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% Macros for Buekenhout diagrams
% Revised version 23 July 1990
% Typical usage is $\node\stroke{c}\node_q\arc\node_q\etc\node^n$
% Note that nodes can have subscripts and superscripts, which will be
%   properly positioned! (I hope)

\mathchardef\tnode="020E % temporary node
% Note : this is an unfilled node. Change to "020F for filled node.
% Someday, I'll rewrite the macros to allow the user to choose!
% (A character superimposed on an unfilled node might be useful.)

\def\arc{% unlabelled stroke for projective plane
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\def\darc{% double arc for GQ
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\def\stroke#1{% labelled stroke for diagrams
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  \kern -0.22em}

\def\dstroke#1{% probably just for triple cover of Sp(4,2) quadrangle
  \kern 0.05em
  \rlap\darc{{\textstyle{#1}}\atop\phantom\darc}
  \kern -0.22em}

\def\etc{\>\>\cdots\>\>} % for ellipsis

\def\centerscript#1{% centres text over or under a node
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  \hbox to \wd0{\hss$\scriptstyle{#1}$\hss}}

% macro for optional superscript or subscript
% adapted from ``Introduction to \TeX'' by Norbert Schwarz

\def\node{%labelled node; usage \node or \node^{label} or \node_{label}
% or \node^{label}_{label} (in either order)
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  \let\sp=^
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  \def\dolabellednode{%
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  \def\getsuper^#1{\def\super{#1}\futurelet\next\dolabellednode}

  \def\donode{%
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\def\varcdn{%vertical arc with node at bottom - use as subscript to node
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\def\an{\node\arc\node\arc\node\etc\node\arc\node\arc\node}
\def\cn{\node\arc\node\arc\node\etc\node\arc\node\darc\node}
\def\dn{\node\arc\node\arc\node\etc\node\arc\node_\varcdn\arc\node}
\def\esix{\node\arc\node\arc\node_\varcdn\arc\node\arc\node}
\def\eseven{\esix\arc\node}
\def\eeight{\eseven\arc\node}
\def\ffour{\node\arc\node\darc\node\arc\node}


\newtheorem{mainth}{Theorem}
\newtheorem{thms}{Theorem}[section]
\newtheorem{thm}{Theorem}[subsection]
\newtheorem{prop}[thms]{Proposition}
\newtheorem{lem}[thms]{Lemma}
\newtheorem{cor}[thms]{Corollary}
\newtheorem{hyp}[thms]{Hypothesis}
\newtheorem{con}[thms]{Conjecture}
\newcommand\pf{\noindent{\it Proof:~~}}
\newcommand{\qed}{\hfill\framebox(6,6){}}



\begin{document}

\phantom{a}

\bigskip

\begin{center}
\begin{large}
A characterization of 3-local geometry of $M(24)$
\end{large}

\bigskip

A.A.~Ivanov and G.~Stroth

\medskip

\end{center}
\begin{center} 
{\it Dedicated to Professor B. Fischer on the occasion of his sixtieth birthday}
\end{center}

\bigskip

\begin{abstract}
The largest Fischer 3-transposition group $M(24)$ acts
flag-transi\-tively on a 3-local incidence geometry ${\cal G}(M(24))$
which is a $c$-extension of the dual polar space associated with the
group $O_7(3)$. The action of the simple commutator subgroup $M(24)'$
is still flag-transitive. We show that ${\cal G}(M(24))$ is characterized
by its diagram under the flag-transitivity assumption. The result implies
in particular that ${\cal G}(M(24))$ is simply connected. The geometry
${\cal G}(M(24))$ appears as a subgeometry in the Buekenhout -- Fischer
3-local geometry ${\cal G}(F_1)$ of the Monster group.
The simple connectedness of ${\cal G}(M(24))$ has played a crucial role
in the characterization of ${\cal G}(F_1)$, achieved recently.
When determining the possible structure of the parabolic subgroups we have
used an unpublished pushing-up result by U.~Meierfrankenfeld.
\end{abstract}

\section{Introduction}
Let $\delta$ be the following diagram:
$$\delta:~\node_1\stroke{c}\node_3\darc\node_3\arc\node_3.$$
An (incidence) geometry ${\cal G}$ with diagram $\delta$ has rank 4;
the types of its elements are 1, 2, 3 and 4 correspond to the nodes from the
left to the right on the diagram. The residue of a flag of type $\{1,2\}$,
$\{1,4\}$ or $\{3,4\}$ is, respectively, a projective plane of order 3,
a generalized quadrangle of order $(3,3)$ or the geometry of vertices and
edges of the complete graph on 5 vertices.

Suppose that ${\cal G}$ is a geometry whose diagram is $\delta$ and that it admits
a flag-transitive automorphism group $G$. Let $\{\alpha_1,\alpha_2,\alpha_3,\alpha_4\}$ be
a flag in ${\cal G}$ where $\alpha_i$ is of type $i$. For a subset $J$ of
$I=\{1,2,3,4\}$ let $G_J$ denote the stabilizer in $G$ of the
flag $F_J=\{\alpha_i \mid i \in J\}$. To simplify the
notation we will write for instance $12$ for the subset $\{1,2\}$. Let $K_J$ denote
the kernel of the action of $G_J$ on the residue ${\cal G}_J$ of $F_J$ in
${\cal G}$ and let
$\bar G_J \cong G_J/K_J$ be the action induced on the residue. For an element $\beta \in {\cal G}$
by $G_{\beta}$ we denote its stabilizer in $G$, so that $G_{\alpha_i}$ is another name for $G_i$.


Let $M(24)$ be the largest sporadic 3-transposition group discovered by
B.~Fischer. It is known \cite{rst} that $M(24)$ acts flag-transitively on
a geometry ${\cal G}(M(24))$ whose diagram is $\delta$ and in which the residue of
an element of type 1 is the dual polar space associated with the group
$O_7(3)$. For the action of $G \cong M(24)$ on ${\cal G}(M(24))$
we have $\bar G_1 \cong O_7(3)$, $K_1 \cong 3^7$. Moreover, $K_1$ is the natural
module for $\bar G_1$ and  $G_1$ does not split over $K_1$. The parabolic
$G_4$ is a split extension of $K_4 \cong 3^{1+10}_+:2$ by
$\bar G_4 \cong U_5(2):2$. Every element $\alpha$ in ${\cal G}(M(24))$ can be identified with
certain elementary abelian 3-subgroup $V(\alpha)$ in $M(24)$ in such a way that
the order of $V(\alpha)$ is $3$, $3^2$, $3^3$ and $3^7$ for $\alpha$ being
of type $1$, $2$, $3$
and $4$, respectively; the stabilizer of $\alpha$ in $G$ coincides with
$N_G(V(\alpha))$ and $\alpha$ is incident to $\beta$ if and only if either $V(\alpha) \le V(\beta)$
or $V(\beta) \le V(\alpha)$. The commutator subgroup $M(24)'$ also acts flag-transitively
on ${\cal G}(M(24))$. With respect to the action of $G \cong M(24)'$ on the
geometry we have $\bar G_1 \cong \Omega_7(3)$, $K_1 \cong 3^7$, $\bar G_4 \cong
U_5(2):2$, $K_4 \cong 3^{1+10}_+$.

A triple cover $3 \cdot M(24)$ of $M(24)$ is the normalizer of a subgroup
of order 3 in the Monster group $F_1$ and ${\cal G}(M(24))$ appears as
a subgeometry in the Buekenhout -- Fischer geometry ${\cal G}(F_1)$ of the
Monster \cite{bf} which has the following diagram:
$$ \node_1\stroke{c}\node_9\darc\node_3\arc\node_3.$$

Both ${\cal G}(M(24))$ and ${\cal G}(F_1)$ belong to the family
of $c$-extended classical dual polar spaces and their characterization in terms
of diagrams is very
important for classification of all flag-transitive extended
dual polar spaces \cite{yosh2}. In the present paper we prove the following.

\begin{mainth}   \label{mainth}

Let ${\cal G}$ be a geometry with the diagram $\delta$ admitting a
flag-transitive automorphism group $G$. Assume that the stabilizer of a maximal flag in ${\cal G}$ is finite. Then ${\cal G} \cong {\cal G}(M(24))$
and $G$ is isomorphic either to $M(24)$ or to $M(24)'$.

\end{mainth}

The geometry ${\cal G}(F_1)$ has been characterized in \cite{im} and
Theorem~\ref{mainth} was crucial for this characterization.

\section{The structure of residues}
For the rest of the paper let ${\cal G}$ be a geometry with the diagram
$\delta$ and $G$ be a flag-transitive automorphism group of ${\cal G}$ with finite stabilizers of maximal flags. We adopt the notation introduced in the previous section.

Consider the residue  ${\cal G}_1$ of $\alpha_1$ in ${\cal G}$. Then ${\cal G}_1$ is
a flag-transitive $C_3$-geometry; its residual projective planes have order 3 and hence
they are Desarguesian. The available results on flag-transitive $C_3$-geometries
imply the following (cf. Section 14 in \cite{pas}).

\begin{lem} \label{residues_1}

${\cal G}_1$ is the (dual) polar space associated with a $7$-dimensional
$GF(3)$-space equipped with a non-singular orthogonal form. $\bar G_1$ is
either the special orthogonal group $O_7(3)$ or its index $2$ simple
commutator subgroup $\Omega_7(3)$.\qed

\end{lem}

Let us recall some well known properties of the dual polar space ${\cal G}_1$.
Let $V=V(\alpha_1)$ be a 7-dimensional $GF(3)$-space
with a non-singular quadratic form $\phi$ on it. Then the elements of type 2, 3 and
4 in ${\cal G}_1$ are the 3-, 2- and 1-dimensional subspaces in $V$ which
are totally singular with respect to $\phi$. Two subspaces are incident if one of them
contains the other one. It is convenient to associate with ${\cal G}_1$ a
graph $\Delta$ called the {\em dual polar space graph}. The
vertices of $\Delta$ are maximal (which means 3-dimensional) totally isotropic
subspaces of $V$ with respect to ${\phi}$ with two such subspaces adjacent if
their intersection is of dimension 2. In other terms the vertices of $\Delta$
are the elements of type 2 in ${\cal G}_1$ and two of them are adjacent if
they are incident
to a common element of type 3. For a vertex $\beta$ of $\Delta$ and an integer $i$ let
$\Delta_i(\beta)$ denote the set of vertices in $\Delta$ which are at distance $i$
from $\beta$ in the natural metric of $\Delta$. It is known \cite{bcn} that $\Delta$ is distance-transitive on 1120 vertices
with the intersection array $\{39,36,27;1,4,13\}$. In particular the valency of
$\Delta$ is 39 and the diameter is 3. Both $O_7(3)$ and $\Omega_7(3)$ act primitively on the vertex set of
$\Delta$.

