\subseteq \langle P_3, P_2, P_5 \rangle$, $P_2 \not\subseteq \langle P_4, P_3, P_5 \rangle,$ and $P_5 \not\subseteq \langle P_3, P_2, P_4 \rangle$. Now by \cite{Str5} such a parabolic system does not exist, a contradiction. $\Box$\\

So we have proved proposition 3.1. In the remainder of this chapter we investigate the two diagrams (1) and (2) in proposition 3.1. 

\begin{lemma}
Let $\{P_1, P_2, P_3, P_4\}$ be a parabolic system of $M_{24}$--type with diagram $\Delta : $  {$
\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(12.5,0.4){\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){1}
\put(11,-2.5){2}
\put(21,-2.5){3}
\put(31,-2.5){4}

\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){$\scriptsize( \scriptstyle{\sim} \scriptsize)$}}
\end{picture}
$}. Then $|K_4| = 2^{11}$ or $2^{12}$, where $K_4$ is the Go-\\[0.3cm] lay code-- or extended Golay code--module, and $G_4^\prime$ 
 does not split over $K_4 \cap G_4^\prime$. Furthermore $G_1$ is an extension of $K_1$, an extraspecial group of order $2^{13}$, by $G_1/K_1$, which  is parabolic isomorphic to a subgroup of $3(\cdot 3) \cdot Aut( U_4(3))$ containing $3(\cdot 3) \cdot U_4(3)$ and the action on $K_1/Z(K_1) $ is uniquely determined.
\end{lemma}

{\bf Proof:} By proposition 3.1 (ASS) holds. Now we have $|K_4| = 2^{11} $ or $2^{12}$ by  \cite{Par1} and \cite{Par2}. If $G_1/K_1$ is not parabolic isomorphic to a subgroup of $3(\cdot 3) \cdot Aut( U_4(3))$, then $K_4 \subseteq K_1$ by lemma 1.7. But this contradicts lemma 1.2. So we  have that  $G_1/K_1 \cong_p K $
{$
\begin{picture}(2,3)
\put(0,0.7){$ <$}
\put(0,-0.7){$\sim$}
\end{picture}
$}
 $3(\cdot 3) \cdot Aut(U_4(3))$. Let $K_4$ be the Todd--module. Then $|C_{K_4}(K_1)| = 2^4$. As $|S| = 2^{21}$ and $2^9 \geq |S : K_1| \geq 2^8$ we see that $2^{12} \leq |K_1| \leq 2^{13}$ and so $|K_1 \cap K_4| \leq 2^7$. Now for the group $3\Sigma_6$ we have a 4--, a 1--, and a 6--dimensional module involved in $K_4$. So we see that $K_1 \cap K_4$ involves the 6--dimensional module. As $C_{K_4}(K_1)$ is 4--dimensional we now get that $|K_1 \cap K_4| \geq 2^{10}$, a contradiction. So we have that $K_4$ involves the Golay--code--module. Now we see  $|K_1K_4/K_4| = 2^6$. The action of $M_{24}$ on the extended Golay--code--module shows that $|[K_1,K_4]| = 2^7$ and that $G_1 \cap G_4$ acts indecomposably on $K_4/[K_1,K_4]$. As $|K_1K_4/K_4| \geq 2^4$, this shows that $K_1 \cap K_4 = [K_1,K_4]$ and $[K_1,K_4,K_1] = \Omega_1(Z(S))$. Hence $|Z(K_1)| = 2$ and so $K_1$ is extraspecial of order $2^{13}$. Application of \cite[Theorem 2]{Wie5} gives that $G_1/C_{G_1}(K_1/Z(K_1)) \cong 3U_4(3)\cdot 2$ or $3U_4(3) \cdot 4$. 

Let us choose notation as in \cite{Wie5}, see also proposition 5.1. Then $\langle s, q, t, x, y \rangle$ $\cong 3A_6$ and is contained in $G_1 \cap G_4$. Using the matrices given there we can see that $K_1 \cap K_4/Z(K_1)$ does not possess a complement in $K_1/Z(K_1)$ invariant under $\langle s, q, t, x, y \rangle$. Hence $K_1\langle s,q,t,x,y \rangle/Z(K_1)$ does not split over $K_1 \cap K_4/Z(K_1)$. In particular a Sylow 2--subgroup of $K_1\langle s,q,t,x,y \rangle$ does not split over $K_1 \cap K_4$.  Then also $S$ does not split over $K_4$. By Gasch\"utz
we get that $G_4$ does not split.$\Box$

\begin{lemma}
Let $\{P_1, P_2, P_3, P_4\}$ be a parabolic system of $M_{24}$--type with diagram $\Delta : $  {$
\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(12.5,1){\line(1,0){8.7}}

\put(2.5,1){\line(1,0){8.7}}
\put(1,-2.5){4}
\put(11,-2.5){3}
\put(21,-2.5){2}
\put(31,-2.5){1}
\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}
$}, then $|K_4|=2^{11}$, $K_4$ is the Todd--module, and\\[0.3cm] $G_4$ splits over $K_4$.
\end{lemma}

{\bf Proof:} By proposition 3.1 (ASS) holds. We have $G_3/K_3 \cong \Sigma_3 \times 3\Sigma_6$ or $(Z_3 \times 3A_6) \cdot 2$. By lemma 1.2 $K_4 \not\leq K_3$ and so $G_3/K_3 \cong \Sigma_3 \times 3\Sigma_6$. Let $x \in P_4 \setminus S$. Then $|(K_4 \cap K_3)^x (K_4 \cap K_3) : K_4 \cap K_3 | = 2^6$, which shows that $2^5 \geq |K_4 \cap K^x_4| \geq 2^4$  As $K_4 \cap K_4^x$ is invariant under $\langle P_1 , P_2 \rangle ^{(\infty)}$, we see that $K_4$ has to be the Todd--module. Then by lemma 1.4 we have $|K_4| = 2^{11}$ and so $|K_4 \cap K_4^x| = 2^4$. 


Let $\rho \in P_4$, $o(\rho) = 3$. Then $\rho$ acts fixed point freely on $K_3/K_4 \cap K_4^x$ and trivially on $K_4 \cap K_4^x$. As $N_{K_4}(\langle \rho \rangle)$ is an indecomposable submodule of $K_4$ of dimension five on which $N_{\langle P_1, P_2 \rangle}(\langle \rho \rangle)$ induces $\Sigma_6$, and any $A_6$ over the fourdimensional module splits, we get that $N_{G_3}(\langle \rho \rangle) \cong 2^43\Sigma_6 \cdot \Sigma_3$ contains a subgroup $3\Sigma_6$.  Furthermore $K_4 $ contains a $3\Sigma_6$ submodule $V$ of order $2^6$. Then $V^{\rho}$ is $3\Sigma_6$-invariant and $V \cap K_4=1$. Hence there is a subgroup $2^63\Sigma_6$ intersecting $K_4$ trivially. This shows by Gasch\"utz that $G_4$ splits over $K_4$. $\Box$

\begin{lemma}
Let $\{P_1, P_2, P_3, P_4\}$ be as in lemma 3.4 and $\rho \in \langle P_1, P_2 \rangle$ of order three with $\rho O_2(\langle P_1, P_2 \rangle \in O_3( \langle P_1, P_2 \rangle / O_2 (\langle P_1, P_2 \rangle ))$.
 Set $P_5 = \langle \rho, S\rangle$. Then $\langle P_2, P_3, P_4, P_5 \rangle $ is an extension of an extraspecial group of order $2^9$ by $\Omega^{+}(8,2)$.
\end{lemma}

{\bf Proof:} The structure of $M_{24}$ shows that $\langle P_3, P_2, P_5 \rangle \cong 2^{11} 2^{4}A_8$. As $P_4 \not\subseteq \langle P_3, P_2, P_1 \rangle $ we see that $\langle P_4, P_3, P_2, P_5 \rangle = : W$ is a parabolic system with diagram  {$
\begin{picture}(23,12)
\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(12.2,1.4){\line(1,1){6.3}}


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

\put(1,-2.5){4}
\put(10,-2.5){3}
\put(21,-6.5){2}
\put(21,8){5}


\end{picture}
$}. By \cite{Tim2} we have that $W/S_W \cong \Omega^+(8,2)$. Furthermore by\\[1cm] lemma 3.4 we have $|S_W| =
 2^9$. Now either $S_W$ is extraspecial or elementary abelian. Suppose the latter. Then $S_W$ involves the natural module $V$. By lemma 1.8 $H_1(V,\Omega^+(8,2))=0$. So $S_W$ is a direct sum of the natural module and the trivial module. Hence $|\Omega_1(Z(S))|=4$. But this contradicts the structure of $G_4$.



