# Two-geodesic transitive graphs of order $p^{n}$ with $n\leq 3$

Jun-Jie Huang, Yan-Quan Feng∗, Jin-Xin Zhou, Fu-Gang Yin  Jun-Jie Huang, Yan-
Quan Feng, Jin-Xin Zhou, Fu-Gang Yin
School of mathematics and statistics
Beijing Jiaotong University
Beijing
100044, P. R. China  20118006@bjtu.edu.cn(J.-J. Huang<EMAIL_ADDRESS>(Y.-Q. Feng<EMAIL_ADDRESS>(J.-X. Zhou), 18118010@bjtu.edu.cn(F.-G. Yin)

###### Abstract.

A vertex triple $(u,v,w)$ of a graph is called a $2$-geodesic if $v$ is
adjacent to both $u$ and $w$ and $u$ is not adjacent to $w$. A graph is said
to be $2$-geodesic transitive if its automorphism group is transitive on the
set of $2$-geodesics. In this paper, a complete classification of $2$-geodesic
transitive graphs of order $p^{n}$ is given for each prime $p$ and $n\leq 3$.
It turns out that all such graphs consist of three small graphs: the complete
bipartite graph $\mathsf{K}_{4,4}$ of order $8$, the Schläfli graph of order
$27$ and its complement, and fourteen infinite families: the cycles
$\mathbf{C}_{p},\mathbf{C}_{p^{2}}$ and $\mathbf{C}_{p^{3}}$, the complete
graphs $\mathsf{K}_{p},\mathsf{K}_{p^{2}}$ and $\mathsf{K}_{p^{3}}$, the
complete multipartite graphs $\mathsf{K}_{p[p]}$, $\mathsf{K}_{p[p^{2}]}$ and
$\mathsf{K}_{p^{2}[p]}$, the Hamming graph $H(2,p)$ and its complement, the
Hamming graph $H(3,p)$, and two infinite families of normal Cayley graphs on
extraspecial group of order $p^{3}$ and exponent $p$.

∗Corresponding author.

1991 MR Subject Classification 20B15, 20B30, 05C25.

This work was partially supported by the National Natural Science Foundation
of China (11731002, 12071023, 12161141005) and the 111 Project of China
(B16002).

keywords. 2-geodesic transitive graph, 2-arc transitive graph, primitive group

##  1\. Introduction

Given a graph $\mathrm{\Gamma}$, let $\mathrm{Aut}(\mathrm{\Gamma})$ be the
automorphism group of $\mathrm{\Gamma}$, and $\mathrm{\Gamma}$ is said to be
vertex-transitive, if $\mathrm{Aut}(\mathrm{\Gamma})$ is transitive on the
vertex set $V(\mathrm{\Gamma})$ of $\mathrm{\Gamma}$. If not otherwise stated,
all graphs consider in this paper are finite, simple, undirected, connected
and vertex-transitive. Let $u,v,w$ be three vertices of $\mathrm{\Gamma}$ such
that $v$ is adjacent to both $u$ and $w$. If $u\neq w$, then the triple
$(u,v,w)$ will be called a $2$-arc, and if, in addition, $u$ is not adjacent
to $w$, then $(u,v,w)$ will be called a $2$-geodesic. An arc of
$\mathrm{\Gamma}$ is an ordered pair of adjacent vertices of
$\mathrm{\Gamma}$. For $G\leq\mathrm{Aut}(\mathrm{\Gamma})$, if $G$ is
transitive on the set of arcs of $\mathrm{\Gamma}$, then $\mathrm{\Gamma}$ is
said to be $G$-arc transitive, and if $G$ is transitive on the set of $2$-arcs
of $\mathrm{\Gamma}$, then $\mathrm{\Gamma}$ is said to be $(G,2)$-arc
transitive. Similarly, if $G$ is transitive on the set of $2$-geodesics of
$\mathrm{\Gamma}$, then $\mathrm{\Gamma}$ is said to be $(G,2)$-geodesic
transitive. When $G=\mathrm{Aut}(\mathrm{\Gamma})$, a $G$-arc transitive
graph, $(G,2)$-arc transitive graph or $(G,2)$-geodesic transitive graph will
be simply called an arc transitive graph, a $2$-arc transitive graph or
$2$-geodesic transitive graph, respectively. Clearly, $2$-arc transitive
graphs are also $2$-geodesic transitive. Following the seminal work of Tutte
[40], the $2$-arc transitive graphs have been received considerable attention,
see [1, 15, 29, 31, 33, 35] and references therein. Note that if
$\mathrm{\Gamma}$ has girth $3$, then the $2$-arcs contained in $3$-cycles are
not $2$-geodesics. There exist $2$-geodesic transitive but not $2$-arc
transitive graphs, and a typical example is the octahedron. The $2$-geodesic
transitive but not $2$-arc transitive graphs have also been extensively
studied in the literature, see, for example, [10, 12, 13, 23, 24, 26, 27].

There are a fair amount of works on $2$-arc transitive graphs with particular
orders in the literature. One of the remarkable results is the classification
of basic $2$-arc transitive graphs of prime-power order ([29]). Later, this
was extended to the class of vertex transitive locally primitive graphs of
prime-power order in [30]. All the $2$-arc transitive graphs of order a prime
or a prime square have also been classified in [33]. Recently, Jin [22]
initiated the study of $2$-geodesic transitive graphs of prime power order,
and he proved that for a $2$-geodesic transitive graph $\mathrm{\Gamma}$ of
prime power, if $\mathrm{Aut}(\mathrm{\Gamma})$ is quasiprimitive on
$V(\mathrm{\Gamma})$, then $\mathrm{\Gamma}$ is isomorphic to some well-known
graphs (see Proposition 2.12), or $\mathrm{Aut}(\mathrm{\Gamma})$ is primitive
on $V(\mathrm{\Gamma})$ of affine type. The following problem was proposed in
[22].

######  Problem 1.1.

([22, Problem 1.4]) Let $\mathrm{\Gamma}$ be a $2$-geodesic transitive graph
of prime power order which is not $2$-arc transitive. Classify such graphs
where $\mathrm{Aut}(\mathrm{\Gamma})$ acts quasiprimitively on
$V(\mathrm{\Gamma})$ of affine type.

Inspired by the facts listed above, in this paper we shall focus on the
classification of 2-geodesic transitive graphs of order $p^{n}$ with $p$ a
prime and $n\leq 3$. The following is our first main result which provides a
partial solution to Problem 1.1.

######  Theorem 1.2.

Let $p$ be a prime and let $\mathrm{\Gamma}$ be a connected $2$-geodesic
transitive but not $2$-arc transitive normal Cayley graph on
$\mathbb{Z}_{p}^{n}$ with $n\leq 3$. Then $\mathrm{Aut}(\mathrm{\Gamma})$ is
primitive on $V(\mathrm{\Gamma})$ if and only if $\mathrm{\Gamma}\cong H(2,3)$
or $H(3,3)$.

Another motivation for us to consider 2-geodesic transitive graphs of order
$p^{n}$ with $p$ a prime and $n\leq 3$ comes from the work in [6, 20]. Before
proceeding, we introduce the following definitions. For two vertices $u$ and
$v$ of a graph $\mathrm{\Gamma}$, the distance $d_{\mathrm{\Gamma}}(u,v)$
between $u$ and $v$ in $\mathrm{\Gamma}$ is the smallest length of paths
between $u$ and $v$, and the diameter $\mathrm{diam}(\mathrm{\Gamma})$ of
$\mathrm{\Gamma}$ is the maximum distance occurring over all pairs of
vertices. A graph $\mathrm{\Gamma}$ is called distance transitive if, for any
vertices $u,v,x,y$ with $d_{\mathrm{\Gamma}}(u,v)=d_{\mathrm{\Gamma}}(x,y)$,
$\mathrm{\Gamma}$ has an automorphism mapping the pair $(u,v)$ to $(x,y)$. For
a positive integer $s$, $\mathrm{\Gamma}$ is said to be $s$-distance
transitive if, for each $1\leq i\leq s$, $\mathrm{Aut}(\mathrm{\Gamma})$ is
transitive on the ordered pairs of form $(u,v)$ with
$d_{\mathrm{\Gamma}}(u,v)=i$. In [6], Chen et al. classified $2$-distance
transitive circulants, and they proposed the following question:

######  Question 1.3.

([6, Question 1.2]) Is there a normal Cayley graph which is $2$-distance
transitive, but neither distance transitive nor $2$-arc transitive?

In [20], we answered this question in positive by constructing an infinite
family of 2-distance transitive Cayley graphs of order a prime cube which are
neither distance transitive nor 2-arc transitive. This family of graphs are
also $2$-geodesic transitive. Stimulated by this, in this paper, we give a
complete classification of the $2$-geodesic transitive graphs of order $p^{n}$
with $p$ a prime and $n\leq 3$. As a result, we find a new family of
$2$-geodesic transitive Cayley graphs which are not $2$-arc transitive.

Note that the $2$-geodesic transitive Cayley graphs on cyclic groups were
classified in [27], from which we find that a $2$-geodesic transitive graph of
order a prime is either a cycle or a complete graph. In our next two main
theorems, we give a complete classification of $2$-geodesic transitive graphs
of order $p^{2}$ or $p^{3}$ with $p$ a prime.

######  Theorem 1.4.

Let $p$ be a prime and let $\mathrm{\Gamma}$ be a connected arc-transitive
graph of order $p^{2}$. Then $\mathrm{\Gamma}$ is $2$-geodesic transitive if
and only if one of the following holds:

  * (1)

If $\mathrm{\Gamma}$ is $2$-arc transitive, then $\mathrm{\Gamma}$ is
isomorphic to $\mathbf{C}_{p^{2}}$ or $\mathsf{K}_{p^{2}}$;

  * (2)

If $\mathrm{\Gamma}$ is not $2$-arc transitive, then $\mathrm{\Gamma}$ is
isomorphic to $\mathsf{K}_{p[p]}$, the Hamming graph $H(2,p)$ or its
complement $\overline{H(2,p)}$, where $p\geq 3$.

######  Theorem 1.5.

Let $p$ be a prime and let $\mathrm{\Gamma}$ be a connected arc transitive
graph of order $p^{3}$. Then $\mathrm{\Gamma}$ is $2$-geodesic transitive if
and only if one of the following holds:

  * (1)

If $\mathrm{\Gamma}$ is $2$-arc transitive, then $\mathrm{\Gamma}$ is
isomorphic to $H(3,2)$, $\mathsf{K}_{4,4}$, $\mathbf{C}_{p^{3}}$ or
$\mathsf{K}_{p^{3}}$;

  * (2)

If $\mathrm{\Gamma}$ is not $2$-arc transitive, then $\mathrm{\Gamma}$ is
isomorphic to one of the following graphs:

    * (i)

the Schläfli graph or its complement;

    * (ii)

$\mathsf{K}_{p^{2}[p]}$ with $p\geq 2$, or $\mathsf{K}_{p[p^{2}]}$ with $p\geq
3$;

    * (iii)

the Hamming graph $H(3,p)$ with $p\geq 3$;

    * (iv)

the normal Cayley graph $\mathrm{Cay}(G,S)$, where $G=\langle a,b,c\mid
a^{p}=b^{p}=c^{p}=1,[a,b]=c,[a,c]=[b,c]=1\rangle$ with $p\geq 3$, and
$S=\\{a^{i},b^{i}\mid i\in\mathbb{Z}_{p}^{*}\\}$ or
$S=\\{a^{i},b^{i},(b^{j}ab^{j})^{i}\mid i,j\in\mathbb{Z}_{p}^{*}\\}$,
respectively.

Remark: All graphs in Theorems 1.2, 1.4 and 1.5 will be explicitly defined in
Sections 3 and 4, and all but graphs in part (iv) of Theorem 1.5 (2) have been
well-studied in the literature. In part (iv) of Theorem 1.5 (2), the graphs
$\mathrm{Cay}(G,S)$ with $S=\\{a^{i},b^{i}\mid i\in\mathbb{Z}_{p}^{*}\\}$ were
first given in [20] while the graphs $\mathrm{Cay}(G,S)$ with
$S=\\{a^{i},b^{i},(b^{j}ab^{j})^{i}\mid i,j\in\mathbb{Z}_{p}^{*}\\}$ are a new
family of $2$-geodesic transitive but not 2-arc transitive graphs.

The layout of this paper is as follows. In Section 2, we list some definitions
and some preliminary results, and in Sections 3 and 4, all $2$-geodesic
transitive graphs in question are listed, where a new infinite family of
$2$-geodesic transitive Cayley graphs on the extraspecial group of order
$p^{3}$ and exponent $p$ is constructed in Section 4. In Section 5, we prove
Theorem 1.2, and in Section 6, we prove Theorems 1.4 and 1.5.

##  2\. Preliminaries

In this section, we give some results which will be used in following
sections. Let $p$ be a prime and let $n$ be a positive integer. We denote by
$n_{p}$ the largest $p$-power dividing $n$, and by $n_{p^{\prime}}$ the
largest divisor of $n$ such that $p\nmid n_{p^{\prime}}$. We use
$\mathbb{Z}_{n}$ and $\mathbb{Z}_{p}^{r}$ to denote the cyclic group of order
$n$ and the elementary abelian group of order $p^{r}$, respectively. Also, we
denote by $\mathbb{Z}_{p}$ the additive group of integers modulo $p$, and by
$\mathbb{Z}_{p}^{*}$ the multiplicative group of all non-zero numbers in
$\mathbb{Z}_{p}$. Denote by $\mathrm{Q}_{8}$, $\mathrm{D}_{2n}$,
$\mathrm{A}_{n}$ and $\mathrm{S}_{n}$, the quaternion group of order $8$, the
dihedral group of order $2n$, the alternating group of degree $n$, and the
symmetric group of degree $n$, respectively. For a subgroup $H$ of a group
$G$, denote by $H^{*}$ the subset of $H$ deleting $1$, that is, $H^{*}=\\{h\
|\ h\in H,h\not=1\\}$. For two groups $A$ and $B$, denote by $A\times B$ the
direct product of $A$ and $B$, by $A\rtimes B$ or $A:B$ a semidirect product
of $A$ by $B$, and by $A.B$ an extension of $A$ by $B$. For a group $G$,
denote by $Z(G)$ the central of $G$, and by $\mathrm{soc}(G)$ the socle of
$G$, that is, the product of all minimal normal subgroup of $G$.

For graph notation, denote by $\mathbf{C}_{n}$ the cycle of length $n$,
$\mathsf{K}_{n}$ the complete graph of order $n$, $\mathsf{K}_{n,n}$ the
complete bipartite graph of order $2n$, $\mathsf{K}_{n,n}-n\mathsf{K}_{2}$ the
subgraph of $\mathsf{K}_{n,n}$ minus a matching, and $\mathsf{K}_{m[b]}$ the
complete multipartite graph consisting $m\geq 3$ parts of size $b\geq 2$,
respectively. Let $\mathrm{\Gamma}$ be a graph. Denote by
$\overline{\mathrm{\Gamma}}$ the complement graph of $\mathrm{\Gamma}$, that
is, the graph with the vertex set $V(\mathrm{\Gamma})$ and two vertices are
adjacent in $\overline{\mathrm{\Gamma}}$ if and only if they are not adjacent
in $\mathrm{\Gamma}$. A clique of a graph $\mathrm{\Gamma}$ is a complete
subgraph and a maximal clique is a clique which is not contained in a larger
clique. For two disjoint subsets $U,W\subset V(\mathrm{\Gamma})$, denote by
$[U]$ the subgraph of $\mathrm{\Gamma}$ induced by $U$, and by $[U,W]$ the
subgraph of $\mathrm{\Gamma}$ with vertex set $U\cup W$ and edge set
consisting of the edges of $\mathrm{\Gamma}$ between $U$ and $W$. Moreover, we
use $\mathrm{\Gamma}_{i}(u)$, with $i\leq\mathrm{diam}(\mathrm{\Gamma})$ and
$u\in V(\mathrm{\Gamma})$, to denote the set of vertices with distance $i$
from $u$ in $\mathrm{\Gamma}$.

A classification of finite groups of order prime cube was given in [17, P. 8,
Theorem 2.14] and [17, P. 203, Theorem 5.1].

######  Proposition 2.1.

Let $p$ be an odd prime and let $G$ be a finite group of order $p^{3}$. Then
$G$ is isomorphic to one of the following groups:
$\mathbb{Z}_{p^{3}},\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p},\mathbb{Z}_{p}^{3},\langle
a,b\mid a^{p^{2}}=b^{p}=1,a^{b}=a^{1+p}\rangle$ or $\langle a,b,c\mid
a^{p}=b^{p}=c^{p}=1,[a,b]=c,[a,c]=[b,c]=1\rangle$.

All non-abelian simple groups with a subgroup of prime-power index were
determined in [19, Theorem 1], and we record it here.

######  Proposition 2.2.

Let $T$ be a nonabelian simple group with a subgroup $H$ of index $p^{n}$,
where $p$ is a prime and $n\geq 1$. Then one of the following holds:

  * (1)

$T\cong\mathrm{A}_{p^{n}}$ and $H\cong\mathrm{A}_{p^{n}-1}$;

  * (2)

$T\cong\mathrm{PSL}(d,q)$ and $H$ is a maximal parabolic subgroup of $T$, with
$p^{n}=(q^{d}-1)/(q-1)$ and $d$ a prime;

  * (3)

$T\cong\mathrm{PSL}(2,11)$ and $H\cong\mathrm{A}_{5}$, with $p^{n}=11$;

  * (4)

$T\cong\mathrm{M}_{11}$ and $H\cong\mathrm{M}_{10}$, with $p^{n}=11$;

  * (5)

$T\cong\mathrm{M}_{23}$ and $H\cong\mathrm{M}_{22}$, with $p^{n}=23$;

  * (6)

$T\cong\mathrm{PSU}(4,2)$ and $H\cong\mathbb{Z}_{2}^{4}\rtimes\mathrm{A}_{5}$,
with $p^{n}=27$.

For a group $G$, $M\lessdot G$ means that $M$ is a maximal subgroup of $G$.
All maximal subgroups of $\mathrm{PSL}(2,q)$ and $\mathrm{PGL}(2,q)$ were
given in [39, P. 392-418] and [5, Theorem 2].

######  Proposition 2.3.

Let $p\geq 5$ be a prime.

  * (1)

If $M\lessdot\mathrm{PGL}(2,p)$, then $M$ is one of the following groups:
$\mathrm{PSL}(2,p)$, $\mathbb{Z}_{p}\rtimes\mathbb{Z}_{p-1}$,
$\mathrm{D}_{2(p-1)}$ with $p\neq 5$, $\mathrm{D}_{2(p+1)}$, or
$\mathrm{S}_{4}$ with $p\equiv\pm 3({\rm mod~{}}8)$;

  * (2)

If $M\lessdot\mathrm{PSL}(2,p)$, then $M$ is one of the following groups:
$\mathbb{Z}_{p}\rtimes\mathbb{Z}_{(p-1)/2}$, $\mathrm{D}_{p-1}$ with $p\geq
13$, $\mathrm{D}_{p+1}$ with $p\neq 7$, $\mathrm{A}_{4}$ with $p\equiv\pm
3({\rm mod~{}}8)$ and $p\equiv\pm 3({\rm mod~{}}10)$, $\mathrm{S}_{4}$ with
$p\equiv\pm 1({\rm mod~{}}8)$, or $\mathrm{A}_{5}$ with $p\equiv\pm 1({\rm
mod~{}}10)$.

By Proposition 2.3, the smallest index of a maximal subgroup of
$\mathrm{PSL}(2,p)$ is $p$, and so every non-trivial transitive action of
$\mathrm{PSL}(2,p)$ has degree at least $p$.

The following can be read off from [3, Tables 8.3 and 8.4], where the maximal
subgroups of $\mathrm{PGL}(3,p)$ corresponding to row 1, row 4 and row 5 of
[3, Table 8.3] should be $\mathbb{Z}_{p}^{2}:\mathrm{GL}(2,p)$,
$(\mathbb{Z}_{p-1}\times\mathbb{Z}_{p-1}):\mathrm{S}_{3}$ and
$\mathbb{Z}_{p^{2}+p+1}:\mathbb{Z}_{3}$ (see [28, Chapter 4]), respectively.
The notation $\frac{1}{d}\mathrm{GL}(2,p)$ is the factor group of
$\mathrm{GL}(2,p)$ by the subgroup of order $d$ in $Z(\mathrm{GL}(2,p))$.

######  Proposition 2.4.

Let $p\geq 5$ be a prime and let $d=(3,p-1)$.

  * (1)

If $M\lessdot\mathrm{PGL}(3,p)$, then $M$ is one of the following groups:
$\mathrm{PSL}(3,p)$, $\mathbb{Z}_{p}^{2}:\mathrm{GL}(2,p)$,
$(\mathbb{Z}_{p-1}\times\mathbb{Z}_{p-1}):\mathrm{S}_{3}$,
$\mathbb{Z}_{p^{2}+p+1}:\mathbb{Z}_{3}$, and
$(\mathbb{Z}_{3}^{2}:\mathrm{Q}_{8}).\mathbb{Z}_{3}$ with $p\equiv 1({\rm
mod~{}}3)$ and $p\not\equiv 1({\rm mod~{}}9)$;

  * (2)

If $M\lessdot\mathrm{PSL}(3,p)$, then $M$ is one of the following groups:
$\mathbb{Z}_{p}^{2}:\frac{1}{d}\mathrm{GL}(2,p)$,
$(\mathbb{Z}_{(p-1)/d}\times\mathbb{Z}_{p-1}):\mathrm{S}_{3}$,
$\mathbb{Z}_{(p^{2}+p+1)/d}:\mathbb{Z}_{3}$,
$(\mathbb{Z}_{3}^{2}:\mathrm{Q}_{8}).\mathbb{Z}_{\frac{(p-1,9)}{3}}$ with
$p\equiv 1({\rm mod~{}}3)$, $\mathrm{PGL}(2,p)$, $\mathrm{PSL}(2,7)$ with
$p\equiv 1,2,4({\rm mod~{}}7)$, and $\mathrm{A}_{6}$ with $p\equiv 1,4({\rm
mod~{}}15)$.

By Proposition 2.4, the smallest index of a maximal subgroup of
$\mathrm{PSL}(3,p)$ is $p^{2}+p+1$, and so every non-trivial transitive action
of $\mathrm{PSL}(3,p)$ has degree at least $p^{2}+p+1$.

From Propositions 2.3 and 2.4, all $2$-transitive actions of subgroups of
$\mathrm{PGL}(3,p)$ can be determined.

######  Lemma 2.5.

Let $p\geq 5$ be a prime. Assume that $X\leq\mathrm{PGL}(3,p)$ has a
$2$-transitive action on $\Omega$ with $n=|\Omega|\geq 4$. Let $K$ be the
kernel of $X$ on $\Omega$. If $\mathrm{soc}(X/K)$ is elementary abelian, then
either

  * (1)

$(n,X/K)=(4,\mathrm{A}_{4})$, $(4,\mathrm{S}_{4})$, $(p,\mathrm{AGL}(1,p))$,
or $(9,L)$ with $L\leq(\mathbb{Z}_{3}^{2}:\mathrm{Q}_{8}).\mathbb{Z}_{3}$ and
$1\equiv({\rm mod~{}}3)$; or

  * (2)

$n=p^{2}$ and $X/K\leq\mathrm{AGL}(2,p)$.

If $\mathrm{soc}(X/K)$ is nonabelian simple, then one of the following holds:

  * (1)

$(n,X/K)=(5,\mathrm{A}_{5})$, $(6,\mathrm{A}_{6})$, $(10,\mathrm{A}_{6})$,
$(8,\mathrm{PSL}(2,7))$ or $(11,\mathrm{PSL}(2,11))$, and if
$X/K=\mathrm{A}_{6}$, $\mathrm{PSL}(2,7)$ or $\mathrm{PSL}(2,11)$ then
$p\equiv 1,4({\rm mod~{}}15)$, $p\equiv 1,2,4({\rm mod~{}}7)$ or $p=11$,
respectively;

  * (2)

$n=p+1$, and $X/K=\mathrm{PSL}(2,p)$ or $\mathrm{PGL}(2,p)$;

  * (3)

$n=p^{2}+p+1$, and $X/K=\mathrm{PSL}(3,p)$ or $\mathrm{PGL}(3,p)$.

