# 2-Arc-transitive Cayley graphs on alternating groups

Jiangmin Pan  J. M. Pan
School of Statistics and Mathematics
Yunnan University of Finance and Economics
Kunming
P. R. China<EMAIL_ADDRESS>,  Binzhou Xia  B. Z. Xia
School of Mathematics and Statistics
The University of Melbourne
Parkville, VIC 3010
Australia<EMAIL_ADDRESS>and  Fugang Yin  F. G. Yin
Department of Mathematics
Beijing Jiaotong University
Beijing 100044, P. R. China<EMAIL_ADDRESS>
###### Abstract.

An interesting fact is that most of the known connected $2$-arc-transitive
nonnormal Cayley graphs of small valency on finite simple groups are
$(\mathrm{A}_{n+1},2)$-arc-transitive Cayley graphs on $\mathrm{A}_{n}$. This
motivates the study of $2$-arc-transitive Cayley graphs on $\mathrm{A}_{n}$
for arbitrary valency. In this paper, we characterize the automorphism groups
of such graphs. In particular, we show that for a non-complete $(G,2)$-arc-
transitive Cayley graph on $\mathrm{A}_{n}$ with $G$ almost simple, the socle
of $G$ is either $\mathrm{A}_{n+1}$ or $\mathrm{A}_{n+2}$. We also construct
the first infinite family of $(\mathrm{A}_{n+2},2)$-arc-transitive Cayley
graphs on $\mathrm{A}_{n}$.

Key words: 2-arc-transitive; Cayley graph; alternating group; automorphism
group

MSC2020: 05E18, 05C25

##  1\. Introduction

All graphs considered in this paper are finite, simple and undirected. For a
positive integer $s$, an _$s$ -arc_ of a graph is a $(s+1)$-tuple of vertices
$(v_{0},v_{1},...,v_{s})$ where $v_{i}$ is adjacent to $v_{i+1}$ for $0\leq
i\leq s-1$ and $v_{i-1}\neq v_{i+1}$ for $1\leq i\leq s-1$. Let $\Gamma$ be a
graph and let $G$ be a subgroup of the full automorphism group
$\mathrm{Aut}(\Gamma)$ of $\Gamma$. The graph $\Gamma$ is said to be _$(G,s)$
-arc-transitive_ if $G$ is transitive on the set of $s$-arcs. For short, we
say that $\Gamma$ is _$s$ -arc-transitive_ if it is
$(\mathrm{Aut}(\Gamma),s)$-arc-transitive, and _arc-transitive_ if it is
$(\mathrm{Aut}(\Gamma),1)$-arc-transitive. Note that for regular graphs
$s$-arc-transitivity with $s\geq 2$ implies $(s-1)$-arc-transitivity. Tutte
[25] in 1947 proved that there exist no finite $s$-arc-transitive cubic graphs
for $s\geq 6$. Since this remarkable result, $s$-arc-transitive graphs have
attracted considerable attention in the literature.

Let $H$ be a group and let $S$ be an inverse-closed nonempty subset of
$H\setminus\\{1\\}$. The _Cayley graph_ $\mathrm{Cay}(H,S)$ on $H$ with
_connection set_ $S$ is defined to be the graph with vertex set $H$ such that
$x,y\in H$ are adjacent if and only if $yx^{-1}\in S$.

Some special classes of $2$-arc-transitive Cayley graphs have been classified
or characterized in the literature. See Alspach, Conder, Marušič and Xu [1]
for $2$-arc-transitive Cayley graphs on cyclic groups, Ivanov and Praeger [14]
and Li and Pan [17] for $2$-arc-transitive Cayley graphs on abelian groups,
Marušič and Du [9] for $2$-arc-transitive Cayley graphs on dihedral groups,
and the recent work of Li and Xia [19] on $2$-arc-transitive Cayley graphs on
solvable groups.

For a Cayley graph $\mathrm{Cay}(H,S)$, denote by $R_{H}(H)$ the subgroup of
$\mathrm{Sym}(H)$ induced by the right multiplication of $H$ on itself.
Clearly, $R_{H}(H)$ is a subgroup of $\mathrm{Aut}(\mathrm{Cay}(H,S))$. The
Cayley graph $\mathrm{Cay}(H,S)$ is said to be _normal_ if $R_{H}(H)$ is a
normal subgroup of $\mathrm{Aut}(\mathrm{Cay}(H,S))$, and is said to be
_nonnormal_ otherwise. Nonnormal Cayley graphs are believed to be rare, and it
is conjectured by Xu [28] that almost all connected Cayley graphs are normal.
Moreover, it is well known that if $\mathrm{Cay}(H,S)$ is a normal Cayley
graph on $H$, then
$\mathrm{Aut}(\mathrm{Cay}(H,S))=R_{H}(H){:}\mathrm{Aut}(H,S)$ where

$\mathrm{Aut}(H,S):=\\{\sigma\in\mathrm{Aut}(H)\mid S^{\sigma}=S\\},$

see Godsil [11, Lemma 2.1]. Hence the symmetries of normal Cayley graphs can
be well understood, and consequently more attention in the literature are on
nonnormal ones while investigating Cayley graphs.

The Jordan-Hölder theorem shows that all finite groups can be viewed as group
extensions of simple groups (namely their compositor factors). This
fundamental theorem increases the interest of the study of Cayley graphs on
simple groups. Fang, Praeger and Wang [10] gave a general description of the
structures of automorphism groups of nonnormal Cayley graphs on nonabelian
simple groups, and $s$-arc-transitive nonnormal Cayley graphs on nonabelian
simple groups of certain valencies have been explicitly characterized. Let
$\Gamma$ be a connected nonnormal $s$-arc-transitive Cayley graph on a
nonabelian simple group. Based on a so called ‘dual action’ given in Li [16],
Xu, Fang, Wang and Xu [29, 30] proved that if the valency
$\mathsf{Val}(\Gamma)=3$ then $\Gamma$ is one of two
$(\mathrm{A}_{48},5)$-arc-transitive Cayley graphs on $\mathrm{A}_{47}$. Du
and Feng [6] proved that if $\mathsf{Val}(\Gamma)=4$ and $s\geq 2$, then
$\Gamma$ is an $\mathrm{A}_{n+1}$-arc-transitive Cayley graph on
$\mathrm{A}_{n}$ for $n\in\\{24,36,72,144,532,14364\\}$ with a unique
exception. Similarly, Du, Feng and Zhou [7] showed that if
$\mathsf{Val}(\Gamma)=5$ then $\Gamma$ is an $\mathrm{A}_{n+1}$-arc-transitive
Cayley graph on $\mathrm{A}_{n}$ where $n$ is among $11$ possible numbers, and
[20] showed that if $\mathsf{Val}(\Gamma)=7$ and the vertex stabilizer is
solvable then $\Gamma$ is an $\mathrm{A}_{n+1}$-arc-transitive Cayley graph on
$\mathrm{A}_{n}$ with $n\in\\{7,21,63,84\\}$. Very recently, Yin, Feng, Zhou
and Chen [31] proved that if $\mathsf{Val}(\Gamma)$ is a prime greater than
$7$ and the vertex stabilizer is solvable, then $\Gamma$ is either an
$\mathrm{A}_{n+1}$-arc-transitive Cayley graph on $\mathrm{A}_{n}$ or one of
the three exceptions.

From the above mentioned results in [6, 7, 20, 29, 30, 31] on $s$-arc-
transitive nonnormal Cayley graphs on nonabelian simple groups of certain
valencies one can observe an interesting phenomenon: most of these graphs turn
out to be Cayley graphs on alternating groups. This motivates a natural
problem as follows.

Problem A. Study $s$-arc-transitive nonnormal Cayley graphs on alternating
groups for arbitrary valency.

In this paper, we give a characterization of the graphs in Problem A with
$s=2$. For a group $G$, the socle $\mathrm{Soc}(G)$ of $G$ is the product of
all minimal normal subgroups of $G$. For a $G$-vertex transitive graph
$\Gamma$ and a normal subgroup $M$ of $G$, the quotient graph $\Gamma_{M}$
associated with $M$ is a graph with the set of orbits of $M$ as vertex set,
and two orbits $O_{1},O_{2}$ are adjacent if there is an edge in $\Gamma$ with
two endpoints in $O_{1},O_{2}$ respectively. The graph $\Gamma$ is called a
_normal cover_ of $\Gamma_{M}$ if the valency
$\mathsf{Val}(\Gamma)=\mathsf{Val}(\Gamma_{M})$.

######  Theorem 1.1.

Let $\Gamma$ be a connected $(G,2)$-arc-transitive Cayley graph on an
alternating group $H\cong\mathrm{A}_{n}$ with $n\geq 5$ and $H\leq G$, and let
$\alpha$ be a vertex of $\Gamma$. Suppose that $H$ is not normal in $G$. Then
one of the following holds:

  1. (a)

$\mathrm{Soc}(G)=\mathrm{A}_{n+1}$, $\mathrm{Soc}(G)_{\alpha}$ is a regular
subgroup of $\mathrm{A}_{n+1}$.

  2. (b)

$G$ has a normal subgroup $N\cong\mathrm{A}_{n+2}$ such that $H<N$, $\Gamma$
is $(N,2)$-arc-transitive, $N_{\alpha}$ is a sharply $2$-transitive subgroup
of $\mathrm{A}_{n+2}$, and $N_{\alpha}$ acts faithfully on $\Gamma(\alpha)$;
in particular, $\mathsf{Val}(\Gamma)=n+2$ is a prime power.

  3. (c)

$\Gamma=\mathsf{K}_{m}$ with $m=n!/2$, and either
$(\mathrm{Soc}(G),\mathrm{Soc}(G)_{\alpha})=(\mathrm{A}_{m},\mathrm{A}_{m-1})$
or $(G,G_{\alpha},n)=(\mathrm{PGL}_{2}(59),\mathrm{AGL}_{1}(11),5)$.

  4. (d)

$G$ has a nontrivial maximal intransitive normal subgroup $M$ such that $G/M$
is almost simple, $\Gamma_{M}$ is $(G/M,2)$-arc-transitive, and $\Gamma$ is a
normal cover of $\Gamma_{M}$; moreover, one of the following is true, where
$v$ is a vertex of $\Gamma_{M}$:

    1. (d.1)

$\mathrm{Soc}(G/M)=\mathrm{A}_{n}$ such that $n\geq 16$ is a power of $2$, and
$\mathrm{Soc}(G/M)_{v}$ contains a Sylow $2$-subgroup of $\mathrm{Soc}(G/M)$.

    2. (d.2)

$\mathrm{Soc}(G/M)=\mathrm{A}_{n+1}$, and $\mathrm{Soc}(G/M)_{v}$ is a
transitive subgroup of $\mathrm{A}_{n+1}$;

    3. (d.3)

$\Gamma_{M}=\mathsf{K}_{m}$, $\mathrm{Soc}(G/M)=\mathrm{A}_{m}$ such that $m$
is the index of a subgroup of $\mathrm{A}_{n}$, and
$\mathrm{Soc}(G/M)_{v}=\mathrm{A}_{m-1}$.

A group $G$ is said to be _almost simple_ if $\mathrm{Soc}(G)$ is a nonabelian
simple group. For Problem A, if we impose that the $2$-arc-transitive
automorphism group of the graph is almost simple, then we have the following
corollary of Theorem 1.1.

######  Corollary 1.2.

Let $\Gamma$ be a connected $(G,2)$-arc-transitive Cayley graph on an
alternating group $H\cong\mathrm{A}_{n}$ with $n\geq 5$ and $H\leq G$, and let
$\alpha$ be a vertex of $\Gamma$. Suppose that $G$ is almost simple with socle
$T$. Then one of the following holds:

  1. (a)

$T=\mathrm{A}_{n+1}$ and $T_{\alpha}$ is a regular subgroup of
$\mathrm{A}_{n+1}$;

  2. (b)

$T=\mathrm{A}_{n+2}$ such that $\Gamma$ is $(T,2)$-arc-transitive,
$T_{\alpha}$ is a sharply $2$-transitive subgroup of $\mathrm{A}_{n+2}$ and
acts faithfully on $\Gamma(\alpha)$, and $\mathsf{Val}(\Gamma)=n+2$ is a prime
power;

  3. (c)

$\Gamma=\mathsf{K}_{n!/2}$.

As introduced above, there are a number of $(\mathrm{A}_{n+1},2)$-arc-
transitive Cayley graphs on $\mathrm{A}_{n}$ known in the literature, which
give rise to examples of case (a) in Corollary 1.2. However, to the best of
our knowledge, there is no example of $(\mathrm{A}_{n+2},2)$-arc-transitive
Cayley graph on $\mathrm{A}_{n}$ in the literature. Note that the vertex
stabilizer $T_{\alpha}$ in case (b) of Corollary 1.2, as a sharply
$2$-transitive group, is an affine group with degree a prime power $p^{d}$,
and is either a subgroup of $\mathrm{A\Gamma L}_{1}(p^{d})$ or with
$p^{d}\in\\{5^{2},7^{2},11^{2},19^{2},23^{2},29^{2},59^{2}\\}$ (see for
example [13, Chapter XII §9]). In this paper, we construct the first infinite
family of $(\mathrm{A}_{n+2},2)$-arc-transitive Cayley graphs on
$\mathrm{A}_{n}$ with vertex stabilizer a subgroup of $\mathrm{A\Gamma
L}_{1}(p^{d})$.

######  Construction 1.3.

Let $q$ be a prime power such that $q\equiv 3\pmod{4}$, and let $\tau\colon
v\mapsto v^{q}$ be the filed automorphism of $\mathbb{F}_{q^{2}}$ of order
$2$. Take $G=\mathrm{Sym}(\mathbb{F}_{q^{2}})$,
$K=\mathrm{AGL}_{1}(q^{2}){:}\langle\tau\rangle\leq\mathrm{A\Gamma
L}_{1}(q^{2})<G$, and $g\in G$ such that $0^{g}=0$ and $v^{g}=v^{-1}$ for all
$v\in\mathbb{F}_{q^{2}}^{\times}$. Then let $\Gamma=\mathrm{Cos}(G,K,g)$ be
the coset graph (see Subsection 2.2 for definition) given by the triple
$(G,K,g)$.

######  Theorem 1.4.

Let $q$, $G$ and $\Gamma$ be as in Construction 1.3. Then $\Gamma$ is a
connected $(\mathrm{A}_{q^{2}},2)$-arc-transitive Cayley graph on
$\mathrm{A}_{q^{2}-2}$ with valency $q^{2}$.

The rest of the paper is organized as follows. After a preliminary section, we
deal with the vertex-quasiprimitive case of Theorems 1.1 in Section 3, which
leads to a proof of Theorems 1.1 in Section 4. Then in Section 5 we prove
Theorem 1.4.

##  2\. Preliminaries

The notations used in this paper are standard. As in [4], we sometimes use $n$
to denote a cyclic group of order $n$, use $[n]$ to denote a group of order
$n$, and use $p^{n}$ with $p$ a prime to denote the elementary abelian group
of order $p^{n}$. For a positive integer $n$ and prime number $p$, let denote
by $n_{p}$ the largest $p$-power dividing $n$. For a graph $\Gamma$, let
$V(\Gamma)$ denote the vertex set of $\Gamma$, and let $\Gamma(v)$ denote the
set of neighborhoods of $v\in V(\Gamma)$.

###  2.1. $2$-arc-transitive graphs

The following result shows that the $s$-arc-transitivity of graph with $s\geq
2$ is inherited by normal quotients, see Praeger [21, Theorem 4.1].

######  Theorem 2.1.

Let $\Gamma$ be a connected $(G,s)$-arc-transitive graph with $s\geq 2$.
Suppose that $G$ has a normal subgroup $M$ with at least three orbits on
$V(\Gamma)$. Then $\Gamma_{M}$ is $(G/M,s)$-arc-transitive with
$(G/M)_{v}\cong G_{\alpha}$, where $v\in V(\Gamma_{M})$ and $\alpha\in
V(\Gamma)$. Moreover, $M$ is semiregular on $V(\Gamma)$, and $\Gamma$ is a
normal cover of $\Gamma_{M}$.

Let $\Gamma$ be a graph and let $G\leq\mathrm{Aut}(\Gamma)$. For adjacent
vertices $\alpha$ and $\beta$ of $\Gamma$, let $G_{\alpha}^{[1]}$ be the
kernel of $G_{\alpha}$ acting on $\Gamma(\alpha)$, and let
$G_{\alpha\beta}^{[1]}=G_{\alpha}^{[1]}\cap G_{\beta}^{[1]}$. The following
theorem is a well-known result of Weiss [26] on local action of 2-arc-
transitive graphs.

######  Theorem 2.2.

Let $\Gamma$ be a connected $(G,2)$-transitive graph with
$G\leq\mathrm{Aut}(\Gamma)$, and let $\alpha$ and $\beta$ be adjacent vertices
of $\Gamma$. Then one of the following holds:

  1. (a)

$G_{\alpha\beta}^{[1]}=1$, and
$G_{\alpha}^{[1]}\cong(G_{\alpha}^{[1]})^{\Gamma(\beta)}\trianglelefteq
G_{\alpha\beta}^{\Gamma(\beta)}\cong G_{\alpha\beta}^{\Gamma(\alpha)}$;

  2. (b)

$G_{\alpha\beta}^{[1]}$ is a nontrivial $p$-group for some prime $p$, and
there exist integers $d\geq 2$ and $f\geq 1$ such that
$G_{\alpha}^{\Gamma(\alpha)}\trianglerighteq\mathrm{PSL}_{d}(p^{f})$ and
$\mathsf{Val}(\Gamma)=(p^{df}-1)/(p^{f}-1)$.

The following lemma is also well known. We include a proof here as it is not
lengthy.

######  Lemma 2.3.

Let $\Gamma$ be a connected graph, let $G$ be a vertex-transitive subgroup of
$\mathrm{Aut}(\Gamma)$, and let $\alpha\in V(\Gamma)$. Then every composition
factor of $G_{\alpha}$ is isomorphic to a section of
$G_{\alpha}^{\Gamma(\alpha)}$.