\section{The $Co_1$ - case}

In this chapter we finish our investigations of the first diagram in proposition 3.1. More precisely we are going to prove the following

\begin{Prop}
If  $\{P_1,P_2,P_3,P_4\}$ is a parabolic system of $M_{24}$--type with diagram $\Delta : $  {$
\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(12.5,1){\line(1,0){8.7}}

\put(2.5,1){\line(1,0){8.7}}
\put(1,-2.5){4}
\put(11,-2.5){3}
\put(21,-2.5){2}
\put(31,-2.5){1}
\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}
$}, then $G \cong Co_1$.
\end{Prop}

{\bf Proof:} By lemma 3.4 we can give generators and relations for $G_4$. For this we just have to construct a split extension of the Todd--module with $M_{24}$. Let $\{
u_1, ..., u_{11}\}$ be a basis of $K_4$ and $\langle v_1, ..., v_6 \rangle$
an elementary abelian subgroup of $M_{24}$ where $\langle s, t, q, p, x, y, z \rangle \cong 3\Sigma_6$, acts on. Then $M_{24}$ is generated by this group and
an element $u$, where $\langle v_1, ...,v_6, p, s, t, q, u, y \rangle \cong 2^{1+6}L_3(2)$ and  $\langle v_1, ...,v_6, p, s, t, q, u, y ,z \rangle \cong
2^4A_8$. Then this group is given by generators and relations as follows. 
\\[0.2cm]
First we have the defining relations for the elementary abelian groups
$\langle u_1, ..., u_{11} \rangle$ and $\langle v_1, ..., v_6 \rangle$. The
action of the $v_i$ on $\langle u_1, ..., u_{11} \rangle$ is given by\\
$
[v_1,u_1]$,
$
u_2^{v_1}   u_2  u_1$,$ [u_3$,$v_1] u_1$,$ [u_4$,$v_1]$,$ [u_5$,$v_1]$,$ u_6^{v_1}  u_2  u_3  u_6 u_1$,$ 
u_7^{v_1} u_2 u_3  u_4  u_7  u_1$,\\ $ [u_8$,$v_1]$,$
u_9^{v_1}  u_4 u_5 u_9$,$ u_{10}^{v_1} u_5 u_{10}$,$ u_{11}^{v_1}  u_4  u_5 u_{11}$,\\
$ 
[u_1$,$v_2]$,$ [u_2$,$v_2] u_1$,$ [u_3$,$v_2]$,$ [u_4$,$v_2] u_1$,$ [u_5$,$v_2] u_1$,$ 
u_6^{v_2}  u_3  u_4  u_5  u_6$,$ u_7^{v_2}  u_4  u_5  u_7  u_1$,\\$
u_8^{v_2}  u_2  u_3  u_4  u_8$,$ u_9^{v_2}  u_2 u_3  u_5  u_9$,$ u_{10}^{v_2}  u_2  u_5  u_{10}$,$ u_{11}^{v_2}  u_2  u_3  u_4  u_{11}$,\\
$
[u_1$,$v_3]$,$ u_2^{v_3}  u_3 u_5$,$ u_3^{v_3} u_2  u_5$,$ [u_4$,$v_3]$,$ [u_5$,$v_3]$,$ 
u_6^{v_3}  u_5  u_7  u_8  u_{10}  u_{11}  u_1$,$
u_7^{v_3}  u_5  u_6  u_9  u_{10}$,\\$ [u_8$,$v_3] u_1$,$ u_9^{v_3}  u_5  u_6  u_7  u_{10}$,$ 
u_{10}^{v_3}  u_5  u_6 u_7  u_8  u_{11}  u_1,u_{11}^{v_3}  u_5  u_6  u_7 u_9  u_{10}  u_{11}  u_1$,\\
$
[u_1$,$v_4]$,$ u_2^{v_4} u_3 u_5$,$ [u_3$,$ v_4]$,$ u_4^{v_4} u_2  u_3  u_4  u_5$,$ 
u_5^{v_4}  u_2  u_3$,$ u_6^{v_4}  u_3  u_5  u_6  u_7  u_8  u_9$,$
u_7^{v_4}  u_2  u_3  u_6  u_9  u_{10}$,\\$ u_8^{v_4}  u_2  u_3  u_6  u_7  u_8  u_9  u_{10}$,$ 
u_9^{v_4}  u_2  u_3  u_5  u_9$,$
u_{10}^{v_4}  u_2  u_5  u_6  u_8$,$ u_{11}^{v_4}  u_2  u_3  u_6  u_7  u_9  u_{10}  u_{11}  u_1$,\\
$
[u_1$,$v_5]$,$ u_2^{v_5}  u_3  u_4  u_1$,$ u_3^{v_5}  u_2  u_4$,$ [u_4$,$v_5]  u_1$,$ 
[u_5$,$v_5]  u_1$,$ u_6^{v_5}  u_2  u_4  u_5  u_7  u_8  u_9  u_{10}$,\\$
{u_7}^{v_5}  u_4  u_5  u_7  u_1$,$ u_8^{v_5}  u_2  u_3  u_4  u_8$,$ 
{u_9}^{v_5}  u_2  u_3  u_4  u_5  u_6  u_7  u_{10}$,\\$
{u_{10}}^{v_5}  u_3  u_4  u_5  u_8  u_{10}$,$ u_{11}^{v_5}  u_2  u_3  u_6  u_7  u_9  u_{10}  u_{11}  u_1$,\\
$
[u_1$,$v_6]$,$ u_2^{v_6}  u_3  u_4$,$ u_3^{v_6}  u_3  u_1$,$ u_4^{v_6}  u_2  u_3  u_1$,$ 
u_5^{v_6}  u_2  u_3  u_4  u_5  u_1$,$ u_6^{v_6}  u_4  u_5  u_6  u_7  u_9  u_{11}$,\\$
u_7^{v_6}  u_5  u_6  u_9  u_{10}$,$ [u_8$,$v_6]  u_1$,$ u_9^{v_6}  u_4  u_6  u_7  u_{10}$,$ u_{10}^{v_6}  u_2  u_3  u_4  u_7  u_9  u_{10}  u_{11}  u_1,u_{11}^{v_6}  u_4  u_5  u_{11}$.
\\[0.4cm]
The relations for $3\Sigma_6$ are\\[0.3cm]
$
s^2$,$ q^2$,$ t^2$,\\
$ (s  t)^2  q$,$ p^2$,$ [p$,$s]$,$ [p$,$t]$,\\
$x^3$, $q^x  s$,$ s^x  q  s$,\\$ (x  t)^2$,$ [p$,$x]$
$
y^3$,$ q^y  t$,$ t^y  q  t$,$ (y  s)^2$,$ p^y  p  q  t$, $(xy)^5$\\
$z^{-1}(xptypqs)^4$, $
[z$,$y]$, $[z,x]$, 
$z^3, [z,s], [z,t], [z,q], (z p)^2.$\\[0.4cm]
To see these relations we use the following identification with elements in $\Sigma_6$ : \\
$q = (12)(34), s = (13)(24), t = (12)(56), \\
x = (123), (y = (153)(264), p = (56).$
\\