Proof. By [14, Theorem 4.1B], either $\mathrm{soc}(X/K)\cong T$, a non-abelian
simple group, or $\mathrm{soc}(X/K)\cong\mathbb{Z}_{q}^{r}$ with $q$ a prime
and $r$ a positive integer. For $\mathrm{soc}(X/K)\cong T$, by Propositions
2.3 and 2.4 we have $T=\mathrm{A}_{5}$, $\mathrm{A}_{6}$, $\mathrm{PSL}(2,p)$
or $\mathrm{PSL}(3,p)$, and then all $2$-transitive actions of $T$ follow from
[31, Table 2.1]. Furthermore, $X/K=\mathrm{A}_{5}$ or $\mathrm{A}_{6}$ for
$T=\mathrm{A}_{5}$ or $\mathrm{A}_{6}$, $X/K=\mathrm{PSL}(2,p)$ or
$\mathrm{PGL}(2,p)$ for $T=\mathrm{PSL}(2,p)$, and $X/K=\mathrm{PSL}(3,p)$ or
$\mathrm{PGL}(3,p)$ for $T=\mathrm{PSL}(3,p)$.

Let $\mathrm{soc}(X/K)\cong\mathbb{Z}_{q}^{r}$. Then $n=|\Omega|=q^{r}$. If
$n=4$, then $X/K=\mathrm{A}_{4}$ or $\mathrm{S}_{4}$. We further assume $n\geq
5$. First we prove a claim.

Claim: Assume $Z_{1}\leq Z(\mathrm{GL}(2,p))$ and $\mathrm{GL}(2,p)/Z_{1}$ has
a subgroup isomorphic to $X/K_{1}$ for some $K_{1}\leq K$. Then $n=p$ and
$X/K\cong\mathrm{AGL}(1,p)$.

Set $\overline{\mathrm{GL}(2,p)}=\mathrm{GL}(2,p)/Z_{1}$ and
$\overline{Z(\mathrm{GL}(2,p))}=Z(\mathrm{GL}(2,p))/Z_{1}$. Then
$\overline{Z(\mathrm{GL}(2,p))}\leq Z(\overline{\mathrm{GL}(2,p)})$ and
$\overline{\mathrm{GL}(2,p)}/\overline{Z(\mathrm{GL}(2,p))}\cong\mathrm{PGL}(2,p)$.
If $\overline{Z(\mathrm{GL}(2,p))}\not=Z(\overline{\mathrm{GL}(2,p)})$, then
$Z(\overline{\mathrm{GL}(2,p)})/\overline{Z(\mathrm{GL}(2,p))}$ is a non-
trivial solvable normal subgroup of
$\overline{\mathrm{GL}(2,p)}/\overline{Z(\mathrm{GL}(2,p))}$, which is
impossible because $\mathrm{soc}(\mathrm{PGL}(2,p))=\mathrm{PSL}(2,p)$. It
follows that $\overline{Z(\mathrm{GL}(2,p))}=Z(\overline{\mathrm{GL}(2,p)})$
and hence
$\overline{\mathrm{GL}(2,p)}/Z(\overline{\mathrm{GL}(2,p)})\cong\mathrm{PGL}(2,p)$.

Set $\overline{X}=X/K_{1}$ and $\overline{K}=K/K_{1}$. Then
$X/K\cong\overline{X}/\overline{K}$ and
$\mathrm{soc}(\overline{X}/\overline{K})\cong\mathrm{soc}(X/K)\cong\mathbb{Z}_{q}^{r}$.
Since $K_{1}\leq K$, $\overline{X}$ has a $2$-transitive action on $\Omega$
with kernel $\overline{K}$, and hence $\overline{X}/\overline{K}$ is a
$2$-transitive permutation group on $\Omega$ with the unique regular subgroup
$\mathrm{soc}(\overline{X}/\overline{K})\cong\mathbb{Z}_{q}^{r}$.

Suppose $Z(\overline{X}/\overline{K})\not=1$. The $2$-transitivity of
$\overline{X}/\overline{K}$ implies that every non-trivial subgroup of
$Z(\overline{X}/\overline{K})$ is transitive on $\Omega$, forcing $|\Omega|=q$
and $Z(\overline{X}/\overline{K})\cong\mathbb{Z}_{q}$. It follows that
$\overline{X}/\overline{K}\cong\mathbb{Z}_{q}\rtimes\mathbb{Z}_{q-1}$, and
since $Z(\overline{X}/\overline{K})\cong\mathbb{Z}_{q}$, we have
$\overline{X}/\overline{K}\cong\mathbb{Z}_{q}\times\mathbb{Z}_{q-1}$,
contradicting the $2$-transitivity of $\overline{X}/\overline{K}$.

Thus, $Z(\overline{X}/\overline{K})=1$, and so
$Z(\overline{X})\leq\overline{K}$. By hypothesis, without loss of any
generality, we assume $X/K_{1}\leq\mathrm{GL}(2,p)/Z_{1}$. Then
$\overline{X}=X/K_{1}\leq\mathrm{GL}(2,p)/Z_{1}=\overline{\mathrm{GL}(2,p)}$.
Furthermore, $\overline{X}\cap Z(\overline{\mathrm{GL}(2,p)})\leq
Z(\overline{X})\leq\overline{K}$, and $\overline{X}/(\overline{X}\cap
Z(\overline{\mathrm{GL}(2,p)}))$ has a $2$-transitive action on $\Omega$.
Since $n=|\Omega|\geq 5$, $\overline{X}/(\overline{X}\cap
Z(\overline{\mathrm{GL}(2,p)}))$ cannot be abelain or dihedral, and since

$\overline{X}/(\overline{X}\cap
Z(\overline{\mathrm{GL}(2,p)}))\cong\overline{X}Z(\overline{\mathrm{GL}(2,p)})/Z(\overline{\mathrm{GL}(2,p)})\leq\overline{\mathrm{GL}(2,p)}/Z(\overline{\mathrm{GL}(2,p)})\cong\mathrm{PSL}(2,p),$

by Proposition 2.3 we obtain $\overline{X}/(\overline{X}\cap
Z(\overline{\mathrm{GL}(2,p)}))\cong\mathbb{Z}_{p}:\mathbb{Z}_{p-1}$, which
implies that $\overline{X}/(\overline{X}\cap Z(\overline{\mathrm{GL}(2,p)}))$
is faithful on $\Omega$, that is, $\overline{X}\cap
Z(\overline{\mathrm{GL}(2,p)})=\overline{K}$. Thus,
$X/K\cong\overline{X}/\overline{K}\cong\mathbb{Z}_{p}:\mathbb{Z}_{p-1}$, and
the $2$-transitivity of $X/K$ implies $X/K\cong\mathrm{AGL}(1,p)$, as claimed.

Since $X\leq\mathrm{PGL}(3,p)$, we have $X\leq M$ with
$M\lessdot\mathrm{PGL}(3,p)$ or $M\lessdot\mathrm{PSL}(3,p)$. We argue by
contradiction and we suppose that
$(n,X/K)\not=(9,\mathbb{Z}_{3}^{2}:\mathrm{Q}_{8})$,
$(9,(\mathbb{Z}_{3}^{2}:\mathrm{Q}_{8}).\mathbb{Z}_{3})$,
$(p,\mathrm{AGL}(1,p)$, or $(p^{2},L)$ with $L\leq\mathrm{AGL}(2,p)$. By
Proposition 2.4, $M=\mathrm{PGL}(2,p)$, $\mathbb{Z}_{p}^{2}:\mathrm{GL}(2,p)$,
$\mathbb{Z}_{p}^{2}:\frac{1}{d}\mathrm{GL}(2,p)$, $\mathrm{A}_{6}$,
$(\mathbb{Z}_{p-1}\times\mathbb{Z}_{p-1}):\mathrm{S}_{3}$,
$(\mathbb{Z}_{(p-1)/d}\times\mathbb{Z}_{p-1}):\mathrm{S}_{3}$,
$\mathbb{Z}_{p^{2}+p+1}:\mathbb{Z}_{3}$ or
$\mathbb{Z}_{(p^{2}+p+1)/d}:\mathbb{Z}_{3}$. We deal with these case by case.

Assume $X\leq\mathrm{PGL}(2,p)$. By taking $K_{1}=1$ and
$Z_{1}=Z(\mathrm{GL}(2,p))$, Claim implies that
$(n,X/K)=(p,\mathrm{AGL}(1,p))$, a contradiction.

Assume $M=\mathbb{Z}_{p}^{2}:\mathrm{GL}(2,p)$ or
$\mathbb{Z}_{p}^{2}:\frac{1}{d}\mathrm{GL}(2,p)$. If
$X\cap\mathbb{Z}_{p}^{2}\not\leq K$, then $K(X\cap\mathbb{Z}_{p}^{2})/K$ is
the unique regular subgroup of $X/K$ on $\Omega$, and since
$K(X\cap\mathbb{Z}_{p}^{2})/K\cong\mathbb{Z}_{p}$ or $\mathbb{Z}_{p}^{2}$,
either $(n,X/K)=(p,\mathrm{AGL}(1,p))$ or $(n,X/K)=(p^{2},L)$ with
$L\leq\mathrm{AGL}(2,p)$, a contradiction. Thus, $X\cap\mathbb{Z}_{p}^{2}\leq
K$, and $X/(X\cap\mathbb{Z}_{p}^{2})\cong
X\mathbb{Z}_{p}^{2}/\mathbb{Z}_{p}^{2}\leq\mathrm{GL}(2,p)$ or
$\frac{1}{d}\mathrm{GL}(2,p)$. By taking $K_{1}=X\cap\mathbb{Z}_{p}^{2}$ and
$Z_{1}=1$ or the subgroup of order $d$ of $Z(\mathrm{GL}(2,p))$ for
$\mathrm{GL}(2,p)$ or $\frac{1}{d}\mathrm{GL}(2,p)$ respectively, Claim
implies that $(n,X/K)=(p,\mathrm{AGL}(1,p))$, a contradiction.

Assume $M=\mathrm{A}_{6}$. By Atlas [8], $X\leq\mathrm{A}_{4}$, which is
impossible because $n\geq 5$.

Assume $M=\mathbb{Z}_{p^{2}+p+1}:\mathbb{Z}_{3}$. Let
$N=\mathbb{Z}_{p^{2}+p+1}$. Since $X$ is not abelian, $X\cap N\not=X$, and
since $X/(X\cap N)\cong XN/N\leq M/N\cong\mathbb{Z}_{3}$, we have $X=(X\cap
N).\mathbb{Z}_{3}$. Thus, $(X\cap N)K=X$ or $(X\cap N)K=X\cap N$. For the
former, $X/K$ is abelian, a contradiction. For the latter, $K\leq X\cap N$,
and $(X\cap N)/K$ is the regular subgroup of the permutation group $X/K$ on
$\Omega$. Since $(X/K)/((X\cap N)/K)\cong X/(X\cap N)\leq\mathbb{Z}_{3}$, we
have $X/K=(X\cap N).\mathbb{Z}_{3}$, forcing $n=4$, a contradiction. A similar
argument gives rise to a contradiction for
$M=\mathbb{Z}_{(p^{2}+p+1)/d}:\mathbb{Z}_{3}$.

Assume $M=(\mathbb{Z}_{p-1}\times\mathbb{Z}_{p-1}):\mathrm{S}_{3}$. Write
$Y=X\cap(\mathbb{Z}_{p-1}\times\mathbb{Z}_{p-1})$. Since $Y$ is abelian,
$Y\not=X$, and since $X/Y\cong
X(\mathbb{Z}_{p-1}\times\mathbb{Z}_{p-1})/(\mathbb{Z}_{p-1}\times\mathbb{Z}_{p-1})\leq
M/(\mathbb{Z}_{p-1}\times\mathbb{Z}_{p-1})\cong\mathrm{S}_{3}$, we have
$Y<X\leq Y.\mathrm{S}_{3}$. Since $n\geq 5$, $Y\not\leq K$, and so $KY/K$ is
the regular subgroup of $X/K$. Since $(X/K)/(KY/K)\cong X/KY$ and
$X/Y\leq\mathrm{S}_{3}$, we have $X/K=(KY/K).\mathbb{Z}_{2}$,
$(KY/K).\mathbb{Z}_{3}$ or $(KY/K).\mathrm{S}_{3}$, of witch the first two are
impossible. Thus, $X/K=(KY/K).\mathrm{S}_{3}$, and since
$\mathrm{soc}(X/K)=KY/K\cong\mathbb{Z}_{q}^{r}$, the $2$-transitivity of $X/K$
on $\Omega$ implies that $(q^{r}-1)\mid 6$. Since $n=q^{r}\geq 5$, we have
$q^{r}=7$ and $X/K\cong\mathbb{Z}_{7}\rtimes\mathbb{Z}_{6}$, which implies
that $(X/K)/(KY/K)\cong\mathbb{Z}_{6}$, a contradiction. A similar argument
gives rise to a contradiction for
$M=(\mathbb{Z}_{(p-1)/d}\times\mathbb{Z}_{p-1}):\mathrm{S}_{3}$. This
completes the proof. ∎

An extension $G=N.H$ is called a central extension if $N$ is a central
subgroup of $G$, and a central extension $G=N.H$ is called a covering group of
$H$ if $G$ is perfect, that is, the derived group $G^{\prime}$ of $G$ is equal
to $G$. Schur [38] showed that a simple group $T$ possesses a universal
covering group $G$ with the property that every covering group of $T$ is a
homomorphic image of $G$, and the center $Z(G)$ is called the Schur multiplier
of $T$, denoted by $\mathrm{Mult}(T)$. The following result was given in [34,
Lemma 2.11].

######  Proposition 2.6.

Let $N$ be a group of order prime or prime-square, and let $T$ be a nonabelian
simple group. Then every extension $G=N.T$ is a central extension.
Furthermore, $G=NG^{\prime}$ and $G^{\prime}=M.T$ with $M\leq
N\cap\mathrm{Mult}(T)$.

For a finite group $G$ and an inverse closed subset $S\subseteq
G\setminus\\{1\\}$, the Cayley graph $\mathrm{Cay}(G,S)$ on $G$ with respect
to $S$ is defined to be the graph with vertex set $G$ and edge set
$\\{\\{g,sg\\}\mid g\in G,s\in S\\}$. For $g\in G$, let $R(g)$ be the
permutation of $G$ defined by $x\mapsto xg$ for all $x\in G$, and then
$R(G):=\\{R(g)\mid g\in G\\}$ is a regular subgroup of automorphisms of
$\mathrm{Cay}(G,S)$. A Cayley graph $\mathrm{Cay}(G,S)$ is called normal if
$R(G)$ is a normal subgroup of $\mathrm{Aut}(\mathrm{Cay}(G,S))$. It is widely
known that a graph $\mathrm{\Gamma}$ is a Cayley graph on $G$ if and only if
$\mathrm{Aut}(\mathrm{\Gamma})$ has a regular subgroup isomorphic to $G$, see
[37]. Moreover, $\mathrm{Aut}(G,S):=\\{\alpha\in\mathrm{Aut}(G)\mid
S^{\alpha}=S\\}$ is also a subgroup of $\mathrm{Aut}(\mathrm{\Gamma})$, which
fixes the vertex $1$. Godsil [16] proved that $R(G)\rtimes\mathrm{Aut}(G,S)$
is the normalizer of $R(G)$ in $\mathrm{Aut}(\mathrm{Cay}(G,S))$, and together
with Xu [42], we have the following proposition.

######  Proposition 2.7.

Let $\mathrm{\Gamma}=\mathrm{Cay}(G,S)$ be a Cayley graph on a finite group
$G$ with respect to $S$, and let $A=\mathrm{Aut}(\mathrm{\Gamma})$. Then
$N_{A}(R(G))=R(G)\rtimes\mathrm{Aut}(G,S)$, and $\mathrm{\Gamma}$ is a normal
Cayley graph if and only if $A_{1}=\mathrm{Aut}(G,S)$, where $A_{1}$ is the
stabilizer of $1$ in $A$.

A Cayley graph $\mathrm{Cay}(G,S)$ is called a circulant if $G$ is a cyclic
group. The 2-arc transitive and 2-geodesic transitive circulants were
classified in [1, 27], respectively. We summarize those results as follows:

######  Proposition 2.8.

Let $\mathrm{\Gamma}$ be a connected $2$-geodesic transitive circulant. Then

  * (1)

If $\mathrm{\Gamma}$ is $2$-arc transitive, then $\mathrm{\Gamma}$ is one of
the following graphs: $\mathsf{K}_{n}$ with $n\geq 1$, $\mathbf{C}_{n}$ with
$n\geq 4$, $\mathsf{K}_{\frac{n}{2},\frac{n}{2}}$ with $n\geq 6$, and
$\mathsf{K}_{\frac{n}{2},\frac{n}{2}}-\frac{n}{2}\mathsf{K}_{2}$ with
$\frac{n}{2}\geq 5$ odd;

  * (2)

If $\mathrm{\Gamma}$ is not $2$-arc transitive, then
$\mathrm{\Gamma}\cong\mathsf{K}_{m[b]}$ for some $m\geq 3$ and $b\geq 2$.

A Cayley graph $\mathrm{Cay}(G,S)$ is called a normal $(X,2)$-geodesic
transitive if it is $(X,2)$-geodesic transitive for a group $X$ such that
$R(G)\leq X\leq R(G)\rtimes\mathrm{Aut}(G,S)$.

######  Proposition 2.9.

([12, Theorem 1.2]) Let $\mathrm{\Gamma}=\mathrm{Cay}(G,S)$ be a connected
normal $(X,2)$-geodesic transitive Cayley graph. Then one of the following
holds:

  * (1)

$\mathrm{\Gamma}\cong\mathbf{C}_{r}$ and $G\cong\mathbb{Z}_{r}$ for some
$r\geq 4$;

  * (2)

$\mathrm{\Gamma}\cong\mathsf{K}_{4[2]}$ and $G\cong\mathrm{Q}_{8}$, the
quaternion group, with $S=G\setminus Z(G)$;

  * (3)

There is a prime $q$ and an integer $m$ such that for all $a\in S$, $a$ has
order $q$ with $\langle a\rangle^{*}\subseteq S$ and $\langle
a\rangle^{*}\not=S$, and $b$ has order $m$ for each $b\in
S^{2}\setminus(S\cup\\{1\\})$.

Let $\mathrm{\Gamma}$ be a graph with $G\leq\mathrm{Aut}(\mathrm{\Gamma})$,
and let $N$ be a normal subgroup of $G$ such that $N$ is intransitive on
$V(\mathrm{\Gamma})$. The normal quotient graph $\mathrm{\Gamma}_{N}$ of
$\mathrm{\Gamma}$ induced by $N$ is defined to be the graph with vertex set
the orbits of $N$ and with two orbits adjacent in $\mathrm{\Gamma}_{N}$ if
$\mathrm{\Gamma}$ has an edge incident to some vertices in the two orbits,
respectively. Furthermore, $\mathrm{\Gamma}$ is called a normal $N$-cover or
simply cover of $\mathrm{\Gamma}_{N}$ if for every vertex $v\in
V(\mathrm{\Gamma})$ and the orbit $O(v)$ of $N$ containing $v$, $v$ and $O(v)$
has the same valency in $\mathrm{\Gamma}$ and $\mathrm{\Gamma}_{N}$,
respectively. The graph $\mathrm{\Gamma}$ is called locally $(G,s)$-distance
transitive with $s\geq 1$, if for every vertex $u\in V(\mathrm{\Gamma})$,
$G_{u}$ is transitive on $\mathrm{\Gamma}_{i}(u)$ for all $i\leq s$, where
$\mathrm{\Gamma}_{i}(u)$ is the set of vertices with distance $i$ from $u$ in
$\mathrm{\Gamma}$. The following proposition gives a reduction for studying
locally $(G,s)$-distance transitive graphs.

######  Proposition 2.10.

([9, Lemma 5.3]) Let $\mathrm{\Gamma}$ be a connected locally $(G,s)$-distance
transitive graph with $s\geq 2$. Let $1\neq N\lhd G$ be intransitive on
$V(\mathrm{\Gamma})$, and let $\mathcal{B}$ be the set of $N$-orbits on
$V(\mathrm{\Gamma})$. Then one of the following holds:

  * (1)

$|\mathcal{B}|=2$;

  * (2)

$\mathrm{\Gamma}$ is bipartite, $\mathrm{\Gamma}_{N}\cong\mathsf{K}_{1,r}$
with $r\geq 2$ and $G$ is intransitive on $V(\mathrm{\Gamma})$;

  * (3)

$s=2$,
$\mathrm{\Gamma}\cong\mathsf{K}_{m[b]},\mathrm{\Gamma}_{N}\cong\mathsf{K}_{m}$,
where $m\geq 3$ and $b\geq 2$;

  * (4)

$N$ is semiregular on $V(\mathrm{\Gamma})$, $\mathrm{\Gamma}$ is a cover of
$\mathrm{\Gamma}_{N}$, $|V(\mathrm{\Gamma}_{N})|<|V(\mathrm{\Gamma})|$ and
$\mathrm{\Gamma}_{N}$ is locally $(G/N,s^{\prime})$-distance transitive, where
$s^{\prime}=\min\\{s,\mathrm{diam}(\mathrm{\Gamma}_{N})\\}$.

The next result illustrates that a connected locally $(G,s)$-distance
transitive with $s\geq 2$, cannot be a cover of $\mathsf{K}_{m[b]}$.

######  Proposition 2.11.

([9, Proposition 4.2]) Let $\mathrm{\Gamma}$ be a connected locally
$(G,s)$-distance transitive graph with $s\geq 2$. Then there exists no
nontrivial $N\lhd G$ such that $\mathrm{\Gamma}$ is a cover of
$\mathrm{\Gamma}_{N}$ and $\mathrm{\Gamma}_{N}\cong\mathsf{K}_{m[b]}$ for some
$m\geq 3$ and $b\geq 2$.

A transitive permutation group $G$ on $\Omega$ is called primitive if it has
only trivial blocks in $\Omega$ (a block is a non-empty subset $\Delta$ of
$\Omega$ such that $\Delta^{g}\cap\Delta\not=\emptyset$ implies
$\Delta^{g}=\Delta$ for every $g\in G$ and a block $\Delta$ is trivial if
$|\Delta|=1$ or $|\Delta|=|\Omega|$), and quasiprimitive if every non-trivial
normal subgroup of $G$ is transitive on $\Omega$. Clearly, primitive group is
quasiprimitive, but the converse is not true. Combining [22, Theorem 1.2] and
[25, Theorem 1.3], we have the following result.

######  Proposition 2.12.

For a prime $p$ and a positive integer $n$, let $\mathrm{\Gamma}$ be a
connected $2$-geodesic but not $2$-arc transitive graph of order $p^{n}$, and
let $\mathrm{Aut}(\mathrm{\Gamma})$ be quasiprimitive on $V(\mathrm{\Gamma})$.
Then $\mathrm{Aut}(\mathrm{\Gamma})$ is primitive on $V(\mathrm{\Gamma})$ and
one of the following holds:

  * (1)

$\mathrm{\Gamma}$ is the Schläfli graph or its complement;

  * (2)

$\mathrm{\Gamma}$ is the Hamming graph $H(s,p^{t})$ with $p^{t}\geq 5$ and
$st=n$, or $\overline{H(2,p^{t})}$ with $p^{t}\geq 5$;

  * (3)

$\mathrm{\Gamma}$ is a normal Cayley graph on $\mathbb{Z}_{p}^{n}$;

Schläfli graph and Hamming graph are defined in Examples 3.2 and 3.3.

Let $p$ be a prime. Marušič [32, Theorem 3.4] proved that a vertex transitive
graph of order $p^{2}$ or $p^{3}$ is a Cayley graph, and $2$-arc transitive
Cayley graphs of order $p^{2}$ were classified by Marušič [33, Corollary 2.3].
This, together with [29, Corollary 1.3 and 3.5], enables us to obtain the
following result.

######  Proposition 2.13.

For an odd prime $p$, a connected $2$-arc transitive graph of order $p^{2}$ is
$\mathbf{C}_{p^{2}}$ or $\mathsf{K}_{p^{2}}$, and a connected $2$-arc
transitive graph of order $p^{3}$ is $\mathbf{C}_{p^{3}}$,
$\mathsf{K}_{p^{3}}$, or a cover of a complete graph.

##  3\. Examples

In this section, we give some 2-geodesic transitive graphs, and these contain
all $2$-geodesic transitive graphs in question except a new construction in
Section 4.

