4.3 we have given an infinite family of examples via a novel group theoretic
construction (see Theorem 4.4). To illustrate that there are many other
examples, we list below the examples for $n=8,9$, which were computed using
Magma.

  * (i)

For
$(X,X_{\alpha},X_{\alpha\beta},G,G_{\alpha},G_{\alpha\beta})=(S_{8},[64],[8],S_{7},D_{8},1)$,
there are several arc-transitive graphs $\Gamma$ of valency $8$ with $630$
vertices such that Hypothesis 2.2 holds, and the groups $([64],[8])$ can have
shape $(D_{8}.D_{8},C_{2}^{3})$ or $(C_{2}\wr C_{2}^{2},D_{8})$.

  * (ii)

For
$(X,X_{\alpha},X_{\alpha\beta},G,G_{\alpha},G_{\alpha\beta})=(S_{8},2\times
S_{4},D_{8},S_{7},D_{6},1)$, there are several arc-transitive graphs $\Gamma$
of valency $6$ with $840$ vertices such that Hypothesis 2.2 holds.

  * (iii)

For
$(X,X_{\alpha},X_{\alpha\beta},G,G_{\alpha},G_{\alpha\beta})=(S_{9},3^{1+2}.2,3^{2},S_{8},D_{6},1)$,
there is an arc-transitive graph $\Gamma$ of valency $6$ with $56\cdot 5!$
vertices such that Hypothesis 2.2 holds.

###  7.6. Case(b) with $k=2$

Here $A_{n-2}\leq G\leq(S_{n-2}\times S_{2})\cap X$ with $X_{\alpha}$
2-homogeneous as in Table 10.2.

####  7.6.1. $X_{\alpha}$ almost simple

Assume that $X_{\alpha}$ is almost simple, with socle $T$. From Table 10.2,
the possibilities are:

$\begin{array}[]{ccc}\hline\cr n&T&G\cap T\leq\\\ \hline\cr
q+1&\operatorname{PSL}_{2}(q)&D_{t(q-1)}\,(t=(2,q))\\\
q^{2}+1&{}^{2}\\!B_{2}(q)&D_{2(q-1)}\\\
q^{3}+1&\operatorname{PSU}_{3}(q)&D_{2(q^{2}-1)}\\\
q^{3}+1&{}^{2}\\!G_{2}(q)&D_{2(q-1)}\\\
11&\operatorname{PSL}_{2}(11)&D_{12}\\\ 7&\operatorname{PSL}_{2}(7)&D_{8}\\\
\hline\cr\end{array}$

For the entries in the first four rows of the table (assuming $n\geq 10$),
[32] shows that there are no factorisations of $X_{\alpha}$ with a factor as
in the third column, contradicting the fact that
$X_{\alpha}=G_{\alpha}X_{\alpha\beta}$.

_Claim 1: For row $5$ of the table above, the graph $\Gamma$ is as in line $5$
of Table 1.2. _

Here $T=\operatorname{PSL}_{2}(11)$ is self-normalising in $S_{11}$ and so
$X_{\alpha}=T<A_{11}$. Then since $|X_{\alpha}:X_{\alpha\beta}|$ divides
$|G_{\alpha}|$, which divides $12$, it follows that
$|X_{\alpha}:X_{\alpha\beta}|=|G_{\alpha}|=12$, and $G_{\alpha}=D_{12}$ by
Theorem 1.1. Also $X_{\alpha\beta}=11.5$, $G_{\alpha\beta}=1$ and $\Gamma$ has
valency $12$. Let $x\in X_{\alpha,\beta}$ of order 11. Then
$N_{X}(X_{\alpha\beta})=N_{X}(\langle x\rangle)\leq 11.10$ must contain an
involution, and hence we must have $X=S_{11}$ and then
$N_{X}(X_{\alpha\beta})=11.10$ does indeed contain an involution $g\in
X\setminus A_{11}$. Further,
$|G|=|X:X_{\alpha}|\cdot|G_{\alpha}|=\frac{11!}{12.11.5}\cdot 12=2\cdot 9!$
and hence $G=S_{9}\times S_{2}$. Note that since the involution $g\in S_{11}$
stabilises a pair in $\\{1,\ldots,11\\}$, it can be taken to lie in $G$, and
since $g$ normalises $X_{\alpha\beta}$ but does not normalise $X_{\alpha}$,
$g$ interchanges $\alpha$ and $\beta$. Thus $\Gamma$ is the graph with edge-
set $\\{\alpha,\beta\\}^{X}$; both $X$ and $G$ act arc-transitively on
$\Gamma$, and $\Gamma$ has $|X:X_{\alpha}|=12\cdot 7!$ vertices. Finally,
$\langle X_{\alpha},g\rangle=X$ as $S_{11}$ is the only insoluble transitive
group of degree 11 containing an odd permutation, and hence $\Gamma$ is
connected. It is the graph in line $4$ of Table 1.2.

We note that $\Gamma$ is bipartite (since $X_{\alpha}<A_{11}<X$), and that $X$
acts $2$-arc-transitively on $\Gamma$ (since $X_{\alpha}$ is $2$-transitive on
the $12$ neighbours of $\alpha$).

_Claim 2: For row $6$ of the table above, the graph $\Gamma$ is as in line $4$
of Table 1.2. _

Here $T=\operatorname{PSL}_{2}(7)$ is self-normalising in $S_{7}$ and so
$X_{\alpha}=T<A_{7}$. Then since $|X_{\alpha}:X_{\alpha\beta}|$ divides
$|G_{\alpha}|$, which divides $8$, it follows that
$|X_{\alpha}:X_{\alpha\beta}|=|G_{\alpha}|=8$, and $G_{\alpha}=D_{8}$ by
Theorem 1.1. Also $X_{\alpha\beta}=7.3$, $G_{\alpha\beta}=1$ and $\Gamma$ has
valency $8$. Let $x\in X_{\alpha,\beta}$ of order 7. Then
$N_{X}(X_{\alpha\beta})=N_{X}(\langle x\rangle)\leq 7.6$ must contain an
involution, and hence we must have $X=S_{7}$ and then
$N_{X}(X_{\alpha\beta})=7.6$ does contain an involution $g\in X\setminus
A_{7}$. Further, $|G|=|X:X_{\alpha}|\cdot|G_{\alpha}|=\frac{7!}{8.7.3}\cdot
8=2\cdot 5!$ and hence $G=S_{5}\times S_{2}$.

Arguing exactly as in Claim 1, we see that there is an arc-transitive
embedding in this case also, with $X_{\alpha\beta}=7.3$ for an edge
$\\{\alpha,\beta\\}$. The graph $\Gamma$ has 30 vertices and valency 8, and in
fact it is isomorphic to the antiflag graph of $\operatorname{PG}_{3}(2)$;
this is a bipartite graph whose vertices are the points and hyperplanes of
$\operatorname{PG}_{3}(2)$, joined if they are non-incident. Its automorphism
group is $\operatorname{PSL}_{4}(2).2=S_{8}$, so as $G=S_{5}\times 2\leq
S_{5}\times S_{3}<S_{8}$, this is also an embedding that occurs under Case (b)
with $k=3$ in Section 7.7 below.

####  7.6.2. $X_{\alpha}$ affine

Assume that $X_{\alpha}$ is affine. From Table 10.2, the possibilities are:

$\begin{array}[]{ccc}\hline\cr n&X_{\alpha}&|G_{\alpha}|\ \hbox{divides }\\\
\hline\cr p^{2}&\leq\operatorname{AGL}_{2}(p)&p(p-1)\\\
q=p^{a}&\leq\operatorname{A\Gamma L}_{1}(q)&2a\\\ p^{2}&\leq
p^{2}.(p-1).S_{4}&\frac{48}{p+1}\\\ (p=5,7,11,23)&&\\\ q^{2}&\leq
q^{2}.(q-1).S_{5}&\frac{240}{q+1}\\\ (q=9,11,\ldots,59)&&\\\
\hline\cr\end{array}$

In all cases we see that the factorisation
$X_{\alpha}=G_{\alpha}X_{\alpha\beta}$ implies that $X_{\alpha\beta}$ contains
the full translation subgroup $T$ of order $n$. But then any involution $g$ as
in Proposition 2.1 normalises $T$, and so $\langle X_{\alpha},g\rangle\neq X$.
So by Proposition 2.1, there are no arc-transitive embeddings in this case.

###  7.7. Case (b) with $k=3$

Here $A_{n-3}\leq G\leq(S_{n-3}\times S_{3})\cap X$ with $n\geq 6$, and
$X_{\alpha}$ is 3-homogeneous as in Table 10.1. In similar fashion to ‘case
(a) with $k=3$’ in Subsection 7.3, the possibilities are:

$\begin{array}[]{ccc}\hline\cr n&X_{\alpha}&|G_{\alpha}|\hbox{ divides}\\\
\hline\cr q+1&{\rm soc}(X_{\alpha})=\operatorname{PSL}_{2}(q)\ (q=p^{f})&6f\\\
12&M_{11}&12\\\
8&\operatorname{AGL}_{3}(2),\,\operatorname{AGL}_{1}(8),\,\operatorname{A\Gamma
L}_{1}(8)&8,1,3\\\ 32&\operatorname{A\Gamma L}_{1}(32)&1\\\
\hline\cr\end{array}$

In the first row, $X_{\alpha\beta}$ does not contain
$\operatorname{PSL}_{2}(q)$ by Hypothesis 2.2(c), and so
$|X_{\alpha}:X_{\alpha\beta}|$ is divisible by the index of a proper subgroup
of $\operatorname{PSL}_{2}(q)$. Since $|X_{\alpha}:X_{\alpha\beta}|$ divides
$|G_{\alpha}|$, which divides $6f$, we must have $q=5$ or 8. If $q=5$ then
$X_{\alpha}=\operatorname{PGL}_{2}(5)<X=S_{6}$ and we have $n=6$,
$\Gamma=K_{6}$ as in line 1 of Table 1.2 of Theorem 7.1. If $q=8$ then
$G_{\alpha}$ must contain a subgroup $3^{2}$, so is not cyclic or dihedral.

Next observe that $M_{11}$ has no factorisation with a factor of order 12, so
$X_{\alpha}\neq M_{11}$. Similarly $X_{\alpha}$ is not
$\operatorname{AGL}_{1}(8)$, $\operatorname{A\Gamma L}_{1}(8)$ or
$\operatorname{A\Gamma L}_{1}(32)$.

Finally, consider the case where $X_{\alpha}=\operatorname{AGL}_{3}(2)$. Here
there is an example of an arc-transitive embedding. In fact this occurs with
the antiflag graph of $\operatorname{PG}_{3}(2)$ in line $4$ of Table 1.2,
which we saw in Section 7.6.1: this graph has automorphism group
$X=\operatorname{PSL}_{4}(2).2=S_{8}$ and vertex-stabiliser a parabolic
subgroup $X_{\alpha}=\operatorname{AGL}_{3}(2)$. The group $G$ for this
embedding is $S_{5}\times 2$.

###  7.8. Case (b) with $k=4$ or $5$

Here $A_{n-k}\leq G\leq(S_{n-k}\times S_{k})\cap X$, and $X_{\alpha}$ is
4-homogeneous as in Table 10.1. However in every case the group $X_{\alpha}$
has no factorisation with one of the factors being $G_{\alpha}$.

