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\begin{document}
\setlength{\unitlength}{1mm}
\begin{center}
\bf $c$ - extensions of the Petersen geometry for $M_{22}$ 
\end{center}
\medskip
\begin{center}{\sc
Anna Fukshansky, Corinna Wiedorn\\
Martin-Luther-Universit\"at Halle-Wittenberg\\
Institut f\"ur Algebra und Geometrie\\
06099 Halle, Germany}
\end{center}
\bigskip
  
In this paper we consider a group $G$ acting flag-transitively on a geometry
with diagram \Diag{1}{2}{3}{4} such that the residue of an element of type 1
is the Petersen geometry for the group $M_{22}$. Explicitely, $G$ is generated by finite subgroups $P_1,P_2,P_3,P_4$ such that
\begin{enumerate}
\item[(a)] $G\neq G_i:=\langle P_j|j\neq i\rangle $ for $1\le i\le 4$.
\item[(b)] For $B:=P_1\cap P_2\cap P_3\cap P_4$ we have $B_G=1$, where $B_G:=\bigcap_{g\in G}B^g$.
\item[(c)] If $K_i=B_{G_i}$, $G_{ij}=\langle P_k|k\neq i,j\rangle $ and $K_{ij}=B_{G_{ij}}$, then $G_1/K_1\cong M_{22}$ or $Aut(M_{22})$ and the structures of the $G_i/K_i$, $i\neq 1$, and the $G_{ij}/K_{ij}$ are as indicated by the diagram.\footnote{The reader unfamiliar with diagram geometries is referred to \cite{Buek} and \cite{Pas}.}
\end{enumerate}
We are going to prove the following\\  
\\
{\bf Theorem.}~~
Let $G$ be a group with a   parabolic system belonging to the diagram
\Diag{1}{2}{3}{4} where $G_1/K_1 
\cong M_{22}$ or $Aut(M_{22})$. Then $G$ is isomorphic to one of the
following groups:\\ 
a) $\quad  2^{10}M_{22}, \; 2^{11}M_{22}, \;  2^{10}Aut(M_{22}),\; 2^{11}Aut(M_{22})$\\
b) $\quad U_6(2),\; 2\!\cdot\! U_6(2),\; U_6(2)\!\cdot\! :,\; 2\!\cdot\! U_6(2)\! :\!2$\\
c) $\quad M_{24}$.\\
\\
{\bf Remark.}~~It follows from the diagram that 
\begin{enumerate}
\item[--] $|P_1:B_{P_1}|=|P_4:B_{P_4}|=2$ and $P_2/B_{P_2}\cong P_3/B_{P_3}\cong\Sigma _3$,
\item[--] $G_2/K_2\cong {\bf Z}_2\times \Sigma _5$, $G_3/K_3\cong \Sigma _4\times {\bf Z}_2$ and $G_4/K_4\cong 2^3L_3(2)$, 
\item[--] $K_{14}=K_4$.
\end{enumerate}
This implies that $K_1K_4/K_1$ and $K_1K_4/K_4$ are 2-groups. Hence, by
\cite{S1} (1.13)-(1.15) $K_1$ and $K_4$ are 2-groups. Since $B/K_1$ is also
a 2-group, $B$ and all $K_i$ are 2-groups.\par
%\bigskip
The following lemma is due to A. Pasini.
%\newpage
\begin{lem}
We have $K_1=1$.
\end{lem}

{\it Proof.}~~Let $\Gamma =\Gamma (G;G_i)$ be the coset geometry belonging
to our parabolic system. 