It is well known that $\mathbf{C}_{n}$, $\mathsf{K}_{n}$, $\mathsf{K}_{n,n}$,
and $\mathsf{K}_{n,n}-n\mathsf{K}_{2}$ are $2$-geodesic and $2$-arc
transitive, and $\mathrm{Aut}(\mathbf{C}_{n})\cong\mathrm{D}_{2n}$,
$\mathrm{Aut}(\mathsf{K}_{n})\cong\mathrm{S}_{n}$,
$\mathrm{Aut}(\mathsf{K}_{n,n})\cong(\mathrm{S}_{n}\times\mathrm{S}_{n})\rtimes\mathbb{Z}_{2}$
and
$\mathrm{Aut}(\mathsf{K}_{n,n}-n\mathsf{K}_{2})\cong\mathrm{S}_{n}\times\mathbb{Z}_{2}$.
However, the complete multipartite graph $\mathsf{K}_{m[b]}$ with $m\geq 3$
and $b\geq 2$, is $2$-geodesic but not $2$-arc transitive, see [6, Example
2.2(1)] and [23, Lemma 2.4].

######  Example 3.1.

Let $\mathrm{\Gamma}=\mathsf{K}_{m[b]}$ with $m\geq 3$ and $b\geq 2$. Then
$\mathrm{Aut}(\mathrm{\Gamma})\cong\mathrm{S}_{b}\wr\mathrm{S}_{m}$ is
imprimitive on $V(\mathrm{\Gamma})$, and $\mathrm{\Gamma}$ is $2$-geodesic but
not $2$-arc transitive.

Let $G$ be a transitive permutation group on $\Omega$ and let $O$ be an orbit
of $G$ on $\Omega\times\Omega$. If $O$ is self-paired, that is, $O=\\{(u,v)\
|\ (v,u)\in O\\}$, then the orbital graph of $G$ corresponding to $O$ has
vertex set $\Omega$ and edge set $\\{\\{u,v\\}\ |\ (u,v)\in O\\}$. The
following is the Schläfli graph, see [8, P.26] and [22, Lemma 3.4].

######  Example 3.2.

Let $T\cong\mathrm{PSU}(4,2)$ and
$H\cong\mathbb{Z}_{2}^{4}\rtimes\mathrm{A}_{5}$. Then $T$ is a primitive
permutation group on $[T:H]$, the set of right cosets of $H$ in $T$, and there
are two orbital graphs of valency $16$ and $10$ of $T$ on $[T:H]$. Both graphs
have order $27$ and are $2$-geodesic but not $2$-arc transitive, with
automorphism groups isomorphic to
$\mathrm{Aut}(T)=\mathrm{PSU}(4,2).\mathbb{Z}_{2}$. The graph with valency
$16$ is called Schläfli graph and the graph with valency $10$ is the
complementary graph of the Schläfli graph in $\mathsf{K}_{27}$.

The following is the Hamming graph, see [4, Section 9.2] and [26, Proposition
2.2].

######  Example 3.3.

Let $d,n$ be positive integers with $d,n\geq 2$. The Hamming graph $H(d,n)$ is
defined to be the Cayley graph $\mathrm{Cay}(G,S)$ with $G=\langle
a_{1}\rangle\times\langle a_{2}\rangle\times\cdots\times\langle
a_{d}\rangle\cong\mathbb{Z}_{n}^{d}$ and $S=\langle
a_{1}\rangle^{*}\cup\langle a_{2}\rangle^{*}\cup\cdots\cup\langle
a_{d}\rangle^{*}$. Then $H(d,n)$ is geodesic transitive and
$\mathrm{Aut}(\mathrm{\Gamma})\cong\mathrm{S}_{n}\wr\mathrm{S}_{d}$.
Furthermore, the complementary graph $\overline{H(2,n)}$ of $H(2,n)$ in the
complete graph $\mathsf{K}_{n^{2}}$, with $n\geq 3$, is $2$-geodesic
transitive.

For an odd prime $p$, the extraspecial group $E(p^{3})$ of exponent $p$ is
defined as follows:

$E(p^{3})=\langle a,b,c\mid a^{p}=b^{p}=c^{p}=1,[a,b]=c,[c,a]=[c,b]=1\rangle.$
(1)

There are two infinite families of $2$-geodesic but not $2$-arc transitive
normal Cayley graphs on $E(p^{3})$, of which one family is given in the
following example which was first constructed in [20]. Another family will be
constructed and studied in the next section.

######  Example 3.4.

Let $p$ be an odd prime and let $S=\\{a^{i},b^{i}\mid
i\in\mathbb{Z}_{p}^{*}\\}$. Then $\mathrm{Cay}(E(p^{3}),S)$ is normal and
$\mathrm{Aut}(\mathrm{Cay}(E(p^{3}),S))\cong
E(p^{3})\rtimes((\mathbb{Z}_{p-1}\times\mathbb{Z}_{p-1})\rtimes\mathbb{Z}_{2})$.
Furthermore, $\mathrm{Cay}(E(p^{3}),S)$ is $2$-geodesic but not $2$-arc
transitive.

Proof. Let $\mathrm{\Gamma}=\mathrm{Cay}(E(p^{3}),S)$ and
$A=\mathrm{Aut}(\mathrm{\Gamma})$. From [20, Theorem 1.2, Lemmas 3.1 and 3.3],
$\mathrm{\Gamma}$ is normal and is not $2$-arc transitive with
$\mathrm{Aut}(\mathrm{\Gamma})\cong
E(p^{3})\rtimes((\mathbb{Z}_{p-1}\times\mathbb{Z}_{p-1})\rtimes\mathbb{Z}_{2})$.
Let $\mathbb{Z}_{p}^{*}=\langle t\rangle$, and let $\beta$ be the automorphism
of $E(p^{3})$ induced by $a\mapsto a$ and $b\mapsto b^{t}$. Then
$\langle\beta\rangle$ is transitive on
$\mathrm{\Gamma}_{2}(1)\cap\mathrm{\Gamma}(a)$, and hence $\mathrm{\Gamma}$ is
$2$-geodesic transitive.∎

To end this section, we give a proposition which lists all $2$-geodesic
transitive graphs of order $4,8,9,25$ or $27$. This was obtained from the list
of all connected symmetric graphs with orders from $2$ to $30$ in [7]. For
convenience, we write

$\begin{array}[]{l}\mathcal{G}_{(27,4)}=\mathrm{Cay}(E(3^{3}),\\{a,b,a^{-1},b^{-1}\\}),\\\
\mathcal{G}_{(27,8)}=\mathrm{Cay}(E(3^{3}),\langle
b\rangle^{*}\cup_{i\in\mathbb{Z}_{3}}\langle
b^{i}ab^{i}\rangle^{*}).\end{array}$

Note that $\mathcal{G}_{27,4}$ is just the graph for $p=3$ in Example 3.4
while $\mathcal{G}_{(27,8)}$ is just the graph for $p=3$ in Theorem 4.3.

######  Proposition 3.5.

Let $\mathrm{\Gamma}$ be a connected $2$-geodesic transitive graph of order
$n$. If $\mathrm{\Gamma}$ is $2$-arc transitive, then

  * (1)

for $n=4$, $\mathrm{\Gamma}\cong\mathbf{C}_{4}$ or $\mathsf{K}_{4}$;

  * (2)

for $n=8$, $\mathrm{\Gamma}\cong\mathbf{C}_{8},H(3,2),\mathsf{K}_{4,4}$ or
$\mathsf{K}_{8}$;

  * (3)

for $n=9$, $\mathrm{\Gamma}\cong\mathbf{C}_{9}$ or $\mathsf{K}_{9}$;

  * (4)

for $n=25$, $\mathrm{\Gamma}\cong\mathbf{C}_{25}$ or $\mathsf{K}_{25}$;

  * (5)

for $n=27$, $\mathrm{\Gamma}\cong\mathbf{C}_{27}$ or $\mathsf{K}_{27}$.

If $\mathrm{\Gamma}$ is not $2$-arc transitive, then $\mathrm{\Gamma}$ cannot
be of order $4$ and

  * (1)

for $n=8$, $\mathrm{\Gamma}\cong\mathsf{K}_{4[2]}$;

  * (2)

for $n=9$, $\mathrm{\Gamma}\cong H(2,3)$ or $\mathsf{K}_{3[3]}$;

  * (3)

for $n=25$, $\mathrm{\Gamma}\cong H(2,5),\overline{H(2,5)}$, or
$\mathsf{K}_{5[5]}$;

  * (4)

for $n=27$,
$\mathrm{\Gamma}\cong\mathcal{G}_{(27,4)},H(3,3),\mathcal{G}_{(27,8)},\mathsf{K}_{9[3]},\mathsf{K}_{3[9]}$,
or the Schläfli graph or its complement.

##  4\. A family of $2$-geodesic transitive Cayley graphs on $E(p^{3})$

The goal of this section is to prove that the Cayley graph
$\mathrm{Cay}(E(p^{3}),S)$ with $S=\langle
b\rangle^{*}\cup_{i\in\mathbb{Z}_{p}}\langle b^{i}ab^{i}\rangle^{*}$ is
$2$-geodesic transitive but not $2$-arc transitive. To do this, we need the
following two lemmas, which will also be used in other Sections.

######  Lemma 4.1.

Let $p\geq 5$ be a prime, and let
$A=\mathbb{Z}_{p}^{2}\rtimes\mathrm{GL}(2,p)$. Then $A\cong\mathrm{AGL}(2,p)$
if and only if every non-identity in the center of $\mathrm{GL}(2,p)$ cannot
commute with every element of order $p$ in $\mathbb{Z}_{p}^{2}$.

Proof. The necessity is well known. To prove the sufficiency, let
$L=\mathbb{Z}_{p}^{2}$. Then $A=L\mathrm{GL}(2,p)$ and
$A/L\cong\mathrm{GL}(2,p)$. To finish the proof, we only need to show that
$\mathrm{soc}(A)=L$. Let $N$ be a minimal normal subgroup of $A$. Then
$N=T^{n}$ for a positive integer $n$ and a simple group $T$. Set
$Z=Z(\mathrm{GL}(2,p))$. Then $LZ\unlhd A$ and $A/LZ\cong\mathrm{PGL}(2,p)$.
Clearly, $\mathrm{soc}(\mathrm{PGL}(2,p))\cong\mathrm{PSL}(2,p)$. By [21, P.
181, (6.10)],
$\mathrm{GL}(2,p)^{\prime}=\mathrm{SL}(2,p)\leq\mathrm{GL}(2,p)$.

Suppose that $T$ is a nonabelian simple group. Then $L\cap N=1$ and $LZ\cap
N=1$. It follows $N\cong N/(LZ\cap N)\cong NLZ/LZ\unlhd
A/LZ\cong\mathrm{PSL}(2,p)$, and then $\mathrm{PSL}(2,p)\cong N\cong N/(N\cap
L)\cong NL/L\unlhd A/L\cong\mathrm{GL}(2,p)$, which is impossible.

Thus, $T$ is abelain, and $N=T^{n}$ with $T\cong\mathbb{Z}_{q}$ for a prime
$q$. Suppose that $q\not=p$. Since $L$ is $p$-group,
$L\cap\mathrm{GL}(2,p)N=1$, and
$\mathrm{GL}(2,p)N\cong\mathrm{GL}(2,p)N/(L\cap\mathrm{GL}(2,p)N)\cong
N\mathrm{GL}(2,p)L/L=A/L\cong\mathrm{GL}(2,p)$, forcing
$N\mathrm{GL}(2,p)=\mathrm{GL}(2,p)$. Thus, $N\leq\mathrm{GL}(2,p)$ and hence
$N\leq Z$. On the other hand, since $N\unlhd A$ and $L\unlhd A$, $N\cap L=1$
implies that $N$ commutes with $L$ pointwise, contradicting the hypothesis.
Thus, $q=p$ and $N=\mathbb{Z}_{p}^{n}$.

If $NL\not=L$, then $NL/L$ is non-trivial $p$-group, and this is impossible
because $NL/L\unlhd A/L\cong\mathrm{GL}(2,p)$. Thus, $NL=L$ and $N\leq L$. It
follows that every minimal normal subgroup of $A$ is a subgroup of $L$, that
is, $\mathrm{soc}(A)\leq L$.

To prove $\mathrm{soc}(A)=L$, the left thing is to show that $L$ is a minimal
normal subgroup of $A$. Suppose that $L_{1}=\langle\sigma\rangle\leq L$ such
that $|L_{1}|=p$ and $L_{1}\unlhd A$. Since
$\mathrm{SL}(2,p)\leq\mathrm{GL}(2,p)$, we consider the conjugate action of
$\mathrm{SL}(2,p)$ on $L_{1}$. By [39, P. 393, (6.3)], $\mathrm{SL}(2,p)$ has
a unique involution, say $\beta$, and $\mathrm{SL}(2,p)$ has an element of
order $4$, say $\gamma$. Then $\gamma^{2}=\beta$. Clearly, $\beta\in Z$. By
assumption, $\sigma\not=\sigma^{\beta}\in\langle\sigma\rangle$, and since
$\beta$ is an involution, $\sigma^{\beta}=\sigma^{-1}$. Then the orbit set of
$\langle\beta\rangle$ on $L_{1}^{*}$ is
$\Delta=\\{\\{\sigma^{i},\sigma^{-i}\\}\mid i\in\mathbb{Z}_{p}^{*}\\}$. Since
$\langle\beta\rangle\unlhd\mathrm{SL}(2,p)$,
$\mathrm{SL}(2,p)/\langle\beta\rangle\cong\mathrm{PSL}(2,p)$ has an induced
natural action on $\Delta$, and then the simplicity of $\mathrm{PSL}(2,p)$
implies the action is trivial because $\mathrm{PSL}(2,p)$ has no non-trivial
action of degree less then $p$ by Proposition 2.3. Thus, $\mathrm{PSL}(2,p)$
fixes $\Delta$ pointwise. In particular, $\gamma$ fixes each
$\\{\sigma^{i},\sigma^{-i}\\}$ for all $i\in\mathbb{Z}_{p}^{*}$. It follows
that $\gamma^{2}=\beta$ fixes $\langle\sigma\rangle$ pointwise, contradicting
$\sigma^{\beta}=\sigma^{-1}$. This means that $L$ is a minimal normal subgroup
of $A$ and hence $\mathrm{soc}(A)=L$, that is, $A\cong\mathrm{AGL}(2,p)$.∎

######  Lemma 4.2.

Let $p\geq 5$ be a prime. Then we have the following:

  * (1)

Every subgroup $G$ of $\mathrm{GL}(2,p)$ with index less than $p$, contains
$\mathrm{SL}(2,p)=\mathrm{GL}(2,p)^{\prime}$, and for every $m\mid(p-1)$,
$\mathrm{GL}(2,p)$ has exactly one subgroup of index $m$.

  * (2)

For each subgroup $G$ with $\mathrm{SL}(2,p)\leq G\leq\mathrm{GL}(2,p)$,
$\mathrm{AGL}(2,p)$ has exactly one conjugate class of subgroups isomorphic to
$G$.

  * (3)

$\mathrm{Aut}(E(p^{3}))\cong\mathrm{AGL}(2,p)$.

Proof. To prove (1), let $G\leq\mathrm{GL}(2,p)$ such that
$|\mathrm{GL}(2,p):G|\bigm{|}<p$. Then $|\mathrm{GL}(2,p)/Z:GZ/Z|<p$, where
$Z=Z(\mathrm{GL}(2,p))$. Since $\mathrm{GL}(2,p)/Z\cong\mathrm{PGL}(2,p)$, by
Proposition 2.3 we have $\mathrm{GL}(2,p)/Z=GZ/Z$, implying
$\mathrm{GL}(2,p)=GZ$. By [21, P. 181, (6.10)],
$\mathrm{GL}(2,p)^{\prime}=\mathrm{SL}(2,p)$, and hence $G\geq
G^{\prime}=(GZ)^{\prime}=\mathrm{SL}(2,p)$. Now let $|\mathrm{GL}(2,p):G|=m$.
Then $m\mid(p-1)$ and $\mathrm{SL}(2,p)\leq G\leq\mathrm{GL}(2,p)$. Since
$\mathrm{GL}(2,p)/\mathrm{SL}(2,p)\cong\mathbb{Z}_{p-1}$, $\mathrm{GL}(2,p)$
has exactly one subgroup of index $m$ for every $m\mid(p-1)$.

To prove (2), we may write $\mathrm{SL}(2,p)\leq G\leq\mathrm{GL}(2,p)\leq
A=\mathrm{AGL}(2,p)$, and let $B$ be a subgroup of $A$ such that $B\cong G$.
We only need to prove that $B$ and $G$ are conjugate in $A$. Let $L$ be the
unique minimal normal subgroup of $A$. Then $L\cong\mathbb{Z}_{p}^{2}$. Note
that
$\mathrm{GL}(2,p)^{\prime}=G^{\prime}=\mathrm{SL}(2,p)^{\prime}=\mathrm{SL}(2,p)$.
Since $B\cong G$, we have $B^{\prime}\cong\mathrm{SL}(2,p)$.

Set $M=\mathrm{SL}(2,p)L$. Then $\mathrm{SL}(2,p)\cong M/L\leq
A/L\cong\mathrm{GL}(2,p)$. Since $B^{\prime}\cong\mathrm{SL}(2,p)$, we have
$L\cap B^{\prime}=1$ and $B^{\prime}L/L\cong B^{\prime}/(B^{\prime}\cap
L)\cong B^{\prime}\cong\mathrm{SL}(2,p)$. Then both $M/L$ and $B^{\prime}L/L$
are subgroups of index $p-1$ in $A/L$. By (1), $M/L=B^{\prime}L/L$, that is,
$M=B^{\prime}L$.

Denote by $\delta$ and $\gamma$ be the center involutions of
$\mathrm{SL}(2,p)$ and $B^{\prime}$, respectively. Then
$L\langle\delta\rangle\unlhd M$ and
$M/L\langle\delta\rangle\cong\mathrm{PSL}(2,p)$. It follows that
$\mathrm{PSL}(2,p)\cong
M/L\langle\delta\rangle=B^{\prime}L/L\langle\delta\rangle\cong
B^{\prime}/(B^{\prime}\cap L\langle\delta\rangle)$, forcing $B^{\prime}\cap
L\langle\delta\rangle=\langle\gamma\rangle$. Thus, $\gamma\in
L\langle\delta\rangle$. By the Sylow Theorem, $\delta$ and $\gamma$ are
conjugate in $L\langle\delta\rangle$. Therefore, the centralizer
$C_{A}(\langle\delta\rangle)$ of $\langle\delta\rangle$ in $A$ is isomorphic
to the centralizer $C_{A}(\langle\gamma\rangle)$. By Lemma 4.1,
$C_{A}(\langle\delta\rangle)=\mathrm{GL}(2,p)$, and hence
$C_{A}(\langle\gamma\rangle)\cong\mathrm{GL}(2,p)$. Thus, $\mathrm{GL}(2,p)$
and $C_{A}(\langle\gamma\rangle)$ are conjugate in $A$.

Since $G\cong B$ and $C_{A}(\langle\gamma\rangle)\cong\mathrm{GL}(2,p)$, we
may write $|\mathrm{GL}(2,p):G|=|C_{A}(\langle\gamma\rangle):B|=m$. By (1),
$\mathrm{GL}(2,p)$ has a unique subgroup of index $m$, that is, $G$, and
$C_{A}(\langle\gamma\rangle)$ has a unique subgroup of index $m$, that is,
$B$. It follows that $G$ and $B$ are conjugate in $A$.

To prove (3), let $A=\mathrm{Aut}(E(p^{3}))$ and $C=\langle c\rangle$. By Eq.
(1), $C=Z(E(p^{3}))$. Then $A$ fixes $C$ and $E(p^{3})\backslash C$ setwise.
Let $x,y\in E(p^{3})\backslash C$ with $x\not\in\langle y\rangle$. Then
$\langle x,y\rangle=E(p^{3})$. It is easy to see that $x$ and $y$ have the
same relations as $a$ and $b$, given in Eq. (1). By the von Dyck s Theorem
(see [36, P. 51, 2.2.1]), the mapping

$\sigma_{x,y}:a\mapsto x,\ b\mapsto y$ (2)

induces an automorphism of $E(p^{3})$, denoted by $\sigma_{x,y}$. Clearly,
$c^{\sigma_{x,y}}=[a,b]^{\sigma_{x,y}}=[x,y]$. By [41, Theorem 1.1],
$\mathrm{Aut}(E(p^{3})\cong\mathbb{Z}_{p}^{2}\rtimes\mathrm{GL}(2,p)$, and we
may assume that $\mathrm{GL}(2,p)\leq A$.

Claim 1. Let $L=\langle\sigma_{a,bc},\sigma_{ac,b}\rangle$. Then
$A=L\rtimes\mathrm{GL}(2,p)\cong\mathrm{AGL}(2,p)$ and $L$ is the unique
minimal normal subgroup of $\mathrm{Aut}(E(p^{3}))$.

It is easy to check that
$\sigma_{ac^{i},bc^{j}}\sigma_{ac^{k},bc^{l}}=\sigma_{ac^{i+k},bc^{j+l}}$ for
all $i,j,k,l\in\mathbb{Z}_{p}$, and so $L=\\{\sigma_{ac^{i},bc^{j}}\ |\
i,j\in\mathbb{Z}_{p}\\}\cong\mathbb{Z}_{p}^{2}$. Clearly, $L\leq A$.

Since $A$ fixes $C$ setwise, $A$ has a natural action on the quotient group
$E(p^{3})/C=\\{a^{i}b^{j}C\ |\ i,j\in\mathbb{Z}_{p}\\}$. Let $K$ be the kernel
of this action. Then $K\unlhd A$. Clearly, $L\leq K$. Take $\sigma\in K$. Then
$\sigma$ fixes $aC$ and $bC$ setwise, respectively, and hence
$\sigma=\sigma_{ac^{i},bc^{j}}$ for some $i,j\in\mathbb{Z}_{p}$, implying
$\sigma\in L$. It follows that $K=L$ and $L$ is a normal subgroup of $A$. In
particular, $A/L$ has a faithful action on
$E(p^{3})/C\cong\mathbb{Z}_{p}^{2}$. Since $L\unlhd A$, we have
$L\cap\mathrm{GL}(2,p)=1$ and so $A=L\rtimes\mathrm{GL}(2,p)$. Then
$\mathrm{GL}(2,p)$ induces the automorphism group of the group $E(p^{3})/C$,
and for convenience, we also write
$\mathrm{GL}(2,p)=\mathrm{Aut}(E(p^{3})/C)$.

Let $Z$ be the center of $\mathrm{GL}(2,p)$. Take a non-identity $\alpha\in
Z$. Let $q$ be the order of $\alpha$. Then $q\neq 1$, and since
$Z\cong\mathbb{Z}_{p-1}$, we have $q\mid(p-1)$. Noting that $\alpha\in
Z\leq\mathrm{GL}(2,p)=\mathrm{Aut}(E(p^{3})/C)$, $\alpha$ maps $aC$ to
$a^{r}C$, and $bC$ to $b^{r}C$, where $r$ is an element of order $q$ in
$\mathbb{Z}_{p}^{*}$. Then $p\nmid r$ and $r\not=1$ as $q\not=1$. It follows
that $a^{\alpha}=a^{r}c_{1}$ and $b^{\alpha}=b^{r}c_{2}$ for some
$c_{1},c_{2}\in C$. In particular, $c^{\alpha}=[a,b]^{\alpha}=c^{r^{2}}$.
Clearly, $c^{r}\not=1$ as $p\nmid r$. For every
$1\not=\sigma_{ac^{s},bc^{t}}\in L$, we have $c^{s}\neq 1$ or $c^{t}\not=1$.
If $c^{t}\not=1$, then
$b^{\alpha\sigma_{ac^{s},bc^{t}}}=b^{r}c_{2}c^{tr}\not=b^{r}c_{2}c^{tr^{2}}=b^{\sigma_{ac^{s},bc^{t}}\alpha}$,
and if $c^{s}\not=1$, then
$a^{\alpha\sigma_{ac^{s},bc^{t}}}=a^{r}c_{1}c^{rs}\not=a^{r}c_{1}c^{r^{2}s}=a^{\sigma_{ac^{s},bc^{t}}\alpha}$.
Thus, $\alpha$ cannot commute with $\sigma_{ac^{s},bc^{t}}$. By Lemma 4.1,
$A\cong\mathrm{AGL}(2,p)$, and hence $L$ is the unique minimal normal subgroup
of $\mathrm{Aut}(E(p^{3}))$. ∎

Now we are ready to prove the main result of this section.