###  7.9. Case (c) with $n=6,8$ or $10$

We performed computations with Magma to find all the possibilities for
$X,G,\Gamma,g$ satisfying Hypothesis 2.2 with ${\rm soc}(X)=A_{n}$, where
$n\in\\{6,8,10\\}$. Many of the possibilities occur in families already
analysed and identified in Tables 1.1 and 1.2, and several that we recorded in
§ 7.5 with $G=A_{n-1}$ or $S_{n-1}$ – the one case where we do not have a
complete classification. There was exactly one new example that did not fit
any of these cases. and we present it below.

####  7.9.1. $n=6$

Here is the new set of subgroups:

$(X,X_{\alpha},G,G_{\alpha},X_{\alpha\beta})=(S_{6}.2,\,S_{5},\,(S_{3}\wr
S_{2}).2,\,D_{12},\,F_{20}).$

Since $S_{5}<S_{6}<S_{6}.2$ it follows that the graph $\Gamma$ is bipartite.
Let $\Delta_{1},\Delta_{2}$ be the parts of the bipartition, where
$\alpha\in\Delta_{1},\beta\in\Delta_{2}$, and let $X^{+}=S_{6}$ fixing each
$\Delta_{i}$ setwise. An involution in $X\setminus S_{6}$ interchanges
$\Delta_{1}$ and $\Delta_{2}$, and acts as a graph automorphism of $X^{+}$.
Thus the actions of $X^{+}$ on $\Delta_{1}$ and $\Delta_{2}$ are not
equivalent, so $X_{\alpha}$ acts transitively on $\Delta_{2}$ and $\Gamma$ is
the complete bipartite graph $K_{6,6}$. Since $S:=O_{5}(X_{\alpha\beta})$ is a
Sylow $5$-subgroup of $X$, it follows that $N_{X}(S)S_{6}/S_{6}\cong C_{2}$,
and by [2], $N_{X}(S)\setminus S_{6}$ contains an involution $g$ such that
$g\in G$ and $g$ interchanges $\alpha$$\beta$. In addition $\langle
X_{\alpha},g\rangle=X$ since $\Gamma$ is connected and so all the conditions
of Proposition 2.1 hold and we have an arc-transitive embedding, as in Line 3
of Table 1.2.

This completes the proof of Theorem 7.1.

##  8\. The map theorem for groups of Lie type

In this section we prove Theorem 1.9. The procedure is to examine each of the
cyclic/dihedral factorisations in Theorem 1.2 to determine which of them
correspond to simple connected graphs admitting arc-transitive embeddings. For
convenience, we restate the theorem here:

######  Theorem 8.1.

Suppose that Hypothesis 2.2 holds for $X,G,\Gamma,g$ with $X$ is an almost
simple group of Lie type (such that ${\rm soc}(X)$ is not an alternating
group). Then one of the lines of Table 1.3 holds.

Proof of Theorem 8.1

Let $X,G,g,$ $\Gamma=(\Omega,E)$ be as in Hypothesis 2.2 where $X_{0}$ is a
simple group of Lie type, not isomorphic to an alternating group. By Theorem
1.2, the possibilities for $X,G,X_{\alpha}$ are as in Tables 10.3 (the
families) and 10.4 \- 10.7 (the exceptional factorisations). Note that
$\\{X_{\alpha},G\\}=\\{A,B\\}$ for $A,B$ in the tables.

###  8.1. Proof for the tables of exceptional factorisations.

Here we assume that $X$ and the pair $\\{X_{\alpha},G\\}$ occurs in one of the
lines of Tables 10.4, 10.5, 10.6 or 10.7. For each possibility
$(X,\\{X_{\alpha},G\\})$ in these tables a computation using Magma [2] was
performed to identify cases where an involution $g$ exists such that
Hypothesis 2.2 holds. This computation yielded exactly two possibilities for
linear groups, and two possibilities for unitary groups, and we consider these
cases below separately.

_Line 1 of Table 10.4 with $X=\operatorname{PSL}_{2}(7).2$,
$X_{\alpha}=S_{4}$, $G=7.6$, $G_{\alpha}=3$, and $G_{\alpha\beta}=1$._
Here $X_{0}=\operatorname{PSL}_{2}(7)$ and $X_{\alpha}<X_{0}<X=X_{0}.2$, and
it follows that $\Gamma$ is bipartite. Let $\Delta_{1},\Delta_{2}$ denote the
two parts of the bipartition of $V\Gamma$ and assume that
$\alpha\in\Delta_{1}$ and $\beta\in\Delta_{2}$. Since $X=X_{\alpha}G=X_{0}G$,
the involution $g\in G\setminus X_{0}$ and so $g$ induces an outer
automorphism of $X_{0}$ and interchanges $\Delta_{1}$ and $\Delta_{2}$, Thus
the $X_{0}$-actions on $\Delta_{1}$ and $\Delta_{2}$ are not equivalent, and
hence $X_{\alpha}$ has orbits of lengths $3,4$ in $\Delta_{2}$. As $\Gamma$
has valency $|G_{\alpha}:G_{\alpha\beta}|=3$, $\beta$ lies in the
$X_{\alpha}$-orbit in $\Delta_{2}$ of length $3$. The most convenient way to
identify $\Gamma$ is to observe that $X=\operatorname{PSL}_{3}(2).2$, so that
we may identify $\Delta_{1}$ and $\Delta_{2}$ with the sets of $7$ points and
lines of the projective plane $\operatorname{PG}_{2}(2)$, respectively, we
have then shown that a projective point $\alpha$ is adjacent to a projective
line $\beta$ if and only if they are incident in $\operatorname{PG}_{2}(2)$.
Thus $\Gamma$ is the incidence graph of $\operatorname{PG}_{2}(2)$, as in line
$1$ of Table 1.3.

_Line 2 of Table 10.4 with $X=\operatorname{PSL}_{2}(11).2$,
$X_{\alpha}=A_{5}$, $G=11.10$, $G_{\alpha}=5$, and $G_{\alpha\beta}=1$._
As in the previous case, $X_{0}=\operatorname{PSL}_{2}(11)$ and
$X_{\alpha}<X_{0}<X=X_{0}.2$, so $\Gamma$ is bipartite, $g\in G\setminus
X_{0}$, $g$ induces an outer automorphism of $X_{0}$, and the $X_{0}$-actions
on the parts $\Delta_{1}$ and $\Delta_{2}$ of the bipartition are not
equivalent, and hence $X_{\alpha}$ (where $\alpha\in\Delta_{1}$) has orbits of
lengths $5,6$ in $\Delta_{2}$, and as $\Gamma$ has valency
$|G_{\alpha}:G_{\alpha\beta}|=5$, $\beta$ lies in the $X_{\alpha}$-orbit in
$\Delta_{2}$ of length $5$. In this case we identify $\Delta_{1}$ and
$\Delta_{2}$ with the sets of $11$ points and blocks of the $2-(11,5,2)$
biplane, respectively, we have then shown that a point $\alpha$ is adjacent to
a block $\beta$ if and only if they are form a flag of the biplane. Thus
$\Gamma$ is the incidence graph of the biplane, as in line $3$ of Table 1.3.

_Line 3 of Table 10.5 or Line 1 of Table 10.7 with
$X=\operatorname{PSU}_{4}(2).2$, $X_{\alpha}=3^{3}.A_{4}.a$, $G=2^{4}.S_{5}$,
$G_{\alpha}=D_{12a}$, and $G_{\alpha\beta}=1$, where $a\leq 2$._
Here
$X_{0}=\operatorname{PSU}_{4}(2)\cong\operatorname{PSp}_{4}(3)\cong\Omega_{5}(3)$
and $X=X_{0}.2=\operatorname{SO}_{5}(3)$, and the two possibilities for
$X_{\alpha}$ are $3^{3}.A_{4}$ and $3^{3}.S_{4}$. Suppose first that
$X_{\alpha}=3^{3}.S_{4}$ (so $a=2$). Here a Magma computation shows that the
group $X_{0}$ is transitive on $\Omega$, and so (see [Atlas, p. 26])
$X_{0}\cap X_{\alpha}$ is the unique index $2$ subgroup $3^{3}.A_{4}$ of the
stabiliser in $X_{0}$ of a singular $1$-space in $V=\mathbb{F}_{3}^{5}$. Thus
$X_{0}\cap X_{\alpha}$ is the stabiliser of a non-zero singular vector of
$V=\mathbb{F}_{3}^{5}$, and so $\Omega$ can be identified with the set of non-
zero singular vectors of $V$. The subgroup $X_{\alpha}$ has exactly one orbit
of length $24=|X_{\alpha}:X_{\alpha\beta}|$ in $\Omega$, namely the set of
singular vectors $\beta$ orthogonal to $\alpha$, so the graph $\Gamma$ is as
in line $4$ of Table 1.3.

Now consider the case where $X_{\alpha}=3^{3}.A_{4}$ (that is, $a=1$), of
index $160$ n $X$. Here the Magma computations show that the group $X_{0}$ has
two orbits in $\Omega$, say $\Omega^{+}$ and $\Omega^{-}$, and that the
$X_{0}$-action on each $\Omega^{\pm}$ is equivalent to its action on the
vertex set of the $80$-vertex graph of the previous paragraph. Thus $\Gamma$
is bipartite with $\Omega^{+},\Omega^{-}$ the parts of the bipartition, and we
identify each $\Omega^{\pm}$ with the set of non-zero singular vectors in
$V=\mathbb{F}_{3}^{5}$. Further, Magma computations also show that
$X_{\alpha}=(X_{0})_{\alpha}$ has two orbits of length $12$ in the set of $24$
neighbours of $\alpha$ in the $80$-vertex graph, and that these suborbits are
paired with each other. Hence the two corresponding $X_{0}$-orbitals $\Delta$
and $\Delta^{\prime}$ in $\Omega^{\pm}\times\Omega^{\pm}$ are paired with each
other. In this case the graph $\Gamma$ has valency
$12=|G_{\alpha}:G_{\alpha\beta}|$, and for $\alpha=v^{+}\in\Omega^{+}$, the
set of edges adjacent to $v^{+}$ must be one of the two orbits $\Delta(v)^{-}$
or $\Delta^{\prime}(v)^{-}$ of $(X_{0})_{\alpha}$ of size $12$ in $\Omega^{-}$
(as the graph is connected). By Lemma 3.1(b), the graphs corresponding to each
of these choices are isomorphic to the graph $\widehat{\Gamma}$ of Example
3.4(c). Thus Line 5 of Table 1.3 holds.

###  8.2. Proof for the infinite families of classical group factorisations.

We now assume that $(X,\\{X_{\alpha},G\\})$ is as in one of the lines of Table
10.3. First we perform Magma computations as described in Subsection 2.2 to
identify any possibilities for $(X,X_{\alpha},G,g)$ satisfying Hypothesis 2.2
for the following small groups in these families:

$\begin{array}[]{l}\operatorname{PSL}_{2}(q)\ (q\leq
11),\operatorname{PSL}_{3}(4),\operatorname{PSL}_{4}(3),\operatorname{PSU}_{3}(3),\
\operatorname{PSU}_{4}(q)\ (q\in\\{2,3\\}),\\\ \operatorname{Sp}_{4}(q)\
(q\in\\{3,4,8\\}),\
\operatorname{Sp}_{6}(2),\operatorname{Sp}_{8}(2),\Omega_{8}^{+}(2).\end{array}$
(8.1)

No examples were found apart from those analysed in Subsection 8.1, and so we
assume that $X_{0}$ is not one of these groups, and $X_{0}$ is not an
alternating group. We examine each of the lines of Table 10.3 separately
below.

