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\title{\bf Local Fusion Graphs and Sporadic Simple Groups}

\author{John Ballantyne \\
\small School of Mathematics\\[-0.8ex]
\small University of Manchester\\[-0.8ex]
\small United Kingdom\\
\small\tt john.ballantyne@manchester.ac.uk\\
\and Peter Rowley\\
\small School of Mathematics\\[-0.8ex]
\small University of Manchester\\[-0.8ex]
\small United Kingdom\\
\small\tt peter.rowley@manchester.ac.uk
}

\date{\dateline{Apr 17, 2014}{Jul 23, 2015}{Jul 31, 2015}\\
\small Mathematics Subject Classifications: 20D08, 05C25}

\begin{document}

\maketitle
  
\begin{abstract}
For a group $G$ with $G$-conjugacy class of involutions $X$, the local fusion graph $\mathcal{F}(G,X)$ has $X$ as its vertex set, with distinct vertices $x$ and $y$ joined by an edge if, and only if, the product $xy$ has odd order. Here we show that, with only three possible exceptions, for all pairs $(G,X)$ with $G$ a sporadic simple group or the automorphism group of a sporadic simple group, $\mathcal{F}(G,X)$ has diameter $2$.

 \bigskip\noindent \textbf{Keywords:} Local Fusion Graph; Sporadic Simple Group; Diameter
\end{abstract}


\section{Introduction}\label{intro}
Suppose that $G$ is a finite group with $X$ a $G$-conjugacy class of
involutions (that is, a $G$-conjugacy class of elements of order $2$). The local fusion graph, $\mathcal{F}(G,X)$, is the
graph whose vertex set is $X$ with distinct vertices $x$ and $y$
joined by an edge whenever $xy$ has odd order. Equivalently, $x$ and
$y$ are joined if $\langle x,y \rangle$ is a dihedral group of order
$2m$, $m$ odd, in which case $x$ and $y$ are conjugate in $\langle
x,y \rangle$, explaining the graph's epithet. It is clear that $G$
induces graph automorphisms (by conjugation) on $\mathcal{F}(G,X)$
and acts transitively on the vertices. Various properties of local
fusion graphs have been investigated in \cite{B} and \cite{BGR}. In \cite{BGR} local fusion graphs for finite symmetric
groups are studied, the main result being that they always have
diameter two, provided that the degree is at least five. The other
finite irreducible Coxeter groups are dealt with in \cite{B}, which also
considers the possible diameters. There, examples are given of groups
which have local fusion graphs whose diameter can be arbitrarily
large. Local fusion graphs have
even being pressed into service \cite{BR2} in the area of
computational algebra, while graphs of a similar nature appear in
\cite{DG}.

Our main result determines the diameter of the local fusion graphs for
(most of) the sporadic simple groups and their automorphism groups. We shall follow the notation
and conventions of the \textsc{Atlas}~\cite{atlas} and also use it as a
source of data on the sporadic simple groups -- $\mathrm{Diam}(\mathcal{F}(G,X))$
will denote the diameter of ${\cal F} (G,X)$.

\begin{thm} \label{Diam=2}
Suppose that $K$ is a sporadic simple group, $G$ a subgroup of $\mathrm{Aut}(K)$ containing $K$, and $X$ is a $G$-conjugacy class of involutions. If $(K,X) \neq (\mathbb{B},2C/2D), (\mathbb{M},2B)$ then $\mathrm{Diam}(\mathcal{F}(G,X)) = 2$.
\end{thm}

In Theorem \ref{Diam=2} we note that $[G:K]=1$ or $2$ (see \cite{atlas}), so $G=K$ or $\mathrm{Aut}(K)$. Before outlining the contents of this paper we introduce some more
notation. Suppose that $G$ is a finite group and $X$ is a
$G$-conjugacy class of involutions. Let $x,y \in \mathcal{F}(G,X)$
and $i \in \mathbb{N}\cup \{0\}$. We shall use $d(x,y)$ to denote
the distance between $x$ and $y$ in $\mathcal{F}(G,X)$, and the
\emph{$i$-th disc} of $\mathcal{F}(G,X)$, $\Xi_i(x)$, is defined by
\begin{displaymath} \Xi_i(x) = \{ y \in X \mid d(x,y)=i \}.
\end{displaymath} So $\Xi_0(x) = \{ x\}$, while $\Xi_1(x)$
consists of all the neighbours of $x$ in $\mathcal{F}(G,X)$. From
now on we fix $t \in X$. For a $G$-conjugacy class $C$ we put
\begin{displaymath} X_C = \{ x \in X \mid tx \in C \},
\end{displaymath} and note that $X_C$ is invariant under the action of $C_G(t)$ by conjugation.
We shall sometimes adapt the \textsc{Atlas}~\cite{atlas} notation for
conjugacy classes by adding a subscript which indicates the group
whose conjugacy class this is. So, for example, $2C_{HS:2}$
indicates that we are considering the $2C$ conjugacy class (as in the \textsc{Atlas}) of
$HS:2$.

For most of the sporadic groups, calculations employing GAP
\cite{GAP4} and \textsc{Magma}~\cite{magma} yield the diameter of
$\mathcal{F}(G,X)$ -- the details of these being given in Section
2.1. When $(G,X) =(\mathbb{B},2A)$ or $(\mathbb{M},2A)$, by
extracting appropriate subgroup information from the \textsc{Atlas}
we demonstrate in Propositions \ref{Baby2A} and \ref{Monster} that
$\mathcal{F}(G,X)$ has diameter $2$. This approach works largely
for the following disparate reasons: the number of $C_G(t)$-orbits of $X$ is small; these orbits are of the form $X_C$ for some conjugacy class $C$ of $G$; and it is possible to identify $G$-conjugacy classes in smaller subgroups. However, in the case of $(G,X) =
(\mathbb{B},2B)$, for example, $X_{2B}$ is not a $C_G(t)$-orbit, so we choose in Proposition \ref{Baby2B}
to investigate $\mathcal{F}(G,X)$ using the detailed description of
the point-line collinearity graph given in \cite{RW1}. As a
by-product, for this case our proof is computer-free. Furthermore, this angle of attack will undoubtedly lead to a sharper picture of the local fusion graph for $(\mathbb{M},2B)$. Indeed, for
$(G,X) = (\mathbb{M},2B)$ it can be shown that the diameter of the
local fusion graph is at most $6$. This follows from \cite{R}, where it is shown that the commuting involution graph of $\mathbb{M}$ on the $2B$ conjugacy class has diameter $3$, when combined with the observation that two commuting $2B$ involutions are distance $2$ apart in $\mathcal{F}(\mathbb{M},2B)$. However, this bound is almost certainly not the best possible. Finally, we remark that for the three graphs not covered by Theorem \ref{Diam=2}
the permutation rank of $G$ on $X$ is very large.