Let's call the elements of $\Gamma $ of type 1 ``points'' and the elements of
type 2 ``lines''. Then each line consists of 2 points and the relation
``having the same points'' is an equivalence relation on the set of lines on
the point $\alpha _1$, where we identify the coset $G_1$ with $\alpha
_1$. Clearly, this equivalence relation  is preserved by $G_1$. So the
action of $G_1/K_1$ 
on $res(\alpha _1)$   implies that there are no two lines with the same
points. In particular, $G_p\cap G_q\le G_l$ for two points $p,q$ on a line
$l$.\par 
Now let $g\in K_1$, $p$ a point collinear to $\alpha _1$ and $l$ the line on
$\alpha _1$ and $p$. Then $g$ fixes $l$ and $\alpha _1$, hence $g\in
G_p$. Let $k$ be a line on $p$ such that $l$ and $k$ are incident
to a common element $x$ of type 3 and $q$ the other point on $k$. Then $x$
is incident to $\alpha _1$ and there is a line through $\alpha _1$ and $q$ in
$res(x)$. So $g$ fixes $q$ and $g\in G_p\cap G_q\le G_k$. This means that
$g$ fixes all lines in $res(p)$ which are coplanar with $l$. The action
$G_p/K_p\cong M_{22}$ resp. $Aut(M_{22})$ on $res(p)$ shows that $K_1\le K_p$, hence $K_1=K_p=1$.  \hfill
$\Box $\\   
\\
So from now on we may assume $G_1\cong M_{22}$ or $Aut(M_{22})$. 
We are going to prove Theorem 1 by determining generators and relations for
$G$ and then applying the Todd-Coxeter-Algorithm in MAGMA. We start with
the following generators for $G_1$:
\[ x_1, x_2, x_3, x_5, x_6, x_7, x_8, x_9, x_{10}, y.\]
All these elements are involutions, $\langle x_1, \ldots ,x_{10}\rangle
\cong M_{22}$, $y$ is  the outer automorphism  and the following 
relations are satisfied:
\begin{eqnarray}
 &[x_1, x_2]=1,
    [x_1, x_3]=1,
    x_1^{-1}   x_5^{-1}   x_1   x_5   x_2=1,
    x_1^{-1}   x_6^{-1}   x_1   x_6   x_3=1,\nonumber \\
   &[x_1, x_7]=1, [x_1, x_8]=1,
    [x_1, x_9]=1,
    [x_2, x_3]=1,
    [x_2, x_5]=1,\nonumber\\
   &[x_2, x_6]=1, x_2^{-1}   x_7^{-1}   x_2   x_7   x_1=1,
    x_2^{-1}   x_8^{-1}   x_2   x_8   x_3=1,
    [x_2, x_9]=1,\nonumber\\
   &[x_3, x_5]=1,
    [x_3, x_6]=1, [x_3, x_7]=1,
    [x_3, x_8]=1,\nonumber\\
   &x_3^{-1}   x_9^{-1}   x_3   x_9   x_2=1,
    [x_5, x_6]=1,
    (x_5   x_7)^3=1, x_5^{-1}   x_8^{-1}   x_5   x_8   x_6=1,\nonumber\\
   &[x_5, x_9]=1,
    x_6^{-1}   x_7^{-1}   x_6   x_7   x_8=1,
    [x_6, x_8]=1,
    x_6^{-1}   x_9^{-1}   x_6   x_9   x_5=1,\label{rel1}\\
    &[x_7, x_8]=1,
    (x_7   x_9)^4=1,
    (x_8   x_9)^3=1,
    [x_{10}, x_3]=1,
    [x_{10}, x_6]=1,\nonumber\\
    &[x_{10}, x_8]=1,
    x_{10}^{-1}   x_1^{-1}   x_{10}   x_1   x_8=1,
    x_{10}^{-1}   x_2^{-1}   x_{10}   x_2   x_6=1,\nonumber\\
    &x_{10}^{-1}   x_5^{-1}   x_{10}   x_5   x_6   x_3=1,
    x_{10}^{-1}   x_7^{-1}   x_{10}   x_7   x_8   x_3=1,
    (x_9   x_{10})^5=1,
    [x_1, y]=1,\nonumber\\
   &[x_2, y]=1, [x_3, y]=1,
    [x_5, y]=1,
    [x_6, y]=1,
    [x_7, y]=1,\nonumber\\
   &[x_8, y]=1, [x_9, y]=1,
    y^{-1}   x_{10}^{-1}   y   x_{10}   x_3=1,
    (x_1   x_{10}   x_7   x_5   x_9)^{11}=1.\nonumber
\end{eqnarray}
Moreover, we have  
  \begin{eqnarray}
    P_2&=\langle x_1,x_2,x_3,x_5,x_6,x_7, x_8,y^i \rangle \nonumber\\
    P_3&=\langle x_1,x_2,x_3,x_5,x_6, x_8,x_9,y^i \rangle \label{rel2}\\
    P_4&=\langle x_1,x_2,x_3,x_5,x_6, x_8,x_{10},y^i \rangle \nonumber
  \end{eqnarray}
and
\begin{equation}
  B=\langle x_1,x_2,x_3,x_5,x_6,x_8,y^i\rangle\nonumber
\end{equation}
where $i=1$  if $G_1\cong Aut(M_{22})$ and $i=0$ otherwise.