####  8.2.1. Lines in Table 10.3 with $X_{0}=\operatorname{PSL}_{n}(q)$,
$n\geq 2$.

Let $V=\mathbb{F}_{q}^{n}$ and $q=p^{f}$ for a prime $p$ and $f\geq 1$. From
the discussion above, if $n=2$ then $q>11$, and $(n,q)\neq(3,4)$ or $(4,3)$.

_Line 1a: $X_{\alpha}\triangleright q^{n-1}.\operatorname{SL}_{n-1}(q)$,
$G\cap X_{0}\leq\frac{q^{n}-1}{(q-1)d}.n$, and $G_{\alpha}\cap X_{0}=r$ where
$r\mid n$ and $d=(n,q-1)$._
Here, applying a graph automorphism of $X_{0}$ if necessary, we may take
$X_{\alpha}\leq P_{1}=X_{\langle v\rangle}$ for some $v\in V$. Suppose first
that $X\leq\operatorname{P\Gamma L}_{n}(q)$. Now $X_{\beta}\leq P_{1}^{g}$,
and if $\langle v\rangle^{g}=\langle v\rangle$ then $\langle
X_{\alpha},g\rangle\leq P_{1}$, which is a contradiction. Hence $\langle
v\rangle^{g}=\langle w\rangle\neq\langle v\rangle$, and as $X_{0}$ is
$2$-transitive on $1$-spaces, this implies that $(q^{n}-q)/(q-1)$ divides
$|X_{\alpha}:X_{\alpha\beta}|=|G_{\alpha}:G_{\alpha\beta}|$. However
$|G_{\alpha}:G_{\alpha\beta}|$ divides $|G_{\alpha}|$ which divides $nf$. This
implies that $(n,q)=(2,4)$, but in this case $X_{0}\cong A_{5}$ which is
excluded.

Now we deal with the case where $X$ contains a graph automorphism so $X=Y.2$
where $Y:=X\cap\operatorname{P\Gamma L}_{n}(q)$ and $n\geq 3$. Since
$X_{\alpha}\leq P_{1}<Y$ it follows that $\Gamma$ is bipartite, and as
$X=X_{\alpha}G=YG$, the subgroup $G$ must also contain a graph automorphism,
while $G_{\alpha}\leq X_{\alpha}<Y$ so $|G_{\alpha}|$ divides $nf$. Let
$\Delta_{1},\Delta_{2}$ denote the two parts of the bipartition of $V\Gamma$,
with $\alpha\in\Delta_{1}$ and $\beta\in\Delta_{2}$. Now $X$ preserves a
partition $\mathcal{B}=\mathcal{B}_{1}\cup\mathcal{B}_{2}$ of $V\Gamma$, where
$\mathcal{B}_{i}$ is a partition of $\Delta_{i}$ ($i=1,2$), such that the
$Y$-actions on $\mathcal{B}_{1},\mathcal{B}_{2}$ are equivalent to its actions
on $1$-spaces and hyperplanes of $V$, respectively. Since $X_{\alpha}\geq
P_{1}^{\prime}$ it follows that $X_{\alpha}$ has exactly two orbits in
$\mathcal{B}_{2}$, with lengths $m_{1}:=q^{n-1}$ and
$m_{2}:=(q^{n-1}-1)/(q-1)$. The point $\beta$ lies in a block of one of these
orbits, and hence $|X_{\alpha}:X_{\alpha\beta}|$ is divisible by $m_{i}$ for
some $i$. This implies that $m_{i}$ divides $|G_{\alpha}|$, so $m_{i}$ divides
$nf$. The only possibilities are $(n,q,m_{i})=(3,2,3)$ and $(3,8,9)$. In the
former case, $X_{0}=\operatorname{PSL}_{3}(2)\cong\operatorname{PSL}_{2}(7)$,
which is excluded. So $(n,q,m_{i})=(3,8,9)$, and we identify the graph
$\Gamma$ as follows. Since $m_{i}=9=q+1=nf$, it follows that $|G_{\alpha}|=9$
and $|V\Gamma|=2(q^{n}-1)/(q-1)=2\cdot 73$, so the partition $\mathcal{B}$ is
trivial. Thus we may identify $\Delta_{1},\Delta_{2}$ with the points and
lines of the projective plane ${\rm PG}_{2}(8)$ such that the point $\alpha$
is joined to the set of $9$ lines incident with it. Thus $\Gamma$ is the
incidence graph of ${\rm PG}_{2}(q)$ as in Line 2 of Table 1.3. The existence
of an involution $g$ satisfying Hypothesis 2.2 is computed using Magma,
completing the proof for this case.

_Line 1b: $X_{\alpha}\cap X_{0}\leq\frac{q^{n}-1}{(q-1)d}.n$, $G\triangleright
q^{n-1}.\operatorname{SL}_{n-1}(q)$, and $G_{\alpha}\cap X_{0}=r$ where $r\mid
n$ and $d=(n,q-1)$._
Let $Y:=X\cap\operatorname{P\Gamma L}_{n}(q)$ so that $a:=|X:Y|\leq 2$, and if
$a=2$ then $n\geq 3$. Note that $G\leq Y$, so $X=X_{\alpha}G=X_{\alpha}Y$ and
hence $a=|X_{\alpha}:X_{\alpha}\cap Y|$ and
$X_{\alpha}\leq\frac{q^{n}-1}{(q-1)d}.nf.a$. Now $|X:P_{1}|=(q^{n}-1)/(q-1)$
divides $|X:G|=|X_{\alpha}:G_{\alpha}|$, and hence $(q^{n}-1)/(q-1)$ divides
$|X_{\alpha}|$. If both $(nf,p)\neq(6,2)$ and $(n,f,p)\neq(2,1,2^{c}-1)$, then
$p^{nf}-1$ has a primitive prime divisor $s$ by [37], and we note that
$s\equiv 1\pmod{nf}$ so in particular $s$ does not divide $nfa$. Thus $s$
divides $|X_{\alpha}\cap X_{0}|$, so $X_{\alpha}$ has an element $x$ of order
$s$ such that $\langle x\rangle$ is normal in $X_{\alpha}$ and irreducible on
$V$, and hence $X_{\alpha}\leq N_{X}(\langle
x\rangle)\leq\frac{q^{n}-1}{(q-1)d}.nf.a$ (see [14, Satz II.7.3]). It follows
from Hypothesis 2.2(c) that $x\not\in X_{\beta}$, and this implies that $s$
divides $|X_{\alpha}:X_{\alpha\beta}|=|G_{\alpha}:G_{\alpha\beta}|$, which in
turn divides $|G_{\alpha}|$. Thus $nf<s\leq|G_{\alpha}|\leq nf$, which is a
contradiction.

If $(nf,p)=(6,2)$ let $s=21$; and if $(n,f,p)=(2,1,2^{c}-1)$ let $s=2^{c-2}$,
and note that in the latter case $c\geq 5$ and $a=1$ (since we assume that
$q>11$ when $n=2$). Then again $X_{\alpha}\cap Y$ has an element $x$ of order
$s$ which is irreducible on $V$ and $X_{\alpha}\leq N_{X}(\langle
x\rangle)\leq\frac{q^{n}-1}{(q-1)d}.nf.a$. If $s=|x|=21$ and $nf=6$, then
$|G_{\alpha}|$ is not divisible by $7$, and hence
$|X_{\alpha}:X_{\alpha\beta}|$ is not divisible by $7$. In this case $\langle
x^{3}\rangle$ is the unique normal subgroup of $X_{\alpha\beta}$ of order $7$,
contradicting Hypothesis 2.2(c). Hence $(n,f,p)=(2,1,2^{c}-1)$, $s=2^{c-2}$,
and $|G_{\alpha}|\leq nf=2$. Thus $|G_{\alpha}|$ is not divisible by $4$, and
hence $|X_{\alpha}:X_{\alpha\beta}|$ is not divisible by $4$, so $\langle
x^{2}\rangle\unlhd X_{\alpha\beta}$. In this case the group
$\frac{q^{n}-1}{(q-1)d}.nf$ is dihedral of order $q+1=2^{c}$ and since $a\geq
5$, $\langle x^{2}\rangle$ is the unique cyclic normal subgroup of
$X_{\alpha\beta}$ of order $2^{c-3}$, contradicting Hypothesis 2.2(c).

_Line 2a: $(n,q)=(4,p)$, $X_{\alpha}\cap X_{0}=\operatorname{PSp}_{4}(p).a$,
$G\cap X_{0}=p^{3}.\frac{p^{3}-1}{d}.e$, and $G_{\alpha}\cap X_{0}=p\times e$,
where $p$ is odd, $e\mid 3$, $d=(4,p-1)$, $ad=4$, and $X\geq X_{0}.2$ if
$p\equiv 1\pmod{4}$._
Here $p>3$ (as $\operatorname{PSL}_{4}(3)$ is excluded by (8.1), and
$X_{\alpha\beta}$ does not contain $\operatorname{PSp}_{4}(p)$ by Hypothesis
2.2(c). Hence, by [25, Table 5.2.A],
$|X_{\alpha}:X_{\alpha\beta}|\geq(p^{4}-1)/(p-1)$ since $p>3$; and
$|X_{\alpha}:X_{\alpha\beta}|$ divides $|G_{\alpha}|\leq(3p)\cdot 2(p-1)$,
which is a contradiction.

_Line 2b: $(n,q)=(4,p)$, $X_{\alpha}\cap X_{0}=p^{3}.\frac{p^{3}-1}{d}.e$,
$G\cap X_{0}=\operatorname{PSp}_{4}(p).a$, and $G_{\alpha}\cap X_{0}=p\times
e$, where $p$ is odd, $e\mid 3$, $d=(4,p-1)$, $ad=4$, and $X\geq X_{0}.2$ if
$p\equiv 1\pmod{4}$._
Again $p>3$ by (8.1), and we have $X_{\alpha}=QR$ where
$Q=[p^{3}]=O_{p}(X_{\alpha})$. Now
$|X_{\alpha}:X_{\alpha\beta}|=|G_{\alpha}:G_{\alpha\beta}|$ divides
$|G_{\alpha}|$, which in turn divides $2(p-1)pe$, and it follows that
$|X_{\alpha\beta}|$ is divisible by $p^{2}(p^{3}-1)/(p-1)$. This implies both
that $X_{\alpha\beta}\cap Q\neq 1$ and that $X_{\alpha\beta}$ contains a
cyclic subgroup of order $(p^{3}-1)/(p-1)$ which acts irreducibly on $Q$.
Hence $Q\leq X_{\alpha\beta}$, and this contradicts Hypothesis 2.2(c).