The action of this group on $\langle u_1,...,u_{11} \rangle$ is given by\\[0.3cm]
$
[u_1$,$s]$,$ [u_2$,$s]$,$ u_3^s  u_2  u_5$,$ [u_4$,$s]$,$ u_5^s  u_2  u_3$,$ 
u_6^s  u_2  u_6  u_7  u_8  u_9  u_1$,$ u_7^s  u_2  u_{11}$,$
u_8^s  u_9  u_{11}$,$ u_9^s  u_2  u_7  u_8$,\\$ u_{10}^s  u_2  u_7  u_{10}  u_{11}$,$ u_{11}^s  u_2  u_7$,\\
$
[u_1$,$t]$,$ [u_2$,$t]  u_1$,$ [u_3$,$t]  u_1$,$ [u_4$,$t]$,$ [u_5$,$t]  u_1$,$ [u_6$,$t]  u_1$,$ 
u_7^t  u_2  u_3  u_7  u_1$,\\$ u_8^t  u_2  u_3  u_4  u_8$,$
u_9^t  u_3  u_5  u_9  u_1$,$ u_{10}^t  u_3  u_4  u_5  u_{10}  u_1$,$ u_{11}^t  u_2  u_5  u_{11}  u_1$,\\
$
[u_1$,$q]$,$ [u_2$,$q]$,$ [u_3$,$q]  u_1$,$ [u_4$,$q]$,$ [u_5$,$q]  u_1$,$ u_6^q  u_3  u_4  u_5  u_6  u_1$,$ 
u_7^q  u_2  u_7$,\\$ u_8^q  u_2  u_3  u_4  u_5  u_8  u_1$,$ u_9^q  u_3  u_4  u_5  u_9  u_1$,$ 
u_{10}^q  u_3  u_5  u_{10}  u_1$,$ u_{11}^q  u_2  u_{11}$,\\
$
[u_1$,$p],u_2^{p}  u_2  u_4  u_1$,$ u_3^{p}  u_3  u_4$,$ u_4^{p}  u_4$,$ [u_5$,$p]  u_1$,$ 
u_6^{p}  u_2  u_4  u_5  u_7  u_8  u_9  u_{10}$,\\$
u_7^{p}  u_3  u_6  u_8  u_9  u_{10}  u_1$,$ u_8^{p}  u_2  u_3  u_4  u_5  u_8$,$ 
u_9^{p}  u_3  u_6  u_7  u_9  u_{10}  u_{11}  u_1$,\\$
u_{10}^{p}  u_3  u_4  u_6  u_7  u_{11}$,$ u_{11}^{p}  u_2  u_4  u_5  u_6  u_7  u_{10}  u_1$,\\
$
[u_1$,$y]$,$ u_2^y  u_2  u_4  u_5$,$ u_3^y  u_2  u_4$,$ [u_4$,$y]$,$ 
u_5^y  u_2  u_3  u_4  u_5$,$ u_6^y  u_2  u_4  u_5  u_6  u_7  u_8  u_9$,$
u_7^y  u_2  u_4  u_6  u_7  u_{10}$,\\$ u_8^y  u_4  u_5  u_9  u_{11}  u_1$,$
u_9^y  u_2  u_4  u_6  u_{10}$,$
u_{10}^y  u_2  u_7  u_8  u_{10}$,$ u_{11}^y  u_2  u_4  u_5  u_7  u_8  u_{11}$,\\
$u_1^xu_1u_2u_3u_5$, $u_2^xu_1u_3u_5u_7u_{11}$, $u_3^xu_1u_2u_6u_8u_{10}$,
$u_4^xu_8u_9u_{11}$, $u_5^xu_1u_2u_3u_6u_7u_8u_{10}u_{11}$, 
$u_6^xu_1u_2u_3u_5u_6$, $u_7^xu_1u_5u_6u_7u_8u_{10}$, $u_8^xu_2u_3u_7u_9$, 
$u_9^xu_1u_2u_4u_6u_{10}u_{11}$, $u_{10}^xu_1u_2u_6u_7u_9$,
$u_{11}^xu_1u_3u_6u_7u_8u_{10}$,\\ 
$
[u_1$,$z]$,$ u_2^z  u_2  u_4  u_1$,$ u_3^z  u_5  u_1$,$ u_4^z  u_3  u_5$,$ u_5^z  u_3  u_4$,$ 
u_6^z  u_2  u_8  u_{10}  u_{11}$,$ 
u_7^z  u_2  u_4  u_5  u_6  u_7  u_{10}  u_1$,\\$ u_8^z  u_3  u_4  u_7  u_8  u_9  u_{11}  u_1$,$ 
u_9^z  u_3  u_4  u_5  u_6  u_{10}$,$
u_{10}^z  u_4  u_5  u_6  u_8  u_9  u_{11}  u_1$,$ u_{11}^z  u_3  u_6  u_8  u_9  u_{10}  u_1$.\\[0.4cm]
The action on $\langle v_1,...,v_6 \rangle $ is given by\\[0.3cm]
$
[v_1$,$q]$,$ [v_2$,$q]$,$ [v_3$,$q]$,$ [v_4$,$q]$,$ v_5^q  v_5  v_1$,$ v_6^q  v_6  v_2$,\\
$
[v_1$,$s]$,$ [v_2$,$s]$,$ [v_3$,$s]$,$ [v_4$,$s]$,$ v_5^s  v_5  v_3$,$ v_6^s  v_6  v_4$,\\
$
[v_1$,$t]$,$ [v_2$,$t]$,$ v_3^t  v_1  v_3$,$ v_4^t  v_4  v_2$,$ [v_5$,$t]$,$ [v_6$,$t]$,\\
$
[v_1$,$p]$,$ v_2^{p}  v_1  v_2$,$ [v_3$,$p]$,$ v_4^{p}  v_4  v_3$,$ [v_5$,$p]$,$ 
v_6^{p}  v_5  v_6$,\\
$
[v_1$,$y]$,$ [v_2$,$y]$,$ v_3^y  v_2  v_5$,$ v_4^y  v_1  v_2  v_6$,$ v_5^y  v_3  v_5$,$ v_6^y  v_4  v_6$,\\
$v_1^xv_3$, $v_2^xv_4$, $v_3^xv_1v_3$, $v_4^xv_2v_4$, $[v_5,x]$, $[v_6,x]$,\\
$
v_1^z  v_1  v_2$,$ v_2^z  v_1$,$ v_3^z  v_3  v_4$,$ v_4^z  v_3$,$ v_5^z  v_5  v_6$,$ v_6^z  v_5$.\\[0.4cm]
These are generators and relations for $2^{11}2^63 \Sigma_6$. Now we
add $u$ with $u^3$ $=1$ and the following relations:\\[0.3cm]
$ 
[u_1,u], u_2^u u_3 u_1, u_3^u u_3 u_4, u_4^u u_2 u_3 u_4 u_1, 
u_5^u u_2 u_3 u_4 u_5 u_1, u_6^u u_4 u_5 u_6 u_7 u_9 u_{11},$\\$
u_7^u u_2 u_3 u_5 u_7 u_{10} u_{11} u_1, u_8^u u_9 u_{11}, u_9^u u_5 u_8 u_{10},
 u_{10}^u u_5 u_7 u_9 u_{10} u_{11},
u_{11}^u u_6 u_7 u_8 u_{11} u_1,$\\$
[v_1,u], v_2^u v_2 q, [v_3,u], v_4^u v_4 s, v_5^u v_5 p,
 (v_6 u)^2,$\\$
q^u v_2, [t,u], p^u v_5, s^u v_4, (y s v_2 u)^3, [x,u] ,(z u)^4.$
\\[0.3cm]
These are the relations for $2^{11}M_{24}$. One easily checks that $\langle v_1, $ $v_2$, $v_3$, $v_4$, $v_5$, $v_6$, $p$, $q$, $s$, $t$, $x$, $y$, $z$, $u \rangle \cong M_{24}$. Maybe the easiest way is to see that $\langle v_1, v_2, v_3, v_4, v_5, v_6, p, q,s,t,x,y,z\rangle $ is an extension of an elementary abelian group of order $2^6$ by $\Sigma_6$, $\langle v_1, v_2, v_3, v_4, v_5, v_6, p, q,s,t,y,u \rangle$ is an extension of a group of order $2^7$ by $L_3(2)$ and $\langle v_1, v_2, v_3, v_4, v_5, v_6, p, q,s,t,x,u \rangle $ is an extension of a group of order $2^8$ by $\Sigma_3 \times \Sigma_3$. Now we have a parabolic system of type $M_{24}$ and rank three and so we may quote \cite{Hei2}. 

As $u_1$ is centralized by $\langle v_1, v_2, v_3, v_4, v_5, v_6, p, q,s,t,x,y,z\rangle $ we see that $\langle u_1$ , $\ldots $, $u_{11} \rangle$ is the module from lemma 3.4. So we have the structure of $G_4$, which is uniquely determined.\\ 

By lemma 3.5 we have $\langle P_4, P_3, P_2,P_5 \rangle \cong 2^{1+8}\Omega^+_8(2)$, where $\langle P_3, P_2,P_5\rangle $ is generated by
$\{u_1,...,u_{11},v_1,...,v_6,s,t,q,p, u,y,z\}$.
Now choose some $v \in P_4$ with $o(v) = 3$ such that $\langle P_2,P_3,P_5,v \rangle = \langle P_2,P_3,P_5,P_4 \rangle$. As $G_3/K_3 \cong \Sigma_3 \times 3\Sigma_6$, we may choose $v$ such that $[v,z] = 1 = [v,y]$. As $(K_4 \cap K_3)^v(K_4 \cap K_3) = K_3$, we get $|C_{K_3}(v)|=2^4$ and so $C_{G_3}(v) \cong 
  \langle v \rangle \times 2^4 3 \Sigma_6$. This implies that we may choose $v$ with $[v,x]=1$.

The extraspecial normal 2-subgroup of $2^{1+8} \Omega_8^+(2) $ is given by 
$X:= \langle u_1,u_2,$ \\ $u_3,u_4, u_5,v_1,v_2,q,t \rangle$, which can be seen inside of $\langle P_2, P_3, P_5 \rangle$. Using the relations obtained so far we see that $v$ acts on $X/\langle u_1 \rangle$ as the following matrix
\begin{center}
$
\left(
\begin{array}{cccccccc}
1 & 0 & 0 & 1 & 1 & 1 & 0 & 0\\
0 & 1 & 1 & 1 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 1 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
\end{array}
\right)
$
\end{center}
where the basis is $\{u_2,u_3,u_4,u_3u_5,v_1,v_2,q,t \}$.\\
In $ \langle P_2, P_3, P_4, P_5 \rangle / X $ we get the following relations:
\\[0.3cm]$
[s,v] ,[p,v] ,[v,y] ,[v,z] ,[u_6  u_7  u_9  u_{10},v],$\\$
v_3^v v_3u_8  u_9  u_{11}, (u_8  u_9  u_{11})^v v_3,
v_4^v u_7  u_8  u_9 v_4, $\\$
v_5^v u_6  u_7 u_8  u_9  u_{10}v_5, (u_7  u_8 u_9)^v v_4, $\\$
(u_6  u_7  u_8  u_9  u_{10})^v v_5, v_6^v v_6 u_7  u_9  u_{11}, (u_7 u_9  u_{11})^v v_6, $ \\ $
v^{u_6 u_7 u_9  u_{11}} v, (v  u )^7, (u^2 v)^4 $.\\

These can easily calculated as we know the action of all elements of $\langle P_2$, $P_3$, $P_4$, $P_5 \rangle /X$ on $X$, so we just have to do matrix multiplications.\\[0.4cm]
As the extension of $X/ \langle u_1 \rangle $ by $\Omega^+_8(2)$ is determined by the extension of the Sylow 2-subgroup and this is known from $G_4$, we can calculate modulo $\langle u_1 \rangle $ :\\[0.3cm]
$
[v,y] ,[v,z] ,(v  u )^7 ,(u^2  v)^4 ,[s,v] ,[p,v] ,$ \\
$
v_3^v   u_8  u_9  u_{11}v_3, 
(u_8  u_9  u_{11})^v v_3u_2  u_3  u_5, 
v_4^v u_2  u_7  u_8  u_9v_4,$\\$
v_5^v u_2  u_4  u_5  u_6  u_7  u_8  u_9 u_{10}v_5, $ 
$
(u_2  u_7  u_8  u_9)^v v_4u_2  u_3  u_5,$\\$
(u_4  u_6  u_7  u_8  u_9  u_{10})^v v_5 v_2 u_2  u_3  u_4,$\\$
v_6^v v_6 u_4  u_5  u_7  u_9  u_{11},
(u_2  u_7  u_9  u_{11})^v v_6 v_2 v_1u_2  u_3  u_4,
v^{u_6  u_7  u_9  u_{11}} v,$\\$
(u_6  u_7  u_9  u_{10})^v  u_4  u_6  u_7  u_9  u_{10} v_1.$\\
\\
As can be check with MAGMA this defines modulo $\langle u_1 \rangle$ a nonsplit extension of an elementary abelian group of order $2^8$ by $\Omega^+(8,2)$.\\