We thank the referee for their careful reading of this paper, and their many helpful suggestions.


\section{Diameter of $\mathcal{F}(G,X)$}\label{Diameters}

Quite a number of the sporadic simple groups and their conjugacy
classes will be dealt with using the next two results. The first is
a consequence of some well-known character theoretic results and the
second is an elementary observation relating to the size of the
first disc of a regular graph. For Lemma \ref{structureconstants}
we require some more notation, so suppose  $G$ is a finite group,
with conjugacy classes $\mathcal{K}_1,\ldots,\mathcal{K}_\ell$ and
corresponding class sums $K_1,\ldots,K_\ell$ in the group algebra
$\mathbb{C}G$. Also let $a_{ijk}$ be defined by
\begin{displaymath} K_iK_j = \sum_{k=1}^\ell a_{ijk} K_k.
\end{displaymath} The $a_{ijk}$ are referred to as the structure
constants of $G$, and may be calculated from the character table of
$G$ (see Chapter 28 of \cite{JL} for further details).



\begin{lemma}\label{structureconstants}
Suppose $G$ is a finite group with $X$ a $G$-conjugacy class of
involutions. Assume that $X = \mathcal{K}_i$. Then for $x \in X$ we
have \begin{displaymath} |\Xi_1(x)| = \sum_j a_{jii},
\end{displaymath} where the sum is over all $j$ such that the
conjugacy class $\mathcal{K}_j$ contains elements of odd order
(excluding the conjugacy class of the identity element).
\end{lemma}

\begin{proof} Let $x \in X$. Then $a_{jii}$ is the number of pairs $(z,y)$ where $z \in \mathcal{K}_j$ and $y \in \mathcal{K}_i=X$ are such that $zy=x$. So, letting $\mathcal{K}_j$ run over all $G$-conjugacy classes of non-trivial odd order elements, $\sum_j a_{jii}$ is the number of $y \in X$ such that $xy$ has odd order, whence the lemma holds.
\end{proof}

\begin{lemma}\label{bigdisc}
Suppose that $\mathcal{X}$ is a finite regular graph with valency $d$. If $d > |\mathcal{X}|/2$, then $\mathcal{X}$ is connected and has diameter at most $2$.
\end{lemma}

\begin{proof}
For $x \in \mathcal{X}$, let $\Delta(x)$ denote the neighbours of $x$ in $\mathcal{X}$. So $|\Delta(x)|=d$. Let $x \in \mathcal{X}$. Since $|\Delta(x)|=d > |\mathcal{X}|/2$, the regularity of $\mathcal{X}$ implies connectedness. Suppose there exists $y \in \mathcal{X}$ such that $x$ and $y$ are distance $3$ apart. Then $\Delta(x) \cap \Delta(y) = \varnothing$. Therefore $$|\Delta(x)| \leq |\mathcal{X}| - |\Delta(y)| = |\mathcal{X}| - |\Delta(x)|$$ by regularity. Hence $|\Delta(x)| \leq |\mathcal{X}|/2$, a contradiction. Thus the diameter of $\mathcal{X}$ is at most $2$.
\end{proof}

Note that Lemma \ref{bigdisc} is best possible, as the example of dumbbell graphs attest.

\subsection{$K$ is not isomorphic to $\mathbb{B}$ or $\mathbb{M}$}

In Table~\ref{spordiscs} we list the first disc sizes for the local fusion
graphs of $G$ where $G=K$ or $\mathrm{Aut}(K)$, $K$ a sporadic simple group. These values have been calculated
using Lemma~\ref{structureconstants} with the aid of GAP~\cite{GAP4}. Note that
for \begin{eqnarray*}
(K,X) &=& (M_{12},2C), (J_1,2A), (M_{22},2C), (J_2,2A), (J_3,2A/2B) \\
 & &(He,2B), (Suz,2A/2B), (O'N, 2B), (Fi_{22},2A), (Ly,2A) \\
& &(Fi_{23},2A), (J_4,2B), (Fi_{24}',2C)  \end{eqnarray*} we have
$|\Xi_1(t)| > |X|/2$, and so by Lemma~\ref{bigdisc} we have
$\mathrm{Diam}(\mathcal{F}(G,X))=2$. Now suppose that $(K,X)$ does
not fall into this category, and also that $K \neq \mathbb{B}$ or
$\mathbb{M}$. Here our strategy is as follows. First we obtain (by
means detailed below) a set of $C_G(t)$-orbit representatives for
$X$. Then for each $C_G(t)$-orbit representative $x$ for which $tx$
has even order, we check that there exists another $C_G(t)$-orbit
representative $y$ for which both $ty$ and $yx$ have odd order, thus
demonstrating that $d(t,x) = 2$. Then by the vertex-transitivity of
$\mathcal{F}(G,X)$ we have $\mathrm{Diam}(\mathcal{F}(G,X))= 2$. For
a number of cases it is straightforward to obtain a set of
$C_G(t)$-orbit representatives using the \textsc{Magma}~\cite{magma}
command \verb'DoubleCosetRepresentatives' to find representatives of
the double cosets $C_G(t)gC_G(t)$, where $g \in G$. However this
command can fail when the index $[G:C_G(t)]$ becomes relatively
large. Therefore, when $(K,X) = (Fi_{22},2C)$, respectively
$(Fi_{23},2B), (Fi_{23},2C), (Fi_{24}',2A)$, $(Fi_{24}',2B)$ and $(Fi_{24}',2D)$, we
use the $C_G(t)$-orbit representatives calculated on page 119,
respectively pages 128, 129, 140, 82 and 83 of~\cite{T} to verify
Theorem~\ref{Diam=2}, while for $K=Co_1$ we make use of the
representatives calculated in~\cite{BR}. Finally, when $K=HN$ and
$Th$ the required representatives are taken from~\cite{RT}.




