# Symmetric graphs of prime valency
with a transitive simple group

Jing Jian Li  J.J. Li
College of Mathematics and Information Science
Guangxi University
Nanning 530004
P. R. China<EMAIL_ADDRESS>and  Zai Ping Lu  Z.P. Lu
Center for Combinatorics
LPMC, Nankai University
Tianjin 300071
P. R. China<EMAIL_ADDRESS>
###### Abstract.

A graph $\Gamma=(V,E)$ is called a Cayley graph of some group $T$ if the
automorphism group $\mathbf{Aut}(\Gamma)$ contains a subgroup $T$ which acts
on regularly on $V$. If the subgroup $T$ is normal in $\mathbf{Aut}(\Gamma)$
then $\Gamma$ is called a normal Cayley graph of $T$. Let $r$ be an odd prime.
Fang et al. [12] proved that, with a finite number of exceptions for finite
simple group $T$, every connected symmetric Cayley graph of $T$ of valency $r$
is normal. In this paper, employing maximal factorizations of finite almost
simple groups, we work out a possible list of those exceptions for $T$.

Keywords. Symmetric graph, Cayley graph, simple group, group factorization.

2010 Mathematics Subject Classification. 05C25, 20B25, 20D40

Supported by the National Natural Science Foundation of China (11861012,
11971248, 11731002) and the Fundamental Research Funds for the Central
Universities.

##  1\. Introduction

All graphs in this paper are assumed to be finite, simple and undirected.

Let $\Gamma=(V,E)$ be a graph with vertex set $V$ and edge set $E$. Denote by
$\mathbf{Aut}(\Gamma)$ the automorphism group of $\Gamma$, and let
$G\leq\mathbf{Aut}(\Gamma)$. The graph $\Gamma$ is called $G$-vertex
transitive if $G$ acts transitively on $V$, and called a Cayley graph of $G$
if $G$ acts regularly on $V$. For $\alpha\in V$, let $G_{\alpha}=\\{g\in
G\mid\alpha^{g}=\alpha\\}$ and $\Gamma(\alpha)=\\{\beta\in
V\mid\\{\alpha,\beta\\}\in E\\}$, called the stabilizer and neighborhood of
$\alpha$ in $G$ and in $\Gamma$, respectively. If $\Gamma$ is $G$-vertex
transitive and $G_{\alpha}$ acts transitively on $\Gamma(\alpha)$ then
$\Gamma$ is called $G$-symmetric.

In the literature, the solution of quite a number of problems on symmetric
graphs have been reduced or partially reduced into the class of graphs arising
from almost simple groups. For example, the reduction for symmetric graphs of
prime valency [34], the reduction for $2$-arc-transitive graphs established in
[38], the Weiss Conjecture [47, Conjecture 3.12] for non-bipartite locally
primitive graphs [6], the normality of Cayley graphs of simple groups [12,
13], the existence and classification of edge-primitive graphs [18, 36], and
so on. Certainly, the class of graphs arising from almost simple groups plays
an important role in the theory of symmetric graphs. In this paper, we will
study those symmetric graphs admitting a transitive simple group.

Our main motivation comes from the study of the automorphism groups or the
normality of symmetric Cayley graphs of simple groups. (Recall that a Cayley
graph $\Gamma$ of a group $T$ is normal [48] if $T$ is a normal subgroup of
$\mathbf{Aut}(\Gamma)$.) Let $\Gamma=(V,E)$ be a Cayley graph of some simple
group $T$ of prime valency $r\geq 3$. Assume that $\Gamma$ is connected and
$G$-symmetric, where $T\leq G\leq\mathbf{Aut}(\Gamma)$. Then the Weiss
Conjecture is true for $\Gamma$, that is, the orders of vertex-stabilizers
have an upper bound depending only on the valency $r$. This ensures that $T$
is normal in $G$ with a finite number of exceptions, see [12, Theorem 1.1].
What interests us are those exceptions for $T$. As proposed in [12], those
exceptions for $T$ are worth figuring out. This problem has been solved for
$r\leq 5$ in several papers, refer to [9, 11, 12, 43]. In a recent paper [44],
the exceptions for $T$ are classified under the assumption that a vertex-
stabilizer $G_{\alpha}$ is solvable. In this paper, we are trying to make some
progress on this problem.

Let $\Gamma=(V,E)$ be a connected $G$-symmetric Cayley graph of a simple group
$T$, where $T\leq G\leq\mathbf{Aut}(\Gamma)$. Assume that $\Gamma$ has odd
prime valency $r$. Take an intransitive normal subgroup $K$ of $G$, and define
a graph $\Gamma_{G/K}$ with vertex set $V_{K}:=\\{\alpha^{K}\mid\alpha\in
V\\}$ and edge set
$E_{K}:=\\{\\{\alpha^{K},\beta^{K}\\}\mid\\{\alpha,\beta\\}\in E\\}$. Let
$\overline{G}$ be the group induced by $G$ on $V_{K}$. Then $\Gamma_{G/K}$ is
a $\overline{G}$-symmetric graph of valency $r$, $K$ is semiregular on $V$ and
$\overline{G}\cong G/K$, see [34, Theorem 9] for example. Note that $T\cap
K=1$. Then $\overline{G}$ has a transitive simple subgroup which is isomorphic
to $T$. These observations suggest us to investigate the symmetric graphs
which have prime valency and admit a transitive simple group.

Let $G^{\infty}$ be the solvable residue of $G$, i.e., the last term of the
derived series of $G$. Let $\mathbf{rad}(G)$ be the solvable radical of $G$,
that is, the maximal solvable normal subgroup of $G$. In Section 3, the
following result is proved.

######  Theorem 1.

Let $\Gamma=(V,E)$ be a connected $G$-symmetric graph of prime valency $r\geq
3$. Assume that $G$ contains a transitive simple subgroup $T$.

  * (1)

If $G$ has a transitive minimal normal subgroup $N$ then either $N$ is simple
and $T\leq N$, or $N\cong\mathbb{Z}_{2}^{3}$, $r=7$ and
$T\cong\mathrm{L}_{3}(2)$; for the former case, $\Gamma$ is $N$-symmetric if
$T\neq N$.

  * (2)

If $T$ is not normal in $G$ then $G$ has a characteristic subgroup $N$ such
that $\Gamma$ is $N$-symmetric, $N/\mathbf{rad}(N)$ is simple,
$|\mathbf{rad}(N)|$ is indivisible by $r$, $\mathbf{rad}(N)T\lneq
N=N^{\infty}$, and either $\mathbf{rad}(N)$ is intransitive on $V$ or
$G=N\cong\mathrm{AGL}_{3}(2)$.

Part (2) of Theorem 1 provides us a method to determine the normality of
symmetric Cayley graphs of prime valency of simple groups, which also suggests
us to consider the symmetric graphs satisfying the following hypothesis.

######  Hypothesis 1.1.

$\Gamma=(V,E)$ is a connected graph of prime valency $r\geq 3$ such that

  * (I)

$\Gamma$ is $G$-symmetric, where $G$ is a nonabelian simple group;

  * (II)

$G=TG_{\alpha}$ for $\alpha\in V$, where $T$ is a simple group with $T\neq G$.

Note that the simple group $T$ in (II) is nonabelian; otherwise, $\Gamma$ has
odd prime order, yielding $r=2$, a contradiction.

Most part of this paper, Sections 4-8, is devoted to analyzing the possibility
of the triples $(G,T,G_{\alpha})$ satisfying Hypothesis 1.1. In Section 4, we
give some limitations on $G_{\alpha}$ or $G$. Then the case where $G_{\alpha}$
is solvable is dwelt in Section 5, the case where $G$ (or $T$) is an
alternating group or a sporadic simple group is dwelt in Section 6, and the
case where $G$ is a classical group is left in Sections 7 and 8. At last, the
following result is proved in Section 9.

######  Theorem 2.

Let $\Gamma=(V,E)$ be a connected $X$-symmetric Cayley graph of a finite
simple group $T$ of prime valency $r\geq 7$, where $T\leq
X\leq\mathbf{Aut}(\Gamma)$. Assume that $T$ is not normal in $X$ and
$X_{\alpha}$ is insolvable, where $\alpha\in V$. Then $T$ is isomorphic to a
proper subgroup of some composition factor $L$ of $X$, and all possible
triples $(L,T,r)$ are listed in Table 1.1.

$\footnotesize\begin{array}[]{|l|l|c|l|l|}\hline\cr\mbox{Line}&L&T&r&\mbox{Conditions}\\\
\hline\cr 1&\mathrm{A}_{n}&\mathrm{A}_{n-1}&\geq
7&r\,\,\big{|}\,\,n,r^{2}\nmid n,n\mbox{ not prime}\\\ \hline\cr
2&\mathrm{A}_{r+1}&|T|=r+1&\geq 7&\Gamma=\mathsf{K}_{r+1}\\\ \hline\cr
3&\mathrm{U}_{2m}(2^{f})&\mathrm{U}_{2m-1}(2^{f})&2^{mf}+1&m\geq
2,mf=2^{e}>2\\\ \hline\cr
4&\Omega_{2m+1}(p^{f})&\mathrm{P\Omega}^{-}_{2m}(p^{f})&{p^{mf}-1\over
p^{{m\over d}f}-1}&m\geq 3,mf=d^{e},\mbox{odd primes }p,d\\\ \hline\cr
5&\mathrm{P\Omega}^{-}_{10}(2^{f})&\mathrm{U}_{5}(2^{f})&2^{4f}+1&f=2^{e}>1\\\
\hline\cr 6&\Omega^{+}_{2m}(2^{f})&\mathrm{Sp}_{2m-2}(2^{f})&{2^{mf}-1\over
2^{mf\over d}-1}&m\geq 4,mf=d^{e},\mbox{prime }d\\\ \hline\cr
7&\Omega^{+}_{2m}(2^{f})&\Omega^{-}_{2m-2}(2^{f})&{2^{mf}-1\over 2^{{m\over
d}f}-1}&m\geq 5,mf=d^{e},\mbox{odd prime }d\\\ \hline\cr
8&\Omega^{+}_{2m}(p^{f})&\mathrm{P\Omega}^{-}_{2m-2}(p^{f})&{p^{mf}-1\over
p^{f}-1}&\mbox{odd prime }m\geq 5,f=m^{e},\\\ \hline\cr
9&\mathrm{P\Omega}^{+}_{2m}(p^{f})&\Omega_{2m-1}(p^{f})&{p^{mf}-1\over
p^{{m\over d}f}-1}&m\geq 5,mf=d^{e},\mbox{odd primes }p,d\\\ \hline\cr
10&\mathrm{Sp}_{4}(2^{f})&\mathrm{SL}_{2}(2^{2f})&2^{f}+1&f=2^{e}>2\\\
\hline\cr
11&\mathrm{Sp}_{6}(2^{f})&\mathrm{G}_{2}(2^{2f}),\mathrm{SL}_{2}(2^{3f})&2^{2f}+1&f=2^{e}\geq
2\\\ \hline\cr
12&\mathrm{Sp}_{2m}(2^{f})&\Omega^{-}_{2m}(2^{f})&{2^{mf}-1\over 2^{{m\over
d}f}-1}&m\geq 4,mf=d^{e},\mbox{prime }d\\\ \hline\cr
13&\mathrm{L}_{n}(p^{f})&\mathrm{L}_{n-1}(p^{f})&{p^{nf}-1\over p^{{n\over
d}f}-1}&n>d,nf=d^{e},\mbox{odd prime }d\\\ \hline\cr
14&\mathrm{L}_{n}(p^{f})&\mathrm{L}_{m}(p^{{n\over m}f})&{p^{(n-1)f}-1\over
p^{f}-1}&1<m<n,\mbox{ and }(*):\,m\,\,\big{|}\,\,n,\\\
&&&&f=(n-1)^{e},\mbox{odd prime }n-1\\\ &&\mathrm{PSp}_{m}(p^{{n\over
m}f})&&(*),\mbox{even }m\geq 4,p=2\mbox{ if }m=n\\\
&&\mathrm{P\Omega}^{-}_{m}(p^{{n\over m}f})&&(*),\mbox{even }{m\over 2}\geq
4,p=2\mbox{ if }m=n\\\ &&\mathrm{G}_{2}(p^{{n\over m}f})&&(*),m=6,p=2\\\
\hline\cr 15&\mathrm{U}_{6}(2)&\mathrm{U}_{5}(2)&7&\\\ \hline\cr
16&\Omega^{+}_{12}(2)&\mathrm{Sp}_{10}(2)&7&\\\ \hline\cr
17&\mathrm{Sp}_{12}(2)&\Omega^{-}_{12}(2)&7&\\\ \hline\cr\end{array}$ Table
1.1. Exceptions for $T$.

##  2\. Preliminaries

In this section, we collects a few of preliminary results which will be used
frequently in the following sections.

###  2.1. Primitive divisors

Let $\Phi_{m}(q)$ be the $m$-th cyclotomic polynomial evaluated at a prime
power $q=p^{f}$. Then

$q^{m}-1=\prod_{d\mid m}\Phi_{d}(q),\,\,q^{m}+1=\prod_{d\mid 2m,d\nmid
m}\Phi_{d}(q).$

Noting that

$\prod_{d\mid mf,d\nmid f}\Phi_{d}(p)={q^{m}-1\over q-1}=\prod_{1<d\mid
m}\Phi_{d}(q),$

the next lemma is easily shown.

######  Lemma 2.1.

If ${q^{d}-1\over q-1}$ is a prime then $d$ is a prime, $(d,q-1)=1$, $f$ is a
power of $d$, and either $d$ is odd or $d=p=2$.

Recall that a primitive prime divisor of $q^{m}-1$ is a prime which divides
$q^{m}-1$ but does not divide $q^{l}-1$ for any $0<l<m$. If $r$ is a primitive
prime divisor of $q^{m}-1$, then $q$ has order $m$ modulo $r$, and thus $m$ is
a divisor of $r-1$; if further $r\,\,\big{|}\,\,(q^{n}-1)$ with $n\geq 1$ then
$m\,\,\big{|}\,\,n$. Let $\Phi_{m}^{*}(q)$ be the product of all primitive
prime divisors of $q^{m}-1$ (counting multiplicities). Then the following
lemma holds.

######  Lemma 2.2.

If $\Phi_{m}^{*}(q)\neq 1$ then every prime divisor of $\Phi_{m}^{*}(q)$ is no
less than $m+1$, and $(\Phi_{m}^{*}(q),q^{n}-1)\neq 1$ yields that
$m\,\,\big{|}\,\,n$.

It is easily shown that $\Phi_{m}^{*}(q)$ is a divisor of $\Phi_{m\over
d}^{*}(q^{d})$, where $d$ is a divisor of $m$. In particular,
$\Phi_{mf}^{*}(p)$ is a divisor of $\Phi_{m}^{*}(q)$. Moreover, by Zsigmondy’s
Theorem, we have the following fact.

######  Lemma 2.3.

$\Phi_{m}^{*}(q)=1$ if and only if $(m,q)=(6,2)$ or $m=2$ and $q=p=2^{s}-1$
for some prime $s\geq 2$.

###  2.2. Factorizations

All groups in this paper are assumed to be finite, and our group-theoretic
notation mostly consistent with [31].

Let $G$ be a group. An expression $G=XY$ is called a factorization of $G$ if
$X$ and $Y$ are proper subgroups of $G$, where $X$ and $Y$ are called factors.
A factorization $G=XY$ is called maximal if both $X$ and $Y$ are maximal
subgroups of $G$. The next lemma collects a few of simple facts, most of them
follow from Lagrange’s Theorem.

######  Lemma 2.4.

Let $G=XY$ be factorization of $G$. Then the following hold.

  * (1)

$YX=G=A^{g}Y^{h}$ for $X\leq A\leq G$ and $g,h\in G$.

  * (2)

$|G:X|=|Y:(X\cap Y)|$ and $|G:Y|=|X:(X\cap Y)|$; in particular, $|X|$ and
$|Y|$ are divisible by $|G:Y|$ and $|G:X|$, respectively.

  * (3)

$|G|_{p}\leq|X|_{p}|Y|_{p}$, $|X|_{p}\geq|G:Y|_{p}$ and
$|Y|_{p}\geq|G:X|_{p}$, where $p$ is a prime and $|\cdot|_{p}$ means the
largest $p$-power dividing $|\cdot|$.

  * (4)

There is a maximal factorization $G=AB$ with $X\leq A$ and $Y\leq B$.

Let $X$ be a subgroup of a group $G$. The intersection $\cap_{g\in G}X^{g}$ is
called the core of $X$ in $G$, denoted by $\mathbf{core}_{G}(X)$. If
$\mathbf{core}_{G}(X)=1$, i.e., $X$ contains no normal subgroup of $G$ other
than $1$, then $X$ is called a core-free subgroup of $G$.

Given a factorization $G=XY$, we have a transitive action of $G$ on
$[G:Y]:=\\{Yg\mid g\\}$ by right multiplication, where the kernel is
$\mathbf{core}_{G}(Y)$. Clearly, $X$ acts transitively on $[G:Y]$. If $Y$ is
core-free in $G$ then $G$ (and hence $X$) acts faithfully on $[G:Y]$, so all
orbits of a normal subgroup of $X$ have equal length greater than $1$. This
implies the next fact.

######  Lemma 2.5.

Let $G$ be a group and $G=XY$ be a factorization. If $Y$ is core-free in $G$,
then $X\cap Y$ is core-free in both $X$ and $G$.

A factorization $G=XY$ is called core-free if $X$ and $Y$ are core-free in
$G$. Let $G=XY$ be a core-free factorization. Assume that $X<A\leq G$. Then we
have a factorization $A=X(A\cap Y)$. By Lemma 2.5, since $G=AY$, we know that
$A\cap Y$ is core-free in $A$. In general, $X$ is not necessarily core-free in
$A$, and the factorization $G=AY$ is not necessarily core-free too.

For a simple group $S$ and a group $G$, by $S\preccurlyeq G$ or $G\succcurlyeq
S$ we means that $G$ has a composition factor (isomorphic to) $S$.

######  Lemma 2.6.

Let $G=XB$ be a core-free factorization and $X\leq A\leq G$. Assume that
$N\unlhd A$, $A/N$ is almost simple, and $X^{\infty}\not\leq N$. Let
$\overline{A}=A/N$, $\overline{X}=XN/N$ and $\overline{Y}=(A\cap B)N/N$. Then
either $\overline{A}=\overline{X}\,\overline{Y}$ is core-free, or one of the
following holds:

  * (1)

$\overline{A}^{\infty}\preccurlyeq X$;

  * (2)

$|G:B|$ is a divisor of $|\overline{A}:\overline{A}^{\infty}||N|$.

If $\mathbf{core}_{G}(A)=1$ and $N=\mathbf{rad}(A)$ normalize $A\cap B$ then
$\overline{A}^{\infty}\not\preccurlyeq A\cap B$.

###### Proof.

Assume that either $\overline{X}$ or $\overline{Y}$ is not core-free in
$\overline{A}$. If $\overline{X}$ is not core-free in $\overline{A}$ then part
(1) of this lemma follows. Suppose that $\overline{Y}$ is not core-free in
$\overline{A}$. Then $NA^{\infty}\leq N(A\cap B)$. Thus $|G:B|=|A:(A\cap
B)|=|A:N(A\cap B)||N(A\cap B):(A\cap B)|$, and so $|G:B|$ is a divisor of
$|A:NA^{\infty}||N|$. Then part (2) of this lemma occurs.

Now let $\mathbf{core}_{G}(A)=1$ and $\mathbf{rad}(A)$ normalize $A\cap B$.
Then $((A\cap B)\mathbf{rad}(A))^{\infty}=(A\cap B)^{\infty}$. Suppose that
$S\preccurlyeq A\cap B$. Then $(A\cap B)\mathbf{rad}(A)\geq
A^{\infty}\mathbf{rad}(A)$, and so $A^{\infty}\leq((A\cap
B)\mathbf{rad}(A))^{\infty}=(A\cap B)^{\infty}$. Thus $A^{\infty}\leq B\cap
A$, which contradicts Lemma 2.5. Then the lemma follows. ∎

######  Lemma 2.7.

Let $G=XY$ be a core-free factorization. Suppose that $G$ is almost simple.
Then there exists a core-free maximal factorization $G_{0}=A_{0}B_{0}$ such
that

  * (1)

$G\geq G_{0}=(A_{0}\cap X)(B_{0}\cap Y)$, $G^{\infty}\leq G_{0}$,
$X^{\infty}\leq A_{0}$ and $Y^{\infty}\leq B_{0}$;

  * (2)

$|X:(A_{0}\cap X)|$ and $|Y:(B_{0}\cap Y)|$ are divisors of $|G:G_{0}|$.

###### Proof.

If $G^{\infty}X=G=G^{\infty}Y$, taking maximal subgroups $A_{0}$ and $B_{0}$
of $G$ with $X\leq A_{0}$ and $Y\leq B_{0}$, then $G=A_{0}B_{0}$ is a desired
factorization. Thus, with out loss of generality, we assume next that $G\neq
G^{\infty}X$.

Let $G_{1}=G^{\infty}X$ . Then $G=G_{1}Y$, and $G_{1}=X(G_{1}\cap Y)$. Let
$Y_{1}=G_{1}\cap Y$. It is easy to see that $X$ and $Y_{1}$ are core-free in
$G_{1}$. Since $G=G_{1}Y$, by Lemma 2.4, $|Y:Y_{1}|=|G:G_{1}|$. Moreover,
$Y^{\infty}\leq G^{\infty}=G_{1}^{\infty}$, and so $Y^{\infty}\leq G_{1}\cap
Y=Y_{1}$, yielding $Y^{\infty}=Y_{1}^{\infty}$. Noting that $|G_{1}|<|G|$, by
induction, there are $G_{0}\leq G_{1}$ and a core-free maximal factorization
$G_{0}=A_{0}B_{0}$ such that $G_{1}^{\infty}\leq G_{0}$, $X^{\infty}\leq
A_{0}$ $Y_{1}^{\infty}\leq B_{0}$, $|X:(A_{0}\cap X)|$ and $|Y_{1}:(B_{0}\cap
Y_{1})|$ are divisors of $|G_{1}:G_{0}|$. Since $Y_{1}\leq Y$, we have
$|Y:(B_{0}\cap Y)||(B_{0}\cap Y):(B_{0}\cap Y_{1})|=|Y:(B_{0}\cap
Y_{1})|=|Y:Y_{1}||Y_{1}:(B_{0}\cap Y_{1})|$, it follows that $|Y:(B_{0}\cap
Y)|$ is a divisor of $|G:G_{0}|$. Clearly, $G^{\infty}=G_{1}^{\infty}\leq
G_{0}$ and $Y^{\infty}=Y_{1}^{\infty}\leq B_{0}$. Then the factorization
$G_{0}=A_{0}B_{0}$ is desired. ∎

###  2.3. Stabilizers

Let $\Gamma=(V,E)$ be a connected $G$-symmetric graph of prime valency $r\geq
3$. Denote by $\mathbf{soc}(G)$ the socle of $G$, and by $\pi(G)$ the set of
prime divisors of $|G|$. By [6, Lemma 1.1], [34, Theorem 9] and [35, Theorem
1], the following lemma holds.

######  Lemma 2.8.

$|G_{\alpha}|_{r}=r=\max\pi(G_{\alpha})$. If $N\unlhd G$ then either $N$ is
semiregular on $V$, or $N_{\alpha}$ is transitive on $\Gamma(\alpha)$ and $N$
has at most two orbits on $V$.

Take $\\{\alpha,\beta\\}\in E$. Then the arc-stabilizer $G_{\alpha\beta}$ is a
Hall $r^{\prime}$-subgroup of $G_{\alpha}$. Since $\Gamma$ is $G$-symmetric,
$(\alpha,\beta)^{x}=(\beta,\alpha)$ for some $x\in G$. Clearly, such an $x$
has even order and lies in the normalizer $\mathbf{N}_{G}(G_{\alpha\beta})$ of
$G_{\alpha\beta}$. Thus the element $x$ may be chosen as a $2$-element.
Moreover, since $\Gamma$ is connected, $\langle x,G_{\alpha}\rangle=G$.

######  Lemma 2.9.

Let $\\{\alpha,\beta\\}\in E$. Then $\langle x,G_{\alpha}\rangle=G$ for some
$2$-element $x\in\mathbf{N}_{G}(G_{\alpha\beta})$ with $x^{2}\in
G_{\alpha\beta}$.

Note that $G_{\alpha}$ induces a transitive permutation group say
$G_{\alpha}^{\Gamma(\alpha)}$ (on $\Gamma(\alpha)$) of degree $r$. Then
$G_{\alpha}^{\Gamma(\alpha)}$ and $r$ are known as in Table 2.2, refer to [7,
pp. 99]. Moreover, if $r={p^{fd}-1\over p^{f}-1}$ for some prime $p$ then, by
Lemma 2.1, $d$ is a prime and $f$ is a power of $d$.