######  Theorem 4.3.

For an odd prime $p$, let $E(p^{3})$ be given as in Eq. (1), and let
$S=\langle b\rangle^{*}\cup_{i\in\mathbb{Z}_{p}}\langle
b^{i}ab^{i}\rangle^{*}$. Then $\mathrm{Aut}(E(p^{3}),S)\cong\mathrm{GL}(2,p)$,
and $\mathrm{Cay}(E(p^{3}),S)$ is $2$-geodesic and distance transitive, but
not $2$-arc transitive. Furthermore,

  * (1)

For every $m\mid(p-1)$, the unique subgroup of index $m$ in
$\mathrm{Aut}(E(p^{3}),S)$ has $2m+2$ orbits: one orbit has length $1$, one
orbit has length $p^{2}-1$, namely $S$, $m$ orbits have length $(p-1)/m$, and
$m$ orbits have length $(p^{2}-1)(p-1)/m$;

  * (2)

$\mathrm{Aut}(E(p^{3}),S)^{\prime}\cong\mathrm{SL}(2,p)$ has orbit-set
$\\{\\{c^{i}\\},c^{i}S\ |\ i\in\mathbb{Z}_{p}\\}$;

  * (3)

$|SS|=p^{2}+(p^{2}-1)(p-1)$, where $SS=\\{s_{1}s_{2}\mid s_{1},s_{2}\in S\\}$.

Proof. When $p=3$, using Magma [2], one may verify that all results are true.
In what follows, we assume that $p\geq 5$.

Let $\mathrm{\Gamma}=\mathrm{Cay}(E(p^{3}),S)$ and $C=\langle c\rangle$. Note
that $E(p^{3})^{\prime}=Z(E(p^{3}))=C$. The following formulae are some basic
facts and very useful, which follow from [17, Theorem 2.1 and Lemma 2.2]. For
$x,y,z\in E(p^{3})$ and $i\in\mathbb{Z}_{p}$, we have $xy=yx[x,y]$,
$[x^{i},y]=[x,y^{i}]=[x,y]^{i}$, $[xy,z]=[x,z][y,z]$, $[x,yz]=[x,y][x,z]$, and
for every integer $k$,

$(xy)^{k}=x^{k}y^{k}[y,x]^{2^{-1}k(k-1)},$

where $2^{-1}$ is the inverse of $2$ in $\mathbb{Z}_{p}^{*}$.

For $i,j\in\mathbb{Z}_{p}$ with $i\not=j$, it is easy to see that $\langle
b^{i}ab^{i}\rangle\cap\langle b^{j}ab^{j}\rangle=1$. Then
$|S|=p-1+p(p-1)=p^{2}-1$. Set

$T=\langle a\rangle^{*}\cup_{i\in\mathbb{Z}_{p}}\langle
a^{i}ba^{i}\rangle^{*}.$

We also have $|T|=p^{2}-1$.

Let $i\in\mathbb{Z}_{p}^{*}$. Then $a^{i}ba^{i}=ba^{2i}c^{i}$ and
$(b^{4^{-1}i^{-1}}ab^{4^{-1}i^{-1}})^{2i}=(b^{2^{-1}i^{-1}}ac^{4^{-1}i^{-1}})^{2i}=ba^{2i}[a,b^{2^{-1}i^{-1}}]^{2^{-1}2i(2i-1)}c^{2^{-1}}=ba^{2i}c^{2^{-1}i^{-1}2^{-1}2i(2i-1)}c^{2^{-1}}=ba^{2i}c^{i}$.
It follows that $a^{i}ba^{i}=(b^{4^{-1}i^{-1}}ab^{4^{-1}i^{-1}})^{2i}$,
implying that $T\subseteq S$. Since $|S|=|T|$, we have $S=T$.

Let $B=\mathrm{Aut}(E(p^{3}),S)$. Then
$B\leq\mathrm{Aut}(\mathrm{\Gamma})_{1}$ fixes $C$ setwise. Since $S=T$, we
have

$S=\langle a\rangle^{*}\cup S_{1},\mbox{ where
}S_{1}=\cup_{i\in\mathbb{Z}_{p}}\langle a^{i}ba^{i}\rangle^{*}=\langle
b\rangle^{*}\cup_{i\in\mathbb{Z}_{p}^{*}}\langle b^{i}ab^{i}\rangle^{*}.$

Let $x,y\in E(p^{3})\backslash C$ with $x\not\in\langle y\rangle$. Then
$\langle x,y\rangle=E(p^{3})$. We still use $\sigma_{x,y}$ to denote the
automorphism of $E(p^{3})$ defined by Eq. (2).

Since $S=\langle a\rangle^{*}\cup S_{1}$, the stabilizer $B_{a}$ of $a$ in $B$
fixes $S_{1}$ setwise. Furthermore, for all $i\in\mathbb{Z}_{p}$ we have
$\sigma_{a,a^{i}ba^{i}}\in B_{a}$ as $S_{1}=\cup_{i\in\mathbb{Z}_{p}}\langle
a^{i}ba^{i}\rangle^{*}$, and for all $j\in\mathbb{Z}_{p}^{*}$ we have
$\sigma_{a,b^{j}}\in B_{a}$ as $S_{1}=\langle
b\rangle^{*}\cup_{i\in\mathbb{Z}_{p}^{*}}\langle b^{i}ab^{i}\rangle^{*}$. It
follows that $B_{a}$ is transitive on $S_{1}$. Since $E(p^{3})$ can be
generated by $a$ and $x$, for every $x\in S_{1}$, $B_{a}$ is regular on
$S_{1}$. Thus, $|B_{a}|=p(p-1)$. Similarly, $B_{b}$ is transitive on
$\cup_{i\in\mathbb{Z}_{p}}\langle b^{i}ab^{i}\rangle^{*}$, and hence $B$ is
transitive on $S$. It follow that
$|B|=(p^{2}-1)|B_{a}|=p(p^{2}-1)(p-1)=|\mathrm{GL}(2,p)|$.

By Lemma 4.2, $\mathrm{Aut}(E(p^{3}))\cong\mathrm{AGL}(2,p)$, and by Claim 1
of the Proof of Lemma 4.2 (3),
$L=\langle\sigma_{a,bc}\rangle\times\langle\sigma_{ac,b}\rangle\cong\mathbb{Z}_{p}^{2}$
is the unique minimal normal subgroup of $\mathrm{Aut}(E(p^{3}))$. Take
$\sigma_{ac^{i},bc^{j}}\in L$ for $i,j\in\mathbb{Z}_{p}$. If
$\sigma_{ac^{i},bc^{j}}\in B$, then $\sigma_{ac^{i},bc^{j}}$ fixes $a$ because
$S=\langle a\rangle^{*}\cup_{i\in\mathbb{Z}_{p}}\langle
a^{i}ba^{i}\rangle^{*}$, and $\sigma_{ac^{i},bc^{j}}$ fixes $b$ because
$S=\langle b\rangle^{*}\cup_{i\in\mathbb{Z}_{p}}\langle
b^{i}ab^{i}\rangle^{*}$. It follows that $\sigma_{ac^{i},bc^{j}}=1$, that is,
$L\cap B=1$. This implies that
$|LB|=|L||B|=p^{3}(p^{2}-1)(p-1)=|\mathrm{AGL}(2,p)|$, and so
$\mathrm{Aut}(E(p^{3}))=L\rtimes B$. In particular,
$B\cong\mathrm{Aut}(E(p^{3}))/L\cong\mathrm{GL}(2,p)$.

Since $S=\langle b\rangle^{*}\cup_{i\in\mathbb{Z}_{p}}\langle
b^{i}ab^{i}\rangle^{*}$, we have $E(p^{3})/C=\\{C,sC\ |\ s\in S\\}$, and since
$|S|=p^{2}-1$, we have $S\cap sC=\\{s\\}$ for all $s\in S$. Since
$B\cong\mathrm{GL}(2,p)$, we have $\mathrm{SL}(2,p)\cong B^{\prime}$, and
since $B$ fixes $C$, $B$ induces the automorphism group of
$E(p^{3})/C\cong\mathbb{Z}_{p}^{2}$. Thus, $B^{\prime}$ is transitive on
$(E(p^{3})/C)^{*}$, and since $S\cap sC=\\{s\\}$ for all $s\in S$,
$B^{\prime}$ is transitive on $S$.

Recall that $|B_{a}|=p(p-1)$ and $\sigma_{a,b^{2}}\in B_{a}$. Then
$B_{a}\cong\mathbb{Z}_{p}\rtimes\mathbb{Z}_{p-1}$, and $B_{a}$ is transitive
on $C\backslash\\{1\\}$ as
$c^{\sigma_{a,b^{2}}}=[a,b]^{\sigma_{a,b^{2}}}=c^{2}$, implying that $B_{a}$
is transitive on $aC\backslash\\{a\\}$. It follows that
$|B_{a}:(B_{a})_{ac}|=p-1$ and $|(B_{a})_{ac}|=p$. On the other hand, $B_{ac}$
fixes $aC\cap S=\\{a\\}$, and so $|B_{ac}|=|(B_{a})_{ac}|=p$. Let $(ac)^{B}$
be the orbit of $B$ containing $ac$. Then
$|(ac)^{B}|=|B|/|B_{ac}|=(p^{2}-1)(p-1)$.

Recall that $B\leq\mathrm{Aut}(\mathrm{\Gamma})_{1}$. Since
$1+(p^{2}-1)+(p^{2}-1)(p-1)+(p-1)=p^{3}$, $B$ has exactly four orbits on
$V(\mathrm{\Gamma})$, that is, $\\{1\\}$, $S$, $(ac)^{B}$ and
$C\backslash\\{1\\}$. Since $S=\langle
b\rangle^{*}\cup_{i\in\mathbb{Z}_{p}}\langle b^{i}ab^{i}\rangle^{*}$, we have
$aS\cap C\backslash\\{1\\}=\emptyset$, and since $B$ is transitive on $S$, we
have $SS\cap C\backslash\\{1\\}=\emptyset$. It follows that
$\mathrm{\Gamma}(1)=S$, $\mathrm{\Gamma}_{2}(1)=(ac)^{B}$, and
$\mathrm{\Gamma}_{3}(1)=C\backslash\\{1\\}$. In particular, $\mathrm{\Gamma}$
has diameter $3$ and is distance transitive. Since $\langle
a\rangle^{*}\subset S$, $\mathrm{\Gamma}$ has girth $3$, and hence it is not
$2$-arc transitive. Recall that $S=\langle a\rangle^{*}\cup S_{1}$ and $B_{a}$
is regular on $S_{1}$. Then $B_{a}$ is regular on $\\{sa\ |\ s\in S_{1}\\}$.
Since $a$ is adjacent to all vertices in $\\{sa\ |\ s\in S_{1}\\}$, we have
$\\{as\ |\ s\in S_{1}\\}\not\subseteq S$ (otherwise $\mathrm{\Gamma}$ is a
complete graph). It follows that $\\{as\ |\ s\in
S_{1}\\}\subseteq\mathrm{\Gamma}(a)\cap\mathrm{\Gamma}_{2}(1)$. Since $\langle
a\rangle^{*}\subseteq S$, we have
$|\mathrm{\Gamma}(a)\cap\mathrm{\Gamma}_{2}(1)|\leq|S_{1}|$, and hence
$\mathrm{\Gamma}(a)\cap\mathrm{\Gamma}_{2}(1)=\\{as\ |\ s\in S_{1}\\}$. Thus,
$\mathrm{\Gamma}$ is $2$-geodesic transitive and $B_{a}$ is regular on
$\mathrm{\Gamma}(a)\cap\mathrm{\Gamma}_{2}(1)$.

To prove part (1), let $m\mid(p-1)$ and let $G$ be the unique subgroup of $B$
of index $m$. Then $G\unlhd B$, and by Lemma 4.2, $B^{\prime}\leq G$. Since
$B^{\prime}$ is transitive on $S$, $G$ is transitive on $S$. Since $B_{a}$ is
regular on $\mathrm{\Gamma}(a)\cap\mathrm{\Gamma}_{2}(1)$, $G_{a}$ is
semiregular, and so has $m$ orbits on
$\mathrm{\Gamma}(a)\cap\mathrm{\Gamma}_{2}(1)$ as $m=|B:G|=|B_{a}:G_{a}|$.
Since $B$ is arc-transitive and $G\unlhd B$, $G$ has $m$ orbits on
$\mathrm{\Gamma}_{2}(1)$ with each length $(p^{2}-1)(p-1)/m$. Since
$C\backslash\\{1\\}$ is an orbit of $B$ on $V(\mathrm{\Gamma})$ and
$B\cong\mathrm{GL}(2,p)$, the kernel of $B$ on $C\backslash\\{1\\}$ is
$B^{\prime}\cong\mathrm{SL}(2,p)$. Then $B/B^{\prime}$ is regular on
$C\backslash\\{1\\}$, and since $|B/B^{\prime}:G/B^{\prime}|=|B:G|=m$,
$G/B^{\prime}$ has $m$ orbits on $C\backslash\\{1\\}$. It follows that $G$ has
$m$ orbits on $C\backslash\\{1\\}$ with each length $(p-1)/m$.

To prove part (2), take $G=B^{\prime}=\mathrm{SL}(2,p)$ in the proof of part
(1). Then $B^{\prime}$ has $2p$ orbits: $p$ orbits have length $1$ and $p$
orbits have length $p^{2}-1$. Furthermore, the $p$ orbits of length $1$ are
$\\{c^{i}\\}$ for all $i\in\mathbb{Z}_{p}$, and the $p$ orbits of length
$p^{2}-1$ are $c^{i}S$ for all $i\in\mathbb{Z}_{p}$.

Note that $\langle s\rangle^{*}\subset S$ for all $s\in S$. Then
$SS=\\{1\\}\cup\mathrm{\Gamma}(1)\cup\mathrm{\Gamma}_{2}(1)$ and
$|SS|=p^{2}+(p^{2}-1)(p-1)$, which is part (3). ∎

##  5\. Proof of Theorem 1.2

Let $p$ be a prime and let $n\geq 2$ be a positive integer. Write
$\mathrm{Aut}(\mathbb{Z}_{p}^{n})=\mathrm{GL}(n,p)$, $Z=Z(\mathrm{GL}(n,p))$,
and $\mathrm{PGL}(n,p)=\mathrm{GL}(n,p)/Z$. For $z\in\mathbb{Z}_{p}^{n}$,
write $\overline{z}=z^{Z}$, the orbit of $Z$ containing $z$, and for
$T\subseteq\mathbb{Z}_{p}^{n}$, set $\overline{T}=\\{\overline{t}\ |\ t\in
T\\}$. Recall that for a subgroup $H$ of a group $G$, $H^{*}$ is the subset of
$H$ deleting $1$. Then $\overline{z}=\langle z\rangle^{*}$ for
$1\not=z\in\mathbb{Z}_{p}^{n}$, but $\overline{1}=\\{1\\}$. The kernel of the
natural action of $\mathrm{GL}(n,p)$ on $\overline{\mathbb{Z}_{p}^{n}}$ is
$Z$, and $\mathrm{PGL}(n,p)$ is a $2$-transitive permutation group on the
point set $\overline{(\mathbb{Z}_{p}^{n})^{*}}=\\{\overline{x}\ |\
1\not=x\in\mathbb{Z}_{p}^{n}\\}$. For $N\leq\mathrm{GL}(n,p)$, write
$\overline{N}=NZ/Z$, and then $\overline{N}\leq\mathrm{PGL}(2,p)$. For points
$\overline{x_{1}},\overline{x_{2}},\cdots,\overline{x_{t}}$ in
$\overline{(\mathbb{Z}_{p}^{n})^{*}}$, denote by
$N_{\overline{x_{1}},\overline{x_{2}},\cdots,\overline{x_{t}}}$ and
$\overline{N}_{\overline{x_{1}},\overline{x_{2}},\cdots,\overline{x_{t}}}$ the
subgroups of $N$ and $\overline{N}$ fixing
$\\{\overline{x_{1}},\overline{x_{2}},\cdots,\overline{x_{t}}\\}$ pointwise,
respectively. First we give some properties on
$\mathrm{Aut}(\mathbb{Z}_{p}^{3})$, which will be used later.

######  Lemma 5.1.

For a prime $p\geq 5$, let $\mathbb{Z}_{p}^{3}=\langle a\rangle\times\langle
b\rangle\times\langle c\rangle$,
$B=\mathrm{Aut}(\mathbb{Z}_{p}^{3})=\mathrm{GL}(3,p)$ and
$\overline{B}=\mathrm{PGL}(3,p)=\mathrm{GL}(3,p)/Z(\mathrm{GL}(3,p))$. Then we
have:

  * (1)

If
$\\{\overline{a},\overline{b},\overline{c},\overline{u}\\}\subseteq\overline{(\mathbb{Z}_{p}^{3})^{*}}$
such that any three points in
$\\{\overline{a},\overline{b},\overline{c},\overline{u}\\}$ generates
$\mathbb{Z}_{p}^{3}$, then
$\overline{B}_{\overline{a},\overline{b},\overline{c}}\cong\mathbb{Z}_{p-1}\times\mathbb{Z}_{p-1}$
and $\overline{B}_{\overline{a},\overline{b},\overline{c},\overline{u}}=1$;

  * (2)

Let $\alpha\in\overline{B}$ be of order a prime $q$ with $q\mid(p-1)$ and
$\beta\in\overline{B}$ be of order $p$. Assume that $\alpha$ commutes with
$\beta$. Then $\alpha$ has a fixed point in
$\overline{(\mathbb{Z}_{p}^{3})^{*}}$. If further
$\alpha\in\overline{B}_{\overline{a},\overline{b},\overline{c}}$ then $\alpha$
has exactly $3$ or $p+2$ fixed points in
$\overline{(\mathbb{Z}_{p}^{3})^{*}}$.

  * (3)

$B_{\overline{c}}\cong\mathrm{AGL}(2,p)\times\mathbb{Z}_{p-1}$ and
$\overline{B}_{\overline{c}}\cong\mathrm{AGL}(2,p)$. Furthermore,
$B_{\overline{c}}$ has exactly one conjugate class of subgroups isomorphic to
$\mathrm{SL}(2,p)$, of which one has orbit set $\\{\\{c^{i}\\},c^{i}\langle
a,b\rangle^{*}\ |\ i\in\mathbb{Z}_{p}\\}$ in $\mathbb{Z}_{p}^{3}$.

Proof. For $i,j,k\in\mathbb{Z}_{p}^{*}$, $\mathbb{Z}_{p}^{3}$ has an
automorphism induced by $a\mapsto a^{i}$, $b\mapsto b^{j}$ and $c\mapsto
c^{k}$, denoted by $\sigma_{a^{i},b^{j},c^{k}}$. Then
$Z=Z(\mathrm{GL}(3,p))=\\{\sigma_{a^{i},b^{i},c^{i}}\ |\
i\in\mathbb{Z}_{p}^{*}\\}$,
$B_{\overline{a},\overline{b},\overline{c}}=\\{\sigma_{a^{i},b^{j},c^{k}}\ |\
i,j,k\in\mathbb{Z}_{p}^{*}\\}\cong\mathbb{Z}_{p-1}\times\mathbb{Z}_{p-1}\times\mathbb{Z}_{p-1}$,
and
$\overline{B}_{\overline{a},\overline{b},\overline{c}}=B_{\overline{a},\overline{b},\overline{c}}/Z\cong\mathbb{Z}_{p-1}\times\mathbb{Z}_{p-1}$
as $Z\leq B_{\overline{x},\overline{y},\overline{z}}$. Let $u=a^{r}b^{s}c^{t}$
with $r,s,t\in\mathbb{Z}_{p}^{*}$. If $\sigma_{a^{i},b^{j},c^{k}}\in
B_{\overline{a},\overline{b},\overline{c}}$ fixes $\overline{u}$, then
$a^{ir}b^{js}c^{kt}=(a^{r}b^{s}c^{t})^{\sigma_{a^{i},b^{j},c^{k}}}=(a^{r}b^{s}c^{t})^{m}$
for some $m\in\mathbb{Z}_{p}^{*}$, implying $i=j=k=m$ and
$\sigma_{a^{i},b^{j},c^{k}}\in Z$. It follows that
$\overline{B}_{\overline{a},\overline{b},\overline{c},\overline{u}}=1$. This
completes the proof of part (1).

To prove part (2), we suppose to the contrary that $\alpha$ has no fixed point
in $\overline{(\mathbb{Z}_{p}^{3})^{*}}$. Since
$|\overline{(\mathbb{Z}_{p}^{3})^{*}}|=p^{2}+p+1$, we have $q\mid(p^{2}+p+1)$,
and hence $q$ is a divisor of $(p^{2}+p+1)-(p-1)=p^{2}+2$. Since
$(p^{2}+2)-p(p-1)=p+2$ and $p+2-(p-1)=3$, we have $q=3$. Since
$p\nmid(p^{2}+p+1)$, $\beta$ has a fixed point in
$\overline{(\mathbb{Z}_{p}^{3})^{*}}$, say $\overline{u}$. Let
$\overline{u}^{\langle\alpha\rangle}=\\{\overline{u},\overline{v},\overline{w}\\}$.
Since $\alpha$ has no fixed point,
$\langle\overline{u},\overline{v},\overline{w}\rangle=\mathbb{Z}_{p}^{2}$ or
$\mathbb{Z}_{p}^{3}$. If
$\langle\overline{u},\overline{v},\overline{w}\rangle=\mathbb{Z}_{p}^{2}$,
then $3\mid(p+1)$, forcing $3$ is a divisor of $(p+1)-(p-1)=2$, a
contradiction. Thus,
$\langle\overline{u},\overline{v},\overline{w}\rangle=\mathbb{Z}_{p}^{3}$.
Since $\alpha$ commutes with $\beta$, $\beta$ fixes $\overline{u}$,
$\overline{v}$ and $\overline{w}$, that is,
$\beta\in\overline{B}_{\overline{u},\overline{v},\overline{w}}$, contradicting
part (1).

Let $\alpha\in\overline{B}_{\overline{a},\overline{b},\overline{c}}$. Then
$\alpha$ fixes $\overline{a},\overline{b}$ and $\overline{c}$, and hence
$\alpha=\sigma_{a^{i},b^{j},c^{k}}Z$ for some $i,j,k\in\mathbb{Z}_{p}^{*}$.
Since $\sigma_{a^{i^{-1}},b^{i^{-1}},c^{i^{-1}}}\in Z$, we have
$\alpha=\sigma_{a^{i},b^{j},c^{k}}Z=\sigma_{a,b^{r},c^{s}}Z$, where
$r=i^{-1}j\in\mathbb{Z}_{p}^{*}$ and $s=i^{-1}k\in\mathbb{Z}_{p}^{*}$. Note
that $o(\alpha)=q$. If $r=1$ then $s\not=1$ and $\alpha$ has the fixed-point
set $\overline{\langle a,b\rangle^{*}}\cup\\{\overline{c}\\}$, and if $s=1$
then $r\not=1$ and $\alpha$ has the fixed-point set $\overline{\langle
a,c\rangle^{*}}\cup\\{\overline{b}\\}$. For $r\not=1$ and $s\not=1$, if $r=s$
then $\alpha$ has the fixed-point set $\\{\overline{a}\\}\cup\overline{\langle
b,c\rangle^{*}}$, and if $r\not=s$ then $\alpha$ has the fixed-point set
$\\{\overline{a},\overline{b},\overline{c}\\}$. Thus, $\alpha$ has exactly $3$
or $p+2$ fixed points in $\overline{(\mathbb{Z}_{p}^{3})^{*}}$.

To prove part (3), let $\mathbb{Z}_{p}^{3}=\langle x,y,z\rangle$ and denote by
$\delta_{x,y,z}$ the automorphism of $\mathbb{Z}_{p}^{3}$ induced by $a\mapsto
x$, $b\mapsto y$ and $c\mapsto z$. Let $L$ be the kernel of $B_{\overline{c}}$
on ${\overline{c}}$. Then $|B_{\overline{c}}/L|\leq p-1$, and since $Z$ is
transitive on ${\overline{c}}$, we have $B_{\overline{c}}=L\times Z$ and
$\overline{B}_{\overline{c}}\cong L$.