In the following, if not stated explicitely anything different, we always
assume $G_1 \cong Aut(M_{22})$. \\
\\
As remarked above, we have $G_2/K_2\cong {\bf Z}_2\times \Sigma _5$. Therefore, we can choose an element $x\in P_1$  such that $o(x)=2$, $G =
\langle G_1, x\rangle$ and $[x,G_{12}]\le K_2$.  

\begin{lem}\label{lem2}
   We have $[K_2,x]=1$.
\end{lem}

{\it Proof.}~~Since  $[x,G_{12}]\le K_2=C_{G_{12}}(K_2)$, $G_{12}$ acts on
$[K_{2},x]$. But $|[K_{2},x]|\le 2^3$ and $K_{2}\cap G_1'$, which is of
order $2^4$, is the only nontrivial $G_{12}/K_{2}$-submodule of
$K_{2}$. So the assertion follows.\hfill $\Box$\\
\\
Let $Q_2:=\langle K_{2},x\rangle $ and $H:=\langle x_1,x_8,x_9,x_{10}\rangle
\cong \Sigma _5$. By Lemma \ref{lem2} $Q_2$ is elementary abelian of order
$2^6$ and 
$[H,Q_2]\le K_2$. With respect to the basis $x_2, x_3, x_5, x_6, y, x$ of
$Q_2$ the elements of $H$ belong to $6 \times 6$-matrices over $GF(2)$ of
the shape
\begin{center}
$
\left(
\begin{array}{cccc}
 &&& 0 \\
 & A & &\vdots \\
 &&& 0 \\
\ast & \cdots & \ast & 1 \\
\end{array}
\right),
$
\end{center}
where the  $5 \times 5$-submatrix $ A$ is determined by the  relations in
(\ref{rel1}). Since $x$ is centralized by an element of order 3 in $H$, we
may assume $[x,x_8x_9]=1$. Using this and the above relations, it is not
difficult to calculate that either $[x,H]=1$ or 
\begin{center}
$x_1= \left(
\begin{array}{cccccc}
1&0&0&0&0&0\\
0&1&0&0&0&0\\
1&0&1&0&0&0\\
0&1&0&1&0&0\\
0&0&0&0&1&0\\
0&0&0&0&a_5&1\\
\end{array}
\right), \qquad
x_8= \left(
\begin{array}{cccccc}
1&1&0&0&0&0\\
0&1&0&0&0&0\\
0&0&1&1&0&0\\
0&0&0&1&0&0\\
0&0&0&0&1&0\\
0&0&0&0&0&1\\
\end{array}
\right),$
\end{center}
\begin{center}
$x_9= \left(
\begin{array}{cccccc}
1&0&0&0&0&0\\
1&1&0&0&0&0\\
0&0&1&0&0&0\\
0&0&1&1&0&0\\
0&0&0&0&1&0\\
0&0&0&0&0&1\\
\end{array}
\right), \qquad
x_{10}= \left(
\begin{array}{cccccc}
1&0&0&1&0&0\\
0&1&0&0&0&0\\
0&1&1&1&0&0\\
0&0&0&1&0&0\\
0&1&0&0&1&0\\
0&d_2&0&1&0&1\\
\end{array}
\right),$
\end{center}
with $a_5,d_2 \in \{0,1\}$. In particular, $[x,x_8]=[x,x_9]=1$ in any case.\\
\\ 
Now look at $G_4=\langle x_1,x_2,x_3,x_5,x_6,x_7,x_8,x_9,y,x\rangle $. 