The following two lemmas come from the standard information on the
actions of $O_7(3)$ and $\Omega_7(3)$ on the associated dual polar space
(compare \cite{atlas}).

\begin{lem} \label{residues_2}

$\bar G_{14} \cong O_5(3) \cong \Omega_5(3):2$ and $O_3(K_{14}/K_1) \cong 3^5$
is the irreducible orthogonal module for $\bar G_{14}$. If $\bar G_1 \cong
\Omega_7(3)$ then $K_{14}/K_1=O_3(K_{14}/K_1)$ and if $\bar G_1 \cong O_7(3)$
then $K_{14}/K_1$ is an extension of $O_3(K_{14}/K_1)$ by a fixed-point free
involution.\qed

\end{lem}

\begin{lem} \label{residues_3}

$\bar G_{12} \cong L_3(3)$ and $O_3(K_{12}/K_1) \cong 3^{3+3}$. If $\bar G_1 \cong \Omega_7(3)$
then $K_{12}/K_1=O_3(K_{12}/K_1)$ and if $\bar G_1 \cong O_7(3)$ then
$K_{12}/K_1$ is an extension of $O_3(K_{12}/K_1)$ by an involution commuting
with a subgroup of order $3^3$ inside $O_3(K_{12}/K_1)$. In any case the elementwise stabilizer
of $\Delta_1(\alpha _2)$ in $K_{12}/K_1$ is elementary abelian of order $3^3$, it acts
faithfully on $\Delta_2(\alpha _2)$ and on $\Delta_3(\alpha _2)$  and it is acted on irreducibly by
$\bar G_{12}$.\qed

\end{lem}

The residue ${\cal G}_4$ of the element $\alpha_4$ in ${\cal G}$ is a flag-transitive
$c$-extension of a classical generalized quadrangle. All such geometries are
known (cf. \cite{yosh2} and references therein). There is
only one flag-transitive
$c$-extension of the generalized quadrangle ${\cal G}_{14}$ of the orthogonal
group $O_5(3)$. This extension has 176 elements of type 1 and it possesses
two flag-transitive automorphism groups: $U_5(2)$ and $U_5(2):2$. In the
former group the stabilizer of an element of type 1 induces on the residual
generalized quadrangle the simple group $\Omega_5(3)$ and this
is in a contradiction with Lemma~\ref{residues_2}. So we have the following :

\begin{lem} \label{residues_4}

$\bar G_4 \cong U_5(2):2$.\qed

\end{lem}

\medskip

Considering the action of $\bar G_4$ on the residual geometry ${\cal G}_4$ (compare \cite{atlas}) we
obtain :

\medskip

\begin{lem} \label{residues_5}

$G_{14}/K_4 \cong (3 \times \Omega_5(3)):2$ and $K_{14}/K_4$ is of order $3$.\qed

\end{lem}

\medskip

Now we are in a position to prove the following :

\begin{lem} \label{residues_6}

$K_1/(K_1 \cap K_4)$ has order $3$, in particular $K_1 \ne 1$.

\end{lem}

\pf By definition $K_1/(K_1 \cap K_4)$ is the action induced by $K_1$ on the
residue ${\cal G}_4$ of $\alpha_4$ and it is contained in the action $K_{14}/K_4$
induced by $K_{14}$ on this residue. By Lemma~\ref{residues_5} the latter action has order 3. Suppose that
$K_1/(K_1 \cap K_4)$ is trivial. Then $O_3(K_{14}/K_1)$ must induce an order 3 action on
${\cal G}_4$ but since $O_3(K_{14}/K_1)$ has order $3^5$ and it is irreducible under $\bar G_{14} \cong O_5(3)$
(cf. Lemma~\ref{residues_2}) this is impossible.\qed


\medskip

Let $\Gamma$ be the collinearity graph of ${\cal G}$ which means that the vertices of
$\Gamma$ are the elements of type 1 in ${\cal G}$ and two of them are
adjacent if and only if they are incident to a common element of type 2.

It follows directly from the diagram $\delta$ of ${\cal G}$ that every element of type 2 is
incident to exactly two elements of type 1. Hence an element
of type 2 determines an edge of $\Gamma$.
Since $G_1$ acts
primitively on the set of elements of type 2 in ${\cal G}_1$,
different elements of type 2 determine different edges of $\Gamma$. In
particular there is a bijective mapping $\psi$ of the vertex set of
the dual polar space graph $\Delta$ associated with $\alpha_1$
onto the set $\Gamma(\alpha_1)$ of vertices in $\Gamma$ adjacent to $\alpha_1$
and $\psi$ commutes with the action of $G_1$.

\begin{lem} \label{residues_7}

Let $\beta$ and $\gamma$ be adjacent vertices of $\Delta$. Then the images
of $\beta$ and $\gamma$ under
$\psi$ are adjacent in $\Gamma$.

\end{lem}

\pf Since $\beta$ and $\gamma$ are adjacent in  $\Delta$, they are incident to a
common element of type 3 in ${\cal G}_1$, say to $\alpha_3$. But the residue of the
flag $\{\alpha_3,\alpha_4\}$ is formed by vertices and edges of a complete
graph on 5 vertices so the result follows. \qed

\medskip

In what follows we will identify the vertices and the edges of
$\Gamma$ with the elements in ${\cal G}$ of type 1 and 2, respectively.

Since the automorphism group
of the residual geometry ${\cal G}_1$ acts faithfully on the
set of elements of type
2 in the residue, we observe that $K_1$ coincides with the
elementwise stabilizer in $G_1$ of the vertices
from $\Gamma(\alpha_1)$.

\begin{lem} \label{residues_8}

$K_1$ is a non-trivial $3$-group.

\end{lem}

\pf $K_1 \ne 1$ by Lemma~\ref{residues_6}. As we observed in the above paragraph,
$K_1$ coincides with the elementwise stabilizer
in $G$ of $\alpha_1$ and of all the vertices adjacent to $\alpha_1$ in $\Gamma$.
On the other hand $\Gamma$ is connected and $G$ acts on
it vertex-transitively. So to prove the lemma it is sufficient to show that $K_1$
induces a $3$-group on $\Gamma(\beta)$ for every
$\beta \in \Gamma(\alpha_1)$. In view of the obvious symmetry
between $\alpha_1$ and $\beta$, by Lemmas~\ref{residues_3} and \ref{residues_7}
the action induced by $K_1$ on $\Gamma(\beta)$ has order $3^3$, so the result follows.\qed

\medskip

In order to identify $K_1$ we apply the following result proved by U.~Meierfrankenfeld using the pushing-up
methods developed in his Ph D. Thesis \cite{meie1} and the classification \cite{meix} of the failure factorization
modules for Lie type groups in odd characteristic.

\begin{prop} \label{residues_9}

{\rm \cite{meie2}} Let $L$ be a finite group and suppose that

\begin{itemize}

\item[{\rm (i)}] for $Q=O_3(L)$ we have $\bar L:= L/Q \cong O_7(3)$ or $\Omega_7(3)$;

\item[{\rm (ii)}] for $V$ being the natural module for $\bar L$ and $P$ being the full preimage of the stabilizer
in $\bar L$ of a maximal totally isotropic subspace in $V$; if $N \le Q$ is normal in $L$ and characteristic in
$P$, then $N=1$.

\end{itemize}
Then $Q$ is either trivial or isomorphic to $V$, that is to the natural module of $\bar L$.\qed

\end{prop}

\begin{lem} \label{residues_10}

$K_1 \cong V \cong 3^7$ is the natural module for $\bar G_1$.\qed

\end{lem}

\pf We are going to apply Proposition~\ref{residues_9} for $L=G_1$, $P=G_{12}$ and $Q=K_1$. Let
$N$ be a subgroup in $K_1$ which is normal in $G_1$ and characteristic in $G_{12}$. Since $G_2$ contains $G_{12}$ as a normal subgroup
(of index 2), $N$ must be normal in $G_2$. Now, $G_1$ and $G_2$ generate $G$ since both ${\cal G}$ and $\Gamma$ are connected. So $N$
fixes a vertex of $\Gamma$ and it is normal in the group $G$ which acts on $\Gamma$ vertex-transitively. This means that $N$ fixes every
vertex of $\Gamma$. Since the action of $G$ on $\Gamma$ is faithful, we have $N=1$ and the result follows.\qed

\medskip

\begin{lem} \label{residues_11}

The following assertions hold:

\begin{itemize}

\item[{\rm (i)}] either $G_1 \cong 3^7.\Omega_7(3)$ or $G_1 \cong 3^7.O_7(3)$;

\item[{\rm (ii)}] if $G_1 \cong 3^7.\Omega_7(3)$ then $G_1$ possesses an automorphism $\tau$ such that $\langle G_1, \tau \rangle \cong 3^7.O_7(3)$.

\end{itemize}

\end{lem}

\pf The assertion (i) follows from Lemmas~\ref{residues_1} and ~\ref{residues_10}.

The assertion (ii) is trivial if $G_1$ splits over $K_1$. By \cite{kus} there is a unique isomorphism type
of non-split extensions of $\Omega_7(3)$ by the natural module. This extension is contained in $M(24)'$ and
the automorphism $\tau$ is realized in $M(24)$.\qed

\medskip

Thus we have established that $K_1$ is canonically isomorphic to the 7-dimensional $GF(3)$-module $V$ equipped with a non-singular quadratic form $\phi$
and the residual geometry ${\cal G}_1$ can be naturally defined in terms of the polar space $(V,\phi)$.
In what follows we will identify $V$ with $K_1$. For an element $\beta \in {\cal G}_1$
let $V(\beta)$ denote the subspace of $V$ which corresponds to $\beta$.
The dimension of $V(\beta)$ is 3, 2 and 1 for $\beta$ being of type 2, 3 and 4, respectively.
Then $G_{1j}$ is the full normalizer of $V(\alpha_j)$ in $G_1$.
For an arbitrary subspace $U$ in $V$ let $U^{\perp}$ denote the orthogonal complement of $U$ with respect to $\phi$.