As $[x,v] = 1$ and $u_1^x = u_1  u_2  u_3  u_5$, we get \\[0.3cm]
$(u_1  u_2  u_3  u_5)^v = (u_1  u_2  u_3  u_5)^{x^{-1}  v  x} = u_1^{v  x} = u_1^x = u_1  u_2  u_3  u_5$.\\[0.3cm]
In this way we can calculate all relations involving $v$, which then yields : \\[0.4cm]
$
v^3, [v,y], [v,z], (v u)^7, (u^{-1} v)^4, $\\$
u_2^v u_2 u_3 u_5 v_1 v_2 u_1, u_3^v u_4 u_5 v_2, u_4^v v_1 u_1,
 (u_3 u_5)^v v_1 v_2 u_1, v_1^v u_4 v_1,$\\$
v_2^v u_3 u_4 u_5 v_2 u_1, [v,q], [t,v], [s,v], [p,v], 
(u_6 u_7 u_9 u_{10})^v v_1 u_4 u_1 u_6 u_7 u_9 u_{10},$\\$
(u_6 u_7 u_9 u_{11} v)^2, v_3^v v_3 u_8 u_9 u_{11} ,
 (u_8 u_9 u_{11})^v u_1 u_2 u_3 u_5 v_3, $\\$
v_4^v v_4 u_2 u_7 u_8 u_9 u_1$, $
 v_5^v v_5 u_2 u_4 u_5 u_6 u_7 u_8 u_9 u_{10},$\\$
(u_2 u_7 u_8 u_9)^v u_2 u_3 u_5 v_4$,
 $(u_4 u_6 u_7 u_8 u_9 u_{10})^v u_2 u_3 u_4 v_2 v_5 ,$\\$
v_6^v u_4 u_5 u_7 u_9 u_{11} v_6$, 
$(u_2 u_7 u_9 u_{11})^v u_1 u_2 u_3 u_4 v_1 v_2 v_6.
$ \\[0.4cm]
This then provides us with generators and relations for $G= \langle P_1, P_2, P_3, P_4 \rangle$, where all these come uniquely from the diagram.\\[0.2cm]

Now we calculate that $ \langle u_1,u_2,u_3,u_4,u_5,u_6 u_7, u_6 u_8, u_6  u_9, u_6  u_{10}, u_{11} \rangle \langle v_1,v_2,$\\$v_3,v_4 v_5,v_6, t,q,ps,yz^{-1}, u^{v_5},y^{x^{-1}}s(y^{-1}s)^{x^{-1}} \rangle$ is an extension of an elementary abelian group of order $2^{10}$ by$Aut(M_{22})$ in $G_4$. We generate this group by two parabolics containing the Sylow 2-subgroup  $ \langle u_1$, $u_2$, $u_3$, $u_4$, $u_5$, $u_6 u_7$, $u_6  u_8$, $u_6  u_9$, $u_6  u_{10}$, $u_{11} \rangle  \langle v_1$, $v_2$, $v_3,v_4 v_5$, $v_6$, $t$, $q$, $ps \rangle =: T$, where $ \tilde{P}_1 = 
\langle T$, $yz^{-1}$, $y^{x^{-1}}s(y^{-1}s)^{x^{-1}} \rangle$ , $\tilde{P}_1/O_2(\tilde{P}_1) \cong \Sigma_5$ and $\tilde{P}_2 = \langle T, u^{v_5} \rangle, \tilde{P}_2/O_2(\tilde{P}_2) \cong \Sigma_3$. Now choose $\tilde{P}_3 = \langle T, v^{v_5} \rangle \subseteq
 \langle P_2, P_3 , P_4, P_5 \rangle $. Then $T \in Syl_2(\tilde{P}_3)$ and $\tilde{P}_3/O_2(\tilde{P}_3) \cong \Sigma_3$. Furthermore 
$\langle \tilde{P}_2, \tilde{P}_3 \rangle / O_2(\langle \tilde{P}_2, \tilde{P}_3 \rangle ) \cong
L_3(2)$ as $\langle u,v \rangle $ generate $L_3(2)$ and $\langle \tilde{P}_3, \tilde{P}_1 \rangle = \tilde{P}_3 \tilde{P}_1$ as $ \langle \tilde{P}_3, \tilde{P}_1 \rangle \subseteq G_3$, 
where $\tilde{P}_3 \subseteq P_4$ and $\tilde{P}_1 \subseteq \langle P_2, P_1 \rangle$. So we have that $\langle \tilde{P}_1,\tilde{P}_2,\tilde{P}_3 \rangle $ possesses a diagram 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(12.5,1){\line(1,0){8.7}}
\put(17,3.5){\makebox(0,0){$\scriptstyle{M_{22}}$}}
\end{picture}
$}
. Application of [Str6, Theorem 4] shows that $\langle \tilde{P}_1,\tilde{P}_2,\tilde{P}_3 \rangle \cong Co_2$. Now using the Todd-Coxeter algorithm we get $|G: \langle \tilde{P}_1,\tilde{P}_2,\tilde{P}_3 \rangle | = 98280$. Hence $|G| = |Co_1|$. By \cite{RoStr} $Co_1$ possesses a geometry of this type, so $G \cong Co_1$.


\section{The $M(24)$-case}

In this chapter we investigate parabolic systems whose diagram is as in proposition 3.1 (2). First we prove that the corresponding amalgam is uniquely determined.

\begin{Prop}
Let $\{P_1, P_2, P_3, P_4\}$ be a parabolic system of $M_{24}$--type with diagram $\Delta$ :  {$
\begin{picture}(35,5)
\put(1,-2.5){1}
\put(11,-2.5){2}
\put(21,-2.5){3}
\put(31,-2.5){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(12.5,0.4){\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){$\scriptsize( \scriptstyle{\sim} \scriptsize) $}}
\end{picture}
$}
, then the amalgam is uniquely determined \\[0.3cm]
and isomorphic to one of $M(24)^\prime $ or $M(24)$, $3M(24)$ or $3M(24)^\prime$.
\end{Prop}

\begin{bf} Proof: \end{bf}
By lemma 3.3 we know that $K_1$ is an extraspecial group of order  $2^{13}$ and $(G_1/C_{G_1}(K_1/Z(K_1)))^{(\infty)} \cong 3U_4(3) $. In particular if $\langle P_3, P_4 \rangle /O_2(\langle P_3, P_4 \rangle) \cong 3\Sigma_6$, then we see that  $C(K_1) \not\leq O_2(\langle P_3, P_4 \rangle)$. Hence we have an element of order 3, which centralizes $K_1$ and so it does not affect the structure of the amalgam, as it also centralizes $K_4 \cap G_4^\prime$. So we may assume that we have $\langle P_3, P_4 \rangle /O_2(\langle P_3, P_4 \rangle) \cong \Sigma_6$. If we can show that this amalgam is as in $M(24)$ or $M(24)^\prime$, then the other is as in the triple cover of these groups.  So we now get  $G_1/C_{G_1}(K_1/Z(K_1)) \cong 3U_4(3) \cdot 2$ or $  3U_4(3) \cdot 4$ depending on $|K_4| = 2^{11}$ or $2^{12}$. 

Let $\{u_1,...,u_{11}\} $ be a basis for $K_4$ in case of $|K_4| = 2^{11}$ and $\{v_1,...,v_6\}$ be an elementary abelian group of order $2^6$ such that $\{u_1,...,u_7,v_1,...,v_6 \} = K_1$ where $[u_{i+1},v_i]=u_1, i=1,...,6$ and 
$[u_j,v_i]=1 $ for $j\ \not= i+1$.
Choose $\langle s,t,q,x,y,p \rangle \cong 3\Sigma_6$ as in proposition 4.1. Hence we have the following relations 
\\
 
$ s^2$ ,$t^2$ ,$q^2$, $(st)^2q$, $q^xs$, $s^xqs$, $x^tx$, $q^yt$, $t^yqt$,
$y^sy$, $(xy)^5$, $x^3$, $y^3$, $p^2$, $[p,s]$, $[p,t]$, $[x,p]$, $p^ypqt$.
\\ 

One should have in mind that these are relations modulo $\langle u_1 \rangle$. As shown in \cite{Wie5} there is a subgroup $3\Sigma_6$ in $G_1/Z(K_1)$, so we may in fact use the relations above for $3\Sigma_6$  in $G_1/Z(K_1)$.
\\
 