We already know that all generators of $G_4$ except $x_7$ and (possibly) $x_1$ 
commute with $x$. Moreover, we have $K_4=\langle x_1,x_2,x_3,y\rangle$,
$H_4:=\langle x_5,x_6,x_7,x_8,x_9\rangle \cong L_3(2)$ and
$Q_4=\langle K_4,x^{H_4}\rangle $, where $Q_4=O_2(G_4)$. Using the relations in (\ref{rel1}), we
calculate that, with respect to the basis $\{ x, x^{x_7}, x^{x_7x_9}\}$, we
get the following matrices for the action of $H_4$ on $Q_4/K_4$:
\begin{center}
$x_{5}= \left(
\begin{array}{ccc}
1&0&0\\
\bar{1}&\bar{1}&\bar{0}\\
\bar{1}&\bar{0}&\bar{1}\\
\end{array}
\right),\quad
x_{6}= \left(
\begin{array}{ccc}
1&0&0\\
0&1&0\\
\bar{1}&\bar{0}&\bar{1}\\
\end{array}
\right),\quad
x_{7}= \left(
\begin{array}{ccc}
0&1&0\\
1&0&0\\
\bar{1}&\bar{1}&\bar{1}\\
\end{array}
\right),$
\end{center}
\begin{center}
$x_{8}= \left(
\begin{array}{ccc}
1&0&0\\
0&1&0\\
\bar{1}&\bar{1}&\bar{1}\\
\end{array}
\right),\quad
x_{9}= \left(
\begin{array}{ccc}
1&0&0\\
0&0&1\\
0&1&0\\
\end{array}
\right).$  
\end{center}
Here we write $\bar{1}$ resp. $\bar{0}$ if we only know that the corresponding
relations hold in the factorgroup $Q_4/K_4$ and we write $1,0$ if we know
that they hold indeed.\par 
Since there exists the involution $x$ in $Q_4\setminus K_4$ and an element of order 7 in $L_3(2)$ acts fixedpointfreely on $Q_4/K_4$, it is  easy to see that $Q_4$ is elementary abelian of order $2^7$ if $[x,H]=1$ or
$a_5=0$, and that $Q_4\cong D_8^3$ is extraspecial with $Z(Q_4)=\langle
y\rangle $ if $a_5=1$. \par
Let's look at $x_5$. We have
$(x^{x_7})^{x_5}=axx^{x_7}$ for some $a\in\langle x_1,x_2,x_3,y\rangle
$. Now  $x^{x_7x_5x_7}=x^{x_5x_7x_5}=x^{x_7x_5}$, hence $[a,x_7]\in\langle
y\rangle $ and 
$a\in C_{K_4}(x_7)=\langle x_1,x_3,y\rangle $.
Furthermore, $x^{x_7x_5x_6}=x^{x_6x_7x_8x_5}=x^{x_7x_5}$, so $a\in
C_{K_4}(x_6)=\langle x_3,y\rangle $.  \par
Suppose $a=by$ with $b\in\langle x_3\rangle $. Let $\tilde{x}=xy$ and
substitute our basis by $\{ \tilde{x}, \tilde{x}^{x_7}, \tilde{x}^{x_7x_9}\}
$. Then we get the same matrices for $x_5,x_6,x_7,x_8,x_9$ and
$\tilde{x}^{x_7x_5}=b\tilde{x}\tilde{x}^{x_7}$. This means that w.l.o.g.
we may assume $a\in\langle x_3\rangle $. \par
It follows from the relations (\ref{rel1}) that the action of $x_5$ on
$x^{x_7}$ already determines the action of $H_4$ on $Q_4$. Now we are going to
use the Todd-Coxeter-Algorithm in MAGMA to calculate $|G:G_1|$ for the different cases. We have to distinguish wether $a=1$ or $a=x_3$, and wether $[x,H]=1$ or $[x,H]\neq 1$, where in the last case we have the open parameters $a_5$, $d_2$. We always
apply MAGMA for both possibilities $G_1\cong M_{22}$ and $G_1\cong Aut(M_{22})$
(except in the case $a_5=1$ which already implies $G_1\cong
Aut(M_{22}$). We get the following results:
\begin{enumerate}
\item If $[x,H]=1$ and $a=1$ then $|G:G_1|=2^{11}$.
\item If $[x,H]=1$ and $a=x_3$ then $|G:G_1|=2^9\cdot 3^4$.