###### Proof.

Let $d$ be the diameter of $\Gamma$. For $i\in\\{0,1,\dots,d\\}$, let $V_{i}$
be the set of vertices of distance $i$ from $\alpha$ in $\Gamma$, let
$U_{i}=V_{0}\cup V_{1}\cup\dots\cup V_{i}$, and let $K_{i}$ be the kernel of
$G_{\alpha}$ acting on $U_{i}$. By the connectivity of $\Gamma$ we have
$U_{d}=V(\Gamma)$, and thus

$1=K_{d}\trianglelefteq K_{d-1}\trianglelefteq\dots\trianglelefteq
K_{1}\trianglelefteq K_{0}=G_{\alpha}.$

Moreover, for $i\in\\{1,\dots,d\\}$,

$K_{i-1}/K_{i}\cong K_{i-1}^{V_{i}}.$

Let $\Delta$ be an orbit of $K_{i-1}$ on $V_{i}$, and let $\gamma\in\Delta$.
Then there exists $\beta\in U_{i-1}$ such that $\gamma\in\Gamma(\beta)$. It
follows that $\gamma^{k}\in\Gamma(\beta^{k})=\Gamma(\beta)$ for each $k\in
K_{i-1}$. Hence $\Delta=\gamma^{K_{i-1}}\subseteq\Gamma(\beta)$, and so
$K_{i-1}^{\Delta}$ is a homomorphic image of $K_{i-1}^{\Gamma(\beta)}$. Note
that $K_{i-1}^{\Gamma(\beta)}\leq G_{\beta}^{\Gamma(\beta)}\cong
G_{\alpha}^{\Gamma(\alpha)}$ since $G$ is vertex-transitive. We then conclude
that $K_{i-1}^{\Delta}$ is (isomorphic to) a section of
$G_{\alpha}^{\Gamma(\alpha)}$. Since $\Delta$ is an arbitrary orbit of
$K_{i-1}$ on $V_{i}$, this implies that $K_{i-1}/K_{i}\cong K_{i-1}^{V_{i}}$
is a section of $G_{\alpha}^{\Gamma(\alpha)}$. Thus every composition factor
of $G_{\alpha}$ is a section of $G_{\alpha}^{\Gamma(\alpha)}$. ∎

Recall that the $2$-transitive groups are known (see for instance [3, Chapter
7]). In particular, a $2$-transitive group is either affine or almost simple.

######  Lemma 2.4.

Let $\Gamma$ be a connected $(G,2)$-arc-transitive graph, let $N$ be a vertex-
transitive normal subgroup of $G$, and let $\alpha\in V(\Gamma)$. Then the
following hold:

  1. (a)

if $N_{\alpha}^{\Gamma(\alpha)}$ is almost simple with socle not isomorphic to
$\mathrm{PSL}_{2}(8)$, then $\Gamma$ is $(N,2)$-arc-transitive;

  2. (b)

if $N_{\alpha}$ is almost simple with socle not isomorphic to
$\mathrm{PSL}_{2}(8)$, then $\Gamma$ is $(N,2)$-arc-transitive.

###### Proof.

Since $\Gamma$ is $(G,2)$-arc-transitive, $G_{\alpha}^{\Gamma(\alpha)}$ is
$2$-transitive. Suppose that $N_{\alpha}^{\Gamma(\alpha)}$ is almost simple
with socle not isomorphic to $\mathrm{PSL}_{2}(8)$. It follows that
$N_{\alpha}^{\Gamma(\alpha)}$ is a normal subgroup of the $2$-transitive group
$G_{\alpha}^{\Gamma(\alpha)}$. Then by the classification of $2$-transitive
groups, $G_{\alpha}^{\Gamma(\alpha)}$ is almost simple with
$\mathrm{Soc}(G_{\alpha}^{\Gamma(\alpha)})=\mathrm{Soc}(N_{\alpha}^{\Gamma(\alpha)})$,
and $N_{\alpha}^{\Gamma(\alpha)}$ is $2$-transitive. Since $N$ is vertex-
transitive, we then conclude that $\Gamma$ is $(N,2)$-arc-transitive. This
proves part (a).

Now suppose that $N_{\alpha}$ is almost simple with socle not isomorphic to
$\mathrm{PSL}_{2}(8)$. Then $N_{\alpha}^{\Gamma(\alpha)}$ is nonsolvable by
Lemma 2.3. Since $N_{\alpha}^{\Gamma(\alpha)}$ is a homomorphic image of the
almost simple group $N_{\alpha}$, this implies that
$N_{\alpha}^{\Gamma(\alpha)}\cong N_{\alpha}$. Thus the conclusion of part (a)
shows that $\Gamma$ is $(N,2)$-arc-transitive, proving part (b). ∎

###  2.2. Coset graphs and orbital graphs

Coset graph and orbital graph are two useful tools to construct and understand
general arc-transitive graphs.

Let $G$ be a group, $K$ a core-free subgroup of $G$ (namely, $K$ contains no
nontrivial normal subgroup of $G$) and $g\in G\setminus K$. Define the _coset
graph_ $\mathrm{Cos}(G,K,g)$ to be a graph with vertex set $[G:K]$ (the set of
right cosets of $K$ in $G$) such that $Kx$ is adjacent to $Ky$ with $x,y\in G$
if and only if $yx^{-1}\in KgK$. The following assertion is due to Sabidussi
[24].

######  Lemma 2.5.

Let $\Gamma$ be a connected $G$-arc-transitive graph of valency $d$, and let
$\alpha\in V(\Gamma)$. Then $\Gamma\cong\mathrm{Cos}(G,K,g)$ for
$K=G_{\alpha}$ and some $2$-element $g$ satisfying:

$g\in\mathbf{N}_{G}(K\cap K^{g}),\ \ g^{2}\in K,\ \ \langle K,g\rangle=G,\ \
|K|/|K\cap K^{g}|=d.$ (1)

In particular, if $\beta\in\Gamma(\alpha)$ then $\langle
G_{\alpha},\mathbf{N}_{G}(G_{\alpha\beta})\rangle=G$.

Conversely, if $H$ is a core-free subgroup of a group $X$ and $x$ is an
element of $X$ such that the triple $(X,H,x)$ $($as $(G,K,g)$ there$)$
satisfies (1), then $\mathrm{Cos}(X,H,x)$ is a connected $X$-arc-transitive
graph of valency $d$.

Let $G$ be a transitive permutation group on a set $\Omega$. Then $G$
naturally acts on $\Omega\times\Omega$ by

$(\alpha,\beta)^{g}=(\alpha^{g},\beta^{g})\ \ \text{for}\ g\in G\ \text{and}\
\alpha,\beta\in\Omega.$

The orbits of $G$ on $\Omega\times\Omega$ are called orbitals of $G$. By the
transitivity of $G$, each orbital of $G$ corresponds to a orbit of
$G_{\alpha}$ on $\Omega$, which is called a _suborbit_ of $G$. Let $\Delta$ be
a suborbit of $G$, and let $\beta\in\Delta$. We say that $\Delta$ is
_nontrivial_ if $\Delta\neq\\{\alpha\\}$, and $\Delta$ is _self-paired_ if
$(\alpha,\beta)^{G}=(\beta,\alpha)^{G}$. If $\Delta$ is nontrivial, then the
associated _orbital graph_ $\Gamma(\Delta)$ of $\Delta$ is the graph with
vertex set $\Omega$ and edge set
$\\{\alpha,\beta\\}^{G}:=\\{\\{\alpha^{g},\beta^{g}\\}:g\in G\\}$. Clearly, if
$\Delta$ is nontrivial and self-paired, then $\Gamma(\Delta)$ is $G$-arc-
transitive. Conversely, every $G$-arc-transitive graph can arise in this way.
Note that if $\Gamma(\Delta)$ is connected, then $\langle
G_{\alpha},g\rangle=G$ for any $g\in G$ such that $\alpha^{g}=\beta$.

The next lemma will be needed in Section 3.

######  Lemma 2.6.

Let $G=\mathrm{A}_{n}$ be a transitive permutation group on a set $\Omega$,
where $n\geq 5$, and let $\alpha\in\Omega$. Suppose that
$G_{\alpha}\cong\mathrm{A}_{n-2}$ or $\mathrm{S}_{n-2}$. Then there is no
connected $(G,2)$-arc-transitive associated orbital graph for any non-trivial
suborbit of $G$ on $\Omega$.

###### Proof.

Suppose for a contradiction that there exists a connected $(G,2)$-arc-
transitive associated orbital graph $\Gamma$ with a suborbit $\Delta$ of $G$
on $\Omega$. Then $\Delta$ is nontrivial and self-paired.

First assume that $G_{\alpha}\cong\mathrm{A}_{n-2}$. Then the action of $G$ on
$\Omega$ can be identified with the natural action of $G$ on $\Phi^{(2)}$, the
set of ordered pairs in the set $\Phi:=\\{1,2,\dots,n\\}$. Identify $\Omega$
with $\Phi^{(2)}$, and assume without loss of generality that $\alpha=(1,2)$.
Let $\beta=(i,j)\in\Delta$. Then $\Delta=\beta^{G_{\alpha}}$, and
$\beta\neq\alpha$ as $\Delta$ is nontrivial. If $\\{i,j\\}=\\{1,2\\}$, then
$\beta=(2,1)$, and so the suborbit $\Delta=\beta^{G_{\alpha}}=\\{\beta\\}$ has
length $1$, contradicting the connectivity of $\Gamma$. If
$\\{i,j\\}\cap=\\{1,2\\}=\emptyset$, then
$G_{\alpha\beta}\cong\mathrm{A}_{n-4}$ is not maximal in
$\mathrm{G}_{\alpha}$, and so the action of $G_{\alpha}$ on
$[G_{\alpha}:G_{\alpha\beta}]$ is not $2$-transitive, contradicting the
$(G,2)$-arc-transitivity of $\Gamma$. Hence $|\\{i,j\\}\cap\\{1,2\\}|=1$.
Assume without loss of generality that $\beta=(1,3)$. Let $g=(2,3)(4,5)$. Then
$g$ maps $\alpha$ to $\beta$, and so $\langle G_{\alpha},g\rangle=G$ by the
connectivity of $\Gamma$. However, $\langle G_{\alpha},g\rangle$ fixes $1$ as
both $G_{\alpha}$ and $g$ fixes $1$, a contradiction.

Next assume that $G_{\alpha}\cong\mathrm{S}_{n-2}$. Then the action of $G$ on
$\Omega$ can be identified with the natural action of $G$ on $\Phi^{\\{2\\}}$,
the set of $2$-subsets of the set $\Phi:=\\{1,2,\dots,n\\}$. Identify $\Omega$
with $\Phi^{\\{2\\}}$, and assume without loss of generality that
$\alpha=\\{1,2\\}$. Let $\beta=\\{i,j\\}\in\Delta$. Then
$\Delta=\beta^{G_{\alpha}}$, and $\beta\neq\alpha$ as $\Delta$ is nontrivial.
If $\\{1,2\\}\cap\\{i,j\\}=\emptyset$, then
$G_{\alpha\beta}\cong\mathrm{S}_{n-4}.\mathrm{C}_{2}$. If
$|\\{i,2\\}\cap\\{i,j\\}|=1$, then $G_{\alpha\beta}\cong\mathrm{A}_{n-3}$. In
either case, $G_{\alpha\beta}$ is not maximal in $G_{\alpha}$, and so the
action of $G_{\alpha}$ on $[G_{\alpha}:G_{\alpha\beta}]$ is not
$2$-transitive, contradicting the $(G,2)$-arc-transitivity of $\Gamma$. ∎

###  2.3. Quasiprimitive permutation groups with a transitive alternating
group

A transitive permutation group is said to be _quasiprimitive_ if all of its
nontrivial normal subgroups are transitive. The following classification of
quasiprimitive permutation groups containing a transitive alternating group
was obtained by the second-named author [27].

######  Proposition 2.7.

Let $G$ be a quasiprimitive permutation group on $\Omega$, and let
$\alpha\in\Omega$. If $G$ contains a transitive subgroup $H=\mathrm{A}_{n}$
with $n\geq 5$, then one of the following holds.

  1. (a)

$G$ is almost simple with socle $L$ such that either $L=H$, or $L=HL_{\alpha}$
satisfies one of the following:

    1. (a.1)

$L=\mathrm{A}_{n+k}$ with $1\leq k\leq 5$, and $L_{\alpha}$ is $k$-transitive
on $n+k$ points;

    2. (a.2)

$L=\mathrm{A}_{m}$ and $L_{\alpha}=\mathrm{A}_{m-1}$, where $m$ is the index
of a subgroup of $A_{n}$;

    3. (a.3)

$(L,n,L_{\alpha})$ lies in Table 1.

  2. (b)

$G$ is primitive with socle $\mathrm{A}_{n}\times\mathrm{A}_{n}$, and $H$ is
regular.

  3. (c)

$n=6$, $G$ is primitive with socle $\mathrm{A}_{6}\times\mathrm{A}_{6}$, and
$G\leq\mathrm{S}_{6}\wr\mathrm{S}_{2}$ by the product action on $6^{2}$
points.

$\begin{array}[]{llll}\hline\cr\text{Row}&L&n&L_{\alpha}\\\ \hline\cr
1&\mathrm{A}_{6}&5&\mathrm{A}_{4},\,\mathrm{S}_{4}\\\
2&\mathrm{A}_{10}&6&\mathrm{A}_{8},\,\mathrm{S}_{8}\\\
3&\mathrm{A}_{15}&7,\,8&\mathrm{A}_{13},\,\mathrm{S}_{13}\\\
4&\mathrm{M}_{12}&5&\mathrm{M}_{11}\\\ 5&\mathrm{PSL}_{2}(11)&5&11,\,11{:}5\\\
6&\mathrm{PSL}_{2}(19)&5&19{:}9\\\
7&\mathrm{PSL}_{2}(29)&5&29{:}7,\,29{:}14\\\
8&\mathrm{PSL}_{2}(59)&5&59{:}29\\\
9&\mathrm{PSL}_{4}(3)&6&3^{3}{:}\mathrm{PSL}_{3}(3)\\\
10&\mathrm{PSU}_{3}(5)&7&5^{1+2}_{+}{:}8\\\
11&\mathrm{PSp}_{4}(3)&6&3^{1+2}_{+}{:}\mathrm{Q}_{8},\,3^{1+2}_{+}{:}2.\mathrm{A}_{4}\\\
12&\mathrm{Sp}_{6}(2)&6,\,7,\,8&\mathrm{PSU}_{3}(3){:}2\\\
13&\mathrm{Sp}_{6}(2)&8&3^{1+2}_{+}{:}8{:}2,\,3^{1+2}_{+}{:}2.\mathrm{S}_{4},\,\mathrm{PSL}_{2}(8),\,\mathrm{PSL}_{2}(8){:}3,\,\mathrm{PSU}_{4}(2){:}2\\\
14&\mathrm{Sp}_{8}(2)&6,\,7,\,8,\,9,\,10&\mathrm{SO}^{-}_{8}(2)\\\
15&\Omega_{7}(3)&8,\,9&3^{3+3}{:}\mathrm{PSL}_{3}(3)\\\
16&\Omega_{7}(3)&9&3^{3}{:}\mathrm{PSL}_{3}(3),\,\mathrm{PSL}_{4}(3),\,\mathrm{PSL}_{4}(3){:}2,\,\mathrm{G}_{2}(3)\\\
17&\Omega^{+}_{8}(2)&6,\,7,\,8,\,9&\mathrm{Sp}_{6}(2)\\\
18&\Omega^{+}_{8}(2)&8&\mathrm{A}_{9}\\\
19&\Omega^{+}_{8}(2)&8,\,9&\mathrm{PSU}_{4}(2),\,\mathrm{PSU}_{4}(2){:}2,\,3\times\mathrm{PSU}_{4}(2),\,(3\times\mathrm{PSU}_{4}(2)){:}2\\\
20&\Omega^{+}_{8}(2)&9&2^{4}{:}15.4,\,2^{6}{:}15,\,2^{6}{:}15.2,\,2^{6}{:}15.4,\,\mathrm{A}_{8},\,\mathrm{S}_{8}\\\
21&\Omega^{+}_{8}(2)&9&2^{4}{:}\mathrm{A}_{5}\leq L_{\alpha}\leq
2^{6}{:}\mathrm{A}_{8}\\\
22&\mathrm{P\Omega}^{+}_{8}(3)&8,\,9&3^{6}{:}\mathrm{PSL}_{4}(3)\\\
23&\mathrm{P\Omega}^{+}_{8}(3)&9&\Omega_{7}(3)\\\
24&\Omega^{-}_{10}(2)&12&2^{8}{:}\Omega^{-}_{8}(2)\\\ \hline\cr\end{array}$
Table 1. Exceptional quasiprimitive groups with a transitive subgroup
$\mathrm{A}_{n}$

Let us outline the proof of Theorem 1.1, in which Proposition 2.7 will play a
crucial role. Let $\Gamma$ be a connected $(G,2)$-arc-transitive Cayley graph
on an alternating group $H\cong\mathrm{A}_{n}$ with $n\geq 5$ and $H\leq G$,
and let $\alpha$ be a vertex of $\Gamma$. Take $M$ to be a maximal
intransitive normal subgroup of $G$, and let $X=G/M$ and $Y=HM/H$. From
Theorem 2.1 we see the quotient graph $\Gamma_{M}$ is $X$-arc-transitive and
$X$-vertex-quasiprimitive, and $X$ contains a vertex-transitive subgroup
$Y\cong H$. Thus the candidates for the triple $(X,Y,X_{v})$ are given in
Proposition 2.7. We investigate those candidates and give a characterization
of such graphs $\Gamma_{M}$ in Section3. We shall see that most of the
candidates for $(X,Y,X_{v})$ satisfy $\mathrm{Soc}(X)=\mathrm{A}_{n+k}$ with
$1\leq k\leq 5$, $Y=\mathrm{A}_{n}$ and that $X_{v}$ is a $k$-transitive group
on $n+k$ points, as in Case (a.1) of Proposition 2.7. Moreover, the
possibility for $k\geq 3$ will be excluded, and the graph $\Gamma$ arising
from this case must will be shown to satisfy part (b) of Theorem 1.1. In
Section 4, we consider the cover of $\Gamma_{M}$ and complete the proof of
Theorem 1.1, where the same technique as in [7] will be used.

###  2.4. Some technical lemmas

Let $G$ be a permutation group on a set $\Omega$. Recall that $G$ is said to
be _semiregular_ on $\Omega$ if $G_{\alpha}=1$ for each $\alpha\in\Omega$. A
nontrivial cyclic subgroup $\langle g\rangle$ of $G$ is semiregular on
$\Omega$ if and only if $g$ can be expressed as a disjoint product of $s$
cycles of length $t$ such that $st=|\Omega|$ and $t>1$.