$\begin{array}[]{|l|l|l|l|l|l|l|}\hline\cr\mathbf{soc}(G_{\alpha}^{\Gamma(\alpha)})&\mathbb{Z}_{r}&\mathrm{A}_{r}&\mathrm{L}_{d}(q)&\mathrm{L}_{2}(11)&\mathrm{M}_{11}&\mathrm{M}_{23}\\\
\hline\cr r&r&r\geq 5&{q^{d}-1\over q-1}&11&11&23\\\ \hline\cr\end{array}$
Table 2.2. Transitive groups of prime degree.

Let $G_{\alpha}^{[1]}$ be the kernel of the action of $G_{\alpha}$ on
$\Gamma(\alpha)$. Then $G_{\alpha}^{\Gamma(\alpha)}\cong
G_{\alpha}/G_{\alpha}^{[1]}$. Take $\beta\in\Gamma(\alpha)$, and set
$G_{\alpha\beta}^{[1]}=G_{\alpha}^{[1]}\cap G_{\beta}^{[1]}$. Then
$G_{\alpha\beta}^{[1]}$ has order a prime power, see [14]. For $r=3$, by [8,
45, 46], $G_{\alpha}$ and $G_{\alpha\beta}^{[1]}$ are listed in Table 2.3.

$\small\begin{array}[]{|l|l|l|l|l|l|}\hline\cr
G_{\alpha}&\mathbb{Z}_{3}&\mathrm{S}_{3}&\mathrm{D}_{12}&\mathrm{S}_{4}&\mathrm{S}_{4}\times\mathbb{Z}_{2}\\\
\hline\cr
G_{\alpha}^{[1]}&1&1&\mathbb{Z}_{2}&\mathbb{Z}_{2}^{2}&\mathbb{Z}_{2}^{3}\\\
\hline\cr G_{\alpha\beta}^{[1]}&1&1&1&\mathbb{Z}_{2}&\mathbb{Z}_{2}^{2}\\\
\hline\cr
G_{\alpha\beta}&1&\mathbb{Z}_{2}&\mathbb{Z}_{2}^{2}&\mathrm{D}_{8}&\mathrm{D}_{8}\times\mathbb{Z}_{2}\\\
\hline\cr\end{array}$ Table 2.3. Stabilizers of symmetric cubic graphs.

In general, we have

(2.1)
$\displaystyle\begin{split}G_{\alpha\beta}/G_{\alpha\beta}^{[1]}\lesssim(G_{\alpha}^{\Gamma(\alpha)})_{\beta}\times(G_{\beta}^{\Gamma(\beta)})_{\alpha},\,\,(G_{\alpha}^{[1]})^{\Gamma(\beta)}\unlhd(G_{\beta}^{\Gamma(\beta)})_{\alpha}\cong(G_{\alpha}^{\Gamma(\alpha)})_{\beta}\\\
G_{\alpha}^{[1]}=G_{\alpha\beta}^{[1]}.(G_{\alpha}^{[1]})^{\Gamma(\beta)},\,\,G_{\alpha}=(G_{\alpha\beta}^{[1]}.(G_{\alpha}^{[1]})^{\Gamma(\beta)}).G_{\alpha}^{\Gamma(\alpha)}.\end{split}$

Then, combining with [15, 47], we have the following result.

######  Lemma 2.10.

  * (1)

$G_{\alpha}$ is soluble if and only if
$G_{\alpha}^{\Gamma(\alpha)}\leq\mathrm{AGL}_{1}(r)$.

  * (2)

Either $\mathrm{L}_{d}(q)\unlhd G_{\alpha}^{\Gamma(\alpha)}$ and
$r={q^{d}-1\over q-1}$, or $G_{\alpha}$ is listed in Table 2.4.

$\footnotesize\begin{array}[]{|l|l|l|l|l|l|l|}\hline\cr
G_{\alpha}^{\Gamma(\alpha)}&\mathbb{Z}_{r}&\mathrm{S}_{r}&\mathrm{A}_{r}&\mathrm{L}_{2}(11)&\mathrm{M}_{11}&\mathrm{M}_{23}\\\
\hline\cr
G_{\alpha}&(\mathbb{Z}_{l^{\prime}}{\times}\mathbb{Z}_{r}){.}\mathbb{Z}_{l}&\mathrm{S}_{r},\mathrm{S}_{r-1}{\times}\mathrm{S}_{r}&\mathrm{A}_{r}&\mathrm{L}_{2}(11)&\mathrm{M}_{11}&\mathrm{M}_{23}\\\
&l^{\prime}\,\,\big{|}\,\,l\,\,\big{|}\,\,r-1&(\mathrm{A}_{r-1}{\times}\mathrm{A}_{r}){.}2&\mathrm{A}_{r-1}{\times}\mathrm{A}_{r}&\mathrm{A}_{5}{\times}\mathrm{L}_{2}(11)&\mathrm{M}_{10}{\times}\mathrm{M}_{11}&\mathrm{M}_{22}{\times\mathrm{M}_{23}}\\\
&&\mathbb{Z}_{2}^{2}{\times}\mathrm{S}_{5}&\mathbb{Z}_{2}^{2}{\times}\mathrm{A}_{5}&&\mathrm{A}_{6}{\times}\mathrm{M}_{11}&\\\
\hline\cr\end{array}$
Table 2.4. Some stabilizers with $G_{\alpha\beta}^{[1]}=1$.

##  3\. The proof of Theorem 1

In this section, $\Gamma=(V,E)$ is a connected graph of prime valency $r\geq
3$.

######  Lemma 3.1.

Let $\Gamma=(V,E)$ be a connected $G$-symmetric graph of prime valency $r\geq
3$. Then $G$ has at most one transitive minimal normal subgroup.

###### Proof.

Suppose that $G$ has two distinct transitive minimal normal subgroups, say $M$
and $N$. Then $M\cap N=1$, and so $M$ and $N$ centralize each other. Thus $M$
and $N$ are nonabelian and regular on $V$, refer to [7, pp.108-109, Lemma 4.2A
and Theorem 4.2A]. Clearly, $X:=MN$ is not regular on $V$. By Lemma 2.8,
$\Gamma$ is $X$-symmetric; in particular, $r\in\pi(X_{\alpha})$ for $\alpha\in
V$. Since $NX_{\alpha}=X=MX_{\alpha}$, we have $M\cong X/N\cong
X_{\alpha}\cong X/M\cong N$. Then $|M|=|N|$ is divisible by $r$ but not by
$r^{2}$. Noting that $M$ and $N$ are direct products of isomorphic simple
groups, it follows that $M\cong T\cong N$ for some nonabelian simple group
$T$. Then the actions of $N$ and $M$ on $V$ are equivalent to the actions of
$T$ on $T$ by right and left multiplications. Identifying $V$ with $T$ and
letting $\alpha=1\in T$, the neighborhood $\Gamma(\alpha)$ is in fact a
conjugacy class of $T$ with length $r$, which is impossible by [22, pp.36,
Theorem 3.7]. This completes the proof. ∎

######  Corollary 3.1.

Let $\Gamma=(V,E)$ be a connected $G$-symmetric graph of prime valency $r\geq
3$. Assume that $G$ has a transitive simple subgroup $T$. If $T$ is normal in
a characteristic subgroup of $G$ then $T$ is normal in $G$.

###### Proof.

Assume $N$ is a characteristic subgroup of $G$, and $T\unlhd N$. Then
$T^{g}\unlhd N$ for each $g\in G$. Suppose that $T\neq T^{g}$ for some $g$.
Then $N$ is not regular on $V$, and so $\Gamma$ is $N$-symmetric. Noting that
$T\cap T^{g}=1$, it follows that $T$ and $T^{g}$ centralize each other, and
hence they are regular on $V$. Thus $N$ has two transitive minimal normal
subgroups, which contradicts Lemma 3.1. Then the corollary follows. ∎

Recall that a permutation group $G$ is quasiprimitive if its minimal normal
subgroups are all transitive.

######  Lemma 3.2.

Let $\Gamma=(V,E)$ be a connected $G$-symmetric graph of prime valency $r\geq
3$, and let $N=\mathbf{soc}(G)$. Assume that $G$ is quasiprimitive on $V$ and
$G$ has a transitive simple subgroup $T$. Then either $r=7$,
$T\cong\mathrm{L}_{3}(2)$ and $G\cong\mathrm{AGL}_{3}(2)$, or $N$ is simple
with $T\leq N$. Moreover, for the latter case, if $T\neq N$ then $\Gamma$ is
$N$-symmetric.

###### Proof.

By Lemma 3.1, $N$ is the unique minimal normal subgroup of $G$. Since $\Gamma$
has odd valency, $|V|$ is even, and then $T$ is a nonabelian simple group.

Assume first that $N$ is abelian. Then $N\cong\mathbb{Z}_{p}^{d}$ and
$G\lesssim\mathrm{AGL}_{d}(p)$ for some prime $p$ and integer $d\geq 1$. In
this case, $T\lesssim\mathrm{GL}_{d}(p)$, and $N$ is regular on $V$. Since
$\Gamma$ has odd valency, $|N|=|V|$ is even, and so $p=2$. Since $T$ is
transitive on $V$, we have $|T:T_{\alpha}|=2^{d}$. By [19], $d\geq 3$ and
either $T\cong\mathrm{A}_{2^{d}}$ or $\mathrm{L}_{n}(q)$ with ${q^{n}-1\over
q-1}=2^{d}$ and $n$ a prime. Note that
$\mathrm{A}_{2^{d}}\not\lesssim\mathrm{GL}_{d}(2)$, see [25, pp. 186,
Proposition 5.3.7]. Then $T\cong\mathrm{L}_{n}(q)$ and ${q^{n}-1\over
q-1}=2^{d}$. By Lemma 2.3, $n=2$ and $q=2^{d}-1$ is a prime. By [25, pp. 188,
Theorem 5.3.9], we have $d\geq{q-1\over(2,q-1)}=2^{d-1}-1$, yielding $d=3$.
Then $N\cong\mathbb{Z}_{2}^{3}$, $|V|=8$ $r=7$, $T\cong\mathrm{L}_{3}(2)$ and
$G\cong\mathrm{AGL}_{3}(2)$.

Next let $N$ be nonabelian. Write $N=T_{1}\times\cdots\times T_{k}$ for
integer $k\geq 1$ and isomorphic nonabelian simple groups
$T_{1},\ldots,T_{k}$. Suppose that $T\not\leq N$. Then $T\cap N=1$, and
$TN/N\cong T$. Since $N$ is the unique minimal normal subgroup of $G$, we have
$\mathbf{C}_{G}(N)=1$, and thus $TN\cong
TN/\mathbf{C}_{TN}(N)\lesssim\mathbf{Aut}(N)$. It follows that $T\cong
TN/N\lesssim\mathbf{Out}(N)$, the outer automorphism group of $N$. Noting that
$\mathbf{Out}(N)\cong(\mathbf{Out}(T_{1})\times\cdots\times\mathbf{Out}(T_{k})){:}\mathrm{S}_{k}$,
it follows that $T\lesssim\mathrm{S}_{k}$. In particular, $|T|$ is a divisor
of $k!$. Since $N$ is transitive on $V$, we have $G=NG_{\alpha}$ for
$\alpha\in V$, and so $T\cong TN/N\leq G/N\cong G_{\alpha}/(G_{\alpha}\cap
N)$. Then $G_{\alpha}$ is insolvable, and so $\Gamma$ is $(G,2)$-arc-
transitive. By [38, Theorem 2], $G$ satisfies III(b)(i) or III(c) given in
[38, pp.229]. It follows that there is a prime divisor $p$ of $|N|$ such that
$|V|$ is divisible by $p^{k}$. Since $T$ is transitive on $V$, it follows that
$p^{k}$ is a divisor of $|T|$. Then $k!$ is divisible by $p^{k}$, and so
$k\leq\sum_{i=1}^{\infty}\lfloor{k\over p^{i}}\rfloor$, which is impossible.
Therefore, $T\leq N$.

Assume that $T\neq N$. Then $N$ is not regular on $V$, and so $\Gamma$ is
$N$-symmetric by Lemma 2.8. Suppose that $k>1$. Then, by Lemma 3.1, every
$T_{i}$ is intransitive on $V$. Noting that $\Gamma$ is not bipartite, by
Lemma 2.8, $|T_{i}|$ is a proper divisor of $|V|$, and so $|T_{i}|$ is a
proper divisor of $|T|$ as $T$ is transitive on $V$, where $1\leq i\leq k$.
Now consider the projections $\phi_{i}:N\rightarrow T_{i},x_{1}\cdots
x_{k}\mapsto x_{i}$, where $x_{j}\in T_{j}$ for $1\leq j\leq k$, and $1\leq
i\leq k$. Then $\phi_{i}(T)\neq 1$ for some $i$, and thus $T\lesssim T_{i}$,
which is impossible. Thus $k=1$, and $N$ is simple. This completes the proof.
∎

The following corollary completes the proof of part (1) of Theorem 1.

######  Corollary 3.2.

Let $\Gamma=(V,E)$ be a connected $G$-symmetric graph of prime valency $r\geq
3$. Assume that $G$ has a transitive minimal normal subgroup $N$ and a
transitive simple subgroup $T$. Then either $r=7$, $T\cong\mathrm{L}_{3}(2)$
and $G\cong\mathrm{AGL}_{3}(2)$, or $T\leq N$ and $N$ is simple.

###### Proof.

Choose a maximal intransitive normal subgroup $M$ of $G$. Then $T\cap M=N\cap
M=1$; in particular, $MN=M\times N$. If $M=1$ then $G$ is quasiprimitive on
$V$, and so the corollary holds by Lemma 3.2. Thus let $M\neq 1$. This implies
that $N$ is nonabelian as $M\leq\mathbf{C}_{G}(N)\neq N$. Write
$N=T_{1}\times\cdots\times T_{k}$ for integer $k\geq 1$ and isomorphic
nonabelian simple groups $T_{1},\ldots,T_{k}$. Then $G$ acts transitively on
$\\{T_{1},\ldots,T_{k}\\}$ by conjugation. Let $\overline{G}=G/M$ and
$\overline{T_{i}}=T_{i}M/M$ for $1\leq i\leq k$. It is easily shown that
$\overline{G}$ acts transitively on
$\\{\overline{T_{1}},\ldots,\overline{T_{k}}\\}$ by conjugation. Let
$\overline{N}=NM/M$. Then $\overline{N}$ is a minimal normal subgroup of
$\overline{G}$. Let $\overline{T}=TM/M$. Note that $\overline{N}\cong N$ and
$\overline{T}\cong T$.

Since $\Gamma$ is not bipartite, $M$ has at least three orbits on $V$. By
Lemma 2.8, $|M|$ is a proper divisor of $|V|$, and so $|M|$ is a proper
divisor of $|T|$. Identify $\overline{G}$ with a subgroup of
$\mathbf{Aut}(\Gamma_{G/M})$. Then Lemma 3.2 works for $\Gamma_{G/M}$. We have
that $\overline{T}\leq\overline{N}$ and $\overline{N}$ is simple. This yields
that $T\leq TM\leq NM$. Suppose that $T\not\leq N$. Then $N\cap T=1$, and so
$T\cong TN/N\leq MN/N\cong M$. In particular, $|T|$ is a divisor of $|M|$, a
contradiction. Therefore, $N\geq T$, and the corollary follows. ∎

######  Corollary 3.3.

Let $\Gamma=(V,E)$ be a connected $G$-symmetric graph of odd prime valency
$r$. Let $K$ be a maximal intransitive normal subgroup of $G$. Assume that $G$
has a transitive simple subgroup $T$. Then either $K=1$, or $T\cong
TK/K\leq\mathbf{soc}(G/K)$.

###### Proof.

Considering $\Gamma_{G/K}$, $TK/K$ and $G/K$, by Lemma 3.2, either the lemma
holds, or $\mathbf{soc}(G/K)\cong\mathbb{Z}_{2}^{3}$. Assume that the latter
case occurs. Then $G/K\cong\mathrm{AGL}_{3}(2)$, $T\cong
TK/K\cong\mathrm{L}_{3}(2)$ and $r=7$. We next show $K=1$.

Let $\mathbf{soc}(G/K)=N/K$, where $N\unlhd G$. Then $N$ is a regular subgroup
of $G$. Noting that $|N|=|V|$, since $T$ is transitive on $V$, we know that
$|V|$ is a divisor of $|T|=168$. It follows that $|K|$ is a divisor of $21$.
Thus $K$ is a subgroup of the cyclic group or the Frobenius group of order
$21$. In particular, $|\mathbf{Aut}(K)|$ is indivisible by $8$. Take a Sylow
$2$-subgroup $P$ of $N$ and consider the conjugation of $P$ on $K$. Noting
that $P\cong\mathbb{Z}_{2}^{3}$, it follows that
$\mathbf{C}_{N}(K)\geq\mathbf{C}_{P}(K)\neq 1$. Clearly,
$\mathbf{C}_{N}(K)\unlhd G$. Then $G/K$ has a (nontrivial) normal subgroup
$\mathbf{C}_{N}(K)K/K$. Recalling that $N/K$ is the unique minimal normal
subgroup of $G/K$, we have $N/K=\mathbf{C}_{N}(K)K/K$, yielding
$N=\mathbf{C}_{N}(K)K$. It follows that $P\leq\mathbf{C}_{N}(K)$ and
$N=K\times P$.

Suppose that $K\neq 1$. Then we may choose a normal Hall subgroup $H$ of $N$
such that $P\leq H$ and $|N:H|\in\\{3,7\\}$. Note that $H$ is a characteristic
subgroup of $N$. This implies that $H\unlhd G$. Then we get a symmetric graph
$\Gamma_{G/H}$ of valency $7$ and order $3$ or $7$, which is impossible. Thus
$K=1$, and the lemma follows. ∎

Now we are ready to prove part (2) of Theorem 1.

######  Lemma 3.3.

Let $\Gamma=(V,E)$ be a connected $G$-symmetric graph of prime valency $r\geq
3$. Assume that $G$ has a transitive simple subgroup $T$, and $T$ is not
normal in $G$. Then either $G\cong\mathrm{AGL}_{3}(2)$, or $G$ has a
characteristic subgroup $N$ such that $\mathbf{rad}(N)$ is intransitive on
$V$, $N/\mathbf{rad}(N)$ is simple, $r\not\in\pi(\mathbf{rad}(N))$ and
$\mathbf{rad}(N)T\lneq N=N^{\infty}$.

###### Proof.

We choose a minimal member $N$ from those characteristic subgroups of $G$
which contain $T$. Note that $T\leq N^{\infty}$, and so $N=N^{\infty}$ by the
choice of $N$. Then $T$ is not normal in $N$ by Corollary 3.1, and $\Gamma$ is
$N$-symmetric by Lemma 2.8.

Assume first that $N$ is quasiprimitive on $V$. Then, by the choice of $N$ and
Lemma 3.2, either $N$ is simple or $N\cong\mathrm{AGL}_{3}(2)$. For the
latter, noting that $\mathbf{rad}(N)$ is normal in $G$, we have
$G\cong\mathrm{AGL}_{3}(2)$ by Corollary 3.2. Then the lemma is true.

Assume that $N$ is not quasiprimitive on $V$. Let $K$ be a maximal
intransitive normal subgroup of $N$. Then $K\neq 1$ and, by Lemma 3.2 and
Corollary 3.3, $T\cong TK/K\leq\mathbf{soc}(N/K)$ and $\mathbf{soc}(N/K)$ is
simple. Recalling that $N=N^{\infty}$, it is easily shown that
$\mathbf{soc}(N/K)=N/K$. By the choice of $K$ and Lemma 2.8, $|K|$ is a proper
divisor of $|V|$, and thus $|K|$ is a proper divisor of $|T|$.

For $\sigma\in\mathbf{Aut}(G)$, we have $K^{\sigma}\unlhd N^{\sigma}=N$, and
so $K^{\sigma}K/K\unlhd N/K$. If $K^{\sigma}\neq K$ then $T\cong TK/K\leq
N/K=K^{\sigma}K/K\cong K^{\sigma}/(K^{\sigma}\cap K)$, and so $|T|$ is a
divisor of $|K|$, a contradiction. Therefore, $K^{\sigma}=K$. It follows that
$K$ is characteristic in $G$.

Set $X=KT$ and let $\alpha\in V$. Noting $K\cap T=1$ and $KT=X=TX_{\alpha}$,
we have $|K|=|X:T|=|X_{\alpha}:T_{\alpha}|$. Take two characteristic subgroups
$I$ and $J$ of $K$ with $I\leq J$. It is easy to see that both $I$ and $J$ are
characteristic in $G$. Then $JT/I=(J/I)(TI/I)$, and $\mathbf{C}_{N/I}(J/I)$ is
characteristic in $G/I$. Set $C/I=\mathbf{C}_{N/I}(J/I)$. Then $C$ is
characteristic in $G$. Choose $J$ and $I$ such $J/I$ is a direct product of
isomorphic simple groups, and write $J/I=S_{1}\times\cdots\times S_{l}$, where
$S_{i}$ are simple.

Suppose that $K$ is insolvable. Then we may choose $J$ and $I$ such that
$S_{1}\cong\cdots\cong S_{l}\cong S$ for some nonabelian simple group $S$. In
this case, $(C/I)\cap(J/I)=1$, and then $C/I$ is properly contained in $N/I$.
By the choice of $N$, we have $l>1$ and $TI/I\not\leq C/I$; otherwise, we have
$N/I=C/I$, yielding that $J/I$ is abelian, a contradiction. Then $T\cong
TI/I\cong(TI/I)(C/I)/(C/I)\lesssim\mathbf{Aut}(J/I)\cong\mathbf{Aut}(S)\wr\mathrm{S}_{l}$.
Recalling that $|K|$ is a proper divisor of $|T|$, it follows that $|S|$ is a
proper divisor of $|T|$, and thus we have $T\lesssim\mathrm{S}_{l}$. Let $p$
be an odd prime of $|S|$. Then we have
$p^{l}\leq|K|_{p}\leq|T|_{p}\leq|\mathrm{S}_{l}|_{p}$, yielding
$l\leq\sum_{i=1}^{\infty}\lfloor{l\over p^{i}}\rfloor$, which is impossible.
Thus $K$ is solvable, and so $K=\mathbf{rad}(N)$.

Suppose that $r\in\pi(K)$. Then we may choose $J$ and $I$ such that
$S_{1}\cong\cdots\cong S_{l}\cong\mathbb{Z}_{r}$. Recalling that
$|K|=|X_{\alpha}:T_{\alpha}|$, we have
$r^{l}\leq|K|_{r}=|X_{\alpha}:T_{\alpha}|_{r}$. This yields that $\Gamma$ is
$X$-symmetric as $r\in\pi(X_{\alpha})$, and so $|X_{\alpha}|_{r}=r$ by Lemma
2.8. Then $|K|_{r}=r$, and so $l=1$ and $J/I\cong\mathbb{Z}_{r}$. Thus
$JT/I=(J/I)(TI/I)=(J/I)\times(TI/I)$, and then $TI/I\leq C/I$. By the choice
of $N$, we have $C=N$ and $N/I=\mathbf{C}_{N/I}(J/I)$. This forces that
$\mathbf{Z}(N/I)\geq J/I\cong\mathbb{Z}_{r}$. Since $|K|_{r}=r$, we know that
$J/I$ is a Sylow $r$-subgroup group of $K/I$. Take a Sylow $r$-subgroup $Q/I$
of $TI/I$. Then $(J/I)\times(Q/I)$ is a Sylow $r$-subgroup of $X/I$ and has
order $r|T|_{r}$. By Lemma 2.8, $|N_{\alpha}|_{r}=r$, since $N=TN_{\alpha}$,
we have $|N|_{r}=r|T|_{r}$. It follows that $(J/I)\times(Q/I)$ is a Sylow
$r$-subgroup of $N/I$. By Gaschütz’ Theorem (refer to [1, pp. 31, (10.4)]),
$N/I$ splits over $J/I$. Let $M/I=(N/I)^{\prime}$. Then $T\leq M$, $M$ is
characteristic in $G$, and $|M|_{r}=|T|_{r}<r|T|_{r}=|N_{r}|$, which
contradicts the choice of $N$. Therefore, $|K|_{r}=1$.