####  8.2.2. Lines in Table 10.3 with $X_{0}=\operatorname{PSp}_{2m}(q)$

Let $V=\mathbb{F}_{q}^{2m}$ be the natural module for $X$, where $m\geq 2$ and
$q=p^{f}$ for a prime $p$ and $f\geq 1$. By (8.1), we are assuming: if $m=2$
then $q\geq 5$ and $q\neq 8$, and also $(2m,q)\neq(6,2),(8,2)$.

_Lines 1–3 a: $q$ even, $\Omega_{2m}^{\epsilon}(q)\unlhd X_{\alpha}$,
$\operatorname{Sp}_{2}(q^{m})\unlhd G\leq\operatorname{Sp}_{2}(q^{m}).(2mf)$,
and $C_{q^{m}-\epsilon}\leq G_{\alpha}\cap X_{0}\leq D_{2(q^{m}-\epsilon)}$,
where $\epsilon=\pm$._
Note that the largest dihedral subgroup of ${\rm
Aut}(\operatorname{SL}_{2}(q^{m}))$ containing $C_{q^{m}-\epsilon}$ is
$D_{2(q^{m}-\epsilon)}$, so $G_{\alpha}\leq D_{2(q^{m}-\epsilon)}$. Thus
$|X_{\alpha}:X_{\alpha\beta}|$ divides $|G_{\alpha}|$ which divides
$2(q^{m}-\epsilon)$. On the other hand, by [15, Theorem 1],
$|X_{\alpha}:X_{\alpha\beta}|$ is divisible by either
$q^{m-1}(q^{m}-\epsilon)$ or $(q^{m}-\epsilon)(q^{m-1}+\epsilon)$, so one of
these quantities must divide $2(q^{m}-\epsilon)$. This implies that $m=2$ and
$q\leq 3$, but these cases are excluded.

_Lines 1–3 b: $q$ even, $\operatorname{Sp}_{2}(q^{m})\unlhd
X_{\alpha}\leq\operatorname{Sp}_{2}(q^{m}).(2mf)$,
$\Omega_{2m}^{\epsilon}(q)\unlhd G$, and $C_{q^{m}-\epsilon}\leq
G_{\alpha}\cap X_{0}\leq D_{2(q^{m}-\epsilon)}$, where $\epsilon=\pm$._
As noted in the previous case, $G_{\alpha}\leq D_{2(q^{m}-\epsilon)}$ and
$|X_{\alpha}:X_{\alpha\beta}|$ divides $2(q^{m}-\epsilon)$. Hence
$|X_{\alpha\beta}|$ is divisible by $\frac{1}{2}q^{m}(q^{m}+\epsilon)$. If
$X_{\alpha\beta}\leq N_{X_{\alpha}}(q^{m}+\epsilon)\leq(q^{m}+\epsilon).4f$
then $q^{m}$ divides $8f$, but this is impossible for the parameter values
being considered. Thus this inclusion does not hold, and so $\epsilon=-$ and
$X_{\alpha\beta}$ is contained in a Borel subgroup $B$ of $X_{\alpha}$.

Now $X_{\alpha}=\operatorname{Sp}_{2}(q^{m}).a$ where $a\mid 2mf$, and
$\frac{1}{2}q^{m}(q^{m}-1)a=|X_{\alpha}:D_{2(q^{m}+1)}|$ divides
$|X_{\alpha}:G_{\alpha}|$ which divides $|X_{\alpha\beta}|$. It follows that
$|B:X_{\alpha\beta}|\leq 2$ and $|D_{2(q^{m}+1)}:G_{\alpha}|\leq 2$. Thus
$X_{\alpha\beta}=RM.b$ where $R=[q^{m}]$, $M=q^{m}-1$, and $b\leq 2$.

_Next we analyse $N_{X_{0}}(X_{\alpha\beta})$._
Now $N:=N_{X_{0}}(X_{\alpha\beta})\leq N_{X}(R)\leq P$ for some parabolic
subgroup $P$. Since $M=q^{m}-1$ acts on the module $V$ as $W\oplus W^{\prime}$
(a direct sum of two totally isotropic $m$-spaces which are not isomorphic as
$M$-modules), the subgroup $P=P_{m}$ is the stabiliser $X_{W}$ or
$X_{W^{\prime}}$. Hence $P=QL$ where $Q=[q^{\frac{1}{2}m(m+1)}]$,
$L=\operatorname{GL}_{m}(q)$, and $R\leq Q$; and we can take $M\leq L$. Now
$Q$ is elementary abelian and, as an $\mathbb{F}_{q}L$-module, we have $Q\cong
S^{2}V_{m}(q)$.

As in the proof of Proposition 6.12, if $m$ is odd then the
$\mathbb{F}_{q}$-composition factors of $Q\downarrow M$ all have dimension
$m$, while if $m$ is even then there is one composition factor of dimension
$m/2$ and the rest have dimension $m$. Hence $R$ is one of the composition
factors of dimension $m$ and we have $Q\downarrow M=R\oplus Q_{1}$.

_Claim 1. $C_{Q}(M)=1$._
As we have seen above, the eigenvalues of $x$ on $S^{2}\overline{V}_{m}$ are
$\lambda^{2q^{i}}$ and $\lambda^{q^{i}+q^{j}}$ for $1\leq i<j\leq m$. None of
these is equal to 1, and Claim 1 follows.

_Claim 2. $N_{Q}(RM)=R$._
Let $rq_{1}\in N_{Q}(RM)$, where $r\in R$ and $q_{1}\in Q_{1}$. Then $q_{1}\in
N_{Q}(RM)$, so as $R$ and $Q_{1}$ are both $M$-invariant, $[q_{1},M]\leq
RM\cap Q_{1}=1$. Hence $q_{1}\in C_{Q}(M)$, and so $q_{1}=1$ by Claim 1. Thus
$N_{Q}(RM)=R$, proving Claim 2.

_Claim 3. $N_{X_{0}}(X_{\alpha\beta})=RN_{L}(M)=RM.m\leq
N_{X_{0}}(X_{\alpha})$._
Note that $N_{X_{0}}(X_{\alpha})=\operatorname{Sp}_{2}(q^{m}).m$ and that
$N_{\operatorname{Sp}_{2}(q^{m}).m}(RM)=RM.m$, so in particular $M.m\leq
N_{X_{0}}(RM)$ and $RM.m\leq N_{X}(X_{\alpha})$. As noted earlier in the
proof, $N_{X_{0}}(X_{\alpha\beta})=N_{X_{0}}(RM)\leq P=QL$, so
$N_{X_{0}}(RM)=N_{P}(RM)$. Now $N_{P}(RM)\leq N_{P}(QM)$ (since $QM=Q(RM)$ and
$Q\unlhd P$), and $N_{P}(QM)/Q\cong M.m$ (the normaliser in
$L=\operatorname{GL}_{m}(q)$ of the Singer cycle $M$). Similarly,
$N_{L}(M)=M.m$ and we just saw that this is contained in $N_{X}(RM)$. It
follows that $N_{P}(QM)=QN_{L}(M)$, and then that
$N_{P}(RM)=N_{Q}(RM)N_{L}(M)$. Using Claim 2, we obtain $N_{P}(RM)=RM.m$,
which is contained in $N_{X_{0}}(X_{\alpha})$. This proves Claim 3.

Now the involution $g$ of Hypothesis 2.2 lies in $N_{G}(X_{\alpha\beta})$ and
$g$ does not normalise $X_{\alpha}$. However, by Claim 3,
$N_{X_{0}}(X_{\alpha\beta})\leq N_{X}(X_{\alpha})$, and hence $g\not\in
X_{0}$. Thus $g$ induces an involutory field automorphism of $X$, and so it
must also induce an involutory field automorphism of $G$. However
$\Omega_{2m}^{-}(q)$ has no involutory field automorphisms (see for example
[25, Prop. 2.8.2]), a final contradiction, completing this case.

_Lines 4–5 a: $q$ even, $X_{\alpha}\cap X_{0}=\Omega_{2m}^{-}(q).a$, $G\cap
X_{0}=[q^{m}].(q^{m}-1).m^{\prime}$, and $G_{\alpha}\cap X_{0}=am^{\prime}$,
where $a\leq 2$ and $m^{\prime}\mid m$._
By Hypothesis 2.2(c), $X_{\alpha\beta}$ does not contain $\Omega_{2m}^{-}(q)$.
Also $|X_{\alpha}:X_{\alpha\beta}|$ divides $|G_{\alpha}|\leq 4mf$, whereas by
[25, Table 5.2.A], $|X_{\alpha}:X_{\alpha\beta}|$ is much larger than $4mf$.

_Lines 4–5 b: $q$ even, $X_{\alpha}\cap X_{0}=[q^{m}].(q^{m}-1).m^{\prime}$,
$G\cap X_{0}=\Omega_{2m}^{-}(q).a$, and $G_{\alpha}\cap X_{0}=am^{\prime}$,
where $a\leq 2$ and $m^{\prime}\mid m$._
Let $X_{\alpha}=RM.m^{\prime}$ where $R=[q^{m}]$ and $M=q^{m}-1$. Now
$|X_{\alpha\beta}|$ is divisible by $|X_{\alpha}:G_{\alpha}|$, which is
divisible by $q^{m}(q^{m}-1)/2fa$. If $R\cap X_{\alpha\beta}=1$ then this
implies that $q^{m}$ divides $2fm$, and hence that $m=q=2$, which is excluded.
Thus $R\cap X_{\alpha\beta}\neq 1$, and as $M$ acts irreducibly on $R$, we
must have $R\leq X_{\alpha\beta}$, contradicting Hypothesis 2.2(c).

_Line 6 a: $q$ even, $X_{\alpha}\cap X_{0}=\Omega_{2m}^{-}(q)$, $G\cap
X_{0}=\operatorname{Sp}_{2}(q^{m/2})\wr S_{2}$, and $G_{\alpha}\cap
X_{0}=D_{2(q^{m}-1)}$, where $m\equiv 2\pmod{4}$._
As usual, $X_{\alpha\beta}$ does not contain $\Omega_{2m}^{-}(q)$ by
Hypothesis 2.2(c). If $m=2$ (so $q\geq 8$ since $q$ is even), then
$X_{\alpha}=\operatorname{PSL}_{2}(q^{2}).a$ with $a\leq 2$ and
$G_{\alpha}=D_{2(q^{2}-1)}$ since there is no larger dihedral subgroup in
$X_{\alpha}$. Thus $|X_{\alpha\beta}|$ is divisible by
$|X_{\alpha}:G_{\alpha}|=q^{2}(q^{2}+1)a/2$, but the only such subgroups of
$X_{\alpha}$ contain $\operatorname{PSL}_{2}(q^{2})$, while $X_{\alpha\beta}$
does not contain $\operatorname{PSL}_{2}(q^{2})$. Thus $m\neq 2$ and so $m\geq
6$. Then by [25, Table 5.2.A],
$|X_{\alpha}:X_{\alpha\beta}|\geq(q^{m}+1)(q^{m-1}-1)/(q-1)$ whereas
$|X_{\alpha}:X_{\alpha\beta}|$ divides $|G_{\alpha}|$, which divides
$2(q^{m}-1)$, giving a contradiction.