\item If $[x,H]\neq 1$, $a_5=1$, $d_2=0$ and $a=1$ then $|G:G_1|=23\cdot 11$.
\end{enumerate}
In all the other cases $G$ is trivial. \par
In the first case $G\cong
2^{11}M_{22}$ resp. $2^{11}Aut(M_{22})$ and $O_2(G)$ is the universal representation group of the $P$-geometry for $M_{22}$ (see \cite[section 3]{IPS}). We have $|Z(G)|=2$ and $G/Z(G)$ as well has a parabolic system of the desired type. So we get case a) of the Theorem.  \par
In the second case, we have
$|G:G_1|=2\cdot |U_6(2):M_{22}|$ and we have shown below (Example 1) that
there exists  a parabolic system for the group $U_6(2)$ satisfying these relations. This means
that, if $G_1\cong M_{22}$, then   
there exists a subgroup $N\unlhd G$ such that $G/N\cong U_6(2)$. Now
$|N|=2$, and since $G=\langle G_1,G_4\rangle =\langle G_1',G_4'\rangle \le
G'$, we get a non-split extension $G\cong 2\!\cdot\! U_6(2)$. If $G_1\cong
Aut(M_{22})$ then $H:=\langle G_1',G_4'\rangle\cong 2\!\cdot\! U_6(2)$ and
$|G:H|=2$. Hence $G\cong 2\!\cdot\! U_6(2)\! :\! 2$. Again in both cases $G/Z(G)$ also has such a parabolic system and we get b).\par
In the third case we can use similar arguments to deduce from Example 2
that $G\cong M_{24}$, i.e., c). \hfill $\Box$\\ 
\\
{\bf Example 1.}
A parabolic system with diagram \Diag{1}{2}{3}{4} for the group $U_6(2)$:\\
\\
We will give two different descriptions for the parabolic system: one by
just giving matrices which satisfy the relations  deduced above, and a
second one which is more geometric and uses the acton of $U_6(2):2\le Co_2$
on the Leech lattice.\par
\bigskip
Let $V$ be a 6-dimensional unitary space over $GF(4)=\{ 0,1,
\omega ,\omega ^2\}$, $G=SU_6(2)$ and  choose a basis 
$\{ v_1,v_2,v_3,w_3,w_2,w_1\} $  for  $V$  such that $\langle v_i,w_i\rangle $ 
 are pairwise orthogonal hyperbolic planes. Then the
 following matrices belong to $G$:
 \begin{center}
$x_{1}= \left(
\begin{array}{cccccc}
1& & & & & \\
 & & &1&0& \\
 & & &0&1& \\
 &1&0& & & \\
 &0&1& & & \\
 & & & & &1\\
\end{array}
\right),$  \quad
$x_{2}= \left(
\begin{array}{cccccc}
1& & & & & \\
1&1& & & & \\
1& &1& & & \\
1& & &1& & \\
1& & & &1& \\
1&1&1&1&1&1\\
\end{array}
\right),$   
 \end{center}
 \begin{center}
$x_{3}= \left(
\begin{array}{cccccc}
1& & & & & \\
0&1& & & & \\
1& &1& & & \\
0& & &1& & \\
1& & & &1& \\
0&1&0&1&0&1\\
\end{array}
\right),$ \quad
$x_{5}= \left(
\begin{array}{cccccc}
1& & & & & \\
\omega &1& & & & \\
\omega & &1& & & \\
\omega^2& & &1& & \\
\omega^2& & & &1& \\
1&\omega &\omega &\omega^2&\omega^2&1\\
\end{array}
\right),$  
 \end{center}
 \begin{center}
   $x_{6}= \left(
\begin{array}{cccccc}
1& & & & & \\
0&1& & & & \\
\omega & &1& & & \\
0& & &1& & \\
\omega^2& & & &1& \\
1&\omega &0&\omega^2&0&1\\
\end{array}
\right),$  \quad
$x_{7}= \left(
\begin{array}{cccccc}
1&1&0&1&0&0 \\
0&0&0&1&0&0 \\
0&1&0&1&1&1 \\
0&1&0&0&0&0 \\
0&1&1&1&0&1 \\
0&0&0&0&0&1\\
\end{array}
\right),$  
 \end{center}
 \begin{center}
$x_{8}= \left(
\begin{array}{cccccc}
1& & & & & \\
0&1&0& & & \\
1&1&1& & & \\