######  Lemma 2.8.

Let $G$ be a permutation group on a set $\Omega$, let $H$ be a subgroup of
$G$, and let $H_{\alpha}$ be the stabilizer of a point $\alpha\in\Omega$ in
$H$. If $H_{\alpha}$ contains an element acting fixed-point-freely on
$\Omega\setminus\\{\alpha\\}$, then $\mathbf{N}_{G}(H_{\alpha})\leq
G_{\alpha}$. In particular, if $H$ is $2$-transitive then
$\mathbf{N}_{G}(H_{\alpha})\leq G_{\alpha}$.

###### Proof.

Let $g\in\mathbf{N}_{G}(H_{\alpha})$, and let $h\in H_{\alpha}$ such that $h$
has no fixed-point on $\Omega\setminus\\{\alpha\\}$. Then
$\mathsf{Fix}(h)=\\{\alpha\\}$, and so $\mathsf{Fix}(h^{g})=\\{\alpha^{g}\\}$.
Since $h^{g}\in(H_{\alpha})^{g}=H_{\alpha}$, we have
$\alpha\in\mathsf{Fix}(h^{g})$. It follows that $\alpha\in\\{\alpha^{g}\\}$.
Hence $\alpha^{g}=\alpha$, that is, $g\in G_{\alpha}$. Thus
$\mathbf{N}_{G}(H_{\alpha})\leq G_{\alpha}$, as required.

Now suppose that $H$ is $2$-transitive. Then $H_{\alpha}$ is transitive on
$\Omega\setminus\\{\alpha\\}$. By a theorem of Jordan, there exits $h\in
H_{\alpha}$ such that $h$ has no fixed-point on $\Omega\setminus\\{\alpha\\}$.
This implies $\mathbf{N}_{G}(H_{\alpha})\leq G_{\alpha}$ by the above
conclusion. ∎

Recall that a _section_ of a group $G$ is a quotient of a subgroup of $G$.

######  Lemma 2.9.

Let $B=M{:}H$ where $H=\mathrm{A}_{n}$ with $n\geq 9$. Suppose that
$|M|_{r}<r^{n-2}$ for each prime divisor $r$ of $|M|$ and that $M$ has no
section isomorphic to $H$. Then $B=M\times H$.

###### Proof.

Suppose for a contradiction that $B\neq M\times H$. Let

$1=M_{0}<M_{1}<\dots<M_{s}=M\lhd B$

be a normal series of $B$ such that $M_{i+1}/M_{i}$ is minimal normal in
$B/M_{i}$ for each $0\leq i\leq s-1$. Since $B\neq M\times H$, there exists
$0\leq j\leq s-1$ such that $M_{j+1}H\neq M_{j+1}\times H$ but
$M_{i}H=M_{i}\times H$ for each $0\leq i\leq j$. Since $M\cap H=1$, we have
$M_{j}H/M_{j}\cong H\cong\mathrm{A}_{n}$ and
$M_{j+1}H/M_{j}=(M_{j+1}/M_{j}){:}(M_{j}H/M_{j})$. If the conjugation action
of $M_{j}H/M_{j}$ on $M_{j+1}/M_{j}$ is trivial, then
$M_{j+1}H/M_{j}=(M_{j+1}/M_{j})\times(M_{j}H/M_{j})$ and so $M_{j}H\lhd
M_{j+1}H$. If this is the case, then since $M_{j}H=M_{j}\times H$ and $M$ has
no section isomorphic to $\mathrm{A}_{n}$, it follows that $H$ is
characteristic in $M_{j}H$ and thus normal in $M_{j+1}H$, which implies that
$M_{j+1}H=M_{j+1}\times H$, a contradiction. Hence the conjugation action of
$M_{j}H/M_{j}$ on $M_{j+1}/M_{j}$ is nontrivial, and is thus faithful by the
simplicity of $M_{j}H/M_{j}$. Therefore, $\mathrm{A}_{n}\cong
M_{j}H/M_{j}\leq\mathrm{Aut}(M_{j+1}/M_{j})$.

Write $M_{j+1}/M_{j}=T_{1}\times\cdots\times T_{m}$ with
$T_{1}\cong\cdots\cong T_{m}\cong T$ for some simple group $T$. If
$T\cong\mathrm{C}_{p}$ for some prime $p$, then
$\mathrm{A}_{n}\leq\mathrm{Aut}(\mathrm{C}_{p}^{m})\cong\mathrm{GL}_{m}(p)$,
and by [15, Proposition 5.3.7] we have $m\geq n-2$, which yields that
$|M|_{p}\geq|M_{j+1}/M_{j}|=p^{m}\geq p^{n-2}$, contradicting the condition of
the lemma. Hence $T$ is a nonabelian simple group. Then since
$\mathrm{A}_{n}\leq\mathrm{Aut}(T^{m})\cong\mathrm{Aut}(T)\wr\mathrm{S}_{m}$
and $M$ has no section isomorphic to $\mathrm{A}_{n}$, we derive that
$\mathrm{A}_{n}\cap\mathrm{Aut}(T)^{m}=1$ and so
$\mathrm{A}_{n}\lesssim\mathrm{S}_{m}$. This implies $n\leq m$. Then for any
prime divisor $r$ of $|T|$ we have $|M|_{r}\geq|T|^{m}\geq r^{m}\geq r^{n}$,
again contradicting the condition of the lemma. ∎

Let $G$ be a $2$-transitive group of degree $n$ not containing
$\mathrm{A}_{n}$. Pyber [23] gave the bound $|G|\leq n^{c\log^{2}n}$ for some
constant $c>0$, and in particular, $|G|\leq n^{8\lceil 4\log n\rceil\log n}$
if $n\geq 400$, where $\log n$ means logarithm to the base $2$. His proof does
not rely on the classification of finite simple groups (CFSG for short), and
he noted that one can prove $|G|\lessapprox n^{(1+o(1))\log n}$ by using CFSG.
In the following lemma we shall prove that $|G|<n(n-1)2^{n-4}$. Computation
shows that

$n(n-1)2^{n-4}<n^{8\lceil 4\log n\rceil\log n}\text{ for }n<168840.$

Thus if we use Pyber’s result, we would still need to investigate
$2$-transitive groups of degree less than $168840$. Hence for the convenience
of the proof we shall use CFSG. Then the list of $2$-transitive groups can be
found in [3, Tables 7.3 and 7.4]. Note that a $2$-transitive group is affine
or almost simple.

######  Lemma 2.10.

Let $G$ be a $2$-transitive permutation group with degree $n\geq 11$ not
containing $\mathrm{A}_{n}$. Then $|G|<n(n-1)2^{n-4}$.

###### Proof.

Computation shows that $n^{1.5\log n}<n(n-1)2^{n-4}$ for $n\geq 31$. Hence for
$n\geq 31$ it suffices to show $|G|\leq n^{1.5\log n}$. By the classification
of $2$-transitive groups, it is straightforward to verify the conclusion for
$n\leq 30$. Thus we assume $n\geq 31$ for the rest of the proof.

Case 1: $G$ is affine. Then $G\leq\mathrm{AGL}_{d}(p)$ for some prime $p$ and
integer $d$ such that $n=p^{d}$, and so $d+1=\log_{p}n+1\leq\log n+1<1.5\log
n$. Consequently,

$|G|\leq|\mathrm{AGL}_{d}(p)|=p^{d}\prod_{i=0}^{d-1}(p^{d}-p^{i})<p^{d}\prod_{i=0}^{d-1}p^{d}=p^{d(d+1)}=n^{d+1}<n^{1.5\log
n}.$

Case 2: $\mathrm{Soc}(G)=\mathrm{PSL}_{d}(q)$ with $n=(q^{d}-1)/(q-1)$, where
$d\geq 2$ and $q=p^{e}$ for some prime $p$ and integer $e$. If $d=2$, then

$|G|\leq e(d,q-1)|\mathrm{PSL}(2,q)|=q(q+1)(q-1)e\leq
q(q+1)2^{q-3}=n(n-1)2^{n-4}.$

Now let $d\geq 3$. Then $|G|\leq 2e(d,q-1)|\mathrm{PSL}_{d}(q)|$. The
candidates of the pair $(d,q)$ such that $31\leq n\leq 63$ are $(5,2)$,
$(6,2)$, $(4,3)$, $(3,5)$, $(3,7)$, and direct calculation shows that the
lemma is true for these candidates. For $n\geq 64$, since $n>q^{d-1}$, we have
$d+2<\log_{q}n+3\leq\log n+3\leq 1.5\log n$ and
$q^{(d-1)(d+2)}<n^{d+2}<n^{1.5\log n}$, which leads to

$|G|\leq
2eq^{d-1}\prod_{i=0}^{d-2}(q^{d}-q^{i})<2eq^{d-1}\prod_{i=0}^{d-2}q^{d}=2eq^{d^{2}-1}\leq
q^{d^{2}}<q^{(d-1)(d+2)}<n^{1.5\log n}.$

Case 3: $G=\mathrm{Sp}_{2d}(2)$ with $n=2^{2d-1}+2^{d-1}$ or
$2^{2d-1}-2^{d-1}$, where $d\geq 3$. If $d=3$, then $G=\mathrm{Sp}_{6}(2)$,
and so $|G|<31^{1.5\log 31}\leq n^{1.5\log n}$ as $n\geq 31$. If $d\geq 4$,
then $n\geq 2^{2d-1}-2^{d-1}>2^{2d-2}$ and $1.5\log n>3d-3>d+2$, which implies
that

$|G|=2^{d^{2}}\prod_{i=1}^{d}(2^{2i}-1)<2^{d^{2}}\prod_{i=1}^{d}2^{2i}=2^{d(2d+1)}\leq
2^{(2d-2)\cdot(d+2)}<n^{d+2}<n^{1.5\log n}.$

Case 4: $\mathrm{Soc}(G)=\mathrm{PSU}_{3}(q)$ with $n=q^{3}+1$, where
$q=p^{e}\geq 4$ for some prime $p$ and integer $e$. In this case, we have

$|G|\leq
2(3,q+1)e|\mathrm{PSU}_{3}(q)|=2eq^{3}(q^{3}+1)(q^{2}-1)<q^{9}<(q^{3}+1)^{3}=n^{3}<n^{1.5\log
n}.$

Case 5: $\mathrm{Soc}(G)=\mathrm{Sz}(q)$ with $n=q^{2}+1$, where
$q=2^{2e+1}\geq 8$ for some integer $e$. Then

$|G|\leq|\mathrm{Sz}(q)|(2e+1)=(2e+1)q^{2}(q^{2}+1)(q-1)<q^{6}<(q^{2}+1)^{3}=n^{3}<n^{1.5\log
n}.$

Case 6: $\mathrm{Soc}(G)=\mathrm{Ree}(q)$ and $n=q^{3}+1$, where
$q=3^{2e+1}\geq 27$ for some integer $e$. In this case,

$|G|\leq|\mathrm{Ree}(q)|(2e+1)=(2e+1)q^{3}(q^{3}+1)(q-1)<q^{8}<(q^{3}+1)^{3}=n^{3}<n^{1.5\log
n}.$

For $G$ not in any of Cases 1–6, by the classification of $2$-transitive
groups, there are finitely many candidates for $G$. For these candidates, one
can directly verify the conclusion of the lemma. ∎

##  3\. Vertex-quasiprimitive case

In this section we prove the following proposition.

######  Proposition 3.1.

Let $\Gamma$ be a connected $(G,2)$-arc-transitive graph, let $\alpha$ be a
vertex of $\Gamma$. Suppose that $G$ is quasiprimitive on $V(\Gamma)$ and has
a vertex-transitive subgroup $H=\mathrm{A}_{n}$ with $n\geq 5$. Then $G$ is
almost simple, and one of the following holds:

  1. (a)

$\mathrm{Soc}(G)=H$;

  2. (b)

$\mathrm{Soc}(G)=\mathrm{A}_{n+1}$, and $\mathrm{Soc}(G)_{\alpha}$ is a
transitive subgroup of $\mathrm{A}_{n+1}$;

  3. (c)

$\mathrm{Soc}(G)=\mathrm{A}_{n+2}$, $\mathrm{Soc}(G)_{\alpha}$ is a
$2$-transitive subgroup of $\mathrm{A}_{n+2}$, the action of $G_{\alpha}$ on
$\Gamma(\alpha)$ is faithful, and $\Gamma$ is $(\mathrm{Soc}(G),2)$-arc-
transitive;

  4. (d)

$\Gamma=\mathsf{K}_{m}$, $\mathrm{Soc}(G)=\mathrm{A}_{m}$ such that $m$ is the
index of a subgroup of $\mathrm{A}_{n}$, and
$\mathrm{Soc}(G)_{\alpha}=\mathrm{A}_{m-1}$;

  5. (e)

$\Gamma=\mathsf{K}_{p+1}$, $H=\mathrm{A}_{5}$, $G=\mathrm{PGL}_{2}(p)$ with
$p\in\\{11,19,29,59\\}$, and $G_{\alpha}=\mathrm{AGL}_{1}(p)$;

  6. (f)

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

Let $\Gamma$, $G$, $\alpha$ and $H$ be as in the assumption of Proposition
3.1, and let $\beta\in\Gamma(\alpha)$. Then $G$ satisfies Proposition 2.7.

######  Lemma 3.2.

The pair $(G,H)$ does not satisfy part _(b)_ or _(c)_ of Proposition 2.7.

###### Proof.

Recall that the O’Nan-Scott-Praeger Theorem divides quasiprimitive permutation
groups into eight types, see [22, Section 5]. Since $\Gamma$ is $(G,2)$-arc-
transitive and $G$ is quasiprimitive on $V(\Gamma)$, it is shown in [21] (see
also [22, Theorem 6.1]) that $G$ has type holomorph affine, almost simple,
twisted wreath product or product action.

If $(G,H)$ satisfies part (b) of Proposition 2.7, then $G$ is of type
holomorph simple or simple diagonal, a contradiction.

Suppose that $(G,H)$ satisfies part (c) of Proposition 2.7. Then
$G\leq\mathrm{S}_{6}\wr\mathrm{S}_{2}$ acts primitively on $V(\Gamma)$ by
product action with $\mathrm{Soc}(G)=\mathrm{A}_{6}\times\mathrm{A}_{6}$ and
$|V(\Gamma)|=36$. Hence

$(\mathrm{A}_{5}\times\mathrm{A}_{5}).2\leq
G_{\alpha}\leq\mathrm{S}_{5}\wr\mathrm{S}_{2},$

and $G_{\alpha}$ has a unique minimal normal subgroup
$\mathrm{A}_{5}\times\mathrm{A}_{5}$. Since $G_{\alpha}^{\Gamma(\alpha)}$ is a
homomorphic image of $G_{\alpha}$, it follows that either
$G_{\alpha}^{\Gamma(\alpha)}\cong G_{\alpha}$ or $G_{\alpha}^{\Gamma(\alpha)}$
is solvable. However, Lemma 2.3 implies that $G_{\alpha}^{\Gamma(\alpha)}$ is
nonsolvable. Thus $G_{\alpha}^{\Gamma(\alpha)}\cong G_{\alpha}$, from which we
deduce that $G_{\alpha}^{\Gamma(\alpha)}$ is not $2$-transitive, contradicting
the $(G,2)$-arc-transitivity of $\Gamma$. ∎

Now assume that $(G,H)$ satisfies part (a) of Proposition 2.7, so that $G$ is
almost simple. Let $L=\mathrm{Soc}(G)$. Then according to Proposition 2.7,
either $L=H$, or $L=HL_{\alpha}$ satisfies one of the following:

  1. (a.1)

$L=\mathrm{A}_{n+k}$ with $1\leq k\leq 5$, and $L_{\alpha}$ is $k$-transitive
on $n+k$ points;

  2. (a.2)

$L=\mathrm{A}_{m}$ and $L_{\alpha}=\mathrm{A}_{m-1}$, where $m$ is the index
of a subgroup in $A_{n}$;

  3. (a.3)

$(L,n,L_{\alpha})$ lies in Table 1.

It is clear that case (a.2) leads to part (d) of Proposition 3.1. Thus we only
need to deal with cases (a.1) and (a.3). These two cases will be treated in
the following two subsections, respectively, after the next lemma.

######  Lemma 3.3.

If $L_{\alpha}$ is nonsolvable, then $\Gamma$ is $(L,2)$-arc-transitive.

###### Proof.

Since $G$ is quasiprimitive on $V(\Gamma)$, the normal subgroup $L$ is
transitive on $V(\Gamma)$. Suppose for a contradiction that $\Gamma$ is not
$(L,2)$-arc-transitive. Then $L_{\alpha}^{\Gamma(\alpha)}$ is not
$2$-transitive, and $L<G$. As $\Gamma$ is $(G,2)$-arc-transitive and
$L_{\alpha}$ is nonsolvable, [18, Corollary 1.2] asserts that
$G_{\alpha}^{\Gamma(\alpha)}$ is one of the groups:

$\mathrm{P\Gamma L}_{2}(8),\ 19^{2}{:}(9\times\mathrm{SL}_{2}(5)),\
29^{2}{:}(7\times\mathrm{SL}_{2}(5)),\ 29^{2}{:}(28\circ\mathrm{SL}_{2}(5)),\
59^{2}{:}(29\times\mathrm{SL}_{2}(5)).$ (2)

Since $L$ is normal in $G$, we see that $L_{\alpha}^{\Gamma(\alpha)}$ is
normal in $G_{\alpha}^{\Gamma(\alpha)}$.

Suppose that $L$ is not an alternating group. Then $(L,n,L_{\alpha})$ lies in
Table 1. As $L_{\alpha}$ has a homomorphic image $L_{\alpha}^{\Gamma(\alpha)}$
that is a normal subgroup of one of the groups in (2), inspecting the
candidates for $L_{\alpha}$ in Table 1 shows that $L=\mathrm{Sp}_{6}(2)$ and
$L_{\alpha}=\mathrm{PSL}_{2}(8)$ or $\mathrm{PSL}_{2}(8){:}3$. However, in
this case we have $\mathrm{Out}(L)=1$, which implies that $G=L$, a
contradiction.