Let us now specify $K_4$. We have that $G_{14}$ is the normalizer in $G_1$ of the
subspace $V(\alpha_4)$ which is 1-dimensional.
It is a standard fact about orthogonal
groups that under the action of $G_{14}/K_1$ the module $V$ is uniserial,
$V(\alpha_4)$ is its unique 1-dimensional submodule and
$V(\alpha_4)^{\perp}$ is the unique submodule of codimension 1. This and Lemma~\ref{residues_6} give :

\begin{lem} \label{residues_12}

$K_1 \cap K_4 = V(\alpha_4)^{\perp}$.\qed

\end{lem}

The action induced by $G_{14}$ on ${\cal G}_4$ is isomorphic to $(3 \times \Omega_5(3)):2$ (cf. Lemma~\ref{residues_5}) and
$K_1$ induces an action of order 3. Hence $G_{14}/ \langle K_1,K_4 \rangle \cong O_5(3) \cong \Omega_5(3):2$. This and
Lemma~\ref{residues_2} imply :

\begin{lem} \label{residues_13}

$\langle K_1 , K_4 \rangle =K_{14}$.\qed

\end{lem}

\begin{lem} \label{residues_14}

$O_3(K_4) \cong 3_+^{1+10}$. If $G_1 \cong 3^7.\Omega_7(3)$ then $K_4=O_3(K_4)$ and if $G_1 \cong
3^7.O_7(3)$ then $K_4$ is an extension of $O_3(K_4)$ by an involution acting fixed-point freely on
$O_3(K_4)/Z(O_3(K_4))$.

\end{lem}

\pf By Lemmas~\ref{residues_12} and \ref{residues_13} $O_3(K_4)$ has
index 3 in $O_3(K_{14})$ and the latter is of order $3^{12}$. Hence $|O_3(K_4)|=3^{11}$.
Clearly $V(\alpha_4)$ is in the center of $K_4$. We have that $O_3(K_{14}/K_1)$ acts faithfully on $V(\alpha_4)^{\perp}$ and centralizes $V(\alpha_4)^{\perp}/V(\alpha_4)$.
Moreover, both $O_3(K_{14}/K_1)$ and $V(\alpha_4)^{\perp}/V(\alpha_4)$ are elementary abelian of order $3^5$ and are acted on irreducibly by
$\bar G_{14} \cong O_3(5)$. Hence $O_3(K_4)$ is extraspecial of plus type.
Suppose that $G_1 \cong 3^7.O_7(3)$. Then there exists an
involution $\tau$ in $K_{14} \setminus O_3(K_{14})$ which
centralizes $V(\alpha_4)$ and acts fixed-point freely both on $O_3(K_{14}/K_4)$ and on
$V(\alpha_4)^{\perp}/V(\alpha_4)$ so the result follows.\qed

\medskip

Suppose that $G_1 \cong 3^7.O_7(3)$ and let $\tau$ be an involution from
$K_4 \setminus O_3(K_4)$. Then $C_{G_4}(\tau)$ is an extension of
$\langle \tau \rangle \times V(\alpha_4)$ by $\bar G_4 \cong U_5(2):2$.
Since the Schur multiplier of $U_5(2)$ is trivial we have $C_{G_4}(\tau) \cong \langle \tau \rangle \times V(\alpha_4) \times
U_5(2):2$ and $G_4$ splits over $O_3(K_4)$.


\begin{prop} \label{residues_15}

Let ${\cal G}$ be a flag-transitive geometry with diagram $\delta$ and let $G$ be the full automorphism group of ${\cal G}$. Then
$G_1 \cong 3^7.O_7(3)$ and $G_4 \cong 3^{1+10}_+:(2 \times U_5(2):2)$.

\end{prop}

\pf By Lemma~\ref{residues_11} and the paragraph before the proposition all we
have to show is that if $G_1 \cong 3^7.\Omega_7(3)$ then $G$ is not the full automorphism group of ${\cal G}$.
By  a
standard argument the elements of  ${\cal G}$ can be identified with the
right cosets of the subgroups $G_i$ in $G$ so that
two elements are incident if and only if the intersection of the cosets is non-empty. So to prove the claim it is sufficient to show that $G$ possesses
a suitable automorphism $\tau$ which normalizes each $G_i$. By Lemma~\ref{residues_11} (ii) there
is an automorphism $\tau$ of $G_1$ such that $\langle G_1, \tau \rangle \cong 3^7.O_7(3)$. We can choose $\tau$ in such a way that it normalizes
$G_{14}$ and centralizes a complement $F$ to $K_{14}=O_3(K_{14})$ in $G_{14}$ (cf. Lemma~\ref{residues_2}). In this case $\tau$ acts fixed-points freely on $O_3(K_{14}/K_1)$ and on $(K_1 \cap K_4)/Z(K_4)$. Hence $\tau$ acts fixed-points freely on $K_4/Z(K_4)$. Then $C_{G_{14}}(\tau)/Z(K_4)$ is a complement to $O_3(K_4)/Z(K_4)$ in $G_{14}/Z(K_4)$. Since $G_{14}$ contains a Sylow 3-subgroup of $G_4$, by Gasch\"utz's theorem and since the Schur multiplier of $U_5(2)$ is trivial, there
is a complement $H$ to $O_3(K_4)$ in $G_4$ which contains $F$. Then we can extend the action of $\tau$ on $G_{14}$ to an automorphism of $G_4$
by declaring that $\tau$ commutes with $H$. The geometry ${\cal G}$ is
connected, hence $G_1$ and $G_4$ generate $G$
and we can extend $\tau$ to an automorphism of $G$. Since $\tau$ fixes the
residue ${\cal G}_4$ elementwise, it clearly normalizes $G_2$ and $G_3$.\qed

\medskip

For the rest of the paper $G$ is assumed to be the full automorphism group of
${\cal G}$. By the above proposition in this case $G_1 \cong 3^7.O_7(3)$ and $G_4 \cong 3^{1+10}_+:(2 \times U_5(2):2)$.

The group $U_5(2)$ has only one faithful 10-dimensional
$GF(3)$-representation (compare \cite{jlpw}). This representation is irreducible which means that
there is a unique invariant
quad\-ratic form on the corresponding module. This observation and Proposition~\ref{residues_15} specify $G_4$ up to isomorphism. By
Proposition~\ref{residues_15} and \cite{kus} the isomorphism type of $G_1$ depends  only on whether or not it splits over $K_1$. On the other hand there is a unique way to specify $G_{14}$ as a subgroup in $G_4$. Namely, let
$U \cong U_5(2)$ be a complement to $O_3(G_4)$ in $O^2(G_4)$ (one can easily
see that there is one class of such complements) and let $\lambda$ be an element of order 3 in $U$ such that $C_U(\lambda) \cong \langle \lambda \rangle \times
\Omega_5(3)$. Then up to conjugacy in $G_4$ we have $K_1=
\langle  \lambda , C_{O_3(G_4)}(\lambda) \rangle$ and $G_{14}=N_{G_4}(K_1)$.
Since $G_{14}$ contains a Sylow 3-subgroup of $G_1$, by Gasch\"utz's theorem $G_1$ splits over $K_1$ if and only if $G_{14}$ does so. By the above argument the
isomorphism type of $G_{14}$ and the location of $K_1$ in $G_{14}$ are uniquely specified and hence the isomorphism type of $G_1$ is also uniquely determined. Clearly, $G_1$ must be isomorphic to the parabolic subgroup associated with
the action of $M(24)$ on ${\cal G}(M(24))$, that is to the non-split extension.
So the amalgam consisting of the subgroups $G_1$ and $G_4$ is isomorphic to the one associated with the action of $M(24)$ on ${\cal G}(M(24))$.

Let $\beta \in \Gamma(\alpha_1)$ and $K(\beta)$ be the elementwise stabilizer in $G$ of the vertices
from $\Gamma(\beta)$ (which is a conjugate of $K_1$). By Lemmas~\ref{residues_3} and \ref{residues_8}
$K(\beta)$ induces on the dual polar space graph $\Delta$ an action of order $3^3$ which is faithful
on $\Delta_2(\beta)$ and on $\Delta_3(\beta)$. This implies the
following two lemmas :

\begin{lem} \label{residues_16}

The intersection of $K(\beta)$ and $K_1$ has order $3^4$ and it coincides with
$V(\beta)^{\perp}$.\qed

\end{lem}

\begin{lem} \label{residues_17}

Let $\psi$ be the mapping of $\Delta$ onto $\Gamma(\alpha_1)$
which commutes with the action of $G_1$.
Then $\beta, \gamma \in \Delta$ are adjacent in $\Delta$ if and only if their
images under $\psi$  are adjacent in $\Gamma$.\qed

\end{lem}

By the above lemma locally $\Gamma$ is the dual polar space graph $\Delta$.
This implies that the subgraph $\Xi=\Xi(\alpha _4)$ in
$\Gamma$ induced by the vertices incident to $\alpha_4$ is strongly regular
with parameters $v=176$, $k=40$, $\lambda=12$, $\mu=8$ and
$\Xi \cap \Gamma(\alpha_1)$ are all the maximal totally isotropic
subspaces in $V$ which contain $V(\alpha_4)$.

\section{Reconstruction of $\Sigma_3 \times D_4(3).\Sigma_3$}

As above let $G$ be the full automorphism group of
a geometry ${\cal G}$ with diagram $\delta$.
To prove Theorem 1 we have to show that $G \cong M(24)$. Recall that $M(24)$ is generated by a conjugacy class of so-called 3-transpositions. If $m$ is a 3-transposition then it is of order 2 and $C_{M(24)}(m) \cong \langle m \rangle \times M(23)$ where $M(23)$ is another 3-transposition group discovered by B.~Fischer. Two distinct 3-transpositions either commute or generate an $\Sigma_3$-subgroup whose
normalizer $T$ in $M(24)$ is isomorphic to
$\Sigma_3 \times D_4(3).\Sigma_3$. The aim of the present section is to reconstruct $T$ inside $G$.

Under the action of $\bar G_1 \cong O_7(3)$ the set of 1-dimensional subspaces of $V$ (the projective points) split into 3 orbits: $S^0$ (isotropic points), $S^+$ (plus points) and $S^-$ (minus points). The orbit $S^0$ naturally corresponds to the elements of type 4 in ${\cal G}_1$.
Let $W \in S^+$. Then $W$ is non-singular, $V=W \oplus W^{\perp}$ and $W^{\perp}$
contains a maximal totally isotropic subspace of $V$.
Let $M=C_{G_1}(W)$ and $\bar M = M/K_1$. Then then the action
of $\bar M \cong O^+_6(3) \cong L_4(3).2_2$ on $W^{\perp}$
is faithful and coincides with the automorphism group of the restriction of $\phi$ to
$W^{\perp}$. Further 
$N_{G_1}(W)$ is an extension of $M$ by an involution $m$ which centralizes $W^{\perp}$ and inverts $W$.