0& & &1&0& \\
1& & &1&1& \\
0&1&0&1&0&1\\
\end{array}
\right),$ \quad
$x_{9}= \left(
\begin{array}{cccccc}
1& & & & & \\
 &0&1& & & \\
 &1&0& & & \\
 & & &0&1& \\
 & & &1&0& \\
 & & & & &1\\
\end{array}
\right),$  
 \end{center}
 \begin{center}
$x_{10}= \left(
\begin{array}{cccccc}
1&0&0& & & \\
0&1&0& & & \\
1&\omega &1& & & \\
 & & &1&0&0 \\
 & & &\omega^2&1&0 \\
 & & &1&0&1\\
\end{array}
\right),$\quad
$x= \left(
\begin{array}{cccccc}
1& & & & & \\
 &1& & & & \\
 & &1& & & \\
 & & &1& & \\
 & & & &1& \\
1& & & & &1\\
\end{array}
\right) .$   
 \end{center}
Moreover, one can calculate that  there images in $\bar{G}:=G/Z(G)\cong U_6(2)$ 
satisfy the relations in (\ref{rel1}) resp. the relations we have deduced
for $x$ in the proof of Theorem 1. In particular, if we set  
\begin{eqnarray*}
    P_1&=\langle x_1,x_2,x_3,x_5,x_6,x_8,x\rangle ,\quad
    P_2=\langle x_1,x_2,x_3,x_5,x_6,x_7, x_8 \rangle ,\\
    P_3&=\langle x_1,x_2,x_3,x_5,x_6, x_8,x_9 \rangle ,\quad
    P_4=\langle x_1,x_2,x_3,x_5,x_6, x_8,x_{10}\rangle ,
\end{eqnarray*}
then $\{ P_1,P_2,P_3,P_4\}$ is a parabolic system of the desired type. So it
remains 
to show that  $G=\langle P_1,P_2,P_3,P_4\rangle $.\par 
Let $G_1=\langle P_2,P_3,P_4\rangle $. Since $\bar{G}_1$ satisfies the relations in (1), we have  $\bar{G}_1\cong M_{22}$. Now by \cite{A} $M_{22}$ is a
maximal subgroup of $U_6(2)$, and   
since $x\not\in G_1$, and the assertion follows. \par
\bigskip
Before starting with the second description we recall the definition of the
$P$-geometry for $Co_2$ as it is given in \cite{Sh1} and some more facts
about the action of this group on the Leech lattice. \par
Let $\Lambda $ denote the Leech lattice with inner
product ( , ) and let $\Lambda _2=\{ v\in \Lambda _2|(v,v)=32\} $. Consider the
group $H:=Co_2$ as the stabilizer in $Aut(\Lambda )$ of a fixed vector $v_0\in 
\Lambda _2$. Let $\Sigma =\{ \{ v,-v\} |v\in \Lambda _2, (v_0,v)=0\} $. Then
$H_{\sigma } \cong 2^{10}.Aut(M_{22})$ for $\sigma\in \Sigma $ and the orbits
of $H_{\sigma }$ on $\Sigma $ are described in \cite{Sh1}.
Define a graph  on $\Sigma $ in the way that $\sigma $ and $\tau
\in \Sigma $ are adjacent iff $H_{\sigma}\cap H_{\tau }\cong 2^9.2^5.\Sigma
_5$. A clique ${\cal C}$ in this graph is called closed iff for each 
$\{ v,-v\} , \{ w,-w\}\in {\cal C}$ there exists $\{ u,-u\}\in {\cal C}$ such
that $v_0+v+w+u\in 2\Lambda $. It is shown in \cite{Sh1} that each closed
clique is of size 1,3,7 or 15, and that, if we take as
objects of type $i$ the closed cliques of size $2^i-1$ and define incidence
by inclusion, than we get the P-geometry for $Co_2$.  \par
Let $\Omega =\{ \{ v,v_0-v\} |v\in\Lambda _2,(v,v_0)=16\} $. Then $H$ acts
transitively on $\Omega $ and $H_{\omega }\!\cong\! U_6(2)\!:\! 2$ for
$\omega\in\Omega $ (see \cite{Sh1}, section 6). By {\sc table} 2 of
\cite{Sh1} $H_{\sigma }$ has 
three orbits $\Omega _1,\Omega _2,\Omega _3$ on $\Omega $ and if $\sigma
=\{u,-u\} ,\omega =\{ w,v_0-w\} $ we may assume $(u,v)=16,0$ resp. $8$.   