Thus $L$ is an alternating group. Since $L$ is transitive on $V(\Gamma)$, we
have $|G|/|G_{\alpha}|=|V(\Gamma)|=|L|/|L_{\alpha}|$ and hence
$|G_{\alpha}/L_{\alpha}|=|G/L|\in\\{2,4\\}$. As a consequence,
$|G_{\alpha}^{\Gamma(\alpha)}/L_{\alpha}^{\Gamma(\alpha)}|\in\\{1,2,4\\}$.
Moreover, since $L_{\alpha}^{\Gamma(\alpha)}$ is not $2$-transitive, we have
$L_{\alpha}^{\Gamma(\alpha)}\neq G_{\alpha}^{\Gamma(\alpha)}$. Therefore,
$|G_{\alpha}^{\Gamma(\alpha)}/L_{\alpha}^{\Gamma(\alpha)}|=2$ or $4$. Now
$G_{\alpha}^{\Gamma(\alpha)}$ is one of the above five groups with a normal
subgroup $L_{\alpha}^{\Gamma(\alpha)}$ of index $2$ or $4$. The only
possibility is that
$G_{\alpha}^{\Gamma(\alpha)}=29^{2}{:}(28\circ\mathrm{SL}_{2}(5))$ and
$L_{\alpha}^{\Gamma(\alpha)}=29^{2}{:}(7\times\mathrm{SL}_{2}(5))$. However,
this implies that $L_{\alpha}^{\Gamma(\alpha)}$ is $2$-transitive, a
contradiction. ∎

###  3.1. Case (a.1)

######  Lemma 3.4.

Let $L=\mathrm{A}_{n+k}$ with $2\leq k\leq 5$. If $L_{\alpha}$ is an affine
$k$-transitive subgroup of $L$, then the action of $G_{\alpha}$ on
$\Gamma(\alpha)$ is faithful.

###### Proof.

Suppose for a contradiction that $L_{\alpha}$ is an affine $k$-transitive
subgroup of $L$ while $G_{\alpha}$ is not faithful on $\Gamma(\alpha)$. Then
$G_{\alpha}$ is an affine $k$-transitive subgroup of $G$ with
$\mathrm{Soc}(G_{\alpha})=\mathrm{C}_{p}^{m}$ for some prime $p$ and integer
$m$, and $G_{\alpha}^{[1]}\trianglerighteq\mathrm{Soc}(G_{\alpha})$. Since
$G_{\alpha}^{\Gamma(\alpha)}\cong G_{\alpha}/G_{\alpha}^{[1]}$, it follows
that $G_{\alpha}^{\Gamma(\alpha)}$ is a quotient of
$G_{\alpha}/\mathrm{Soc}(G_{\alpha})$.

Suppose $G_{\alpha\beta}^{[1]}=1$. Then by Theorem 2.2(a), the group
$G_{\alpha}^{[1]}$ is isomorphic to a normal subgroup of
$G_{\alpha\beta}^{\Gamma(\alpha)}$. In particular, $|G_{\alpha}^{[1]}|$
divides $|G_{\alpha\beta}^{\Gamma(\alpha)}|$ and thus divides
$|G_{\alpha}^{\Gamma(\alpha)}|$, which implies that $p^{2m}$ divides
$|G_{\alpha}^{[1]}.G_{\alpha}^{\Gamma(\alpha)}|=|G_{\alpha}|$. Checking the
candidates (for example, in [3, Table 7.3]) of affine $2$-transitive groups
$G_{\alpha}$ for the condition that $|\mathrm{Soc}(G_{\alpha})|^{2}=p^{2m}$
divides $|G_{\alpha}|$, we obtain the following possibilities:

  1. (i)

$G_{\alpha}/\mathrm{Soc}(G_{\alpha})\trianglerighteq\mathrm{SL}_{d}(q)$ with
$d\geq 2$ and $q^{d}=p^{m}$;

  2. (ii)

$G_{\alpha}/\mathrm{Soc}(G_{\alpha})\trianglerighteq\mathrm{Sp}_{2d}(q)$ with
$d\geq 2$ and $q^{2d}=p^{m}$;

  3. (iii)

$G_{\alpha}/\mathrm{Soc}(G_{\alpha})\trianglerighteq\mathrm{G}_{2}(q)$ with
$q^{6}=p^{m}$.

Notice that $G_{\alpha}^{\Gamma(\alpha)}$ is $2$-transitive and is a quotient
group of $G_{\alpha}/\mathrm{Soc}(G_{\alpha})$. We see that (iii) is not
possible because there is no 2-transitive permutation group with socle
$\mathrm{G}_{2}(q)$. Moreover, if (i) occurs, then
$G_{\alpha}^{\Gamma(\alpha)}$ is almost simple with socle
$\mathrm{PSL}_{d}(q)$, and the largest normal $p$-subgroup of
$G_{\alpha\beta}^{\Gamma(\alpha)}$ is of order $q^{d-1}$, contradicting the
condition that $G_{\alpha}^{[1]}$ is isomorphic to a normal subgroup of
$G_{\alpha\beta}^{\Gamma(\alpha)}$ with
$G_{\alpha}^{[1]}\trianglerighteq\mathrm{Soc}(G_{\alpha})=\mathrm{C}_{p}^{m}$.
Now assume (ii) occurs. Then the 2-transitivity of
$G_{\alpha}^{\Gamma(\alpha)}$ implies that it is almost simple with socle
$\mathrm{Sp}_{2d}(q)$, where $q=2$, and $G_{\alpha\beta}^{\Gamma(\alpha)}$ is
almost simple with socle $\mathrm{P\Omega}_{2d}^{+}(2)$ or
$\mathrm{P\Omega}_{2d}^{-}(2)$. This also contradicts the condition that
$G_{\alpha}^{[1]}$ is isomorphic to a normal subgroup of
$G_{\alpha\beta}^{\Gamma(\alpha)}$ with
$G_{\alpha}^{[1]}\trianglerighteq\mathrm{Soc}(G_{\alpha})=\mathrm{C}_{p}^{m}$.

Therefore, $G_{\alpha\beta}^{[1]}\neq 1$. By Theorem 2.2, there exist prime
$r$ and integers $e\geq 2$ and $f\geq 1$ such that $G_{\alpha\beta}^{[1]}$ is
a nontrivial $r$-group,
$G_{\alpha}^{\Gamma(\alpha)}\trianglerighteq\mathrm{PSL}_{e}(r^{f})$ and
$\mathsf{Val}(\Gamma)=(r^{ef}-1)/(r^{f}-1)$. If $r\neq p$, then

$\mathrm{C}_{p}^{m}=\mathrm{Soc}(G_{\alpha})\cong\mathrm{Soc}(G_{\alpha})G_{\alpha\beta}^{[1]}/G_{\alpha\beta}^{[1]}\leq
G_{\alpha}^{[1]}/G_{\alpha\beta}^{[1]}=(G_{\alpha}^{[1]})^{\Gamma(\beta)}\trianglelefteq(G_{\beta})^{\Gamma(\beta)}\cong(G_{\alpha})^{\Gamma(\alpha)},$

and so $p^{2m}$ divides
$|G_{\alpha}^{[1]}.G_{\alpha}^{\Gamma(\alpha)}|=|G_{\alpha}|$, which is not
possible by the same argument as in the previous paragraph. Hence $r=p$, and
so $\mathrm{Soc}(G_{\alpha}^{\Gamma(\alpha)})=\mathrm{PSL}_{e}(p^{f})$. Then
since $G_{\alpha}^{\Gamma(\alpha)}$ is a quotient of
$G_{\alpha}/\mathrm{Soc}(G_{\alpha})$, the only possibility for the affine
$2$-transitive group $G_{\alpha}$ is

$\mathrm{ASL}_{e}(p^{f})\trianglelefteq G_{\alpha}\leq\mathrm{A\Gamma
L}_{e}(p^{f})\ \text{ with }\ ef=m.$ (3)

Suppose $(e,p^{f})\in\\{(2,3),(2,4),(3,2),(3,3)\\}$. Then viewing (3) and
$G=G_{\alpha}\mathrm{A}_{n}$ we have

$\displaystyle(G,G_{\alpha})\in\\{$
$\displaystyle(\mathrm{A}_{9},\mathrm{ASL}_{2}(3)),\,(\mathrm{S}_{9},\mathrm{AGL}_{2}(3)),\,(\mathrm{A}_{16},\mathrm{ASL}_{2}(4)),$
$\displaystyle(\mathrm{A}_{16},\mathrm{AGL}_{2}(4)),\,(\mathrm{A}_{16},\mathrm{A\Sigma
L}_{2}(4)),\,(\mathrm{A}_{16},\mathrm{A\Gamma L}_{2}(4)),$
$\displaystyle(\mathrm{A}_{8},\mathrm{ASL}_{3}(2)),\,(\mathrm{A}_{27},\mathrm{ASL}_{3}(3)),\,(\mathrm{S}_{27},\mathrm{AGL}_{3}(3))\\}.$

Moreover, $G_{\alpha\beta}$ is a maximal subgroup of $G_{\alpha}$ such that
the action of $G_{\alpha}$ on $[G_{\alpha}:G_{\alpha\beta}]$ is
$2$-transitive. However, for all the candidates computation in Magma [2] shows
that $\langle\mathbf{N}_{G}(G_{\alpha\beta}),G_{\alpha}\rangle<G$,
contradicting the connectivity of $\Gamma$.

Thus $(e,p^{f})\notin\\{(2,3),(2,4),(3,2),(3,3)\\}$. Let $V$ be an
$e$-dimensional vector space over $\mathbb{F}_{p^{f}}$ such that
$\mathrm{Alt}(V)\leq G\leq\mathrm{Sym}(V)$ and $\mathrm{ASL}(V)\trianglelefteq
G_{\alpha}\leq\mathrm{A\Gamma L}(V)$, and let $N=\mathrm{Soc}(G_{\alpha})$.
Since $G_{\alpha\beta}\geq G_{\alpha}^{[1]}\geq N$, we have
$G_{\alpha\beta}=N{:}(G_{\alpha\beta})_{0}$. Moreover, $(G_{\alpha\beta})_{0}$
is the stabilizer in $(G_{\alpha})_{0}$ of a $1$-dimensional subspace or a
$(e-1)$-dimensional subspace of $V$, as the action of $G_{\alpha}$ on
$[G_{\alpha}:G_{\alpha\beta}]$ is $2$-transitive. Hence
$(G_{\alpha\beta})_{0}=(P{:}Q).\mathcal{O}$ with $P=\mathrm{C}_{p}^{(e-1)f}$,
$Q=\mathrm{GL}_{e-1}(p^{f})$ and
$\mathcal{O}=(G_{\alpha})_{0}/\mathrm{SL}(V)\leq\mathrm{\Gamma
L}(V)/\mathrm{SL}(V)$.

By Lemma 2.5, there exits $g\in\mathbf{N}_{G}(G_{\alpha\beta})$ such that
$\Gamma\cong\mathrm{Cos}(G,G_{\alpha},g)$ and $\langle G_{a},g\rangle=G$.
Since $N\trianglelefteq G_{\alpha\beta}$ and
$g\in\mathbf{N}_{G}(G_{\alpha\beta})$, we have $N^{g}\trianglelefteq
G_{\alpha\beta}$. Then $NN^{g}\leqslant NP$, as $NP$ is the largest normal
$p$-subgroup of $G_{\alpha\beta}$. If $N^{g}=N$, then $N$ is normal in
$\langle G_{\alpha},g\rangle=G$, contradicting that $G$ is $\mathrm{Alt}(V)$
or $\mathrm{Sym}(V)$. Hence $N^{g}\neq N$, and so $NN^{g}/N$ is a nontrivial
normal subgroup of $G_{\alpha\beta}/N\cong(P{:}Q).\mathcal{O}$. Since $P$ is a
minimal normal subgroup of $(P{:}Q).\mathcal{O}$, it follows that
$NN^{g}/N\cong P$, which yields $NN^{g}=NP$. For $h\in\mathrm{GL}(V)$ and
$w\in V$, define $t_{h,w}\in\mathrm{AGL}(V)$ by letting

$t_{h,w}\colon v\mapsto v^{h}+w\quad\text{for }\,v\in V,$

and identify $t_{h,0}$ with $h$. For each $h\in P$, since $t_{h,0}\in
NP=NN^{g}$, there exists $x\in N$ such that $t_{h,0}x\in N^{g}$, and so there
exists $w\in V$ such that $t_{h,w}=t_{h,0}t_{1,w}\in N^{g}$.

Case 1: $(G_{\alpha\beta})_{0}$ is the stabilizer in $(G_{\alpha})_{0}$ of a
$(e-1)$-dimensional subspace of $V$. In this case, $PQ=\mathrm{SL}(V)_{U}$ for
some $(e-1)$-dimensional subspace $U$ of $V$. Let $v_{1},v_{2},\dots,v_{e}$ be
a basis of $V$ such that $U=\langle v_{2},\dots,v_{e}\rangle$. Since
$|N|=|N^{g}|=(p^{f})^{e}$ and $|NN^{g}|=|NP|=(p^{f})^{2e-1}$, we have $|N\cap
N^{g}|=p^{f}$. Moreover, since both $N$ and $N^{g}$ are normal in
$G_{\alpha\beta}$, their intersection $N\cap N^{g}$ is normalized by $PQ$.
Then since the orbits of $PQ=\mathrm{SL}(V)_{U}$ on $V\setminus\\{0\\}$ are
$U\setminus\\{0\\}$ and $V\setminus U$, we conclude that $N\cap
N^{g}=\\{t_{1,u}\mid u\in U\\}$ with $e=2$. From $p^{ef}=p^{m}=n+k\geq n+2\geq
7$ and $(e,p^{f})\notin\\{(2,3),(2,4)\\}$ we deduce that $p^{f}\geq 5$. Hence
there exists $c\in\mathbb{F}_{p^{f}}^{\times}$ with $c\neq c^{-2}$. Let
$h,\ell\in\mathrm{SL}(V)$ defined by

$v_{1}^{h}=v_{1}+v_{2},\quad v_{2}^{h}=v_{2},\quad v_{1}^{\ell}=cv_{1},\quad
v_{2}^{\ell}=c^{-1}v_{2}.$

Then $h\in P$ and $\ell\in Q$. As shown above, there exists $w\in V$ such that
$t_{h,w}\in N^{g}$. Write $w=av_{1}+bv_{2}$ with $a,b\in\mathbb{F}_{p^{f}}$.
Since $N^{g}$ is normalized by $Q$, we have
$t_{\ell,0}^{-1}t_{h,w}t_{\ell,0}\in N^{g}$. Then since $N^{g}$ is abelian, we
obtain
$t_{\ell,0}^{-1}t_{h,w}t_{\ell,0}t_{h,w}=t_{h,w}t_{\ell,0}^{-1}t_{h,w}t_{\ell,0}$,
which implies that

$(ac+a)v_{1}+(ac+bc^{-1}+b)v_{2}=0^{t_{\ell,0}^{-1}t_{h,w}t_{\ell,0}t_{h,w}}=0^{t_{h,w}t_{\ell,0}^{-1}t_{h,w}t_{\ell,0}}=(ac+a)v_{1}+(ac^{-2}+bc^{-1}+b)v_{2}$

by direct calculation. Since $c\neq c^{-2}$, this yields $a=0$, that is, $w\in
U$. Consequently, $t_{1,w}\in\\{t_{1,u}\mid u\in U\\}=N\cap N^{g}$, and so
$t_{h,0}=t_{h,w}t_{1,w}^{-1}\in N^{g}$, contradicting the fact that $N^{g}$ is
regular on $V$.

Case 2: $(G_{\alpha\beta})_{0}$ is the stabilizer in $(G_{\alpha})_{0}$ of a
$1$-dimensional subspace of $V$. In this case, $PQ=\mathrm{SL}(V)_{\langle
v_{1}\rangle}$ for some $v_{1}\in V$. Extend $v_{1}$ to a basis
$v_{1},v_{2},\dots,v_{e}$ of $V$. Then the linear transformation $h$ defined
by

$v_{2}^{h}=v_{1}+v_{2}\quad\text{and}\quad v_{i}^{h}=v_{i}\ \text{ for }\
i\in\\{1,3,4,\dots,e\\}$

lies in $P$, and so there exists $w\in V$ such that $t_{h,w}\in N^{g}$. Then
for any $\ell\in(PQ)_{w}$, since

$t_{h\ell h^{-1}\ell^{-1},0}=t_{h,w}t_{\ell
h\ell^{-1},w}^{-1}=t_{h,w}(t_{\ell,0}t_{h,w}^{-1}t_{\ell,0}^{-1})\in N^{g}$

and $N^{g}$ acts regularly on $V$, we obtain $h\ell h^{-1}\ell^{-1}=1$, that
is, $\ell\in\mathbf{C}_{PQ}(h)$. Hence

$(PQ)_{w}\leq\mathbf{C}_{PQ}(h).$ (4)

If there exists $z\in\mathrm{Fix}(h)\setminus\langle v_{1},v_{2},w\rangle$,
then the linear transformation $\ell$ such that $v_{1}^{\ell}=v_{1}$,
$w^{\ell}=w$ and $z^{\ell}=v_{2}+z$ satisfies $\ell\in\mathrm{SL}(V)_{\langle
v_{1}\rangle}=PQ$, $w^{\ell}=w$ and $z^{h\ell}=v_{2}+z\neq
v_{1}+v_{2}+z=z^{\ell h}$, contradicting (4). Thus $\mathrm{Fix}(h)\leq\langle
v_{1},v_{2},w\rangle$. As $v_{2}\notin\mathrm{Fix}(h)$, we obtain

$\mathrm{Fix}(h)<\langle v_{1},v_{2},w\rangle.$ (5)

In particular, $e-1=\dim(\mathrm{Fix}(h))<3$, which means $e<4$. The
possibility of $e=2$ has already been ruled out in Case 1. Therefore, $e=3$.
Then it follows from $(e,p^{f})\notin\\{(3,2),(3,3)\\}$ that $p^{f}\geq 4$.
Hence there exists $a\in\mathbb{F}_{p^{f}}^{\times}$ with $a\neq a^{-1}$.
Since $\mathrm{Fix}(h)=\langle v_{1},v_{3}\rangle$, we derive from (5) that
$w\notin\langle v_{1}\rangle$ and $v_{2}\notin\langle v_{1},w\rangle$. Let
$\ell$ be the transformation defined by $v_{1}^{k}=av_{1}$, $w^{k}=w$ and
$v_{2}^{k}=a^{-1}v_{2}$. Then $\ell\in\mathrm{SL}(V)_{\langle
v_{1}\rangle}=PQ$, $w^{\ell}=w$, and $v_{2}^{h\ell}=av_{1}+a^{-1}v_{2}\neq
a^{-1}v_{1}+a^{-1}v_{2}=v_{2}^{\ell h}$. This contradicts (4). ∎

######  Lemma 3.5.

Let $L=\mathrm{A}_{n+k}$ with $1\leq k\leq 5$. If $L_{\alpha}$ is
$k$-transitive on $n+k$ points, then $k\leq 2$.