\subsection{$K$ is isomorphic to $\mathbb{B}$}

\begin{prop} \label{Baby2A}
Suppose that $K \cong \mathbb{B}$ and $X = 2A$. Then $\mathrm{Diam}(\mathcal{F}(G,X))=2$.
\end{prop}

\begin{proof}
Here we have $G=K\cong \mathbb{B}$. By page 216 of~\cite{atlas}, $C_G(t)$ has five orbits on $X$, and they are $\{t \}, X_{2B}, X_{2C}, X_{3A}$
and $X_{4B}$.
Clearly $\Xi_1(t) = X_{3A}$.
Let $\xi \in 11A$. Then $C_G(\xi) = \langle \xi \rangle \times L$
with $L \cong S_5$. Considering elements of order $22$ and using
\cite{atlas} we infer that $L \cap 2A \ne \varnothing \ne L \cap 2B$.
Since there are involutions in $L$ of cycle type $2^2$ whose product
has cycle type $2^2$ and $X_{2A} = \varnothing$, we must have $L \cap
2A = (1,2)^L$ and $L \cap 2B = (1,2)(3,4)^L$. Without loss we may
take $t = (1,2) \in L$ and then for $x = (3,4) \in L$ we have $x \in
X_{2B}$. With $y = (2,3) \in L$ we obtain a path $(t,y,x)$
in $\mathcal{F}(G,X)$. Since $X_{2B}$ is a $C_G(t)$-orbit, this gives $X_{2B} \subseteq \Xi_2(t)$.

Moving on to examine $X_{2C}$, this time we choose $\xi \in 13A$. So
$C_G(\xi) = \langle \xi \rangle \times L$ with $L \cong S_4$.
Looking at elements of order $26$ and employing \cite{atlas} again
we see that $L \cap 2A \ne \varnothing \ne L \cap 2C$. Since $X_{2A} =
\varnothing$, we deduce that $L \cap 2A = (1,2)^L$ and $L \cap 2C =
(1,2)(3,4)^L$. Again, taking $t = (1,2) \in L$ and $x= (3,4) \in L$
we get $x \in X_{2C}$
and then arguing as in the case of $X_{2B}$ gives $X_{2C} \subseteq \Xi_2(t)$.

From \cite{atlas}, looking at elements of order $10$ we see $X \cap
C_G(\xi) \ne \varnothing$ for $\xi \in 5A$. Now, by \cite{atlas}
$C_G(\xi) = \langle \xi \rangle \times L$ where $L \cong HS:2$. So
we may suppose that $t \in L$. Looking at products of conjugate
involutions in $HS:2$ we see that $t \in 2C_{HS:2}$ and there exists
$x \in X_{4B} \cap L$ (see, for example \cite{BBHR}, Table 2).
Employing \textsc{Magma} \cite{magma}, and using the $100$ degree permutation
representation of $HS:2$, we check that $d(t,x) = 2$ (note that $t$
has $30$ fixed points in this permutation representation), which
completes the proof. % of Proposition \ref{Baby2A}.
\end{proof}

We shall use $\mathcal{G}$ to denote the point-line collinearity
graph of $\Gamma$, the maximal $2$-local geometry, for $G \cong
\mathbb{B}$. The data arrayed in \cite{RW1} and \cite{RW2} which
describes the structure of the graph $\mathcal{G}$ will form the
backbone of the proof of Proposition \ref{Baby2B}, and we recommend that the reader has these sources to hand as they are referenced extensively. The vertex set of
$\mathcal{G}$ is $X = 2B$. For $ x \in X$, the edges of
$\mathcal{G}$ joined to $x$ are encoded by the lines in $\Gamma_x$,
the residue geometry at $x$. Now the lines in $\Gamma_x$ correspond
to certain type-$2$ vectors in the Leech lattice (see  \cite{RW1}
and \cite{RW2} again). We shall display these vectors by writing
their co-ordinates on a $24$-element set which we denote by
$\Omega_x$ (the subscript $x$ is to indicate that we are working in
$\Gamma_x$, as we will be considering a number of different vertices
of $\mathcal{G}$). Blank entries mean the co-ordinate is zero.
Further, the Steiner system $S(24,8,5)$ on $\Omega_x$ plays an
important role and, just as in \cite{RW1} and \cite{RW2}, we employ
Curtis's MOG (see \cite{C} and also \cite{CS}) to describe this Steiner system.

\begin{prop} \label{Baby2B}
Suppose that $K \cong \mathbb{B}$ and $X = 2B$. Then $\mathrm{Diam}(\mathcal{F}(G,X))=2$.
\end{prop}

\begin{proof} Again we note that $G=K\cong \mathbb{B}$. By Theorem 1(ii) of \cite{RW1} $G$ has
permutation rank $10$ on $X$, and the $C_G(t)$-orbits are $\{t\},
\Delta_1(t), \Delta_2^j(t) \;(1 \le j \le 3), \Delta_3^j \; (1 \le j
\le 4)$ and $\Delta_4(t)$. We proceed by examining each of these $C_G(t)$-orbits. \begin{equation}\label{2B1} \Xi_1(t) = \Delta_3^4(t) \cup
\Delta_4(t).\end{equation} From calculation of structure constants and the sizes of the $C_G(t)$-orbits (Table 1 of \cite{RW1}) we see
that $\Delta_3^4(t) = X_{3A}$, $\Delta_4(t) = X_{5A}$ and for $x
\in X \setminus (X_{3A} \cup X_{5A} \cup \{t\})$, $tx$ has even
order, so giving (\ref{2B1}).
\begin{equation}\label{2B2}\Delta_1(t) \subseteq \Xi_2(t). \end{equation}
Let $y \in \Delta_4(t)$. By Theorem $10$ of \cite{RW1} we may choose
$x \in \Delta_1(t) \cap \Delta_3^4(t)$. Then, by (\ref{2B1}),
$(y,t,x)$ is a path in $\mathcal{F}(G,X)$. Since $G$ acts
transitively on $X$, there exists $g \in G$ such that $y^g = t$.
Then $x^g \in \Delta_1(t)$ with $d(t,x^g) = 2$. Because
$\Delta_1(t)$ is a $C_G(t)$-orbit, it follows that $\Delta_1(t)
\subseteq \Xi_2(t)$ and (\ref{2B2}) holds.
\begin{equation}\label{2B3} \Delta_2^1(t) \cup \Delta_2^2(t) \cup \Delta_2^3(t) \subseteq \Xi_2(t).\end{equation}
Choose $y \in \Delta_3^4(t)$. Our aim, for each of $i = 1,2,3$, is
to find $x_1,x_2 \in \Delta_1(y) \cap \Delta_4(t)$ so as $x_2 \in
\Delta_2^i(x_1)$. This would then imply that $(x_1,t,x_2)$ is a path
in $\mathcal{F}(G,X)$ with $x_2 \in \Delta_2^i(x_1)$. The
transitively of $G$ on $X$ and the fact that $\Delta_2^i(t)$ is a
$C_G(t)$-orbit will then
yield~(\ref{2B3}).