Finally, if $TK=N$ then
$r|T|_{r}=|T|_{r}|N_{\alpha}|_{r}=|N|_{r}=|T|_{r}|K|_{r}$, yielding
$|K|_{r}=r$, a contradiction. Thus $TK\neq N$, and the lemma follows. ∎

##  4\. Some limitations on the vertex-stabilizers

######  Lemma 4.1.

Let $\Gamma$, $G$, $T$ and $r$ be as in Hypothesis 1.1. Then $G$ is not an
exceptional simple group of Lie type, and $G_{\alpha}$ has at most one
insolvable composition factor.

###### Proof.

Suppose that $G_{\alpha}$ has at least two insolvable composition factors. By
(2.1) and Lemma 2.10, $G_{\alpha}$ has exactly two insolvable composition
factors, one is $S=\mathbf{soc}(G_{\alpha}^{\Gamma(\alpha)})$ and the other
one, say $S_{1}$, is a composition factor of
$(G_{\alpha}^{\Gamma(\alpha)})_{\beta}$ for $\beta\in\Gamma(\alpha)$. We list
them as follows.

$\small\begin{array}[]{l|l|l|l|l|l}S&\mathrm{A}_{r}&\mathrm{L}_{d}(q)&\mathrm{L}_{2}(11)&\mathrm{M}_{11}&\mathrm{M}_{23}\\\
\hline\cr
S_{1}&\mathrm{A}_{r-1}&\mathrm{L}_{d-1}(q)&\mathrm{A}_{5}&\mathrm{A}_{6}&\mathrm{M}_{22}\end{array}$

Checking the factorizations given in [28, Theorem 1.1], we conclude that there
is no a simple group $T$ such that $G=TG_{\alpha}$, a contradiction. Then
$G_{\alpha}$ has at most one insolvable composition factor. Noting the
factorization $G=TG_{\alpha}$, our lemma follows from [21, Theorem 1]. ∎

In view of Lemma 4.1, we make some assumptions for the following sections.

######  Hypothesis 4.1.

$\Gamma=(V,E)$ is a connected graph of prime valency $r\geq 3$ such that

  * (I)

$\Gamma$ is $G$-symmetric, where $G$ is a nonabelian simple group;

  * (II)

$G=TG_{\alpha}$ for $\alpha\in V$, where $T$ is a nonabelian simple group with
$T\neq G$;

  * (III)

$G$ is not an exceptional simple group of Lie type, and $G_{\alpha}$ has at
most one insolvable composition factor.

By [41], (2.1) and Lemmas 2.1, 2.10 and 4.1, we have the following lemma.

######  Lemma 4.2.

Assume that Hypothesis 4.1 holds, and $G_{\alpha}$ is insolvable. Let
$S=\mathbf{soc}(G_{\alpha}^{\Gamma(\alpha)})$. Then $\pi(G_{\alpha})=\pi(S)$,
and one of the following holds.

  * (1)

$(S,G_{\alpha},r)$ is one of $(\mathrm{A}_{r},\mathrm{A}_{r},r)$,
$(\mathrm{A}_{r},\mathrm{S}_{r},r)$,
$(\mathrm{L}_{2}(11),\mathrm{L}_{2}(11),11)$,
$(\mathrm{M}_{11},\mathrm{M}_{11},11)$ and
$(\mathrm{M}_{23},\mathrm{M}_{23},23)$.

  * (2)

$S\cong\mathrm{L}_{d}(q)$ and $(r,|\mathbf{O}_{p}(G_{\alpha})|,d,q)$ is listed
as in Table 4.5. In this case, if $\mathbf{O}_{p}(G_{\alpha})=1$ then
$G_{\alpha}\lesssim\mathrm{P\Gamma L}_{d}(q)$, if
$\mathbf{O}_{p}(G_{\alpha})\neq 1$ then
$|G_{\alpha}:\mathbf{O}_{p}(G_{\alpha})|$ is a divisor of
$lf^{2}|\mathrm{GL}_{d}(q)|$, where $l=1$ or $(l,d,q)$ is one of $(6,3,2)$ and
$(24,3,3)$.

$\small\begin{array}[]{|c|c|l|l|}\hline\cr
r&\mathbf{O}_{p}(G_{\alpha})&d&q=p^{f}\\\ \hline\cr
q+1&1,\,q,\,q^{2},\,q^{3}&2&p=2,f={2^{t}}\\\ \hline\cr{q^{d}-1\over
q-1}&1,\,q^{d-1},\,q^{d(d-1)\over 2}&\mbox{prime }d\geq 5&f=d^{t}\\\
\hline\cr{q^{d}-1\over q-1}&1,\,q^{d-1},\,q^{d}&\mbox{prime }d\geq
3&f=d^{t}\\\ \hline\cr 2^{d}-1&1,\,2^{d-1},\,2^{d+1}&\mbox{prime }d\geq
3&q=2\\\ \hline\cr q^{2}+q+1&1,\,q^{2},\,q^{6}&3&q\in\\{2,3\\}\\\ \hline\cr
7&1,\,4,\,2^{20}&3&q=2\\\ \hline\cr 13&1,\,9,\,3^{6}&3&q=3\\\ \hline\cr
31&1,\,2^{4},\,2^{30}&5&q=2\\\ \hline\cr\end{array}$ Table 4.5.

######  Remark 4.1.

Given Hypothesis 4.1, we consider the (faithful) transitive action of
$G_{\alpha}$ on $[G:T]$ by right multiplication. Let $n=|G:T|$. Since $G$ is
simple, we view $G$ as a transitive subgroup of the alternating group
$\mathrm{A}_{n}$, and thus we may let $T=\mathrm{A}_{n-1}\cap G$. Then there
is a general construction based on the coset graphs for connected graphs
satisfying the above (I) and (II), shown below. Let $H$ be a group which is a
possible vertex-stabilizer of some connected symmetric graph of valency $r$.
For any given faithful transitive permutation representation of $H$ on
$\Delta=\\{1,2,\ldots,n\\}$ and a Hall $r^{\prime}$-subgroup $K$ of $H$, let
$\overline{H}$ and $\overline{K}$ be the images of $H$ and $K$ respectively.
Determine the normalizer $\mathbf{N}_{\mathrm{A}_{n}}(\overline{K})$ of
$\overline{K}$ in $\mathrm{A}_{n}$, and then seek the elements $x$ in
$\mathbf{N}_{\mathrm{A}_{n}}(\overline{K})\setminus\overline{H}$ which have
the property that $x^{2}\in\overline{K}$ and $\langle x,\overline{H}\rangle$
is a simple group. Note that the simplicity of $\langle x,\overline{H}\rangle$
ensures that $\overline{H}\leq\mathrm{A}_{n}$. Suppose that there is such an
$x$, and let $\overline{G}=\langle x,\overline{H}\rangle$. Then $\overline{G}$
has a factorization
$\overline{G}=(\mathrm{A}_{n-1}\cap\overline{G})\overline{H}$. If further
$\mathrm{A}_{n-1}\cap\overline{G}$ is simple then the coset graph
$\mathbf{Cos}(\overline{G},\overline{H},x)$ is a desired graph. In view of
this, we sometimes deal with $G$ and $G_{\alpha}$ as two transitive subgroups
of $\mathrm{A}_{n}$. ∎

######  Lemma 4.3.

Let $\Gamma$, $G$, $T$ and $r$ be as in Hypothesis 4.1. Let $n=|G:T|$ and view
$G$ as a transitive subgroup of $\mathrm{A}_{n}$ acting on
$\Delta=\\{1,2,\ldots,n\\}$. Then either $G_{\alpha}$ is imprimitive on
$\Delta$, or one of following holds.

  * (1)

$G$, $T$ and $G_{\alpha}$ are listed in Table 4.6 up to isomorphism.

  * (2)

$G=\mathrm{A}_{n}$, $T=\mathrm{A}_{n-1}$, and $G_{\alpha}$ is almost simple
but not $2$-transitive on $\Delta$.

  * (3)

$n=r\geq 7$, $G=\mathrm{A}_{r}$, $T=\mathrm{A}_{r-1}$, $|G:T|_{r}=r$ and
$G_{\alpha}=\mathbb{Z}_{r}{:}\mathbb{Z}_{l}$, where $l$ is a divisor of
$r-1\over 2$ with $(r,l)\neq(7,3)$.

$\tiny\begin{array}[]{|l|l|c|c|c|l|l|}\hline\cr\mbox{Line}&r&G_{\alpha}&G&T&|G:T|_{r}&\Gamma\\\
\hline\cr
1&11&\mathrm{A}_{11}&\mathrm{A}_{12}&\mathrm{M}_{12}&1&\mathsf{K}_{12}\\\
\hline\cr
2&23&\mathrm{A}_{23}&\mathrm{A}_{24}&\mathrm{M}_{24}&1&\mathsf{K}_{24}\\\
\hline\cr 3&r=2^{2d-1}\pm 2^{d-1}-1,d\geq
3&\mathrm{A}_{r}&\mathrm{A}_{r+1}&\mathrm{Sp}_{2d}(2)&1&\mathsf{K}_{r+1}\\\
\hline\cr
4&r\not\in\\{7,11,17,23\\}&\mathrm{A}_{r}&\mathrm{A}_{r+1}&\mathrm{L}_{2}(r)&1&\mathsf{K}_{r+1}\\\
\hline\cr
5&11&\mathrm{M}_{11}&\mathrm{M}_{12}&\mathrm{M}_{11}&1&\mathsf{K}_{12}\\\
\hline\cr
6&11&\mathrm{L}_{2}(11)&\mathrm{A}_{12}&\mathrm{A}_{11}&1&\mbox{unique}\\\
\hline\cr
7&11&\mathbb{Z}_{11}&\mathrm{L}_{2}(11)&\mathrm{A}_{5}&11&\mbox{Cayley}\\\
\hline\cr
8&11&\mathbb{Z}_{11}{:}\mathbb{Z}_{5}&\mathrm{L}_{2}(11)&\mathrm{A}_{5}&11&\mathsf{K}_{12}\\\
\hline\cr
9&23&\mathbb{Z}_{23}&\mathrm{M}_{23}&\mathrm{M}_{22}&23&\mbox{Cayley}\\\
\hline\cr\end{array}$ Table 4.6.

###### Proof.

Suppose that $G_{\alpha}$ is primitive on $\Delta$. Since $G_{\alpha}$ has at
most one insolvable composition factor, $G_{\alpha}$ is almost simple or of
affine type.

Case 1. Assume that $G_{\alpha}$ is almost simple. Since $G$ is primitive on
$\Delta$ with a simple point-stabilizer $T$, either $G=\mathrm{A}_{n}$ or
$(G,G_{\alpha})$ can be read out from [30, Tables III-VI] by [37, Proposition
6.1]. Suppose that $G\neq\mathrm{A}_{n}$. Then, combining Lemma 4.2, either
$(G,T,G_{\alpha})$ is desired as at Lines 1-5 of Table 4.6 and
$\Gamma\cong\mathsf{K}_{r+1}$, or $r=11$, $G_{\alpha}\cong\mathrm{L}_{2}(11)$,
$G\cong\mathrm{M}_{12}$ and $T\cong\mathrm{M}_{11}$. The latter is excluded by
Lemma 2.9 and calculation of $\mathbf{N}_{G}(G_{\alpha\beta})$, where
$\beta\in\Gamma(\alpha)$.

Now let $G=\mathrm{A}_{n}$. Then $T=\mathrm{A}_{n-1}$ by the assumption in
this lemma. We assume that $G_{\alpha}$ is $2$-transitive on $\Delta$, and
check $2$-transitive subgroups of $\mathrm{A}_{n}$, refer to[4, pp. 197, Table
7.3]. Combining Lemma 4.2, $n$, $\mathbf{soc}(G_{\alpha})$ and $r$ are listed
as follows:

$\tiny\begin{array}[]{l|l|l|l}n&\mathbf{soc}(G_{\alpha})&r&\\\ \hline\cr
6,5&\mathrm{L}_{2}(4)&5&\\\ \hline\cr
8,7&\mathrm{L}_{3}(2)&7&G_{\alpha}=\mathrm{L}_{3}(2)\\\ \hline\cr
12,11&\mathrm{L}_{2}(11)&11&G_{\alpha}=\mathrm{L}_{2}(11)\\\ \hline\cr
12,11&\mathrm{M}_{11}&11&G_{\alpha}=\mathrm{M}_{11}\\\ \hline\cr
23&\mathrm{M}_{23}&23&G_{\alpha}=\mathrm{M}_{23}\\\ \hline\cr
15&\mathrm{A}_{7}&7&G_{\alpha}=\mathrm{A}_{7}\\\ \hline\cr{q^{d}-1\over
q-1}&\mathrm{L}_{d}(q)&{q^{d}-1\over q-1}>7&d\mbox{ a prime}\\\ \end{array}$

For all lines but the last one, since $\Gamma$ has odd valency,
$|V|=|G:G_{\alpha}|$ is even, we have $(n,r)\neq(5,5)$, $(7,7)$. By
calculation using GAP [40], we conclude that either $(n,r)=(6,5)$ or $n=12$
and $G_{\alpha}\cong\mathrm{L}_{2}(11)$. Noting that
$\mathrm{A}_{5}\cong\mathrm{L}_{2}(5)$, the former case is included at Line 4
of Table 4.6. The latter case implies Line 6 of Table 4.6.

We suppose next that $n=r={q^{d}-1\over q-1}>7$ and
$\mathbf{soc}(G_{\alpha})=\mathrm{L}_{d}(q)$, and deduce a contradiction. Let
$q=p^{f}$ for some prime $p$, and let $\sigma$ be the Frobenius automorphism
of the field of order $q$. Note that, by Lemma 2.1, $d$ is a prime,
$(d,q-1)=1$ and $f$ is a power of $d$. Thus, by Lemma 2.10, we may let
$G_{\alpha}=\mathrm{SL}_{d}(q){:}\langle\sigma^{l}\rangle\leq\mathrm{\Gamma
L}_{d}(q)\cap\mathrm{A}_{n}$, where $l$ is a divisor of $f$. Without loss of
generality, we identify $\Delta$ with the point-set of the projective geometry
$\mathrm{PG}_{d-1}(q)$, and let
$G_{\alpha\beta}=q^{d-1}{:}\mathrm{GL}_{d-1}(q){:}\langle\sigma^{l}\rangle$.
Fix $\beta\in\Gamma(\alpha)$, and choose a $2$-element
$x\in\mathbf{N}_{\mathrm{A}_{n}}(G_{\alpha\beta})$ with
$(\alpha,\beta)^{x}=(\beta,\alpha)$ and $\mathrm{A}_{n}=\langle
x,G_{\alpha}\rangle$.

Suppose that $d=2$, in this case, $p=2$ and $f=2^{s}$ for some $s\geq 2$. Then
$G_{\alpha\beta}$ is the stabilizer of some projective point $i$, and thus $x$
fixes $i$. Then $G_{\alpha\beta}\langle x\rangle$ is a solvable $2$-transitive
group of degree $q$. It follows that $G_{\alpha\beta}\langle
x\rangle\leq\mathrm{A\Gamma
L}_{1}(q)=\mathrm{AGL}_{1}(q){:}\langle\sigma\rangle$. Thus $\langle
x,G_{\alpha}\rangle\leq\mathrm{SL}_{2}(q){:}\langle\sigma\rangle$, yielding
$\langle x,G_{\alpha}\rangle\neq\mathrm{A}_{n}=G$, a contradiction.

Now $d$ is an odd prime, and so $f$ is odd. Then
$G_{\alpha}=\mathrm{SL}_{d}(q){:}\langle\sigma^{l}\rangle\leq\mathrm{\Gamma
L}_{d}(q)\leq\mathrm{A}_{n}$, and
$\mathbf{N}_{\mathrm{A}_{n}}(G_{\alpha\beta})$ has a subgroup
$P{:}K\cong\mathrm{A\Gamma L}_{d-1}(q)$, where
$P=\mathbf{O}_{p}(G_{\alpha\beta})$ and $K\cong\mathrm{\Gamma L}_{d-1}(q)$
with $G_{\alpha\beta}\leq PK$. Let $C=\mathbf{C}_{\mathrm{A}_{n}}(P)$ and
$N=\mathbf{N}_{\mathrm{A}_{n}}(P)$ . Then $P\leq C$, $C\cap K=1$, $N\geq
CK=C{:}K$, and so $K\lesssim
N/C\lesssim\mathbf{Aut}(P)=\mathrm{GL}_{(d-1)f}(p)$. Noting that
$\mathbf{N}_{\mathrm{A}_{n}}(G_{\alpha\beta})\leq N$, we have
$\mathbf{N}_{\mathrm{A}_{n}}(G_{\alpha\beta})=\mathbf{N}_{N}(G_{\alpha\beta})$.
Then $K\cong
PKC/C\leq\mathbf{N}_{\mathrm{A}_{n}}(G_{\alpha\beta})C/C=\mathbf{N}_{N}(G_{\alpha\beta})C/C\leq\mathbf{N}_{N/C}(G_{\alpha\beta}C/C)\lesssim\mathbf{N}_{\mathrm{GL}_{(d-1)f}(p)}(\mathrm{GL}_{d-1}(q){:}\langle\sigma^{l}\rangle)\cong\mathrm{\Gamma
L}_{d-1}(q)\cong K$. It follows
$C{:}K=\mathbf{N}_{N}(G_{\alpha\beta})C\geq\mathbf{N}_{N}(G_{\alpha\beta})$.
Write $G_{\alpha\beta}=P{:}H$ with $H\leq K$. Recalling that
$K\leq\mathbf{N}_{N}(G_{\alpha\beta})\leq C{:}K$, we have
$\mathbf{N}_{N}(G_{\alpha\beta})=(C\cap\mathbf{N}_{N}(G_{\alpha\beta}){:}K=P(C\cap
H)K=PK$. Then $x=yz\sigma^{t}$, where $y\in P$, $t$ is a divisor of $f$ and
$z$ is contained in a subgroup of $H$ isomorphic to $\mathrm{GL}_{d-1}(q)$.
Since $f$ is odd, the order of $x$ is not a $2$-power unless $t=f$. Then
$x=yz\in G_{\alpha\beta}$, a contradiction.

Case 2. Suppose that $G_{\alpha}$ is not almost simple. Then $4<n=p^{m}$ for
some prime $p$ and integer $m\geq 1$, and $G_{\alpha}\leq\mathrm{AGL}_{m}(p)$.
By [37, Proposition 6.2], one of the following holds: $G=\mathrm{A}_{n}$; or
$n=p=r=11$, $G=\mathrm{L}_{2}(11)$, $T\cong\mathrm{A}_{5}$ and
$\mathbb{Z}_{11}\lesssim G_{\alpha}\lesssim\mathbb{Z}_{11}{:}\mathbb{Z}_{5}$;
or $n=p=r=23$, $G=\mathrm{M}_{23}$, $T=\mathrm{M}_{22}$ and
$\mathbb{Z}_{23}\lesssim G_{\alpha}\lesssim\mathbb{Z}_{23}{:}\mathbb{Z}_{11}$.
Calculation shows that the last two case give Lines 7-9 of Table 4.6.

Assume that $G=\mathrm{A}_{n}$. Then $T=\mathrm{A}_{n-1}$. If $G_{\alpha}$ is
solvable, then $n=r=p$ and $G_{\alpha}=\mathbb{Z}_{r}{:}\mathbb{Z}_{l}$, where
$l$ is a divisor of $r-1\over 2$. In this case, by calculation, we exclude the
case $(r,l)=(7,3)$, and part (3) of this lemma follows. Thus we assume further
that $G_{\alpha}$ is insolvable. In particular, $m>1$, and $n\geq 8$.

Clearly, $\mathbf{soc}(G_{\alpha})\leq\mathbf{O}_{p}(G_{\alpha})\neq 1$, and
so $\mathbf{Z}(\mathbf{O}_{p}(G_{\alpha}))\neq 1$. It follows that
$\mathbf{Z}(\mathbf{O}_{p}(G_{\alpha}))$ is a regular normal subgroup of
$G_{\alpha}$ (acting on $\Delta$), and thus
$\mathbf{soc}(G_{\alpha})=\mathbf{Z}(\mathbf{O}_{p}(G_{\alpha}))$. Then, by
Lemma 2.10, we may let
$G_{\alpha}=\mathbf{O}_{p}(G_{\alpha}){:}((\mathbb{Z}_{l}\times\mathrm{SL}_{d}(q)){:}\langle\sigma^{e}\rangle)$
and $r={q^{d}-1\over q-1}$, where $l$ is a divisor of $q-1$, $\sigma$ is the
Frobenius automorphism of the field of order $q$, and $q$ is a power of the
prime $p$. (Recall that, since $r$ is a prime, $d$ is a prime and
$(d,q-1)=1$.) Replacing $T$ by its a conjugation in $G$, we let
$(\mathbb{Z}_{l}\times\mathrm{SL}_{d}(q)){:}\langle\sigma^{e}\rangle\leq T$.
Then $\Gamma$ is $T$-symmetric and
$T_{\alpha}=(\mathbb{Z}_{l}\times\mathrm{SL}_{d}(q)){:}\langle\sigma^{e}\rangle$,
and thus $T_{\alpha}^{[1]}\cong\mathbb{Z}_{l}$ and $T_{\alpha\beta}^{[1]}=1$,
where $\beta\in\Gamma(\alpha)$. By (2.1), it is easily shown that $l=1$. Then
we have
$G_{\alpha}=\mathbf{O}_{p}(G_{\alpha}){:}(\mathrm{SL}_{d}(q){:}\langle\sigma^{e}\rangle)$.
It follows that $p^{m}=|\mathbf{O}_{p}(G_{\alpha})|=q^{d}$, and then
$\mathrm{ASL}_{d}(q)\leq G_{\alpha}\leq\mathrm{A\Gamma L}_{d}(q)$. In
particular, $G_{\alpha}$ is $3$-transitive on $\Delta$. Then, noting that
$\mathrm{A}_{n}=\mathrm{A}_{n-2}G_{\alpha}$, we have a factorization
$T=\mathrm{A}_{n-2}T_{\alpha}$ with
$T_{\alpha}=\mathrm{SL}_{d}(q){:}\langle\sigma^{e}\rangle$. Moreover,
$T_{\alpha}$ is a $2$-transitive subgroup of $\mathrm{A}_{n-1}$. Recalling
that $n\geq 8$, we have $n-2\geq 6$. Then the argument in Case 1 works for the
triple $(n-1,n-2,r)$. Thus, noting that $n=q^{d}$, the only possibility is
that $n=8$, $r=7$, and $T_{\alpha}\cong\mathrm{L}_{3}(2)$. Then
$G_{\alpha}=2^{3}{:}\mathrm{L}_{3}(2)$, which has odd index in
$\mathrm{A}_{8}$, a contradiction. ∎

##  5\. Graphs with solvable vertex-stabilizers

In this section, we deal with the graphs described as in Hypothesis 4.1 under
the further assumption that $G_{\alpha}$ is solvable. We always let $n=|G:T|$,
and view $G$ as a transitive subgroup of $\mathrm{A}_{n}$ (acting on
$\Delta=\\{1,2,\ldots,n\\}$).

By [27, Theorem 1.1], combining Table 2.3 and Lemma 2.10, $G$, $T$ and
$G_{\alpha}$ are listed as in Table 5.7 up to isomorphism.

$\footnotesize\begin{array}[]{|l|l|c|c|c|l|}\hline\cr\mbox{Line}&G&T&G_{\alpha}&r&\\\
\hline\cr
1&\mathrm{L}_{2}(r)&\mathrm{A}_{5}&\mathbb{Z}_{r}{:}\mathbb{Z}_{r-1\over
2}&r\in\\{11,19,29,59\\}&\\\ \hline\cr
2&\mathrm{L}_{2}(11)&\mathrm{A}_{5}&\mathbb{Z}_{11}&11&\\\ \hline\cr
3&\mathrm{L}_{2}(19)&\mathrm{A}_{5}&\mathbb{Z}_{19}{:}\mathbb{Z}_{3}&19&\\\
\hline\cr
4&\mathrm{L}_{2}(29)&\mathrm{A}_{5}&\mathbb{Z}_{29}{:}\mathbb{Z}_{7}&29&\\\
\hline\cr 5&\mathrm{A}_{c}&\mathrm{A}_{c-1}&&&c\geq 6\\\ \hline\cr
6&\mathrm{A}_{6}&\mathrm{A}_{5}&&r=3&\\\ \hline\cr
7&\mathrm{M}_{c}&\mathrm{M}_{c-1}&&&c\in\\{12,24\\}\\\ \hline\cr
8&\mathrm{M}_{23}&\mathrm{M}_{22}&\mathbb{Z}_{23},\mathbb{Z}_{23}{:}\mathbb{Z}_{11}&23&\\\
\hline\cr\end{array}$ Table 5.7.

Note that, since $G_{\alpha}$ is solvable, if $r\geq 5$ then $G_{\alpha}$ has
a unique Sylow $r$-subgroup, which is semiregular on $\Delta$.