###### Proof.

Let $\Omega=\\{1,\dots,n+k\\}$ be the set that $L=\mathrm{A}_{n+k}$ naturally
acts on. Suppose for a contradiction that $k\geq 3$. Then $L_{\alpha}$ is a
$3$-transitive subgroup of $\mathrm{A}_{n+k}$, and so by [3, Table 7.4], the
triple $(L_{\alpha},L_{\alpha 1},n+k)$ lies in Table 2, where $L_{\alpha 1}$
is the stabilizer of the point $1\in\Omega$ in $L_{\alpha}$. In particular,
$L_{\alpha}$ is nonsolvable.

$\begin{array}[]{lllll}\hline\cr\text{Row}&L_{\alpha}&L_{\alpha
1}&n+k&\text{Remarks}\\\ \hline\cr
1&\mathrm{AGL}_{d}(2)&\mathrm{GL}_{d}(2)&2^{d}&d\geq 3\\\
2&\mathrm{M}_{11}&\mathrm{M}_{10}&11&\text{sharply $4$-transitive}\\\
3&\mathrm{M}_{11}&\mathrm{PSL}_{2}(11)&12&\\\
4&\mathrm{M}_{12}&\mathrm{M}_{11}&12&\text{sharply $5$-transitive}\\\
5&\mathrm{M}_{22}&\mathrm{PSL}_{3}(4)&22&\\\
6&\mathrm{M}_{23}&\mathrm{M}_{22}&23&\text{$4$-transitive}\\\
7&\mathrm{M}_{24}&\mathrm{M}_{23}&24&\text{$5$-transitive}\\\
8&\mathrm{PSL}_{2}(p^{f}).\mathcal{O}&\Big{(}p^{f}{:}\frac{p^{f}-1}{(2,p-1)}\Big{)}.\mathcal{O}&p^{f}+1&p\text{
prime, }(2,p-1)\leq\mathcal{O}\leq(2,p-1)\times f,\\\ &&&&\mathcal{O}\text{ is
not Frobenius if }p>2\\\ \hline\cr\end{array}$ Table 2. $3$-transitive
permutation groups on $n+k$ points

Since $L_{\alpha}$ is $2$-transitive on $\Omega$, it follows from Lemma 2.8
that

$\mathbf{N}_{L}(L_{\alpha 1})=\mathbf{N}_{L_{1}}(L_{\alpha 1}),$ (6)

where $L_{1}$ is the stabilizer of $1\in\Omega$ in $L$. Take
$\beta\in\Gamma(\alpha)$. As $L_{\alpha}$ is nonsolvable, Lemma 3.3 asserts
that $\Gamma$ is $(L,2)$-arc-transitive. Hence $L_{\alpha}$ is $2$-transitive
on $\Gamma(\alpha)$, which can be identified with
$[L_{\alpha}:L_{\alpha\beta}]$. In particular, $L$ has a suborbit of length
$|L_{\alpha}|/|L_{\alpha\beta}|=\mathsf{Val}(\Gamma)$ on $[L:L_{\alpha}]$. Let
$M=\mathbf{N}_{L}(L_{\alpha\beta})$. Then by Lemma 2.5 we have $\langle
L_{\alpha},M\rangle=L$. In the following we consider each row of Table 2
separately.

Row 1. For this row we have $L=\mathrm{A}_{2^{d}}$,
$L_{\alpha}=\mathrm{AGL}_{d}(2)$ and $L_{\alpha
1}=\mathrm{GL}_{d}(2)\cong\mathrm{PSL}_{d}(2)$, where $d\geq 3$. Then Lemma
3.4 asserts that $L_{\alpha}$ acts faithfully on $\Gamma(\alpha)$. Since
$L_{\alpha}=\mathrm{AGL}_{d}(2)$ has a unique faithful $2$-transitive
permutation presentation, it follows that $L_{\alpha\beta}$ is conjugate to
$L_{\alpha 1}$ in $L_{\alpha}$, that is, $L_{\alpha\beta}=(L_{\alpha 1})^{x}$
for some $x\in L_{\alpha}$. Since $L_{\alpha 1}\cong\mathrm{PSL}_{d}(2)$ is
$2$-transitive on $\Omega\setminus\\{1\\}$ and $\mathrm{PSL}_{d}(2)$ is the
only $2$-transitive permutation group with socle $\mathrm{PSL}_{d}(2)$, we
have $\mathbf{N}_{L_{1}}(L_{\alpha 1})=L_{\alpha 1}$. This together with (6)
leads to $\mathbf{N}_{L}(L_{\alpha 1})=L_{\alpha 1}$. Hence

$M=\mathbf{N}_{L}(L_{\alpha\beta})=\mathbf{N}_{L}((L_{\alpha
1})^{x})=(\mathbf{N}_{L}(L_{\alpha 1}))^{x}=(L_{\alpha 1})^{x}\leq
L_{\alpha},$

which implies that $\langle L_{\alpha},M\rangle=L_{\alpha}<L$, a
contradiction.

Row 2. Then $L=\mathrm{A}_{11}$ and $L_{\alpha}=\mathrm{M}_{11}$. Since the
$2$-transitive permutation representations of $L_{\alpha}=\mathrm{M}_{11}$
have degree $11$ or $12$, we have $\mathsf{Val}(\Gamma)=11$ or $12$. However,
computation in Magma [2] shows that $L=\mathrm{A}_{11}$ acting on
$[\mathrm{A}_{11}:\mathrm{M}_{11}]$ has no suborbit of length $11$ or $12$, a
contradiction.

Row 3. Then $L=\mathrm{A}_{12}$ and $L_{\alpha}=\mathrm{M}_{11}$. For the same
reason as Row 2, we have $\mathsf{Val}(\Gamma)=11$ or $12$. Then since
computation in Magma [2] shows that $L=\mathrm{A}_{12}$ has no suborbit of
length $12$ on $[\mathrm{A}_{12}:\mathrm{M}_{11}]$, we obtain
$|L_{\alpha}|/|L_{\alpha\beta}|=\mathsf{Val}(\Gamma)=11$. Thus
$L_{\alpha\beta}=\mathrm{M}_{10}$. It then follows from the Atlas [4] that
$M=\mathbf{N}_{L}(L_{\alpha\beta})=\mathrm{M}_{10}.2$. However, this yields
$\langle L_{\alpha},M\rangle=\mathrm{M}_{12}$, contradicting the condition
$\langle L_{\alpha},M\rangle=L$.

Row 4. For this row we have $L=\mathrm{A}_{12}$ and
$L_{\alpha}=\mathrm{M}_{12}$. Then
$|L_{\alpha}|/|L_{\alpha\beta}|=\mathsf{Val}(\Gamma)=12$ as the $2$-transitive
permutation representations of $\mathrm{M}_{12}$ have degree $12$. However,
computation in Magma [2] shows that $L=\mathrm{A}_{12}$ has no suborbit of
length $12$ on $[\mathrm{A}_{12}:\mathrm{M}_{12}]$, a contradiction.

Row 5. Then $L=\mathrm{A}_{22}$ and $L_{\alpha}=\mathrm{M}_{22}$. Since the
unique $2$-transitive permutation representation of
$L_{\alpha}=\mathrm{M}_{22}$ is of degree $22$, we have
$\mathsf{Val}(\Gamma)=22$ and $L_{\alpha\beta}=\mathrm{PSL}_{3}(4)$. Then
computation in Magma [2] shows that
$M=\mathbf{N}_{L}(L_{\alpha\beta})=L_{\alpha\beta}.3$. This implies that every
$2$-element $g$ of $M$ lies in $L_{\alpha\beta}$ and hence $\langle
L_{\alpha},g\rangle=L_{\alpha}<L$, contradicting Lemma 2.5.

Row 6. Then $L=\mathrm{A}_{23}$ and $L_{\alpha}=\mathrm{M}_{23}$. Since the
unique $2$-transitive permutation representation of
$L_{\alpha}=\mathrm{M}_{23}$ is of degree $23$, we have
$\mathsf{Val}(\Gamma)=23$ and $L_{\alpha\beta}=\mathrm{M}_{22}$. Then
computation in Magma [2] shows that
$M=\mathbf{N}_{L}(L_{\alpha\beta})=L_{\alpha\beta}$, which implies $\langle
L_{\alpha},M\rangle=L_{\alpha}<L$, a contradiction.

Row 7. Then $L=\mathrm{A}_{24}$ and $L_{\alpha}=\mathrm{M}_{24}$. Similarly as
Row 6, we derive that $\mathsf{Val}(\Gamma)=24$ and
$L_{\alpha\beta}=\mathrm{M}_{23}$, and then computation in Magma [2] shows
$M=L_{\alpha\beta}$, so $\langle L_{\alpha},M\rangle=L_{\alpha}<L$, giving a
contradiction.

Row 8. Then $L_{\alpha}$ is an almost simple $3$-transitive group on $\Omega$
with socle $\mathrm{PSL}_{2}(p^{f})$. Since $L_{\alpha}$ is $2$-transitive on
$[L_{\alpha}:L_{\alpha\beta}]$, the classification of $2$-transitive groups
shows that either $L_{\alpha\beta}$ is conjugate to $L_{\alpha 1}$ in
$L_{\alpha}$, or
$(L_{\alpha},L_{\alpha\beta})=(\mathrm{PSL}_{2}(11),\mathrm{A}_{5})$ or
$(\mathrm{P\Gamma L}_{2}(8),9{:}6)$. As $L_{\alpha}$ is $3$-transitive group
on $\Omega$, we have $L_{\alpha}\neq\mathrm{PSL}_{2}(11)$. If
$(L_{\alpha},L_{\alpha\beta})=(\mathrm{P\Gamma L}_{2}(8),9{:}6)$, then
computation in Magma [2] shows that
$M=\mathbf{N}_{L}(L_{\alpha\beta})=L_{\alpha\beta}$, which implies $\langle
L_{\alpha},M\rangle=L_{\alpha}<L$, a contradiction.

Thus $L_{\alpha\beta}$ is conjugate to $L_{\alpha 1}$ in $L_{\alpha}$, that
is, $L_{\alpha\beta}=(L_{\alpha 1})^{x}$ for some $x\in L_{\alpha}$. Let
$K=\mathrm{P\Gamma L}_{2}(p^{f})$ be an overgroup of $L_{\alpha}$ in
$\mathrm{Sym}(\Omega)=\mathrm{S}_{n+k}$. Then the stabilizer
$K_{1}=\mathrm{A\Gamma L}_{1}(p^{f})$ is $2$-transitive on
$\Omega\setminus\\{1\\}$. Also, $L_{\alpha 1}$ is $2$-transitive on
$\Omega\setminus\\{1\\}$, and so is its overgroup
$\mathbf{N}_{L_{1}}(L_{\alpha 1})$. Now $\mathbf{N}_{L_{1}}(L_{\alpha 1})$ is
a $2$-transitive group on $\Omega\setminus\\{1\\}$ and has a normal
$2$-transitive subgroup $L_{\alpha 1}$ on $\Omega\setminus\\{1\\}$ with
$L_{\alpha 1}\leq K_{1}=\mathrm{A\Gamma L}_{1}(p^{f})$. We conclude from the
classification of $2$-transitive groups that $\mathbf{N}_{L_{1}}(L_{\alpha
1})\leq K_{1}=\mathrm{A\Gamma L}_{1}(p^{f})$. This together with (6) implies
that $(\mathbf{N}_{L}(L_{\alpha 1}))^{x}=(\mathbf{N}_{L_{1}}(L_{\alpha
1}))^{x}\leq(K_{1})^{x}\leq K$ as $x\in L_{\alpha}\leq K$. Hence

$\langle L_{\alpha},M\rangle=\langle
L_{\alpha},\mathbf{N}_{L}(L_{\alpha\beta})\rangle=\langle
L_{\alpha},\mathbf{N}_{L}((L_{\alpha 1})^{x})\rangle=\langle
L_{\alpha},(\mathbf{N}_{L}(L_{\alpha 1}))^{x}\rangle\leq K,$

contradicting the condition $\langle L_{\alpha},M\rangle=L$. ∎

By the above lemma, we have either $k=1$ or $k=2$. If $k=1$ then part (b) of
Proposition 3.1 holds, and for $k=2$, to complete the statement of part (c) of
Proposition 3.1, we remain to show the following:

######  Lemma 3.6.

Let $L=\mathrm{A}_{n+2}$. If $L_{\alpha}$ is $2$-transitive on $n+2$ points,
then $\Gamma$ is $(L,2)$-arc-transitive.

###### Proof.

Recall $L_{\alpha}$ is either almost simple or affine. The lemma is true for
the case $L_{\alpha}$ is almost simple by Lemma 3.3. So we assume that
$L_{\alpha}$ is an affine 2-transitive subgroup of $\mathrm{A}_{n+2}$. Then
$G_{\alpha}$ is an affine 2-transitive subgroup of $\mathrm{S}_{n+2}$. By
Lemma 3.4, $G_{\alpha}$ acts faithfully on $\Gamma(\alpha)$. Thus
$G_{\alpha}^{\Gamma(\alpha)}$ is equivalent to the $2$-transitive permutation
representation of $G_{\alpha}$ naturally on $n+2$ points. This implies
$L_{\alpha}^{\Gamma(\alpha)}$ is 2-transitive as $L_{\alpha}$ is 2-transitive
on $n+2$ points. Thus $\Gamma$ is $(L,2)$-arc-transitive. ∎

###  3.2. Case (a.3)

Next we consider $(L,n,L_{\alpha})$ in Table 1, as in case (a.3). We deal with
Row 21 of Table 1 separately in the following lemma.

######  Lemma 3.7.

The triple $(L,n,L_{\alpha})$ cannot lie in Row $21$ of Table 1.

###### Proof.

Suppose for a contradiction that $L=\Omega_{8}^{+}(2)$, $n=9$ and
$2^{4}{:}\mathrm{A}_{5}\leq L_{\alpha}\leq 2^{6}{:}\mathrm{A}_{8}$, as in Row
21 of Table 1. Then computation in Magma [2] shows that there are 36
candidates for $L_{\alpha}$ (up to $L$-conjugate) giving rise to a
factorization $L=\mathrm{A}_{n}L_{\alpha}$. In Table 3, we list these
candidates together with the lengths of their orbits on
$V(\Gamma)\setminus\\{\alpha\\}$, where each row corresponds to two isomorphic
conjugacy classes, and the expression $64^{1}70^{1}$ (in the last row) etc.
means that there are exactly 1 orbit with length 64 and 1 orbit with length 70
of $L_{\alpha}$ in the corresponding row etc.

$\begin{array}[]{ll}\hline\cr L_{\alpha}&\text{Orbits on
}V(\Gamma)\setminus\\{\alpha\\}\\\ \hline\cr
2^{4}{:}\mathrm{A}_{5}&1^{23}10^{132}16^{96}80^{441}160^{72}240^{1}320^{216}480^{98}960^{16}\\\
2^{4}{:}\mathrm{S}_{5}&1^{1}2^{5}10^{6}16^{4}20^{30}32^{22}40^{1}80^{24}120^{1}160^{114}240^{6}320^{32}480^{22}640^{43}960^{18}1920^{1}\\\
2^{4}{:}3{:}\mathrm{A}_{5}&1^{1}3^{2}10^{2}16^{2}30^{14}48^{10}80^{1}160^{2}240^{49}320^{1}480^{10}960^{26}1440^{10}2880^{1}\\\
2^{4}{:}3{:}\mathrm{S}_{5}&3^{1}10^{1}16^{1}30^{5}40^{1}48^{3}60^{1}96^{1}120^{1}160^{2}240^{16}480^{15}720^{4}960^{4}1440^{4}1920^{3}\\\
2^{5}{:}\mathrm{A}_{5}&1^{3}2^{4}5^{4}10^{16}20^{24}32^{24}40^{9}80^{24}120^{1}160^{100}240^{2}320^{20}480^{16}640^{52}960^{16}1920^{4}\\\
2^{5}{:}\mathrm{S}_{5}&1^{1}4^{1}5^{2}20^{9}32^{4}40^{14}60^{1}64^{4}80^{5}160^{12}240^{9}320^{28}480^{4}640^{15}960^{8}1280^{6}1920^{2}\\\
2^{5}{}^{\boldsymbol{.}}\mathrm{S}_{5}&1^{1}4^{1}5^{2}20^{5}40^{8}60^{1}64^{6}80^{1}160^{6}240^{1}320^{27}480^{2}640^{4}960^{3}1280^{13}1920^{4}3840^{1}\\\
2^{6}{:}\mathrm{A}_{5}&1^{5}5^{18}20^{9}40^{6}60^{1}64^{6}120^{2}160^{18}320^{27}480^{6}640^{9}960^{9}1280^{9}3840^{1}\\\
\text{[}2^{6}\text{]}{:}\mathrm{S}_{5}&2^{1}5^{1}10^{5}20^{2}30^{1}40^{2}64^{1}80^{3}120^{1}128^{1}160^{5}240^{2}320^{6}480^{3}640^{8}960^{2}1280^{3}1920^{2}2560^{1}\\\
2^{6}{:}3{:}\mathrm{A}_{5}&1^{1}15^{2}20^{1}40^{4}60^{1}64^{2}480^{4}960^{3}1280^{1}1920^{1}2880^{1}3840^{1}\\\
2^{6}{:}3{:}\mathrm{S}_{5}&10^{1}15^{1}30^{1}40^{2}64^{1}240^{4}480^{1}640^{1}960^{2}1440^{1}1920^{1}\\\
\text{[}2^{5}\text{]}{:}\mathrm{A}_{6}&1^{3}12^{4}32^{8}60^{5}120^{4}192^{16}480^{8}640^{4}720^{1}960^{4}\\\
2^{5}{:}\mathrm{S}_{6}&1^{1}24^{1}30^{5}32^{4}240^{1}360^{1}384^{4}480^{8}640^{2}\\\
\text{[}2^{5}\text{]}{}^{\boldsymbol{.}}\mathrm{S}_{6}&1^{1}24^{1}30^{1}64^{2}120^{3}360^{1}384^{4}640^{2}960^{2}1920^{1}\\\
\text{[}2^{6}\text{]}{:}\mathrm{A}_{6}&1^{1}6^{4}30^{1}64^{2}120^{1}240^{1}360^{1}384^{4}480^{4}640^{2}1920^{1}\\\
\text{[}2^{6}\text{]}{:}\mathrm{S}_{6}&12^{1}15^{1}60^{1}64^{1}120^{1}180^{1}480^{2}640^{1}768^{1}960^{1}\\\
2^{6}{:}\mathrm{A}_{7}&7^{1}64^{1}280^{2}448^{1}\\\
2^{6}{:}\mathrm{A}_{8}&64^{1}70^{1}\\\ \hline\cr\end{array}$ Table 3.
Candidates for $L_{\alpha}$ in Row $21$ of Table 1

Since $L_{\alpha}$ is insolvable, Lemma 3.3 asserts that $\Gamma$ is
$(L,2)$-arc-transitive, and so $L_{\alpha}^{\Gamma(\alpha)}$ is
$2$-transitive. Thus $L_{\alpha}^{\Gamma(\alpha)}$ is affine or almost simple.