Write $\mathrm{Aut}(\langle a,b\rangle)=\mathrm{GL}(2,p)$ and
$\mathrm{GL}(2,p)^{\prime}=\mathrm{SL}(2,p)$. Set
$G=\\{\delta_{a^{\alpha},b^{\alpha},c}\ |\ \alpha\in\mathrm{GL}(2,p)\\}$ and
$H=\\{\delta_{a^{\alpha},b^{\alpha},c}\ |\ \alpha\in\mathrm{SL}(2,p)\\}$. Then
$H\leq G\leq L$ with $H\cong\mathrm{SL}(2,p)$ and $G\cong\mathrm{GL}(2,p)$.
Clearly, $G$ and $H$ have the same orbit set on $\mathbb{Z}_{p}^{3}$, that is,
$\\{\\{c^{i}\\},c^{i}\langle a,b\rangle^{*}\ |\ i\in\mathbb{Z}_{p}\\}$.

Let $M\cong\mathrm{SL}(2,p)$ be a subgroup of $B_{\overline{c}}$. Then
$M=M^{\prime}\leq B_{\overline{c}}^{\prime}=L^{\prime}\leq L$. Note that Lemma
4.2 (2) means that $\mathrm{AGL}(2,p)$ has one conjugate class of subgroups
isomorphic to $\mathrm{SL}(2,p)$. Thus, to finish the proof of part (3), it
suffices to show that $L\cong\mathrm{AGL}(2,p)$.

Let $C=\langle c\rangle$. Since $L$ fixes $C$ pointwise, $L$ has a natural
action on $\mathbb{Z}_{p}^{3}/C$, and let $K$ be the kernel of the action of
$L$ on the quotient group $\mathbb{Z}_{p}^{3}/C$. Then $K$ fixes $aC$ and
$bC$, and hence $K=\\{\delta_{ac_{1},bc_{2},c}\ |\ c_{1},c_{2}\in
C\\}\cong\mathbb{Z}_{p}^{2}$. Since $K\unlhd L$, we have $G\cap K=1$, and
since $|L/K|\leq|\mathrm{Aut}(\mathbb{Z}_{p}^{3}/C)|=|G|$, we have $L=K\rtimes
G$.

Note that $Z(G)=\\{\delta_{a^{r},b^{r},c}\ |\ r\in\mathbb{Z}_{p}^{*}\\}$. Take
$1\not=\delta_{ac_{1},bc_{2},c}\in K$ and $1\not=\delta_{a^{r},b^{r},c}\in
Z(G)$. Then $r\not=1$, and either $c_{1}\not=1$ or $c_{2}\not=1$. If
$c_{1}\not=1$ then
$a^{\delta_{ac_{1},bc_{2},c}\delta_{a^{r},b^{r},c}}=a^{r}c_{1}\not=a^{r}(c_{1})^{r}=a^{\delta_{a^{r},b^{r},c}\delta_{ac_{1},bc_{2},c}}$,
and if $c_{2}\not=1$ then
$b^{\delta_{ac_{1},bc_{2},c}\delta_{a^{r},b^{r},c}}=b^{r}c_{2}\not=b^{r}(c_{2})^{r}=b^{\delta_{a^{r},b^{r},c}\delta_{ac_{1},bc_{2},c}}$.
Thus,
$\delta_{ac_{1},bc_{2},c}\delta_{a^{r},b^{r},c}\neq\delta_{a^{r},b^{r},c}\delta_{ac_{1},bc_{2},c}$.
By Lemma 4.1, $L\cong\mathrm{AGL}(2,p)$, as required. ∎

We also need some properties on cliques of certain normal Cayley graphs on
$\mathbb{Z}_{p}^{n}$.

######  Lemma 5.2.

Let $\mathrm{\Gamma}=\mathrm{Cay}(\mathbb{Z}_{p}^{n},S)$ be a connected
$2$-geodesic but not $2$-arc transitive normal Cayley graph on
$\mathbb{Z}_{p}^{n}$. Then

  * (1)

$\mathrm{Aut}(G,S)$ contains the center of $\mathrm{Aut}(\mathbb{Z}_{p}^{n})$,
and for every positive integer $m$, if $x\in\mathrm{\Gamma}_{m}(1)$ then
$\langle x\rangle^{*}\subseteq\mathrm{\Gamma}_{m}(1)$;

  * (2)

Let $K_{1}$ and $K_{2}$ be subgroups of $\mathbb{Z}_{p}^{n}$. Assume that the
induced subgroups $[K_{1}]$ and $[K_{2}]$ of $\mathrm{\Gamma}$ are complete
graphs and every vertex in $K_{1}$ is adjacent to every vertex in $K_{2}$.
Then the induced subgraph $[K_{1}K_{2}]$ of $\mathrm{\Gamma}$ is a complete
graph;

  * (3)

$G$ has a subgroup $H$ such that $H^{*}\subseteq S$ and $[H]$ is a maximal
clique of $\mathrm{\Gamma}$.

Proof. For $r\in\mathbb{Z}_{p}^{*}$, let $\sigma_{r}$ be the automorphism of
$\mathbb{Z}_{p}^{n}$ which maps every element $x\in\mathbb{Z}_{p}^{n}$ to
$x^{r}$. Then the center $Z(\mathrm{Aut}(\mathbb{Z}_{p}^{n}))=\\{\sigma_{r}\
|\ r\in\mathbb{Z}_{p}^{*}\\}$. By Proposition 2.9, $\langle
s\rangle^{*}\subset S=\mathrm{\Gamma}_{1}(1)$ for each $s\in S$, and hence
$\sigma_{r}$ fixes $S$ setwise. Thus,
$Z(\mathrm{Aut}(\mathbb{Z}_{p}^{n}))\leq\mathrm{Aut}(\mathbb{Z}_{p}^{n},S)\leq\mathrm{Aut}(\mathrm{\Gamma})$,
which implies that for every positive integer $m$, if
$x\in\mathrm{\Gamma}_{m}(1)$ then $\langle
x\rangle^{*}\subseteq\mathrm{\Gamma}_{m}(1)$.

To prove part (2), assume that $[K_{1}]$ and $[K_{2}]$ are complete graphs and
every vertex in $K_{1}$ is adjacent to every vertex in $K_{2}$. Clearly,
$K_{1}K_{2}\leq\mathbb{Z}_{p}^{n}$. Take two distinct vertices
$x_{1}y_{1},x_{2}y_{2}\in K_{1}K_{2}$ with $x_{1},x_{2}\in K_{1}$ and
$y_{1},y_{2}\in K_{2}$. Then $x_{1}x_{2}^{-1}\not=y_{1}^{-1}y_{2}$. Since
$x_{1}x_{2}^{-1}\in K_{1}$ is adjacent to $y_{1}^{-1}y_{2}\in K_{2}$, we have
$x_{1}x_{2}^{-1}(y_{1}^{-1}y_{2})^{-1}\in S$, that is,
$x_{1}y_{1}(x_{2}y_{2})^{-1}\in S$. Thus, $x_{1}y_{1}$ is adjacent to
$x_{2}y_{2}$, and so $[K_{1}K_{2}]$ is a complete graph.

To prove part (3), assume that $H$ is a maximal subgroup of $G$ such that
$[H]$ is a complete graph. Then $H^{*}\subseteq S$. If $[H]$ is not a maximal
clique of $\mathrm{\Gamma}$, then there is $x\not\in H$ such that $x$ is
adjacent to all elements of $H$. In particular, $x\in S$ as $x$ is adjacent to
$1$. By (1), $\langle x\rangle^{*}\subseteq S=\mathrm{\Gamma}_{1}(1)$ and
hence $\langle x\rangle$ is a complete graph. Since
$Z(\mathrm{Aut}(\mathbb{Z}_{p}^{n}))\leq\mathrm{Aut}(\mathrm{\Gamma})$, every
vertex in $\langle x\rangle$ is adjacent to every vertex in $H$, and by (2),
$[\langle x\rangle H]$ is a complete graph, contradicting the maximality of
$H$. It follows that $[H]$ is a maximal clique of $\mathrm{\Gamma}$. ∎

In the next lemma, we consider $2$-geodesic but not $2$-arc transitive normal
Cayley graphs on $\mathbb{Z}_{p}^{2}$.

######  Lemma 5.3.

Let $p\geq 5$ be a prime and let
$\mathrm{\Gamma}=\mathrm{Cay}(\mathbb{Z}_{p}^{2},S)$ be a connected
$2$-geodesic but not $2$-arc transitive normal Cayley graph. Then
$\mathrm{Aut}(\mathrm{\Gamma})$ cannot be primitive on $V(\mathrm{\Gamma})$.

Proof. Let $B=\mathrm{GL}(2,p)=\mathrm{Aut}(\mathbb{Z}_{p}^{2})$ and
$A=\mathrm{Aut}(\mathrm{\Gamma})$ . Since $\mathrm{\Gamma}$ is normal,
Propositions 2.7 implies
$A=R(\mathbb{Z}_{p}^{2})\rtimes\mathrm{Aut}(\mathbb{Z}_{p}^{2},S)$ and
$A_{1}=\mathrm{Aut}(\mathbb{Z}_{p}^{2},S)\leq B$, where $A_{1}$ is the
stabilizer of $1$ in $A$. In particular,
$\mathrm{\Gamma}\not=\mathsf{K}_{p^{2}}$. By Lemma 5.2 (1),
$Z=Z(\mathrm{GL}(2,p))\leq A_{1}$. Denote by $\overline{\mathrm{\Gamma}}$ the
quotient graph of $\mathrm{\Gamma}$ corresponding to $Z$, that is, the graph
with the orbits of $Z$ as vertices and with two orbits adjacent if and only if
there is an edge of $\mathrm{\Gamma}$ between the two orbits. Then
$V(\overline{\mathrm{\Gamma}})=\overline{\mathbb{Z}_{p}^{2}}$. Since $Z\leq
A_{1}\leq B$, the kernel of the action of $A_{1}$ on
$\overline{\mathrm{\Gamma}}$ is $Z$. Then
$\overline{A_{1}}\leq\mathrm{Aut}(\overline{\mathrm{\Gamma}})_{\overline{1}}$,
and $\overline{A_{1}}$ is a permutation group on
$\overline{(\mathbb{Z}_{p}^{2})^{*}}$. By Lemma 5.2 (1),
$\overline{\mathrm{\Gamma}}_{m}(\overline{1})=\overline{\mathrm{\Gamma}_{m}(1)}$
for every $m\geq 1$.

Suppose to the contrary that $A$ is primitive on $V(\mathrm{\Gamma})$. Since
$\mathrm{\Gamma}$ is connected, $\langle S\rangle=\mathbb{Z}_{p}^{2}$, and so
we may let $a,b\in S$ with $\mathbb{Z}_{p}^{2}=\langle a,b\rangle$. Then
$\overline{a}\subseteq S$ and $\overline{b}\subseteq S$. Since
$\mathbb{Z}_{p}^{2}=\langle a\rangle\langle b\rangle\subseteq SS$,
$\mathrm{\Gamma}$ has diameter $2$. Note that
$|\overline{(\mathbb{Z}_{p}^{2})^{*}}|=p+1$. Suppose $|\overline{S}|=p+1$. By
Lemma 5.2 (1), $S=(\mathbb{Z}_{p}^{2})^{*}$ and $\mathrm{\Gamma}$ is a
complete graph, a contradiction. Suppose $|\overline{S}|=p$. Then
$|\overline{\mathrm{\Gamma}}_{2}(\overline{1})|=1$ and hence
$\langle\mathrm{\Gamma}_{2}(1)\rangle\cong\mathbb{Z}_{p}$. Therefore, $A_{1}$
fixes a subgroup of order $p$ in $\mathbb{Z}_{p}^{2}$, contrary to the
primitivity of $A$ on $V(\mathrm{\Gamma})$. Suppose $|\overline{S}|=2$. Then
$S=\overline{a}\cup\overline{b}$ and $\mathrm{\Gamma}=H(2,p)$. By Example 3.3,
$\mathrm{Aut}(\mathrm{\Gamma})\cong(\mathrm{S}_{p}\times\mathrm{S}_{p})\rtimes\mathbb{Z}_{2}$,
which is impossible because $\mathrm{Aut}(\mathrm{\Gamma})$ has no normal
subgroup isomorphic to $\mathbb{Z}_{p}^{2}$. Suppose $|\overline{S}|=p-1$.
Then $|\overline{\mathrm{\Gamma}}_{2}(\overline{1})|=(p+1)-(p-1)=2$ and
$\mathrm{\Gamma}=\overline{H(2,p)}$, the complement of $H(2,p)$ in the
complete graph $\mathsf{K}_{p^{2}}$, which is also impossible because
$\mathrm{Aut}(\overline{H(2,p)})\cong\mathrm{Aut}(H(2,p))$. It follows that
$3\leq|\overline{S}|\leq p-2$.

Clearly,
$\overline{(\mathbb{Z}_{p}^{2})^{*}}=\\{\overline{a},\overline{ba^{i}}\ |\
i\in\mathbb{Z}_{p}\\}$. Since $\mathrm{\Gamma}$ has diameter $2$,
$\overline{\mathrm{\Gamma}}_{2}(\overline{1})=\\{\overline{ba^{i}}\ |\
\overline{ba^{i}}\not\in\overline{S},i\in\mathbb{Z}_{p}^{*}\\}$, and for
$\overline{ba^{i}}\in\overline{\mathrm{\Gamma}}_{2}(\overline{1})$, we have
$\overline{ba^{i}}\in\overline{\mathrm{\Gamma}}_{1}(\overline{b})$ because $b$
is adjacent $ba^{i}$. Thus,

$\overline{\mathrm{\Gamma}}_{2}(\overline{1})\subseteq\overline{\mathrm{\Gamma}}_{1}(\overline{b}).$

Let
$\overline{u},\overline{v}\in\overline{\mathrm{\Gamma}}_{2}(\overline{1})$.
Then
$\overline{u},\overline{v}\in\overline{\mathrm{\Gamma}}_{1}(\overline{b})$,
and since $Z\leq\mathrm{Aut}(\mathrm{\Gamma})$, $b$ is adjacent to some vertex
$u^{i}\in\overline{u}$ and $v^{j}\in\overline{v}$. This implies that
$(1,b,u^{i})$ and $(1,b,v^{j})$ are two 2-geodesics of $\mathrm{\Gamma}$, and
since $\mathrm{\Gamma}$ is $2$-geodesic transitive, the stabilizer $A_{1b}$ of
$b$ in $A_{1}$ has an element mapping $(1,b,u^{i})$ to $(1,b,v^{j})$. It
follows that $\overline{A_{1b}}$ is transitive on
$\overline{\mathrm{\Gamma}}_{2}(\overline{1})$, and hence
$(\overline{A_{1}})_{\overline{b}}$ is transitive on
$\overline{\mathrm{\Gamma}}_{2}(\overline{1})$. Since $\mathrm{\Gamma}$ is
arc-transitive, $A_{1}$ is transitive on $\mathrm{\Gamma}_{1}(1)=S$, and so
$\overline{A_{1}}$ is transitive on $\overline{S}$. Thus,
$|\overline{A_{1}}|=|\overline{S}||(\overline{A_{1}})_{\overline{b}}|$, and
the transitivity of $(\overline{A_{1}})_{\overline{b}}$ on
$\overline{\mathrm{\Gamma}}_{2}(\overline{1})$ implies that
$|\overline{S}||\overline{\mathrm{\Gamma}}_{2}(\overline{1})|\mid|\overline{A_{1}}|$.
Since $\mathrm{\Gamma}$ has diameter $2$,
$|\overline{\mathrm{\Gamma}}_{2}(\overline{1})|=(p+1)-|\overline{S}|$, and
therefore, $\ell(p+1-\ell)\mid|\overline{A_{1}}|$, where
$\ell=|\overline{S}|$. Since $3\leq\ell\leq p-2$, we have $3\leq p+1-\ell\leq
p-2$. Thus, $(\ell(p+1-\ell),p)=1$ and hence
$\ell(p+1-\ell)\mid|\overline{A_{1}}|_{p^{\prime}}$. Note that the function
$f(t)=t(p+1-t)$ has the minimal value $3(p-2)$ when $3\leq t\leq p-2$. It
follows that

$3\leq\ell\leq p-2,\ 3(p-2)\leq\ell(p+1-\ell)\mbox{ and }\
\ell(p+1-\ell)\mid|\overline{A_{1}}|_{p^{\prime}}.$

Suppose $\mathrm{PSL}(2,p)\leq\overline{A_{1}}$. By Proposition 2.3, every
transitive action of $\overline{A_{1}}$ has degree at least $p$, and so
$|\overline{S}|=\ell\geq p$, a contradiction. Thus, $\overline{A_{1}}\leq M$,
where $\mathrm{PSL}(2,p)\nleq M$ and $M$ is a maximal subgroup of
$\mathrm{PSL}(2,p)$ or $\mathrm{PGL}(2,p)$. By Proposition 2.3,
$M\cong\mathbb{Z}_{p}\rtimes\mathbb{Z}_{p-1},\mathbb{Z}_{p}\rtimes\mathbb{Z}_{(p-1)/2},\mathrm{D}_{2(p-1)},\mathrm{D}_{p-1},\mathrm{D}_{2(p+1)},\mathrm{D}_{p+1},\mathrm{A}_{4},\mathrm{S}_{4}$
or $\mathrm{A}_{5}$, and since
$\ell(p+1-\ell)\mid|\overline{A_{1}}|_{p^{\prime}}$, $\ell(p+1-\ell)$ is
divisor of $2(p+1)$, $2(p-1)$, $24$ or $60$.

Suppose $\ell(p+1-\ell)\mid 2(p-1)$. Since $\ell(p+1-\ell)\geq 3(p-2)$ and
$3(p-2)=2(p-1)+p-4$, we have $p\leq 4$, contradicting the hypothesis $p\geq
5$.

Suppose $\ell(p+1-\ell)\mid 24$. Since $\ell(p+1-\ell)\geq 3(p-2)$, we have
either $p=5$ and $\ell=3$, or $p=7$ and $3\leq\ell\leq 5$, of which each is
impossible.

Suppose $\ell(p+1-\ell)\mid 60$. Then $\overline{A_{1}}\leq\mathrm{A}_{5}$. If
$\overline{A_{1}}<\mathrm{A}_{5}$, then $\overline{A_{1}}$ is a subgroup of a
maximal subgroup of $\mathrm{A}_{5}$, and by Atlas [8], $\ell(p+1-\ell)$ is a
divisor of $12$ or $20$, which are impossible by a similar argument to the
above case $\ell(p+1-\ell)\mid 24$. Thus, $\overline{A_{1}}=\mathrm{A}_{5}$,
and since $\overline{S}$ and $\overline{\mathrm{\Gamma}}_{2}(\overline{1})$
are orbits of $\overline{A_{1}}$, we have $\ell=|\overline{S}|\geq 5$ and
$p+1-\ell=|\overline{\mathrm{\Gamma}}_{2}(\overline{1})|\geq 5$. It follows
that $p\geq 9$, and since $3(p-2)\leq\ell(p+1-\ell)\leq 60$, we have $p\leq
22$. Thus, $p=11,13,17$ or $19$, and for each case, $\ell(p+1-\ell)\nmid 60$,
a contradiction.

Suppose $\ell(p+1-\ell)\mid 2(p+1)$. Then $\ell(p+1-\ell)\geq
3(p+1-3)=2(p+1)+p-8$. If $p\geq 9$ then $\ell(p+1-\ell)$ is not a divisor of
$2(p+1)$. Thus, $p=5$ or $7$, and $\ell(p+1-\ell)\mid 2(p+1)$ implies that
$p=7$ and $\ell=4$. It follows $\overline{A_{1}}=M\cong\mathrm{D}_{16}$, and
$|\overline{\mathrm{\Gamma}}_{1}(\overline{1})|=|\overline{\mathrm{\Gamma}}_{2}(\overline{1})|=4$.
Since $\overline{\mathrm{\Gamma}}_{1}(\overline{1})$ and
$\overline{\mathrm{\Gamma}}_{2}(\overline{1})$ are orbits of
$\overline{A_{1}}$, the unique involution of $\overline{A_{1}}$ fixes every
point in $\overline{(\mathbb{Z}_{7}^{2})^{*}}$, contradicting that
$\overline{A_{1}}$ is a permutation group on
$\overline{(\mathbb{Z}_{7}^{2})^{*}}$. ∎

Now we consider $2$-geodesic but not $2$-arc transitive normal Cayley graphs
on $\mathbb{Z}_{p}^{3}$.

######  Lemma 5.4.

Let $p\geq 5$ be a prime and let
$\mathrm{\Gamma}=\mathrm{Cay}(\mathbb{Z}_{p}^{3},S)$ be a connected
$2$-geodesic but not $2$-arc transitive normal Cayley graph. Then
$\mathrm{Aut}(\mathrm{\Gamma})$ cannot be primitive on $V(\mathrm{\Gamma})$.

Proof. Let $B=\mathrm{Aut}(\mathbb{Z}_{p}^{3})=\mathrm{GL}(3,p)$,
$Z=Z(\mathrm{GL}(3,p))$ and $\overline{B}=\mathrm{PGL}(2,p)$. Let
$A=\mathrm{Aut}(\mathrm{\Gamma})$. By Proposition 2.7,
$A=R(\mathbb{Z}_{p}^{3})\rtimes\mathrm{Aut}(\mathbb{Z}_{p}^{3},S)$, and
$A_{1}=\mathrm{Aut}(\mathbb{Z}_{p}^{3},S)\leq B$, implying
$\mathrm{\Gamma}\not=\mathsf{K}_{p^{3}}$. By Lemma 5.2 (1), $Z\leq A_{1}$. Let
$\overline{\mathrm{\Gamma}}$ be the quotient graph of $\mathrm{\Gamma}$
corresponding to $Z$. Then
$V(\overline{\mathrm{\Gamma}})=\overline{\mathbb{Z}_{p}^{3}}$, and since
$Z\leq A_{1}\leq B$, we have
$\overline{A_{1}}\leq\mathrm{Aut}(\overline{\mathrm{\Gamma}})_{\overline{1}}$.
Furthermore, $\overline{A_{1}}\leq\overline{B}=\mathrm{PGL}(3,p)$ and
$\overline{\mathrm{\Gamma}}_{m}(\overline{1})=\overline{\mathrm{\Gamma}_{m}(1)}$
for every $m\geq 1$.

Suppose, by way of contradiction, that $A$ is primitive on
$V(\mathrm{\Gamma})$. Let
$\mathrm{O}=\\{\mathrm{O}_{1},\cdots,\mathrm{O}_{t}\\}$ be the set of orbits
of $\overline{A_{1}}$ on $\overline{(\mathbb{Z}_{p}^{3})^{*}}$. The
primitivity of $A$ implies that
$\langle\mathrm{O}_{i}\rangle=\mathbb{Z}_{p}^{3}$ for each $1\leq i\leq t$,
forcing $|\overline{\mathrm{O}_{i}}|\geq 3$ for each $1\leq i\leq t$. In
particular, $|\overline{\mathrm{\Gamma}}_{m}(\overline{1})|\geq 3$ and
$\langle\overline{\mathrm{\Gamma}}_{m}(\overline{1})\rangle=\mathbb{Z}_{p}^{3}$,
for every $1\leq m\leq\mathrm{diam}(\mathrm{\Gamma})$.

Denote by $O_{p}(\overline{A_{1}})$ the maximal normal $p$-subgroup of
$\overline{A_{1}}$.

Claim: $O_{p}(\overline{A_{1}})=1$.

Suppose $O_{p}(\overline{A_{1}})\not=1$. If $O_{p}(\overline{A_{1}})$ has no
fixed point in $\overline{(\mathbb{Z}_{p}^{3})^{*}}$, then $p$ is a divisor of
$|\overline{(\mathbb{Z}_{p}^{3})^{*}}|=p^{2}+p+1$, which is impossible. Now
assume that $O_{p}(\overline{A_{1}})$ has a fixed point in $\mathrm{O}_{j}$
for some $1\leq j\leq m$. Since
$O_{p}(\overline{A_{1}})\unlhd\overline{A_{1}}$, $O_{p}(\overline{A_{1}})$
fixes $\mathrm{O}_{j}$ pointwise. Since
$\langle\mathrm{O}_{j}\rangle=\mathbb{Z}_{p}^{3}$, we may assume that
$\overline{x},\overline{y},\overline{z}\in\mathrm{O}_{j}$ and
$\mathbb{Z}_{p}^{3}=\langle x,y,z\rangle$. Then $O_{p}(\overline{A_{1}})$
fixes $\overline{x},\overline{y}$ and $\overline{z}$, and hence
$O_{p}(\overline{A_{1}})\leq\overline{B}_{\overline{x},\overline{y},\overline{z}}$,
contradicting Lemma 5.1 (1). Thus, $O_{p}(\overline{A_{1}})=1$, as claimed.