######  Theorem 5.1.

Assume that Hypothesis 4.1 holds and $G_{\alpha}$ is solvable. Let $n=|G:T|$.
Then $n$ is divisible by $r$, and one of the following holds.

  * (1)

$\Gamma\cong\mathsf{K}_{r+1}$ for $r\in\\{11,19,29,59\\}$,
$G=\mathrm{L}_{2}(r)$, $T\cong\mathrm{A}_{5}$ and
$G_{\alpha}\cong\mathbb{Z}_{r}{:}\mathbb{Z}_{r-1\over 2}$.

  * (2)

$n=11$, $r=11$, $G=\mathrm{L}_{2}(11)$, $T\cong\mathrm{A}_{5}$ and
$G_{\alpha}\cong\mathbb{Z}_{11}$.

  * (3)

$n=203$, $r=29$, $G=\mathrm{L}_{2}(29)$, $T\cong\mathrm{A}_{5}$ and
$G_{\alpha}\cong\mathbb{Z}_{29}{:}\mathbb{Z}_{7}$.

  * (4)

$n=r=23$, $G=\mathrm{M}_{23}$, $T=\mathrm{M}_{22}$ and
$G_{\alpha}\cong\mathbb{Z}_{23}$.

  * (5)

$G=\mathrm{A}_{n}$, $T=\mathrm{A}_{n-1}$, $n\geq 7$, $r\geq 5$ and $r^{2}$ is
not a divisor of $n$.

  * (6)

$G=\mathrm{A}_{48}$, $T=\mathrm{A}_{47}$, $r=3$ and
$G_{\alpha}\cong\mathbb{Z}_{2}\times\mathrm{S}_{4}$.

###### Proof.

First, Lines 1, 2 and 4 of Table 5.7 give parts (1), (2) and (3) of this
lemma, respectively. For Line 3 of Table 5.7, $G$ has a unique conjugacy class
of elements of order $3$. This yields that $T\cap G_{\alpha}$ contains an
element of order $3$, see [17, Lemma 2.2] for example. Then
$3420=|G|={|T||G_{\alpha}|\over|T\cap G_{\alpha}|}\leq 1140$, a contradiction.
For Line 8 of Table 5.7, if $G_{\alpha}$ has order $253$, then
$G_{\alpha\beta}\cong\mathbb{Z}_{11}$ for $\beta\in\Gamma(\alpha)$, and
$|\mathbf{N}_{G}(G_{\alpha\beta})|$ has order $55$ (confirmed by GAP), which
contradicts Lemma 2.9. Thus $G_{\alpha}\cong\mathbb{Z}_{23}$, and part (4) of
this lemma follows.

Suppose that Line 6 of Table 5.7 occurs. Noting that $G_{\alpha}$ is known as
in Table 2.3, we have $G_{\alpha}\cong\mathrm{S}_{3}$ or $\mathrm{S}_{4}$, and
$|V|=60$ or $15$ respectively. Since $\Gamma$ has odd valency, $|V|$ is even,
and so $G_{\alpha}\cong\mathrm{S}_{3}$. Then $|G|=|T||G_{\alpha}|$, and so
$T\cap G_{\alpha}=1$. Thus $G_{\alpha}$ acts regularly on $\Delta$. This
yields that $G_{\alpha}$ contains odd permutations of $\mathrm{S}_{6}$, a
contradiction.

Suppose that Line 7 of Table 5.7 occurs. Then $n=c\in\\{12,24\\}$. If
$r\geqslant 5$ then the Sylow $r$-subgroup of $G_{\alpha}$ is semiregular on
$\Delta$, and so $r$ is a divisor of $n$, which is impossible. Thus we have
$r=3$. If $T\cap G_{\alpha}=1$ then $\Gamma$ is isomorphic to a symmetric
Cayley graph of $T$, and so $T$ is normal in $G$ by [29], which is impossible.
Therefore, $T\cap G_{\alpha}\neq 1$. Recalling that
$n=|G:T|=|G_{\alpha}:(T\cap G_{\alpha})|$, it follows that $n$ is a proper
divisor of $|G_{\alpha}|$. By Table 2.3, $(n,G_{\alpha})$ is one of
$(12,\mathrm{S}_{4})$, $(12,\mathrm{S}_{4}\times\mathbb{Z}_{2})$ and
$(24,\mathrm{S}_{4}\times\mathbb{Z}_{2})$. With the help of GAP, up to
conjugation, $\mathrm{A}_{n}$ has at most three transitive subgroups
isomorphic to $G_{\alpha}$, and none of them has a Sylow $2$-subgroup $P$ such
that $|\langle G_{\alpha},x\rangle|=|\mathrm{M}_{n}|$ for some
$x\in\mathbf{N}_{\mathrm{A}_{n}}(P)$. Noting that $P=G_{\alpha\beta}$ for some
$\beta\in\Gamma(\alpha)$, we get a contradiction, see Lemma 2.9.

Finally, suppose that Line 5 of Table 5.7 occurs. Then $n=c$, and
$n=|G_{\alpha}:(T\cap G_{\alpha})|$ as $G_{\alpha}$ is a transitive subgroup
of $\mathrm{A}_{n}$. If $r\geq 5$ then the Sylow $r$-subgroup of $G_{\alpha}$
is semiregular on $\Delta$, and so $n$ is divisible by $r$, yielding part (5)
of this lemma. Assume that $r=3$. Then
$G_{\alpha}\cong\mathrm{S}_{3},\,\mathrm{D}_{12}$, $\mathrm{S}_{4}$ or
$\mathrm{S}_{4}\times\mathbb{Z}_{2}$. Suppose that $T\cap G_{\alpha}\neq 1$.
Recalling that $6\leq n=|G_{\alpha}:(T\cap G_{\alpha})|$, we have
$G_{\alpha}\cong\mathrm{D}_{12},\,\mathrm{S}_{4}$ or
$\mathrm{S}_{4}\times\mathbb{Z}_{2}$. It follows that $(n,G_{\alpha})$ is one
of $(6,\mathrm{D}_{12})$, $(6,\mathrm{S}_{4})$, $(8,\mathrm{S}_{4})$,
$(8,\mathrm{S}_{4}\times\mathbb{Z}_{2})$,
$(16,\mathrm{S}_{4}\times\mathbb{Z}_{2})$, $(12,\mathrm{S}_{4})$,
$(12,\mathrm{S}_{4}\times\mathbb{Z}_{2})$ and
$(24,\mathrm{S}_{4}\times\mathbb{Z}_{2})$. With the help of GAP, a similar
argument as in the above paragraph leads to a contradiction. Thus $T\cap
G_{\alpha}=1$. Then $\Gamma$ is a Cayley graph of $\mathrm{A}_{n-1}$, and so
part (6) of this lemma holds by [29]. ∎

######  Remark 5.1.

For $r\in\\{5,7\\}$, there is a graph satisfying Theorem 5.1 (5) if and only
if $(n,r,|G_{\alpha}|)$ is one of $(20,5,40)$, $(40,5,40)$, $(40,5,80)$,
$(80,5,80)$, $(7,7,7)$, $(7,7,21)$, $(21,7,21)$, $(21,7,63)$, $(63,7,63)$ and
$(84,7,84)$. This is confirmed by GAP.

######  Corollary 5.1.

Assume that Hypothesis 4.1 holds, and that $G$ is isomorphic to a linear
simple group of prime dimension. Then one of the following holds.

  * (1)

$\Gamma\cong\mathsf{K}_{r+1}$ for $r\in\\{11,19,29,59\\}$,
$G=\mathrm{L}_{2}(r)$, $T\cong\mathrm{A}_{5}$ and
$G_{\alpha}\cong\mathbb{Z}_{r}{:}\mathbb{Z}_{r-1\over 2}$.

  * (2)

$n=11$, $r=11$, $G=\mathrm{L}_{2}(11)$, $T\cong\mathrm{A}_{5}$ and
$G_{\alpha}\cong\mathbb{Z}_{11}$.

  * (3)

$n=203$, $r=29$, $G\cong\mathrm{L}_{2}(29)$, $T\cong\mathrm{A}_{5}$ and
$G_{\alpha}\cong\mathbb{Z}_{29}{:}\mathbb{Z}_{7}$.

###### Proof.

Since $G=TG_{\alpha}$ and $T$ is nonabelian simple, by [27, Corollary 3.4],
$G_{\alpha}$ is solvable. By Theorem 5.1, either one of parts (1)-(3) follows,
or $G=\mathrm{A}_{n}$, $T=\mathrm{A}_{n-1}$, $r\geq 5$ and $r$ is a divisor of
$n$. For the latter, we have $G\cong\mathrm{L}_{2}(9)$ but $n=6$, a
contradiction. ∎

######  Corollary 5.2.

Assume that Hypothesis 4.1 holds. Then $G$ is not (isomorphic to) a simple
unitary group of odd dimension.

###### Proof.

Suppose that $G\cong\mathrm{U}_{2m+1}(q)$. Then, by Theorem 5.1, $G_{\alpha}$
is insolvable. Thus $G=AB$ for maximal subgroups $A$ and $B$ of $G$ such that
one of $A$ and $B$ contains $T$ and the other one contains $G_{\alpha}$. Then,
checking the pairs $(A,B)$ given in [31, pp. 13, Table 3], the only
possibility is that $G\cong\mathrm{U}_{9}(2)$, $A\cong\mathrm{J}_{3}$ and
$B\cong[2^{15}]{:}\mathrm{U}_{7}(2)$. If $T\leq B$ then $G=AT$; however,
$|A|_{2}|T|_{2}\leq 2^{28}<2^{36}=|G|_{2}$, a contradiction. Thus $T\leq A$
and $G_{\alpha}\leq B$. It is easy to check that $|B|$ is indivisible by $19$
or $17$, and $|A|$ is indivisible by $43$ or $11$. Check the maximal subgroups
of $A$ and $\mathrm{U}_{7}(2)$, refer to the Atlas [5]. We conclude that $T=A$
and $\mathrm{U}_{7}(2)\preccurlyeq G_{\alpha}$, which contradicts Lemma 4.2.
This completes the proof. ∎

##  6\. Graphs arising from alternating groups

In this section, using some known factorizations of simple groups, we
investigate the graphs in Hypothesis 4.1 under the further assumption that
$G_{\alpha}$ is insolvable and either $G$ or $T$ is isomorphic to an
alternating group. Let $n=|G:T|$, view $G$ as a transitive subgroup of
$\mathrm{A}_{n}$ acting on $\Delta=[G:T]$.

######  Lemma 6.1.

Assume that Hypothesis 4.1 holds. Let $n=|G:T|$. Assume that
$|\Gamma(\alpha)|=r\geq 5$ and $G_{\alpha}\cong\mathrm{A}_{r}$. Then one of
the following holds.

  * (1)

$G=\mathrm{A}_{6}$, $T=\mathrm{A}_{5}$, $G_{\alpha}=\mathrm{L}_{2}(4)$ and
$\Gamma\cong\mathsf{K}_{6}$.

  * (2)

$n>{r\choose 2}$, and either $n\geq{r\choose{r-1\over 2}}$ or $r\geq 11$ and
$n$ is divisible by $r\choose l$ for some $l$ with $2\leq l<{r-1\over 2}$.
Moreover, one the following holds:

    * (i)

$\Gamma\cong\mathsf{K}_{r+1}$ for $r\geq 7$, $G\cong\mathrm{A}_{r+1}$, and $T$
is isomorphic to a transitive subgroup of $\mathrm{A}_{r+1}$;

    * (ii)

$G=\mathrm{A}_{n}$, $T=\mathrm{A}_{n-1}$ and $G_{\alpha}$ is a transitive
subgroup of $\mathrm{A}_{n}$.

###### Proof.

Note that $n>r$, $2n$ is a divisor of $r!$ and $\mathrm{A}_{r}$ has a subgroup
$H$ of index $n$. Let $k={r-1\over 2}$. Suppose that $n<{r\choose k}$. By [7,
Theorem 5.2A], either $\mathrm{A}_{r-l}\lesssim
H\lesssim(\mathrm{S}_{l}\times\mathrm{S}_{r-l})\cap\mathrm{A}_{r}$ for some
$l$ with $2\leq l<k$, or $(r,k,n,H)$ is one of $(5,2,6,\mathrm{D}_{10})$ and
$(7,2,15,\mathrm{L}_{3}(2))$. For $(r,k,n,H)=(5,2,6,\mathrm{D}_{10})$, we have
$G=\mathrm{A}_{6}$, $T=\mathrm{A}_{5}$, $G_{\alpha}=\mathrm{L}_{2}(4)$ and
$\Gamma\cong\mathsf{K}_{6}$ as in part (1) of this lemma. Assume that
$(r,k,n,H)=(7,2,15,\mathrm{L}_{3}(2))$. Up to equivalence, $\mathrm{A}_{7}$
has two permutation representations of degree $15$. Given such a
representation of $G_{\alpha}$, using GAP, we search the $2$-elements $x$ in
$\mathrm{A}_{15}\setminus G_{\alpha}$ which normalize a subgroup
$\mathrm{A}_{6}$ of $G_{\alpha}$. Computation shows that $\langle
x,G_{\alpha}\rangle\cong\mathrm{A}_{8}$ for each a desired $x$. We have
$G\cong\mathrm{A}_{8}$; however, a subgroup of index $15$ in $\mathrm{A}_{8}$
is isomorphic to $2^{3}{:}\mathrm{L}_{3}(2)$, which can not play the role of
$T$, a contradiction. Thus we assume next that either $n\geq{r\choose{r-1\over
2}}$, or $n<{r\choose{r-1\over 2}}$ and $n$ is divisible by $r\choose l$ for
some $l$ with $2\leq l<{r-1\over 2}$. Moreover, for the latter case, $r\geq
11$. In fact, if $r\in\\{5,7\\}$ and $n<{r\choose{r-1\over 2}}$ then
$(n,r)=(21,7)$, which is excluded by calculation in a similar way as above.

Suppose that $n={r\choose 2}$. Then either $r=5$ or $r\geq 11$. Let $M$ be a
maximal subgroup of $G_{\alpha}$ with $H\leq M$. Assume that $M$ is a
primitive subgroup of $\mathrm{A}_{r}$. Then $|H|\leq|M|<4^{r}$ by [39], and
so ${r\choose 2}=n>{r!\over 4^{r}2}$, yielding $r\leq 13$. Then
$(n,r)=(10,5)$, $(21,7)$, $(55,11)$ and $(78,13)$. However, for such an $r$,
every primitive subgroup of $\mathrm{A}_{r}$ has index indivisible by
${r\choose 2}$, a contradiction. Thus $M$ is an intransitive or imprimitive
subgroup of $\mathrm{A}_{r}$. It follows from [30] that $M$ is the stabilizer
of some $2$-subset of $\\{1,2,\ldots,r\\}$, and thus
$H=M\cong\mathrm{S}_{r-2}$. Then the action of $G_{\alpha}$ on $\Delta=[G:T]$
is equivalent that on the $2$-subsets of an $r$-set. Thus, for
$\beta\in\Gamma(\alpha)$, the arc-stabilizer $G_{\alpha\beta}$ has two orbits
on $\Delta$, say $\Delta_{1}$ and $\Delta_{2}$, which have length $r-1$ and
$(r-1)(r-2)\over 2$ respectively. Moreover, it is easily shown that
$G_{\alpha\beta}$ acts primitively on both $\Delta_{1}$ and $\Delta_{2}$.
Noting that $|\Delta_{1}|\neq|\Delta_{2}|$, we know that
$\mathbf{C}_{G}(G_{\alpha\beta})$ fixes both $\Delta_{1}$ and $\Delta_{2}$
set-wise. By [7, Theorem 4.2A], $\mathbf{C}_{G}(G_{\alpha\beta})$ acts
trivially on both $\Delta_{1}$ and $\Delta_{2}$, yielding
$\mathbf{C}_{G}(G_{\alpha\beta})=1$. Recalling $r\neq 7$, we have
$\mathbf{N}_{G}(G_{\alpha\beta})\lesssim\mathrm{S}_{r-1}$. Note that the
transposition $(1\,2)$ induces a product of $r-1$ transpositions on $\Delta$.
This implies that $\mathrm{S}_{r}\lesssim\mathbf{N}_{G}(G_{\alpha})$, and thus
$\mathbf{N}_{G_{\alpha}}(G_{\alpha\beta})\gtrsim\mathrm{S}_{r}$. Then
$\mathbf{N}_{G}(G_{\alpha\beta})\leq G_{\alpha}$, a contradiction. Thus
$n>{r\choose 2}$.

Since $G_{\alpha}\cong\mathrm{A}_{r}$ and $G=TG_{\alpha}$, by [42, Theorem
1.1], one of the following occurs:

  * (i)

$G\cong\mathrm{A}_{r+1}$, $G_{\alpha}\cong\mathrm{A}_{r}$, $T$ is isomorphic
to a transitive subgroup of $\mathrm{A}_{r+1}$;

  * (ii)

$G=\mathrm{A}_{n}$, $G_{\alpha}\cong\mathrm{A}_{r}$, $T=\mathrm{A}_{n-1}$ and
$G_{\alpha}$ is a transitive subgroup of $\mathrm{A}_{n}$;

  * (iii)

$G\cong\mathrm{A}_{r+k}$, $G_{\alpha}\cong\mathrm{A}_{r}$, and $T$ is
isomorphic to a $k$-transitive subgroup of $\mathrm{A}_{r+k}$, where $2\leq
k\leq 5$;

  * (vi)

the triple $(G,r,T)$ is one of $(\mathrm{M}_{12},5,\mathrm{M}_{11})$,
$(\Omega_{8}^{+}(2),7,\mathrm{Sp}_{6}(2))$,
$(\mathrm{A}_{15},13,\mathrm{A}_{13})$, $(\mathrm{A}_{15},13,\mathrm{A}_{8})$
and $(\mathrm{A}_{15},7,\mathrm{A}_{13})$.

For the case (i) above, $\Gamma$ is the complete graph $\mathsf{K}_{r+1}$. If
$r=5$ then $G\cong\mathrm{A}_{6}$, yielding $T\cong\mathrm{A}_{5}$, and so
$6=n>{5\choose 2}=10$, which is impossible. Then $r\geq 7$, desired as in part
(i) of this lemma. For the case (ii) above, we have part (ii) of this lemma.
To complete the proof, we next exclude the left two cases.

Case 1. Assume that case (iii) occurs. Consider the action of $G$ on an
$(r+k)$-set, and take a $G_{\alpha}$-orbit with maximal length, say $m$. Then
$m\geq r$, and $G_{\alpha}$ has a subgroup $K$ of index $m$. Clearly, $m\leq
r+k\leq 2r\leq{r\choose 2}$. Suppose that $m={r\choose 2}$. Then $r(r-1)=2m$,
the only possibility is that $r=5$ and $m=10$. This implies that $k=r=5$. Thus
$T$ is (isomorphic to) a $5$-transitive subgroup of $\mathrm{A}_{10}$. This
yields that $T=\mathrm{A}_{10}=G$, a contradiction. Therefore,
$|G_{\alpha}:K|=m<{r\choose 2}$ and $m\leq r+5$.

By [7, Theorem 5.2A], either $(r,m)=(5,6)$, or $K\lesssim\mathrm{A}_{r-1}$.
For the latter case, $r$ is a divisor of $m=|G_{\alpha}:K|$, since $m\leq r+k$
and $2\leq k\leq 5$, we have $r=k=5$, which yields that $T$ is a
$5$-transitive subgroup of $\mathrm{A}_{10}$, a contradiction. If $r+k\geq 9$
and $r=5$ then $k\geq 4$, and we have a similar contradiction as above. All in
all, $(r,m)=(5,6)$ and $7\leq r+k<9$. Computation using GAP shows that, for
$\beta\in\Gamma(\alpha)$, there is no $2$-element
$x\in\mathbf{N}_{G}(G_{\alpha\beta})$ with $G=\langle G_{\alpha},x\rangle$,
which contradicts Lemma 2.9.

Case 2. Assume that case (iv) occurs. Suppose that
$(G,r,T)=(\Omega_{8}^{+}(2),7,\mathrm{Sp}_{6}(2))$. Then, as a permutation
group on $\Delta$, the group $G$ has $2$ transitive subgroups $\mathrm{A}_{7}$
up to conjugacy, confirmed by GAP. Taking $G_{\alpha}$ as any one of them,
further calculation shows that there is no $2$-element
$x\in\mathbf{N}_{G}(G_{\alpha\beta})$ with $x^{2}\in G_{\alpha\beta}$ and
$G=\langle G_{\alpha},x\rangle$, which contradicts Lemma 2.9. Suppose that
$(G,r,T)=(\mathrm{M}_{12},5,\mathrm{M}_{11})$. Then $G_{\alpha}$ is a
transitive subgroup of $\mathrm{A}_{12}$ acting on a $12$-set. Note that, up
to equivalence, $G_{\alpha}$ has a unique permutation representation of degree
$12$, and $G_{\alpha\beta}$ is regular under this representation, where
$\beta\in\Gamma(\alpha)$. Computation shows that there is no
$x\in\mathbf{N}_{\mathrm{A}_{12}}(G_{\alpha\beta})$ with $x^{2}\in
G_{\alpha\beta}$ and $|\mathrm{M}_{12}|=|\langle G_{\alpha},x\rangle|$, a
contradiction.

For the left three triples, we conclude that $G_{\alpha}\cong\mathrm{A}_{r}$
has a $2$-transitive permutation representation of degree $15$. It follows
that $r=7$. Considering the transitive permutation representations of
$\mathrm{A}_{7}$ of degree $15$, we get a similar contradiction by
calculation. This completes the proof. ∎

######  Theorem 6.1.

Let $\Gamma$, $G$, $T$ and $r$ be as in Hypothesis 4.1, and let $n=|G:T|$.
Assume that $G$ or $T$ is isomorphic to an alternating group, and that
$G_{\alpha}$ is insolvable, where $\alpha\in V$. Then one of the following
holds.

  * (1)

$G=\mathrm{A}_{6}$, $T=\mathrm{A}_{5}$, $G_{\alpha}=\mathrm{L}_{2}(4)$, $r=5$
and $\Gamma\cong\mathsf{K}_{6}$.

  * (2)

$G\cong\mathrm{M}_{12}$, $T\cong\mathrm{A}_{5}$,
$G_{\alpha}\cong\mathrm{M}_{11}$, $r=11$ and $\Gamma\cong\mathsf{K}_{12}$.

  * (3)

$G$, $T$ and $G_{\alpha}$ are given in Table 4.6.

  * (4)

$G=\mathrm{A}_{n}$, $T=\mathrm{A}_{n-1}$, $G_{\alpha}$ is a transitive but not
$2$-transitive subgroup of $\mathrm{A}_{n}$.

  * (5)

$n>{r\choose 2}$ for $r\geq 7$, $\Gamma\cong\mathsf{K}_{r+1}$,
$G\cong\mathrm{A}_{r+1}$, $G_{\alpha}\cong\mathrm{A}_{r}$, $T$ is isomorphic
to a transitive subgroup of $\mathrm{A}_{r+1}$, and either
$n\geq{r\choose{r-1\over 2}}$ or $r\geq 11$ and $n$ is divisible by $r\choose
l$ for some $l$ with $2\leq l<{r-1\over 2}$.

###### Proof.

If $G_{\alpha}\cong\mathrm{A}_{r}$ then, by Lemma 6.1 one of parts (1), (4)
and (5) occurs. Thus we assume next that $G_{\alpha}\not\cong\mathrm{A}_{r}$,
and discuss in two cases: $T\cong\mathrm{A}_{c}$ for some $c\geq 5$;
$G\cong\mathrm{A}_{c+1}$ for some $c\geq 5$, but $T$ is not isomorphic to an
alternating group.

Case 1. Assume that $T\cong\mathrm{A}_{c}$ for some $c\geq 5$. By [42, Theorem
1.1], combining Lemma 4.2, one line of the following table holds.