Consulting Theorem 9 of \cite{RW1} we see that choosing $x_1, x_2
\in X$ with $y + x_j \in U_2 \; (j = 1,2)$ will ensure that $x_1,
x_2 \in \Delta_1(y) \cap \Delta_4(t)$. By (4.7) of \cite{RW2},
$(\Delta_2^1,+) \subseteq U_2$. The set $(\Delta_2^1,+)$ is defined
on page 278 of \cite{RW2}, and we recall its definition here. Using
the labelling of the MOG \cite{C} on the $24$-element set $\Omega$
as in \cite{C}, $(\Delta_2^1,+)$ is the set of all type-$2$ vectors
of the Leech lattice whose underlying $\mathcal{C}$-set is an octad
contained in $\Omega \setminus \{14,\infty \}$ and which contains
$0$. So we fix
$$y+x_1 = \begin{tabular}{|p{.1in}p{.1in}|p{.1in}p{.1in}|p{.1in}p{.1in}|}
  \hline
  \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} \\
    \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} \\\hline
    \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} \\
    \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} \\
    \hline
\end{tabular} \in (\Delta_2^1,+).$$\\
If we take $$y + x_2 =
\begin{tabular}{|p{.1in}p{.1in}|p{.1in}p{.1in}|p{.1in}p{.1in}|}
  \hline
  \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} \\
    \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} \\\hline
    \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} \\
    \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} \\
    \hline
\end{tabular},$$ then $y + x_2 \in (\Delta_2^1,+)$ and the inner product
$(y+x_1) \cdot (y+x_2) =16$, whence $x_2 \in \Delta_2^2(x_1)$ by
(3.2)(iii) of \cite{RW2} and the definition of $\Delta_2^2(x_1)$. On
the other hand, selecting $$y+x_2 =
\begin{tabular}{|p{.1in}p{.1in}|p{.1in}p{.1in}|p{.1in}p{.1in}|}
  \hline
  \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{} \\
    \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{} \\\hline
    \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{~2} \\
    \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{} \\
    \hline
\end{tabular}  \in (\Delta_2^1,+)$$
gives that $(y+x_1) \cdot (y+x_2) =8.$ Hence, using (3.2)(iv) of
\cite{RW2} and the definition
of $\Delta_2^3(x_1)$ we have $x_2 \in \Delta_2^3(x_1)$.