There are 80 maximal totally isotropic subspaces in $W^{\perp}$ which form a single $M$-orbit
and
under the action of $O^2(M) \cong 3^7.L_4(3)$ they split into
two orbits of length 40 each. If the elements from different
orbits are assumed to be of different types, then together with
the set of 1-dimensional singular subspaces contained in
$W^{\perp}$ they form the parabolic geometry of $O^2(M)/K_1 \cong L_4(3)$
with respect to the natural incidence relation. This is the geometry of type $A_3$ (or $D_3$) over $GF(3)$ and we will denote it by ${\cal H}(\alpha_1,W)$.

We assume that $V(\alpha_2)$ is contained in $W^{\perp}$.
Then $V(\alpha_3)$ is also in $W^{\perp}$ and there is
exactly one maximal totally isotropic subspace $Z$ in $W^{\perp}$
such that $V(\alpha_2) \cap Z =V(\alpha_3)$. Moreover, $V(\alpha_2)$ and $Z$
are in different $O^2(M)$-orbits and $\{V(\alpha_2),V(\alpha_4),Z\}$
is a maximal flag in the
geometry ${\cal H}(\alpha_1,W)$.

Let $\beta$ and $\gamma$ be the vertices from $\Gamma(\alpha_1)$ such that
$\{\alpha_1,\beta\}=\alpha_2$,
$V(\{\alpha_1,\gamma\})=Z$ (here $\{\alpha_1,\beta\}$
and $\{\alpha_1,\gamma\}$ are edges of $\Gamma$ also considered as elements of type 2 in ${\cal G}$).
Since ${\rm dim}~(V(\alpha_2) \cap Z)=2$, $\alpha_1$, $\beta$ and $\gamma$ form a triangle in $\Gamma$ by Lemma~\ref{residues_17}.


\begin{lem} \label{triality_1}

The setwise stabilizer in $C_{G_4}(W)$ of the triangle $\{\alpha_1,\beta,\gamma\}$ induces the symmetric group
$\Sigma_3$ on the vertices of the triangle.

\end{lem}


\pf Let $\Theta$ be the set of vertices of $\Gamma$ such that for $\lambda \in \Theta$, $W$ is contained
in $O_3(G_{\lambda})$ and
is of plus type with respect to the unique quadratic form on $O_3(G_{\lambda})$ preserved
by $G_{\lambda}$. Since $G_1$ acts transitively on $S^+$, $C_G(W)$ acts transitively on
$\Theta$. Since $O^2(M)$ acts flag-transitively
on ${\cal H}(\alpha_1,W)$ and $M/O^2(M)$ induces a diagram automorphism of the geometry,
the result follows.\qed

\medskip

Let $a_1=\beta$, $a_2=\alpha_4$, $a_3=\gamma$, $a_4=\alpha_1$. Let $P_i$ be the intersection
of the centralizers
$C_{G_{a_j}}(W)$ for all $1 \le j \le 4$, $j \ne i$, in other words $P_i$ is the
largest subgroup in $G$ which centralizes $W$ and stabilizes $a_j$ for every
$1 \le j \le 4$, $j \ne i$.
Put $\bar P_i=P_i/O_3(P_i)$ and
$\bar P_{ij}= \langle P_i,P_j \rangle /O_3(\langle P_i,P_j \rangle)$, $1 \le i,j \le 4$.

\begin{lem} \label{triality_2}

$\bar P_i \cong GL_2(3)$ for $1 \le i \le 4$. If $i \ne 2$ and $j \ne 2$ then
$\bar P_{ij} \cong (SL_2(3)*SL_2(3)).2$ and $\bar P_{2i} \cong L_3(3)$
for $i=1$, $3$ and $4$.

\end{lem}

\pf Let $1 \le i < j \le 4$. By Lemma~\ref{triality_1} we can assume that
$i \ne 4$ and $j \ne 4$. In this case
$P_i,P_j \le O^2(M)$ where $M=C_{G_1}(W)$ as above.
Then $P_i$ and $P_j$ are minimal parabolic subgroups associated with the (unfaithful) action of $O^2(M)$ on ${\cal H}(\alpha_1,W)$ and the result follows
from the standard facts about
the action of $L_4(3)$ on its parabolic geometry (that is on the projective space $PG(3,3)$).\qed

\medskip

Notice that for every $1 \le i \le 4$, $P_i$ is contained either in $C_{G_1}(W)$ or in
$C_{G_4}(W)$ or in both.

\begin{lem} \label{triality_3}

Let $T$ be the subgroup in $G$ generated by $N_{G_1}(W)$ and
$N_{G_4}(W)$.
Then $T \cong \Sigma_3 \times D_4(3).\Sigma_3$.

\end{lem}

\pf Let $P=\langle P_i \mid 1 \le i \le 4 \rangle$. It is easy to see that
$W$ is the largest subgroup which is contained and normal in every $P_i$ for
$1 \le i \le 4$.
This and Lemma~\ref{triality_2} imply that $\{P_i \mid 1 \le i \le 4 \}$ is a parabolic system of type $D_4$ over
$GF(3)$ and by \cite{tit} $P/W \cong D_4(3)$. Since the
3-part of the Schur multiplier of $D_4(3)$ is trivial, $P \cong 3 \times D_4(3)$. There is an element
$\tau_{13} \in M \setminus O^2(M)$ which normalizes $P_2$ and $P_4$  and permutes $P_1$ and $P_3$.
By Lemma~\ref{triality_1} there
are similar elements $\tau_{14}$ and $\tau_{34}$ and these three elements generate $\Sigma_3$ modulo $P$.
This shows that the centralizers of $W$ in $G_1$ and $G_4$ generate in $G$ a subgroup
$T_1 \cong 3 \times D_4(3).\Sigma_3$. Finally $T$ contains $T_1$ with index 2 and since the outer automorphism group of $D_4(3)$ (isomorphic to $\Sigma_4$) does not contain $\Sigma_3 \times 2$, the result follows.\qed

\medskip

\section{Reconstruction of $2 \times M(23)$}
In this section we reconstruct in $G$ a subgroup $Q$ isomorphic to the
centralizer $2 \times M(23)$ of a 3-transposition in $M(24)$. The reconstruction is based on
enumeration by a computer of the cosets of a subgroup in
a group given in terms of generators and relations.

We start with the Steinberg presentation for
$D \cong D_4(3)$ to obtain generators and relations for
$T \cong \Sigma_3 \times D_4(3).\Sigma_3$.

We use the numbering for the
fundamental system $\Pi=\{1,2,3,4\}$ in the root system of type
$D_4$ as in the previous section, so that 2 is the central node in the
Dynkin diagram. Then the positive roots correspond to multisets from the
following family (here, for instance, $12234$ stays for the multiset
$\{1,2,2,3,4\}$):
%
$${\cal J}=\{1,2,3,4,12,23,24,123,124,234,1234,12234\}.$$
%
With every $I \in {\cal J}$ a subgroup $X_I$ of order 3 in
$D \cong D_4(3)$ (a root subgroup)
is associated so that such subgroups taken for all $I \in {\cal J}$ generate
a Sylow 3-subgroup of $D$. The root subgroups
satisfy the following commutator relations:
%
$$ [X_I,X_J]=\left\{\begin{array}{ll}
                       X_{I \cup J}, & \mbox{if $I \cup J \in {\cal J}$}; \\
                       1, & \mbox{otherwise.}
\end{array}
\right.
$$
%
Here $\cup$ is understood as the union operation on multisets.

For every fundamental root $i \in \Pi$ we have a Weyl element $w_i$ in $D$,
which is of order 4 and together with
$X_i$ generate in $D$ a subgroup isomorphic to $SL_2(3)$. In addition
for $I \in {\cal J}$ and $I \ne \{i\}$ we have
%
$$ X_I^{w_i}=\left\{\begin{array}{ll}
                       X_{I \cup \{i\}}, & \mbox{if $I \cup \{i\} \in {\cal J}$}; \\
                       X_{I \setminus \{i\}}, & \mbox{if $I \setminus \{i\} \in {\cal J}$}; \\
                       X_{I}, & \mbox{otherwise.}
\end{array}
\right.
$$
%
The product of two Weyl elements $w_i$ and $w_j$ is an element of order 3 if $i$ and
$j$ are adjacent in the Dynkin diagram of $D_4$ and they commute otherwise.

To obtain a concrete realization of the Steinberg presentation for $D_4(3)$,
in each root subgroup $X_I$ for $I \in {\cal J}$  we have to choose a
generator $x_I$ (a root element) and the concrete form of the relations depends on this choice.
We have chosen the generators so that the non-trivial commutator relations
among the root elements are determined by the following ones:
%
$$[x_{1},x_{2}]x_{12},~~[x_{3},x_{2}]x_{23},~~[x_{1},x_{23}]x_{123},~~
[x_{1},x_{24}]x_{124},~~[x_{1},x_{234}]x_{1234},~~
[x_{2},x_{4}]x_{24},$$
$$[x_{2},x_{1234}]x_{12234},~~
[x_{3},x_{24}]x_{234},~~[x_{3},x_{124}]x_{1234},~~
[x_{4},x_{23}]x_{234}^{-1},~~
[x_{4},x_{123}]x_{1234}^{-1},$$
$$[x_{234},x_{12}]x_{12234},~~[x_{124},x_{23}]x_{12234},~~
[x_{123},x_{23}]x_{12234},~~[x_{12},x_{4}]x_{124}~~, [x_{12},x_{3}]x_{123}^{-1}.$$
%
For $1 \le i \le 4$ the $SL_2(3)$-subgroup generated by $x_i$ and
$w_i$ is given by the following relations:
%
$$(x_{i}^{w_{i}}x_{i}^{-1})^4,~~[(x_{i}^{w_{i}}x_{i}^{-1})^2,x_{i}],~~
(x_{i}^{w_{i}}x_{i}^{-1})^2w_{i}^2,~~w_{i}^{-1}x_{i}x_{i}^{w_{i}}x_{i}.$$
%
The center of $\langle x_i, w_i \rangle$ is generated by $w_i^2$.