$\tiny\begin{array}[]{|l|l|l|l|}\hline\cr G&T&G_{\alpha}&r\\\
\hline\cr\mathrm{M}_{12}&\mathrm{A}_{5}&\mathrm{M}_{11}&11\\\
\hline\cr\mathrm{A}_{15}&\mathrm{A}_{7},\mathrm{A}_{8}&\mathrm{S}_{13}&13\\\
\hline\cr\mathrm{L}_{4}(3)&\mathrm{A}_{6}&3^{3}{:}\mathrm{L}_{3}(3)&13\\\
\hline\cr\Omega_{7}(3)&\mathrm{A}_{8},\mathrm{A}_{9}&[3^{6}]{:}\mathrm{L}_{3}(3)&13\\\
\hline\cr\Omega_{7}(3)&\mathrm{A}_{9}&3^{3}{:}\mathrm{L}_{3}(3)&13\\\
\hline\cr\Omega^{+}_{8}(2)&\mathrm{A}_{9}&[2^{4}]{:}\mathrm{SL}_{2}(4)\leq
G_{\alpha}\leq[2^{6}]{:}\mathrm{\Gamma L}_{2}(4)\mbox{ (see
\cite[cite]{[\@@bibref{}{567}{}{}]})}&5\\\
\hline\cr\mathrm{A}_{c+k}&\mathrm{A}_{c}&k\mbox{-transitive subgroup of
}\mathrm{A}_{c+k},1\leq k\leq 5&\\\ \hline\cr\end{array}$

If $G\cong\mathrm{M}_{12}$ then $\Gamma$ is the complete graph
$\mathsf{K}_{12}$, and part (2) of this lemma follows.

For $G\cong\mathrm{A}_{15}$ and $G_{\alpha}\cong\mathrm{S}_{13}$, the action
of $G$ on $V$ is equivalent to that on the $2$-subsets of a $15$-set, and thus
$G_{\alpha}$ has no orbit of length $13$ on $V$, a contradiction. For
$G\cong\mathrm{L}_{4}(3)$ and $G_{\alpha}\cong 3^{3}{:}\mathrm{L}_{3}(3)$, the
action of $G$ on $V$ is $2$-transitive, yielding $\Gamma\cong\mathsf{K}_{40}$,
which is not the case.

Suppose that $G\cong\Omega_{7}(3)$. If
$G_{\alpha}\cong[3^{6}]{:}\mathrm{L}_{3}(3)$ then $G$ is a primitive group on
$V$ with subdegrees $1$, $39$, $351$ and $729$, and so $G_{\alpha}$ has no
orbit of length $13$ on $V$, a contradiction. Let $G_{\alpha}\cong
3^{3}{:}\mathrm{L}_{3}(3)$. Then
$G_{\alpha\beta}\cong[3^{5}]{:}2\mathrm{S}_{4}$ for $\beta\in\Gamma(\alpha)$.
Checking the maximal subgroups of $G$ in the Atlas [5], $G_{\alpha}$ is
contained in a maximal subgroup $[3^{6}]{:}\mathrm{L}_{3}(3)$ or
$\mathrm{L}_{4}(3).2$. Computation shows that $|\langle
x,G_{\alpha}\rangle|\leq 2|\mathrm{L}_{4}(3)|<|G|$ for
$x\in\mathbf{N}_{G}(G_{\alpha\beta})$ with $x^{2}\in G_{\alpha\beta}$, a
contradiction. For $G\cong\Omega^{+}_{8}(2)$, we know that $G_{\alpha}$ is one
of $[4^{2}]{:}\mathrm{SL}_{2}(4)$, $[4^{2}]{:}\mathrm{GL}_{2}(4)$,
$[4^{2}]{:}\mathrm{\Gamma L}_{2}(4)$, $[4^{3}]{:}\mathrm{GL}_{2}(4)$ and
$[4^{3}]{:}\mathrm{\Gamma L}_{2}(4)$, see [26, Theorem 3.4] for example.
Confirmed by GAP, all subgroups of $\Omega^{+}_{8}(2)$ isomorphic to
$G_{\alpha}$ are conjugate under $\mathbf{Aut}(G)$. Then calculation gives a
similar contradiction as above.

Now $G\cong\mathrm{A}_{c+k}$, $T\cong\mathrm{A}_{c}$ and $G_{\alpha}$ is
(isomorphic to) a $k$-transitive subgroup of $\mathrm{A}_{c+k}$, where $1\leq
k\leq 5$. If $k=1$ then $n=c+1$, and thus either part (3) of this lemma occurs
by Lemma 4.3 when $G_{\alpha}$ is a $2$-transitive subgroup of
$\mathrm{A}_{n}$ or, otherwise we have part (4) of this lemma. Now let $k>1$.
Then Lemma 4.3 works for the triple
$(\mathrm{A}_{c+k},\mathrm{A}_{c+k-1},G_{\alpha})$. In this case, the only
possibility is that $c+k=12$ and $G_{\alpha}\cong\mathrm{L}_{2}(11)$, and so
$\Gamma$ is desired as in part (3) of this lemma.

Case 2. Assume that $G\cong\mathrm{A}_{c+1}$ and $T$ is not (isomorphic to) an
alternating group. By [31, Theorem D], combining Lemma 4.2, $T$ is a
$k$-homogeneous subgroup of $\mathrm{A}_{c+1}$, and either $r>5$ and
$G_{\alpha}\cong\mathrm{S}_{r}$ or $r=5$ and $c+1<9$, where $1\leq k\leq 5$
and $c+1=r+k$. If $k=1$ then $c=r$, and so
$\mathrm{S}_{5}\lesssim\mathrm{A}_{6}$, which is impossible. Then $k\geq 2$.

Consider the action of $G$ on a $(c+1)$-set, and take a $G_{\alpha}$-orbit
with maximal length, say $m$. Then $m\geq r$, and $G_{\alpha}$ has a subgroup
$K$ of index $m$. Clearly, $m\leq c+1\leq 2r\leq{r\choose 2}$. Suppose that
$m={r\choose 2}$. Then $r(r-1)=2m$, the only possibility is that $r=5$ and
$m=10$. This implies that $k=r=5$, and $c+1=10$. Thus $T$ is (isomorphic to) a
$5$-transitive subgroup of $\mathrm{A}_{10}$, yielding
$T\cong\mathrm{A}_{10}\cong G$, a contradiction. Therefore,
$|G_{\alpha}:K|=m<{r\choose 2}$ and $m\leq r+5$.

By [7, Theorem 5.2B], $(r,m)=(5,6)$ or $K\lesssim\mathrm{S}_{r-1}$. For the
latter case, $r$ is a divisor of $m=|G_{\alpha}:K|$, since $m\leq c+1$ and
$2\leq k\leq 5$, we have $r=k=5$, which yields that $T$ is a $5$-transitive
subgroup of $\mathrm{A}_{10}$, a contradiction. Thus we have $(r,m)=(5,6)$ and
$7\leq r+k<9$, yielding $c+1=8$. Computation shows that, for
$\beta\in\Gamma(\alpha)$, there is no $x\in\mathbf{N}_{G}(G_{\alpha\beta})$
with $x^{2}\in G_{\alpha\beta}$ and $G=\langle G_{\alpha},x\rangle$, which
contradicts Lemma 2.9. ∎

In the following theorem, we consider the case where $G$ is a sporadic simple
group.

######  Theorem 6.2.

Assume that Hypothesis 4.1 holds, and let $n=|G:T|$. Assume that $G$ is
isomorphic to a sporadic simple group. Then one of the following holds.

  * (1)

$n=r=23$, $G=\mathrm{M}_{23}$, $T=\mathrm{M}_{22}$,
$G_{\alpha}\cong\mathbb{Z}_{23}$.

  * (2)

$n=12$, $r=11$, $G=\mathrm{M}_{12}$, $T=\mathrm{M}_{11}$,
$G_{\alpha}\cong\mathrm{M}_{11}$, and $\Gamma\cong\mathsf{K}_{12}$.

  * (3)

$r=11$, $G\cong\mathrm{M}_{12}$, $T\cong\mathrm{A}_{5}$,
$G_{\alpha}\cong\mathrm{M}_{11}$, and $\Gamma\cong\mathsf{K}_{12}$.

  * (4)

$r=11$, $G\cong\mathrm{M}_{12}$, $T\cong\mathrm{L}_{2}(11)$,
$G_{\alpha}\cong\mathrm{M}_{11}$, and $\Gamma\cong\mathsf{K}_{12}$.

  * (5)

$r=23$, $G\cong\mathrm{M}_{24}$, $T\cong\mathrm{L}_{3}(2)$,
$G_{\alpha}\cong\mathrm{M}_{23}$, and $\Gamma\cong\mathsf{K}_{24}$.

  * (6)

$r=23$, $G\cong\mathrm{M}_{24}$, $T\cong\mathrm{L}_{2}(23)$,
$G_{\alpha}\cong\mathrm{M}_{23}$, and $\Gamma\cong\mathsf{K}_{24}$.

  * (7)

$n=24$, $r=7$, $G=\mathrm{M}_{24}$, $T=\mathrm{M}_{23}$,
$G_{\alpha}\cong\mathrm{L}_{3}(2)$, and $\Gamma$ is unique up to isomorphism.

###### Proof.

By Lemmas 4.3 and 6.1 and Theorems 5.1 and 6.1, one of parts (1)-(3) holds if
one of the following cases occurs: (i) $G_{\alpha}$ is primitive on $\Delta$;
(ii) $G_{\alpha}$ is solvable; (iii) one of $T$ and $G_{\alpha}$ is an
alternating group. Thus we suppose that none of (i)-(iii) occurs. By [17,
Theorem 1.1], combining Lemma 4.2, the following cases are left:

  * (iv)

$r=c-1$, $G\cong\mathrm{M}_{c}$, $G_{\alpha}\cong\mathrm{M}_{c-1}$,
$c\in\\{12,24\\}$, and $T$ is a transitive subgroup of $M_{c}$ acting on $c$
points.

  * (v)

$G=\mathrm{M}_{n}$, $T=\mathrm{M}_{n-1}$, $n\in\\{12,24\\}$, and $G_{\alpha}$
is a transitive subgroup of $M_{n}$.

Assume that (iv) occurs. Then $\Gamma\cong\mathsf{K}_{c}$ and, in this case,
$n>c$ as $G\leq\mathrm{A}_{n}$. By [16], $(c,T)$ is one of
$(12,\mathrm{L}_{2}(11))$, $(24,\mathrm{L}_{2}(7))$ and
$(24,\mathrm{L}_{2}(23))$, which gives one of parts (4)-(6) of this lemma,
respectively.

Assume that (v) occurs. Combining Lemma 4.2, calculation using GAP shows that
either $n=12$ and $G_{\alpha}\cong\mathrm{S}_{5}$, or $n=24$ and $G_{\alpha}$
is isomorphic to one of the following groups: $\mathrm{S}_{5}$ (twice up to
the conjugacy in $\mathrm{M}_{24}$), $2^{2}\times\mathrm{A}_{5}$ (twice),
$(2^{2}\times\mathrm{A}_{5}){:}2$ (twice),
$\mathrm{A}_{4}\times\mathrm{A}_{5}$, $2^{4}\times\mathrm{A}_{5}$,
$(\mathrm{A}_{4}\times\mathrm{A}_{5}){:}2$, $2^{6}{:}\mathrm{GL}_{2}(4)$,
$2^{6}{:}\mathrm{\Gamma L}_{2}(4)$, $\mathrm{L}_{3}(2)$,
$2^{6}{:}(3\times\mathrm{L}_{3}(2))$ and
$2^{6}{:}(3\times\mathrm{L}_{3}(2)){:}2$. Consider the normalizer $N$ of a
Hall $r^{\prime}$-subgroup of $G_{\alpha}$ in $\mathrm{M}_{n}$. Since $N$
contains a $2$-element which together with $G_{\alpha}$ generate
$\mathrm{M}_{n}$, it follows from a GAP calculation that all cases but
$G_{\alpha}\cong\mathrm{L}_{3}(2)$ are excluded. The group $\mathrm{L}_{3}(2)$
has two subgroups of index $7$ up to conjugation, which are isomorphic to
$\mathrm{S}_{4}$. Embedding $\mathrm{L}_{3}(2)$ in $\mathrm{M}_{24}$ as a
transitive subgroup, one $\mathrm{S}_{4}$ of $\mathrm{L}_{3}(2)$ is self-
normalized in $\mathrm{M}_{24}$, and the second one has normalizer
$\mathbb{Z}_{2}\times\mathrm{S}_{4}$. Then we have a unique $\Gamma$ arising
from the second $\mathrm{S}_{4}$, and thus part (7) of this lemma follows. ∎

By the argument in Section 5 and this section, our task is reduced to those
graphs which satisfy the following assumptions.

######  Hypothesis 6.1.

$\Gamma=(V,E)$, $G$, $T$ and $r$ satisfy Hypothesis 4.1, and

  * (IV)

$G$ is a classical simple group over some finite field, $G_{\alpha}$ has a
unique insolvable composition factor, say
$S=\mathbf{soc}(G_{\alpha}^{\Gamma(\alpha)})$, where $\alpha\in V$;

  * (V)

$G\not\cong\mathrm{A}_{5},\,\mathrm{A}_{6},\,\mathrm{A}_{8}$,
$\mathrm{L}_{m}(q)$ (with $m$ a prime) or $\mathrm{U}_{2m+1}(q)$, and neither
$T$ nor $G_{\alpha}$ is isomorphic to an alternating group.

##  7\. Some technique lemmas

The following two lemmas are derived from [2, Theorem 3.1], which bring a lot
of convenience for us to determine the composition factor $S$ of $G_{\alpha}$
in Hypothesis 6.1.

######  Lemma 7.1.

Let $S$ be a nonabelian simple group with a subgroup of prime index $r$. Let
$q=p^{f}$ for a prime $p$ and integer $f\geq 1$, and let
$H\leq\mathrm{GL}_{m}(q)$ for some $m\geq 2$. Suppose that $H$ has a unique
insolvable composition factor $S$. If $\Phi^{*}_{mf}(p)\neq 1$ and $|H|$ is
divisible by $\Phi^{*}_{mf}(p)$ then one of the following holds.

  * (1)

$S=\mathrm{L}_{d}(q^{m\over d})$, $|H|_{p}$ is a divisor of ${m\over
d}|S|_{p}$, and either $(m,d,q)=(2,2,11)$ or $d$ is a prime and $mf$ is a
power of $d$.

  * (2)

$S$, $m$ and $q$ are listed in Table 7.8.

$\tiny\begin{array}[]{|l|l|l|l|l|l|l|l|l|l|}\hline\cr
S&\mathrm{A}_{5}&\mathrm{A}_{7}&\mathrm{A}_{11}&\mathrm{A}_{13}&\mathrm{A}_{19}&\mathrm{M}_{11}&\mathrm{L}_{3}(2)&\mathrm{L}_{2}(11)&\mathrm{L}_{3}(3)\\\
\hline\cr(m,q)&(4,2)&(6,3)&(10,2)&(12,2)&(18,2)&(10,2)&(6,3)&(10,2)&(12,2)\\\
&(4,3)&(6,5)&&&&&(6,5)&(5,4)&\\\ &&(4,2)&&&&&(3,9)&&\\\
&&(4,7)&&&&&(3,25)&&\\\ &&(3,25)&&&&&&&\\\ \hline\cr\end{array}$ Table 7.8.

###### Proof.

Assume that $\Phi^{*}_{mf}(p)\neq 1$ and $|H|$ has a divisor
$\Phi^{*}_{mf}(p)$. By [2, Theorem 3.1], ignoring those $H$ which does not
produce an suitable $S$, one of the following holds:

  * (1.1)

$H$ has a normal subgroup $\mathrm{SL}_{m}(q)$, $\mathrm{Sp}_{2}(q)$ (with
$m=2$) or $\Omega^{-}_{4}(q)$ (with $m=4$);

  * (1.2)

$H\leq\mathrm{GL}_{m\over b}{.}[b]$, where $b$ is a proper divisor of $m$;

  * (1.3)

$S$, $m$ and $q$ are listed in Table 7.8.

We only need to deal with cases (1.1) and (1.2).

Assume that (1.1) holds. Then either $S=\mathrm{L}_{2}(11)$, or $r$ is one of
${q^{m}-1\over q-1}$, $q+1$ and $q^{2}+1$ respectively. By Lemma 2.1, for the
latter case, $m$ is a prime and $mf$ is a power of $m$, or $m=4$ and $2f$ is a
power of $2$. Moreover, it is easy to see that $|H|_{p}=|S|_{p}$. Then part
(1) of this lemma occurs.

Assume that case (1.2) occurs. Let $H_{1}=H\cap\mathrm{GL}_{m\over b}(q^{b})$.
Then $S\preccurlyeq H_{1}$, and $|H|_{p}$ is a divisor of $|H_{1}|_{p}b$. By
[31, pp. 38,Proposition B], $\Phi^{*}_{mf}(p)$ is a divisor of $|H_{1}|$.
Applying [2, Theorem 3.1] to $H_{1}$ and $\Phi^{*}_{{m\over b}bf}(p)$, we have

  * (2.1)

$H_{1}$ has a normal subgroup $\mathrm{SL}_{m\over b}(q^{b})$,
$\mathrm{Sp}_{2}(q^{b})$ (with $b={m\over 2}$) or $\Omega^{-}_{4}(q^{b})$
(with $b={m\over 4}$); or

  * (2.2)

$H_{1}\leq\mathrm{GL}_{m\over ab}{.}[a]$, where $a$ is a proper divisor of
${m\over b}$; or

  * (2.3)

$(S,{m\over b},q^{b})$ is one of $(\mathrm{A}_{7},3,25)$,
$(\mathrm{L}_{3}(2),3,9)$, $(\mathrm{L}_{3}(2),3,25)$,
$(\mathrm{L}_{2}(11),5,4)$.

First, all triples in (2.3) are included in Table 7.8.

Assume that (2.1) occurs. Then $q^{b}\neq 11$ as $b>1$, and
$|H_{1}|_{p}=|S|_{p}$. Then $|H|_{p}$ is a divisor of $b|S|_{p}$. By Lemma
2.1, $S$ is described as in part (1) of this lemma.

Assume that case (2.2) occurs. By [2, Theorem 3.1] and an induction on $m$, we
conclude that $S=\mathrm{L}_{d}(p^{m\over d})$, $\mathrm{PSp}_{2}(q^{m\over
2})$ or $\mathrm{P\Omega}^{-}_{4}(q^{m\over 4})$. Then, by Lemma 2.1, $S$ is
described as in part (1) of this lemma. ∎

######  Lemma 7.2.

Let $S$ be a nonabelian simple group with a subgroup of prime index $r$. Let
$q=p^{f}$ for a prime $p$ and integer $f\geq 1$, and let
$H\leq\mathrm{GL}_{m}(q)$ for some $m\geq 3$. Suppose that $H$ has a unique
insolvable composition factor $S$. If $\Phi^{*}_{(m-1)f}(p)\neq 1$ and $|H|$
is divisible by $\Phi^{*}_{(m-1)f}(p)$ then one of the following holds.

  * (1)

$S=\mathrm{L}_{m}(q)$, $m$ is a prime and $f$ is a power of $m$.

  * (2)

$S=\mathrm{L}_{d}(q^{m-1\over d})$, $|H|_{p}$ is a divisor of ${m-1\over
d}q^{m-1}|S|_{p}$, and either $(m-1,q)=(2,11)$ or $d$ is a prime and $(m-1)f$
is a power of $d$.

  * (3)

$S$, $m$ and $q$ are listed in Table 7.9.

$\tiny\begin{array}[]{|l|l|l|l|l|l|l|l|l|l|l|}\hline\cr
S&\mathrm{A}_{5}&\mathrm{A}_{7}&\mathrm{A}_{11}&\mathrm{A}_{13}&\mathrm{A}_{19}&\mathrm{M}_{11}&\mathrm{M}_{23}&\mathrm{L}_{3}(2)&\mathrm{L}_{2}(11)&\mathrm{L}_{3}(3)\\\
\hline\cr(m,q)&(5,2)&(7,3)&(11,2)&(13,2)&(19,2)&(5,3)&(11,2)&&&\\\
&(5,3)&(7,5)&&(11,2)&&&&&&\\\ &&(5,2)&&&&&&&&\\\ &&(5,3)&&&&&&&&\\\
&&(4,2)&&&&&&&&\\\ &&(4,9)&&&&&&&&\\\ &&(4,25)&&&&&&&&\\\
\hline\cr(m-1,q)&(4,2)&(6,3)&(10,2)&(12,2)&(18,2)&(10,2)&&(6,3)&(10,2)&(12,2)\\\
&(4,3)&(6,5)&&&&&&(6,5)&(5,4)&\\\ &&(4,2)&&&&&&(3,9)&&\\\
&&(4,7)&&&&&&(3,25)&&\\\ &&(3,25)&&&&&&&&\\\ &&(3,2)&&&&&&&&\\\
\hline\cr\end{array}$ Table 7.9.

###### Proof.

Assume that $\Phi^{*}_{(m-1)f}(p)\neq 1$ and $|H|$ has a divisor
$\Phi^{*}_{(m-1)f}(p)$. By [2, Theorem 3.1], one of the following holds:

  * (1.1)

$H$ has a normal subgroup $\mathrm{SL}_{m}(q)$;

  * (1.2)

$H\leq G_{1}:=q^{m-1}{:}((q-1)\times\mathrm{GL}_{m-1}(q))$;

  * (1.3)

$S$, $m$ and $q$ are listed in Table 7.9.

Case (1.1) gives part (1) of this lemma occurs. If (1.3) occurs then $(S,m,q)$
is given as in part (3) of this lemma. For case (1.2), by Lemma 7.1, either
$S$ is described as in part (2) of this lemma, or $(S,m-1,q)$ is listed in
Table 7.9. ∎

Under Hypothesis 6.1, let $A$ and $B$ be maximal subgroups of $G$ such that
one of them contains $T$ and the other one contains $G_{\alpha}$. Then we have
a maximal factorization $G=AB$ with two factors $A$ and $B$ insolvable.

Employing [31, Theorem A], we will work on the possible candidates for the
triple $(G,T,G_{\alpha})$. An essential part of our approach is determine the
inclusion of $G_{\alpha}$ in $A$ or $B$. Suppose that $G_{\alpha}\leq B$ and
$B$ has an almost simple quotient $B/N=L.O$, where $L$ is simple. Assume
further that $|G:A|$ has a prime divisor which does not divide $|O||N|$. By
Lemma 2.6, either $S\preccurlyeq B$, or we have a core-free maximal
factorization of $B/N$ with a factor admitting $S$ as a composition factor.
Then, by Lemma 2.7 and [31, Theorem A], we may determine $S$ in a recursive
process. We use some notation defined in [31, pp.5], such as
$\mathsf{P}_{i},\,\mathsf{N}_{i},\,\mathsf{N}_{i}^{-},\,\mathsf{N}_{i}^{+}$.

######  Lemma 7.3.

Let $S$ be a nonabelian simple group with a subgroup of prime index $r$. Let
$q=p^{f}$ for a prime $p$ and integer $f\geq 1$, $L=\mathrm{U}_{2m}(q)$ and
$\mathbf{soc}(G_{0})=L$ for some $m\geq 2$. Suppose that $G_{0}=A_{0}B_{0}$ is
a core-free maximal factorization, $A_{0}\cap L=\mathsf{N}_{1}$ and
$G_{0}=A_{0}Y$ for $Y\leq B_{0}$. Suppose that $S$ is the unique insolvable
composition factor of $Y$, and $\pi(S)=\pi(Y)$. Then one of the following
holds.

  * (1)

$S=\mathrm{L}_{2}(q^{m})$, $p=2$, $r=q^{m}+1$ and $mf$ is a power of $2$.

  * (2)

$S$, $2m$ and $q$ are listed in Table 7.10.

$\tiny\begin{array}[]{|l|l|l|l|l|l|l|l|l|l|}\hline\cr
S&\mathrm{A}_{5}&\mathrm{A}_{7}&\mathrm{A}_{11}&\mathrm{A}_{13}&\mathrm{A}_{19}&\mathrm{M}_{11}&\mathrm{L}_{3}(2)&\mathrm{L}_{2}(11)&\mathrm{L}_{3}(3)\\\
\hline\cr(2m,q)&&&&&&&(6,2)&&(6,2)\\\
\hline\cr(2m,q)&(4,2)&(6,3)&(10,2)&(12,2)&(18,2)&(10,2)&(6,3)&(10,2)&(12,2)\\\
&(4,3)&(6,5)&&&&&(6,5)&&\\\ &&(4,7)&&&&&&&\\\
\hline\cr(m,q^{2})&&(3,25)&&&&&(3,25)&(5,4)&\\\ &&&&&&&(3,9)&&\\\
\hline\cr\end{array}$ Table 7.10.