First assume that $L_{\alpha}^{\Gamma(\alpha)}$ is affine. Inspecting Table 3
we see that the homomorphic image $L_{\alpha}^{\Gamma(\alpha)}$ of
$L_{\alpha}$ has socle a $2$-group, and has a unique nonsolvable composition
factor, which is one of $\mathrm{A}_{5}$, $\mathrm{A}_{6}$, $\mathrm{A}_{7}$
or $\mathrm{A}_{8}$. By the classification of $2$-transitive groups, a
$2$-transitive affine group with socle a $2$-group and the nonsolvable
composition factor $\mathrm{A}_{5}$, $\mathrm{A}_{6}$, $\mathrm{A}_{7}$ or
$\mathrm{A}_{8}$ is one of the following:

$2^{4}{:}\mathrm{A}_{6},\ 2^{4}{:}\mathrm{S}_{6},\ 2^{4}{:}\mathrm{A}_{7},\ \
2^{4}{:}\mathrm{SL}_{2}(4),\ 2^{4}{:}\mathrm{GL}_{2}(4),\
2^{4}{:}\mathrm{\Gamma L}_{2}(4),\ \ 2^{4}{:}\mathrm{SL}_{4}(2),\ \
2^{4}{:}\mathrm{Sp}_{4}(2).$

Since these permutation groups all have degree $16$, we conclude that
$|\Gamma(\alpha)|=16$. Thus $L_{\alpha}$ has an orbit of length $16$. Then by
Table 3, the possibilities for $L_{\alpha}$ are:

$2^{4}{:}\mathrm{A}_{5},\ \ 2^{4}{:}\mathrm{S}_{5},\ \
2^{4}{:}3{:}\mathrm{A}_{5},\ \ 2^{4}{:}3{:}\mathrm{S}_{5}.$

However, for each possible $L_{\alpha}$ and each orbit of length $16$,
computation in Magma [2] shows that the action of $L_{\alpha}$ on this orbit
is not $2$-transitive, contradicting the $2$-transitivity of
$L_{\alpha}^{\Gamma(\alpha)}$.

Next assume that $L_{\alpha}^{\Gamma(\alpha)}$ is almost simple. Then from
Table 3 we see that $\mathrm{Soc}(L_{\alpha}^{\Gamma(\alpha)})=\mathrm{A}_{t}$
with $t\in\\{5,6,7,8\\}$. By the classification of almost simple
$2$-transitive groups, either $|\Gamma(\alpha)|=t$ or
$(t,|\Gamma(\alpha)|)\in\\{(5,6),(7,15),(8,15)\\}$. As $L_{\alpha}$ has an
orbit of length $|\Gamma(\alpha)|$, by checking Table 3 we obtain the
following possibilities for $(L_{\alpha},|\Gamma(\alpha)|)$:

$(2^{5}{:}\mathrm{A}_{5},5)\ \ (2^{5}{:}\mathrm{S}_{5},5),\ \
(2^{5}{}^{\boldsymbol{.}}\mathrm{S}_{5},5)\ \ (2^{6}{:}\mathrm{A}_{5},5)\ \
([2^{6}]{:}\mathrm{S}_{5},5),\ \ ([2^{6}]{:}\mathrm{A}_{6},6),\ \
(2^{6}{:}\mathrm{A}_{7},7).$

However, for each possible $L_{\alpha}$ in these pairs of
$(L_{\alpha},|\Gamma(\alpha)|)$ and for each orbit of length
$|\Gamma(\alpha)|$, computation in Magma [2] shows that the action of
$L_{\alpha}$ on this orbit is not $2$-transitive, contradicting the
$2$-transitivity of $L_{\alpha}^{\Gamma(\alpha)}$. ∎

We are now in a position to completely determine $(G,H,\Gamma)$ for case
(a.3), which will finish the proof of Proposition 3.1.

######  Lemma 3.8.

Let $(L,n,L_{\alpha})$ be in Table 1. Then $n=5$, and one of the following
holds:

  1. (a)

$G=\mathrm{M}_{12}$, $G_{\alpha}=\mathrm{M}_{11}$ and
$\Gamma=\mathsf{K}_{12}$;

  2. (b)

$G=\mathrm{PGL}_{2}(p)$ with $p\in\\{11,19,29,59\\}$,
$G_{\alpha}=\mathrm{AGL}_{1}(p)$ and $\Gamma=\mathsf{K}_{p+1}$.

###### Proof.

By Lemma 3.7, Row 21 of Table 1 cannot occur. Thus we only need to deal with
Rows 1–20 and 22–24 of Table 1.

Row 1. For this row we have $L=\mathrm{A}_{6}$, $n=5$ and
$L_{\alpha}=\mathrm{A}_{4}$ or $\mathrm{S}_{4}$. Then
$\mathrm{A}_{6}\cong\mathrm{PSL}_{2}(9)\leq G\leq\mathrm{P\Gamma L}_{2}(9)$,
and $G=HG_{\alpha}$ with $H=\mathrm{A}_{5}$ and $G_{\alpha}\cap
L=\mathrm{A}_{4}$ or $\mathrm{S}_{4}$. Searching for such a group
factorization in Magma [2] shows that $(G,G_{\alpha})$ is the one of the
following pairs:

$(\mathrm{A}_{6},\,\mathrm{A}_{4}),\ \ (\mathrm{A}_{6},\,\mathrm{S}_{4}),\ \
(\mathrm{S}_{6},\,\mathrm{S}_{4}),\ \ (\mathrm{S}_{6},\,\mathrm{A}_{4}\times
2),\ \ (\mathrm{S}_{6},\,\mathrm{S}_{4}\times 2).$

By Lemma 2.6, the first two candidates for $(G,G_{\alpha})$ are not possible.
Hence $G=\mathrm{S}_{6}$ and $G_{\alpha}=\mathrm{S}_{4}$,
$\mathrm{A}_{4}\times 2$ or $\mathrm{S}_{4}\times 2$. Then since the action of
$G_{\alpha}$ on $[G_{\alpha}:G_{\alpha\beta}]$ is $2$-transitive, we conclude
that $|G_{\alpha}|/|G_{\alpha\beta}|=3$ or $4$. For the three possibilities of
$G_{\alpha}$, computation in Magma [2] shows that $G=\mathrm{S}_{6}$ has a
suborbit of length $3$ or $4$ only if $G_{\alpha}=\mathrm{S}_{4}$ and the
suborbit length is $4$. However, in this case computation in Magma [2] shows
that there is no element $g\in G$ satisfying:

$g\in\mathbf{N}_{G}(G_{\alpha}\cap G_{\alpha}^{g}),\ \ g^{2}\in G_{\alpha},\ \
\langle G_{\alpha},g\rangle=G,\ \ |G_{\alpha}|/|G_{\alpha}\cap
G_{\alpha}^{g}|=4.$

This contradicts Lemma 2.5.

Rows 2–3. For these two rows we have $L=\mathrm{A}_{m}$ with $m\in\\{10,15\\}$
and $L_{\alpha}=\mathrm{A}_{m-2}$ or $\mathrm{S}_{m-2}$. Since $L_{\alpha}$ is
insolvable, Lemma 3.3 asserts that $\Gamma$ is $(L,2)$-arc-transitive. This
contradicts Lemma 2.6.

Row 4. Here $L=\mathrm{M}_{12}$, $n=5$ and $L_{\alpha}=\mathrm{M}_{11}$. Then
$L$ is 2-transitive on $V(\Gamma)|$ with $|V(\Gamma)|=|L|/|L_{\alpha}|=12$, it
follows that $\Gamma=\mathsf{K}_{12}$. According to [4],
$\mathrm{Out}(\mathrm{M}_{12})=2$, and $\mathrm{Aut}(\mathrm{M}_{12})$ has no
subgroup of index $12$. Hence $G\neq\mathrm{Aut}(\mathrm{M}_{12})$, and so
$G=\mathrm{M}_{12}$. This leads to part (a) of the lemma.

Rows 5–8. Here $L=\mathrm{PSL}_{2}(p)$ with $p\in\\{11,19,29,59\\}$, $n=5$ and
$p\leq L_{\alpha}\leq p{:}(p-1)/2$. Since $G_{\alpha}$ is $2$-transitive on
$\Gamma(\alpha)$, we have $G_{\alpha}=\mathrm{AGL}_{1}(p)=p{:}(p-1)$ and
$G=\mathrm{PGL}_{2}(p)$. Hence $|\Gamma(\alpha)|=p$ and
$|V(\Gamma)|=|G|/|G_{\alpha}|=|\mathrm{PGL}_{2}(p)|/|\mathrm{AGL}_{1}(p)|=p+1$.
This implies that $\Gamma\cong\mathsf{K}_{p+1}$, leading to part (b) of the
lemma.

Row 9. For this row, $L=\mathrm{PSL}_{4}(3)$ and
$L_{\alpha}=3^{3}{:}\mathrm{PSL}_{3}(3)$. Then $L$ and $G$ are $2$-transitive
but not 3-transitive on $V(\Gamma)$, which implies that $\Gamma$ is a complete
graph. Notice that an automorphism group acting 2-arc-transitively on a
complete graph should be 3-transitive on the vertices, contradicting that $G$
is not 3-transitive on $V(\Gamma)$.

Row 10. For this row, $L=\mathrm{PSU}_{3}(5)$ and
$L_{\alpha}=5_{+}^{1+2}{:}8$. Then $L$ and $G$ are $2$-transitive but not
3-transitive on $V(\Gamma)$, hence $\Gamma$ is a complete graph, a
contradiction occurs by the same reason as in Row 9.

Row 11. Here $L=\mathrm{PSp}_{4}(3)$ and
$L_{\alpha}=3_{+}^{1+2}{:}\mathrm{Q}_{8}$ or $3_{+}^{1+2}{:}2.\mathrm{A}_{4}$.
Since $\mathrm{Out}(L)=2$, we have $G_{\alpha}=L_{\alpha}$ or $L_{\alpha}.2$.
Then since $G_{\alpha}$ is $2$-transitive on $\Gamma(\alpha)$, we see that
$(G,G_{\alpha},|\Gamma(\alpha)|)$ is one of the following triples:

$(\mathrm{PSp}_{4}(3),\,3_{+}^{1+2}{:}\mathrm{Q}_{8},\,9),\ \
(\mathrm{PSp}_{4}(3),\,3_{+}^{1+2}{:}2.\mathrm{A}_{4},\,4),$
$(\mathrm{PSp}_{4}(3).2,\,3_{+}^{1+2}{:}\mathrm{Q}_{8}.2,\,9),\ \
(\mathrm{PSp}_{4}(3).2,\,3_{+}^{1+2}{:}2.\mathrm{A}_{4}.2,\,4).$

However, for each of the possible triples $(G,G_{\alpha},|\Gamma(\alpha)|)$,
computation in Magma [2] shows that $G$ acting on $[G:G_{\alpha}]$ has no
suborbit of length $|\Gamma(\alpha)|$, a contradiction.

Rows 12–13. For these two rows, $L=\mathrm{Sp}_{6}(2)$, and $L_{\alpha}$ is
one of the following groups:

$\mathrm{PSU}_{3}(3){:}2,\ \ 3_{+}^{1+2}{:}8{:}2,\ \
3_{+}^{1+2}{:}2{.}\mathrm{S}_{4},\ \ \mathrm{PSL}_{2}(8),\ \
\mathrm{PSL}_{2}(8){:}3,\ \ \mathrm{PSU}_{4}(2){:}2.$

Since $\mathrm{Out}(L)=1$, we have $G=L$, and so $\Gamma$ is $(L,2)$-arc-
transitive. Then since $G_{\alpha}$ is $2$-transitive on $\Gamma(\alpha)$, we
see that $(G_{\alpha},|\Gamma(\alpha)|)$ is one of the following pairs:

$(\mathrm{PSU}_{3}(3){:}2,\,28),\ \ (3_{+}^{1+2}{:}8{:}2,\,9),\ \
(3_{+}^{1+2}{:}2{.}\mathrm{S}_{4},\,9),\ \
(3_{+}^{1+2}{:}2{.}\mathrm{S}_{4},\,4),$
$(3_{+}^{1+2}{:}2{.}\mathrm{S}_{4},\,3),\ \ (\mathrm{PSL}_{2}(8),\,9),\ \
(\mathrm{PSL}_{2}(8){:}3,\,9),\ \ (\mathrm{PSL}_{2}(8){:}3,\,28).$

However, for each of the possible pairs $(G_{\alpha},|\Gamma(\alpha)|)$,
computation in Magma [2] shows that $G$ acting on $[G:G_{\alpha}]$ has no
suborbit of length $|\Gamma(\alpha)|$, a contradiction.

Row 14. For this row, $L_{\alpha}=\mathrm{SO}_{8}^{-}(2)$ is nonsolvable, and
so $\Gamma$ is $(L,2)$-arc-transitive by Lemma 3.3. However,
$\mathrm{SO}_{8}^{-}(2)$ has no $2$-transitive permutation representation, a
contradiction.

Rows 15–16. For these two rows, $L=\Omega_{7}(3)$, and $L_{\alpha}$ is one of
the following groups:

$3^{3+3}{:}\mathrm{PSL}_{3}(3),\ \ 3^{3}{:}\mathrm{PSL}_{3}(3),\ \
\mathrm{PSL}_{4}(3),\ \ \mathrm{PSL}_{4}(3){:}2,\ \ \mathrm{G}_{2}(3).$

In particular, $L_{\alpha}$ is nonsolvable. Hence Lemma 3.3 implies that
$\Gamma$ is $(L,2)$-arc-transitive. Then as $L_{\alpha}$ is $2$-transitive on
$\Gamma(\alpha)$, we see that $(L_{\alpha},|\Gamma(\alpha)|)$ is one of the
following pairs:

$(3^{3+3}{:}\mathrm{PSL}_{3}(3),\,27),\ \
(3^{3+3}{:}\mathrm{PSL}_{3}(3),\,13),\ \ (3^{3}{:}\mathrm{PSL}_{3}(3),\,27),$
$(3^{3}{:}\mathrm{PSL}_{3}(3),\,13),\ \ (\mathrm{PSL}_{4}(3),\,40),\ \
(\mathrm{PSL}_{4}(3){:}2,\,40).$

However, for each of the possible pairs $(L_{\alpha},|\Gamma(\alpha)|)$,
computation in Magma [2] shows that there is no subgroup $K$ of $L_{\alpha}$
with $|L_{\alpha}|/|K|=|\Gamma(\alpha)|$ and $\langle
L_{\alpha},\mathbf{N}_{L}(K)\rangle=L$, contradicting Lemma 2.5.

Rows 17–20. Here $L=\Omega_{8}^{+}(2)$, and $L_{\alpha}$ is one of the
following groups:

$\mathrm{Sp}_{6}(2),\ \ \mathrm{A}_{9},\ \ \mathrm{PSU}_{4}(2),\ \
\mathrm{PSU}_{4}(2){:}2,\ \ 3\times\mathrm{PSU}_{4}(2),\ \
(3\times\mathrm{PSU}_{4}(2)){:}2,$ $2^{4}{:}15.4,\ \ 2^{6}{:}15,\ \
2^{6}{:}15.2,\ \ 2^{6}{:}15.4,\ \ \mathrm{A}_{8},\ \ \mathrm{S}_{8}.$

Suppose that $L_{\alpha}$ is nonsolvable. Then $\Gamma$ is $(L,2)$-arc-
transitive by Lemma 3.3, and so $L_{\alpha}$ is $2$-transitive on
$\Gamma(\alpha)$, and $L_{\alpha}^{\Gamma(\alpha)}$ is almost simple by Lemma
2.3, which implies that $(L_{\alpha},|\Gamma(\alpha)|)$ is one of the
following pairs:

$(\mathrm{Sp}_{6}(2),\,28),\ \ (\mathrm{Sp}_{6}(2),\,36),\ \
(\mathrm{A}_{9},\,9),\ \ (\mathrm{A}_{8},\,8),\ \ (\mathrm{A}_{8},\,15),\ \
(\mathrm{S}_{8},\,8).$

However, for each of these pairs $(L_{\alpha},|\Gamma(\alpha)|)$, computation
in Magma [2] shows that $L$ acting on $[L:L_{\alpha}]$ has no suborbit of
length $|\Gamma(\alpha)|$, a contradiction. Thus $L_{\alpha}$ is solvable,
that is, $L_{\alpha}=2^{4}{:}15.4$, $2^{6}{:}15$, $2^{6}{:}15.2$ or
$2^{6}{:}15.4$. These candidates for $L_{\alpha}$ only occur in Row 20 of
Table 1, where it is shown that $n=9$. As $L\leq G\leq\mathrm{Aut}(L)$ and
$G_{\alpha}\cap
L=L_{\alpha}\in\\{2^{4}{:}15.4,2^{6}{:}15,2^{6}{:}15.2,2^{6}{:}15.4\\}$,
computation in Magma [2] shows that such pairs $(G,G_{\alpha})$ that give rise
to a factorization $G=\mathrm{A}_{9}G_{\alpha}$ are the following:

$(\Omega_{8}^{+}(2),\,2^{4}{:}15.4),\ \ (\Omega_{8}^{+}(2),\,2^{6}{:}15),\ \
(\Omega_{8}^{+}(2),\,2^{6}{:}15.2),\ \ (\Omega_{8}^{+}(2),\,2^{6}{:}15.4),$
$(\mathrm{O}_{8}^{+}(2),\,(2^{4}{:}15.4){:}2),\ \
(\mathrm{O}_{8}^{+}(2),\,(2^{6}{:}15){:}2),\ \
(\mathrm{O}_{8}^{+}(2),\,(2^{6}{:}15.2).2),\ \
(\mathrm{O}_{8}^{+}(2),\,(2^{6}{:}15.4){:}2).$

Then as $G_{\alpha}$ is $2$-transitive on $\Gamma(\alpha)$, we conclude that
$|\Gamma(\alpha)|=16$. However, for each of the above pairs $(G,G_{\alpha})$,
computation in Magma [2] shows that $G$ acting on $[G:G_{\alpha}]$ has no
suborbit of length $16$ on which $G_{\alpha}$ acts $2$-transitively, a
contradiction.