For $1 \le i \le 4$ and $I \in {\cal J}$ the Weyl element $w_i$ commutes with
the root element $x_I$ if $I \cup \{i\} \notin {\cal J}$ and
$I \setminus \{i\} \notin {\cal J}$ and the non-trivial conjugation relations
are determined by the following:
%
$$x_2^{w_{1}}x_{12}^{-1},~~x_{23}^{w_{1}}x_{123}^{-1},~~
x_{24}^{w_{1}}x_{124}^{-1},~~x_{234}^{w_{1}}x_{1234}^{-1},~~
x_{1}^{w_{2}}x_{12},~~x_{3}^{w_{2}}x_{23},~~x_{1234}^{w_{2}}x_{12234}^{-1},~~
x_{4}^{w_{2}}x_{24}^{-1},$$
%
$$x_{12}^{w_{3}}x_{123}^{-1},~~
x_{2}^{w_{3}}x_{23}^{-1},~~x_{24}^{w_{3}}x_{234}^{-1},~~
x_{124}^{w_{3}}x_{1234}^{-1},~~
x_{2}^{w_{4}}x_{24},~~x_{23}^{w_{4}}x_{234},~~
x_{12}^{w_{4}}x_{124},~~
x_{123}^{w_{4}}x_{1234}.$$
%

The above relations (where it is also assumed that $x_I^3=w_i^4=1$ for $I \in
{\cal J}$ and $1 \le i \le 4$) constitute the Steinberg presentation for
the universal Lie group of type $D_4$ over $GF(3)$, that is of
an extension of $D_4(3)$ by a center of order 4. In order to obtain
a presentation for the simple group we put the additional
relations
%
$$w_1^2=w_3^2=w_4^2.$$
%

Now to obtain a presentation for $T \cong \Sigma_3 \times D_4(3).\Sigma_3$
we have to add automorphisms of $D$ inducing $\Sigma_3$ and
elements generating another copy of
$\Sigma_3$ as a direct factor. It is clear that $D= \langle x_i,~w_j \mid
1 \le i \le 4,~1 \le j \le 4 \rangle$ so it is sufficient to define
the automorphisms on the $x_i$ and $w_j$. The automorphisms
generating $\Sigma_3$ correspond to symmetries of the Dynkin diagram and
can be generated by
elements $t$ and $\nu$ satisfying the relations:
%
$$t^2,~~w_{1}^tw_{3}^{-1},~~[t,w_2],~~[t,w_4],~~
x_{1}^tx_{3}^{-1},~~[t,x_2],~~[t,x_4]$$
%
and
%
$$\nu^3,~~(t\nu)^2,~~w_{1}^\nu w_{3}^{-1},~~w_{3}^\nu w_{4},~~
x_{1}^\nu x_{3}^{-1},~~x_{3}^\nu x_{4},~~x_{2}^{\nu}x_2^{-1},~~
w_{2}^{\nu}w_2^{-1}.$$
%

The $\Sigma_3$ direct factor comes by adjoining the
generators $m$ and $\varrho$ satisfying the following
obvious relations, where $1 \le i \le 4$ and $I \in {\cal J}$:
%
$$m^2,~~[m,w_i],~~[m,x_I],~~[m,t],~~[m,\nu];$$
$$\varrho^3,~~(\varrho m)^2,~~
[\varrho,w_i],~~[\varrho,x_I],~~
[\rho,t],~~[\varrho,\nu].$$
%

If $P_i$ is the minimal parabolic subgroup of
$D \cong D_4(3)$ as in the previous section, then $O^2(P_i)$
is generated by the root elements $x_{I}$ for all $I \in {\cal J}$ together
with the Weyl element $w_i$, and the root elements other
than $x_i$ generate $O_3(P_i)$, $1 \le i \le 4$. The elements $t$ and $\nu$ normalize
$P_2$ and conjugate the parabolics $P_1$, $P_3$ and $P_4$ in the way
they conjugate the corresponding Weyl elements.



We are going to produce $Q \cong 2 \times M(23)$ as the subgroup
in $G$ generated by
$C_T(m)$ and $C_{G_4}(m)$. The former of the centralizers is isomorphic to
$2 \times D_4(3).\Sigma_3$; it is generated by all the above generators
of $T$ except for $\varrho$ subject to all the defining relations of $T$,
except for the relations which involve $\varrho$.

Let us discuss $C_{G_4}(m)$. First of all, it follows directly from the construction of $T$ in the
previous section that $T \cap G_4$ contains the normalizer $T_2$ in $T$
of the maximal parabolic subgroup of $D_4(3)$ generated by
$P_1$, $P_3$ and $P_4$.
Since $T_2$ is maximal
in $T$ we have $T_2=T \cap G_4$. In order to modify the above presentation for
$T$ into a presentation for $T_2$ we have to drop the generator $w_2$ together
with all the relations which involve $w_2$ and to introduce a new generator $h$ (which corresponds
to the square of $w_2$ in $T$) and the following set of relations:
$$h^2;~(w_{i}h)^2~{\rm for}~i=1,3,4;~~[h,t],~[h,\nu],~[h,m],~[h,\rho];$$
$$(x_Ih)^2,~{\rm if}~I \cup \{2\} \in {\cal J}~
{\rm or}~I \setminus \{2\} \in {\cal J}~{\rm and}~
[x_I,h],~{\rm otherwise}.$$
%

Similarly by dropping the generator $\varrho$ together with the relations
which involve $\varrho$ we obtain a
presentation for $C_{T_2}(m)$.

One can see in \cite{atlas} that $C_{T_2}(m) \cong \langle m \rangle \times
3_+^{1+8}2^{1+6}_-3_+^{1+2}(2 \times \Sigma_3)$,
in particular $O_3(C_{T_2}(m))$ is extraspecial of order $3^9$. Let us
locate $m$ and $C_{T_2}(m)$ in $G_4$. Comparing the 3-parts in the
orders of $C_{T_2}(m)$ and $U_5(2):2 \times 2 \cong G_4/O_3(G_4)$ we see
that $O_3(C_{T_2}(m))$ is contained in $O_3(G_4)$ which means that $m$ has 8-dimensional
centralizer in $O_3(G_4)/Z(O_3(G_4)) \cong 3^{10}$. Comparing this
with \cite{jlpw}, we conclude that $m$ is the product of an involution from
$O_{3,2}(G_4)$ (acting fixed-point freely on $O_3(G_4)/Z(O_3(G_4))$ and a
central involution from $O^2(G_4) \cong 3^{1+10}_+.U_5(2)$. This shows
(compare \cite{atlas}) that $C_{G_4}(m) \cong \langle m \rangle.3^{1+8}_+2^{1+6}_-3_+^{1+2}GL_2(3)$,
which means that $C_{G_4}(m)$ contains $C_{T_2}(m)$ with index 4. In more
details the situation is the following. 
Let $R=O_{3,2}(C_{G_4}(m)) \cong \langle m \rangle \times 
3^{1+8}_+2^{1+6}_-$. Then
$C_{G_4}(m)/R$ and $C_{T_2}(m)/R$ are, respectively, a maximal 3-local parabolic subgroup
and the
corresponding Borel subgroup in ${\rm Out}(R/O_3(R)) \cong Sp_4(3)$.

Thus $[C_{G_4}(m):C_{T_2}(m)]=4$ and $C_{G_4}(m) \cong   
\langle m \rangle.
3_+^{1+8}2_-^{1+6}3_+^{1+2}GL_2(3)$ while
$C_{T_2}(m) \cong  \langle m \rangle \times 
3_+^{1+8}2_-^{1+6}3_+^{1+2}(2 \times \Sigma_3)$. Let us
introduce some subgroups in $C_{T_2}(m)$:
$$X_1 = \langle x_{12234}, x_{1234}, x_{123}, x_{124}, x_{234}, x_{23},
x_{12}, x_{24}, x_2 \rangle,$$
$$X_2 = \langle w_1,w_3,w_4, w_1^{x_1}, w_3^{x_3},
w_4^{x_4} \rangle,$$
$$X_3 = \langle x_1x_3x_4^{-1}, \nu, x_1x_4 \rangle,~~
X_4=\langle t,h,x_4 \rangle.
$$
%
Then $X_1=O_3(C_{T_2}(m)) \cong 3_+^{1+8}$, $X_2$ is a complement
to $X_1 \times \langle m \rangle$ in $O_{3,2}(C_{T_2}(m))$, which is isomorphic to the central
product of three copies of the quaternion group, that is to $2_-^{1+6}$,
$X_3 \cong 3_+^{1+2}$ and $X_4 \cong (2 \times \Sigma_3)$. So that
$C_{T_2}(m) = \langle m \rangle \times X_1X_2X_3X_4$.
Our
intention is to obtain a presentation for $C_{G_4}(m)$ by adjoining to
$C_{T_2}(m)$  an element $t_1$ so that
$\langle t_1,x_4,h \rangle (\langle m \rangle X_1X_2X_3)/
(\langle m \rangle X_1X_2X_3) \cong GL_2(3)$.
To do so we have to determine the action of $t_1$ on $X_3$, $X_2$ and $X_1$.
After that,
we will find a relation between $t_1$ and $w_2$.
Eventually we obtain generators and relations for the subgroup $Q$ in
$G$ generated by $C_T(m)$ and $C_{G_4}(m)$ and identify $Q$ with $2 \times M(23)$ using coset
enumeration.