###### Proof.

By [31, Theorem A, Tables 1 and 3], $|G_{0}:A_{0}|=q^{2m-1}{q^{2m}-1\over
q+1}$ and $L\cap B_{0}$ is described as follows:

$\tiny\begin{array}[]{l|l|l|l|l|l|l}L\cap
B_{0}&\mathsf{P}_{m}&\mathrm{PSp}_{2m}(q){.}a&{}^{\bf\hat{}}\mathrm{SL}_{m}(4).2&{}^{\bf\hat{}}\mathrm{SL}_{m}(16).[6]&\mathrm{U}_{4}(3){.}2,\mathrm{M}_{22}&\mathrm{Suz}\\\
\hline\cr&&a\leq 2&q=2,m\geq 3&q=4,G_{0}\geq L{.}4&q=2,m=3&q=2,m=6\\\
\end{array}$

Noting that $Y^{\infty}\unlhd L\cap Y$, it follows that $S\preccurlyeq L\cap
Y$. Since $LY\leq G_{0}$, it follows that $|Y:(L\cap Y)|$ is a divisor of
$o:=|G_{0}:L|$. By Lemma 2.4, $|Y|$ is divisible by $|G_{0}:A_{0}|$. Then
$|L\cap Y|$ has a divisor $q^{2m-1}{q^{2m}-1\over o(q+1)}$.

Assume that $L\cap B_{0}\cong\mathrm{Suz}$. Then $|(L\cap B_{0})\cap Y|$ is
divisible by $2^{11}\cdot 5\cdot 7\cdot 13$. By the Atlas [5], we have $(L\cap
B_{0})\cap Y\lesssim\mathrm{G}_{2}(4)$. Again by the Atlas [5],
$\mathrm{G}_{2}(4)$ has no maximal subgroup of order divisible $2^{11}\cdot
5\cdot 7\cdot 13$. It follows that $(L\cap B_{0})\cap Y=\mathrm{G}_{2}(4)$,
and then $S\cong\mathrm{G}_{2}(4)$, which is impossible.

Assume that $(m,q)=(3,2)$. Then $|L\cap Y|$ is divisible by $2^{5}\cdot 7$,
and $L\cap B_{0}\cong 2^{9}{:}\mathrm{L}_{3}(4)$, $\mathrm{Sp}_{6}(2)$,
$\mathrm{L}_{3}(4){.}2$, $\mathrm{U}_{4}(3){.}2$ or $\mathrm{M}_{22}$. Noting
that $S\preccurlyeq(L\cap Y)\mathbf{rad}(L\cap B_{0})/\mathbf{rad}(L\cap
B_{0})$, with the help of GAP, we search the subgroups of $(L\cap
B_{0})/\mathbf{rad}(L\cap B_{0})$ with order divisible by $2^{5}\cdot
7\over(2^{5},|\mathbf{rad}(L\cap B_{0})|)$. It follows that
$S\cong\mathrm{L}_{3}(2)$ or $\mathrm{L}_{3}(3)$, desired as in part (2) of
this lemma.

For the left candidates for $L\cap B_{0}$, we have $\Phi^{*}_{2m}(q)\neq 1$,
and $|L\cap Y|$ has a divisor $q^{2m-1}\Phi^{*}_{2m}(q)$. Let
$\overline{B}=(L\cap B_{0})/\mathbf{rad}(L\cap B_{0})$ and
$\overline{Y}=(L\cap Y)\mathbf{rad}(L\cap B_{0})/\mathbf{rad}(L\cap B_{0})$.
Then $S\preccurlyeq\overline{Y}$, and $|\overline{Y}|$ is divisible by
$\Phi^{*}_{2m}(q)$. By [31, pp. 38, Proposition B], we conclude that both
$|\overline{B}^{\infty}|$ and $|\overline{B}^{\infty}\cap\overline{Y}|$ are
divisible by $\Phi^{*}_{2m}(q)$. Thus Lemma 7.1 works for the preimages of
$\overline{B}^{\infty}\cap\overline{Y}$ and $\overline{B}^{\infty}$ in
$\mathrm{GL}_{m}(q^{2})$ (or in $\mathrm{GL}_{2m}(q)$ when
$\overline{B}^{\infty}\cong\mathrm{PSp}_{2m}(q)$). Then the lemma follows. ∎

######  Lemma 7.4.

Let $S$ be a nonabelian simple group with a subgroup of prime index $r$. Let
$q=p^{f}$ for a prime $p$ and integer $f\geq 1$,
$L=\mathrm{P\Omega}^{+}_{8}(q)$ and $\mathbf{soc}(G_{0})=L$. Suppose that
$G_{0}=A_{0}B_{0}$ is a core-free maximal factorization, $A_{0}\cap
L\cong\Omega_{7}(q)$ and $G_{0}=A_{0}Y$ for $Y\leq B_{0}$. Suppose that $S$ is
the unique insolvable composition factor of $Y$, and $\pi(S)=\pi(Y)$. Then
either $S\cong\mathrm{L}_{2}(q^{2})$ and $r=q^{2}+1$ with $p=2$ and $f$ a
power of $2$, or $(S,q)$ is one of $(\mathrm{A}_{5},3)$, $(\mathrm{A}_{7},2)$,
$(\mathrm{A}_{7},3)$ and $(\mathrm{A}_{7},7)$.

###### Proof.

By [31, Table 4], $L\cap B_{0}$ is described as follows:

$\tiny\begin{array}[]{l|l|l|l|l|l}L\cap B_{0}&(L\cap
A_{0})^{\tau}&\mathsf{P}_{i}&{}^{\bf\hat{}}({q\mp 1\over
e}{\times}\Omega^{\pm}_{6}(q)){.}2^{e}&(\mathrm{PSp}_{2}(q)\otimes\mathrm{PSp}_{4}(q)){.}2&\Omega^{-}_{8}(\sqrt{q})\\\
\hline\cr&\mbox{triality }\tau&i\in\\{1,3,4\\}&e=(2,q-1)&q\mbox{ odd}&q\mbox{
square}\\\ \hline\cr L\cap
B_{0}&\mathrm{A}_{9}&(\mathrm{L}_{2}(4)\times\mathrm{L}_{2}(4)){.}2^{2}&(\mathrm{L}_{2}(16)\times\mathrm{L}_{2}(16)){.}2^{2}&\Omega^{+}_{8}(2)\\\
\hline\cr&q=2&q=2&q=4,G_{0}\geq L{.}2&q=3\end{array}$

It is easily shown that $S\preccurlyeq L\cap Y$, and $|L\cap Y|$ has a divisor
${1\over o}|G_{0}:A_{0}|$, where $o=|G_{0}:L|$. Note that
$|G_{0}:A_{0}|={1\over(2,q-1)}q^{3}(q^{4}-1)$.

Suppose that $q=3$ and $L\cap B_{0}\cong\Omega^{+}_{8}(2)$. Then $|L\cap Y|$
has a divisor $2^{3}\cdot 3^{2}\cdot 5$. Checking the orders of maximal
subgroups of $\Omega^{+}_{8}(2)$, it follows that $L\cap
Y\lesssim\mathrm{Sp}_{6}(2)$, $2^{6}{:}\mathrm{A}_{8}$, $\mathrm{A}_{9}$,
$(3\times\mathrm{U}_{4}(2)){:}2$ or
$(\mathrm{A}_{5}\times\mathrm{A}_{5}){.}2^{2}$. Inspecting the subgroups of
$\mathrm{Sp}_{6}(2)$, $\mathrm{A}_{9}$, $\mathrm{A}_{8}$, $\mathrm{U}_{4}(2)$
and $\mathrm{A}_{5}\times\mathrm{A}_{5}$, we conclude that
$S\cong\mathrm{A}_{5}$ or $\mathrm{A}_{7}$. If $L\cap
B_{0}\cong\mathrm{A}_{9},\,(\mathrm{L}_{2}(4)\times\mathrm{L}_{2}(4)){.}2^{2}$
or $(\mathrm{L}_{2}(16)\times\mathrm{L}_{2}(16)){.}2^{2}$ then we have
$(S,q)=(\mathrm{A}_{7},2)$, $(\mathrm{L}_{2}(q^{2}),2)$ or
$(\mathrm{L}_{2}(q^{2}),4)$.

Next we deal with the left factorizations $G_{0}=A_{0}B_{0}$.

Case 1. Assume that $L\cap B_{0}$ is one of
$\mathsf{P}_{i},\,{}^{\bf\hat{}}({q\mp 1\over
2}{\times}\Omega^{\pm}_{6}(q)){.}2^{e},\,(\mathrm{PSp}_{2}(q)\otimes\mathrm{PSp}_{4}(q)){.}2$
and $\Omega^{-}_{8}(\sqrt{q})$. Let $\overline{B}=(L\cap
B_{0})/\mathbf{rad}(L\cap B_{0})$ and $\overline{Y}=(L\cap
Y)\mathbf{rad}(L\cap B_{0})/\mathbf{rad}(L\cap B_{0})$. Then
$S\preccurlyeq\overline{Y}$, and $|\overline{Y}|$ is divisible by ${1\over
a}|G_{0}:A_{0}|$, where $a$ is a divisor of $o|\mathbf{rad}(L\cap B_{0})|$.
Then $S\preccurlyeq\overline{B}^{\infty}\cap\overline{Y}$, and it is easily
shown that $|\overline{B}^{\infty}\cap\overline{Y}|$ is divisible by
$\Phi^{*}_{4}(q)$ (and by $\Phi^{*}_{8}(\sqrt{q})$ if $q$ is a square). If
$\overline{B}^{\infty}$ is neither $\mathrm{U}_{4}(q)$ nor
$\mathrm{PSp}_{2}(q)\times\mathrm{PSp}_{4}(q)$ then, applying Lemma 7.1 to the
preiamge of $\overline{B}^{\infty}\cap\overline{Y}$ in $\mathrm{GL}_{4}(q)$
(or $\mathrm{GL}_{8}(\sqrt{q})$), we conclude that either
$S\cong\mathrm{L}_{2}(q^{2})$, $p=2$ and $f$ is a power of $2$, or $(S,q)$ is
one of $(\mathrm{A}_{5},2)$, $(\mathrm{A}_{5},3)$, $(\mathrm{A}_{7},2)$,
$(\mathrm{A}_{7},3)$ and $(\mathrm{A}_{7},7)$.

Assume that $\overline{B}^{\infty}\cong\mathrm{U}_{4}(q)$. Then
$|\overline{B}^{\infty}\cap\overline{Y}|$ has a divisor
${1\over(2,p)}q^{3}\Phi^{*}_{4}(q)$. Checking maximal subgroups of
$\mathrm{U}_{4}(q)$ (refer to [3, Tables 8.10 and 8.11]), we conclude that
$\overline{B}^{\infty}\cap\overline{Y}\leq[q^{4}]{:}[a_{1}]{.}\mathrm{L}_{2}(q^{2}){.}[a_{2}]$,
where $a_{1}a_{2}={(q-1)(2,q^{2}-1)\over(4,q+1)}$. Applying Lemma 7.1 to a
suitable quotient of $\overline{B}^{\infty}\cap\overline{Y}$, we have
$S\cong\mathrm{L}_{2}(q^{2})$, $p=2$ and $f$ is a power of $2$.

Now let
$\overline{B}^{\infty}\cong\mathrm{PSp}_{2}(q)\times\mathrm{PSp}_{4}(q)$ with
odd $q>3$. Then $|\overline{B}^{\infty}\cap\overline{Y}|$ is divisible by
$q^{3}\Phi^{*}_{4}(q)$. Let $N$ be the direct factor of
$\overline{B}^{\infty}$ with $N\cong\mathrm{PSp}_{2}(q)$. Noting that
$S\preccurlyeq\overline{B}^{\infty}\cap\overline{Y}$, we have
$\pi(S)=\pi(\overline{B}^{\infty}\cap\overline{Y})$. It follows that
$S\not\preccurlyeq N$, and so
$S\preccurlyeq(\overline{B}^{\infty}\cap\overline{Y})N/N$. Clearly,
$(\overline{B}^{\infty}\cap\overline{Y})N/N$ has order divisible by
$\Phi^{*}_{4}(q)$. Since $q^{2}+1$ is not a prime as $q$ is odd, it follows
from Lemma 7.1 that $q=7$ and $S\cong\mathrm{A}_{7}$.

Case 2. Assume that $L\cap A_{0}\cong\Omega_{7}(q)\cong L\cap B_{0}$. Then
$B_{0}=A_{0}^{\tau}$ for some triality $\tau$. By [24, Proposition 3.1.1],
$L\cap A_{0}\cap B_{0}\cong\mathrm{G}_{2}(q)$. Considering the limitations on
$S$, we know that $\mathrm{G}_{2}(q)\not\preccurlyeq Y$. Since $G_{0}=A_{0}Y$,
we have $B_{0}=(A_{0}\cap B_{0})Y$. By Lemma 2.7, we have a maximal core-free
factorization $G_{1}=A_{1}B_{1}$ with $B_{0}^{\infty}\leq G_{1}$, $(A_{0}\cap
B_{0})^{\infty}\leq A_{1}$ and $Y^{\infty}\leq B_{1}$. By [31, Theorem A],
$B_{1}$ can be read out from [31, Tables 2 and 3]:

$\tiny\begin{array}[]{l|l|l|l|l|l}B_{1}\cap
B_{0}^{\infty}&[q^{5}]{:}({q-1\over
2}{\times}\mathrm{PSp}_{4}(q)){.}2&\Omega^{\pm}_{6}(q){.}2&({q\pm 1\over
2}{\times}\mathrm{PSp}_{4}(q)){.}2^{2}&\mathrm{Sp}_{6}(2),q=3&\mathrm{S}_{9},q=3\\\
\hline\cr B_{1}\cap
B_{0}^{\infty}&\Omega^{\pm}_{6}(q){.}2&[q^{5}]{:}({q-1\over
2}{\times}\mathrm{PSp}_{4}(q))&\mathrm{Sp}_{2}(q){\times}\mathrm{PSp}_{4}(q)\end{array}$

For $\mathrm{Sp}_{6}(2)$ and $\mathrm{S}_{9}$, we have $S\cong\mathrm{A}_{5}$
or $\mathrm{A}_{7}$ by checking their subgroups. For the left candidates for
$B_{1}$, by a similar argument as in Case 1, either
$S\cong\mathrm{L}_{2}(q^{2})$, $p=2$ and $f$ is a power of $2$, or $(S,q)$ is
one of $(\mathrm{A}_{5},2)$, $(\mathrm{A}_{5},3)$, $(\mathrm{A}_{7},2)$,
$(\mathrm{A}_{7},3)$ and $(\mathrm{A}_{7},7)$. ∎

######  Lemma 7.5.

Let $S$ be a nonabelian simple group with a subgroup of prime index $r$. Let
$q=p^{f}$ for a prime $p$ and integer $f\geq 1$,
$L=\mathrm{P\Omega}^{+}_{2m}(q)$ and $\mathbf{soc}(G_{0})=L$ for some $m\geq
5$. Suppose that $G_{0}=A_{0}B_{0}$ is a core-free maximal factorization,
$A_{0}\cap L=\mathsf{N}^{-}_{2}$ and $G_{0}=A_{0}Y$ for $Y\leq B_{0}$. Suppose
that $S$ is the unique insolvable composition factor of $Y$, and
$\pi(S)=\pi(Y)$. Then one of the following holds.

  * (1)

$S=\mathrm{L}_{m}(q)$, $r={q^{m}-1\over q-1}$, $m$ is a prime and $f$ is a
power of $m$.

  * (2)

$(m,q)=(6,2)$, $S\cong\mathrm{L}_{5}(2)$ and $r=31$.

###### Proof.

By [31, Tables 1-3], $|G_{0}:A_{0}|={q^{2m-2}(q^{m}-1)(q^{m-1}-1)\over
2(q+1)}$, and $B_{0}$ is of type $\mathsf{P}_{m}$, $\mathsf{P}_{m-1}$,
${}^{\bf\hat{}}\mathrm{GL}_{m}(2){.}2$ or
${}^{\bf\hat{}}\mathrm{GL}_{m}(4){.}2$. Let
$\overline{B_{0}}=B_{0}^{\prime}/\mathbf{O}_{p}(B_{0}^{\prime})$ and
$\overline{Y}=(Y\cap
B^{\prime})\mathbf{O}_{p}(B_{0}^{\prime})/\mathbf{O}_{p}(B_{0}^{\prime})$.
Then $S\preccurlyeq\overline{Y}$ and $|\overline{Y}|$ is divisible by
$\Phi^{*}_{m}(q)\Phi^{*}_{m-1}(q)$. Consider the preimage of $\overline{Y}$ in
$\mathrm{GL}_{m}(q)$, and note that $\pi(S)=\pi(\overline{Y})$. If
$\Phi^{*}_{m}(q)=1$ then $(m,q)=(6,2)$, and thus $S\cong\mathrm{L}_{5}(2)$ by
Lemma 7.2. Assume that $\Phi^{*}_{m}(q)\neq 1$. Then $S$ is described as in
Lemma 7.1. For those $S$ in Table 7.8, we have $\Phi^{*}_{m-1}(q)\neq 1$ and
$(|S|,\Phi^{*}_{m-1}(q))=1$. Thus $L\cong\mathrm{L}_{d}(q^{m\over d})$, where
$d$ is a prime and $mf$ is a power of $d$. The only possibility is that $m=d$,
and the lemma follows. ∎

######  Lemma 7.6.

Let $S$ be a nonabelian simple group with a subgroup of prime index $r$. Let
$q=p^{f}$ for a prime $p$ and integer $f\geq 1$,
$L=\mathrm{P\Omega}^{+}_{2m}(q)$ and $\mathbf{soc}(G_{0})=L$ for some $m\geq
5$. Suppose that $G_{0}=A_{0}B_{0}$ is a core-free maximal factorization,
$A_{0}\cap L=\mathsf{N}_{1}$ and $G_{0}=A_{0}Y$ for $Y\leq B_{0}$. Suppose
that $S$ is the unique insolvable composition factor of $Y$, and
$\pi(S)=\pi(Y)$. Then one of the following holds.

  * (1)

$S=\mathrm{L}_{d}(q^{m\over d})$, $r={q^{m}-1\over q^{m\over d}-1}$, $d$ is a
prime and $mf$ is a power of $d$.

  * (2)

$(S,m,q)$ is one of the following triples:
$(\mathrm{A}_{7},6,3)$, $(\mathrm{A}_{7},6,5)$, $(\mathrm{A}_{11},10,2)$,
$(\mathrm{A}_{13},12,2)$, $(\mathrm{A}_{19},18,2)$, $(\mathrm{M}_{11},10,2)$,
$(\mathrm{L}_{3}(2),6,3)$, $(\mathrm{L}_{3}(2),6,5)$,
$(\mathrm{L}_{2}(11),10,2)$, $(\mathrm{L}_{2}(11),5,4)$,
$(\mathrm{L}_{3}(3),12,2)$,
$(\mathrm{A}_{7},6,2)$, $(\mathrm{L}_{3}(2),6,2)$ and
$(\mathrm{L}_{3}(3),6,2)$.

###### Proof.

By [31, Tables 1-3], $L\cap B_{0}$ is described as follows:

$\tiny\begin{array}[]{l|l|l|l|l|l|l}L\cap B_{0}&\mathsf{P}_{m}\mbox{ or
}\mathsf{P}_{m-1}&{}^{\bf\hat{}}\mathrm{GU}_{m}(q){.}2&(\mathrm{PSp}_{2}(q){\times}\mathrm{PSp}_{m}(q)){.}c&{}^{\bf\hat{}}\mathrm{GL}_{m}(q){.}2&\Omega^{+}_{m}(4).2^{2}&\Omega^{+}_{m}(16).2^{2}\\\
\hline\cr&&m\mbox{ even}&q>2,m\mbox{ even},c=(2,q-1)&G_{0}\geq L.2\mbox{
if}&q=2&q=4,m\mbox{ even}\\\ &&&&m\mbox{ odd}&m\mbox{ even}&G_{0}\geq L.2\\\
\hline\cr L\cap B_{0}&\Omega_{9}(q).a&&\mathrm{Co}_{1}\\\ \hline\cr&m=8,a\leq
2&&m=12,q=2\par\end{array}$

Note that $|Y|$ is divisible by
$|G_{0}:A_{0}|={1\over(2,q-1)}q^{m-1}(q^{m}-1)$, $S\preccurlyeq L\cap Y$, and
$|L\cap Y|$ has a divisor ${1\over o}|G_{0}:A_{0}|$, where $o=|G_{0}:L|$.

Assume that $L\cap B_{0}\cong\mathrm{Co}_{1}$. Checking the maximal subgroups
of $\mathrm{Co}_{1}$ with order divisible by $2^{10}\cdot 3^{2}\cdot 5\cdot
7\cdot 13$, we have $L\cap Y\leq B_{1}\cong 3{.}\mathrm{Suz}{.}2$ or
$(\mathrm{A}_{4}{\times}\mathrm{G}_{2}(4)){.}2$. Clearly, $S\preccurlyeq(L\cap
Y)\mathbf{rad}(B_{1})/\mathbf{rad}(B_{1})$. Checking the subgroups of
$B_{1}/\mathbf{rad}(B_{1})$ with order divisible by $2^{8}\cdot 3\cdot 5\cdot
7\cdot 13$, we conclude that $(L\cap
Y)\mathbf{rad}(B_{1})/\mathbf{rad}(B_{1})\succcurlyeq\mathrm{G}_{2}(4)$ or
$\mathrm{Suz}$. This implies that $S\cong\mathrm{G}_{2}(4)$ or $\mathrm{Suz}$,
a contradiction.

Case 1. Assume that $L\cap B_{0}\cong\Omega_{9}(q).a$,
$\mathsf{P}_{m},\,\mathsf{P}_{m-1},\,(\mathrm{PSp}_{2}(q){\times}\mathrm{PSp}_{m}(q)){.}c$
or ${}^{\bf\hat{}}\mathrm{GL}_{m}(q){.}2$. Let $B_{1}=(L\cap B_{0})^{\prime}$
and $Y_{1}=Y\cap B_{1}$. Then $S\preccurlyeq Y_{1}$ and $|Y_{1}|$ is divisible
by $\Phi^{*}_{m}(q)$. If $L\cap B_{0}\cong\Omega_{9}(q).a$ then Lemmas 7.1 or
7.2 work for $B_{1}$ and $Y_{1}$, and thus $S\cong\mathrm{L}_{2}(q^{4})$ and
$r=q^{4}+1$ with $p=2$ and $f$ a power of $2$, desired as in part (1) of this
lemma with $d=p=2$ and $m=8$. (Note $\Omega_{9}(q)=\mathrm{Sp}_{8}(q)$ for
even $q$.) For the left cases, choose $N\unlhd B_{1}$ such that $B_{1}/N$ is
almost simple, $S\preccurlyeq Y_{1}N/N$ and $|Y_{1}N/N|$ is divisible by
$\Phi^{*}_{m}(q)$. Then Lemma 7.1 works for the preimages of $B_{1}/N$ and
$Y_{1}N/N$ in $\mathrm{GL}_{m}(q)$. It follows that $S$ is desired as in part
(1) or (2) of this lemma.

Case 2. Assume that $L\cap B_{0}\cong\mathrm{GU}_{m}(q){.}2$ for even $m$.
Noting that $B_{0}=(A_{0}\cap B_{0})Y$, by Lemma 2.6, we have a core-free
factorization $\overline{B_{0}}=\overline{(A_{0}\cap B_{0})}\,\overline{Y}$,
where $\overline{B_{0}}=B_{0}/\mathbf{rad}(B_{0})$, $\overline{(A_{0}\cap
B_{0})}=(A_{0}\cap B_{0})\mathbf{rad}(B_{0})/\mathbf{rad}(B_{0})$ and
$\overline{Y}=Y\mathbf{rad}(B_{0})/\mathbf{rad}(B_{0}))$. It is easily shown
that $|\overline{(A_{0}\cap B_{0})}|$ is divisible by $\Phi^{*}_{2m-2}(q)$. By
Lemma 2.7, we may choose a core-free maximal factorization $G_{1}=A_{1}B_{1}$
such that $\mathbf{soc}(G_{1})\cong\mathrm{U}_{m}(q)$, $|A_{1}|$ is divisible
by $\Phi^{*}_{2m-2}(q)$, $\overline{Y}^{\infty}\leq B_{1}$ and
$G_{1}=A_{1}(B_{1}\cap\overline{Y})$. By [31, Theorem A], either $A_{1}$ is a
subgroup of $G_{1}$ with type $\mathsf{N}_{1}$, or $(m,q)=(6,2)$,
$A_{1}\cap\mathbf{soc}(G_{1})\cong\mathrm{M}_{22}$ and
$B_{1}\cap\mathbf{soc}(G_{1})\cong\mathrm{U}_{5}(2)$. By Lemma 7.3, we
conclude that $S$ is desired as in part (1) (with $d=p=2$) or (2) of this
lemma.