_Line 6 b: $q$ even, $X_{\alpha}\cap X_{0}=\operatorname{Sp}_{2}(q^{m/2})\wr
S_{2}$, $G\cap X_{0}=\Omega_{2m}^{-}(q)$, and $G_{\alpha}\cap
X_{0}=D_{2(q^{m}-1)}$, where $m\equiv 2\pmod{4}$._
Now $X_{\alpha}=(\operatorname{Sp}_{2}(q^{m/2})\wr S_{2}).a$ for some $a$
dividing $f$, and the base group of $X_{\alpha}\cap X_{0}$ is $T_{1}\times
T_{2}$ with each $T_{i}=\operatorname{Sp}_{2}(q^{m/2})$ (a nonabelian simple
group). If $X_{\alpha\beta}$ contains $T_{i}$, then
$G_{\alpha\beta}=G_{\alpha}\cap X_{\alpha\beta}\geq D_{2(q^{m}-1)}\cap
T_{i}\geq(q^{m/2}\pm 1)$, contradicting Hypothesis 2.2(a). Thus
$X_{\alpha\beta}\cap T_{i}$ is a proper subgroup of $T_{i}$ for each $i$. On
the other hand, the index $|X_{\alpha}:G_{\alpha}|=q^{m}(q^{m}-1).a$, and this
divides $|X_{\alpha\beta}|$. Hence, for each $i$, $|X_{\alpha\beta}\cap
T_{i}|$ is divisible by $q^{m/2}/2$; and for some $i$, say $i=1$, a primitive
prime divisor $r$ of $2^{fm}-1$ (which exists since $(2m,q)\neq(4,8),(6,2)$)
divides $|X_{\alpha\beta}\cap T_{1}|$. The latter condition implies that
$X_{\alpha\beta}\cap T_{1}\leq N_{T_{1}}(C_{r})\leq(q^{m/2}+1).2f$, and hence
$q^{m/2}/2$ divides $2f$. This implies that $m=2$ and $q=16$. However,
$\operatorname{Sp}_{2}(16)$ has no proper subgroup of order divisible by
$17.8$, and we have a contradiction.

_Line 7 a: $q$ even, $X_{\alpha}\cap X_{0}=\Omega_{2m}^{-}(q)$, $G\cap
X_{0}=\operatorname{Sp}_{4}(q^{m/4}).2$, and $G_{\alpha}\cap
X_{0}=D_{2(q^{m/2}-1)}$, where $m\equiv 4\pmod{8}$ and $f$ is odd._
Arguing as in the case of Line 6.a, $X_{\alpha\beta}$ does not contain
$\Omega_{2m}^{-}(q)$, and $|X_{\alpha}:X_{\alpha\beta}|$ divides
$|G_{\alpha}|$, which divides $2(q^{m/2}-1).f$; whereas (since $m\geq 4$) by
[25, Table 5.2.A],
$|X_{\alpha}:X_{\alpha\beta}|\geq(q^{m}+1)(q^{m-1}-1)/(q-1)$, which is a
contradiction.

_Line 7 b: $q$ even, $X_{\alpha}\cap X_{0}=\operatorname{Sp}_{4}(q^{m/4}).2$,
$G\cap X_{0}=\Omega_{2m}^{-}(q)$, and $G_{\alpha}\cap X_{0}=D_{2(q^{m/2}-1)}$,
where $m\equiv 4\pmod{8}$ and $f$ is odd._
As usual, $X_{\alpha\beta}$ does not contain
$\operatorname{Sp}_{4}(q^{m/4})^{\prime}$ and $|X_{\alpha}:X_{\alpha\beta}|$
divides $|G_{\alpha}|$, which divides $2(q^{m/2}-1).f$, whereas (since $m\geq
4$, and $(2m,q)\neq(8,2)$) by [25, Table 5.2.A],
$|X_{\alpha}:X_{\alpha\beta}|\geq(q^{m}-1)/(q^{m/4}-1)$, giving a
contradiction.

_Line 8 a: $q$ even, $X_{\alpha}\cap X_{0}=\Omega_{2m}^{+}(q).2$, $G\cap
X_{0}=\operatorname{Sz}(q^{m/2})$, and $G_{\alpha}\cap
X_{0}=D_{2(q^{m/2}-1)}$, where $m\equiv 2\pmod{4}$ and $f$ is odd, $f>1$._
If $m\geq 4$, we obtain a contradiction by an exactly similar argument to that
for Line 7 a, so we may assume that $m=2$, and hence that $q>8$ (as $q$ is
even). Then $X_{\alpha}\cap
X_{0}=O_{4}^{+}(q)=\operatorname{Sp}_{2}(q^{m/2})\wr S_{2}$ with base group
$T_{1}\times T_{2}$ (where each $T_{i}=\operatorname{Sp}_{2}(q)$), as in Line
6 b. Arguing as for Line 6 b, for some $i$, $X_{\alpha\beta}\cap T_{i}$ is a
proper subgroup of $T_{i}$ with order divisible by $rq/2$, where $r$ is a
primitive prime divisor of $2^{2f}-1$ (note that $r$ exists since $q>8$), and
this implies that $q/2$ divides $2f$. This in turn implies that $q=16$, and
hence that $r=17$. However, $\operatorname{Sp}_{2}(16)$ has no proper subgroup
of order divisible by $17.8$, and we have a contradiction.

_Line 8 b: $q$ even, $X_{\alpha}\cap X_{0}=\,^{2}\\!B_{2}(q^{m/2})$, $G\cap
X_{0}=\Omega_{2m}^{+}(q).2$, and $G_{\alpha}\cap X_{0}=D_{2(q^{m/2}-1)}$,
where $m\equiv 2\pmod{4}$ and $f$ is odd, $f>1$._
In this case $X_{\alpha\beta}$ does not contain ${}^{2}\\!B_{2}(q^{m/2})$ and
$|X_{\alpha}:X_{\alpha\beta}|$ divides $|G_{\alpha}|$, which divides
$(q^{m/2}-1).f$. Also, see [3, Theorem 7.3.3],
$|X_{\alpha}:X_{\alpha\beta}|\geq q^{m}+1$, which is strictly larger than
$(q^{m/2}-1).f$, a contradiction.

_Line 9 a: $m=2,f=1$, $X_{\alpha}\cap X_{0}=\operatorname{PSp}_{2}(p^{2}).a$,
$G\cap X_{0}=[p^{1+2}](p^{2}-1)$, and $G_{\alpha}\cap X_{0}=p$ or $D_{2p}$,
where $a\mid 2$, and $X\geq X_{0}.2$ if $p\equiv 1\pmod{4}$._
Here $p\geq 5$, and by Hypothesis 2.2(c), $X_{\alpha\beta}$ does not contain
$\operatorname{Sp}_{2}(p^{2})$. Also $|X_{\alpha}:X_{\alpha\beta}|$ divides
$|G_{\alpha}|$ which divides $4p$, whereas by [25, Table 5.2.A],
$|X_{\alpha}:X_{\alpha\beta}|\geq p^{2}+1$, which gives a contradiction.

_Line 9 b: $m=2,f=1$, $X_{\alpha}\cap X_{0}=[p^{1+2}].(p^{2}-1)$, $G\cap
X_{0}=\operatorname{PSp}_{2}(p^{2}).a$, and $G_{\alpha}\cap X_{0}=p$ or
$D_{2p}$, where $a\mid 2$, and $X\geq X_{0}.2$ if $p\equiv 1\pmod{4}$._
Here $p\geq 5$, and $X_{\alpha}=QR$ with $Q=[p^{1+2}]$. Also
$p^{2}(p^{2}-1)/2$ divides $|X_{\alpha}:G_{\alpha}|$ which divides
$|X_{\alpha\beta}|$. Thus $|X_{\alpha\beta}\cap Q|\geq p^{2}$ and as
$(p^{2}-1)/2$ acts irreducibly on $Q/Q^{\prime}$ (since $p\geq 5$), it follows
that $X_{\alpha\beta}$ contains $Q$, which is normal in $X_{\alpha}$, and we
have a contradiction to Hypothesis 2.2(c).

####  8.2.3. Line in Table 10.3 with $X_{0}=\operatorname{PSU}_{2m}(q)$.

Let $V=\mathbb{F}_{q^{2}}^{2m}$ be the natural module for $X$, where $m\geq 2$
and $q=p^{f}$ for a prime $p$ and $f\geq 1$. There is only one Line to
consider, and it has $q=p$ or $q=4$. Moreover, from our assumptions, if $m=2$
then $q\geq 4$, and in particular $\operatorname{PSL}_{2m-1}(q)$ is a
nonabelian simple group.

_Line 1 a: $\operatorname{SU}_{2m-1}(q)\leq X_{\alpha}\cap X_{0}\leq N_{1}$,
$G\cap X_{0}=[q^{2m}].\frac{q^{2m}-1}{(q+1)d}.m$, and $G_{\alpha}\cap
X_{0}=[q]$, where $d=(2m,q+1)$ and $q=p$ or $q=4$._
The index $|X_{\alpha}:X_{\alpha\beta}|$ divides $|G_{\alpha}|$ which divides
$qf$. On the other hand, by Hypothesis 2.2(c), $X_{\alpha\beta}$ does not
contain $\operatorname{SU}_{2m-1}(q)$, and by [25, Table 5.2.A], therefore
$|X_{\alpha}:X_{\alpha\beta}|$ is much larger than $qf$.

_Line 1 b: $X_{\alpha}\cap X_{0}=[q^{2m}].\frac{q^{2m}-1}{(q+1)d}.m$,
$\operatorname{SU}_{2m-1}(q)\leq G\cap X_{0}\leq N_{1}$, $G_{\alpha}\cap
X_{0}=[q]$, where $d=(2m,q+1)$ and $q=p$ or $q=4$._
Let $Q=[q^{2m}]=O_{p}(X_{\alpha})$. Since $|G_{\alpha}|\leq qf$ and
$|X_{\alpha}:G_{\alpha}|$ divides $|X_{\alpha\beta}|$, it follows that
$f\cdot|X_{\alpha\beta}|$ is divisible by $q^{2m-1}(q^{2m}-1)/(q+1)d$. This
implies firstly that $Q\cap X_{\alpha\beta}\neq 1$. It also implies that
$X_{\alpha\beta}$ acts irreducibly on $Q$, and hence that $Q\leq
X_{\alpha\beta}$, which contradicts Hypothesis 2.2(c).

####  8.2.4. Line in Table 10.3 with
$X_{0}=\operatorname{P\Omega}_{2m}^{+}(q)$.

Let $V=\mathbb{F}_{q}^{2m}$ be the natural module for $X$. There is only one
line to consider. Moreover, from our assumptions, if $m=4$ then $q\geq 3$.

_Line 1 a: $\Omega_{2m-1}(q)\leq X_{\alpha}\cap X_{0}$, $G\cap
X_{0}\leq[q^{m}].\frac{q^{m}-1}{d}.m$, $G_{\alpha}\cap X_{0}=[q]$ or $[2q]$,
where $m\geq 4$, $q=p$ or $q=4$, and $d=(4,q^{m}-1)$._
This case gives a contradiction by the same argument as for Line 1a in
Subsection 8.2.3, using [25, Table 5.2.A] for
$\operatorname{P\Omega}_{2m-1}(q)$ with $m\geq 4$.