Then $ \langle u_1,...,u_{11} \rangle \langle v_1,...,v_6 \rangle
\langle s,t,q,x,y,p \rangle / 
\langle u_1,...,u_{11} \rangle$ is a subgroup isomorphic to $2^63 \Sigma_6$ in $ M_{24}$. The action of $v_1,...,v_6,s,t,q,x,y,p $ on $ \langle u_1,...
u_{11} \rangle $ is given by \\[0.5cm]
$
u_8^{v_1} u_5 u_8, $\\$
u_9^{v_1} u_1 u_3 u_6 u_7 u_9, u_{10}^{v_1} u_5 u_7 u_{10}, u_{11}^{v_1} u_7 u_{11},$\\$ u_8^{v_2} u_4 u_8, u_9^{v_2} u_1 u_2 u_7 u_9,$\\$ u_{10}^{v_2} u_4 u_6 u_{10}, u_{11}^{v_2} u_6 u_{11},$\\$
u_8^{v_3} u_1 u_3 u_5 u_6 u_7 u_8, $\\$ u_9^{v_3} u_1 u_5 u_6 u_7 u_9, u_{10}^{v_3} u_1 u_3 u_5 u_6 u_{10}, u_{11}^{v_3} u_{11},$\\$
u_8^{v_4} u_1 u_2 u_4 u_7 u_8, $\\$ u_9^{v_4} u_1 u_4 u_7 u_9, u_{10}^{v_4} u_1 u_2 u_4 u_6 u_7 u_{10},
u_{11}^{v_4} u_{11},$\\$
u_8^{v_5} u_4 u_8, $\\$ u_9^{v_5} u_2 u_4 u_9, u_{10}^{v_5} u_3 u_4 u_5 u_{10}, u_{11}^{v_5} u_3 u_{11},$\\$
u_8^{v_6} u_4 u_5 u_8, $\\$ u_9^{v_6} u_2 u_3 u_4 u_5 u_9, u_{10}^{v_6} u_2 u_5 u_{10}, u_{11}^{v_6} u_2 u_{11},$\\$
[q,u_1], u_2^q u_2 u_6, u_3^q u_3 u_7, [u_4,q], [u_5,q], [u_6,q], [u_7,q], 
u_8^q u_{10} u_{11},$\\$ u_9^q u_1 u_9 u_{11}, u_{10}^q u_8 u_{11},
[u_{11},q],$\\$
[u_1,s], [u_2,s], [u_3,s], u_4^s u_4 u_6, u_5^s u_5 u_7, [u_6, s], [u_7,s], 
u_8^s u_1 u_{10},$\\$  u_9^s u_1 u_8 u_9 u_{10} u_{11}, 
u_{10}^s u_1 u_8, [u_{11},s],$\\$
[u_1,t], u_2^t u_2 u_4, u_3^t u_3 u_5, [u_4,t], [u_5,t], [u_6,t], [u_7,t],
[u_8,t], $\\$ u_9^t u_8 u_9, u_{10}^t u_8 u_{11},
u_{11}^t u_8 u_{10},$\\$
[p,u_1], u_2^{p} u_2 u_3, [u_3,p], u_4^{p} u_4 u_5, [u_5,p],
u_6^{p} u_6 u_7, [u_7,p],u_8^{p} u_{11} u_{10}, $\\$
u_9^{p} u_8 u_{10} u_9, 
u_{10}^{p}u_{11} u_8, [u_{11},p], $\\$
[u_1,x], u_2^x u_2 u_4, u_3^x u_3 u_5, u_4^x u_2, u_5^x u_3, [u_6,x],
[u_7,x], u_8^x u_8 u_9, $\\$ u_9^x u_8, u_{10}^x u_9 u_{10} u_{11},
u_{11}^x u_8 u_{10},$\\$
[y,u_1], [y,u_2] u_5 u_7, [y,u_3] u_4 u_5 u_6 u_7,[y,u_4] u_6,[y,u_5] u_7,u_6^y u_4, u_7^y u_5,
u_8^y u_1 u_{10},$\\$  [u_9,y], u_{10}^y u_{11}, u_{11}^y u_1 u_8.$\\[0.4cm]
As the action of $ \langle s,t,q,x,y,p \rangle $ on $K_1/Z(K_1)$ is uniquely 
determined by lemma 3.3 we get, using the description in �\cite{Wie5}, that
 modulo $
\langle u_1 \rangle $ we have \\[0.4cm]
$
[v_1,q], [v_2,q], [v_3,q] u_7, [v_4,q] u_6, v_5^q v_5 v_1 u_5 u_7, 
v_6^q v_2 v_6 u_6 u_4,$\\$
[v_1,s], [v_2,s], [v_3,s], [v_4,s], [v_5,s] v_3, [v_6,s] v_4,$\\$
[v_1,t], [v_2,t], v_3^t v_1 v_3 u_5 u_7, v_4^t v_2 v_4 u_4 u_6, [v_5,t] u_5, 
[v_6,t] u_4,$\\$
[v_1,p], [v_2,p] v_1, [v_3,p], [v_4,p] v_3 u_7, 
[v_5,p] u_7, v_6^p u_5 u_6 u_7 v_5 v_6,$\\$
v_1^x v_3 u_3, v_2^x v_4 u_2 u_6, v_3^x v_1 v_3 u_5 u_7, v_4^x v_2 v_4 u_4 u_6, [v_5,x] u_5, [v_6,x] u_2,$\\$
[y,v_1], [y,v_2], v_3^y v_2 v_5, v_4^y v_1 v_2 v_6, v_5^y v_3 v_5, 
v_6^y v_4 v_6.
$ \\[0.4cm]
All these relations also hold in $\langle u_1,...,u_{11},v_1,...,v_6,s,t,q,x,y,
p  \rangle $ besides possibly the relations involving $s,t,q,x,y,p$. If $s^2 = u_1$, then we see that $s\langle u_1 , \ldots , u_{11} \rangle$ does not contain  involutions. But in $G_4$ we have that $s\langle u_1 , \ldots , u_{11} \rangle$ is conjugate to some element in $\langle u_1, \ldots, u_{11}, v_1, \ldots , v_6 \rangle / \langle u_1 , \ldots , u_{11} \rangle$. So we must have a subgroup $3A_6$ in $G_1$ and then all relations not involving $p$ are left unchanged. From the remaining relations there are four which have to be
changed to $[v_2, p] = v_1u_1, [v_4,p] = v_3v_7u_1, [v_5,p] = u_7u_1,
v_6^{p} = v_6v_5u_1u_5u_6u_7$.
So we have defining relations for $\langle u_1,...,u_{11},v_1,...,v_6,s,t,q,x,y,
p  \rangle $. \\