Assume that $(m,q)=(6,2)$, $A_{1}\cap\mathbf{soc}(G_{1})\cong\mathrm{M}_{22}$
and $B_{1}\cap\mathbf{soc}(G_{1})\cong\mathrm{U}_{5}(2)$. In this case,
$|G_{1}:A_{1}|=2^{8}\cdot 3^{4}$. It is easily shown that
$S\preccurlyeq\overline{Y}\cap B_{1}\cap\mathbf{soc}(G_{1})$, and
$|\overline{Y}\cap B_{1}\cap\mathbf{soc}(G_{1})|$ is divisible by $2^{7}\cdot
3^{3}$. Checking the maximal subgroups of $\mathrm{U}_{5}(2)$, we conclude
that $\overline{Y}\cap B_{1}\cap\mathbf{soc}(G_{1})$ is solvable, a
contradiction.

Case 3. Assume that $L\cap B_{0}\cong\Omega^{+}_{m}(q^{2}).2^{2}$ for
$q\in\\{2,4\\}$ and even $m$. Suppose that $m=6$. Then $(L\cap
B_{0})^{\prime}\cong\mathrm{L}_{4}(q^{2})$. It is easily shown that
$S\preccurlyeq Y\cap(L\cap B_{0})^{\prime}$, and $|Y\cap(L\cap
B_{0})^{\prime}|$ is divisible by ${1\over 8}q^{5}(q^{6}-1)$. Assume that
$q=4$. Then $S$ is given in Lemma 7.2, and the only possibility is that
$S\cong\mathrm{L}_{3}(16)$. Then $r={16^{3}-1\over 16-1}$, which is not a
prime. Thus $q\neq 4$, and so $q=2$. Checking the subgroups of maximal
subgroups of $\mathrm{L}_{4}(4)$, we get $S\cong\mathrm{L}_{3}(2)$ or
$\mathrm{A}_{7}$, which is included in part (2) of this lemma.

Next let $m\geq 8$. Similarly as in Case 2, we have a core-free maximal
factorization $G_{1}=A_{1}B_{1}$ such that
$\mathbf{soc}(G_{1})\cong\Omega^{+}_{m}(q^{2})$, $|A_{1}|$ is divisible by
$\Phi^{*}_{m-2}(q^{2})$, $S\preccurlyeq\overline{Y}^{\infty}\leq B_{1}$ and
$G_{1}=A_{1}(B_{1}\cap\overline{Y})$, where
$\overline{Y}=Y\mathbf{rad}(B_{0})/\mathbf{rad}(B_{0}))$.

Assume thta $m=8$. If $A_{1}$ has a normal subgroup isomorphic to
$\mathrm{Sp}_{6}(q)$ then $S\cong\mathrm{L}_{2}(q^{4})$ by Lemma 7.4, desired
as in part (1) of this lemma. Thus, by [31, Table 4], we may let
$|G_{1}:A_{1}|={1\over 2}q^{12}(q^{6}-1)(q^{4}+1)(q^{2}-1)$, and
$B_{1}\cap\mathbf{soc}(G_{1})\cong[q^{12}]{:}\mathrm{GL}_{4}(q^{2})$ or
$\mathrm{GL}_{4}(4){.}2$. Using Lemmas 7.1 and 7.2, by a similar argument as
in the proof of Lemma 7.5, we conclude that $S\cong\mathrm{L}_{2}(q^{4})$ or
$\mathrm{L}_{3}(q^{2})$. Noting that $\pi(S)=\pi(B_{1}\cap\overline{Y})$ and
$|B_{1}\cap\overline{Y}|$ is divisible by $|G_{1}:A_{1}|$, it follows that
$q=2$ and $S\cong\mathrm{L}_{2}(16)$, desired as in part (1) of this lemma.

Let $m\geq 10$. Then, by [31, Theorem A and Proposition 2.5], $A_{1}$ has type
$\mathsf{N}_{1}$, $\mathsf{N}^{-}_{2}$ or ${}^{\bf\hat{}}\mathrm{GU}_{m\over
2}(q^{2}){.}2$ (with ${m\over 2}$ even). For $\mathsf{N}^{-}_{2}$, by Lemma
7.5, $S\cong\mathrm{L}_{m\over 2}(q^{2})$ with ${m\over 2}$ a prime and $mf$ a
power of ${m\over 2}$, yielding $m=2$, a contradiction. Assume that
$A_{1}\cap\mathbf{soc}(G_{1})\cong{}^{\bf\hat{}}\mathrm{GU}_{m\over
2}(q^{2}){.}2$. By [31, Table 1], $B_{1}$ has type $\mathsf{P}_{1}$ or
$\mathsf{N}^{+}_{2}$. It follows that $(|B_{1}|,\Phi^{*}_{m}(q))=1$. Clearly,
$\Phi^{*}_{m}(q)\neq 1$. Recalling that $\pi(S)=\pi(Y)$ and $|Y|$ is divisible
by $\Phi^{*}_{m}(q)$, since $S\preccurlyeq B_{1}\cap\overline{Y}$, it follows
that $(|B_{1}|,\Phi^{*}_{m}(q))\neq 1$, a contradiction. Thus $A_{1}$ is of
type $\mathsf{N}_{1}$. Then, by an induction on $m$, the lemma follows. ∎

######  Lemma 7.7.

Let $S$ be a nonabelian simple group with a subgroup of prime index $r$. Let
$q=2^{f}$, $L=\mathrm{Sp}_{2m}(q)$ and $\mathbf{soc}(G_{0})=L$, where $m\geq
3$ and $mf>3$. Suppose that $G_{0}=A_{0}B_{0}$ is a core-free maximal
factorization, $A_{0}\cap L\cong\mathrm{GO}_{2m}^{-}(q)$ and $G_{0}=A_{0}Y$
for $Y\leq B_{0}$. Suppose that $S$ is the unique insolvable composition
factor of $Y$, and $\pi(S)=\pi(Y)$. Then $B_{0}$ is given as in [31, Table 1],
and one of the following holds.

  * (1)

$S\cong\mathrm{L}_{d}(q^{m\over d})$, where $d$ is a prime and $mf$ is a power
of $d$.

  * (2)

$(S,m,q)$ is one of $(\mathrm{A}_{7},6,2)$, $(\mathrm{L}_{3}(2),6,2)$ and
$(\mathrm{A}_{13},12,2)$.

  * (3)

$B_{0}=\mathsf{P}_{m}$, $(S,m,q)$ is one of $(\mathrm{A}_{7},4,2)$ and
$(\mathrm{L}_{5}(2),6,2)$.

###### Proof.

Assume that $B_{0}$ is described as in [31, Table 3]. Then $(m,q)=(4,2)$,
$G_{0}=\mathrm{Sp}_{8}(2)$, $B_{0}\cong\mathrm{S}_{10}$ and $A_{0}\cap
B_{0}\cong\mathrm{S}_{3}\times\mathrm{S}_{7}$, see [31, pp.97, 5.1.9]. By [31,
Theorem D], $Y$ is a primitive group of degree $10$. Noting that
$\mathrm{A}_{10}\not\preccurlyeq Y$, it follows that
$S=\mathbf{soc}(Y)\cong\mathrm{A}_{6}$ or $\mathrm{A}_{5}$. Noting the
limitations on $S$, we have $Y\cong\mathrm{A}_{5}$ or $\mathrm{S}_{5}$. Then
$Y\cong\mathrm{S}_{5}$ as $|Y|$ is divisible by $|G_{0}:A_{0}|=2^{3}\cdot
3\cdot 5$, and part (1) of this lemma follows.

Assume that $B_{0}$ is described as in [31, Table 2]. Then $L\cap
B_{0}\cong\mathrm{G}_{2}(q)$. Noting that $B_{0}=(A_{0}\cap B_{0})Y$, by Lemma
2.6, we have a core-free factorization $B_{1}=X_{1}Y_{1}$ with
$\mathbf{soc}(B_{1})\cong\mathrm{G}_{2}(q)$, and $S\preccurlyeq Y_{1}$. It
follows from [21] that $q=4$ and $S\cong\mathrm{J}_{2}$, $\mathrm{U}_{3}(4)$
or $\mathrm{U}_{3}(3)$, which is impossible.

Assume next that $B_{0}$ is given as in [31, Table 1]. We discuss in several
cases according the structure of $B_{0}$. Note that $|Y|$ is divisible by
$|G_{0}:A_{0}|={1\over 2}q^{m}(q^{m}-1)$.

Case 1. Let $B_{0}=\mathsf{P}_{m}$,
$\overline{B_{0}}=B_{0}/\mathbf{O}_{2}(B_{0})$ and
$\overline{Y}=Y\mathbf{O}_{2}(B_{0})/\mathbf{O}_{2}(B_{0})$. Then
$\overline{B_{0}}\cong\mathrm{GL}_{m}(q)$, $S\preccurlyeq\overline{Y}$ and
$|\overline{Y}|$ is divisible by $q^{m}-1$. Assume that $(m,q)\neq(6,2)$.
Then, by Lemma 7.1, either part (1) of this lemma occurs, or $(S,m,q)$ is one
of $(\mathrm{A}_{5},4,2)$, $(\mathrm{A}_{7},4,2)$, $(\mathrm{A}_{11},10,2)$,
$(\mathrm{A}_{13},12,2)$, $(\mathrm{A}_{19},18,2)$, $(\mathrm{M}_{11},10,2)$,
$(\mathrm{L}_{2}(11),10,2)$, $(\mathrm{L}_{2}(11),5,4)$ and
$(\mathrm{L}_{3}(3),12,2)$. For the latter case, considering $\pi(S)=\pi(Y)$,
only $(\mathrm{A}_{5},4,2)$, $(\mathrm{A}_{7},4,2)$ and
$(\mathrm{A}_{13},12,2)$ are left. Note the triple $(\mathrm{A}_{5},4,2)$ is
included in part (1) of this lemma. Now let $(m,q)=(6,2)$. By Lemma 2.7, there
is a factorization of $\mathrm{SL}_{6}(2)$ with a factor having a composition
factor $S$. Checking the subgroups of $\mathrm{SL}_{6}(2)$, we have
$S\cong\mathrm{A}_{7}$, $\mathrm{SL}_{3}(2)$ or $\mathrm{SL}_{5}(2)$, as in
part (2) or (3) of this lemma.

Case 2. Let $B_{0}=\mathrm{Sp}_{m}(q)\wr\mathrm{S}_{2}$. Note that $A_{0}\cap
B_{0}\cong(\mathrm{GO}^{-}_{m}(q)\times\mathrm{GO}^{+}_{m}(q))$, see [31, pp.
50, 3.2.4(b)]. We have $A_{0}\cap B_{0}\leq B_{0}^{\prime}$ and
$B_{0}^{\prime}=(A_{0}\cap B_{0})(Y\cap B_{0}^{\prime})$. Let $N$ be a direct
factor of $B_{0}^{\prime}$ such that $N$ contains the direct factor
$\mathrm{GO}^{-}_{m}(q)$ of $A_{0}\cap B_{0}$. Let
$\overline{B_{0}}=B_{0}^{\prime}/N$, $\overline{Y}=(Y\cap B_{0}^{\prime})N/N$
and $\overline{H}=(A_{0}\cap B_{0})N/N$. Then
$\overline{B}\cong\mathrm{Sp}_{m}(q)$,
$\overline{H}\cong\mathrm{GO}^{+}_{m}(q)$ and
$\overline{B}=\overline{H}\,\overline{Y}$, and thus $|\overline{Y}|$ is
divisible by $|\overline{B}:\overline{H}|={1\over 2}q^{m\over 2}(q^{m\over
2}+1)$.

Assume that $\overline{Y}$ is insolvable. Then $S\preccurlyeq\overline{Y}$.
Applying Lemma 7.1 to the pair $(\overline{B},\overline{Y})$, by a similar
argument as in Case 1, we have that either $S\cong\mathrm{L}_{2}(q^{m\over
2})$ with $mf$ a power of $2$, or $(S,m,q)$ is one of $(\mathrm{A}_{5},4,2)$,
$(\mathrm{A}_{7},6,2)$, $(\mathrm{L}_{3}(2),6,2)$, $(\mathrm{A}_{7},6,2)$ and
$(\mathrm{A}_{13},12,2)$. Assume that $\overline{Y}$ is solvable. It follows
from [27, Theorem 1.1] that $(m,q)=(4,2)$ or $(6,2)$, and
$N\cong\mathrm{S}_{6}$ or $\mathrm{Sp}_{6}(2){.}2$, respectively. Noting that
$\overline{Y}=(Y\cap B_{0}^{\prime})N/N\cong(Y\cap B_{0}^{\prime})/(Y\cap N)$,
we have $S\preccurlyeq Y\cap N$. Since $\pi(Y\cap N)\subseteq\pi(S)=\pi(Y)$,
checking the subgroups of $N$, we have that $(S,m,q)$ is one of
$(\mathrm{A}_{5},4,2)$, $(\mathrm{A}_{7},6,2)$ and $(\mathrm{L}_{3}(2),6,2)$.
Then our lemma holds in this case.

Case 3. Let $B_{0}\cong\mathrm{GO}^{+}_{2m}(2)$, $m\geq 4$ and $q=2$. Then
$A_{0}\cap B_{0}=\mathrm{Sp}_{2m-2}(2)\times 2$ is maximal in both $A_{0}$ and
$B_{0}$, see [31, pp. 52, 3.2.4(e)]. Noting that $B_{0}=(A_{0}\cap
B_{0})Y_{0}$ is a core-free factorization, by Lemma 2.7, choose a core-free
maximal factorization $G_{1}=A_{1}B_{1}$ and a core-free factorization
$G_{1}=A_{1}Y_{1}$ such that $\mathbf{soc}(G_{1})\cong\Omega^{+}_{2m}(2)$,
$A_{1}\gtrsim\mathrm{Sp}_{2m-2}(2)$ and $S\preccurlyeq Y_{1}\leq B_{1}$. Then
$S$ is described as in Lemma 7.6. Recalling $|G_{\alpha}|$ is divisible by
$2^{m-1}(2^{m}-1)$, except for $S\cong\mathrm{L}_{3}(2)$ and $m=6$, all the
triples in Lemma 7.6 (2) are easily excluded. Thus
$S\cong\mathrm{L}_{2}(2^{m\over 2})$, or $m=6$ and $S\cong\mathrm{L}_{3}(2)$
or $\mathrm{A}_{7}$, desired as part (1) or (2) of this lemma.

Case 4. Let $B_{0}\cong\mathrm{Sp}_{2a}(q^{b}){.}b$, where $b$ is a prime and
$m=ab$. In this case, $A_{0}\cap B_{0}\cong\mathrm{GO}^{-}_{2a}(q^{b}){.}b$,
which is maximal in $A_{0}$ (if $a\neq 1$) and $B_{0}$. Noting that
$B_{0}=(A_{0}\cap B_{0})Y$, if $a\geq 3$ then part (1) of this lemma occurs by
an induction on $m$.

Assume that $a=1$. Then $m$ is an odd prime and
$B_{0}\cong\mathrm{L}_{2}(q^{m}){.}m$. Checking the subgroups of
$\mathrm{L}_{2}(q^{m})$ with order divisible by $q^{m}-1$, we conclude that
$S\cong\mathrm{L}_{2}(q^{m})$. Then $r=q^{m}+1$, which is not a prime, a
contradiction. Thus $a=2$, ${m\over 2}$ is a prime and
$B_{0}\cong\mathrm{Sp}_{4}(q^{m\over 2}){.}{m\over 2}$. By Lemmas 2.7 and 2.6,
since $B_{0}=(A_{0}\cap B_{0})Y$, choose a core-free maximal factorization
$G_{1}=A_{1}B_{1}$ and a core-free factorization $G_{1}=A_{1}Y_{1}$ such that
$\mathbf{soc}(G_{1})\cong\mathrm{Sp}_{4}(q^{m\over 2})$,
$A_{1}\gtrsim\Omega^{-}_{4}(q^{m\over 2})$ and $S\preccurlyeq Y_{1}\leq
B_{1}$. By [31, Theorem A], we conclude that
$B_{1}^{\infty}\mathbf{rad}(B_{1})/\mathbf{rad}(B_{1})\cong\mathrm{L}_{2}(q^{m})$,
$\mathrm{L}_{2}(q^{m\over 2})$ or $\mathrm{L}_{2}(q^{m\over
2})\times\mathrm{L}_{2}(q^{m\over 2})$. It follows that
$S\cong\mathrm{L}_{2}(q^{m})$ or $\mathrm{L}_{2}(q^{m\over 2})$. This implies
that $r=q^{m}+1$ or $q^{m\over 2}+1$, and so $r$ is not a prime, a
contradiction. ∎

##  8\. The classical case

In this section, we will work under Hypothesis 6.1. Let $A$ and $B$ be maximal
subgroups of $G$ such that one of them contains $T$ and the other one contains
$G_{\alpha}$. Then $G=AB$, and both $A$ and $B$ are insolvable. Without loss
of generality, we let $A$ and $B$ be described as in the third and the fourth
columns of Tables 1-4 in [31], respectively. For convenience, we define $X$
and $Y$ as follows:

$\\{X,Y\\}=\\{T,G_{\alpha}\\},\,X\leq A\mbox{ and }Y\leq B.$

Recall that $S=\mathbf{soc}(G_{\alpha}^{\Gamma(\alpha)})$.

###  8.1. Unitary case

Let $G=\mathrm{U}_{2m}(q)$, where $m\geq 2$ and $q=p^{f}$ for some prime $p$.
Then $A$ and $B$ are given in [31, Tables 1 and 3].

######  Lemma 8.1.

$A$ and $B$ are described as in [31, Table 1].

###### Proof.

Suppose that $A$, $B$ are given as in [31, Table 3], and we list $A$ and $B$
as follows:

$\tiny\begin{array}[]{l|l|l|l|l}(m,q)&A&|G:A|&B&|G:B|\\\
\hline\cr(2,3)&\mathrm{L}_{3}(4)&2\cdot 3^{4}&\mathrm{PSp}_{4}(3)&2\cdot
3^{2}\cdot 7\\\ \hline\cr(3,2)&\mathrm{U}_{5}(2)&2^{5}\cdot 3\cdot
7&\mathrm{U}_{4}(3).2&2^{7}\cdot 11\\\ &\mathrm{U}_{5}(2)&2^{5}\cdot 3\cdot
7&\mathrm{M}_{22}&2^{8}\cdot 3^{4}\\\
\hline\cr(6,2)&\mathrm{Suz}&\Phi_{22}^{*}(2)\Phi_{18}^{*}(2)\mid|G:A|&\mathrm{U}_{11}(2)&2^{11}\cdot
3\cdot 5\cdot 7\cdot 13\end{array}$

Inspect the insolvable groups contained in $A$ and $B$. For $(m,q)=(2,3)$,
combining Lemma 2.4, we conclude that $\\{T,G_{\alpha}\\}=\\{A,B\\}$; however,
neither $\mathrm{L}_{3}(4)$ nor $\mathrm{PSp}_{4}(3)$ has a subgroup of prime
index. Similarly, the pair $(3,2)$ is excluded for $(m,q)$. Then only
$(m,q)=(6,2)$ is left. By [32, Table 10.1], $\mathrm{U}_{11}(2)$ has no proper
subgroup with order divisible by $\Phi_{22}^{*}(2)\Phi_{18}^{*}(2)$. By the
Atlas [5], $\mathrm{Suz}$ has no maximal subgroup with order divisible by
$2^{11}\cdot 3\cdot 5\cdot 7\cdot 13$. It follows that
$\\{T,G_{\alpha}\\}=\\{A,B\\}$. By Lemma 4.2, none of $\mathrm{U}_{11}(2)$ and
$\mathrm{Suz}$ is a possible vertex-stabilizer of some symmetric graph of
prime degree, a contradiction. This completes the proof. ∎

In the following we let $A=\mathsf{N}_{1}$. Then $B$ and $|G:B|$ are listed as
follows:

$\tiny\begin{array}[]{l|l|l|l}B&\mathsf{P}_{m}&\mathrm{PSp}_{2m}(q){.}{(2,q-1)(m,q+1)\over(2m,q+1)}&{}^{\bf\hat{}}\mathrm{SL}_{m}(4).2,q=2\\\
\hline\cr|G:B|&\prod_{i=1}^{m}(q^{2i-1}+1)&{1\over(m,q+1)}q^{m(m-1)}\prod_{i=2}^{m}(q^{2i-1}+1)&{1\over
2}q^{m^{2}}\prod_{i=1}^{m}(q^{2i-1}+1)\\\ \end{array}$

By Lemma 2.4, $|X|$ is divisible by $|G:B|$. Since $A$ is insolvable,
$(m,q)\neq(2,2)$.

######  Lemma 8.2.

If $(m,q)=(3,2)$ then $T\cong\mathrm{U}_{5}(2)$, $\mathrm{SL}_{3}(2)\unlhd
G_{\alpha}^{\Gamma(\alpha)}$ and $\mathbf{O}_{2}(G_{\alpha})\neq 1$.

###### Proof.

Let $(m,q)=(3,2)$. Then $A\cong\mathrm{U}_{5}(2)$, $|G:A|={2^{5}\cdot 3\cdot
7}$, and $B$ is listed as follows:

$\tiny\begin{array}[]{l|l|l|l}B&2^{9}{:}\mathrm{L}_{3}(4)&\mathrm{PSp}_{6}(2)&\mathrm{L}_{3}(4).2\\\
\hline\cr|G:B|&3^{4}\cdot 11&2^{6}\cdot 3^{2}\cdot 11&2^{8}\cdot 3^{4}\cdot
11\end{array}$

Searching the subgroups of $\mathrm{U}_{5}(2)$ with order divisible by
$|G:B|$, we get $X=A\cong\mathrm{U}_{5}(2)$. Then $T=X\leq A$, and so
$G_{\alpha}=Y\leq B$. Thus $|G_{\alpha}|$ is divisible $|G:A|$. For
$B=\mathrm{PSp}_{6}(2)$, by calculation using GAP, we have $G_{\alpha}\cong
2^{3}{:}\mathrm{SL}_{3}(2)$ or $2^{6}{:}\mathrm{L}_{3}(2)$.

Assume that $B\cong\mathrm{L}_{3}(4).2$ or $2^{9}{:}\mathrm{L}_{3}(4)$. Let
$\overline{B}=B^{\prime}/\mathbf{rad}(B^{\prime})$ and $\overline{Y}=(Y\cap
B^{\prime})\mathbf{rad}(B^{\prime})/\mathbf{rad}(B^{\prime})$. Then
$\overline{B}\cong\mathrm{L}_{3}(4)$, $S\preccurlyeq\overline{Y}$, and
$|\overline{Y}|$ has a divisor $21$. Checking the subgroups of
$\mathrm{L}_{3}(4)$, we get $S\cong\overline{Y}\cong\mathrm{L}_{3}(2)$. By
Lemma 4.2, $|G_{\alpha}|$ is a divisor of
$6|\mathbf{O}_{2}(G_{\alpha})||\mathrm{GL}_{3}(2)|$. Then
$2^{5}\leq|G_{\alpha}|_{2}\leq 2^{4}|\mathbf{O}_{2}(G_{\alpha})|$, yielding
$\mathbf{O}_{2}(G_{\alpha})\neq 1$, and the lemma follows. ∎

######  Lemma 8.3.

$A\geq T\cong\mathrm{U}_{2m-1}(q)$.

###### Proof.

By Lemma 8.2, our lemma holds if $(m,q)=(3,2)$. Thus assume $(m,q)\neq(3,2)$.
Let $\overline{A}=A/\mathbf{rad}(A)$ and
$\overline{X}=X\mathbf{rad}(A)/\mathbf{rad}(A)$. Then
$\overline{A}\cong\mathrm{PGU}_{2m-1}(q)$, and $|\overline{X}|$ is divisible
by $\Phi_{2(2m-1)}^{*}(q)$. By Lemma 2.3, $\Phi_{2(2m-1)}^{*}(q)\neq 1$ as
$(2(2m-1),q)\neq(6,2)$ and $2(2m-1)\geq 6$.

Assume that $m\geq 3$. Then $\Phi_{2(2m-3)}^{*}(q)\neq 1$, and it is easily
shown that $\overline{X}\cap\mathbf{soc}(\overline{A})$ has order divisible by
$\Phi_{2(2m-1)}^{*}(q)\Phi_{2(2m-3)}^{*}(q)$. By [32, Theorem 4 and Table
10.1], we conclude that
$\overline{X}\cap\mathbf{soc}(\overline{A})\cong\mathrm{U}_{2m-1}(q)$. This
implies that $T=X\cong\mathrm{U}_{2m-1}(q)$.