Since $\mathrm{\Gamma}$ is connected, $\langle S\rangle=\mathbb{Z}_{p}^{3}$,
and so we may assume $\mathbb{Z}_{p}^{3}=\langle a,b,c\rangle$ with
$\\{\overline{a},\overline{b},\overline{c}\\}\subseteq\overline{S}$. If
$S=\overline{a}\cup\overline{b}\cup\overline{c}$, then
$\mathrm{\Gamma}=H(3,p)$, and by Example 3.3,
$\mathrm{Aut}(\mathrm{\Gamma})\cong(\mathrm{S}_{p}\times\mathrm{S}_{p}\times\mathrm{S}_{p})\rtimes\mathrm{S}_{3}$,
which is impossible because $\mathrm{Aut}(\mathrm{\Gamma})$ has no normal
subgroup isomorphic to $\mathbb{Z}_{p}^{3}$. Thus, we may assume that
$\ell:=|\overline{S}|\geq 4$. By Lemma 5.2 (3), $\mathbb{Z}_{p}^{3}$ has a
subgroup $H$ such that $H^{*}\subseteq S$ and $[H]$ is a maximal clique of
$\mathrm{\Gamma}$. Since $\mathrm{\Gamma}$ is not a complete graph,
$H\cong\mathbb{Z}_{p}^{2}$ or $\mathbb{Z}_{p}$.

Case 1: $H\cong\mathbb{Z}_{p}^{2}$.

Without loss of any generality, we may let $H=\langle a,b\rangle$. Then
$\overline{H^{*}}\subseteq\overline{S}$ and $\overline{c}\in\overline{S}$,
implying $|\overline{S}|\geq p+2$, and since $\mathbb{Z}_{p}^{3}=H\langle
c\rangle\subseteq SS$, $\mathrm{\Gamma}$ has diameter $2$. Note that
$|\overline{(\mathbb{Z}_{p}^{3})^{*}}|=p^{2}+p+1$ and
$\overline{(\mathbb{Z}_{p}^{3})^{*}}=\overline{H^{*}}\cup\\{\overline{ch}\ |\
h\in H\\}$, where $|\overline{H^{*}}|=p+1$ and $|\\{\overline{ch}\ |\ h\in
H\\}|=p^{2}$. Then
$\overline{\mathrm{\Gamma}}_{2}(\overline{1})=\\{\overline{ch}\ |\ 1\not=h\in
H,\overline{ch}\not\in\overline{S}\\}$. Since $c$ is adjacent $ch$, we have

$\overline{\mathrm{\Gamma}}_{2}(\overline{1})\subseteq\overline{\mathrm{\Gamma}}_{1}(\overline{c}).$

Let
$\overline{u},\overline{v}\in\overline{\mathrm{\Gamma}}_{2}(\overline{1})$.
Then
$\overline{u},\overline{v}\in\overline{\mathrm{\Gamma}}_{1}(\overline{c})$,
and since $Z\leq\mathrm{Aut}(\mathrm{\Gamma})$, $c$ is adjacent to some vertex
$u^{i}\in\overline{u}$ and $v^{j}\in\overline{v}$. Since $\mathrm{\Gamma}$ is
$2$-geodesic transitive, the stabilizer $A_{1c}$ of $c$ in $A_{1}$ has an
element mapping the $2$-geodesic $(1,c,u^{i})$ to $(1,c,v^{j})$. Thus,
$\overline{A_{1c}}$ is transitive on
$\overline{\mathrm{\Gamma}}_{2}(\overline{1})$, and hence
$(\overline{A_{1}})_{\overline{c}}$ is transitive on
$\overline{\mathrm{\Gamma}}_{2}(\overline{1})$. The arc-transitivity of
$\mathrm{\Gamma}$ implies that $A_{1}$ is transitive on
$\mathrm{\Gamma}_{1}(1)=S$, and so $\overline{A_{1}}$ is transitive on
$\overline{S}$. It follows that
$|\overline{S}||\overline{\mathrm{\Gamma}}_{2}(\overline{1})|\mid|\overline{A_{1}}|$.
Since $\mathrm{\Gamma}$ has diameter $2$, we have
$|\overline{S}|+|\overline{\mathrm{\Gamma}}_{2}(\overline{1})|=p^{2}+p+1$.
Then $\ell(p^{2}+p+1-\ell)\mid|\overline{A_{1}}|$, where
$p+2\leq\ell=|\overline{S}|\leq p^{2}+p$. If $\ell\geq p^{2}+p+1-2$, then
$|\overline{\mathrm{\Gamma}}_{2}(\overline{1})|=p^{2}+p+1-\ell\leq 2$,
contradicting $|\overline{\mathrm{\Gamma}}_{2}(\overline{1})|\geq 3$. If
$\ell=p^{2}+p+1-3$, then $|\overline{\mathrm{\Gamma}}_{2}(\overline{1})|=3$
and $\mathrm{\Gamma}=\overline{H(3,p)}$, which is impossible because
$\mathrm{Aut}(\mathrm{\Gamma})\cong(\mathrm{S}_{p}\times\mathrm{S}_{p}\times\mathrm{S}_{p})\rtimes\mathrm{S}_{3}$
has no normal subgroup isomorphic to $\mathbb{Z}_{p}^{3}$. Therefore,

$7\leq p+2\leq\ell\leq p^{2}+p-3,\ \mbox{ where }\ell=|\overline{S}|.$

For every $m\geq 1$, it is easy to see that

$f(\ell,p):=\ell(p^{2}+p+1-\ell)\geq m(p^{2}+p+1-m),\ \mbox{ when
}m\leq\ell\leq p^{2}+p+1-m.$

In particular, $f(\ell,p)\geq 4(p^{2}+p-3)$. Note that
$f(\ell,p)\mid|\overline{A_{1}}|$.

By Proposition 2.4 (2), $\mathrm{PSL}(3,p)$ has no non-trivial transitive
action with degree less than least $p^{2}+p+1$. This implies that if
$\mathrm{PSL}(3,p)\leq\overline{A_{1}}$, then $|\overline{S}|=\ell=p^{2}+p+1$,
a contradiction. Thus, $\overline{A_{1}}\leq M$, where $\mathrm{PSL}(3,p)\nleq
M$ and $M$ is a maximal subgroup of $\mathrm{PGL}(3,p)$ or
$\mathrm{PSL}(3,p)$. By Claim, $O_{p}(\overline{A_{1}})=1$, and by Proposition
2.4, $|\overline{A_{1}}|$ is a divisor of $p(p+1)(p-1)^{2}$, $6(p-1)^{2}$,
$3(p^{2}+p+1)$, $3^{3}\cdot 8$, $168$ or $360$, where the last three cases
correspond to $M=(\mathbb{Z}_{3}^{2}:\mathrm{Q}_{8}).\mathbb{Z}_{3}$ with
$p\equiv 1({\rm mod~{}}3)$, $M=\mathrm{PSL}(2,7)$ with $p\equiv 1,2,4({\rm
mod~{}}7)$, or $M=\mathrm{A}_{6}$ with $p\equiv 1,4({\rm mod~{}}15)$,
respectively.

Suppose $|\overline{A_{1}}|\mid 6(p-1)^{2}$. Then $f(\ell,p)\mid 6(p-1)^{2}$.
If $7\leq\ell\leq p^{2}+p-6$ then $f(\ell,p)\geq 7(p^{2}+p-6)>6(p-1)^{2}$,
which is impossible. Since $7\leq\ell\leq p^{2}+p-3$, we have
$\ell=p^{2}+p-5,p^{2}+p-4$ or $p^{2}+p-3$, implying $f(\ell,p)\nmid
6(p-1)^{2}$, a contradiction.

Suppose $|\overline{A_{1}}|\mid 3(p^{2}+p+1)$. Then $f(\ell,p)\mid
3(p^{2}+p+1)$, which is impossible because $f(\ell,p)\geq
4(p^{2}+p-3)>3(p^{2}+p+1)$.

Suppose $|\overline{A_{1}}|$ is a divisor of $3^{3}\cdot 8$, $168$ or $360$.
Then $f(\ell,p)\mid 3^{3}\cdot 8$, $168$ or $360$, of which each is impossible
for $p\geq 11$, because otherwise $f(\ell,p)\geq 4(p^{2}+p-3)\geq 516$. Thus,
$p=7$ or $5$. For $3^{3}\cdot 8$,
$M=(\mathbb{Z}_{3}^{2}:\mathrm{Q}_{8}).\mathbb{Z}_{3}$ with $p\equiv 1({\rm
mod~{}}3)$, and for $168$ or $360$, $M=\mathrm{PSL}(2,7)$ with $p\equiv
1,2,4({\rm mod~{}}7)$ or $M=\mathrm{A}_{6}$ with $p\equiv 1,4({\rm
mod~{}}15)$, of which the latter two are impossible because $p\geq 11$. Thus,
$M=(\mathbb{Z}_{3}^{2}:\mathrm{Q}_{8}).\mathbb{Z}_{3}$ and $p=7$, which is
also impossible because $f(\ell,7)=\ell(57-\ell)$ is not a divisor of
$3^{3}\cdot 8$ for every $7\leq\ell\leq 53$.

Suppose $|\overline{A_{1}}|\mid p(p+1)(p-1)^{2}$. Then $f(\ell,p)\mid
p(p+1)(p-1)^{2}$. By Proposition 2.4, $M=\mathbb{Z}_{p}^{2}:\mathrm{GL}(2,p)$,
$\mathbb{Z}_{p}^{2}:\frac{1}{d}\mathrm{GL}(2,p)$ or $\mathrm{PGL}(2,p)$. Since
$O_{p}(\overline{A_{1}})=1$, we have
$\overline{A_{1}}=\overline{A_{1}}/(\overline{A_{1}}\cap\mathbb{Z}_{p}^{2})\cong\overline{A_{1}}\mathbb{Z}_{p}^{2}/\mathbb{Z}_{p}^{2}\leq
M/\mathbb{Z}_{p}^{2}\cong\mathrm{GL}(2,p)$, $\frac{1}{d}\mathrm{GL}(2,p)$, or
$\mathrm{PGL}(2,p)$. Thus, we may assume $\overline{A_{1}}\leq N$, where
$N=\mathrm{GL}(2,p)$, $\frac{1}{d}\mathrm{GL}(2,p)$, or $\mathrm{PGL}(2,p)$.
Then $|Z(N)|\mid(p-1)$ and $\overline{A_{1}}/(\overline{A_{1}}\cap
Z(N))\cong\overline{A_{1}}Z(N)/Z(N)\leq\mathrm{PGL}(2,p)$.

Assume $p\nmid|\overline{A_{1}}|$. Then
$\overline{A_{1}}/(\overline{A_{1}}\cap Z(N))\not=\mathrm{PGL}(2,p)$. By
Proposition 2.3, $|\overline{A_{1}}|$ and so $f(\ell,p)$ is a divisor of
$2(p-1)^{2}$, $2(p^{2}-1)$, $24(p-1)$ or $60(p-1)$, of which the last one
corresponds to $\mathbb{Z}_{p-1}.\mathrm{A}_{5}$ with $p\equiv\pm 1({\rm
mod~{}}10)$. The former two are impossible because $f(\ell,p)\geq
4(p^{2}+p-3)>2(p^{2}-1)$, and the latter two are impossible for $p\geq 17$
because $f(\ell,p)\geq 4(p^{2}+p-3)>4(p+1)(p-1)>72(p-1)$. Furthermore,
$24(p-1)$ cannot happen because $f(\ell,p)>4(p+1)(p-1)\geq 24(p-1)$, and for
$60(p-1)$, we have $p=11$ or $13$ as $p\equiv\pm 1({\rm mod~{}}10)$. If
$7\leq\ell\leq(p^{2}+p-6)$ then $f(\ell,p)\geq
7(11^{2}+11-6)=882>60(13-1)>60(11-1)$, which is impossible. It follows that
$p=11$ or $13$ with $\ell=p^{2}+p-5,p^{2}+p-4$ or $p^{2}+p-3$, which are also
impossible because otherwise $f(\ell,p)\nmid 60(p-1)$.

Assume $p\mid|\overline{A_{1}}|$. If $Z(N)\cap\overline{A_{1}}$ has an element
$\alpha$ of order a prime $q$, then $q\mid(p-1)$, and by Lemma 5.1 (2),
$\alpha$ has a fixed point in $\overline{(\mathbb{Z}_{p}^{3})^{*}}$, which is
contained in $\overline{S}$ or $\overline{\mathrm{\Gamma}}_{2}(\overline{1})$.
Since $\langle\alpha\rangle\unlhd\overline{A_{1}}$, $\alpha$ fixes
$\overline{S}$ or $\overline{\mathrm{\Gamma}}_{2}(\overline{1})$ pointwise,
and so $\alpha$ has fixed-point set $\overline{S}$ or
$\overline{\mathrm{\Gamma}}_{2}(\overline{1})$. Since
$\langle\overline{S}\rangle=\langle\overline{\mathrm{\Gamma}}_{2}(\overline{1})\rangle=\mathbb{Z}_{p}^{3}$,
$\alpha$ fixe $\overline{x},\overline{y},\overline{z}$ such that
$\mathbb{Z}_{p}^{3}=\langle x,y,x\rangle$. Since $7\leq\ell\leq p^{2}+p-3$,
again by Lemma 5.1 (2) we have either $|\overline{S}|=p+2$ or
$|\overline{\mathrm{\Gamma}}_{2}(\overline{1})|=p+2$. If follows that
$f(\ell,p)=(p+2)(p^{2}-1)$ is a divisor of $p(p+1)(p-1)^{2}$, which is clearly
impossible. Thus, $\overline{A_{1}}\cap Z(N)=1$, and hence
$\overline{A_{1}}\cong\overline{A_{1}}Z(N)/Z(N)\leq\mathrm{PGL}(2,p)$.

Thus, $|\overline{A_{1}}|\mid p(p+1)(p-1)$ and $f(\ell,p)\mid p(p+1)(p-1)$. If
$p+1\leq\ell\leq p^{2}+p+1-(p+1)=p^{2}$ then $f(\ell,p)\geq
p^{2}(p+1)>p(p+1)(p-1)$, which is impossible. It follows that $7\leq\ell\leq
p$ or $p^{2}+1\leq\ell\leq p^{2}+p-2$, which implies that either
$\ell=p,p^{2}+1$, or $(f(\ell,p),p)=1$. If $\ell=p$ or $p^{2}+1$ then
$f(\ell,p)=p(p^{2}+1)\mid p(p+1)(p-1)$, which is impossible. If
$(f(\ell,p),p)=1$ then $\ell(p^{2}+p+1-\ell)\mid(p+1)(p-1)$, which is also
impossible because $\ell(p^{2}+p+1-\ell)\geq 4(p^{2}+p-3)>(p+1)(p-1)$.

Case 2: $H\cong\mathbb{Z}_{p}$.

Recall that $\overline{a}\cup\overline{b}\cup\overline{c}\subseteq S$. Since
$\mathbb{Z}_{p}^{3}=\langle a\rangle\langle b\rangle\langle c\rangle\leq SSS$,
$\mathrm{\Gamma}$ has diameter at most $3$. Take
$\overline{x},\overline{y}\in\overline{S}$ with
$\overline{x}\not=\overline{y}$. If $\overline{x}$ is adjacent to every vertex
in $\overline{y}$, then Lemma 5.2 implies that $\langle x,y\rangle$ is a
maximal clique of order $p^{2}$, a contradiction. Thus, if
$\overline{c}\not=\overline{x}$ and $\overline{c}\not=\overline{y}$, then $c$
is not adjacent to some $x^{i}\in\overline{x}$ and $y^{j}\in\overline{y}$, and
since $\mathrm{\Gamma}$ is $2$-geodesic transitive, $A_{1c}$ has an element
mapping the $2$-geodesic $(c,1,x^{i})$ to $(c,1,y^{j})$. Thus,
$(\overline{A_{1}})_{\overline{c}}$ is transitive on
$\overline{S}\backslash\\{\overline{c}\\}$, and the transitivity of
$\overline{A_{1}}$ on $\overline{S}$ means that $\overline{A_{1}}$ is
$2$-transitive on $\overline{S}$.

Similarly, if $x$ is not adjacent to $y^{i}$ and $y^{j}$ in $\overline{y}$,
then $(x,1,y^{i})$ and $(x,1,y^{j})$ are two $2$-geodesics, and hence $A_{1x}$
is transitive on $\overline{y}_{x-}$, where

$\overline{y}_{x-}=\\{y^{k}\ |\ y^{k}\in\overline{y},\ x\mbox{ is not adjacent
to }y^{k}\\}.$

Since $A_{1}\leq B=\mathrm{Aut}(\mathbb{Z}_{p}^{3})$, the block stabilizer
$(A_{1x})_{\overline{y}}$ of $\overline{y}$ in $A_{1x}$ is transitive on
$\overline{y}_{x-}$. Then the restriction
$(A_{1x})_{\overline{y}}^{\overline{y}}$ of $(A_{1x})_{\overline{y}}$ on
$\overline{y}$ is a subgroup of $\mathbb{Z}_{p-1}$, and hence regular on
$\overline{y}_{x-}$, implying
$|(A_{1x})_{\overline{y}}^{\overline{y}}|=|\overline{y}_{x-}|$. By the
$2$-transitivity of $\overline{A_{1}}$ on $\overline{S}$, we may assume that

$|\overline{y}_{x-}|=r\ \mbox{ for all }x,y\in S\mbox{ with
}\overline{x}\not=\overline{y},\ \mbox{ where }r\mid(p-1).$

Denote by $\overline{y}_{x+}$ the set of vertices in $\overline{y}$ that are
adjacent to $x$ in $\mathrm{\Gamma}$. Then

$|\overline{y}_{x+}|=(p-1)-r=(p-1)(1-\frac{r}{p-1}).$

Recall that $\overline{A_{1}}$ is $2$-transitive on $\overline{S}$ and
$\overline{A_{1}}\leq\mathrm{PGL}(3,p)$. Let $K$ be the kernel of
$\overline{A_{1}}$ on $\overline{S}$. Since $\langle
S\rangle\cong\mathbb{Z}_{p}^{3}$, by Lemma 5.1 (1) we have
$K\leq\mathbb{Z}_{p-1}\times\mathbb{Z}_{p-1}$.

Suppose $O_{p}(\overline{A_{1}}/K)\not=1$. Let
$O_{p}(\overline{A_{1}}/K)=O/K$. Then $O\unlhd\overline{A_{1}}$. Since
$K\leq\mathbb{Z}_{p-1}\times\mathbb{Z}_{p-1}$, we have $O=KO_{p}$, where
$O_{p}\not=1$ is a Sylow $p$-subgroup of $O$. We prove that $O_{p}$ has
trivial conjugate action on $K$ by induction on $|K|$. Let $q\mid|K|$ be a
prime. Then $q\mid(p-1)$, and the subgroup $L$ generated by all elements of
order $q$ in $K$ is characteristic in $K$. Since $K\unlhd O$, we have $L\unlhd
O$, and $L\cong\mathbb{Z}_{q}$ or $\mathbb{Z}_{q}^{2}$. Since $p\geq 5$, we
have $p\nmid|\mathrm{Aut}(L)|$, and hence $O_{p}$ has a trivial conjugate
action on $L$. Then $O_{p}$ has a natural action on $K/L$, and by inductive
hypothesis, we may assume that the action is trivial. Since $(|O_{p}|,|L|)=1$,
$O_{p}$ has a trivial conjugate action on $K$ (see [21, P. 131, 18.6]), that
is, $O=K\times O_{p}$. Then $O_{p}$ is characteristic in $O$, and since
$O\unlhd\overline{A_{1}}$, we have $O_{p}\unlhd\overline{A_{1}}$,
contradicting Claim. Thus, $O_{p}(\overline{A_{1}}/K)=1$.

If $\mathrm{PSL}(3,p)\leq\overline{A_{1}}/K$, then $|\overline{S}|=p^{2}+p+1$
because $\mathrm{PSL}(3,p)$ has the minimum transitive degree $p^{2}+p+1$ by
Proposition 2.4, contradicting that $\mathrm{\Gamma}\not=\mathsf{K}_{p^{3}}$.
Since $\ell=|\overline{S}|\geq 4$ and $O_{p}(\overline{A_{1}}/K)=1$, by
Proposition 2.5 we have the following lists:

  * List A:

$(\ell,\overline{A_{1}}/K)=(4,\mathrm{A}_{4})$, $(4,\mathrm{S}_{4})$ or
$(9,L)$ with $L\leq(\mathbb{Z}_{3}^{2}:\mathrm{Q}_{8}).\mathbb{Z}_{3}$ and
$p\equiv 1({\rm mod~{}}3)$;

  * List B:

$(\ell,\overline{A_{1}}/K)=(5,\mathrm{A}_{5})$, $(6,\mathrm{A}_{6})$,
$(10,\mathrm{A}_{6})$, $(8,\mathrm{PSL}(2,7))$, $(11,\mathrm{PSL}(2,11))$,
$(p+1,L)$ with $L=\mathrm{PSL}(2,p)$, or $\mathrm{PGL}(2,p)$, and if
$X/K=\mathrm{A}_{6}$, $\mathrm{PSL}(2,7)$ or $\mathrm{PSL}(2,11)$ then
$p\equiv 1,4({\rm mod~{}}15)$, $p\equiv 1,2,4({\rm mod~{}}7)$ or $p=11$,
respectively.

The $2$-transitivity of $\overline{A_{1}}$ on $\overline{S}$ implies that the
induced subgraph $[\overline{S}]$ of $\overline{S}$ in
$\overline{\mathrm{\Gamma}}$ is a complete graph or an empty graph.

Assume that $[\overline{S}]$ is a complete graph. Then $r\not=p-1$, and for
$\overline{x},\overline{y}\in\overline{S}$ with
$\overline{x}\not=\overline{y}$ we have
$|\overline{y}_{x+}|=(p-1)(1-r/(p-1))\geq(p-1)/2$ as $r\mid(p-1)$. Thus,
$|\overline{\langle x,y\rangle}\cap\overline{S}|\geq(p-1)/2+2$. Since
$\\{\overline{a},\overline{b},\overline{c}\\}\subseteq\overline{S}$, we have
$|\overline{\langle a,b\rangle}\cap\overline{S}|\geq(p-1)/2+2$,
$|\overline{\langle a,c\rangle}\cap\overline{S}|\geq(p-1)/2+2$ and
$|\overline{\langle b,c\rangle}\cap\overline{S}|\geq(p-1)/2+2$. Since $\langle
a,b\rangle\cap\langle a,c\rangle=\langle a\rangle$, $\langle
a,b\rangle\cap\langle b,c\rangle=\langle b\rangle$ and $\langle
a,c\rangle\cap\langle b,c\rangle=\langle c\rangle$, we have
$\ell=|\overline{S}|\geq 3(p-1)/2+6-3=3(p+1)/2$. In particular, $\ell\geq 9$
as $p\geq 5$, and $\ell>p+1$. By Lists A and B, either
$(\ell,\overline{A_{1}}/K)=(9,L)$ with
$L\leq(\mathbb{Z}_{3}^{2}:\mathrm{Q}_{8}).\mathbb{Z}_{3}$ and $p\equiv 1({\rm
mod~{}}3)$, or $(\ell,\overline{A_{1}}/K)=(11,\mathrm{PSL}(2,11))$ with
$p=11$, of which both are impossible because otherwise $p\geq 7$ and $\ell\geq
3(p+1)/2\geq 12$.