To complete the proof of (\ref{2B3}) it remains to find $y + x_2 \in U_2$
such that $x_2 \in \Delta_2^1(x_1)$. Now we also have that
$(\Delta_2^2, -) \subseteq U_2$ by (4.7) of \cite{RW2}, where
$(\Delta_2^2, -)$ consists of all type-$2$ vectors in the Leech
lattice whose underlying octad of $\Omega$ contains $\{14, \infty
\}$ but not $0$. We shall select $y + x_2$ from $(\Delta_2^2, -)$.
Taking
$$y + x_2 = \begin{tabular}{|p{.1in}p{.1in}|p{.1in}p{.1in}|p{.1in}p{.1in}|}
  \hline
  \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{} \\
    \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{~2} \\\hline
    \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{~2} \\
    \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{~2} \\
    \hline
\end{tabular} \in (\Delta_2^2, -)$$ we have $(y + x_1) \cdot (y + x_2) = 0$. By (3.2) of \cite{RW2}
we now need to show that $(y + x_1)^\circ \cap (y + x_2)^\circ =
\varnothing$. The sets $(y + x_i)^\circ$ are defined prior to (3.2) of
\cite{RW2} -- they are certain subsets of $\mathcal{G}^\circ$, where
$\mathcal{G}^\circ$ consists of the following $2$-element sets of
type-$2$ vectors (the numbers in brackets give the number of each
type of $2$-element set).
\begin{center}
\begin{tabular}{lll}
 $\{ \{ 4v_\infty \pm 4v_j, -4v_{14} \mp 4v_j \mid j \in \Omega \setminus \{\infty, 14\} \}$  & &(44)\\
 $\{ \{ (3,-1,\pm1^{22}), (1,-3,\pm1^{22})\} \}$ & &(1024)\\
  $\{ \{ (2,-2,(\pm2)^6,0^{16}), (2,-2,(\pm2)^6,0^{16})\} \}$  & &(1232).
                  \end{tabular}
\end{center}
Then $$(y + x_i)^\circ = \{ \{z_1,z_2 \} \in \mathcal{G}^\circ \; |
\; (y + x_i) \cdot z_1 = \pm16 = (y + x_i) \cdot z_2 \}.$$ First we
consider which elements of $\mathcal{G}^\circ$ of the form $\{
4v_{\infty} \pm 4v_j, -4v_{14} \mp 4v_j \}$ are in $(y+x_2)^\circ$.
Let $Y_i$ denote the octad of $\Omega$ underlying $y + x_i$, $i =
1,2$. If $j \notin Y_2$, then $(y + x_2) \cdot (4v_{\infty} \pm
4v_j) = 8$, so to be in $(y + x_2)^\circ$ we must have $j \in Y_2$.
But for $j \in Y_2$, $(y + x_1) \cdot (4v_{\infty} \pm 4v_j) = 0$
and so $\{ 4v_{\infty} \pm 4v_j, -4v_{14} \mp 4v_j \} \notin (y +
x_1)^\circ$. Now we look at elements of $\mathcal{G}^\circ$ of the
form $\{(2,-2,\pm2^6,0^{16}), (2,-2,\pm2^6,0^16) \}$. For such
elements to be in $(y + x_2)^\circ$ their non-zero co-ordinates must
all be located in $Y_2$ and hence are not in $(y + x_1)^\circ$.
Finally, considering the last type of element of $\mathcal{G}^\circ$
we see that those in $(y + x_2)^\circ$ must look like $$\left\{ z_1
=
\begin{tabular}{|p{.1in}p{.1in}|p{.1in}p{.1in}|p{.1in}p{.1in}|}
  \hline
  \makebox[0.1in][r]{~3} & \makebox[0.1in][r]{-1} & \makebox[0.1in][r]{~1} & \makebox[0.1in][r]{~1} & \makebox[0.1in][r]{~1} & \makebox[0.1in][r]{} \\
    \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{~1} \\\hline
    \makebox[0.1in][l]{$(\pm1)^{16}$} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{~1} \\
    \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{~1} \\
    \hline
\end{tabular}, z_2 = \begin{tabular}{|p{.1in}p{.1in}|p{.1in}p{.1in}|p{.1in}p{.1in}|}
  \hline
  \makebox[0.1in][r]{~1} & \makebox[0.1in][r]{-3} & \makebox[0.1in][r]{-1} & \makebox[0.1in][r]{-1} & \makebox[0.1in][r]{-1} & \makebox[0.1in][r]{} \\
    \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{-1} \\\hline
    \makebox[0.1in][l]{$(\pm1)^{16}$} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{-1} \\
    \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{-1} \\
    \hline
\end{tabular} \right\}.$$ If $\{z_1, z_2\} \in (y + x_1)^\circ$, then we must have $z_1
= z_1^\prime$ or $z_1^{\prime\prime}$ where $$ z_1^\prime = \begin{tabular}{|p{.1in}p{.1in}|p{.1in}p{.1in}|p{.1in}p{.1in}|}
  \hline
  \makebox[0.1in][r]{~3} & \makebox[0.1in][r]{-1} & \makebox[0.1in][r]{~1} & \makebox[0.1in][r]{~1} & \makebox[0.1in][r]{~1} & \makebox[0.1in][r]{} \\
    \makebox[0.1in][r]{~1} & \makebox[0.1in][r]{~1} & \makebox[0.1in][r]{~1} & \makebox[0.1in][r]{~1} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{~1} \\\hline
    \makebox[0.1in][r]{~1} & \makebox[0.1in][r]{~1} & \makebox[0.1in][r]{~1} & \makebox[0.1in][r]{~1} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{~1} \\
    \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{~1} \\
    \hline
\end{tabular} \textrm{ and }
z_1^{\prime\prime} = \begin{tabular}{|p{.1in}p{.1in}|p{.1in}p{.1in}|p{.1in}p{.1in}|}
  \hline
  \makebox[0.1in][r]{~3} & \makebox[0.1in][r]{-1} & \makebox[0.1in][r]{~1} & \makebox[0.1in][r]{~1} & \makebox[0.1in][r]{~1} & \makebox[0.1in][r]{} \\
    \makebox[0.1in][r]{-1} & \makebox[0.1in][r]{-1} & \makebox[0.1in][r]{-1} & \makebox[0.1in][r]{-1} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{~1} \\\hline
    \makebox[0.1in][r]{-1} & \makebox[0.1in][r]{-1} & \makebox[0.1in][r]{-1} & \makebox[0.1in][r]{-1} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{~1} \\
    \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{~1} \\
    \hline
\end{tabular}$$ (the blank entries being $\pm1^8$). Since $z_1$
is required to be a type-$2$ vector in the Leech lattice, the $3$
and $-1$ co-ordinates positions of $z_1$ must be a $\mathcal{C}$-set
of $\Omega$. So if $z_1 = z_1^\prime$, then there must be an octad
of $\Omega$ containing $\{ \infty, 14 \}$ and being contained in
\[
\{\infty, 14,15,18,10,2,19,1,12,21 \}.
\]
From the MOG we see there
are no such octads. Thus $z_1 \ne z_1^\prime$ and so $z_1 =
z_1^{\prime\prime}$. Then $\Pi = \{\infty, 14,0,8,3,20,4,13,16,17
\}$ must be contained in a $\mathcal{C}$-set (which is either a
dodecad or a $16$-ad), with $$\Omega \setminus \Pi \supseteq \Psi =
\{17,11,22,9,5,6 \}.$$ Since the octad containing $\Psi$ is not
contained in $\Omega \setminus \Pi$, we must have that $\Pi$ is contained
in a dodecad. But this is impossible as $Y_1 \subseteq \Pi$ and a
dodecad cannot contain an octad. Thus we have verified that $(y +
x_1)^\circ \cap (y + x_2)^\circ = \varnothing$ and therefore $x_2 \in
\Delta_2^1(x_1)$ by (3.2) of \cite{RW2}. Hence (\ref{2B3}) is
established.

There are three $C_G(t)$-orbits remaining which require our
attention, namely $\Delta_3^1(t)$, $\Delta_3^2(t)$ and
$\Delta_3^3(t)$. In dealing with these we shall first prove the following.