Let $m=2$. Then $A=[{q+1\over(4,q+1)}]{.}\mathrm{PGU}_{3}(q)$, and
$|G:A|={q^{3}(q-1)(q^{2}+1)}$. Suppose that
$\overline{X}\cap\mathbf{soc}(\overline{A})\not\cong\mathrm{U}_{3}(q)$. Let
$M$ be a maximal subgroup of $\mathbf{soc}(\overline{A})$ with
$\overline{X}\cap\mathbf{soc}(\overline{A})\leq M$. Noting that
$\Phi_{6}^{*}(q)\geq 7$, $|M|$ is divisible by $\Phi_{6}^{*}(q)$ and $M$ is
insolvable, by [3, Tables 8.5 and 8.6], $q=p$ and either
$M\cong\mathrm{L}_{3}(2)$ with $p\neq 5$ or $M\cong\mathrm{A}_{7}$ with $p=5$.
In both cases, $\Phi_{6}^{*}(q)=7$, and so $q\in\\{3,5\\}$ by [20, Theorem
3.9]. Assume that $q=3$. Then $A\cong\mathrm{U}_{3}(3)$,
$X\cong\mathrm{L}_{3}(2)$, and $|G:X|=2^{4}\cdot 3^{5}\cdot 5$. Moreover,
$B\cong 3^{4}{:}\mathrm{A}_{6}$ or $\mathrm{PSp}_{4}(3)$. Since $G=XY$, by
Lemma 2.4, $|Y|$ is divisible by $2^{4}\cdot 3^{5}\cdot 5$. It follows that
$B\cong 3^{4}{:}\mathrm{A}_{6}$ and $Y\cong 3^{4}{:}\mathrm{A}_{5}$. This
forces $G_{\alpha}=Y\cong 3^{4}{:}\mathrm{A}_{5}$, which is impossible by
Lemma 4.2. Assume that $q=5$. Then $A\cong 3{.}\mathrm{U}_{3}(5){.}3$,
$X\lesssim 3{.}\mathrm{A}_{7}{.}3$, and $|G:X|$ is divisible by $2^{4}\cdot
5^{5}\cdot 13$. In this case, $|Y|$ is divisible by $2^{4}\cdot 5^{5}\cdot
13$. It follows that $B\cong 5^{4}{:}\mathrm{L}_{2}(25){:}[4]$, and then
$Y\cong 5^{4}{:}\mathrm{L}_{2}(25){:}[c]$, where $c$ is a divisor of $4$. This
implies that $G_{\alpha}=Y$, and so $r=25+1$, which contradicts that $r$ is a
prime. Therefore,
$\overline{X}\cap\mathbf{soc}(\overline{A})\cong\mathrm{U}_{3}(q)$. Then
$T=X\cong\mathrm{U}_{3}(q)$, and the lemma holds. ∎

######  Theorem 8.1.

Assume that Hypothesis 6.1 holds, and let $G=\mathrm{U}_{2m}(q)$, where $m\geq
2$, $(m,q)\neq(2,2)$ and $q=p^{f}$ for some prime $p$. Then
$T\cong\mathrm{U}_{2m-1}(q)$, $p=2$, $\mathbf{O}_{2}(G_{\alpha})\neq 1$ and
one of the following holds.

  * (1)

$(m,q)=(3,2)$ and $\mathrm{SL}_{3}(2)\unlhd G_{\alpha}^{\Gamma(\alpha)}$.

  * (2)

$mf$ is a power of $2$ and $\mathrm{SL}_{2}(q^{m})\unlhd
G_{\alpha}^{\Gamma(\alpha)}$.

###### Proof.

By Lemmas 8.1 and 8.3, $\mathrm{U}_{2m-1}\cong T\leq A=\mathsf{N}_{1}$, and so
$G_{\alpha}=Y\leq B$. Thus $S=\mathbf{soc}(G_{\alpha}^{\Gamma(\alpha)})$ is
described as in Lemma 7.3. If $(m,q)=(3,2)$ then part (1) holds by Lemma 8.2.
Next let $m>3$ if $q=2$. Note that
$|G:T|={(2m-1,q+1)\over(2m,q+1)}q^{2m-1}(q^{2m}-1)$.

Since $|G_{\alpha}|$ is divisible by $|G:T|$, we have
$|G_{\alpha}|_{p}\geq|G:T|_{p}$. Suppose that $S$ is given in Table 7.10. Then
$(S,2m,q)$ is one of $(\mathrm{A}_{5},4,3)$, $(\mathrm{A}_{7},4,7)$,
$(\mathrm{A}_{7},6,3)$, $(\mathrm{A}_{7},6,5)$, $(\mathrm{A}_{11},10,2)$,
$(\mathrm{A}_{13},12,2)$, $(\mathrm{A}_{19},18,2)$, $(\mathrm{M}_{11},10,2)$,
$(\mathrm{L}_{3}(2),6,3)$, $(\mathrm{L}_{3}(2),6,5)$,
$(\mathrm{L}_{2}(11),10,2)$ and $(\mathrm{L}_{3}(3),12,2)$. For these triples,
by Lemma 4.2, calculation shows that $|G_{\alpha}|_{p}<|G:T|_{p}$, a
contradiction. Therefore, $S\cong\mathrm{L}_{2}(q^{m})$, $r=q^{m}+1$, $p=2$
and $mf$ is a power of $2$. Suppose that $\mathbf{O}_{p}(G_{\alpha})=1$. Then,
by Lemma 4.2, $|G_{\alpha}|_{2}$ is a divisor of $mfq^{m}$. Then
$q^{2m-1}=|G:T|_{p}\leq|G_{\alpha}|_{2}|\leq mfq^{m}$, yielding $(m,q)=(2,2)$,
a contradiction. Thus $\mathbf{O}_{p}(G_{\alpha})\neq 1$, and then our result
follows. ∎

###  8.2. Orthogonal case I

Let $G=\Omega_{2m+1}(q)$, where $m\geq 3$ and $q=p^{f}$ for some odd prime
$p$. Then $A$ and $B$ are given in [31, Tables 1-3]. Clearly,
$\Phi_{2m}^{*}(q)\neq 1\neq\Phi_{2m-2}^{*}(q)$.

######  Lemma 8.4.

$A$ and $B$ are described as in [31, Table 1].

###### Proof.

Suppose that the lemma is false. We deduce the contradiction in two cases.

Case 1. Let $A$ and $B$ be as in [31, Table 2]. We have the following three
cases.

1.1. $p=3$, $m=12$, $A\cong\mathrm{F}_{4}(q)$ and
$B\cong\Omega_{24}^{-}(q){.}2$. In this case,
$|G:A|={q^{120}(q^{4}-1)(q^{10}-1)\over 2}\prod_{i=7}^{m}(q^{2i}-1)$, and
$|G:B|={1\over 2}q^{12}(q^{12}-1)$. Clearly, $\Phi_{2m-4}^{*}(q)\neq 1$.

Recall that $X\leq A$ and $Y\leq B$. Then $|Y|$ is divisible by
$\Phi_{2m}^{*}(q)\Phi_{2m-2}^{*}(q)\Phi_{2m-4}^{*}(q)$. By [32, Theorem 4 and
Table 10.1], since $q$ is odd, $\mathrm{P\Omega}^{-}_{2m}(q)$ has no proper
subgroup with order divisible by
$\Phi_{2m}^{*}(q)\Phi_{2m-2}^{*}(q)\Phi_{2m-4}^{*}(q)$. It follows that
$Y\succcurlyeq\mathrm{P\Omega}^{-}_{2m}(q)$. Then $Y\neq G_{\alpha}$ by Lemma
4.2. Therefore, $T=Y$ and $G_{\alpha}=X\leq A$. Noting that $G=BG_{\alpha}$,
we have $A=(A\cap B)G_{\alpha}$. By Lemma 4.2, $G_{\alpha}\neq A$. Then we
have a factorization of $\mathrm{F}_{4}(q)$, which is impossible by [21,
Theorem 1] as $q$ is odd here.

1.2. $p=3$, $m=6$, $A\cong\mathrm{PSp}_{6}(q).a$ and
$B\cong\Omega_{12}^{-}(q){.}2$, where $a\leq 2$. In this case, $|G:A|={1\over
2a}q^{27}\prod_{i=4}^{m}(q^{2i}-1)$, and $|G:B|={1\over 2}q^{6}(q^{6}-1)$.

Similarly as in 1.1, we have $T\leq B$, $G_{\alpha}\leq A$, and so
$|G_{\alpha}|$ is divisible $q^{6}(q^{6}-1)$. By Lemma 4.2,
$S\not\cong\mathrm{PSp}_{6}(q)$. It is easily shown that $S\preccurlyeq
G_{\alpha}\cap A^{\prime}$, and $|G_{\alpha}\cap\mathrm{A}^{\prime}|$ is
divisible by $q^{6}\Phi_{6}^{*}(q)$, where $A^{\prime}$ is the derived
subgroup of $A$. Clearly, $\Phi_{6}^{*}(q)\neq 1$. Noting that $q$ is odd, it
follows from Lemma 7.1 that $q\in\\{3,5\\}$ and $S\cong\mathrm{A}_{7}$ or
$\mathrm{L}_{3}(2)$. By Lemma 4.2, $|G_{\alpha}|$ is indivisible by $3^{3}$;
however, here, $|G_{\alpha}|$ is divisible by $q^{6}$, a contradiction.

1.3. $m=3$, $A\cong\mathrm{G}_{2}(q)$ and $B$ is one of $[q^{5}]{:}({q-1\over
2}\times\Omega_{5}(q)).2$, $\Omega_{6}^{+}(q){.}2$, $\Omega_{6}^{-}(q){.}2$,
$({q-1\over 2}\times\Omega_{5}(q)).2^{2}$ (with $q>3$) and $({q+1\over
2}\times\Omega_{5}(q)).2^{2}$.

Suppose that $X=G_{\alpha}$. Then $G=G_{\alpha}B$, $A=G_{\alpha}(A\cap B)$,
and $|G_{\alpha}|$ is divisible $|G:B|$; in particular, $|G_{\alpha}|$ is
divisible by ${1\over 2}q^{3}(q^{3}\pm 1)$ or ${q^{6}-1\over 2(q\pm 1)}$.
Assume that $|G_{\alpha}|$ is divisible by ${1\over 2}q^{3}(q^{3}\pm 1)$. By
[21, Theorem 1], combining Lemma 4.2, we have $\mathrm{SL}_{3}(q)\leq
G_{\alpha}\leq\mathrm{SL}_{3}(q).2$ and $p=3$. It follows that
$r={q^{3}-1\over q-1}$, and so $f$ is a power of $3$ by Lemma 2.1. In
particular, $(4,q+1)=4$. Moreover, in this case, $|T|$ is divisible by
$|G:G_{\alpha}|={1\over 2a}q^{6}(q^{4}-1)(q^{3}+1)$, where $a\leq 2$. Then
$T\leq B=\Omega_{6}^{-}(q){.}2$. By [32, Tables 10.3 and 10.4],
$\mathrm{P\Omega}_{6}^{-}(q)$ has no proper subgroup of order divisible by
$q^{6}(q^{4}-1)(q^{3}+1)\over q-1$. It follows that
$T=\mathrm{P\Omega}_{6}^{-}(q)$. This forces that $B$ has trivial center, and
then $(4,q+1)\leq 2$, a contradiction. Assume that $|G_{\alpha}|$ is divisible
by ${q^{6}-1\over 2(q\pm 1)}$. Then $|G_{\alpha}|$ has a divisor
$\Phi_{6}^{*}(q)\Phi_{3}^{*}(q)$. By [32, Table 10.5], we have
$G_{\alpha}\lesssim\mathrm{L}_{2}(13)$, forcing
$G_{\alpha}\cong\mathrm{L}_{2}(13)$ and $r=13+1$, which contradicts that $r$
is a prime.

Now $G_{\alpha}=Y\leq B$ and $T=X\leq A$. Then $G=AG_{\alpha}$, yielding
$|G_{\alpha}:(A\cap G_{\alpha})|=|G:A|={1\over 2}q^{3}(q^{4}-1)$. Assume that
$B\succcurlyeq\mathrm{PSp}_{4}(q)$ or $\mathrm{L}_{4}(q)$. Let
$\overline{B}=B^{\prime}\mathbf{rad}(B)/\mathbf{rad}(B)$. Then
$\overline{B}\cong\mathrm{PSp}_{4}(q)$ or $\mathrm{L}_{4}(q)$. Note that
$\Phi^{*}_{4}(q)\neq 1$. Considering the preimage of $(G_{\alpha}\cap
B^{\prime})\mathbf{rad}(B)/\mathbf{rad}(B)$ in $\mathrm{GL}_{4}(q)$, by Lemma
7.1, we conclude that $q=3$ or $7$, and $S\cong\mathrm{A}_{5}$ or
$\mathrm{A}_{7}$, respectively. By Lemma 4.2, $|G_{\alpha}|$ is indivisible by
$3^{3}$, which is impossible as $|G_{\alpha}|$ has a divisor $q^{3}$.

The left case is that $G_{\alpha}\leq B\cong\Omega_{6}^{-}(q){.}2$ and $T\leq
A\cong\mathrm{G}_{2}(q)$. In particular, $|G_{\alpha}:(A\cap
G_{\alpha})|=|G:A|={1\over 2}q^{3}(q^{4}-1)$ and $|T:(B\cap T)|=|G:B|={1\over
2}q^{3}(q^{3}-1)$. It is easily shown that $S\preccurlyeq G_{\alpha}\cap
B^{\prime}$. Since $|B:B^{\prime}|=2$, either $G_{\alpha}\leq B^{\prime}$ or
$B=G_{\alpha}B^{\prime}$, yielding $|G:(G_{\alpha}\cap B^{\prime})|=1$ or $2$,
respectively. It follows that $|G_{\alpha}\cap B^{\prime}|$ is divisible by
$q^{3}\Phi_{4}^{*}(q)$. Take a maximal subgroup $M$ of $B^{\prime}$ with
$G_{\alpha}\cap B^{\prime}\leq M$. Then $|M|$ is divisible by
$q^{3}\Phi_{4}^{*}(q)$. Checking the maximal subgroups of $\Omega_{6}^{-}(q)$,
refer to [3, Tables 8.33 and 8.34], we conclude that
$M\cong\Omega_{5}(q){.}2\cong\mathrm{PSp}_{4}(q){.}2$. This gives a similar
contradiction as in the above paragraph.

Case 2. Let $A$ and $B$ be as in [31, Table 3]. Then $G=\Omega_{7}(3)$ and, up
to isomorphism, $A$ and $B$ are listed as follow:

$\tiny\begin{array}[]{l|l|l|l|l|l|l|l}\hline\cr
A&\mathrm{G}_{2}(3)&\mathrm{G}_{2}(3)&\mathrm{S}_{9}&\mathrm{S}_{9}&\mathrm{Sp}_{6}(2)&\mathrm{Sp}_{6}(2)&2^{6}{.}\mathrm{A}_{7}\\\
\hline\cr|G:A|&2^{3}\cdot 3^{3}\cdot 5&2^{3}\cdot 3^{3}\cdot 5&2^{2}\cdot
3^{5}\cdot 13&2^{2}\cdot 3^{5}\cdot 13&3^{5}\cdot 13&3^{5}\cdot 13&3^{7}\cdot
13\\\ \hline\cr\hline\cr
B&\mathrm{Sp}_{6}(2)&\mathrm{S}_{9}&\mathrm{L}_{4}(3).2&[3^{6}]{:}\mathrm{L}_{3}(3)&\mathrm{L}_{4}(3).2&[3^{6}]{:}\mathrm{L}_{3}(3)&[3^{6}]{:}\mathrm{L}_{3}(3)\\\
\hline\cr|G:B|&3^{5}\cdot 13&2^{2}\cdot 3^{5}\cdot 13&2\cdot 3^{3}\cdot
7&2^{5}\cdot 5\cdot 7&2\cdot 3^{3}\cdot 7&2^{5}\cdot 5\cdot 7&2^{5}\cdot
5\cdot 7\\\ \hline\cr\end{array}$

Note that $|X|$ and $|Y|$ are divisible by $|G:B|$ and $|G:A|$, respectively.
Inspecting simple subgroups contained in $A$ and $B$, since neither $X$ nor
$Y$ is an alternating group, we conclude that either
$T=A\cong\mathrm{G}_{2}(3)$ or $\mathrm{U}_{4}(2)\cong T\leq
B\cong\mathrm{Sp}_{6}(2)$. Suppose that $T\cong\mathrm{G}_{2}(3)$. Then
$G_{\alpha}\leq B\cong\mathrm{Sp}_{6}(2)$ or $\mathrm{S}_{9}$ and, since
$G=TG_{\alpha}$, we know that $|G_{\alpha}|$ is divisible by $2^{3}\cdot
3^{3}\cdot 5$. By Lemma 4.2, $G_{\alpha}\neq B$. Let $M$ be a maximal subgroup
of $B$ with $G_{\alpha}\leq M$. Then, by the Atlas [5],
$M\cong\mathrm{U}_{4}(2).2$, $\mathrm{S}_{3}\times\mathrm{S}_{6}$ or
$(3\times\mathrm{A}_{6}){:}2$. However, such an $M$ does not contain an
insolvable subgroup satisfying Lemma 4.2 and having order divisible by
$2^{3}\cdot 3^{3}\cdot 5$. Thus $\mathrm{U}_{4}(2)\cong T\leq
B\cong\mathrm{Sp}_{6}(2)$, and $G_{\alpha}\leq A\cong\mathrm{G}_{2}(3)$. Then
$|G:T|=2^{3}\cdot 3^{5}\cdot 7\cdot 13$. Noting that $|G_{\alpha}|$ is
divisible by $|G:T|$, it follows that $G_{\alpha}=A\cong\mathrm{G}_{2}(3)$,
which is impossible by Lemma 4.2. ∎

######  Theorem 8.2.

Assume that Hypothesis 6.1 holds and $G=\Omega_{2m+1}(q)$, where $m\geq 3$ and
$q=p^{f}$ for some odd prime $p$. Then $mf$ is a power of some odd prime $d$,
$T\cong\mathrm{P\Omega}^{-}_{2m}(q)$,
$\mathbf{soc}(G_{\alpha}^{\Gamma(\alpha)})\cong\mathrm{L}_{d}(q^{m\over d})$,
$p\equiv 1\,({\mathrm{mod}~{}}4)$ and $r={q^{m}-1\over q^{m\over d}-1}$.

###### Proof.

By Lemma 8.4, we let $A$ and $B$ be as in [31, Table 1]. Then
$A\cong\Omega^{-}_{2m}(q).2$, $B\cong[q^{m(m+1)\over 2}]{:}{1\over
2}\mathrm{GL}_{m}(q)$ and $|G:B|=\prod_{i=1}^{m}(q^{i}+1)$. In particular,
$|G:B|$ and hence $|X|$ is divisible by
$\Phi_{2m}^{*}(q)\Phi_{2m-2}^{*}(q)\Phi_{2m-4}^{*}(q)$.

Case 1. Assume that $\Phi_{2m-4}^{*}(q)=1$. Then $m=3$ and $q=p=2^{s}-1$ for
some prime $s\geq 2$. In this case, we have $A\cong\Omega_{6}^{-}(q){.}2\cong
2{.}\mathrm{U}_{4}(q){.}2$.

Suppose that $T=X\leq A$. Then $T\lesssim\Omega_{6}^{-}(q)$. Noting that
$(4,q+1)=4$, we have $T\not\cong\mathrm{U}_{4}(q)$. Recall that $|X|$ is
divisible by $(q+1)(q^{2}+1)(q^{3}+1)$, and hence $|T|$ is divisible by
$\Phi_{2m}^{*}(q)\Phi_{2m-2}^{*}(q)$. If $q\geq 5$ and $|T|$ is divisible by
$p$ then $T\cong\mathrm{U}_{4}(q)$ by [32, Table 10.3], which is impossible.
If $q=3$ then $|T|$ is divisible by $35$, and so $T\cong\mathrm{U}_{4}(3)$ by
[32, Table 10.4], a contradiction. Thus we have $q\geq 5$ and $|T|$ is
indivisible by $p$. Note that $|T|$ is divisible by
$\Phi_{6f}^{*}(p)\Phi_{4}^{*}(q)$. Considering the preimage of $T$ in
$\mathrm{GL}_{6}(q)$, by [2, Theorem 3.1], we conclude that
$T\leq\mathrm{GU}_{3}(q)$. Then $T\leq\mathrm{U}_{3}(q)$, and thus $|T|$ is
indivisible by $(q+1)(q^{2}+1)(q^{3}+1)$, a contradiction.

Now $T\leq B\cong[q^{6}]{:}{1\over 2}\mathrm{GL}_{3}(q)$ and $G_{\alpha}\leq
A\cong\Omega_{6}^{-}(q){.}2$. Then $G=G_{\alpha}B$, and $|G_{\alpha}:(B\cap
G_{\alpha})|=|G:B|=\prod_{i=1}^{3}(q^{i}+1)$. Thus $|G_{\alpha}|$ is divisible
$\Phi^{*}_{6}(q)$. By Lemma 7.1, we conclude that $q=3$ and
$S\cong\mathrm{A}_{7}$ or $\mathrm{L}_{3}(2)$. By Lemma 4.2, calculation shows
that $|G_{\alpha}|_{3}\leq 3^{3}$. Then
$3^{9}=|G|_{3}\leq|T|_{3}|G_{\alpha}|_{3}\leq 3^{6}$, a contradiction.

Case 2. Assume that $\Phi_{2m-4}^{*}(q)\neq 1$. By [32, Theorem 4 and Table
10.1], since $q$ is odd, $\mathrm{P\Omega}^{-}_{2m}(q)$ has no proper subgroup
with order divisible by
$\Phi_{2m}^{*}(q)\Phi_{2m-2}^{*}(q)\Phi_{2m-4}^{*}(q)$. It follows that
$X\succcurlyeq\mathrm{P\Omega}^{-}_{2m}(q)$. Therefore,
$T=X\cong\mathrm{P\Omega}^{-}_{2m}(q)$, and $G_{\alpha}=Y\leq B$. This implies
that $(4,q^{m}+1)=2$ and $|G:T|=q^{m}(q^{m}-1)$. In particular,
$q^{m}(q^{m}-1)$ is a divisor of $|G_{\alpha}|$, and so $|G_{\alpha}|$ is
divisible by $\Phi_{mf}^{*}(p)$. Clearly, $\Phi_{mf}^{*}(p)\neq 1$.

Let $\overline{G_{\alpha}}=G_{\alpha}\mathbf{O}_{p}(B)/\mathbf{O}_{p}(B)$.
Then $\overline{G_{\alpha}}\leq{1\over 2}\mathrm{GL}_{m}(q)$, and
$|\overline{G_{\alpha}}|$ is divisible by $q^{m}-1$. Note that
$S\preccurlyeq\overline{G_{\alpha}}$. By Lemma 7.1, either
$S\cong\mathrm{L}_{d}(q^{m\over d})$ for some divisor $d$ of $m$, or $(S,m,q)$
is one of $(\mathrm{A}_{5},4,3)$, $(\mathrm{A}_{7},6,3)$,
$(\mathrm{A}_{7},6,5)$, $(\mathrm{A}_{7},4,7)$, $(\mathrm{A}_{7},3,25)$,
$(\mathrm{L}_{3}(2),6,3)$, $(\mathrm{L}_{3}(2),6,5)$,
$(\mathrm{L}_{3}(2),3,9)$ and $(\mathrm{L}_{3}(2),3,25)$. Since $|G_{\alpha}|$
is divisible by $q^{m}$, the latter case is easily excluded by calculating
$|G_{\alpha}|_{p}$. Assume that the former case occurs. If $(d,q^{m\over
d},r)=(2,11,11)$ then we have $|G_{\alpha}|_{11}=11$ by Lemma 4.2, which
contradicts that $|G_{\alpha}|$ is divisible by $q^{m}$. Thus we have
$r={q^{m}-1\over q^{m\over d}-1}$, and so $d$ is an odd prime and $mf$ is a
power of $d$, see Lemma 2.1. Recalling $(4,q^{m}+1)=2$, we have $p\equiv
1\,({\mathrm{mod}~{}}4)$. ∎

###  8.3. Orthogonal case II

Let $G=\mathrm{P\Omega}^{-}_{2m}(q)$, where $m\geq 4$ and $q=p^{f}$ for some
prime $p$. Then, by [31, Theorem A], $m$ is odd and one of the following
holds:

  * (1)

$m=5$, $q=2$, $A\cong\mathrm{A}_{12}$ and $B\cong 2^{8}{:}\Omega^{-}_{8}(2)$;

  * (2)

$A\cong[q^{2m-2}]{:}[{q-1\over a}]{.}\mathrm{P\Omega}^{-}_{2m-2}(q){.}a$ and
$B\cong{q+1\over(4,q+1)}{.}\mathrm{PGU}_{m}(q)$, where $a=2$ if $q\equiv
1\,({\mathrm{mod}~{}}4)$, or $a=1$ otherwise;

  * (3)