Row 22. For this row, $L=\mathrm{P\Omega}_{8}^{+}(3)$, and
$L_{\alpha}=3^{6}{:}\mathrm{PSL}_{4}(3)$ is nonsolvable. Hence $\Gamma$ is
$(L,2)$-arc-transitive by Lemma 3.3. This implies that $L_{\alpha}$ is
$2$-transitive on $\Gamma(\alpha)$, and so $|\Gamma(\alpha)|=40$. However,
computation in Magma [2] shows that $L=\mathrm{P\Omega}_{8}^{+}(3)$ acting on
$[L:L_{\alpha}]$ has no suborbit of length $40$, a contradiction.

Rows 23–24. Here $L_{\alpha}=\Omega_{7}(3)$ or $2^{8}{:}\Omega_{8}^{-}(2)$.
Then Lemma 3.3 asserts that $\Gamma$ is $(L,2)$-arc-transitive since
$L_{\alpha}$ is nonsolvable. However, neither $\Omega_{7}(3)$ nor
$2^{8}{:}\Omega_{8}^{-}(2)$ has a $2$-transitive permutation representation, a
contradiction. ∎

##  4\. Proof of Theorem 1.1

Let $\Gamma$ be a connected $(G,2)$-arc-transitive Cayley graph on an
alternating group $H\cong\mathrm{A}_{n}$ with $H\leq G$ and $n\geq 5$, and let
$\alpha$ be a vertex of $\Gamma$. Let $M$ be a maximal intransitive normal
subgroup of $G$. Suppose that $H$ is not normal in $G$. We divides the proof
into several lemmas.

######  Lemma 4.1.

If $M=1$, then part (a), (b) or (c) of Theorem 1.1 holds.

###### Proof.

Since $M=1$, the group $G$ is quasiprimitive on $V(\Gamma)$, and so $\Gamma$
is described in Proposition 3.1. It follows that $G$ is almost simple, and one
of (b)–(f) of Proposition 3.1 occurs as $H$ is not normal in $G$ by our
assumption. Let $L=\mathrm{Soc}(G)$. Since $H$ is nonabelian simple and
$H/(H\cap L)\cong HL/L\leq G/L\leq\mathrm{Out}(L)$ is solvable, we deduce
$H\cap L=H$, that is, $H\leq L$. Note that both $L$ and $G$ are transitive on
$V(\Gamma)$. Since $H$ is regular on $V(\Gamma)$, we have
$|L|=|H||L_{\alpha}|$ and $|G|=|H||G_{\alpha}|$.

First assume that case (b) of Proposition 3.1 occurs. Then
$L=\mathrm{A}_{n+1}$, and $L_{\alpha}$ is a transitive subgroup of
$\mathrm{A}_{n+1}$. Since $|L_{\alpha}|=|L|/|H|=n+1$, we see that $L_{\alpha}$
is a regular subgroup of $\mathrm{A}_{n+1}$, as in part (a) of Theorem 1.1.

Next assume that case (c) of Proposition 3.1 occurs. Then $\Gamma$ is
$(L,2)$-arc-transitive with $L=\mathrm{A}_{n+2}$, and $L_{\alpha}$ is a
$2$-transitive subgroup of $\mathrm{A}_{n+2}$ acting faithfully on
$\Gamma(\alpha)$. Since $|L_{\alpha}|=|L|/|H|=(n+2)(n+1)$, it follows that
$L_{\alpha}$ is a sharply $2$-transitive subgroup of $\mathrm{A}_{n+2}$. By
[13, Chapter XII: Theorem 9.1], sharply $2$-transitive groups are affine
groups with degree a prime power. Since $L_{\alpha}$ acts faithfully on
$\Gamma(\alpha)$ and $L_{\alpha}$ has only one $2$-transitive permutation
representation, we derive that $\mathsf{Val}(\Gamma)=n+2$ is a prime power.
Hence part (b) of Theorem 1.1 holds (with $N=L$).

Now assume that case (d) of Proposition 3.1 occurs. Then $L=\mathrm{A}_{m}$,
where $m$ is the index of a subgroup of $\mathrm{A}_{n}$, and
$L_{\alpha}=\mathrm{A}_{m-1}$. It follows that
$|\mathrm{A}_{m-1}|=|L_{\alpha}|=|L|/|H|=|\mathrm{A}_{m}|/|\mathrm{A}_{n}|$,
which implies $m=|\mathrm{A}_{n}|$. This lies in part (c) of Theorem 1.1.

Finally, for cases (e) and (f) of Proposition 3.1, since
$|G|=|H||G_{\alpha}|$, computation shows that there is only one possibility,
namely, $\Gamma=\mathsf{K}_{60}$, $H=\mathrm{A}_{5}$, $G=\mathrm{PGL}_{2}(59)$
and $G_{\alpha}=\mathrm{AGL}_{1}(59)$. This also lies in part (c) of Theorem
1.1. ∎

For the rest of this section we assume that $M\neq 1$. If $M$ has exactly two
orbits on $V(\Gamma)$, then $\Gamma$ is a bipartite graph, and so the
stabilizer in $H$ of one part has index $2$ in $H$, which is not possible as
$H\cong\mathrm{A}_{n}$ is simple. Thus $M$ has at least three orbits on
$V(\Gamma)$. Then by Theorem 2.1, $\Gamma_{M}$ is $(G/M,2)$-arc-transitive,
$(G/M)_{v}\cong G_{\alpha}$ for any $v\in V(\Gamma_{M})$, and $\Gamma$ is a
normal cover of $\Gamma_{M}$. Moreover, $M$ is semiregular but not transitive
on $V(\Gamma)$, and so $|M|$ properly divides $|V(\Gamma)|=|H|$. By the
maximality of $M$, $G/M$ is quasiprimitive on $V(\Gamma)$. Since $H$ is
regular on $V(\Gamma)$, the group $HM/M\cong H\cong\mathrm{A}_{n}$ is
transitive on $V(\Gamma_{M})$. Hence the $(G/M,2)$-arc-transitive graph
$\Gamma_{M}$ is described in Proposition 3.1.

######  Lemma 4.2.

$\Gamma_{M}$ does not satisfy (e) or (f) of Proposition 3.1.

###### Proof.

First suppose that $\Gamma_{M}$ satisfies (e) of Proposition 3.1. Then
$\Gamma_{M}=\mathsf{K}_{p+1}$ and $H=\mathrm{A}_{5}$, where
$p\in\\{11,19,29,59\\}$. Then
$|M|=|V(\Gamma)|/|V(\Gamma_{M})|=60/(p+1)\in\\{1,2,3,4,5\\}$. By our
assumption, $|M|\neq 1$. By [8, Theorem 1.1], $|M|\neq 3$ or $5$. If $|M|=2$,
then $p=29$ and by [8, Theorem 1.1],
$\Gamma=\mathsf{K}_{30,30}-30\mathsf{K}_{2}$ is bipartite, which is not
possible as $H$ is simple.

Next suppose that $\Gamma_{M}$ satisfies (f) of Proposition 3.1. Then
$\Gamma_{M}=\mathsf{K}_{12}$ and $H=\mathrm{A}_{5}$. It follows that
$|M|=|V(\Gamma)|/|V(\Gamma_{M})|=60/12=5$ and so $M\cong\mathrm{C}_{5}$. Note
that $\Gamma$ is a normal $M$-cover of $\Gamma_{M}=\mathsf{K}_{12}$, and $G/M$
lifts to a $2$-arc-transitive automorphism group $G$ of $\Gamma$. We conclude
from [8, Theorem 1.1] that such a graph $\Gamma$ dose not exist, a
contradiction. ∎

######  Lemma 4.3.

If $\Gamma_{M}$ satisfies (a), (b) or (d) of Proposition 3.1, then part (d.1),
(d.2) or (d.3) of Theorem 1.1 holds, respectively.

###### Proof.

It is clear that if $\Gamma_{M}$ satisfies (b) or (d) of Proposition 3.1 then
part (d.2) or (d.3) of Theorem 1.1 holds, respectively. Suppose that
$\Gamma_{M}$ satisfies (a) of Proposition 3.1. Then
$\mathrm{Soc}(G/M)=HM/M\cong\mathrm{A}_{n}$. Let $B=HM=M{:}H$. Then
$B/M=\mathrm{Soc}(G/M)$ and so $B\unlhd G$. Since $B\geq H$ is transitive on
$V(\Gamma)$, we have $|M|=|B|/|H|=|B|/|V(\Gamma)|=|B_{\alpha}|$ and
$G=BG_{\alpha}$, which implies that $G/B\cong G_{\alpha}/B_{\alpha}$. Since
$\Gamma$ is a normal cover of $\Gamma_{M}$, it follows that
$|B_{\alpha}|=|(B/M)_{v}|$ divides $|B/M|=|H|=|\mathrm{A}_{n}|$, and

$|M|=|B_{\alpha}|=|(B/M)_{v}|=|(\mathrm{Soc}(G/M))_{v}|.$

Recall that $|M|$ properly divides $|H|=|\mathrm{A}_{n}|$. If $B=M\times H$,
then $H$ is characteristic in $B$ and hence normal in $G$, a contradiction.
Hence $B\neq M\times H$. Then as in the proof of Lemma 2.9, there are two
normal subgroups $M_{j}$ and $M_{j+1}$ of $B$ such that

$1\leq M_{j}<M_{j+1}\leq M,\quad M_{j}H=M_{j}\times H,\quad M_{j+1}H\neq
M_{j+1}\times H,$

$M_{j+1}/M_{j}$ is a minimal normal subgroup of $B/M_{j}$, and
$M_{j}H/M_{j}\cong\mathrm{A}_{n}$ acts faithfully on $M_{j+1}/M_{j}$ by
conjugation.

Suppose that $M_{j+1}/M_{j}\cong T^{m}$ for some nonabelian simple group $T$.
Then $2^{m}\leq|T^{m}|_{2}\leq|M|_{2}<|\mathrm{A}_{n}|_{2}<2^{n}$, and so
$m<n$. Since $M_{j}H/M_{j}\cong\mathrm{A}_{n}$ acts faithfully on
$M_{j+1}/M_{j}$, we have
$\mathrm{A}_{n}\lesssim\mathrm{Aut}(T^{m})=\mathrm{Aut}(T)\wr\mathrm{S}_{m}\cong(T^{m}.\mathrm{Out}(T)^{m}){:}\mathrm{S}_{m}$.
As $\mathrm{Out}(T)$ is solvable and $m<n$, we then conclude that
$\mathrm{A}_{n}\leq T^{m}$. This is a contradiction as
$|T^{m}|\leq|M|<|A_{n}|$.

Thus we conclude that $M_{j+1}/M_{j}\cong\mathrm{C}_{p}^{m}$ for some prime
$p$. Consequently, $\mathrm{A}_{n}\lesssim\mathrm{GL}_{m}(p)$. Since
$|M|=|(\mathrm{Soc}(G/M))_{v}|$, we have
$p^{m}\leq|M|_{p}\leq|(\mathrm{Soc}(G/M))_{v}|_{p}$. Suppose $5\leq n\leq 8$.
Recall that $\Gamma_{M}$ is a connected $(G/M,2)$-arc-transitive graph with
$\mathsf{Val}(\Gamma_{M})=\mathsf{Val}(\Gamma)$ and
$|(G/M)_{v}|=|G_{\alpha}|$. Inspecting the results in [29, 6, 7] regarding
$2$-arc-transitive nonnormal Cayley graphs on nonabelian simple groups of
valency $3$, $4$ and $5$, respectively, we conclude that
$\mathsf{Val}(\Gamma)\geq 6$. With the help of Magma [2], we can easily find
all the pairs $(G/M,(G/M)_{v})$ such that $\mathrm{Soc}(G/M)=\mathrm{A}_{n}$
with $5\leq n\leq 8$ and the action of $G/M$ on $[G/M:(G/M)_{v}]$ admits a
connected non-bipartite $(G/M,2)$-arc-transitive orbital graph of valency at
least $6$. Then for each pair $(G/M,(G/M)_{v})$, we check whether there exist
prime $p$ and integer $m$ such that $p^{m}\leq|\mathrm{Soc}(G/M)_{v}|_{p}$ and
$\mathrm{Soc}(G/M)\lesssim\mathrm{GL}_{m}(p)$. It turns out that no such prime
$p$ and integer $n$ exist for any pair $(G/M,(G/M)_{v})$, a contradiction.

Therefore, $n\geq 9$. By [15, Proposition 5.3.7], $m\geq n-2$. Note that
$(n!)_{p}<p^{n/(p-1)}$ (see [5, Example 2.6.1] for example). If $p\geq 3$,
then $p^{n-2}\leq
p^{m}\leq|M|_{p}\leq|\mathrm{A}_{n}|_{p}<p^{n/(p-1)}<p^{n-2}$, a
contradiction. Hence $p=2$, and $2^{n-2}\leq
2^{m}\leq|M|_{2}\leq|\mathrm{A}_{n}|_{2}$. Note that
$|\mathrm{A}_{n}|_{2}=2^{n-2}$ if $n$ is a $2$-power and
$|\mathrm{A}_{n}|_{2}<2^{n-2}$ otherwise. We conclude that $n\geq 16$ is a
2-power, and $|M|_{2}=|\mathrm{A}_{n}|_{2}$. Thus
$|\mathrm{Soc}(G/M)_{v}|_{2}=|M|_{2}=|\mathrm{A}_{n}|_{2}=|\mathrm{Soc}(G/M)|_{2}$,
and so $\mathrm{Soc}(G/M)_{v}$ contains a Sylow $2$-subgroup of
$\mathrm{Soc}(G/M)$, leading to part (d.1) of Theorem 1.1. ∎

######  Lemma 4.4.

If $\Gamma_{M}$ satisfies (c) of Proposition 3.1, then part (b) of Theorem 1.1
holds.

###### Proof.

Let $X=G/M$ and $Y=HM/M\cong\mathrm{A}_{n}$. Then from Proposition 3.1(c) we
see that $\mathrm{Soc}(X)\cong\mathrm{A}_{n+2}$, $\mathrm{Soc}(X)_{v}$ is
$2$-transitive on $n+2$ points, the action of $(\mathrm{Soc}(X))_{v}$ on
$\Gamma_{M}(v)$ is faithful, and $\Gamma_{M}$ is $(\mathrm{Soc}(X),2)$-arc-
transitive. Since both $X$ and $Y$ are transitive on $V(\Gamma_{M})$, we have
$|X:Y|=|X_{v}:Y_{v}|$, and so

$|M|=\frac{|V(\Gamma)|}{|V(\Gamma_{M})|}=\frac{|Y|}{|V(\Gamma_{M})|}=|Y_{v}|=\frac{|X_{v}|}{|X:Y|}\
\text{ divides }\ \frac{|X_{v}|}{(n+2)(n+1)}.$ (7)

Suppose that $n\leq 8$. As $\mathrm{Soc}(X)_{v}$ is a subgroup of
$\mathrm{A}_{n+2}$ with a $2$-transitive permutation representation of degree
$n+2$, by the classification of $2$-transitive groups, the candidates for
$(n+2,\mathrm{Soc}(X)_{v})$ are

$(7,\mathrm{PSL}_{3}(2)),(8,\mathrm{ASL}_{3}(2)),(9,\mathrm{PSL}_{2}(8)),(9,\mathrm{P\Gamma
L}_{2}(8)),$ $(9,\mathrm{A\Gamma
L}_{1}(9)\cap\mathrm{A}_{9}),(10,\mathrm{PSL}_{2}(9)),(10,\mathrm{M}_{10}).$

For these candidates, computation in Magma [2] shows that such a
$(\mathrm{Soc}(X),2)$-arc-transitive graph $\Gamma_{M}$ does not exist except
when $(n+2,\mathrm{Soc}(X)_{v})=(9,\mathrm{A\Gamma
L}_{1}(9)\cap\mathrm{A}_{9})$. However, for this exception, we have
$(X,X_{v})=(\mathrm{A}_{9},\mathrm{A\Gamma L}_{1}(9)\cap\mathrm{A}_{9})$ or
$(\mathrm{S}_{9},\mathrm{A\Gamma L}_{1}(9))$, which yields that
$|M|=|X_{v}|/|X:Y|=1$, contradicting our assumption.

Therefore, $n\geq 9$. Since $X_{v}$ is has a $2$-transitive permutation
representation of degree $n+2$, we derive from (7) and Lemma 2.10 that
$|M|<2^{n-2}$. Let $B=MH=M{:}H$. Recall that $|M|$ is a proper divisor of
$|H|$. Then $M$ has no section isomorphic to $H=\mathrm{A}_{n}$, and so Lemma
2.9 implies that $B=M\times H$. Let $L$ be a normal subgroup of $G$ containing
$M$ such that $L/M=\mathrm{Soc}(X)\cong\mathrm{A}_{n+2}$, and let
$C=\mathbf{C}_{L}(M)$. Then $H\leq C$ and $C\cap M\leq\mathbf{Z}(C)$. Since
$CM/M\unlhd L/M\cong\mathrm{A}_{n+2}$ and $CM/M\geq HM/M\cong\mathrm{A}_{n}$,
we have $CM/M=L/M$. Hence

$\mathbf{Z}(C)/(C\cap M)\unlhd C/(C\cap M)\cong
CM/M=L/M\cong\mathrm{A}_{n+2}.$

Since $\mathbf{Z}(C)/(C\cap M)$ is abelian, this implies that
$\mathbf{Z}(C)/(C\cap M)=1$. Accordingly,

$C/\mathbf{Z}(C)=C/(C\cap M)\cong\mathrm{A}_{n+2},$

and so

$C/\mathbf{Z}(C)=(C/\mathbf{Z}(C))^{\prime}=C^{\prime}\mathbf{Z}(C)/\mathbf{Z}(C)\cong
C^{{}^{\prime}}/(C^{\prime}\cap\mathbf{Z}(C))=C^{\prime}/\mathbf{Z}(C^{\prime}).$

It follows that $C^{\prime}/\mathbf{Z}(C^{\prime})\cong\mathrm{A}_{n}$ and
$C=C^{\prime}\mathbf{Z}(C)$, which yields
$C^{\prime}=(C^{\prime}\mathbf{Z}(C))^{\prime}=C^{\prime\prime}$. Hence
$C^{\prime}$ is a covering group of $\mathrm{A}_{n+2}$.