\addtocounter{equation}{1}\hspace{1cm}\\
(5.4) Let $x_1, x_2, x_3 \in X$ where $x_2 \in \Delta_1(x_1)$ and
$x_3 \in \Delta_1(x_2)$. Further assume that $x_2 \in \Delta_3^4(t)$, $x_3 \in \Delta_4(t)$
and $x_3 \in \Delta_2^3(x_1)$. Then there exists $g \in C_G(t)$ and a line $\ell \in \Gamma_1(x_3)$
(whose points are $x_3, x_4, x_5$) such that
\begin{enumerate}
\item[(i)] $x_4 \in \Delta_4(t)$ and $x_5 \in \Delta_3^2(t)$; and
\item[(ii)] $x_4, x_5 \in \Delta_3^4(x_1^g)$.
\end{enumerate}
By Theorem 10 of \cite{RW1} we have that $C_G(t) \cap C_G(x_2) \cap
C_G(x_3) \sim M_{22}:2$ and $x_3 + x_2 \in \mathcal{HS}_{x_3}$ (the
subscript $x_3$ telling us that this set of lines are to be viewed
in $\Gamma_{x_3}$, the residue of $x_3$). Using the explicit
description of $\mathcal{HS}$ given in (3.8) of \cite{RW2} we may,
without loss, assume that $x_3 + x_2 = 4v_{\infty} + 4v_{14}$. By
hypothesis $x_3 \in \Delta_2^3(x_1)$, and so, relative to $x_1$, the
$C_G(x_1) \cap C_G(x_3)$ orbits of lines incident with $x_3$ are
listed in Theorem 5 of \cite{RW1}. The description of such $C_G(x_1)
\cap C_G(x_3)$ orbits revolves around a certain element of
$\Omega_{x_3} \setminus \{\infty, 14 \}$. Since $C_G(t) \cap
C_G(x_2) \cap C_G(x_3)$ acts transitively on $\Omega_{x_3} \setminus
\{\infty, 14 \}$, we may suppose this element is $0$ (and replace
$x_1$ by $x_1^g$, for some $g \in C_G(t) \cap C_G(x_2) \cap
C_G(x_3)$). Consulting Theorem 5 of \cite{RW1} again (applied with
$x_1^g = t$ and $x_3 = x$) we see that the lines in $(\alpha_3, x_3
+ x_2, \pm3)$ have one point in $\Delta_2^3(x_1^g)$ and the other
two are in $\Delta_3^4(x_1^g)$. Let $\ell \in \Gamma_1(x_3)$
correspond to the following type-2 vector
$$ v=\begin{tabular}{|p{.1in}p{.1in}|p{.1in}p{.1in}|p{.1in}p{.1in}|}
  \hline
  \makebox[0.1in][r]{~1} & \makebox[0.1in][r]{~1} & \makebox[0.1in][r]{~1} & \makebox[0.1in][r]{~1} & \makebox[0.1in][r]{~1} & \makebox[0.1in][r]{~1} \\
    \makebox[0.1in][r]{~3} & \makebox[0.1in][r]{-1} & \makebox[0.1in][r]{-1} & \makebox[0.1in][r]{-1} & \makebox[0.1in][r]{~1} & \makebox[0.1in][r]{~1} \\\hline
    \makebox[0.1in][r]{-1} & \makebox[0.1in][r]{-1} & \makebox[0.1in][r]{-1} & \makebox[0.1in][r]{-1} & \makebox[0.1in][r]{~1} & \makebox[0.1in][r]{~1} \\
    \makebox[0.1in][r]{~1} & \makebox[0.1in][r]{~1} & \makebox[0.1in][r]{~1} & \makebox[0.1in][r]{~1} & \makebox[0.1in][r]{~1} & \makebox[0.1in][r]{~1} \\
    \hline
\end{tabular},$$
and let $x_3, x_4, x_5$ be the three points collinear with $\ell$.
Then, as $\ell \in (\alpha_3, x_3 + x_2, \pm3)$, we have $x_4, x_5
\in \Delta_3^4(x_1^g)$. We now wish to determine the $C_G(t)$-orbits
to which $x_4$ and $x_5$ belong. This can be done by pinning down
which $C_G(t) \cap C_G(x_3)$ orbit $\ell$ belongs to and applying
Theorem 10 of \cite{RW1}. Note that $\ell \notin
\mathcal{HS}_{x_3}$. The inner product of $v$ with the following six
type-2 vectors in $\mathcal{HS}_{x_3}$
$$ \begin{tabular}{|p{.1in}p{.1in}|p{.1in}p{.1in}|p{.1in}p{.1in}|}
  \hline
  \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} \\
    \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} \\\hline
    \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} \\
    \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} \\
    \hline
\end{tabular} \hspace{1cm} \begin{tabular}{|p{.1in}p{.1in}|p{.1in}p{.1in}|p{.1in}p{.1in}|}
  \hline
  \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} \\
    \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} \\\hline
    \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} \\
    \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} \\
    \hline
\end{tabular}$$ $$
\begin{tabular}{|p{.1in}p{.1in}|p{.1in}p{.1in}|p{.1in}p{.1in}|}
  \hline
  \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{} \\
    \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{~2} \\\hline
    \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{~2} \\
    \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{~2} \\
    \hline
\end{tabular} \hspace{1cm} \begin{tabular}{|p{.1in}p{.1in}|p{.1in}p{.1in}|p{.1in}p{.1in}|}
  \hline
  \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} \\
    \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} \\\hline
    \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} \\
    \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} \\
    \hline
\end{tabular}$$ $$
\begin{tabular}{|p{.1in}p{.1in}|p{.1in}p{.1in}|p{.1in}p{.1in}|}
  \hline
  \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} \\
    \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} \\\hline
    \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} \\
    \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} \\
    \hline
\end{tabular} \hspace{1cm} \begin{tabular}{|p{.1in}p{.1in}|p{.1in}p{.1in}|p{.1in}p{.1in}|}
  \hline
  \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} \\
    \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} \\\hline
    \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} \\
    \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} \\
    \hline
\end{tabular}$$
is $16$. Thus for at least six of the lines $k$ in
$\mathcal{HS}_{x_3}$, $\ell \in \alpha_2^2(x_3,k)$. Surveying the
$C_G(t) \cap C_G(x_3)$-orbits on lines at $x_3$ we see the only
possibility is that $\ell \in [0,8,28,64]_{\mathcal{HS}_{x_3}}$.
Hence by Theorem 10 of \cite{RW1}, one of $x_4$ and $x_5$ is in
$\Delta_3^2(t)$ and the other is in $\Delta_4(t)$, which proves (5.4).
\begin{equation}\label{2B5} \Delta_3^1(t) \cup \Delta_3^3(t) \subseteq \Xi_2(t).\end{equation}
Choose $x_2 \in \Delta_3^4(t)$. Let $x_1, x_3 \in \Delta_1(x_2)$ be
chosen so as $x_2 + x_1$ corresponds to $4v_3 + 4v_{15}$ and $x_2 +
x_3$ to the type-$2$ vector
$$w=\begin{tabular}{|p{.1in}p{.1in}|p{.1in}p{.1in}|p{.1in}p{.1in}|}
  \hline
  \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} \\
    \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} \\\hline
    \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} \\
    \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} \\
    \hline
\end{tabular}.$$
Then $x_2 + x_1 \in (\Delta_1, -) \subseteq U_3$ (at $x_2$) and $x_2
+ x_3 \in (\Delta_2^1, +) \subseteq U_2$ (at $x_2$) (see (4.7) of
\cite{RW2}), where $(\Delta_1,-)$ and $(\Delta_2^1,+)$ are relative
to $0 \in \Omega_{x_2}$. Let $\Gamma_0(x_2 + x_1) =
\{x_1,x_1^{\prime},x_2 \}$. Consequently, without loss, we have $x_1
\in \Delta_3^1(t)$, $x_1^{\prime} \in \Delta_3^3(t)$ and $x_3 \in
\Delta_4^1(t)$ by Theorem 9 of \cite{RW1}. Further the inner product
of $4v_3 + 4v_{15}$ and $w$ is $8$, whence $x_3 \in \Delta_2^3(x_1)$
and $x_3 \in \Delta_2^3(x_1^{\prime})$ by (3.2) of \cite{RW2}.
Applying (2.4.4) to $x_1, x_2, x_3$ yields that there exists $x_4
\in \Delta_1(x_3) \cap \Delta_4(t) \cap \Delta_3^4(x_1^g)$ for some
$g \in C_G(t)$. So, by (\ref{2B1}), $(x_1^g,x_4,t)$ is a path of
length $2$ in $\mathcal{F}(G,X)$. Since $\Delta_3^1(t)$ is a
$C_G(t)$-orbit, this forces $\Delta_3^1(t) \subseteq \Xi_2(t)$. A
similar argument, with $x_1^{\prime}$ in place of $x_1$, proves that
$\Delta_3^3(t) \subseteq \Xi_2(t)$ also, whence (\ref{2B5}) follows.
\begin{equation}\label{2B6} \Delta_3^2(t) \subseteq \Xi_2(t).\end{equation}
We start with $x_2 \in \Delta_3^4(t)$, and again let $0$ be the
element of $\Omega_{x_2}$ relative to $t$ which encodes the line
orbits at $x_2$. Choose $x_1 \in \Delta_1(x_2)$ and $x_3 \in
\Delta_1(x_2)$ such that
$$x_2 + x_1 = \begin{tabular}{|p{.1in}p{.1in}|p{.1in}p{.1in}|p{.1in}p{.1in}|}
  \hline
  \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} \\
    \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} \\\hline
    \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} \\
    \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} \\
    \hline
\end{tabular}$$ and $$x_2 + x_3 = \begin{tabular}{|p{.1in}p{.1in}|p{.1in}p{.1in}|p{.1in}p{.1in}|}
  \hline
  \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} \\
    \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} \\\hline
    \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} \\
    \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{} & \makebox[0.1in][r]{~2} & \makebox[0.1in][r]{~2} \\
    \hline
\end{tabular}.$$
Then both of these type-$2$ vectors are in $(\Delta_2^1, +)$ and
hence (see (4.7) of \cite{RW2}) in $U_2$ (based at $x_2$).
Therefore, by Theorem 9 of \cite{RW1}, $x_1,x_3 \in \Delta_4(t)$.
Moreover, as $(x_2 + x_1) \cdot (x_2 + x_3) = 8$, $x_3 \in
\Delta_2^3(x_1)$. So we may apply (2.4.4) to conclude there is an
$x_5 \in \Delta_1(x_3)$ with $x_5 \in \Delta_3^2(t)$ and $x_5 \in
\Delta_3^4(x_1^g)$ for some $g \in C_G(t)$. Hence, as $x_1^g \in
\Delta_4(t)$, (\ref{2B1}) implies that $(t, x_1^g, x_5)$ is a path
in $\mathcal{F}(G,X)$ of length $2$. Thus $x_5 \in \Xi_2(t)$, whence
$\Delta_3^2(t) \subseteq \Xi_2(t)$, so giving (\ref{2B6}).