Assume that $[\overline{S}]$ is an empty graph. Then $r=p-1$, and for
$\overline{x},\overline{y}\in\overline{S}$ with
$\overline{x}\not=\overline{y}$, we have $|\overline{y}_{x-}|=p-1$ and so
$\overline{\langle
x,y\rangle}\cap\overline{\mathrm{\Gamma}}_{2}(\overline{1})=p-1$. Let
$\overline{c}\not=\overline{x}$ and $\overline{c}\not=\overline{y}$. Then
$|\overline{\langle
c,x\rangle}\cap\overline{\mathrm{\Gamma}}_{2}(\overline{1})|=|\overline{\langle
c,y\rangle}\cap\overline{\mathrm{\Gamma}}_{2}(\overline{1})|=p-1$, and since
$\langle c,x\rangle\cap\langle c,y\rangle=\langle c\rangle$, we have
$|(\overline{\langle c,x\rangle}\cup\overline{\langle
c,y\rangle})\cap\overline{\mathrm{\Gamma}}_{2}(\overline{1})|=2(p-1)$. Note
that $(\overline{\langle c,x\rangle}\cup\overline{\langle
c,y\rangle})\cap\overline{\mathrm{\Gamma}}_{2}(\overline{1})\subseteq\overline{\mathrm{\Gamma}}_{1}(\overline{c})$.
Since $\ell=|\overline{S}|$, the arbitrariness of $\overline{x}$ and
$\overline{y}$ implies that
$|\overline{\mathrm{\Gamma}}_{1}(\overline{c})\backslash\\{\overline{1}\\}|\geq(\ell-1)(p-1)$.
Conversely, for
$\overline{z}\in\overline{\mathrm{\Gamma}}_{1}(\overline{c})\backslash\\{\overline{1}\\}$,
$c$ is adjacent to $z^{i}$ with $i\in\mathbb{Z}_{p}^{*}$, implying
$\overline{cz^{-i}}\in\overline{S}$ and $\overline{z}\in\overline{\langle
c,cz^{-i}\rangle}$. Noting that $|\overline{\langle
a,b\rangle}\cap\overline{\mathrm{\Gamma}}_{2}(\overline{1})|=p-1$ and
$\overline{\langle
a,b\rangle}\cap\overline{\mathrm{\Gamma}}_{1}(\overline{c})=\\{\overline{1}\\}$,
we have

$|\overline{\mathrm{\Gamma}}_{1}(\overline{c})\backslash\\{\overline{1}\\}|=(\ell-1)(p-1)\mbox{
and
}|\overline{\mathrm{\Gamma}}_{2}(\overline{1})|\geq(\ell-1)(p-1)+(p-1)=\ell(p-1).$

Note that
$\overline{\mathrm{\Gamma}}_{1}(\overline{c})\backslash\\{\overline{1}\\}\subseteq\overline{\mathrm{\Gamma}}_{2}(\overline{1})$.
Then $(\overline{A_{1}})_{\overline{c}}$ is transitive
$\overline{\mathrm{\Gamma}}_{1}(\overline{c})\backslash\\{\overline{1}\\}$. It
follows that

$(\ell-1)(p-1)\mid|(\overline{A_{1}})_{\overline{c}}|\ \mbox{ and
}\ell(\ell-1)(p-1)\mid|\overline{A_{1}}|.$

Since $\ell=|\overline{S}|\geq 4$, we may let $\overline{u}\in\overline{S}$
with $\overline{u}\not\in\\{\overline{a},\overline{b},\overline{c}\\}$. If
$u\in\langle a,b\rangle$ then $\overline{u}\in\overline{\langle a,b\rangle}$,
which is impossible because $|\overline{\langle
a,b\rangle}\cap\overline{\mathrm{\Gamma}}_{2}(\overline{1})|=p-1$. Then
$u\not\in\langle a,b\rangle$, and similarly, $u\not\in\langle a,c\rangle$ and
$u\not\in\langle b,c\rangle$. By Lemma 5.1 (1),
$\overline{B}_{\overline{u},\overline{a},\overline{b},\overline{c}}=1$ and so
$K=1$. Thus, Lists A and B hold for $K=1$. Note that $p\geq 5$ and
$|\overline{A_{1}}|\geq\ell(\ell-1)(p-1)$. If $\ell=4$ then
$|\overline{A_{1}}|\geq 3\cdot 4^{2}>|\mathrm{S}_{4}|$; if $\ell=5$ then
$|\overline{A_{1}}|\geq 80>|\mathrm{A}_{5}|$; if $\ell=6$ or $10$ then
$p\equiv 1,4({\rm mod~{}}15)$ and $|\overline{A_{1}}|\geq 30\cdot
18>|\mathrm{A}_{6}|$; if $\ell=8$ then $p\equiv 1,2,4({\rm mod~{}}7)$ and
$|\overline{A_{1}}|\geq 560>|\mathrm{PSL}(2,7)$; if $\ell=9$ then $p\equiv
1({\rm mod~{}}3)$ and $|\overline{A_{1}}|\geq 9\cdot 8\cdot
6>|(\mathbb{Z}_{3}^{2}:\mathrm{Q}_{8}).\mathbb{Z}_{3}|$; and if $\ell=11$ then
$p=11$ and $|\overline{A_{1}}|\geq 11\cdot 10^{2}>|\mathrm{PSL}(2,11)|$. By
Lists A and B, we have $(\ell,\overline{A_{1}})=(p+1,\mathrm{PSL}(2,p)$) or
$(p+1,\mathrm{PGL}(2,p))$. Then $p^{2}\nmid|\overline{A_{1}}|$ and
$|\overline{\mathrm{\Gamma}}_{2}(\overline{1})|\geq\ell(p-1)=p^{2}-1$. Since
$\overline{S}=\ell=p+1$ and $|\overline{(\mathbb{Z}_{p}^{3})^{*}}|=p^{2}+p+1$,
we have $|\overline{\mathrm{\Gamma}}_{2}(\overline{1})|=p^{2}$ or $p^{2}-1$:
for the former, $p^{2}\mid|\overline{A_{1}}|$, a contradiction, and for the
latter, $\overline{\mathrm{\Gamma}}_{3}(\overline{1})=1$, contradicting
$\overline{\mathrm{\Gamma}}_{3}(\overline{1})\geq 3$. ∎

Proof of Theorem 1.2. Let $\mathrm{\Gamma}$ be a 2-geodesic transitive but not
$2$-arc transitive Cayley graph on $\mathbb{Z}_{p}^{n}$ with $p$ a prime and
$n\leq 3$. By [27], every $2$-geodesic transitive graph of order $p$ is
isomorphic to $\mathbf{C}_{p}$ or $\mathsf{K}_{p}$, which are also $2$-arc
transitive. So there do not exist 2-geodesic transitive but not 2-arc
transitive graphs of order $p$.

In what follows, assume that $n\geq 2$. If $p\geq 5$, then by Lemmas 5.3 and
5.4, $\mathrm{Aut}(\mathrm{\Gamma})$ is imprimitive on $V(\mathrm{\Gamma})$.
If $p\leq 3$, then $p^{n}=4,8,9$ or $27$ and then by Proposition 3.5,
$\mathrm{\Gamma}$ is isomorphic to $H(2,3)$, $\mathsf{K}_{3[3]}$,
$\mathsf{K}_{4[2]}$, $\mathcal{G}_{(27,4)}$, $H(3,3)$, $\mathcal{G}_{(27,8)}$,
$\mathsf{K}_{9[3]}$, $\mathsf{K}_{3[9]}$, the Schläfli graph or its
complement. Among these ten graphs, by using Magma [2], it is easy to verify
that $\mathrm{\Gamma}$ is a Cayley graph on $\mathbb{Z}_{p}^{2}$ or
$\mathbb{Z}_{p}^{3}$ and $\mathrm{Aut}(\mathrm{\Gamma})$ is primitive on
$V(\mathrm{\Gamma})$ if and only if $\mathrm{\Gamma}\cong H(2,3)$ or $H(3,3)$.
This completes the proof.∎

##  6\. Proofs of Theorems 1.4 and 1.5

In this section, we prove Theorems 1.4 and 1.5. First we prove the following
lemma.

######  Lemma 6.1.

For an odd prime $p$, let $\mathrm{\Gamma}$ be a connected $2$-geodesic but
not $2$-arc transitive graph of order $p^{2}$ or $p^{3}$, and let
$A=\mathrm{Aut}(\mathrm{\Gamma})$. Let $1\neq N\lhd A$ be intransitive on
$V(\mathrm{\Gamma})$ and let $\mathrm{\Gamma}$ be a cover of
$\mathrm{\Gamma}_{N}$. Then
$\mathrm{\Gamma}_{N}{~{}\not\cong~{}}\mathsf{K}_{p}$.

Proof. Suppose, by way of contradiction, that
$\mathrm{\Gamma}_{N}\cong\mathsf{K}_{p}$. Then $\mathrm{\Gamma}$ has valency
$p-1$, and $|N|=p$ or $p^{2}$ as $|V(\mathrm{\Gamma})|=p^{2}$ or $p^{3}$. Note
that the complete multipartite graphs $\mathsf{K}_{p[p]}$,
$\mathsf{K}_{p^{2}[p]}$ and $\mathsf{K}_{p[p^{2}]}$ cannot be a cover of
$\mathsf{K}_{p}$, because non of the valencies of them is $p-1$. Since $p$ is
odd, only (2) or (4) of Proposition 2.10 can occur. For (2), $\mathrm{\Gamma}$
is bipartite, and since $\mathrm{\Gamma}$ is $2$-geodesic transitive, it is
$2$-arc transitive, a contradiction. For (4), $N$ is semiregular, and
$\mathrm{\Gamma}_{N}$ is $A/N$-arc transitive but not $(A/N,2)$-arc
transitive. Since $\mathrm{\Gamma}_{N}\cong\mathsf{K}_{p}$, $A/N$ is
2-transitive but not 3-transitive on $V(\mathrm{\Gamma}_{N})$, and by [14, P.
99], $A/N$ is one of the following groups:

  * (a)

$A/N=\mathrm{AGL}(1,p)\cong\mathbb{Z}_{p}\rtimes\mathbb{Z}_{p-1}$;

  * (b)

$A/N=\mathrm{A}_{p}$ or $\mathrm{S}_{p}$;

  * (c)

$A/N=\mathrm{M}_{11}$ and $p=11$, or $\mathrm{M}_{23}$ and $p=23$;

  * (d)

$A/N=\mathrm{PSL}(2,11)$ and $p=11$;

  * (e)

$\mathrm{PSL}(d,q)\leq A/N\leq\mathrm{P}\Gamma L(d,q)$, where $d\geq 2$ and
$\frac{q^{d}-1}{q-1}=p$.

Assume that case (a) occurs. Then $A=N.\mathrm{AGL}(1,p)$. Let $G$ be a Sylow
$p$-subgroup of $A$. Then $N\leq G$ and $G/N$ is the unique Sylow $p$-subgroup
of $\mathrm{AGL}(1,p)$. In particular, $|G|=p^{2}$ or $p^{3}$, $G\lhd A$, and
$G$ is regular on $V(\mathrm{\Gamma})$. Thus, $\mathrm{\Gamma}$ is a normal
Cayley graph on $G$, say $\mathrm{Cay}(G,S)$. By Proposition 2.9, $\langle
s\rangle^{*}\subseteq S$ for all $s\in S$ and $s$ has order $p$. Since
$\mathrm{\Gamma}$ has valency $p-1$, we have $S=\langle s\rangle^{*}$ for some
$s\in S$, and by the connectivity of $\mathrm{\Gamma}$, $G=\langle
S\rangle=\langle s\rangle\cong\mathbb{Z}_{p}$, contradicting that $|G|=p^{2}$
or $p^{3}$.

Cases (b) and (c) cannot occur, because otherwise $A/N$ is $3$-transitive on
$V(\mathrm{\Gamma}_{N})$.

Assume that case (d) or (e) occurs. Then
$T:=\mathrm{soc}(A/N)=\mathrm{PSL}(d,q)$. Clearly, $p^{2}\nmid|T|$.
Furthermore, $T=X/N$ for some $X\unlhd A$, and hence $X=N.T\unlhd A$. Since
$|N|=p$ or $p^{2}$, by Proposition 2.6 we have that $X$ is a central
extension, $X=NX^{\prime}$ and $X^{\prime}\cong M.T$ with
$M\leq\mathrm{Mult}(T)\cap N$.

Now we claim $M=1$. Since $M\leq N$, $M$ is a $p$-group. For case (d),
$A/N=\mathrm{PSL}(2,11)$, $p=11$, and $T\cong\mathrm{PSL}(2,11)$. By [18, P.
302, Table 4.1], $\mathrm{Mult}(\mathrm{PSL}(2,11))\cong\mathbb{Z}_{2}$, and
since $M$ is a $p$-group, $M=1$, as claimed. For case (e),
$T\cong\mathrm{PSL}(d,q)$. If $d=2$, then $\frac{q^{d}-1}{q-1}=p$ implies that
$q$ is a power of $2$ and $T$ is $3$-transitive on $V(\mathrm{\Gamma}_{N})$,
forcing that $A/N$ is $3$-transitive on $V(\mathrm{\Gamma}_{N})$, a
contradiction. Thus, $d\geq 3$, and $\frac{q^{d}-1}{q-1}=p$ implies that $d$
is an odd prime and $d<p$. If $(d,q)\neq(3,2)$ and $(3,4)$, by [18, P. 302,
Table 4.1] we have
$\mathrm{Mult}(\mathrm{PSL}(d,q))\cong\mathbb{Z}_{(d,q-1)}$, and since $M$ is
a $p$-group, $M=1$, as claimed. If $(d,q)=(3,4)$, then $\frac{q^{d}-1}{q-1}$
is not a prime, a contradiction. If $(d,q)=(3,2)$, then $p=7$, and again by
[18, P. 302, Table 4.1],
$\mathrm{Mult}(\mathrm{PSL}(3,2))\cong\mathbb{Z}_{2}$, forcing $M=1$, as
claimed.

It follows that $M=1$ and $X^{\prime}\cong T$. Since $X^{\prime}$ is
characteristic in $X$, we have $X^{\prime}\unlhd A$. Since $p^{2}\nmid|T|$,
$X^{\prime}$ is intransitive on $V(\mathrm{\Gamma})$, and since $p$ is odd and
$\mathrm{\Gamma}$ has valency $p-1$, Proposition 2.10 implies that
$X^{\prime}\cong T$ is semiregular on $V(\mathrm{\Gamma})$, contradicting that
$|V(\mathrm{\Gamma})|=p^{2}$ or $p^{3}$. ∎

Proof of Theorem 1.4. By Examples in Section 3, all graphs in Theorem 1.4 are
2-geodesic transitive of order $p^{2}$.

Assume that $\mathrm{\Gamma}$ is a connected $2$-geodesic transitive graph of
order $p^{2}$ and let $A=\mathrm{Aut}(\mathrm{\Gamma})$. If $\mathrm{\Gamma}$
is 2-arc transitive, Proposition 2.13 implies that
$\mathrm{\Gamma}\cong\mathbf{C}_{p^{2}}$ or $\mathsf{K}_{p^{2}}$, which is
Theorem 1.4 (1). Suppose that $\mathrm{\Gamma}$ is not 2-arc transitive. If
$p\leq 5$, then Proposition 3.5 implies that $\mathrm{\Gamma}\cong
H(2,3),H(2,5),\overline{H(2,5)},\mathsf{K}_{3[3]}$ or $\mathsf{K}_{5[5]}$.
Suppose that $p\geq 7$.

Assume that $A$ is quasiprimitive on $V(\mathrm{\Gamma})$. By Proposition
2.12, either $\mathrm{\Gamma}\cong H(2,p)$ or $\overline{H(2,p)}$ that is
listed in Theorem 1.4 (2); or $A$ is primitive on $V(\mathrm{\Gamma})$ and
$\mathrm{\Gamma}=\mathrm{Cay}(\mathbb{Z}_{p}^{2},S)$ is a normal Cayley graph,
which is impossible by Lemma 5.3.

Assume that $\mathrm{\Gamma}$ is not quasiprimitive on $V(\mathrm{\Gamma})$.
Then $A$ has a nontrivial intransitive normal subgroup $N$ on
$V(\mathrm{\Gamma})$. By Proposition 2.10, either
$\mathrm{\Gamma}\cong\mathsf{K}_{p[p]}$ with $p\geq 3$, or $N$ is semiregular
on $V(\mathrm{\Gamma})$ and $\mathrm{\Gamma}$ is a cover of
$\mathrm{\Gamma}_{N}$. For the former, $\mathrm{\Gamma}$ is listed in Theorem
1.4 (2). For the latter, $|V(\mathrm{\Gamma}_{N})|=p$ and
$\mathrm{\Gamma}_{N}$ is $(A/N,2)$-geodesic transitive. Then
$\mathrm{\Gamma}_{N}$ is circulant, and by Proposition 2.8,
$\mathrm{\Gamma}_{N}\cong\mathsf{K}_{p}$, which is impossible by Lemma 6.1. ∎

To prove Theorem 1.5, we first classify $2$-arc transitive graphs of order
prime-cube.

######  Lemma 6.2.

Let $p$ be a prime and let $\mathrm{\Gamma}$ be a connected $2$-arc transitive
graph of order $p^{3}$. Then $\mathrm{\Gamma}$ is isomorphic to $H(3,2)$,
$\mathsf{K}_{4,4}$, $\mathbf{C}_{p^{3}}$ or $\mathsf{K}_{p^{3}}$.

Proof. If $p=2$ or 3, then Proposition 3.5 implies that
$\mathrm{\Gamma}\cong\mathbf{C}_{8},H(3,2),\mathsf{K}_{4,4},\mathsf{K}_{8},\mathbf{C}_{27}$
or $\mathsf{K}_{27}$. Let $p\geq 5$. By Proposition 2.13, either
$\mathrm{\Gamma}\cong\mathbf{C}_{p^{3}}$ or $\mathsf{K}_{p^{3}}$, or
$\mathrm{\Gamma}$ is a cover of a complete graph. Let
$A=\mathrm{Aut}(\mathrm{\Gamma})$. If $\mathrm{\Gamma}$ is a cover of a
complete graph, then $A$ has a nontrivial normal subgroup $N$ such that $N$ is
semiregular on $V(\mathrm{\Gamma})$ and $\mathrm{\Gamma}$ is a $2$-arc
transitive cover of $\mathrm{\Gamma}_{N}$ with
$\mathrm{\Gamma}_{N}=\mathsf{K}_{p}$ or $\mathsf{K}_{p^{2}}$. Clearly,
$N\cong\mathbb{Z}_{p}$, $\mathbb{Z}_{p^{2}}$ or
$\mathbb{Z}_{p}\times\mathbb{Z}_{p}$. By [15, Theorem 1.1], $p=2$,
contradicting that $p\geq 5$. It follows that
$\mathrm{\Gamma}\cong\mathbf{C}_{p^{3}}$ or $\mathsf{K}_{p^{3}}$. ∎

Proof of Theorem 1.5. By Examples in Section 3 and Theorem 4.3, all graphs in
Theorem 1.5 are $2$-geodesic transitive of order $p^{3}$.

Let $\mathrm{\Gamma}$ be a $2$-geodesic transitive graph of order $p^{3}$ and
let $A=\mathrm{Aut}(\mathrm{\Gamma})$. If $\mathrm{\Gamma}$ is $2$-arc
transitive, by Lemma 6.2, Theorem 1.5 (1) holds. In what follows we assume
that $\mathrm{\Gamma}$ is not $2$-arc transitive. If $p=2$ or $3$, Proposition
3.5 implies that $\mathrm{\Gamma}\cong\mathsf{K}_{4[2]}$,
$\mathcal{G}_{(27,4)}$, $H(3,3)$, $\mathcal{G}_{(27,8)}$, $\mathsf{K}_{3[9]}$,
$\mathsf{K}_{9[3]}$, or the Schläfli graph or its complement. Thus, we let
$p\geq 5$.

Assume that $A$ is quasiprimitive on $V(\mathrm{\Gamma})$. By Proposition
2.12, either $\mathrm{\Gamma}$ is the Hamming graph $H(3,p)$ that is Theorem
1.5 (2)-(iii), or $A$ is primitive on $V(\mathrm{\Gamma})$ and
$\mathrm{\Gamma}$ is a normal Cayley graph on $\mathbb{Z}_{p}^{3}$, which is
impossible by Lemma 5.4.

Assume that $A$ is not quasiprimitive on $V(\mathrm{\Gamma})$. Then $A$ has a
nontrivial maximal intransitive normal subgroup $N$ and so $A/N$ is
quasiprimitive on $V(\mathrm{\Gamma}_{N})$. By Proposition 2.10, either
$\mathrm{\Gamma}\cong\mathsf{K}_{p^{2}[p]}$ or $\mathsf{K}_{p[p^{2}]}$ that is
Theorem 1.5 (2)-(ii), or $N$ is semiregular on $V(\mathrm{\Gamma})$ and
$\mathrm{\Gamma}$ is a cover of $\mathrm{\Gamma}_{N}$. To finish the proof,
assume that $\mathrm{\Gamma}$ is a cover of $\mathrm{\Gamma}_{N}$, and we only
need to show that Theorem 1.5 (2) (iv) occurs. Thus, $N$ is semiregular on
$V(\mathrm{\Gamma})$, and $\mathrm{\Gamma}$ and $\mathrm{\Gamma}_{N}$ have the
same valency.

Note that $|V(\mathrm{\Gamma}_{N})|=p$ or $p^{2}$, and $\mathrm{\Gamma}_{N}$
is $(A/N,2)$-geodesic transitive. If $|V(\mathrm{\Gamma}_{N})|=p$, then
$\mathrm{\Gamma}_{N}$ is circulant, and by Proposition 2.8,
$\mathrm{\Gamma}_{N}\cong\mathbf{C}_{p}$ or $\mathsf{K}_{p}$, of which the
latter is impossible by Lemma 6.1. For the former,
$\mathrm{\Gamma}\cong\mathbf{C}_{p^{3}}$, contradicting that $\mathrm{\Gamma}$
is not $2$-arc-transitive. Thus, we may let $|V(\mathrm{\Gamma}_{N})|=p^{2}$
and $N\cong\mathbb{Z}_{p}$.

Let $M/N$ be a minimal normal subgroup of $A/N$. Then $M\unlhd A$ and
$M/N=T_{1}\times T_{2}\times\cdots\times T_{t}$, where $T_{i}\cong T$ is a
simple group. Since $A/N$ is quasiprimitive on $V(\mathrm{\Gamma}_{N})$, $M/N$
is transitive on $V(\mathrm{\Gamma}_{N})$, and since $\mathrm{\Gamma}_{N}$ is
$2$-geodesic transitive and $|V(\mathrm{\Gamma}_{N})|=p^{2}$,
$\mathrm{\Gamma}_{N}$ is listed in Theorem 1.4. By Proposition 2.11,
$\mathrm{\Gamma}_{N}{~{}\not\cong~{}}\mathsf{K}_{p[p]}$. Since
$\mathrm{\Gamma}$ is 2-geodesic but not 2-arc transitive, $\mathrm{\Gamma}$
has girth 3, and hence $\mathrm{\Gamma}_{N}$ has girth $3$, which implies
$\mathrm{\Gamma}_{N}{~{}\not\cong~{}}\mathbf{C}_{p^{2}}$. It follows from
Theorem 1.4 that

$\mathrm{\Gamma}_{N}\cong\mathsf{K}_{p^{2}},H(2,p)\text{~{}or~{}}\overline{H(2,p)}.$
(3)

Claim 1: $M/N$ is abelian.

Suppose that $M/N$ is non-abelian and we will obtain a contradiction. In this
case, $T$ is a non-abelian simple group and $M/N=T_{1}\times
T_{2}\times\cdots\times T_{t}$ with $T_{i}\cong T$ for all $i$. We first prove
$t=1$, that is, $M/N$ is non-abelian simple.

Suppose $t\geq 2$. Recall that $M/N$ is transitive on $V(\mathrm{\Gamma}_{N})$
and $|V(\mathrm{\Gamma}_{N})|=p^{2}$. If some $T_{i}$ is transitive on
$V(\mathrm{\Gamma}_{N})$, say $T_{1}$, then a stabilizer $(T_{2})_{\alpha}$
with $\alpha\in V(\mathrm{\Gamma}_{N})$ fixes every vertex in
$V(\mathrm{\Gamma}_{N})$, because $T_{1}$ commutes with $T_{2}$ pointwise.
Thus, $T_{2}$ is semiregular on $V(\mathrm{\Gamma}_{N})$ and hence $|T_{2}|$
is a divisor of $p^{2}=|V(\mathrm{\Gamma}_{N})|$, contradicting that $T_{1}$
is non-abelian simple. It follows that each orbit of $T_{i}$ with $1\leq i\leq
p$, has length $p$. Let

$\Delta=\\{\Delta_{1},\cdots,\Delta_{p}\\}$

be the orbit set of $T_{1}$ on $V(\mathrm{\Gamma}_{N})$. Then $|\Delta_{j}|=p$
for all $1\leq j\leq p$, and $\Delta$ is a complete imprimitive block system
of $M/N$.