To define $G_4$ we need some further element $u$ of order three.
 We choose $u$ such that $\langle S,u \rangle = P_1.$ We get the following 
action of $u$ on $K_4$:\\[0.4cm]
$
u_1^u u_1 u_7, u_2^u u_2 u_{11}, [u_3,u], u_4^u u_4 u_8 u_{10} u_{11}, [u_5,u], [u_6,u],
u_7^u u_1, u_8^u u_8 u_4,u_9^u u_2 u_4 u_9, $\\$
u_{10}^u u_2 u_8,u_{11}^u u_2. $\\[0.4cm]
Modulo $ \langle u_1,...,u_{11} \rangle$ we have, using the matrices for the action on the 11-dimen\-sional module obtained before, that \\[0.4cm]
$[v_1,u]=1=[t,u]=[v_3,u],
u^{v_6}=u^{-1},
v_2^u = qv_2, q^u = v_2, v_4^u = v_4s, s^u = v_4, [x,u]=1,
(v_1v_2v_5)^u = v_2v_5qp, (v_1qp)^u = v_1v_2v_5,
(ysv_2u)^3 = 1.$\\[0.3cm]
We may choose $u$ such that $1=[x,u]$ holds in $G_4$. Now we have 
$v_1^u = v_1r_1, t^u = tr_2$, where $r_1,r_2 \in [K_4,u]$.
Furthermore $r_2 \in C_{[K_4,u]}(x)$, which shows 
\begin{center} $r_2 \in \langle
u_1, u_7 \rangle $.\end{center}
Now $t^{ysv_2u}=tr_2$. Let $(ysv_2u)^3 = r \in K_4$.
Then $t^r = tr_2^{(ysv_2u)^2} r_2^{ysv_2u} r_2$. \\
This yields 
\begin{center}$ t^r = u_5t$ and $r_2 \in \langle u_7 \rangle u_1$, or $t^r = t$
and $r_2 \in \langle u_7 \rangle $.�\end{center}
�
Next we determine $r_4$ with $u^{v_6u_2u_4}=u^{-1}r_4$.
As $[v_6u_2u_4,x] = 1 $ , we see $ r_4 \in \langle u_1, u_7 \rangle$.
Now $u=u^{(v_6u_2u_4)^2}=r_4ur_4^{v_6u_2u_4}$ implies $r_4 \in \langle u_1u_7 
\rangle$ .
If $r_4 �\not= 1$ we may replace $u$ by $uu_1u_7$. So we may assume
\begin{center}$r_4 =1$.\end{center}
Then $t^{v_6u_2u_4uv_6u_2u_4} = t^{u^2}$ implies
$r_2r_2^u = r_2^{v_6u_2u_4}$ and so $ r_2 \in \langle u_7 \rangle$.
In particular
$r \in C_{K_4}(t) = \langle u_1,u_4,u_5,u_6,u_7,u_8,u_{10}u_{11} \rangle $.
We have 
$v_1^u = v_1r_1$ with $r_1 \in C_{K_4 }(v_1) \cap [K_4,u] = 
\langle u_1,u_7,u_4,u_8u_{10}u_{11} \rangle$. Then
$v_1^{v_6u_2u_4uv_6u_2u_4} = v_1^{u^2}$
 implies \begin{center}$r_1 \in \langle u_7 \rangle u_1$.\end{center}
Now $[x,u]=1$ shows $ v_3^u = v_3r_1$.
Then $v_3^{ut} = v_1v_3u_5u_7r_1^t = v_3^{tur_2} = v_1v_3u_1u_5$ and so 
$r_1 = u_1u_7$.
Set $v_2^u = qv_2r_5$,$ q^u = v_2r_6,$ where $r_5 \in C_{K_4}(qv_2)$.
Then $v_2^{ut}=v_2^{tur_2} $ and $v_2^{uv_3}=v_2^{v_3uu_1u_7} $
show $r_5^{v_3}= u_7r_5 $ and  $r_5 \in \langle u_1,u_4,u_5,u_6,u_7 \rangle
u_8u_{10}u_{11}$. 
From $ v_2^{v_6u_2u_4uv_6u_2u_4} = v_2^{u^2} $ we get
$r_6=r_5^u$ and $ r_5 \in \langle u_1,u_4,u_5,u_6 \rangle u_8u_{10}u_{11}$
or
$r_6=r_5^uu_1$ and $ r_5 \in \langle u_1,u_4,u_5,u_6 \rangle u_7u_8u_{10}u_{11}$.
From $v_2 = v_2^{u^3}$ we get $r_6 = r_5^u$.
As $r \in C_{K_4}(t)$ we see $ v_2^{(ysv_2u)^3} = v_2$ or $v_2u_4$.
As $ v_2^{(ysv_2u)^3}=v_2r_5^ur_2^{(ysv_2u)}r_5^{u(ysv_2u)}r_5^{(ysv_2u)^2}$
we get by checking the possibilities for $r_5$ and $r_2$ that $r_2=1$ and
\begin{center}
$r_5 \in \langle u_1,u_4,u_5 \rangle u_8u_{10}u_{11} $ if $ v_2^{ysv_2u} = v_2$ or
\end{center}
\begin{center}
$r_5 \in \langle u_1,u_4,u_5 \rangle u_6u_8u_{10}u_{11} $ if $ v_2^{ysv_2u} = v_2u_4$.
\end{center}
Now $q^{v_6u_2u_4uv_6u_2u_4} = q^{u^2}$ implies
\begin{center}
$r_5 \in \langle u_1,u_5 \rangle u_8u_{10}u_{11} $ or 
$r_5 \in \langle u_1,u_5 \rangle u_6u_8u_{10}u_{11}$.
\end{center}
As $[x,u]=1$ we get 
$v_4^u = v_4su_{11}r_5^x$ and $s^u = v_4r_6^xu_2u_6$. 
As $s^{ut} = s^{tu}$ we then get 
$r_5 \in \langle u_5 \rangle u_6u_8u_{10}u_{11} $ .
The relation $v_4^{ut} = v_4^{tu} $ then shows 
\begin{center} $r_5 = u_6u_8u_{10}u_{11}$. \end{center}
Set $ (v_1v_2v_5)^u = v_2v_5qpr_9$ and $
(v_1qp)^u = r_{10}v_1v_2v_5$.
We have $r_{10} \in \langle u_1, u_2u_3,$\\$u_2u_6,u_4,u_5,u_7,u_{10}u_{11} \rangle
= C_{K_4}(v_1v_2v_5)$.
As $[u,t]=1 $ we get $r_{10}^t=r_{10}u_5$ and so 
 $r_{10} \in \langle u_1,u_4,u_5,u_7,u_{10}u_{11} \rangle u_3u_6$
.
As $[u,v_1] = u_1u_7$, we get 
\begin{center} $r_{10} \in \langle u_1,u_4,u_5,u_7 \rangle u_3u_6$
. 
\end{center}
As  $[u,v_3] = u_1u_7$, we get 
 $r_{10}^{v_3} = u_1r_{10}$ 
and so \begin{center} $r_{10} \in
 \langle u_1,u_5,u_7 \rangle u_3u_4u_6$. \end{center}
 As $[u,x] = 1$, we get \begin{center} $r_{10} \in \langle u_1,u_7 \rangle u_3u_4u_6$.\end{center}
Applying the same relations to $(v_1v_2v_5)^u$ , we see
 \begin{center}
$r_9 \in \langle u_1,u_7 \rangle u_8u_{10}u_{11}$.
\end{center}
From $v_1qp = (v_1qp)^{u^3} $ we get 
$r_{10}^{u^2}r_9r_{10}r_9^u = 1$ and so we have the following pairs for $(r_9,r_{10})$:\\[0.3cm]
$(u_8u_{10}u_{11},u_3u_4u_6);
 (u_7u_8u_{10}u_{11},u_1u_3u_4u_6);
 (u_1u_7u_8u_{10}u_{11},u_7u_3u_4u_6)$ and \\ $
 (u_1u_8u_{10}u_{11},u_1u_7u_3u_4u_6)$.\\

As $(v_1qp)^{v_6u_2u_4u} = (v_1qp)^{u^2v_6u_2u_4}$, we get by checking
 these pairs that just $(u_7u_8u_{10}u_{11},u_1u_3u_4u_6)$ and
$(u_1u_7u_8u_{10}u_{11},u_7u_3u_4u_6)$ are possible.\\
Now finally $(v_1qp)^{(ysv_2u)^3}=(v_1qp)^r$ gives 
\begin{center}
$r_9 = u_7u_8u_{10}u_{11}$ and $r_{10} = u_1u_3u_4u_6$.
\end{center}
Now also $r$ is determined and we have \begin{center} $r=u_4u_8$.\end{center}
So we have the following relations for $u$:\\[0.4cm]
$
v_1^u u_7 u_1 v_1,[t,u], v_3^u u_7 u_1 v_3, u^{v_6} u_{10} u_8 u_4 u_2 u,v_2^u u_6 u_{11} u_{10} u_8 q v_2, $\\$
q^u v_2 u_4 u_6,v_4^u v_4 s u_6,s^u v_4,[x,u],
(v_1 v_2 v_5)^u u_7 u_8 u_{10} u_{11} p q v_5 v_2,$\\$
(v_1 q p)^u v_1 v_2 v_5 u_3 u_4 u_6 u_1,
(y s v_2 u)^3 u_8 u_4.
$\\[0.4cm]
So we have generators and relations for $G_4$ in case $|K_4|=2^{11}$.
If $|K_4|=2^{12}$, we have a further basis element $u_{12}$ where\\[0.3cm]
$
u_{12}^{v_1}  =u_3u_6u_{12},\quad
u_{12}^{v_2}  =u_1u_2u_6u_7u_{12},\quad
u_{12}^{v_3}  =u_{12},\quad
u_{12}^{v_4}  =u_{12},$\\$
u_{12}^{v_5}  =u_2u_3u_7u_{12},\quad
u_{12}^{v_6}  =u_3u_6u_{12},\quad
u_{12}^s  =u_{12},\quad
u_{12}^t  =u_4u_{10}u_{11}u_{12},$\\$
u_{12}^q  =u_1u_6u_{11}u_{12},\quad
u_{12}^{p}  =u_1u_3u_7u_{11}u_{12},\quad
u_{12}^x  =u_4u_{10}u_{11}u_{12},$\\$
u_{12}^y  =u_1u_4u_6u_8u_{11}u_{12},\quad
u_{12}^u  =u_{12}.
$\\[0.3cm]

By lemma 3.3 we know that the action of $G_1$ on $K_1/Z(K_1)$ is uniquely determined. We choose $v \in P_4$, $o(v) = 3$. Adopting the notation of \cite{Wie5}
to our basis $\{u_2,...,u_7,v_1,...,v_6\}$ of $K_1/Z(K_1)$ we may choose 
$v$ such that\\[0.3cm]
$
[u_6,v]=1=[u_7,v]=[v,u_3v_5v_2]=[u_5,v]=[v,v_1u_2v_2v_6]=[v_3,v]=[u_4,v]
=[v_4,v],\quad
u_2^v=v_2u_6u_7, \quad 
v_2^v=v_2u_2u_7,\quad
v_5^v=u_6u_7v_5u_2v_1u_3, \quad
v_6^v=u_7v_6v_2u_3.$\\[0.3cm]
We have that $\langle S, v, x \rangle \cong \langle P_3,P_4 \rangle $
and $S/Z(K_1)$ splits over $K_1/Z(K_1)$. So we have a complement in $
\langle P_3,P_4 \rangle$ to $K_1/Z(K_1)$. Hence we may assume that we have the following
further relations: \\[0.3cm]$
[u_8u_{10}u_{11},v]=1,\quad
u_{11}^v=qu_{11}u_{10}u_8,\quad
q^v=qu_{11}, \quad
s^v=s,\quad
u_8^v=t,\quad
t^v=u_8t,\quad
(u_9v)^2 = 1,\quad
p^v= u_{11}p,\quad
(xv)^4=1$.\\[0.3cm]
All these relations are modulo $Z(K_1)=\langle u_1 \rangle$.
We have $\langle S, y, v \rangle = \langle P_2,P_4 \rangle$. Calculating
the action of $[v,y] $ on $K_1/Z(K_1)$ we see that this is the same as the action of $u_{10}qt$. As $ \langle P_2,P_4 \rangle /K_1$ acts faithfully on $K_1/Z(K_1)$ we have $[v,y]qtu_{10} \in K_1$.
Applying $[v,y]qtu_{10}$ to $X:=\langle u_8u_{10}u_{11},u_{11},q,s,u_8,t,u_9,p,v,y \rangle$ 
we see that $[v,y]qtu_{10}$ centralizes $X$. So $[v,y]qtu_{10}$ is contained in 
$C_{K_1/Z(K_1)}(X)=1$.
Now we have defining relations for $G_1/Z(K_1)$.