_Line 1 b: $X_{\alpha}\cap X_{0}\leq[q^{m}].\frac{q^{m}-1}{d}.m$,
$\Omega_{2m-1}(q)\leq G\cap X_{0}$, $G_{\alpha}\cap X_{0}=[q]$ or $[2q]$,
where $m\geq 4$, $q=p$ or $q=4$, and $d=(4,q^{m}-1)$._
This case gives a contradiction by the same argument as for Line 1b in
Subsection 8.2.3.

####  8.2.5. Lines in Table 10.3 with $X_{0}=\Omega_{7}(q)$.

There are two lines to consider, and in both lines $q=3^{f}$ for some $f\geq
1$.

_Line 1 a: $X_{\alpha}\cap X_{0}=\Omega_{6}^{\epsilon}(q)$, $G\cap
X_{0}=\operatorname{SL}_{3}^{-\epsilon}(q)$, $G_{\alpha}\cap X_{0}=q^{2}-1$,
with $\epsilon=\pm$._
The index $|X_{\alpha}:X_{\alpha\beta}|$ divides $|G_{\alpha}|$ which divides
$(q^{2}-1)f$. On the other hand, by Hypothesis 2.2(c), $X_{\alpha\beta}$ does
not contain $\Omega_{6}^{\epsilon}(q)$, and by [25, Table 5.2.A], therefore
$|X_{\alpha}:X_{\alpha\beta}|$ is much larger than $(q^{2}-1)f$.

_Line 1 b: $X_{\alpha}\cap X_{0}=\operatorname{SL}_{3}^{-\epsilon}(q)$, $G\cap
X_{0}=\Omega_{6}^{\epsilon}(q)$, $G_{\alpha}\cap X_{0}=q^{2}-1$, with
$\epsilon=\pm$._
If $\epsilon=+$ then this case gives a contradiction by the same argument as
for Line 1a, with $\operatorname{SL}_{3}^{-\epsilon}(q)$ in place of
$\Omega_{6}^{\epsilon}(q)$. If $\epsilon=-$, then $|X_{\alpha\beta}|$ is
divisible by $|X_{\alpha}:G_{\alpha}|$, which is divisible by
$q^{3}(q^{3}-1)$. However there is no proper subgroup of
$\operatorname{SL}_{3}(q).2f$ with this property (see [3, Tables 8.3,8.4]),
and hence $X_{\alpha\beta}$ contains $\operatorname{SL}_{3}(q)$, contradicting
Hypothesis 2.2(c).

_Line 2 a: $X_{\alpha}\cap X_{0}=\Omega_{6}^{+}(q).a$, $G\cap
X_{0}={}^{2}G_{2}(q)$, $G_{\alpha}\cap X_{0}=q-1$ or $D_{2(q-1)}$, where
$q=3^{f}$ with $f$ odd and $a\leq 2$._
This case gives a contradiction by the same argument as for Line 1a, with
$2(q-1)f$ in place of $(q^{2}-1)f$.

_Line 2 b: $X_{\alpha}\cap X_{0}={}^{2}G_{2}(q)$, $G\cap
X_{0}=\Omega_{6}^{+}(q).a$, $G_{\alpha}\cap X_{0}=q-1$ or $D_{2(q-1)}$, where
$q=3^{f}$ with $f$ odd and $a\leq 2$._
This case gives a contradiction by the same argument as for Line 1a, with
$2(q-1)f$ in place of $(q^{2}-1)f$, using the fact that each proper subgroup
of ${}^{2}G_{2}(q)$ has index at least $q^{3}+1$ for $q>3$, see [24, Theorem
C], and at least $9$ for $f=3$ as
${}^{2}G_{2}(q)\cong\operatorname{PSL}_{2}(8)$.

####  8.2.6. Lines in Table 10.3 with $X_{0}=G_{2}(q)$.

There are two lines to consider, and in both cases $p=3$.

_Lines 1 a and 1 b: $X_{\alpha}\cap
X_{0}=\operatorname{SL}_{3}^{\epsilon}(q)$, $G\cap
X_{0}=\operatorname{SL}_{3}^{-\epsilon}(q)$, $G_{\alpha}\cap X_{0}=q^{2}-1$,
with $\epsilon=\pm$._
The index $|X_{\alpha}:X_{\alpha\beta}|$ divides $|G_{\alpha}|$ which divides
$(q^{2}-1)f$. On the other hand, by Hypothesis 2.2(c), $X_{\alpha\beta}$ does
not contain $\operatorname{SL}_{3}^{\epsilon}(q)$. If $\epsilon=-$ then by
[25, Table 5.2.A], $|X_{\alpha}:X_{\alpha\beta}|\geq q^{3}+1$ which is much
larger than $(q^{2}-1)f$. So $\epsilon=+$. In this case the same argument as
in Line 1b of Subsection 8.2.5 gives a contradiction.

_Lines 2 a and 2 b: $\\{X_{\alpha}\cap X_{0},G\cap
X_{0}\\}=\\{{}^{2}G_{2}(q),\operatorname{SL}_{3}(q).a\\}$, $G_{\alpha}\cap
X_{0}=(q-1)$ or $D_{2(q-1)}$, with $q=3^{f}$, $f$ odd, and $a\leq 2$._
By Hypothesis 2.2(c), $X_{\alpha\beta}$ does not contain ${}^{2}G_{2}(q)$ or
$\operatorname{SL}_{3}(q)$, and by [24, Theorem C] or [25, Table 5.2.A],
$|X_{\alpha}:X_{\alpha\beta}|$ is at least $q^{3}+1$ or $q^{2}+q+1$,
respectively. On the other hand $|X_{\alpha}:X_{\alpha\beta}|$ divides
$|G_{\alpha}|$ which is at most $2(q-1)$, and we have a contradiction.

This completes the proof of Theorem 8.1.

##  9\. Sporadic groups

Proof of Theorems 1.4 and 1.10

All factorisations of the sporadic simple groups and their automorphism groups
are given in [9]. The simple sporadic groups that factorise are the Mathieu
groups (apart from $M_{22}$), together with $J_{2}$, $HS$, $Ru$, $Suz$,
$Fi_{22}$ and $Co_{1}$; in addition, there are factorisations of the almost
simple groups $M_{22}.2$, $J_{2}.2$, $HS.2$, $Suz.2$ and $He.2$ that do not
intersect the simple group in a factorisation. Given the results of [9], for
the largest groups $Fi_{22}$ and $Co_{1}$, we can read off from [6] that the
factorisations are not cyclic/dihedral. For the other groups, it is a routine
matter to compute using Magma (as discussed in Subsection 2.2) the
factorisations with cyclic or dihedral intersections, and to determine which
of these satisfy Hypothesis 2.2.

##  10\. Tables of cyclic/dihedral factorisations

This section contains the tables for the Factorisation Theorems 1.1, 1.2 and
1.4.

There is some notation that we use throughout the tables:

  * •

for $r$ a positive integer, $a_{r}$ and $b_{r}$ denote divisors of $r$

  * •

as is standard group-theoretic notation, a positive integer $n$ denotes a
cyclic group of order $n$, and the symbol $[n]$ denotes a (possibly non-
cyclic) group of order $n$

  * •

in Table 10.3 we just give the possibilities for $(X_{0},A\cap X_{0},B\cap
X_{0})$; whereas in Tables 10.4-10.7, we give the possibilities $(X,A,B)$
where $X$ is an almost simple group which is minimal subject to having a
factorisation $X=AB$ with cyclic or dihedral intersection.

  * •

in Table 10.8, if the socle factorises as $X_{0}=A_{0}B_{0}$, we do not
include factorisations of $X=X_{0}.2$ of the form $X=AB$, where $A\cap
X_{0}=A_{0},B\cap X_{0}=B_{0}$.