The additional generator $t_1$ can
be taken so that $\langle t_1,x_4,h\rangle \cong GL_2(3)$ and
the following relations hold:
%
$$(t_1^{x_4})^2t_1^2,~~[t_1,t_1^{x_4}]t_1^2,~~t_1^{x_4^{-1}}t_1t_1^{x_4},~~
[t_1,t],~~t_1^ht_1.$$
%
Then $t_1$ is of order 4 and $t_1^2$ must be equal to $t$, $tm$, $w_1^2t$, or $w_1^2tm$. Since $w_1^2$ is in the
center of $X_2X_3$, the action of $t_1$ on $X_2X_3$ is independent on the exact form of $t_1^2$.
%
Since $\langle t_1,x_4,h \rangle \cong GL_3(2)$ is the full outer automorphism group of $X_3$,
the action of $t_1$ on $X_3$ is uniquely determined and we get
the following relations :
%
$$[x_1x_3x_4^{-1}, t_1],~~ (x_1x_4)^{t_1}\nu^{-1}x_1x_3x_4^{-1}.$$
%
Now we determine the action of $t_1$ on $X_2$. One can see
that $C_{X_2}(\langle t, h \rangle)/\langle w_1^2 \rangle = \langle w_1w_3, w_1^2 \rangle/\langle w_1^2 \rangle$, which shows that $\langle t_1, t_1^{x_4} \rangle$
has to centralize $w_1w_3$. In addition $[t_1,x_1x_3x_4^{-1}] = 1$,
$t_1^h = t_1^{-1}$ and $t_1^2 \in \{t, tm, w_1^2t, w_1^2tm\}$
which imply the following relations :
%
$$[w_1w_3,t_1],~~ [w_1^{x_1}w_3^{x_3}, t_1],~~ [w_1^2,t_1],~~
[t_1^{x_1x_3},w_1w_3],~~ [t_1^{x_1x_3},w_1^{x_1}w_3^{x_3}],$$
$$
[t_1^{x_1x_3},w_1^2],~~
(w_3^{t_1})^{-1}(w_1w_3w_3^{x_3}w_4^{-1}(w_4^{-1})^{x_4})^{x_1^{-1}x_3^{-1}x_4}w_1^2.$$
%

Now it is the right time to decide whether $t_1^2 = t$, $tm$, $w_1^2t$, or $w_1^2tm$. We can
calculate the action of $t$ and $w_1^2t$ on $Y = X_1/\langle  x_{12234} \rangle$.
We get that $|C_Y(w_1^2t)| = 3^2$ while $|C_Y(t)| = 3^6$.
This shows that the images  of $w_1^2$ and $t$ in $U:=O^2(G_4)/O_3(G_4) \cong U_5(2)$
are not conjugated.
On the other hand we have the following :

\begin{lem}
In the above notation the images of $t_1^2$ and $w_1^2$ in $U$ are conjugated.

\end{lem}

\pf We identify $t_1^2$ and $w_1^2$ with their images in $U$ and follow notation
from \cite{atlas}. Then $w:=w_1^2$ is a central involution (that is of type
$2a$) in $U$ and $t_1$ is an element of order 4 in $C_U(w)$ which commutes with
the element $x:=x_1x_3x_4^{-1}$ which is also contained in $C_U(w)$ and acts
fixed-point freely on $O_2(C_U(w))/ \langle w \rangle \cong 2^6$. Then $C_U(wx)$ has
order 1\,296 and hence $wx$ is of type $6cd$. This means that $x$ is of type $3cd$.
The square of $t_1^2x$ is an element of type $3cd$, hence $t_1^2x$ is of type
$6cd$, $6ef$ or $6jk$. In the former two cases the cube is of type $2a$ and in the
latter case it is of type $2b$. But the elements of type $6jk$ are not squares in
$U$. Hence $t_1^2$ is of type $2a$ and the result follows.\qed

\medskip

Since $|C_Y(w_1^2)| = 1$, we see
that $w_1^2$ must be conjugate to $w_1^2t$ or $w_1^2tm$. As $|C_{O_3(G_{4})/Z(O_3(G_{4}))}(w_1^2)| = 3^2$ and $[\rho, w_1^2t] = 1$, we see that $|C_{O_3(G_{4})/Z(O_3(G_{4}))}(w_1^2t)| \geq 3^3$. This gives that $w_1^2$ is conjugate to $w_1^2tm$ and then
%
$$t_1^2 = w_1^2tm.$$
%
Next we determine the action of $t_1$ on $X_1$. Set $Y_1 = [X_1,w_1^2t]$. Then
%
$$Y_1 = \langle x_{1234}, x_{123}, x_{124}x_{234}, x_{12}x_{23}, x_{24}, x_2 \rangle.$$
%
Since $t_1^2 = w_1^2tm$, $h^{t_1} = hw_1^2tm$, $[x_1x_3x_4^{-1}, t_1] = 1$, and
$t_1^{x_4^{-1}}t_1t_1^{x_4} = 1$ we get the relations:
$$x_{12234}^{t_1}x_{12234}^{-1},~~ x_{1234}^{t_1}x_{123}x_{234}x_{124},~~
(x_{124}x_{234})^{t_1}(x_{1234}(x_{12}x_{23})^{-1}x_{24}^{-1})^{-1},$$
$$(x_{12}x_{23}x_{24})^{t_1}x_{123},~~x_{24}^{t_1}(x_{124}x_{234}x_2^{-1})^{-1},~~
x_2^{t_1}(x_{1234}x_{23}^{-1}x_{12}^{-1})^{-1}.$$
%
Now using the fact that $O_2(\langle t_1,x_4,h \rangle )$
acts trivially on $C_{X_1}(w_1^2tm)$,
we see that
%
$$[t_1, x_{124}x_{234}^{-1}] = 1 = [t_1, x_{12}x_{23}^{-1}].$$
%
This completes the set of relations presenting $C_{G_4}(m)$.

To generate the subgroup $Q=\langle C_T(m),C_{G_4}(m) \rangle$ we
need all the generators of $T$ except for $\varrho$ together with the
element $t_1$. Nevertheless, the relations given above still define
an infinite group, namely the free amalgamated product of $C_T(m)$ and $C_{G_4}(m)$.
In fact, for each of the above relations,
all the generators involved in the relation
are contained either in $T$ or in $G_4$ and none of the relations involve both
$t_1$ (which is not in $T$) and $w_2$ (which is not in $G_4$). To find
the missing relation we are going to analyse the parabolic subgroup $G_3$
associated with the action of $G$ on $\cal G$.


First we are going to identify the intersections $G_i \cap T$ for $1 \le i \le 4$.
It follows directly from the construction of $T$ in Section 3, that
$G_1 \cap T$ is the normalizer in $T$ of the subgroup generated by the
minimal parabolics $P_1$, $P_2$ and $P_3$. This means that $G_1 \cap T$ is
generated by all the generators of $T$ except for $w_4$ and $\nu$. Moreover,
$O_3(G_1)$ is contained in $G_1 \cap T$ and it is generated by the
root elements $x_{I}$ such that $I$ contains $4$, and $\varrho$. Similarly, directly from the construction of $T$ we see that
$G_2 \cap T$ is the normalizer in $T$ of the subgroup generated by $P_2$
and $P_3$; $O_3(G_2) \cap T$ is generated by the root
elements $x_{I}$ such that $I$ contains either $1$ or $4$ or both of them.
This means that $O_3(G_2) \cap T$ is of order $3^9$ and hence its index
in $O_3(G_2)$ is $3^4$. The element $\alpha_2$ is identified with
the edge $\{\alpha_1,\beta\}$ of the collinearity graph $\Gamma$ of
$\cal G$. Clearly $G_{\beta} \cap T$ is the normalizer in $T$ of the
subgroup generated by the minimal parabolics $P_2$, $P_3$ and $P_4$,
so that $O_3(G_{\beta})$ is generated by the root elements $x_I$ for
$I$ containing $1$. This and Lemma~\ref{residues_16} imply that
$E:=\langle x_{124},x_{1234},x_{12234},\varrho \rangle$ is the intersection
of $O_3(G_1)$ and $O_3(G_{\beta})$, hence $N_{G_i}(E)=G_i \cap G_2$
for $i=1,~3$ and 4.
Moreover it is easy to see, that in terms introduced in Section 2
we have $V(\alpha_2)=
\langle x_{124}, x_{1234}, x_{12234} \rangle$.

We have already observed that $G_4 \cap T$ is the normalizer in $T$ of the
subgroup generated by $P_1$, $P_3$ and $P_4$, in particular $Z(O_3(G_4)) =
\langle x_{12234} \rangle$.

Let us consider the intersection $G_3 \cap T$. We know that 
$G_3$ is the setwise stabilizer in $G$ of
the subgraph $\Phi$ in the collinearity graph $\Gamma$ of $\cal G$,
induced by the vertices (elements of type 1)
incident to $\alpha_3$. It follows directly from the diagram of ${\cal G}$
that $\Phi$ is a complete subgraph on 5 vertices and that it contains
$\alpha_1$ and $\beta$.
Since $G_{13}$ is the full preimage in $G_1$ of the stabilizer
in $\bar G_1 \cong O_7(3)$ of an isotropic 2-dimensional subspace in
$V$, one can see (compare \cite{atlas}) that $G_{13}/O_3(G_{13}) \cong
\Sigma_4 \times GL_2(3)$ and that $G_{13}$
induces the natural action of $\Sigma_4$ on $\Phi \setminus
\{\alpha_1\}$. This implies that $G_3/O_3(G_3) \cong \Sigma_5 \times GL_2(3)$ and that $G_3$ induces
on $\Xi$ the natural action of $\Sigma_5$.
By Lemma~\ref{residues_17} $\Phi$ is a maximal clique in $\Gamma$ and
each triangle in $\Gamma$ is contained in a unique clique of size 5
which is an image of $\Phi$ under $G$. So without loss of generality
we assume that the triangle $\{\alpha_1,\beta,\gamma\}$ is contained in
$\Phi$ (cf. the terms introduced
before Lemma~3.1).
Moreover this triangle is the intersection of $\Phi$ with
the subgraph $\Theta$ defined in the proof of Lemma~3.1.
This means that $G_3 \cap T$ is the setwise stabilizer
in $G_3$ of the triangle $\{\alpha_1,\beta,\gamma\}$ and
$(G_3 \cap T)O_3(G_3)/O_3(G_3)
\cong GL_2(3) \times \Sigma_3 \times 2$.
Similarly $G_3 \cap T$ is
the setwise stabilizer of $\{\alpha_1,\beta,\gamma\}$ in
$T$, which means that $G_3 \cap T$ is
the normalizer of $P_2$ in $T$. This shows that
$\langle x_2,w_2 \rangle$ is contained in $G_3 \cap T$ and it
maps isomorphically into the
$GL_2(3)$-direct
factor of $G_3/O_3(G_3)$. Notice that $\nu$ is not in $O_3(G_3)$ since
it rotates the vertices $\alpha_1$, $\beta$ and $\gamma$.
Thus $O_3(G_3) \cap T$ is generated by the root
elements $x_I$ for $I \in {\cal J} \setminus \{2\}$ together with $\varrho$.
In particular $O_3(G_3) \cap T$ is of index $3^2$ in $O_3(G_3)$.