Since $M/N$ is transitive on $V(\mathrm{\Gamma}_{N})$, some $T_{i}$ is
transitive on $\Delta$, say $T_{2}$. Let $\alpha_{1}\in\Delta_{1}$. Since
$T_{1}$ commutes with $T_{2}$ pointwise, the transitivity of $T_{2}$ on
$\Delta$ implies that $(T_{1})_{\alpha_{1}}$ fixes a vertex $\alpha_{i}$ in
$\Delta_{i}$ for all $1\leq i\leq p$. Thus,
$(T_{1})_{\alpha_{1}}=(T_{1})_{\alpha_{2}}=\cdots=(T_{1})_{\alpha_{p}}$.

Consider the natural action of $M/N$ on $\Delta$ and let $K$ be the kernel of
the action. Clearly, $T_{1}\leq K$ and $T_{2}\nleq K$. If $T_{1}\not=K$, then
some $T_{i}$ with $i\geq 3$ is a subgroup of $K$ because $K\unlhd M/N$, says
$T_{3}\leq K$. Then $\Delta$ is also the orbit set of $T_{3}$. Since $T_{1}$
and $T_{3}$ commutes pointwise, $(T_{1})_{\alpha_{1}}$ fixes every vertex in
$\Delta_{1}$, and similarly, $(T_{1})_{\alpha_{i}}$ fixes every vertex in
$\Delta_{i}$ for each $1\leq i\leq p$. Since
$(T_{1})_{\alpha_{1}}=(T_{1})_{\alpha_{2}}=\cdots=(T_{1})_{\alpha_{p}}$,
$T_{1}$ is semiregular on $V(\mathrm{\Gamma}_{N})$, a contradiction. Thus,
$K=T_{1}$.

It follows that $(M/N)/T_{1}$ has a faithful action on $\Delta$, and since
$|\Delta|=p$, $(M/N)/T_{1}$ is isomorphic to a subgroup of the symmetric group
$S_{p}$. Thus, $p^{2}\nmid(M/N)/T_{1}$. It follows that $t=2$ and so
$M/N=T_{1}\times T_{2}$ with $p^{2}\nmid|T_{2}|$. On the other hand, the
simplicity of $T_{1}$ implies that $T_{1}$ is isomorphic to a subgroup of the
symmetric group on $\Delta_{1}$, and similarly, $p^{2}\nmid|T_{1}|$. Thus,
$p^{3}\nmid|T_{1}\times T_{2}|$.

Since $M/N=T_{1}\times T_{2}$, we may let $X/N=T_{2}$ with $N\leq X\unlhd M$.
Since $T_{2}$ is transitive on $\Delta$, it has a subgroup of index $p$. By
Proposition 2.2, we have

$T_{2}\cong\mathrm{M}_{11},\mathrm{M}_{23},\mathrm{A}_{p},\mathrm{PSL}(2,11)\text{~{}with~{}}p=11,\text{~{}or~{}}\mathrm{PSL}(d,q)\text{~{}with~{}}(q^{d}-1)/(q-1)=p.$

Recall that $p\geq 5$. By [18, P.302, Table 4.1],
$\mathrm{Mult}(M_{11})=\mathrm{Mult}(M_{23})=1$,
$\mathrm{Mult}(\mathrm{A}_{7})=\mathbb{Z}_{6}$,
$\mathrm{Mult}(\mathrm{A}_{p})=\mathbb{Z}_{2}$ for $p\not=7$, and
$\mathrm{Mult}(\mathrm{PSL}(2,11))=\mathbb{Z}_{2}$. Furthermore,
$\mathrm{Mult}(\mathrm{PSL}(d,q))=\mathbb{Z}_{(d,q-1)}$ for
$(d,q)\neq(2,4),(2,9),(3,2)$ or $(3,4)$, and for $(d,q)=(2,4),(2,9),(3,2)$ or
$(3,4)$, we have $(d,q)=(2,4)$ or $(3,2)$ as $(q^{d}-1)/(q-1)=p$ and
$\mathrm{Mult}(\mathrm{PSL}(2,4))=\mathrm{Mult}(\mathrm{PSL}(3,2))=\mathbb{Z}_{2}$.
In all cases, $p\nmid|\mathrm{Mult}(T_{2})|$.

Since $N\cong\mathbb{Z}_{p}$ and $X=N.T_{2}$, by Proposition 2.6, $X$ is a
central extension of $N$ by $T_{2}$, $X=NX^{\prime}$ and $X^{\prime}=H.T_{2}$,
where $H\leq\mathrm{Mult}(T_{2})\cap N$. Since $p\nmid|\mathrm{Mult}(T_{2})|$,
$\mathrm{Mult}(T_{2})\cap N=1$ and hence $H=1$. It follows that
$H^{\prime}=T_{2}$ and $X=N\times H^{\prime}$. Thus, we may write $X=N\times
T_{2}$.

Let $Y/N=T_{1}$ with $N\leq Y\unlhd M$. Since $T_{1}$ is faithful on
$\Delta_{1}$ and $|\Delta_{1}|=p$, a similar argument to $T_{2}$ above gives
rise to $Y=N\times T_{1}$. It follows that $M=N\times T_{1}\times T_{2}$.

Clearly, $T_{1}\times T_{2}$ is characteristic in $M$, an hence $T_{1}\times
T_{2}\unlhd A$ as $M\unlhd A$. Since $p^{3}\nmid|T_{1}\times T_{2}|$,
$T_{1}\times T_{2}$ is intransitive on $V(\mathrm{\Gamma})$, and by
Proposition 2.10, $T_{1}\times T_{2}$ is semiregular, which is impossible.
Thus, $t=1$, as required.

Now we may let $M/N=T$ for a non-abelian simple group $T$. Since $M/N$ is
transitive on $V(\mathrm{\Gamma}_{N})$, $T$ has a subgroup of index $p^{2}$.
Again by Proposition 2.2, we have

$T\cong\mathrm{A}_{p^{2}},\mbox{ or }\mathrm{PSL}(d,q)\mbox{ with
}p^{2}=(q^{d}-1)/(q-1)\mbox{ and }d\mbox{ a prime}.$

For the latter, it is easy to see that $(d,q-1)\not=p$. By [18, P.302, Table
4.1], we have $p\nmid|\mathrm{Mult}(T)|$, and by a similar argument to $T_{2}$
above, we may write $M=N\times T$.

Clearly, $T$ is characteristic in $M$ and hence $T\unlhd A$. Since
$|V(\mathrm{\Gamma})|=p^{3}$, $T$ cannot be semiregualr on
$V(\mathrm{\Gamma})$, and by Proposition 2.10, $T$ is transitive on
$V(\mathrm{\Gamma})$. Thus, $p^{3}\mid|T|$.

Assume $T\cong\mathrm{PSL}(d,q)$ with $p^{2}=(q^{d}-1)/(q-1)$ and $d$ a prime.
Clearly, $p\nmid q$. If $p\mid(q^{d}+1)$ then $p$ is divisor of
$(q^{d}-1,q^{d}+1)$, forcing $p=2$, a contradiction. Then $p\nmid(q^{d}+1)$.
In particular, $p^{3}\nmid|\mathrm{PSL}(2,q)|$, and since $p^{3}\mid|T|$, $d$
is an odd prime. Since $p^{2}=(q^{d}-1)/(q-1)=q^{d-1}+\cdots+q+1>q^{d-1}$, we
have $p>q^{\frac{d-1}{2}}$, forcing $p\nmid(q^{i}-1)$ for every $1\leq
i\leq\frac{d-1}{2}$.

Let $p\mid(q^{j}-1)$ for some $1+\frac{d-1}{2}\leq j\leq d-1$. Since
$p\mid(q^{d}-1)$, $p$ is a divisor of $(q^{d}-1)-(q^{j}-1)=q^{j}(q^{d-j}-1)$,
and since $p\nmid q^{j}$, $p$ is a divisor of $(q^{d-j}-1)$, which is
impossible because $d-j\leq d-(1+\frac{d-1}{2})=\frac{d-1}{2}$.

It follows that $p\nmid(q^{i}-1)$ for every $1\leq i\leq d-1$. Since
$p^{2}=(q^{d}-1)/(q-1)$, we have $p^{3}\nmid|\mathrm{PGL}(d,q)|$, and in
particular, $p^{3}\nmid|\mathrm{PSL}(d,q)|$, contradicting $p^{3}\mid|T|$.

Thus, $T\cong\mathrm{A}_{p^{2}}$. Furthermore, $|T:T_{\alpha}|=p^{2}$ for
$\alpha\in V(\mathrm{\Gamma}_{N})$, and $T_{\alpha}\cong\mathrm{A}_{p^{2}-1}$,
where $\alpha$ is an orbit of $N$ on $V(\mathrm{\Gamma})$ and $|\alpha|=p$.
Then $T_{\alpha}$ has a subgroup of index $p$, which is impossible. This
completes the proof of Claim 1.

By Claim 1, $M/N$ is abelian, and so $M/N$ is regular on
$V(\mathrm{\Gamma}_{N})$. It follows that $M/N\cong\mathbb{Z}_{p}^{2}$ and
$|M|=p^{3}$. Since $M$ is transitive on $V(\mathrm{\Gamma})$, $M$ is regular
on $V(\mathrm{\Gamma})$. Then we may let $\mathrm{\Gamma}=\mathrm{Cay}(M,S)$
be a normal Cayley graph, where $R(M)$ is identified with $M$. By Proposition
2.7, $A_{1}=\mathrm{Aut}(M,S)$, and by Proposition 2.9, $S=\cup_{s\in
S}\langle s\rangle^{*}$, implying $(p-1)\mid|S|$. Since $N\unlhd A$, $A_{1}$
fixes $N$ setwise. Set $\ell=|S|/(p-1)$. Since $\mathrm{\Gamma}$ is a cover of
$\mathrm{\Gamma}_{N}$, we have
$\mathrm{\Gamma}_{N}=\mathrm{Cay}(M/N,\overline{S})$, where
$\overline{S}=\\{sN\ |\ s\in S\\}$, and $M/N\unlhd A/N$. Thus, the stabilizer
$(A/N)_{N}\leq\mathrm{Aut}(M/N,\overline{S})\leq\mathrm{Aut}(M/N)\cong\mathrm{GL}(2,p)$,
and noting that $A_{1}\cong(A/N)_{N}$, we may write

$A_{1}\leq\mathrm{GL}(2,p),\ \mbox{ and }A/N\leq
N_{\mathrm{Aut}(\mathrm{\Gamma}_{N})}(M/N),$

where $N_{\mathrm{Aut}(\mathrm{\Gamma}_{N})}(M/N)$ is the normalizer of $M/N$
in $\mathrm{Aut}(\mathrm{\Gamma}_{N})$. Since $\mathrm{\Gamma}$ is connected,
$\langle S\rangle=M$, and by Proposition 2.9, $M$ is generated by elements of
order $p$. By Proposition 2.1, $M=\mathbb{Z}_{p}^{3}$ or $\langle a,b,c\ |\
a^{p}=b^{p}=c^{p}=[a,c]=[b,c]=1,c=[a,b]\rangle$.

Claim 2: Assume that for all $x,y\in S$ with $\langle x\rangle\not=\langle
y\rangle$, $x$ is not adjacent to some vertex in $\langle y\rangle^{*}$. Then
$\ell(\ell-1)(p-1)n\mid|A_{1}|$, where $n\geq 1$ and $\ell=|S|/(p-1)$.

Denote by $\overline{y}_{x+}$ the set of vertices of $\langle y\rangle^{*}$
that are adjacent to $x$ in $\mathrm{\Gamma}$, and by $\overline{y}_{x-}$ the
set of vertices of $\langle y\rangle^{*}$ that are not adjacent to $x$ in
$\mathrm{\Gamma}$. Then $|\overline{y}_{x+}|+|\overline{y}_{x-}|=p-1$. Let
$\overline{S}=\\{\langle u\rangle^{*}\ |\ u\in S\\}$. Then
$|\overline{S}|=\ell$. Let $z\in S$ with $\langle z\rangle\not=\langle
x\rangle$. By hypothesis, we may take $y^{i}\in\overline{y}_{x-}$ and
$z^{j}\in\overline{z}_{x-}$. Then $(x,1,y^{i})$ and $(x,1,z^{j})$ are two
2-geodesics, and since $\mathrm{\Gamma}$ is $2$-geodesic transitive, $A_{1}$
is $2$-transitive on $\overline{S}$. Thus, we may assume that

$|\overline{y}_{x-}|=n>0,\ \mbox{ for all }x,y\in S\mbox{ with
}x\not\in\langle y\rangle.$

Clearly, $\mathrm{\Gamma}(x)\cap\mathrm{\Gamma}(1)=\cup_{y\in
S,y\not\in\langle x\rangle}\overline{y}_{x+}$, and
$|S|=1+|\mathrm{\Gamma}(x)\cap\mathrm{\Gamma}(1)|+|\mathrm{\Gamma}(x)\cap\mathrm{\Gamma}_{2}(1)|$
as $\mathrm{\Gamma}$ has valency $|S|$. Thus,
$|\mathrm{\Gamma}(x)\cap\mathrm{\Gamma}(1)|=(|\overline{S}|-1)|\overline{y}_{x+}|+p-2=(\ell-1)(p-1-n)+p-2$
and
$(p-1)\ell=1+(\ell-1)(p-1-n)+p-2+|\mathrm{\Gamma}(x)\cap\mathrm{\Gamma}_{2}(1)|$,
that is, $|\mathrm{\Gamma}_{2}(1)\cap\mathrm{\Gamma}(x)|=(\ell-1)n$. Since
$\mathrm{\Gamma}$ is $2$-geodesic transitive, $A_{1}$ is transitive on $S$ and
$A_{1x}$ is transitive on $\mathrm{\Gamma}(x)\cap\mathrm{\Gamma}_{2}(1)$,
which imply that $\ell(\ell-1)(p-1)n\mid|A_{1}|$, as claimed.

Recall that $M=\mathbb{Z}_{p}^{3}$ or $\langle a,b,c\ |\
a^{p}=b^{p}=c^{p}=[a,c]=[b,c]=1,c=[a,b]\rangle$.

Case 1: $M=\mathbb{Z}_{p}^{3}$.

Let $\mathrm{Aut}(\mathbb{Z}_{p}^{3})=\mathrm{GL}(3,p)$ and
$Z=Z(\mathrm{GL}(3,p))$. Since $A_{1}=\mathrm{Aut}(M,S)\leq\mathrm{GL}(3,p)$
and $S=\cup_{s\in S}\langle s\rangle^{*}$, we have $Z\leq
A_{1}\leq\mathrm{GL}(3,p)$. Let $x,y\in S$ with $\langle x\rangle\not=\langle
y\rangle$. If $x$ is adjacent to all vertices of $\langle y\rangle$, then
every vertex in $\langle x\rangle$ is adjacent to every vertex in $\langle
y\rangle$ because $Z\leq A_{1}$. By Lemma 5.2, $[\langle x\rangle\langle
y\rangle]$ is a complete graph, and hence $\langle x,y\rangle^{*}\subseteq S$.
Since $\mathrm{\Gamma}$ and $\mathrm{\Gamma}_{N}$ have the same valency, Eq.
(3) implies that $\mathrm{\Gamma}$ has valency at most $p^{2}-1$, and hence
$S=\langle x,y\rangle^{*}$, contradicting $\langle S\rangle=M$. Thus, $x$ is
not adjacent to some vertex in $\langle y\rangle^{*}$. By Claim 2,
$\ell(\ell-1)(p-1)n\mid|A_{1}|$.

Since $M=\mathbb{Z}_{p}^{3}=\langle S\rangle$, we have $\ell=|S|/(p-1)\geq 3$.
If $\ell=3$, by Example 3.3 we have $\mathrm{\Gamma}=H(3,p)$, and since
$\mathrm{\Gamma}$ and $\mathrm{\Gamma}_{N}$ have the same valency,
$\mathrm{\Gamma}$ has valency $3(p-1)$, which is impossible by Eq. (3). Thus,
we may assume that $\ell\geq 4$, and again by Eq. (3),
$\mathrm{\Gamma}_{N}\cong\mathsf{K}_{p^{2}}$ or $\overline{H(2,p)}$, and
$\mathrm{\Gamma}$ has valency $p^{2}-1$ or $(p-1)^{2}$, respectively.

Suppose $\mathrm{\Gamma}_{N}\cong\overline{H(2,p)}$. By Example 3.3,
$\mathrm{Aut}(\mathrm{\Gamma}_{N})\cong\mathrm{Aut}(H(2,p))\cong(\mathrm{S}_{p}\times\mathrm{S}_{p})\rtimes\mathbb{Z}_{2}$,
and since $M/N\cong\mathbb{Z}_{p}^{2}$, we have $A/N\leq
N_{\mathrm{Aut}(\mathrm{\Gamma}_{N})}(M/N)\lesssim(\mathrm{AGL}(1,p)\times\mathrm{AGL}(1,p))\rtimes\mathbb{Z}_{2}$.
It follows that $|(A/N)_{N}|\mid 2(p-1)^{2}$ and hence $|A_{1}|\mid
2(p-1)^{2}$. Since $\overline{H(2,p)}$ has valency $(p-1)^{2}$, we have
$|S|=(p-1)^{2}$ and $\ell=p-1$. Then $(p-1)^{2}(p-2)n\mid|A_{1}|$ and so
$(p-1)^{2}(p-2)n\mid 2(p-1)^{2}$, a contradiction.

Suppose $\mathrm{\Gamma}_{N}\cong\mathsf{K}_{p^{2}}$. Then $|S|=p^{2}-1$ and
$\ell=p+1$. Since $p(p^{2}-1)\mid|A_{1}|$ and $A_{1}\leq\mathrm{GL}(2,p)$, we
have $|\mathrm{GL}(2,p):A_{1}|\leq p-1$, and by Lemma 4.2 (1), $A_{1}$
contains $\mathrm{GL}(2,p)^{\prime}\cong\mathrm{SL}(2,p)$. Since
$A_{1}=\mathrm{Aut}(M,S)$ fixes $N$ setwise, $A_{1}$ is a subgroup of
$\mathrm{Aut}(M)_{N}$, the block stabilizer of $N$ in $\mathrm{Aut}(M)$. We
may assume that $M=\mathbb{Z}_{p}^{3}=\langle a\rangle\times\langle
b\rangle\times\langle c\rangle$ and $N=\langle c\rangle$. By Lemma 5.1 (3),
$\mathrm{Aut}(M)_{N}$ has one conjugate class of subgroup isomorphic to
$\mathrm{SL}(2,p)$, of which one has orbit set $\\{\\{c^{i}\\},c^{i}\langle
a,b\rangle^{*}\mid i\in\mathbb{Z}_{p}\\}$. Thus, each orbit of $A_{1}$ on
$V(\mathrm{\Gamma})$ is union of some orbits of $\mathrm{SL}(2,p)$. Since
$A_{1}$ is transitive on $S$, $S$ is an orbit of $A_{1}$ with length
$p^{2}-1$, and hence $S=c^{i}\langle a,b\rangle^{*}$ for some
$i\in\mathbb{Z}_{p}$. Since $\langle S\rangle=M$, $c^{i}\neq 1$ and $\langle
ch\rangle^{*}\subseteq S$ for all $h\in\langle a,b\rangle^{*}$. Thus,
$|S|>p^{2}-1$, a contradiction.

Case 2: $M=\langle a,b,c\ |\ a^{p}=b^{p}=c^{p}=[a,c]=[b,c]=1,c=[a,b]\rangle$.

In this case, $N=\langle c\rangle\unlhd A$. By Eq. (3), $\mathrm{\Gamma}$ has
valency at least $2(p-1)$. Then $|S|\geq 2(p-1)$, and since
$A_{1}=\mathrm{Aut}(M,S)$ fixes $\langle c\rangle$ setwise, the arc-
transitivity of $A$ implies $\langle c\rangle\cap S=\emptyset$. Thus, $\langle
x,y\rangle=M$ and $[x,y]\not=1$, for all $x,y\in S$ with $x\not\in\langle
y\rangle$.

Suppose that $x$ is adjacent to all elements in $\langle y\rangle^{*}$. Then
$xy^{i}\in S$ for all $i\in\mathbb{Z}_{p}^{*}$. If $x^{j}\in\langle
x\rangle^{*}$ is not adjacent to $y^{k}\in\langle y\rangle^{*}$, then $A_{1}$
has an element $\alpha$ interchanging $x^{j}$ and $y^{k}$ and hence $\langle
x\rangle^{*}$ and $\langle y\rangle^{*}$ because $(x^{j},1,y^{k})$ and
$(y^{k},1,x^{j})$ are $2$-geodesics, implying that $x^{\alpha}\in\langle
y\rangle^{*}$ is adjacent to all vertices in $\langle x\rangle^{*}$. Thus, we
may always assume that $\langle x\rangle^{*}$ has an element $x^{s}$ adjacent
to some $y^{t}\in\langle y\rangle^{*}$ with $1\not=s\in\mathbb{Z}_{p}^{*}$ and
$t\in\mathbb{Z}_{p}^{*}$. Thus, $x^{s}y^{-t}\in S$ and so $\langle
x^{s}y^{-t}\rangle^{*}\in S$. Then
$(x^{s}y^{-t})^{s^{-1}}=xy^{-s^{-1}t}[y^{-t},x^{s}]^{2^{-1}s^{-1}(s^{-1}-1)}=xy^{-s^{-1}t}c_{1}\in
S$, where $c_{1}=[y^{-t},x^{s}]^{2^{-1}s^{-1}(s^{-1}-1)}\in\langle c\rangle$.
Since $s\not=1$ and $[x,y]\not=1$, we have $c_{1}\not=1$. Recall that
$xy^{-s^{-1}t}\in S$. Then we have two edges $\\{1,xy^{-s^{-1}t}\\}$ and
$\\{1,xy^{-s^{-1}t}c_{1}\\}$, implying the subgraph $[N,xy^{-s^{-1}t}N]$ of
$\mathrm{\Gamma}$ is not a matching, contradicting that $\mathrm{\Gamma}$ is a
cover of $\mathrm{\Gamma}_{N}$. Thus, $x$ is not adjacent to some vertex in
$\langle y\rangle^{*}$. By Claim 2,
$|\mathrm{\Gamma}_{2}(1)\cap\mathrm{\Gamma}(x)|=(\ell-1)n$ and
$\ell(\ell-1)(p-1)n\mid|A_{1}|$, and by Eq. (3), $\mathrm{\Gamma}_{N}\cong
H(2,p)$, $\mathsf{K}_{p^{2}}$ or $\overline{H(2,p)}$.

Let $\mathrm{\Gamma}_{N}\cong H(2,p)$. Then $\mathrm{\Gamma}$ has valency
$2(p-1)$, and since $S=\cup_{s\in S}\langle s\rangle^{*}$, we may let
$S=\langle x\rangle^{*}\cup\langle y\rangle^{*}$. Thus, $M=\langle
S\rangle=\langle x,y\rangle$, and so $M$ has an automorphism mapping $x$ to
$a$ and $y$ to $b$. It follows that
$\mathrm{\Gamma}\cong\mathrm{Cay}(M,\langle a\rangle^{*}\cup\langle
b\rangle^{*})$, which corresponds to the first graph of Theorem 1.5 (2)-(iv).

Let $\mathrm{\Gamma}_{N}\cong\overline{H(2,p)}$. Then
$\ell=(p-1)^{2}/(p-1)=p-1$, and a similar argument to Case 1 yields that
$|A_{1}|\mid 2(p-1)^{2}$. Since $(p-1)^{2}(p-2)n\mid|A_{1}|$, we have
$(p-1)^{2}(p-2)n\mid 2(p-1)^{2}$, a contradiction.

Let $\mathrm{\Gamma}_{N}\cong\mathsf{K}_{p^{2}}$. Then $|S|=p^{2}-1$ and
$\ell=p+1$. Since $\ell(\ell-1)(p-1)n\mid|A_{1}|$, we have
$p(p^{2}-1)\mid|A_{1}|$ and $|\mathrm{GL}(2,p):A_{1}|\leq p-1$ as
$A_{1}\leq\mathrm{GL}(2,p)$. By Lemma 4.2, $\mathrm{Aut}(M)$ has one conjugate
class of subgroup isomorphic to $A_{1}$, and
$\mathrm{GL}(2,p)^{\prime}=\mathrm{SL}(2,p)\leq A_{1}$. By Theorem 4.3, we may
let

$A_{1}\leq\mathrm{Aut}(M,R)\cong\mathrm{GL}(2,p),\mbox{ where }R=\langle
b\rangle^{*}\cup_{i\in\mathbb{Z}_{p}}\langle b^{i}ab^{i}\rangle^{*}\mbox{ with
}|R|=p^{2}-1.$