Suppose $\mathbf{Z}(C^{\prime})\not=1$. Then as the Schur multiplier of
$\mathrm{A}_{n}$ is $\mathrm{C}_{2}$ (see for instance [15, Theorem 5.1.4]),
we have $\mathbf{Z}(C^{\prime})=\mathrm{C}_{2}$ and $C^{\prime}\cong
2.\mathrm{A}_{n}$. Since $H\leq C$ and $C/C^{\prime}$ is abelian, we have
$H\leq C^{\prime}$. Thus $C^{\prime}$ has a subgroup
$H\times\mathbf{Z}(C^{\prime})\cong\mathrm{A}_{n}\times\mathrm{C}_{2}$.
However, this is not possible by [7, Proposition 2.6].

Therefore, $\mathbf{Z}(C^{\prime})=1$ and so
$C^{\prime}\cong\mathrm{A}_{n+2}$. Since $C^{\prime}$ is characteristic in $C$
and $C$ is normal in $L$, the group $C^{\prime}$ is normal in $L$, and so
$L\geq M\times C^{\prime}$. This together with $L/M\cong\mathrm{A}_{n+2}\cong
C^{\prime}$ gives $L=M\times C^{\prime}$. Since
$|M|<2^{n-2}<(n+2)!/2=|C^{\prime}|$, the group $C^{\prime}$ is characteristic
in $L$ and hence normal in $G$. Since $H/(H\cap C^{\prime})\cong
HC^{\prime}/C^{\prime}\leq C/C^{\prime}$ is abelian and $H=\mathrm{A}_{n}$, we
derive that $H=H\cap C^{\prime}$. Thus $C^{\prime}\geq H$ is transitive on
$V(\Gamma)$. Let $N=C^{\prime}$. Then $N=HN_{\alpha}$ with
$N\cong\mathrm{A}_{n+2}$ and $H=\mathrm{A}_{n}$, and

$N_{\alpha}=\frac{|N|}{|V(\Gamma)|}=\frac{|N|}{|H|}=\frac{|\mathrm{A}_{n+2}|}{|\mathrm{A}_{n}|}=(n+2)(n+1).$

This implies that $N_{\alpha}$ is a sharply $2$-transitive subgroup of
$\mathrm{A}_{n+2}$. By [13, Chapter XII, Theorem 9.1] we then have
$\mathrm{Soc}(N_{\alpha})=\mathrm{C}_{p}^{d}$ and $n+2=p^{d}$ for some prime
$p$ and integer $d$. Since $\mathrm{Soc}(N_{\alpha})$ is characteristic in
$N_{\alpha}$ and $N_{\alpha}\unlhd L_{\alpha}\cong(L/M)_{v}$, the group
$(L/M)_{v}$ has a normal subgroup $\mathrm{C}_{p}^{d}$. Recall that
$\Gamma_{M}$ is $(L/M,2)$-arc-transitive, $(L/M)_{v}$ is a $2$-transitive
subgroup of $L/M=\mathrm{A}_{n+2}$, and $(L/M)_{v}$ acts faithfully on
$\Gamma_{M}(v)$. Thereby we conclude that
$\mathrm{Soc}((L/M)_{v})\cong\mathrm{C}_{p}^{d}$,
$\mathsf{Val}(\Gamma)=\mathsf{Val}(\Gamma_{M})=p^{d}=n+2$, the action of
$N_{\alpha}$ on $\Gamma(\alpha)$ is faithful, and
$N_{\alpha}^{\Gamma(\alpha)}$ is a sharply $2$-transitive permutation group of
degree $p^{d}$. As a consequence, $\Gamma$ is $(N,2)$-arc-transitive, and part
(b) of Theorem 1.1 holds. ∎

##  5\. Proof of Theorem 1.4

Let $q$, $G$, $K$ and $\Gamma$ be as in Construction 1.3, and let
$M=\mathrm{AGL}_{1}(q^{2})$. Then $K=M{:}\langle\tau\rangle$, and the
stabilizer $M_{0}$ of $0\in\mathbb{F}_{q^{2}}$ in $M$ is
$\mathrm{GL}_{1}(q^{2})$ and hence generated by a single element $\omega\in
M_{0}$. Since both $g\tau$ and $\tau g$ fix $0$ and send $v$ to $v^{-q}$ for
$v\in\mathbb{F}_{q^{2}}^{\times}$, we have $g\tau=\tau g$. Moreover,
$\omega^{g}=\omega^{-1}$, and so $g\in\mathbf{N}_{G}(M_{0})$. Thus $g$
normalizes $M_{0}{:}\langle\tau\rangle=K_{0}$, the stabilizer of
$0\in\mathbb{F}_{q^{2}}$ in $K$.

Let $N=\mathbf{N}_{G}(K)$. Then
$\mathrm{Soc}(K)=\mathrm{Soc}(M)\cong\mathbb{F}_{q^{2}}^{+}$ is a normal
subgroup of $N$, and $N$ is $2$-transitive on $\mathbb{F}_{q^{2}}$ as $K\geq
M$ is $2$-transitive on $\mathbb{F}_{q^{2}}$. By the classification of
$2$-transitive groups, this implies that $N\leq\mathrm{A\Gamma L}_{1}(q^{2})$.
Then since $\mathrm{A\Gamma L}_{1}(q^{2})\leq\mathbf{N}_{G}(K)=N$, we obtain
$N=\mathrm{A\Gamma L}_{1}(q^{2})$. Similarly, we have
$\mathbf{N}_{G}(M)=\mathrm{A\Gamma L}_{1}(q^{2})=N$. Write $q=p^{f}$ with
prime $p$ and integer $f$.

Suppose that $g\in N$. Then as $g$ fixes both $0$ and $1$, we have $g\in
N_{0,1}=\mathrm{Gal}(\mathbb{F}_{q^{2}}/\mathbb{F}_{p})$. Since
$\mathrm{Gal}(\mathbb{F}_{q^{2}}/\mathbb{F}_{p})$ is cyclic and $g$ has order
$2$, the same order as
$\tau\in\mathrm{Gal}(\mathbb{F}_{q^{2}}/\mathbb{F}_{p})$, it follows that
$g=\tau$. However, this leads to
$\omega^{-1}=\omega^{g}=\omega^{\tau}=\omega^{q}$, which implies that $\omega$
has order dividing $q+1$, contradicting
$\langle\omega\rangle=M_{0}=\mathrm{GL}_{1}(q^{2})$.

Thus we conclude that $g\notin N$. As a consequence, $K^{g}\neq K$, and so
$K\cap K^{g}<K$. From $g\in\mathbf{N}_{G}(K_{0})$ we deduce that $K\cap
K^{g}\geq K_{0}$. Then since $K_{0}$ is maximal in $K$, we conclude that
$K\cap K^{g}=K_{0}$. Hence the action of $K$ on $[K:K\cap K^{g}]$ is
permutation isomorphic to the natural action of $K$ on $\mathbb{F}_{q^{2}}$ as
a subgroup of $\mathrm{A\Gamma L}_{1}(q^{2})$, which is $2$-transitive. In
particular, $\Gamma$ has valency $|K:K\cap K^{g}|=|K:K_{0}|=q^{2}$.

Since $\tau$ is an involution with $\mathrm{Fix}(\tau)=\mathbb{F}_{q}$, it is
a product of $(q^{2}-q)/2$ transpositions. From the condition $q\equiv
3\pmod{4}$ we derive that $(q^{2}-q)/2$ is odd. Thus
$\tau\notin\mathrm{Alt}(\mathbb{F}_{q^{2}})$, and so
$|K\cap\mathrm{Alt}(\mathbb{F}_{q^{2}})|=|K|/2=q^{2}(q^{2}-1)$. Since $\omega$
is a $(q^{2}-1)$-cycle with $q$ odd, we have
$\omega\notin\mathrm{Alt}(\mathbb{F}_{q^{2}})$. Hence
$K\cap\mathrm{Alt}(\mathbb{F}_{q^{2}})\geq\mathrm{Soc}(K){:}\langle\omega^{2},\omega\tau\rangle$,
which in conjunction with
$|K\cap\mathrm{Alt}(\mathbb{F}_{q^{2}})|=q^{2}(q^{2}-1)=|\mathrm{Soc}(K){:}\langle\omega^{2},\omega\tau\rangle|$
implies that

$K\cap\mathrm{Alt}(\mathbb{F}_{q^{2}})=\mathrm{Soc}(K){:}\langle\omega^{2},\omega\tau\rangle.$

Since $1^{\omega}$ is a generator of $\mathbb{F}_{q^{2}}^{\times}$ and $q$ is
odd, the orbits of $1^{\omega}$ and $(1^{\omega})^{q}$ under
$\langle\omega^{2}\rangle$, respectively, are the orbits of
$\langle\omega^{2}\rangle$ on $\mathbb{F}_{q^{2}}^{\times}$. Then since
$(1^{\omega})^{q}=1^{\omega\tau}$, we obtain
$\mathbb{F}_{q^{2}}^{\times}=1^{\langle\omega^{2}\rangle}\cup(1^{\omega\tau})^{\langle\omega^{2}\rangle}$,
which implies that $\langle\omega^{2},\omega\tau\rangle$ is transitive on
$\mathbb{F}_{q^{2}}^{\times}$. Therefore, the action of
$\mathrm{Soc}(K){:}\langle\omega^{2},\omega\tau\rangle$ on
$\mathbb{F}_{q^{2}}$ is $2$-transitive, and so is the action of
$K\cap\mathrm{Alt}(\mathbb{F}_{q^{2}})$ on $[K:K\cap K^{g}]$.

Since $G=K\mathrm{Alt}(\mathbb{F}_{q^{2}})$, the right multiplication action
of $\mathrm{Alt}(\mathbb{F}_{q^{2}})$ on $[G:K]$ is transitive. This together
with the $2$-transitivity of $K\cap\mathrm{Alt}(\mathbb{F}_{q^{2}})$ on
$[K:K\cap K^{g}]$ implies that $\Gamma$ is
$(\mathrm{Alt}(\mathbb{F}_{q^{2}}),2)$-arc-transitive. Since
$K_{0,1}=\langle\tau\rangle$, we have

$K\cap\mathrm{Alt}(\mathbb{F}_{q^{2}})_{0,1}=\langle\tau\rangle\cap\mathrm{Alt}(\mathbb{F}_{q^{2}})_{0,1}=1.$

Observe that
$|K||\mathrm{Alt}(\mathbb{F}_{q^{2}})_{0,1}|=2q^{2}(q^{2}-1)|\mathrm{A}_{q^{2}-2}|=|G|$.
We then conclude that $\mathrm{Alt}(\mathbb{F}_{q^{2}})_{0,1}$ acts regularly
on $[G:K]$, and so $\Gamma$ is a Cayley graph on
$\mathrm{Alt}(\mathbb{F}_{q^{2}})_{0,1}\cong\mathrm{A}_{q^{2}-2}$.

Now we prove that $\Gamma$ is connected, that is, $\langle K,g\rangle=G$. Let
$X=\langle K,g\rangle$. Then $X$ is $2$-transitive on $\mathbb{F}_{q^{2}}$.
Since $\tau\notin\mathrm{Alt}(\mathbb{F}_{q^{2}})$ and $\tau\in K\leq X$, the
group $X$ is not contained in $\mathrm{Alt}(\mathbb{F}_{q^{2}})$. Then by the
classification of $2$-transitive groups, either
$X=\mathrm{Sym}(\mathbb{F}_{q^{2}})$, or $X\leq Y:=\mathrm{AGL}_{2f}(p)$.
Suppose for a contradiction that the latter occurs. Then $g\in X_{0}\leq
Y_{0}$ and $K_{0}\leq X_{0}\leq Y_{0}$. As $g$ normalizes $M_{0}$, it follows
that $g\in\mathbf{N}_{Y_{0}}(M_{0})$. However, since $M_{0}$ is a Single cycle
in $Y_{0}=\mathrm{GL}_{2f}(p)$, we see from [12, Theorem 7.3] that
$\mathbf{N}_{Y_{0}}(M_{0})=\mathrm{\Gamma L}_{1}(q^{2})=N_{0}$, and so $g\in
N_{0}$, contradicting the conclusion that $g\notin N$. Thus
$X=\mathrm{Sym}(\mathbb{F}_{q^{2}})=G$, which implies that $\Gamma$ is
connected. This completes the proof of Theorem 1.4.

## Acknowledgements

This work was partially supported by NSFC 12061092.

## References

  * [1] B. Alspach, M. Conder, D. Marušič, M.-Y. Xu, A classification of 2-arc-transitive circulants, _J. Algebraic Combin._ 5 (1996), 83–86.
  * [2] W. Bosma, J. Cannon, C. Playoust, The Magma algebra system I: The user language, _J. Symbolic Comput._ 24 (1997), 235–265.
  * [3] P. J. Cameron, _Permutation Groups_ , London Math. Soc. Stud. Texts, vol.45, Cambridge Univ. Press, Cambridge, 1999.
  * [4] J. H. Conway, R. T. Curtis, S. P. Norton, R. A. Parker, R. A. Wilson, _Atlas of Finite Groups_ , Oxford Univ. Press, London/New York, 1985.
  * [5] J. D. Dixon, B. Mortimer, _Permutation Groups_ , Springer, New York, 1996.
  * [6] J. L. Du, Y.-Q. Feng, Tetravalent 2-arc-transitive Cayley graphs on non-abelian simple groups, _Comm. Algebra_ 47 (2019), 4565–4574.
  * [7] J. L. Du, Y.-Q. Feng, J.-X. Zhou, Pentavalent symmetric graphs admiting a vertex-transitive non-abelian simple groups, _European J. Combin._ 63 (2017), 134–145.
  * [8] S. F. Du, D. Marušič, A. O. Waller, On 2-arc-transitive covers of complete graphs, _J. Combin. Theory Ser. B_ 74 (1998), 276–290.
  * [9] S. F. Du, D. Malnič, D. Marušič, Classification of 2-arc-transitive dihedrants, _J. Combin. Theory Ser. B_ 98 (2008), 1349–1372.
  * [10] X. G. Fang, C. E. Praeger, J. Wang, On the automorphism groups of Cayley graphs of finite simple groups, _J. Lond. Math. Soc. (2)_ 66 (2002), 563–578.
  * [11] C. D. Godsil, On the full automorphism group of a graph, _Combinatorica_ 1 (1981), 243–256.
  * [12] B. Huppert, _Endliche Gruppen. I_ , Die Grundlehren der Mathematischen Wissenschaften, Band 134, Springer-Verlag, Berlin-New York, 1967.
  * [13] B. Huppert, N. Blackburn, _Finite Groups III_ , Springer-Verlag, New York, 1982.
  * [14] A. A. Ivanov, C. E. Praeger, On finite affine 2-arc transitive graphs, _European J. Combin._ 14 (1993), 421–444.
  * [15] P. B. Kleidman, M. W. Liebeck, _The subgroup structure of the finite classical groups_ , London Math. Soc. Lecture Note Series 129, Cambridge Univ. Press, 1990.
  * [16] C. H. Li, Isomorphisms of finite Cayley graphs, Ph. D. Thesis, The University of Western Australia, 1996.
  * [17] C. H. Li, J. M. Pan, Finite 2-arc-transitive abelian Cayley graphs, _European J. Combin._ 29 (2007), 148–158.
  * [18] C. H. Li, Á. Seress, S. J. Song, $s$-Arc-transitive graphs and normal subgroups, _J. Algebra_ 421 (2015), 331–348.
  * [19] C. H. Li, B. Z. Xia, Factorizations of almost simple groups with a solvable factor, and Cayley graphs of solvable groups, accepted by Memoirs of the American Mathematical Society, available online at arXiv:1408.0350.
  * [20] J. M. Pan, F. G. Yin, B. Ling, Arc-transitive Cayley graphs on non-abelian simple groups with soluble vertex stabilizers and valency seven, _Discrete Math._ 342 (2019), 689–696.
  * [21] C. E. Praeger, An O’Nan-Scott theorem for finite quasiprimitive permutation groups and an application to 2-arc-transitive graphs, _J. Lond. Math. Soc. (2)_ 47 (1992), 227–239.
  * [22] C. E. Praeger, Finite quasiprimitive graphs, _Surveys in combinatorics, 1997 (London)_ , 65–85, London Math. Soc. Lecture Note Ser., 241, Cambridge Univ. Press, Cambridge, 1997.
  * [23] L. Pyber, On the orders of doubly transitive permutation groups, elementary estimates, _J. Combin. Theory Ser. A_ 62 (1993), 361–366.
  * [24] G. Sabidussi, Vertex-transitive graphs, _Monatsh. Math._ 68 (1964), 426–438.
  * [25] W. T. Tutte, A family of cubical graphs, _Math. Proc. Cambridge Philos. Soc._ 43 (1947), 459–474.
  * [26] R. Weiss, $s$-Transitive graphs, Algebraic methods in graph theory, Vol. I, II (Szeged, 1978), 827–847, Colloq. Math. Soc. János Bolyai, 25, North-Holland, Amsterdam, New York, 1981.
  * [27] B. Z. Xia, Quasiprimitive groups containing a transitive alternating group, _J. Algebra_ 490 (2017), 555–567.
  * [28] M. Y. Xu, Automorphism groups and isomorphisms of Cayley digraphs, _Discrete Math._ 182 (1998), 309–319.
  * [29] S. J. Xu, X. G. Fang, J. Wang, M. Y. Xu, On cubic $s$-arc-transtive Cayley graphs of finite simple groups, _European J. Combin._ 26 (2005), 133–143.
  * [30] S. J. Xu, X. G. Fang, J. Wang, M. Y. Xu, $5$-Arc-transitive Cayley graphs on finite simple groups, _European J. Combin._ 28 (2007), 1023–1036.
  * [31] F.-G. Yin, Y.-Q. Feng, J.-X. Zhou, S.-S. Chen, Arc-transitive Cayley graphs on nonabelian simple groups with prime valency, _J. Combin. Theory Ser. A_ 177 (2021), 105303.