Moreover, for $\omega\in \Omega _3$ we have $H_{\sigma }\cap H_{\omega
}\cong Aut(M_{22})$. Since the orbits of $H_{\sigma }$ on $\Omega $
bijectively correspond to the orbits of $H_{\omega }$ on $\Sigma $, for a
fixed $\omega _0=\{ v,v_0-v\}\in\Omega $, $H_{\omega _0}$ has exactly one
orbit, say $\Sigma _3$, on $\Sigma $ of length $|\Sigma
_3|=|(U_6(2)\! :\! 2):Aut(M_{22})|=20736$     which consists of the pairs 
$\{ v,-v\}\in\Sigma $ such that w.l.o.g. $(v,w)=8$. Now, for $i=1,2,3,4$, let
$$\Gamma _i=\{ {\cal C}\cap \Sigma _3| {\cal C} \hbox{ is a closed
clique of size $2^i-1$ in $\Sigma $ }\}  $$
and define an incidence relation between the $\Gamma _i$ by inclusion.
Considering the shapes of vectors from $\Lambda _2$ as given in \cite{Sh1} it
is easy to see that $|x|=2^{i-1}$ for $x\in\Gamma _i$ and that this really
defines a geometry of the desired type with automorphism group $H_{\omega
_0}\cong U_6(2)\!:\! 2$.\\
\\
{\bf Example 2.} A geometry with diagram \Diag{1}{2}{3}{4} for the
group $M_{24}$:\\
\\
Let ${\cal S}={\cal S}(24,8,5)$ be the Steiner system for $G=M_{24}$
with underlying set $\Omega $. Let ${\cal O}$ denote the set of octads and
${\cal T}$ the set of trios. We assume that the reader is familiar with the
action of $G$ on $\cal{S}$, particularly we need the following property:
\begin{quotation}\noindent
{\it
Let $O\in\cal{O}$ and $G_O$ the stabilizer of $O$ in $G$. Then there is
defined the structure of an affine space on the complement
$\Omega\setminus O$ and $G_O\cong 2^4A_8$ acts as affine group $AGL(4,2)$ 
on $\Omega\setminus O$.}
\end{quotation}
Now we define a rank-4-geometry $\Gamma =\Gamma _1\cup\Gamma _2\cup\Gamma
_3\cup\Gamma _4$ by
\begin{itemize}
\item[--] $\Gamma _1=\{ \{ p,q\} |p,q\in\Omega , p\neq q\} $
\item[--] $\Gamma _2=\{ \{ a,b\} |a,b\in \Gamma _1, a\cap b=\emptyset \}$
\item[--] $\Gamma _3=\{ (O_1,P,\{ O_2,O_3\} )| O_i\in{\cal O}, \{
O_1,O_2,O_3\}\in{\cal T}$, $P$ is a partition of $O_1$ into four parallel
lines in the affine geometry  on $\Omega\setminus O_2\} $
\item[--] $\Gamma _4=\{ (O,P) |O\in{\cal O}$, $P$ is a partiton of
$\Omega\setminus O$ into parallel lines $\}$.
\end{itemize}
We define incidences between the $\Gamma _i$ as follows:\par
Let $x\in \Gamma _1$, $\{ a,b\}\in \Gamma _2$ and $y\in \Gamma _3$ or $\Gamma
_4$. Then $x$ is incident to $\{ a,b\}$ iff $x\in\{ a,b\}$, $x$ is
incident to $y$ iff $x\in P$ and
$\{ a,b\}$ is incident to $y$ iff $a$ and $b$ are incident to $y$.
Finally, $(O_1,P_1,\{ O_2,O_3\} )\in\Gamma _3$ and $(O,P)\in
\Gamma _4$ are incident iff $O=O_2$ or $O_3$ and $P_1\subset P$. 