Let $V(\alpha_3)$ be the subspace of dimension 2 in
$V=O_3(G_1)$ as introduced in Section 2, such that
$G_1 \cap G_3=N_{G_1}(V(\alpha_3))$. Since $O_3(G_3)$ is contained in
$G_1$, we can calculate in the latter group that $V(\alpha_3)$ is the center of
$O_3(G_3)$. Using the relations for $T$ one can see that
$\langle x_{12234}, x_{1234} \rangle$ is the center of
$O_3(G_3) \cap T$. This shows that
%
$$Z_3:=Z(O_3(G_3))=V(\alpha_3)=\langle x_{12234},x_{1234}\rangle$$
%
and that $\langle x_2,w_2\rangle$ acts faithfully on $Z_3$.

Put $S=C_{O_3(G_3)}(\langle x_2,w_2\rangle)Z_3$.
One can calculate in $G_1$ that $S$ is of order
$3^6$ and that $S/Z_3=Z(O_3(G_3)/Z_3)$. This and direct calculations with
the relations for $T$, show that
$S=\langle x_{12234},x_{1234},x_{234},x_{123},x_{124},\varrho
\rangle$ and that $[S,O_3(G_3)]=Z_3$.

We assume that $\Phi=\{\alpha_1, \beta, \gamma, \delta, \epsilon \}$. Notice
that every pair of vertices from $\Phi$ form an edge of $\Gamma$ and hence
can be considered as an element of type 2 in $\cal G$
incident to $\alpha_3$. Clearly an automorphism from $O_3(G_2) \setminus O_3(G_3)$
stabilizes $\alpha_1$ and $\beta$, permuting $\gamma$, $\delta$ and
$\epsilon$. This means that
$\langle \nu \rangle O_3(G_3)/O_3(G_3)$ and
$O_3(G_2)O_3(G_3)/O_3(G_3)$  both are Sylow 3-subgroups in the
$\Sigma_5$-direct factor  of $G_3/O_3(G_3) \cong \Sigma_5 \times GL_2(3)$.

Let
$W:=C_{S/Z_3}(\langle \nu \rangle)Z_3$. Then direct calculations with the relations for $T$ show that
%
$$W=C_{S/Z_3}(\langle \nu \rangle)Z_3=\langle x_{12234}, x_{1234}, x_{234}x_{123}x_{124}, x_{124}^{-1}x_{234} \rangle.$$
%
As above let $E=O_3(G_1) \cap O_3(G_{\beta})$.
Then $E$ is contained $S$, contains
$Z_3$ and $O_3(G_3)$ centralizes $E/Z_3$. This and order reason give
%
$$E=C_{S/Z_3}(O_3(G_2))Z_3=\langle x_{124}, x_{1234}, x_{12234}, \varrho \rangle.$$
%
Since $\langle \nu \rangle O_3(G_3)/O_3(G_3)$ and $O_3(G_2)O_3(G_3)/O_3(G_3)$
are Sylow 3-subgroups in the $\Sigma_5$-direct factor of
$G_3/O_3(G_3)$, there is an element $y \in G_3 \cap G_4$
which conjugates $W$ to $E$
and by Frattini argument
we can choose $y$ so that $[y,w_2]=1$.
Since $[t_1,W]\subseteq W$ and $G_4 \cap G_2=N_{G_4}(E)$ we see
$t_1^y \in G_2$ and $o(t_1^yw_2)=o(t_1w_2)$.


The relations for $T$ and $C_{G_4}(m)$ describe the actions of
$w_2$ and $t_1$ on $W$. Comparing these actions we observe
that the product $w_2t_1$ induces on $W$ an action of order
$3$ which means that $(w_2t_1)^3$ centralizes $W$.
Hence $(w_2t_1^y)^3$ is in $C_{G_2}(E)$.  The
structure of $G_2$ shows that $C_{G_2}(E)$ is
contained in $O_3(G_2)$ and has order $3^{10}$. On
the other hand the relations for $T$ show that the subgroup
%
$$U_1=\langle x_{12234},x_{1234},x_{234},\varrho,x_{124},x_{123},x_{24},x_{23},
x_3,x_4\rangle$$
%
has order $3^{10}$ and centralizes $W$. Hence
$U_1 = C_{G_2^{y^{-1}}}(W)$ and so $(w_2t_1)^3 \in U_1$. We know that $\langle w_{1}, w_{2} \rangle \cong \Sigma_4$ and so ${w_{1}^2}^{w_2} = w_1^2w_2^2 = w_1^2h$. Further $(w_2t_1)^h = w_2t_1w_1^2tm$. As $t$ and $m$ both centralize $w_2$ and $t_1$, we see $((w_2t_1)^3)^h = (w_2t_1w_1^2)^3tm$. Now 
$$w_2t_1w_1^2w_2t_1w_1^2w_2t_1w_1^2 = w_2t_1w_2w_1^2ht_1w_1^2w_2t_1w_1^2 =$$ 
$$w_2t_1w_2t_1w_1^2tmhw_2t_1w_1^2 = (w_2t_1)^3tm.$$
 So we have that $[h,(w_2t_1)^3]=1$.
%
Hence $(w_2t_1)^3 \in C_{U_1}(h)=\langle x_{234},x_{123},x_{124},
\varrho \rangle$. As both $w_2$ and $t_1$ commute
with $m$ we see that
$(w_2t_1)^3\in \langle x_{234},x_{123},x_{124}\rangle$. As $w_2t_1$
centralizes $(w_2t_1)^3$, we have $(w_2t_1)^3\in \langle x_{124}^{-1}x_{234}
\rangle$. But $t$ centralizes $w_2t_1$ and inverts $x_{124}^{-1}x_{234}$,
so we finally get
%
$$(w_2t_1)^3=1.$$
%

Now the subgroup $Q=\langle C_{T}(m), C_{G_4}(m) \rangle$ is generated
by $t_1$ and all the generators of $T$ except for $\varrho$.
These generators satisfy the relations given above.
The coset enumeration with a computer program implementing the
Todd-Coxeter algorithm has shown that the presentation defines a group
whose order is equal to the order of $2 \times M(23)$ (we enumerated
the cosets of $C_{T}(m)$ in $Q$ and have obtained the index $137\,632$).
Since the relations are uniquely
determined by the conditions on the geometry and since $M(24)$ acts
as the full automorphism group of a geometry of the considered type,
we have the following result.

\begin{lem}

Let $Q$ be the subgroup in $G$ generated by the subgroups
$C_T(m)$ and $C_{G_4}(m)$. Them $Q \cong 2 \times M(23)$.\qed

\end{lem}

\section{The transposition graph}
In this section we complete the proof of Theorem 1 by showing that the
full automorphism group $G$ of the geometry $\cal G$ under consideration
is isomorphic to the Fischer group $M(24)$. We will achieve this by
considering a graph $\Lambda$ on the set of cosets in $G$ of the subgroup
$Q \cong 2 \times M(23)$. We will establish the isomorphism between
$\Lambda$ and the graph on the set of $3$-transpositions of $M(24)$ in
which two transpositions are adjacent if they do not commute (i.e. if
their product is of order 3).

As above let $W = \langle \rho \rangle$ be the subgroup of order 3 in $V=O_3(G_1)$ which is of
plus type with respect to the orthogonal form on $V$ preserved by
$G_1/V$ and $T=\langle N_{G_1}(W),N_{G_4}(W) \rangle$. Then $T \cong
\Sigma_3 \times D_4(3).\Sigma_3$ by Lemma~3.3. Let $m$ be an involution
in $T$ which commutes with the subgroup $D \cong D_4(3)$ in $T$ and
let $Q=\langle C_T(m), C_{G_4}(m) \rangle$. Then $Q \cong 2 \times M(23)$
by Lemma~4.2 and it is straightforward that $Q \cap T=C_T(m) \cong
2 \times D_4(3).\Sigma_3$, so that $[T:T \cap Q]=3$.

Let $R=O_3(G_4)C_{G_4}(m)$. Then directly from the construction of $Q$ in the
previous section we have the following :

\begin{lem} \label{trans_1} The following assertions hold.

\begin{itemize}

\item[{\rm (i)}]
$R \cap T \cong \Sigma_3 \times 3^{1+8}_+2_-^{1+6}3_+^{1+2}(2 \times \Sigma_3)$
and $[R:R \cap T]=12$;

\item[{\rm (ii)}]
$R \cap Q \cong 2 \times 3^{1+8}_+2^{1+6}_-3^{1+2}_+GL_2(3)$ and
$[R:R \cap Q]=9$;

\item[{\rm (iii)}]
$W \le O_3(G_4)$ and $|W \cap Q|=1$.\qed

\end{itemize}

\end{lem}

Define $\Lambda$ to be a graph of the set of (right) cosets of $Q$ in $G$ in which
two cosets are adjacent if and only if they intersect a common
coset of $T$ in $G$. Let $\lambda \in \Lambda$ be the coset containing the
identity element so that $G_{\lambda}=Q$. Let $\Theta$ be the orbit of $T$ on the vertex set of $\Lambda$ which
contains $\lambda$. Since
$[T:T \cap Q]=3$, $\Theta$ contains besides $\lambda$ 
exactly two vertices, say $\mu$ and $\nu$. Then $G_{\lambda} \cap G_{\mu}$ contains the subgroup $D \cong D_4(3)$
from $T$. As one can check with \cite{atlas},
every proper subgroup in $Q$ containing $D$ is contained in $T \cap Q$. This
shows that $G_{\lambda} \cap G_{\mu} \le T$
and we have the following.

\begin{lem} \label{trans_2}

\begin{itemize}

\item[{\rm (i)}] The valency of $\Lambda$ is $275\,264=2 \cdot [Q:Q \cap T]$;

\item[{\rm (ii)}] $\Theta$ induces a triangle in
$\Lambda$ and $T$ is its full stabilizer in $G$;

\item[{\rm (iii)}] $\langle W,m \rangle \cong \Sigma_3$ acts faithfully
on $\Theta$.\qed

\end{itemize}

\end{lem}


Let $\Omega$ be the orbit of $R$ on the vertex set of $\Lambda$ which contains
$\lambda$. By Lemma~\ref{trans_1} (iii) and Lemma~\ref{trans_2} (iii), $\Omega$ contains
$\Theta$ and by Lemma~\ref{trans_1} (ii) $|\Omega|=9$. Let $L$ be the kernel of the
action of $R$ on $\Omega$. By Lemma~\ref{trans_1} (ii) we see that
$O_3(R)/(L \cap O_3(R))$ is elementary abelian of order 9 and acts
regularly in $\Omega$. Hence $R/O_3(R)L$ is a subgroup in
$GL_2(3) \cong Aut(O_3(R)/(L \cap O_3(R)))$. On the other hand $L$ is contained
and normal in both $R \cap T$ and $R \cap Q$, by Lemma~\ref{trans_1} this gives
the following :

\begin{lem} \label{trans_3}

Let $N=R/O_3(R)L$. Then $N \cong GL_2(3)$ and $\langle m \rangle$ maps
onto the center of $N$.\qed