Combining (\ref{2B1}) -- (\ref{2B6}) completes the proof of Proposition
\ref{Baby2B}.
\end{proof}

\subsection{$K$ is isomorphic to $\mathbb{M}$}

\begin{prop} \label{Monster}
Suppose that $K \cong \mathbb{M}$ and $X = 2A$. Then $\mathrm{Diam}(\mathcal{F}(G,X))=2$.
\end{prop}

\begin{proof} Here we have $G=K\cong\mathbb{M}$. By Table 2 of~\cite{norton}, $C_G(t)$ has $9$ orbits on $X$, namely
$\{t\}$, $X_{2A}$, $X_{2B}$, $X_{3A}$, $X_{3C}$, $X_{4A}$, $X_{4B}$, $X_{5A}$ and  $X_{6A}$. Hence
$\Xi_1(t) = X_{3A} \cup X_{3C} \cup X_{5A}$. By \cite{atlas} the eleventh power of any element of $G$ of order
$44$ is in $4A$. Also from \cite{atlas}, $G$ has only one conjugacy
class of elements of order $11$. Let $g$ be an element of $G$ of
order $11$. Then $C_G(g) = \langle g \rangle \times M$ with $M \cong
M_{12}$, again by \cite{atlas}. So $M \cap 4A \ne \varnothing$. Moreover,
looking at elements of order $22$ and using \cite{atlas} once more we deduce
that $M \cap 2A \ne \varnothing \ne M \cap 2B$. Since elements of
order $4$ in $M_{12}$ square to the class $2B_{M_{12}}$ (the $2B$
class in $M_{12}$) and, in $G$, $4A$ elements square into $2B$, we see
that $2A \cap M = 2A_{M_{12}}$ and $2B \cap M = 2B_{M_{12}}$. Hence,
if $x \in X_{2A} \cup X_{2B} \cup X_{4A}$, we may without loss
suppose that $\langle t, x \rangle \le M$ (see, for example, Table
$2$, line $2$ of \cite{BBHR}). Then, by Section 2.1, $d(t,x) = 2$.

Now suppose that $x \in X_{4B}$. Consulting page $234$ of
\cite{atlas} we see that $G$ contains a subgroup $H$ where $H \cong
A_6$, the involutions of $H$ are in $2A$ and the order four elements
of $H$ are in $4B$. Thus, without loss of generality, $\langle t,x
\rangle \le H$, whence $d(t,x) = 2$
by Theorem 1.1 of \cite{BGR}.