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\bibitem{A} J.H. Conway, R.T. Curtis, S.P. Norton, R.A. Parker,
R.A. Wilson, Atlas of Finite Groups, Oxford, 1985.
\bibitem{Buek} F. Buekenhout (Ed.), Handbook of Incidence Geometries, North Holland, 1995.
\bibitem{IPS} A. Ivanov, D. Pasechnik, S. Shpectorov, Non-abelian Representations of Some Sporadic Geometries, J. Algebra 181, 1996, 523 - 557.
\bibitem{Pas} A. Pasini, Diagram Geometries, Oxford, 1994.
\bibitem{Sh1} S. Shpectorov, The universal 2-cover of the P-geometry for
$Co_2$, Europ. J. Comb. 13, 1992, 291-312.
\bibitem{S1} G. Stroth, The nonexistence of certain Tits geometries with affine diagram, Geom. Dedicata 28, 1988, 277 - 319.
\end{thebibliography}

\end{document}

{\bf Proof.} Suppose false. Since $|P_1:B|=2$, we have $\Omega _1(Z(B))\unlhd P_1$. Therefore  $\ozb \not\!\!\unlhd G_1$.  If $C_{G_1}(K_1)\le K_1$ this implies that 
$Z_1:=\langle \ozb ^{G_1}\rangle $ is a nontrivial module for
$G_1/K_1$. Since $M_{22}$ does not possess an $F$-module, we have $J(B)\le
K_1\le K_4$ (where $J(B)$ denotes the Thompson-subgroup of $B$). So $J(B)=J(K_1)=J(K_4)\unlhd \langle G_1,G_4\rangle =G$, a
contradiction. \par 
So we have $C_{G_1}(K_1)\not\le K_1$ and $G_1=C_{G_1}(K_1)K_1\langle
y\rangle $, for some $y\in G_1\setminus G_1'K_1$ if $G_1/K_1\cong
Aut(M_{22})$ and $y=1$ if $G_1/K_1\cong M_{22}$. In particular,
$[O^2(G_1),K_1]=1$. Hence also $[O^2(G_{14}),K_1]=1$, which implies that
$K_4$ involves exactly one nontrivial composition factor for
$G_{14}/K_4\cong G_4/Q_4\cong
L_3(2)$. Since $G=\langle G_1,Q_4\rangle $, $K_1\not\!\!\unlhd G$ and $K_1\le
K_4$, we have $[K_4,Q_4]\not\le K_1$. So the irreducible action of $G_4/Q_4$
on $K_4/K_1$ implies that $K_4=[K_4,Q_4]K_1$ and that $G_4/Q_4$ acts
nontrivially $W:=[K_4,Q_4]$. Now choose $W_1\le W$ minimal and $W_2\le W_1$
maximal such that $W_i\unlhd G_4$, $[W_1,K_4]\le W_2$, $[W_1,Q_4,Q_4]\le
W_2$ and $[W_1,Q_4]/W_2$ is the irreducible 3-dimensional $G_4/Q_4$-module.   
Let $w\in W_1\setminus [W_1,Q_4]$ and define
\begin{eqnarray*}
  \varphi: & Q_4/K_4\to [W_1,Q_4]/W_2\\
           & qK_4\to [w,q]W_2.
\end{eqnarray*}
It is easy to see that $\varphi $ is a
$G_4/Q_4$-isomorphism. This means that $Q_4/K_4$ and $[W_1,Q_4]/W_2\cong
K_4/K_1$ are isomorphic as $G_4/Q_4$-modules. But on the one hand, it
follows from the action of $G_4/K_4\cong 2^3L_3(2)$ on the 
affine geometry that $C_{G_4}(qK_4)$ is conjugate to $G_{34}$ for $q\in
Q_4\setminus K_4$. On the other hand, one can see in $G_1$ (see  the
relations for $M_{22}$ below) that, for $x\in
K_4\setminus K_1$, $C_{G_{14}}(xK_1)$ is conjugate to $G_{14}\cap
G_{12}=P_3$.  So $K_4/K_1$ must be the dual of $Q_4/K_4$ and we have again a
contradiction.\hfill $\Box$\\