Table 10.1. Cyclic/dihedral factorisations of $A_{n}$ and $S_{n}$ $k$ | $X$ | $A$ | Conditions on $B$
---|---|---|---
2 |  $S_{n}$ or $A_{n}$ | $(S_{n-2}\times S_{2})\cap X$ |  $B^{\Omega}$ $2$-homog., $B<X$, $B_{\\{x,y\\}}$ as in Table 10.2
|  $S_{n}$ or $A_{n}$ | $(S_{n-2}\times 1)\cap X$ |  $B^{\Omega}$ $2$-trans., $B<X$, $B_{xy}$ as in Table 10.2
| $S_{n}$ | $(S_{n-2}\times S_{2})\cap A_{n}$ |  $B$ contains an odd perm., $(B\cap A_{n})^{\Omega}$ $2$-homog.,
|  |  |  $(B\cap A_{n})_{\\{x,y\\}}$ as in Table 10.2
| $S_{n}$ | $A_{n-2}\times 1$ |  $B$ contains an odd perm., $(B\cap A_{n})^{\Omega}$ $2$-trans.,
|  |  |  $(B\cap A_{n})_{xy}$ as in Table 10.2
| $S_{n}$ | $A_{n-2}\times S_{2}$ |  $(B\cap A_{n})^{\Omega}$ $2$-trans., $(B\cap A_{n})_{xy}$ as in Table 10.2
3 |  $S_{n}$ or $A_{n}$ | $(S_{n-3}\times S)\cap X$ |  $\operatorname{PSL}_{2}(q)\leq B\leq\widehat{B}:=\operatorname{P\Gamma L}_{2}(q)$ with $n=q+1$,
|  |  |  $S\leq S_{3}$; see Remark 5.2(a)
|  $S_{12}$ or $A_{12}$ | $(S_{9}\times 1)\cap X$ |  $B=M_{11}$, $A\cap B=B_{xyz}=D_{6}$
|  $S_{8}$ or $A_{8}$ | $(S_{5}\times S)\cap X$ |  $B=\operatorname{AGL}_{3}(2)$ and $A\cap B=D_{8},2^{2}$ for $S=S_{2},1$, resp.
|  $S_{8}$ or $A_{8}$ | $(S_{5}\times S_{3})\cap X$ |  $B=\operatorname{AGL}_{1}(8)$ and $A\cap B=B_{\\{x,y,z\\}}=1$
|  $S_{8}$ or $A_{8}$ | $(S_{5}\times S)\cap X$ |  $B=\operatorname{A\Gamma L}_{1}(8)$ and $A\cap B=C_{3},1$ for $S=S_{3},S_{2}$, resp.
|  $S_{32}$ or $A_{32}$ | $(S_{29}\times S_{3})\cap X$ |  $B=\operatorname{A\Gamma L}_{1}(32)$ and $A\cap B=B_{\\{x,y,z\\}}=1$
4 |  $S_{11}$ or $A_{11}$ | $(S_{7}\times S)\cap X$ |  $B=M_{11}$, $B_{xyzw}=1$, and $A\cap B=S\leq S_{4}$ cyclic/dihedral
|  $S_{9}$ or $A_{9}$ | $(S_{5}\times S)\cap X$ |  $B=\operatorname{PGL}_{2}(8)$ and $(S,A\cap B)=(S_{4},2^{2})$ or $(S_{3},1)$
|  $S_{9}$ or $A_{9}$ | $(S_{5}\times S)\cap X$ |  $B=\operatorname{P\Gamma L}_{2}(8)$ and $(S,A\cap B)$ as in Remark 5.2(b)
|  $S_{33}$ or $A_{33}$ | $(S_{29}\times S)\cap X$ |  $B=\operatorname{P\Gamma L}_{2}(32)$ and $(S,A\cap B)=(S_{4},2^{2})$ or $(S_{3},1)$
5 |  $S_{12}$ or $A_{12}$ | $(S_{7}\times S)\cap X$ |  $B=M_{12}$, $B_{xyzwv}=1$, $A\cap B=S\leq S_{5}$ cyclic/dihedral
Table 10.2. $2$-homogeneous groups $B$ with $B_{xy}$ or $B_{\\{x,y\\}}$ cyclic or dihedral: notation of Lemma 5.1 $n$ | $B^{\circ}$ | $B^{\circ}_{xy}$ | $B^{\circ}_{\\{x,y\\}}$ |  $\exists$ odd perm.
---|---|---|---|---
$q^{2}+1$ | ${}^{2}\\!B_{2}(q),\,q=2^{2a+1}$ | $q-1$ | $D_{2(q-1)}$ | $\times$
$q^{3}+1$ | ${}^{2}\\!G_{2}(q),\,q=3^{2a+1}$ | $q-1$ | $D_{2(q-1)}$ | $\times$
$q^{3}+1$ | $\operatorname{PSU}_{3}(q)$ | $(q^{2}-1)/(3,q+1)$ | $D_{2(q^{2}-1)/u}$ | $\times$
$q+1$ | $\operatorname{PSL}_{2}(q)$ | $(q-1)/(2,q-1)$ | $D_{2(q-1)/u}$ |  $\checkmark$ iff $q$ odd
$p^{2}$ | $\operatorname{ASL}_{2}(p)$ | $p$ | $D_{2p}$ |  $\checkmark$ iff $p\equiv 3\,(4)$
$q$ | $B\leq\operatorname{A\Gamma L}_{1}(q)$ | $B_{xy}\leq\langle\sigma\rangle$ | $B_{\\{x,y\\}}\leq$ |  $\checkmark$ iff $q$ odd
|  | ($\sigma$ field aut.) | $\langle\sigma\rangle\times 2$ |
$7$ | $\operatorname{PSL}_{2}(7)$ | $2^{2}$ | $D_{8}$ | $\times$
$11$ | $\operatorname{PSL}_{2}(11)$ | $D_{6}$ | $D_{12}$ | $\times$
$3^{4}$ | $3^{4}.2^{1+4}.5$ | 1 | 2 | $\times$
| $3^{4}.2A_{5}.2$ | 3 | 6 | $\times$
$3^{6}$ | $3^{6}.\operatorname{SL}_{2}(13)$ | 3 | 6 | $\times$
$5^{2}$ | $5^{2}.2A_{4}$ | $1$ | $2$ | $\checkmark$
$7^{2}$ | $7^{2}.2S_{4}$ | 1 | 2 | $\times$
$11^{2}$ | $11^{2}.(2A_{4}\times 5)$ | 1 | 2 | $\checkmark$
| $11^{2}.2A_{5}$ | 1 | 2 | $\times$
$19^{2}$ | $19^{2}.(2A_{5}\times 9)$ | 3 | 6 | $\times$
$23^{2}$ | $23^{2}.(2S_{4}\times 11)$ | 1 | 2 | $\times$
$29^{2}$ | $29^{2}.(2A_{5}\times 7)$ | 1 | 2 | $\times$
$59^{2}$ | $59^{2}.(2A_{5}\times 29)$ | 1 | 2 | $\times$
Table 10.3. Cyclic/dihedral factorisations of groups of Lie type: families
$\begin{array}[]{|l|l|l|l|l|}\hline\cr X_{0}&A\cap X_{0}&B\cap X_{0}&A\cap
B\cap X_{0}&\hbox{Conditions}\\\
\hline\cr\operatorname{PSL}_{n}(q),&A\triangleright
q^{n-1}.\operatorname{SL}_{n-1}(q)&\leq\frac{q^{n}-1}{(q-1)d}.n&r&r|n,\,d=(n,q-1)\\\
n\geq 2&&&&\\\
\operatorname{PSL}_{4}(p)&\operatorname{PSp}_{4}(p).a&p^{3}.\frac{p^{3}-1}{d}.e&p\times
e&p\geq 3\hbox{ prime},\,e|3,\\\ &&&&d=(4,p-1),\,ad=4,\\\ &&&&X\geq
X_{0}.2\hbox{ if }p\equiv 1\,(4)\\\
\hline\cr\operatorname{Sp}_{2m}(q),&\operatorname{O}_{2m}^{\epsilon}(q)&\operatorname{Sp}_{2}(q^{m})&D_{2(q^{m}-\epsilon)}&\epsilon=\pm
1\\\ q\hbox{
even},&\Omega_{2m}^{\epsilon}(q)&\operatorname{Sp}_{2}(q^{m})&q^{m}-\epsilon&m\hbox{
odd}\\\ m\geq
2&\Omega_{2m}^{-}(q)&\operatorname{Sp}_{2}(q^{m}).2&D_{2(q^{m}+1)}&m\hbox{
even},\,m_{2}=2\\\
&\operatorname{O}_{2m}^{-}(q)&q^{m}.(q^{m}-1).m^{\prime}&2m^{\prime}&m^{\prime}|m\\\
&\Omega_{2m}^{-}(q)&q^{m}.(q^{m}-1).m^{\prime}&m^{\prime}&m^{\prime}|m\,^{*}\\\
&\Omega_{2m}^{-}(q)&\operatorname{Sp}_{2}(q^{m/2})\wr
S_{2}&D_{2(q^{m}-1)}&\frac{m}{2}\hbox{ odd}\\\
&&\operatorname{Sp}_{4}(q^{m/4}).2&D_{2(q^{m/2}-1)}&\frac{m}{4}\hbox{
odd},\,q=2^{2f+1}\\\
&\operatorname{O}_{2m}^{+}(q)&{}^{2}\\!B_{2}(q^{m/2})&D_{2(q^{m/2}-1)}&\frac{m}{2}\hbox{
odd}\\\ &&&&\\\
\operatorname{PSp}_{4}(p)&\operatorname{PSp}_{2}(p^{2}).a&p^{1+2}.(p^{2}-1)&p,\,D_{2p}&a|2,\,X\geq
X_{0}.2\hbox{ if }p\equiv 1\,(4)\\\
\hline\cr\operatorname{PSU}_{2m}(q),&A\triangleright\operatorname{SU}_{2m-1}(q)&\leq
q^{2m}.\frac{q^{2m}-1}{(q+1)d}.m&[q]&q=p\hbox{ or }q=4,\\\ m\geq
2&&&&d=(2m,q+1)\\\ \hline\cr
P\Omega_{2m}^{+}(q),&A\triangleright\Omega_{2m-1}(q)&\leq
q^{m}.\frac{q^{m}-1}{d}.m&[q],\,[2q]&q=p\hbox{ or }q=4,\\\ m\geq
4&&&&d=(4,q^{m}-1)\\\
\hline\cr\Omega_{7}(q)&\Omega_{6}^{\epsilon}(q)&SL_{3}^{-\epsilon}(q)&q^{2}-1&q=3^{f},\,\epsilon=\pm
1\\\
&\Omega_{6}^{+}(q).a&{}^{2}\\!G_{2}(q)&q-1,\,D_{2(q-1)}&q=3^{2f+1},\,a|2\\\
\hline\cr G_{2}(q)&SL_{3}(q)&SU_{3}(q)&q^{2}-1&q=3^{f}\\\
&{}^{2}\\!G_{2}(q)&SL_{3}(q).a&q-1,\,D_{2(q-1)}&q=3^{2f+1},\,a|2\\\
\hline\cr\end{array}$

$*$ Only know an example if $(m^{\prime})_{2}=m_{2}$, where $m_{2}$ denotes
the $2$-part of $m$