 Next we determine the relations involving $v$ in $G_1$. If for some $r\in G_1$ we have $[v,r]\in Z(K_1)$ then
 $[v,r]=1$. This shows that we just have to treat $u_2^v, v_2^v, v_5^v, v_6^v, u_{11}^v, q^v, u_8^v, t^v, p^v, [v,y]$. Working in $\langle P_2, P_4\rangle$ we check all possiblities by computer. This shows that we have the following relations: \\[0.4cm]
$
v^3$, $[u_1,v]$, $[u_6,v]$, $[u_7,v]$, $[v,u_3 v_5 v_2]$, $[u_5,v]$, $[v,v_1 u_2 v_2 v_6]$, $[v_3,v]$,
$u_2^v u_6 u_7 v_2 u_1$, $[u_4,v]$, $v_2^v u_7 u_2 v_2 u_1$, $[v_4,v]$, 
$v_5^v u_3 v_1 u_2 v_5 u_7 u_6 u_1$, 
$v_6^v u_3 v_2 v_6 u_7$, $[u_8 u_{10} u_{11},v]$,\\ $u_{11}^v u_8 u_{10} u_{11} q u_1$, $q^v u_{11} q u_1$, $[s,v]$, $u_8^v t$, $t^v u_8 t$, $(u_9 v)^2$, $p^v p u_{11} u_1$, \\ $[v,u_1 u_2 u_6 u_7 u_8 u_9 u_{10} u_{12}]$, 
$[v,y] q t u_{10} u_1$, $(x v)^4. 
$\\[0.4cm]
To get some relation between $u$ and $v$ we have to move to $G_2$. We have 
$(G_2/K_2)^{(\infty)}\cong Z_3 \times A_6$ and so $[v,u] \in K_2$. We calculate $[v,u] \in C_{K_2}(K_4 \cap K_2) = \langle u_1,u_2,u_3,u_5,u_6,u_7,u_{11}
\rangle$. 

Now, as $[v,u] \in [K_2,v] \cap [K_2,u] $ we have $[v,u] \in \langle u_1u_7u_2, u_1 u_{11} \rangle$. Checking the order of $G_2$ by Todd-Coxeter we get
$[v,u] = u_1u_2u_7$.

So we have shown that the amalgam is uniquely determined if $|K_4|=2^{11}$.\\[0.4cm] 
So let $|K_4|=2^{12}$. To prove uniqueness we just have to calculate $u_{12}^v$. From the known action on $K_1/Z(K_1)$ we see that $(vu_{12})^2 = qu_8u_{10}u_{11} $ modulo $K_1$. Now again by applying the word $(vu_{12})^2u_{11}u_{10}u_8q $ to the elements of $X$, we see that $(vu_{12})^2u_{11}u_{10}u_8q \in C_{K_1/Z(K_1)}(\langle u_8u_{10}u_{11}, q,s,t,y \rangle) = \langle v_1, v_2 \rangle Z(K_1)/Z(K_1) $. As $((vu_{12})^u)^2 = (vu_1u_7u_2u_{12})^2$ we get 
 $[(vu_{12})^2,u] = [v_2qu_8u_{10}u_{11}, u]$. But we can distinguish the elements in $\langle u_1, v_1, v_2 \rangle$ by their commutator with $u$, so
\begin{center} $(vu_{12})^2 = v_2qu_{11}u_{10}u_8$. 
\end{center}
Hence proposition 5.1 is proved.	$\Box $\\[0.4cm]


The following proposition has been shown by A. Ivanov \cite{Iv}. We present a proof using our generators and relations.

\begin{Prop}
The geometry for $M(24)$ with diagram {$
\begin{picture}(35,10)
\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}}
\end{picture}
$}
is simply connected, i.e. if the maximal parabolic subgroups of $G$ are as in $M(24)$ then $G \cong M(24)$.
\end{Prop}



{\bf Proof:} It is sufficient to prove it for the $M(24)^\prime$ parabolic system. Now the additional information is 
$(G_1/K_1)^{(\infty)} \cong 3U_4(3)$. We add a further relation 
$(s^{(xyx^{-1}sxy^{-1}sx^{-1})^{-1}}vu_9)^u=1$. This is the relation, which comes from the Weyl group of $U_4(2)$ inside of $U_4(3)$. Then we can show that
$G_1^{\infty} \cong 2^{1+12}3U_4(3)$ just by using the Todd-Coxeter
algorithm, and the  group generated by 
$ \{u_1,...,u_{11}\} \{v_1,v_2,q,t,v_3v_4v_5, v_3v_6, 
ps,u^{v_5},y,xyx^{-1}sxy^{-1}sx^{-1}\}$ is a sub\-group isomorphic to
$2^{11}M_{23}$ in $G_4$.
Furthermore the group generated by
$ \{u_1,...,u_{11}\} $ $\{v_1,v_2,q,t,v_3v_4v_5, v_3v_6,
ps,y,xyx^{-1}sxy^{-1}sx^{-1},v\}$ is a subgroup isomorphic to
 $2^22^{1+8}(U_4(2) \times Z_3) �\cdot 2$ in $G_1$.
Hence these two groups generate a group $H$ which belongs to the following
 diagram
\begin{center}
{$
\begin{picture}(35,10)

\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(17,3.5){\makebox(0,0){{$\scriptstyle M_{22}$}}}

\put(22.5,1){\line(1,0){8.7}}
\put(27,3.5){\makebox(0,0){{$\scriptstyle U_4(2) \! \cdot \! 2$}}}

\end{picture}
$}.\end{center}

Application of [Str6, Theorem 9] shows $H \cong M(23)$. Now using Todd-Coxeter we get
$|G:H| = |M(24)^\prime :M(23)|$ and so $G\cong M(24)^\prime$. $\Box$\\[0.4cm]
The following lemma is due to A. Pasini.

\begin{lemma}
The geometry 
{$
\begin{picture}(35,10)

\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}}
\end{picture}
$}
for $M(24)$ is not 2-simply connected.
\end{lemma}


{\bf Proof:}
We have to show that the universal 2-cover of the given diagram is infinite. The diagram supplies us with the information about 2-residues, and they are like those in $M(24)$. Let $ \Gamma $ be such a geometry. Let us choose the type
 numbers like this:  $\Delta$ = {$
\begin{picture}(35,5)
\put(1,-2.5){1}
\put(11,-2.5){2}
\put(21,-2.5){3}
\put(31,-2.5){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(12.5,0.4){\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}}

\end{picture}
$}. Note, that all the
 3-residues\\[0.4cm] which contain type 1, these are 
 {$
\begin{picture}(30,5)
\put(4,-2.5){1}
\put(14,-2.5){3}
\put(24,-2.5){4}

\put(5,1){\makebox(0,0){$\circ$}}
\put(15,1){\makebox(0,0){$\circ$}}
\put(25,1){\makebox(0,0){$\circ$}}



\put(15.5,0.4){\line(1,0){8.9}}
\put(15.5,1.7){\line(1,0){8.9}}

\end{picture}
$}
,
 {$
\begin{picture}(30,5)
\put(4,-2.5){1}
\put(14,-2.5){2}
\put(24,-2.5){3}
\put(5,1){\makebox(0,0){$\circ$}}
\put(15,1){\makebox(0,0){$\circ$}}
\put(25,1){\makebox(0,0){$\circ$}}

\put(5.5,1){\line(1,0){8.7}}

\put(15.5,0.4){\line(1,0){8.9}}
\put(15.5,1.7){\line(1,0){8.9}}
\put(20,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\end{picture}
$}, and  {$
\begin{picture}(30,5)
\put(4,-2.5){1}
\put(14,-2.5){2}
\put(24,-2.5){4}

\put(5,1){\makebox(0,0){$\circ$}}
\put(15,1){\makebox(0,0){$\circ$}}
\put(25,1){\makebox(0,0){$\circ$}}

\put(5.5,1){\line(1,0){8.7}}


\end{picture}
$} are simply connected, so by [Pas, theorem 1] the universal\\[0.4cm]
 cover of the 1-truncation of $\Gamma$ (that is, the geometry, that is left, if we drop all elements of $\Gamma$ of type 1) is equal to the 1-truncation of the universal
cover of $\Gamma$: $ \widetilde{Tr_1(\Gamma)} = Tr_1(\tilde{\Gamma})$.
So we are done if we can show that  $ \widetilde{Tr_1(\Gamma)}$ is infinite.

Now we want to get information about the diagram $\Delta_1$ of $Tr_1(\Gamma)$. Since the diagram of $Res(3)$ in $\Gamma$ is {$
\begin{picture}(30,5)
\put(4,-2.5){1}
\put(14,-2.5){2}
\put(24,-2.5){4}

\put(5,1){\makebox(0,0){$\circ$}}
\put(15,1){\makebox(0,0){$\circ$}}
\put(25,1){\makebox(0,0){$\circ$}}

\put(5.5,1){\line(1,0){8.7}}


\end{picture}
$}, we have $\Delta_1$ =  {$
\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(1,-2.5){2}
\put(11,-2.5){3}
\put(21,-2.5){4}
\put(2.5,0.9){.........}
\put(12.5,0.9){.........}


\end{picture}
$}.\\[0.3cm]
 $Res(2)$ in $\Gamma$ has the diagram 
{$
\begin{picture}(27,5)
\put(4,-2.5){1}
\put(14,-2.5){3}
\put(24,-2.5){4}

\put(5,1){\makebox(0,0){$\circ$}}
\put(15,1){\makebox(0,0){$\circ$}}
\put(25,1){\makebox(0,0){$\circ$}}



\put(15.5,0.4){\line(1,0){8.9}}
\put(15.5,1.7){\line(1,0){8.9}}

\end{picture}
$}, and thus  $\Delta_1$ =  {$
\begin{picture}(23,5)
\put(2,1){\makebox(0,0){$\circ$}}
\put(12,1){\makebox(0,0){$\circ$}}
\put(22,1){\makebox(0,0){$\circ$}}
\put(1,-2.5){2}
\put(11,-2.5){3}
\put(21,-2.5){4}
\put(2.5,0.9){.........}
\put(12.5,0.4){\line(1,0){8.9}}
\put(12.5,1.7){\line(1,0){8.9}}