Finally we assume that $x \in X_{6A}$. Put $z = tx$ and $H =
N_G(\langle z^2 \rangle)$. By \cite{atlas}, $z^2 \in 3A$ and hence $H
\sim 3^. Fi_{24}$. Set $\overline{H} = H/\langle z^2 \rangle$. In
$\overline{H} \setminus \overline{H}^{\prime}$ there are two
$\overline{H}$ involution conjugacy classes, namely $2C_{Fi_{24}}$
and $2D_{Fi_{24}}$. Looking at the possible product orders of
involutions we see that $\overline{X \cap H} = 2C_{Fi_{24}}$ (the
Fischer transpositions). So $\overline{t}$ and $\overline{x}$ are
transpositions in $Fi_{24}$. Thus we may find $y \in X \cap H$ for
which $\overline{t}\overline{y}$ and $\overline{y}\overline{x}$ both
have order $3$. Consequently $ty$ and $yx$ have odd order (in fact
order $3$) and so $d(t,x) = 2$, whence Proposition \ref{Monster} holds.
\end{proof}

Section 2.1 and Propositions \ref{Baby2A}, \ref{Baby2B} and
\ref{Monster} together prove Theorem \ref{Diam=2}.

\section{Discs of $\mathcal{F}(G,X)$}\label{Discs}






The disc sizes for the local fusion graphs featuring in Theorem
\ref{Diam=2} are given in Table~\ref{spordiscs}.



%\begin{table}[H]\caption{First disc sizes for $\mathcal{F}(G,X)$} \label{spordiscs}
\begin{center}
\begin{small}
\begin{longtable}{ccrr} \caption{First disc sizes for $\mathcal{F}(G,X)$}\label{spordiscs}\\
$K$ & Class & Class size & $|\Xi_1(t)|$ \\ \hline
$M_{11}$ & 2A & $165$ & $80$  \\
$M_{12}$ & 2A & $396$ & $180$  \\
 & 2B & $495$ & $176$  \\
 & 2C & $792$ & $460$ \\
$J_1$ & 2A & $1,463$ & $1,072$ \\
$M_{22}$ & 2A & $1,155$ & $576$  \\
 & 2B & $330$ & $112$ \\
 & 2C & $1,386$ & $1,040$ \\
$J_2$ & 2A & $315$ & $224$  \\
 & 2B & $2,520$ & $1,212$  \\
 & 2C & $1,800$ & $532$ \\
$M_{23}$ & 2A & $3,795$ & $1,344$  \\
$HS$ & 2A & $5,775$ & $2,304$  \\
& 2B & $15,400$ & $7,152$  \\
& 2C & $1,100$ & $336$ \\
 & 2D & $23,100$ & $10,704$ \\
$J_3$ & 2A & $26,163$ & $16,832$  \\
 & 2B & $20,520$ & $12,716$ \\
$M_{24}$ & 2A & $11,385$ & $2,816$   \\
 & 2B & $31,878$ & $10,880$  \\
$McL$ & 2A & $22,275$ & $10,304$  \\
 & 2B & $113,400$ & $47,564$ \\
$He$ & 2A & $24,990$ & $4,992$  \\
 & 2B & $187,425$ & $119,552$  \\
 & 2C & $266,560$ & $104,796$ \\
$Ru$ & 2A & $593,775$ & $149,504$   \\
 & 2B & $1,252,800$ & $570,752$  \\
$Suz$ & 2A & $135,135$ & $69,632$  \\
 & 2B & $2,779,920$ & $1,454,432$  \\
 & 2C & $370,656$ & $137,960$ \\
 & 2D & $2,358,720$ & $757,592$ \\
$O'N$ & 2A & $2,857,239$ & $1,079,168$  \\
 & 2B & $2,624,832$ & $1,435,412$ \\
$Co_3$ & 2A & $170,775$ & $59,264$ \\
 & 2B & $2,608,200$ & $904,112$  \\
$Co_2$ & 2A & $56,925$ & $14,336$ \\
 & 2B & $1,024,650$ & $379,904$ \\
 & 2C & $28,690,200$ & $5,084,672$ \\
$Fi_{22}$ & 2A & $3,510$ & $2,816$  \\
 & 2B & $1,216,215$ & $484,352$  \\
 & 2C & $36,468,450$ & $12,015,872$  \\
 & 2D & $61,776$ & $22,400$ \\
 & 2E & $19,459,440$ & $7,102,592$ \\
& 2F & $22,239,360$ & $10,969,856$ \\
$HN$ & 2A & $1,539,000$ & $391,424$ \\
 & 2B & $74,064,375$ & $26,906,624$ \\
 & 2C & $75,240,000$ & $28,083,824$ \\
$Ly$ & 2A & $1,296,826,875$ & $659,509,424$   \\
$Th$ & 2A & $976,841,775$ & $377,298,944$ \\
$Fi_{23}$ & 2A & $31,671$ & $28,160$ \\
 & 2B & $55,582,605$ & $15,234,560$  \\
 & 2C & $12,839,581,755$ & $3,308,650,496$  \\
$Co_1$ & 2A & $46,621,575$ & $13,451,264$  \\
 & 2B & $2,065,694,400$ & $902,774,912$  \\
 & 2C & $10,680,579,000$ & $3,014,586,368$ \\
$J_4$ & 2A & $3,980,549,947$ & $1,112,555,520$  \\
 & 2B & $47,766,599,364$ & $26,545,360,896$  \\
$Fi_{24}'$ & 2A & $4,860,485,028$ & $1,504,701,440$ \\
 & 2B & $7,819,305,288,795$ & $3,351,534,645,248$  \\
 & 2C & $306,936$ & $275,264$ \\
 & 2D & $5,686,767,482,760$ & $1,780,551,713,600$ \\
$\mathbb{B}$ & 2A & $13,571,955,000$ & $2,370,830,336$  \\
 & 2B & $11,707,448,673,375$ & $4,010,408,935,424$  \\
 & 2C & $156,849,238,149,120,000$ & $56,546,114,902,065,152$  \\
 & 2D & $355,438,141,723,665,000$ & $94,228,887,171,497,984$  \\
$\mathbb{M}$ & 2A & $97,239,461,142,009,186,000$ & $30,528,114,911,948,570,624$  \\
 & 2B & $5,791,748,068,511,982,636,944,259,375$ & \begin{minipage}{5.5cm}$1,486,325,429,210,105,899,724,$\\$570,624$\end{minipage}  \\
 \hline
\end{longtable}
\end{small}
\end{center}
%\end{table}















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\end{document}