Table 10.4. Exceptional cyclic/dihedral factorisations,
$X_{0}=\operatorname{PSL}_{n}(q)$ $\begin{array}[]{|l|l|l|l|}\hline\cr
X&A&B&A\cap B\\\ \hline\cr\operatorname{PSL}_{2}(7)&S_{4}&7,\,7.3&1,\,3\\\
\operatorname{PSL}_{2}(11)&A_{5}&11,\,11.5&1,\,5\\\ &A_{4}&11.5&1\\\
\operatorname{PSL}_{2}(16).4&A_{5}.4&D_{34}.4&2\\\
\operatorname{PSL}_{2}(19)&A_{5}&19.9&3\\\
\operatorname{PSL}_{2}(23)&S_{4}&23.11&1\\\
\operatorname{PSL}_{2}(29)&A_{5}&29.7,\,29.14&1,\,2\\\
\operatorname{PSL}_{2}(59)&A_{5}&59.29&1\\\
\hline\cr\operatorname{PSL}_{3}(3)&13.3&3^{2}.8&1\\\
\operatorname{PSL}_{3}(4).2&A_{6}.2&\operatorname{PSL}_{3}(2).2&D_{6}\\\
\operatorname{PSL}_{3}(4).S_{3}&7.3.S_{3}&2^{4}.(3\times D_{10}).2&1\\\
\operatorname{PSL}_{3}(8).3&73.9&2^{3+6}.7^{2}.3&1\\\
\operatorname{PSL}_{4}(3)&S_{5},\,4.A_{5},\,4.S_{5}&3^{3}.\operatorname{SL}_{3}(3)&3,\,D_{6},\,D_{12}\\\
&2^{4}.5.a_{4}&3^{3}.\operatorname{SL}_{3}(3)&2,\,4,\,8\\\
\operatorname{PSL}_{4}(3).2&(4\times A_{6}).2^{2}&3^{3}.13.3.2&1\\\
\operatorname{SL}_{4}(4)&\operatorname{Sp}_{4}(4)&2^{6}.63&2^{2}\\\
\operatorname{SL}_{4}(4).2&(a_{5}\times\operatorname{SL}_{2}(16)).4&\operatorname{SL}_{3}(4).[2a_{3}]&1,\,3,\,5,\,15\\\
&(5\times\operatorname{SL}_{2}(16)).4&2^{6}.63.6&1\\\
\operatorname{SL}_{5}(2)&31.5&2^{6}.(\operatorname{SL}_{3}(2)\times
S_{3})&1\\\
\operatorname{SL}_{6}(2)&G_{2}(2)&\operatorname{SL}_{5}(2)&D_{6}\\\
\operatorname{PSL}_{6}(3)&\operatorname{PSL}_{2}(13)&3^{5}.\operatorname{SL}_{5}(3)&3\\\
\hline\cr\end{array}$ Table 10.5. Exceptional cyclic/dihedral factorisations,
$X_{0}=\operatorname{PSU}_{n}(q)$. Note: exceptional factorisations of
$\operatorname{PSU}_{4}(2)\cong\operatorname{PSp}_{4}(3)$ given in Table 10.7
$\begin{array}[]{|l|l|l|l|}\hline\cr X&A&B&A\cap B\\\
\hline\cr\operatorname{PSU}_{3}(3)&\operatorname{PSL}_{2}(7)&3^{1+2}.8&D_{6}\\\
\operatorname{PSU}_{3}(8).3^{2}&57.9&2^{3+6}.63.3&1\\\
\operatorname{PSU}_{4}(3)&\operatorname{PSL}_{3}(4)&[2^{3}.3^{4}]&4\\\
\operatorname{PSU}_{4}(3).2&\operatorname{PSL}_{3}(4).2&[2.3^{4}.a_{8}]&1,\,2,\,2^{2},\,D_{8}\\\
&&[2.3^{5}.a_{2}]&3,\,D_{6}\\\ &&3^{4}.[10],\,3^{4}.[20]&5,\,D_{10}\\\
\operatorname{PSU}_{4}(3).[4]&\operatorname{PSL}_{2}(7).[4]&3^{4}.A_{6}.[4]&D_{6}\\\
\operatorname{SU}_{4}(4).4&(a\times\operatorname{SL}_{2}(16).4,\,a\leq
3&(a_{5}\times\operatorname{SU}_{3}(4)).4&1,\,2,\,3,\,5,\,15,\,D_{10}\\\
&(D_{6}\times\operatorname{SL}_{2}(16)).4&(a_{5}\times\operatorname{SU}_{3}(4)).4&D_{6},\,D_{30}\\\
\operatorname{SU}_{4}(8).3&513.3.3&2^{12}.\operatorname{SL}_{2}(64).7.3&1\\\
\hline\cr\end{array}$ Table 10.6. Exceptional cyclic/dihedral factorisations,
$X_{0}$ orthogonal $\begin{array}[]{|l|l|l|l|}\hline\cr X&A&B&A\cap B\\\
\hline\cr\Omega_{7}(3)&A_{9},\,S_{9}&3^{3}.\operatorname{SL}_{3}(3)&D_{6},\,D_{12}\\\
&\operatorname{Sp}_{6}(2)&[3^{6}].13.a_{3}&3,\,9\\\
\hline\cr\Omega_{8}^{+}(2)&\operatorname{Sp}_{6}(2)&A_{5}.[2a_{4}]&1,\,2,\,2^{2}\\\
&&A_{6}.[a_{4}]&3,\,D_{6},\,D_{12}\\\ &&[2^{4}.15.a_{4}]\hbox{
(sol.)}&2,\,4,\,8\\\ &A_{9}&2^{4}.A_{5}.[a_{8}]&1,\,2,\,2^{2},\,4,\,D_{8}\\\
&&2^{4}.A_{5}.[3a_{2}]&3,\,D_{6}\\\ &&[2^{6}.15.a_{4}]\hbox{
(sol.)}&1,\,2,\,4\\\
&A_{8},\,S_{8}&\operatorname{SU}_{4}(2),\,\operatorname{SU}_{4}(2).2&3,\,6,\,D_{6},\,D_{12}\\\
P\Omega_{8}^{+}(3).a_{2}&\Omega_{7}(3).b_{2}&3^{4}.[40c_{4}]&3,\,6,\,12\\\
\Omega_{8}^{+}(4).2&\operatorname{Sp}_{6}(4).2&\operatorname{SL}_{2}(16).4,\,2^{2}.\operatorname{SL}_{2}(16).4&1,\,2^{2}\\\
&\Omega_{8}^{-}(2).2&\Omega_{6}^{-}(4).4&D_{12}\\\
\Omega_{12}^{+}(2)&\operatorname{Sp}_{10}(2)&G_{2}(2)&D_{6}\\\
\hline\cr\end{array}$ Table 10.7. Exceptional cyclic/dihedral factorisations,
$X_{0}$ symplectic $\begin{array}[]{|l|l|l|l|}\hline\cr X&A&B&A\cap B\\\
\hline\cr\operatorname{PSp}_{4}(3)&2^{4}.A_{5}&[3^{3}a_{12}]&a_{12},\,D_{a_{12}}\\\
&S_{5}&3^{1+2}.Q_{8}.a_{3}&a_{3}\\\ &A_{6},\,S_{6}&3^{1+2}.Q_{8}&3,\,D_{6}\\\
&2^{4}.5.a_{2}&3^{1+2}.2A_{4}&2,\,4\\\
\operatorname{PSp}_{4}(3).2&2^{4}.5.4&[3^{3}].S_{3}&1\\\
&&3^{3}.A_{4}.2&2^{2}\\\
\operatorname{Sp}_{4}(4)&\operatorname{SL}_{2}(16)&2^{6}.15&2^{2}\\\
\operatorname{Sp}_{4}(4).2&\operatorname{SL}_{2}(16).4&A_{5}.[2a_{2}]&a_{2}\\\
&&A_{6}.[2a_{2}]&D_{6a_{2}}\\\
\operatorname{PSp}_{4}(5)&\operatorname{PSL}_{2}(25).a_{2}&5^{1+2}.Q_{8}.3b_{2}&5,\,D_{10},\,D_{20}\\\
\operatorname{PSp}_{4}(7)&\operatorname{PSL}_{2}(49).a_{2}&7^{1+2}.[48]&7,\,D_{14}\\\
\operatorname{PSp}_{4}(11)&\operatorname{PSL}_{2}(121).a_{2}&11^{1+2}.10A_{4}&11,\,D_{22}\\\
&&11^{1+2}.2A_{5}&11,\,D_{22}\\\
\operatorname{PSp}_{4}(23)&\operatorname{PSL}_{2}(23^{2}).a_{2}&23^{1+2}.22S_{4}&23,\,D_{46}\\\
\operatorname{PSp}_{4}(29)&\operatorname{PSL}_{2}(29^{2}).a_{2}&29^{1+2}.14b_{2}.A_{5}&29,\,D_{58}\\\
\operatorname{PSp}_{4}(59)&\operatorname{PSL}_{2}(59^{2}).a_{2}&59^{1+2}.(2A_{5}\times
29)&59,\,D_{118}\\\
\operatorname{Sp}_{6}(2)&S_{5}.[a_{6}]&\operatorname{SU}_{3}(3).b_{2}&1,\,2,\,3,\,D_{6}\\\
&A_{6}.[a_{4}]&\operatorname{SU}_{3}(3).b_{2}&3,\,6,\,D_{6},\,D_{12}\\\
&\operatorname{SL}_{3}(2).a_{2}&\operatorname{SU}_{4}(2).b_{2}&D_{6},\,D_{12}\\\
&2^{4}.A_{5}.[a_{4}]&\operatorname{SL}_{2}(8).3&1,\,2,\,2^{2}\\\
&2^{4}.A_{5}.a_{2}&\operatorname{SU}_{3}(3)&4,\,8\\\
&2^{5}.A_{6}&\operatorname{SL}_{2}(8)&2^{2}\\\
&A_{8}.a_{2}&[3^{3}].[8b_{2}]&D_{6},\,D_{12}\\\
\operatorname{PSp}_{6}(3)&\operatorname{PSL}_{2}(13)&P_{1}&3\\\
&\operatorname{PSL}_{2}(27).3&3^{1+4}.2^{1+4}.A_{5}&3\\\
\operatorname{PSp}_{6}(3).2&\operatorname{PSL}_{2}(27).6&3^{1+4}.2^{1+4}.D_{10}.2&1\\\
\operatorname{Sp}_{6}(4).2&\operatorname{O}_{6}^{-}(4).2&\operatorname{SU}_{3}(3).2&D_{6}\\\
&G_{2}(4).2&\operatorname{SL}_{2}(16).4&1\\\
\operatorname{Sp}_{8}(2)&\operatorname{O}_{8}^{+}(2)&\operatorname{PSL}_{2}(17)&D_{18}\\\
&\Omega_{8}^{-}(2)&A_{6}.2^{2}&D_{6}\\\
&\operatorname{O}_{8}^{-}(2)&A_{5}.[2a_{4}]&1,\,2,\,2^{2}\\\
&&A_{5}.[a],\,a=6,12,20&3,\,D_{6},\,D_{10}\\\
&&A_{6}.[a_{4}]&3,\,6,\,D_{6},\,D_{12}\\\
&&2^{4}.A_{5}.a_{2}&D_{8},\,D_{16}\\\
\operatorname{Sp}_{12}(2)&\operatorname{O}_{12}^{-}(2)&\operatorname{SU}_{3}(3).2&D_{6}\\\
\hline\cr\end{array}$ Table 10.8. Cyclic/dihedral factorisations of sporadic
groups $\begin{array}[]{|l|l|l|l|}\hline\cr X&A&B&A\cap B\\\ \hline\cr
M_{11}&M_{10}&11,\,11.5&1,\,5\\\
&M_{9}.2&\operatorname{PSL}_{2}(11),\,11.5&D_{12},\,1\\\
&M_{9},\,9.8&\operatorname{PSL}_{2}(11)&D_{6}\\\ \hline\cr
M_{12}&M_{11}&\hbox{many poss.}&C_{i}\,(1\leq i\leq 6),\\\ &&&D_{2j}\,(3\leq
j\leq 6)\\\
&\operatorname{PSL}_{2}(11)\,\hbox{(max.)}&M_{10},\,M_{10}.2&5,\,D_{10}\\\
&&M_{9}.2,\,M_{9},S_{3}&1,\,3\\\ &&S_{6}&5\\\ \hline\cr
M_{23}&M_{22}&23,\,23.11&1,\,11\\\ &M_{21}&23.11&1\\\ &2^{4}.A_{7}&23.11&1\\\
\hline\cr M_{24}&M_{23}&\hbox{many poss.}&C_{i}\,(1\leq i\leq 8),\\\
&&&D_{2j}\,(j=2,4,5,6)\\\
&\operatorname{PSL}_{2}(23)&M_{22},\,M_{22}.2&11,\,D_{22}\\\
&&\operatorname{PSL}_{3}(4).2,\,\operatorname{PSL}_{3}(4).S_{3}&1,\,3\\\
&&2^{4}.A_{7},\,2^{4}.A_{8}&1,\,D_{8}\\\ \hline\cr
J_{2}&\operatorname{PSU}_{3}(3)&A_{5}\times 5,\,A_{5}\times D_{10}&3,\,6\\\
J_{2}.2&\operatorname{PSU}_{3}(3).2&5^{2}.[4a_{6}],\,(5\times
A_{5}).[a_{4}]&1,\,2,\,3,\,6,\,12\\\ \hline\cr HS&M_{22}&A_{5}\times
5,\,A_{5}\times D_{10}&3,\,6\\\
HS.2&M_{22}.2&5^{2}.[4a],\,a=1,2,4,5&1,\,2,\,4,\,5\\\ &&5^{1+2}.(2\times
4)&D_{10}\\\ &&(5\times A_{5}).[a_{4}]&3,\,6,\,12,\,D_{6},\,D_{12}\\\
\hline\cr
He&\operatorname{Sp}_{4}(4).a_{2}&7^{2}.\operatorname{SL}_{2}(7)&4,\,8\\\
He.2&\operatorname{Sp}_{4}(4).4&7^{1+2}.[6a_{6}]&1,\,2,\,3,\,D_{6}\\\
&&7^{2}.\operatorname{SL}_{2}(7).a_{2}&8,\,16\\\ \hline\cr
Ru&{}^{2}\\!F_{4}(2).2&\operatorname{PSL}_{2}(29)&3\\\ \hline\cr
Suz.2&G_{2}(4).2&3^{5}.11.2a_{5}&3,\,15\\\ \hline\cr\end{array}$

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