\end{picture}
$}.\\[0.3cm]
 We now want to determine the gonality $g$ of the residue of type 4 in $Tr_1(\Gamma)$.\\[0.3cm]

(1) $g \ge 3$.\\[0.2cm]
Suppose $ g = 2$. Then there are $e_1, e_2 $ of type 3 and $l_1, l_2$ of type 2, such that $e_1$ and $e_2$ are incident with both, $l_1$ and $l_2$.\\
\begin{center}
{$
\begin{picture}(15,15)
\put(2,0){\makebox(0,0){$\circ$}}
\put(12.5,0){\makebox(0,0){$\circ$}}
\put(2,10){\makebox(0,0){$\circ$}}
\put(12.5,10){\makebox(0,0){$\circ$}}

\put(1,12){$l_1$}
\put(1,-3){$l_2$}
\put(11.5,12){$e_1$}
\put(11.5,-3){$e_2$}

\put(2.5,0.2){\line(1,0){9.1}}
\put(2.5,10.2){\line(1,0){9.1}}

\put(2.5,0.2){\line(1,1){9.4}}
\put(2.5,10.2){\line(1,-1){9.4}}

\end{picture}
$}
\end{center}

Consider $Res(4)$ in $\Gamma$ with diagram 
{$
\begin{picture}(28,5)
\put(4,-2.5){1}
\put(14,-2.5){2}
\put(24,-2.5){3}
\put(5,1){\makebox(0,0){$\circ$}}
\put(15,1){\makebox(0,0){$\circ$}}
\put(25,1){\makebox(0,0){$\circ$}}

\put(5.5,1){\line(1,0){8.7}}

\put(15.5,0.4){\line(1,0){8.9}}
\put(15.5,1.7){\line(1,0){8.9}}
\put(20,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\end{picture}
$}. We find $p$ of type 1\\[0.3cm]
 in the residue of $e_2$, such that $p$ is incident with $l_1$ and $l_2$, since the residue is a projective plane.
\begin{center}
{$
\begin{picture}(25,15)
\put(12,0){\makebox(0,0){$\circ$}}
\put(22.5,0){\makebox(0,0){$\circ$}}
\put(12,10){\makebox(0,0){$\circ$}}
\put(22.5,10){\makebox(0,0){$\circ$}}
\put(2,4.9){\makebox(0,0){$\circ$}}

\put(11,12){$l_1$}
\put(11,-3){$l_2$}
\put(21.5,12){$e_1$}
\put(21.5,-3){$e_2$}
\put(1,1.9){$p$}

\put(12.5,0.2){\line(1,0){9.1}}
\put(12.5,10.2){\line(1,0){9.1}}

\put(12.5,0.2){\line(1,1){9.4}}
\put(12.5,10.2){\line(1,-1){9.4}}

\put(2.5,5.1){\line(2,1){9.1}}
\put(2.5,5.1){\line(2,-1){9.1}}

\end{picture}
$}
\end{center}
But now there is a generalized 2-gon in the residue of $p$. This is not possible in $Res(4)$, since the gonality of 
{$
\begin{picture}(15,5)
\put(1,-2.5){2}
\put(11,-2.5){3}

\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 $5$. So we have $g \ge 3$.\\[0.2cm]
Let us define a property of diagram geometries {\bf (PL)}:\\[0.1cm]

{\bf (PL)} is true in a geometry of rank 3, with the types 1,2,3, if for any $p$ of type 1, $l$ of type 2, and $u,v$ of type 3, where $l$ and $p$ are
incident with $u$ and $v$, then necessarily $p$ is incident with
$l$:
\begin{center}
{$
\begin{picture}(25,15)

\put(22.5,0){\makebox(0,0){$\circ$}}

\put(22.5,10){\makebox(0,0){$\circ$}}
\put(12,4.9){\makebox(0,0){$\circ$}}
\put(2,4.9){\makebox(0,0){$\circ$}}

\put(21.5,12){$u$}
\put(21.5,-3){$v$}
\put(11,1.9){$l$}
\put(1,1.9){$p$}

\put(2.5,5.1){\line(4,1){19}}
\put(2.5,5.1){\line(4,-1){19}}

\put(12.5,5.1){\line(2,1){9.1}}
\put(12.5,5.1){\line(2,-1){9.1}}

\put(2.5,4.9){.........}


\end{picture}
$}
\end{center}

(2) {\bf (PL)} holds in the geometry of $M_{24}$ with diagram
{$
\begin{picture}(30,5)
\put(4,-2.5){1}
\put(14,-2.5){2}
\put(24,-2.5){3}
\put(5,1){\makebox(0,0){$\circ$}}
\put(15,1){\makebox(0,0){$\circ$}}
\put(25,1){\makebox(0,0){$\circ$}}

\put(5.5,1){\line(1,0){8.7}}

\put(15.5,0.4){\line(1,0){8.9}}
\put(15.5,1.7){\line(1,0){8.9}}
\put(20,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\end{picture}
$}.\\[0.2cm]
In this geometry for $M_{24}$ in terms of the 11- or 12-dimensional module over $GF(2)$ the points, lines, and planes are certain 1-, 2-, and 3-spaces, where two different spaces are incident if one is included in the other (See \cite{RoStr} or the action of $G_4$ on $K_4$ in proposition 5.1). So if we have two different 3-spaces, which intersect in a 2-space and a 1-space, the 1-space must lie inside the 2-space.
Hence (2) is proved.\\[0.2cm]

(3) $g \ge 4$. \\[0.2cm]
Suppose $g=3$. Then we can find elements $m,n,l$ of type 2 and elements $u,v,w$ of type 3, such that their incidence graph is a generalized 3-gon:
\begin{center}
{$
\begin{picture}(15,25)
\put(2,0){\makebox(0,0){$\circ$}}
\put(12.5,0){\makebox(0,0){$\circ$}}
\put(2,10){\makebox(0,0){$\circ$}}
\put(12.5,10){\makebox(0,0){$\circ$}}
\put(2,20){\makebox(0,0){$\circ$}}
\put(12.5,20){\makebox(0,0){$\circ$}}

\put(-2,10){$n$}
\put(-2,0){$l$}
\put(13.5,0){$v$}
\put(13.5,10){$w$}
\put(-2,20){$m$}
\put(13.5,20){$u$}

\put(2.5,0.2){\line(1,0){9.1}}
\put(2.5,20.2){\line(1,0){9.1}}

\put(2.5,10.2){\line(1,1){9.4}}
\put(2.5,10.2){\line(1,-1){9.4}}

\put(2.5,0.2){\line(1,1){9.4}}
\put(2.5,20.2){\line(1,-1){9.4}}

\end{picture}
$} \qquad \qquad \qquad \qquad
{$
\begin{picture}(25,25)
\put(12,0){\makebox(0,0){$\circ$}}
\put(22.5,0){\makebox(0,0){$\circ$}}
\put(12,10){\makebox(0,0){$\circ$}}
\put(22.5,10){\makebox(0,0){$\circ$}}
\put(12,20){\makebox(0,0){$\circ$}}
\put(22.5,20){\makebox(0,0){$\circ$}}
\put(2,10){\makebox(0,0){$\circ$}}


\put(8,10){$n$}
\put(8,0){$l$}
\put(23.5,0){$v$}
\put(23.5,10){$w$}
\put(8,20){$m$}
\put(23.5,20){$u$}
\put(-2,10){$p$}


\put(12.5,0.2){\line(1,0){9.1}}
\put(12.5,20.2){\line(1,0){9.1}}

\put(12.5,10.2){\line(1,1){9.4}}
\put(12.5,10.2){\line(1,-1){9.4}}

\put(12.5,0.2){\line(1,1){9.4}}
\put(12.5,20.2){\line(1,-1){9.4}}

\put(2.5,10.2){\line(1,1){9.4}}
\put(2.5,10.2){\line(1,-1){9.4}}

\end{picture}
$}

\end{center}
Consider $Res(4)$ again. Here we can find an element $p$ of type 1 in the residue of $w$ such that $p$ is incident with $m$ and $l$. Now $p$ is incident with $u$ and $v$, since it is in the residue of $m$ and $l$.

But because {\bf (PL)} holds in the $M_{24}$ - geometry, $p$ is incident with $n$ and we have found a generalized 3-gon in the residue of $p$ in the 
geometry with diagram 
{$
\begin{picture}(30,5)
\put(4,-2.5){1}
\put(14,-2.5){2}
\put(24,-2.5){3}
\put(5,1){\makebox(0,0){$\circ$}}
\put(15,1){\makebox(0,0){$\circ$}}
\put(25,1){\makebox(0,0){$\circ$}}

\put(5.5,1){\line(1,0){8.7}}

\put(15.5,0.4){\line(1,0){8.9}}
\put(15.5,1.7){\line(1,0){8.9}}
\put(20,3.5){\makebox(0,0){$\scriptstyle{\sim}$}}

\end{picture}
$}, but this is a contradiction. So $g \ge 4$.
We have seen, that the gonality graph of $Tr_1(\Gamma)$ looks like this:
{$
\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){\line(1,0){8.7}}

\put(6,3.5){\makebox(0,0){$\ge 4$}}
\put(17,3.5){\makebox(0,0){$ 4$}}

\end{picture}
$}. Note that $Tr_1(\Gamma)$ is thick, since it is a truncation of a thick geometry. Now we apply [Pas, theorem 2] and $\widetilde{Tr_1(\Gamma)}$ is 
infinite.$\Box$









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