\end{lem}

Let $Q'$ be the commutator subgroup of $Q$, so that
$Q' \cong M(23)$ and $Q= \langle m \rangle \times Q'$.
As a direct consequence of the above lemma we have the
following :

\begin{figure}[t]
\begin{center}
\setlength{\unitlength}{0.0099in}%
\begin{picture}(580,327)(-50,597)
\thicklines
\put( 80,760){\circle{40}}
\put(190,760){\circle{80}}
\put(190,660){\circle{40}}
\put(308,660){\circle{80}}
\put(308,857){\circle{80}}
\put(101,760){\line( 1, 0){ 60}}
\put(190,731){\line( 0,-1){ 50}}
\put(210,780){\line( 1, 1){ 70}}
\put(210,740){\line( 1,-1){ 70}}
\put(308,830){\line( 0,-1){141}}
\put( 75,755){\makebox(0,0)[lb]{\raisebox{0pt}[0pt][0pt]{\twlrm 1}}}
\put(166,755){\makebox(0,0)[lb]{\raisebox{0pt}[0pt][0pt]{\small 218\,400}}}
\put(184,656){\makebox(0,0)[lb]{\raisebox{0pt}[0pt][0pt]{\twlrm 1}}}
\put(288,655){\makebox(0,0)[lb]{\raisebox{0pt}[0pt][0pt]{\small 28\,431}}}
\put(288,852){\makebox(0,0)[lb]{\raisebox{0pt}[0pt][0pt]{\small 28\,431}}}
\put(149,764){\makebox(0,0)[lb]{\raisebox{0pt}[0pt][0pt]{\tiny 1}}}
\put(181,719){\makebox(0,0)[lb]{\raisebox{0pt}[0pt][0pt]{\tiny 1}}}
\put(295,892){\makebox(0,0)[lb]{\raisebox{0pt}[0pt][0pt]{\tiny 22\,400}}}
\put(295,620){\makebox(0,0)[lb]{\raisebox{0pt}[0pt][0pt]{\tiny 22\,400}}}
\put(175,792){\makebox(0,0)[lb]{\raisebox{0pt}[0pt][0pt]{\tiny 173\,200}}}
\put(208,708){\makebox(0,0)[lb]{\raisebox{0pt}[0pt][0pt]{\tiny 22\,599}}}
\put(230,680){\makebox(0,0)[lb]{\raisebox{0pt}[0pt][0pt]{\tiny 173\,600}}}
\put(208,806){\makebox(0,0)[lb]{\raisebox{0pt}[0pt][0pt]{\tiny 22\,599}}}
\put(230,834){\makebox(0,0)[lb]{\raisebox{0pt}[0pt][0pt]{\tiny 173\,600}}}
\put(319,815){\makebox(0,0)[lb]{\raisebox{0pt}[0pt][0pt]{\tiny 22\,400}}}
\put(319,705){\makebox(0,0)[lb]{\raisebox{0pt}[0pt][0pt]{\tiny 22\,400}}}
\end{picture}
\end{center}
\caption{Distribution diagram of $M(23)$ acting of $D_4(3).3$}
\end{figure}

\begin{lem} \label{trans_4}

\begin{itemize}

\item[{\rm (i)}] The subgraph in $\Lambda$ induced by $\Omega$
is complete and $R$ induces a doubly transitive action on $\Omega$;

\item[{\rm (ii)}] both $Q \cap R$ and $Q' \cap R$ induce
$GL_2(3)$ on $\Omega \setminus \{\lambda\}$;

\item[{\rm (iii)}] $Q'$ acts transitively on the set
$\Lambda(\lambda)$ of vertices
adjacent to $\lambda$ and the stabilizer in $Q'$ of $\mu \in \Lambda(\lambda)$
is isomorphic to $D_4(3).3$;

\item[{\rm (iv)}] if $\sigma \in \Omega \setminus \Theta$, then
the order of $Q'_{\mu} \cap Q'_{\sigma}$ is divisible
by $3^{12} \cdot 2^7$.\qed

\end{itemize}

\end{lem}

\medskip

\begin{lem} \label{trans_5}

$G_{\lambda} \cap G_{\mu}$ acting on
$\Lambda(\lambda) \setminus \{\mu\}$ has exactly $4$
orbits: $\Lambda^1(\lambda,\mu)=\{\nu\}$,
$\Lambda^2(\lambda,\mu)$,
$\Lambda^3(\lambda,\mu)$ and
$\Lambda^4(\lambda,\mu)$ with lengths $1$, $218\,400$, $28\,431$ and $28\,431$,
respectively. The intersection diagram of the graph on $\Lambda(\lambda)$
associated with $\Lambda^2(\lambda,\mu)$ is given on Figure~$1$ and
$\Omega \setminus \Theta \subseteq \Lambda^2(\lambda,\mu)$.

\end{lem}


\pf By Lemma~\ref{trans_2} (i) $G_{\lambda}=Q$ acts on $\Lambda(\lambda)$ as
it acts on the cosets of an index 2 subgroup in its maximal subgroup
isomorphic to $2 \times D_4(3).\Sigma_3$ and by Lemma~\ref{trans_4} (ii) $Q'$
acts transitively on $\Lambda(\lambda)$. This specifies the action
under consideration up to similarity. The subdegrees of this action
and the intersection diagrams can be found in \cite{ps}. It is clear that
$\nu$ is stabilized by $G_{\lambda} \cap G_{\mu}$. If $\sigma \in
\Omega \setminus \Theta$ then Lemma~\ref{trans_4} (iv) and straightforward
congruences show that $\sigma \in \Lambda^2(\lambda,\mu)$.\qed

\medskip



\begin{lem} \label{trans_6}

Let $\Xi$ be the graph on the set of $3$-transpositions of $M(24)$
in which two transpositions are adjacent if they are different and
commute. Then $\Lambda$ is isomorphic to the complement of $\Xi$
and $G \cong M(24)$.

\end{lem}

\pf For $\alpha \in \Lambda$ let $m_{\alpha}$ be the unique non-trivial
element in the center of $G_{\alpha} \cong 2 \times M(23)$, so that
$m=m_{\lambda}$. If $\beta \in \Lambda(\alpha)$ then the product of
$m_{\alpha}$ and $m_{\beta}$ has order 3. On the other hand it is clear that
$Q$ contains elements conjugated to $m$ and different from $m$. Hence
$m_{\alpha}$ and $m_{\beta}$ commute for some vertices $\alpha$ and $\beta$
of $\Lambda$ and the diameter of $\Lambda$ is at least 2. Let $\alpha$ be
a vertex adjacent to $\mu$ and at distance 2 from $\lambda$ and let
$\Phi$ the orbit of $Q$ containing $\alpha$. Then by
Lemma~\ref{trans_5} $\Phi$ contains
$\Lambda^3(\lambda,\mu)$ or $\Lambda^4(\lambda,\mu)$ or both.
The intersection diagram
of the graph on $\Lambda(\lambda)$ associated with
$\Lambda^2(\lambda,\mu)$ shows that $\alpha$ is adjacent to
173\,600 vertices from
$\Lambda^2(\mu,\lambda) \subseteq \Lambda(\lambda)$. Hence
$\Phi$ contains at most
$275\,264 \cdot (28\,431+28\,431)/173\,600 < 90\,162$. It is clear that
$Q$ induces a non-trivial action on $\Phi$. Comparing this with the indices
of maximal subgroups in
$Q' \cong M(23)$ \cite{atlas} we conclude that $Q'$ acting on
$\Phi$ has one or two
orbits of length $31\,671$ and the action on each of the orbits is similar to
the action of $M(23)$ on the cosets of its subgroup $2 \cdot M(22)$. Suppose that
$|\Phi|=2 \cdot 31\,671$. Since $\alpha$ is adjacent to at least
$173\,600$ vertices from $\Lambda(\lambda)$, in this case
$\Lambda^3(\mu,\lambda) \cup \Lambda^4(\mu,\lambda) \subseteq \Phi$ and
$\alpha$ is adjacent to
$247\,104$ vertices from $\Lambda(\lambda)$. We may assume that
$\alpha \in \Lambda^3(\mu,\lambda)$. Then the intersection diagram shows
that $\alpha$ is adjacent to $22\,400$ vertices from $\Lambda^3(\mu,\lambda)$
and to $22\,400$ vertices from $\Lambda^4(\mu,\lambda)$. Since
$247\,104+2 \cdot 22\,400=291\,904$ is larger than the valency
of $\Lambda$, which is $275\,264$, this is impossible. Hence
$|\Phi|=31\,671$ and $Q'$ acts transitively on $\Phi$,
$\Lambda(\mu) \cap \Phi =\Lambda^3(\mu,\lambda)$ and $\alpha$ is
adjacent to $247\,104$ vertices from $\Lambda(\lambda)$. Comparing
this with the valency of $\Lambda$ we see that $\alpha$ is adjacent to
$28\,160$ vertices outside $\Lambda(\lambda)$. The
subdegrees of the action of $M(23)$ on the cosets of
$2 \cdot M(22)$ are well known to be $1$, $3\,510$  and
$28\,160$. Since  $\alpha$ is adjacent to $22\,400$ vertices from
$\Lambda^3(\mu,\lambda)$ we see that $\Lambda=\{\lambda\}
\cup \Lambda(\lambda) \cup \Phi$. Now the complement of $\Lambda$
has valency $31\,671$ and locally it is the commuting graph of transpositions
in $M(23)$. Also for any two vertices $\alpha$ and $\beta$
in $\Lambda$ the product $m_{\alpha}m_{\beta}$ has order at most 3, which
means that $\{m_{\alpha} \mid \alpha \in \Lambda\}$ is a class of
$3$-transpositions in $G$ and to complete the proof we can apply one of
the available characterizations of $M(24)$, for instance \cite{bw}.\qed


\bigskip

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A.A.~Ivanov,

Department of Mathematics,

Imperial College,

180 Queen's Gate, London,

SW7 2BZ, UK


\bigskip
G.~Stroth,

Institut f\"ur Algebra und Geometrie,

Fachbereich Mathematik u. Informatik,

Martin-Luther Universit\"at Halle-Wittenberg,

06099 Halle, Germany


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