On edge-primitive graphs with soluble edge-stabilizers

TL;DR

This paper classifies edge-primitive, 2-arc-transitive graphs with soluble edge-stabilizers, advancing understanding of their symmetry properties and automorphism groups.

Contribution

It provides a complete classification of a specific class of highly symmetric graphs with soluble edge-stabilizers.

Findings

Classification of edge-primitive, 2-arc-transitive graphs with soluble edge-stabilizers

Identification of automorphism group actions on these graphs

Characterization of their symmetry and structural properties

Abstract

A graph is edge-primitive if its automorphism group acts primitively on the edge set, and 2-arc-transitive if its automorphism group acts transitively on the set of 2-arcs. In this paper, we present a classification for those edge-primitive graphs which are 2-arc-transitive and have soluble edge-stabilizers.

Tables

Table 1. Table 1 . Graphs.

Table 2. Table 2 .

Table 3. Table 3 .

Table 4. Table 4 .

Table 5. Table 5 .

Equations

$$|G_{\{u,v\}}:G_{uv}|=2\text{mbox}and(G_v^{\Gamma (v)}{)}_u\cong (G_u^{\Gamma (u)}{)}_v.$$

$${\mathrm{\Gamma L}}_1(p^f)=⟨\tau ,\sigma \mid {\tau }^{p^f-1}=1={\sigma }^f,{\sigma }^{-1}\tau \sigma ={\tau }^p⟩.$$

$$(G_v^{\Gamma (v)}{)}_u/⟨{\tau }^m⟩\cong ⟨\tau ⟩(G_v^{\Gamma (v)}{)}_u/⟨\tau ⟩\lesssim ⟨\sigma ⟩.$$

$$(G_v^{\Gamma (v)}{)}_u\cong {\mathbb{Z}}_{\frac{p^f-1}{m}}.{\mathbb{Z}}_{\frac{f}{e}}.$$

$$(G_v^{\Gamma (v)}{)}_u=⟨{\tau }^m⟩⟨{\tau }^l{\sigma }^e⟩.$$

$$X_v^{[1]}⊴(X_u^{\Gamma (u)}{)}_v\cong (X_v^{\Gamma (v)}{)}_u,X_{uv}\lesssim (X_u^{\Gamma (u)}{)}_v\times (X_v^{\Gamma (v)}{)}_u.$$

$$(X_v^{\Gamma (v)}{)}_u/((X_v^{\Gamma (v)}{)}_u\cap ⟨{\tau }^m⟩)\cong (X_v^{\Gamma (v)}{)}_u⟨{\tau }^m⟩/⟨{\tau }^m⟩\lesssim {\mathbb{Z}}_{\frac{f}{e}},$$

$$\begin{array}[]{l|l|l}X_{v}^{{\it\Gamma}(v)}&(X_{v}^{{\it\Gamma}(v)})_{u}&\\ \hline\cr{\rm Sz}(q).e&p^{f{+}f}{:}(q-1).e&e\mbox{ a divisor of }f,\,p=2,\,\mbox{odd }f>1\\ \hline\cr{\rm Ree}(q).e&p^{f{+}2f}{:}(q-1).e&e\mbox{ a divisor of }f,\,p=3,\,\mbox{odd }f>1\end{array}$$

$$\begin{array}[]{l|l|l|l|l}X_{v}^{[1]}&1&2^{2}&{\rm A}_{4}&{\rm S}_{4}\\ \hline\cr X_{\{u,v\}}&2^{2}{:}{\rm S}_{3}.2&2^{4}.{\rm S}_{3}.2&2^{4}{:}3^{2}.[4]&2^{4}{:}{\rm S}_{3}^{2}.2\\ \end{array}$$

$$\begin{array}[]{c|l|l|l|l|l|l}X_{v}^{[1]}&1&3^{2}&3^{2}{:}2&3^{2}{\rm Q}_{8}&3^{2}{:}2{\rm A}_{4}&3^{2}{:}2{\rm S}_{4}\\ \hline\cr X_{\{u,v\}}&3^{2}{:}2{\rm S}_{4}{.}2&3^{4}{:}2{\rm S}_{4}{.}2&3^{4}{:}([4].{\rm S}_{4}){.}2&3^{4}{:}{\rm Q}_{8}^{2}{.}{\rm S}_{3}{.}2&3^{4}{:}(2{\rm A}_{4})^{2}{.}[4]&3^{4}{:}(2{\rm S}_{4})^{2}{.}2\\ \end{array}$$

$$\begin{array}[]{c|c|l|l|l}X_{v}&X_{\{u,v\}}&s&d&p\\ \hline\cr 2^{6}{.}({\rm S}_{3}{\times}{\rm SL}_{3}(2))&[2^{8}]{.}{\rm S}_{3}^{2}.2&3&7&2\\ \hline\cr[2^{20}]{.}({\rm S}_{3}{\times}{\rm SL}_{3}(2))&[2^{22}]{.}{\rm S}_{3}^{2}.2&3&7&2\\ \hline\cr 2^{3}{.}{\rm SL}_{3}(2)&[2^{5}].{\rm S}_{3}.2&2&7&2\\ \hline\cr 2^{4}{:}{\rm SL}_{3}(2)&[2^{6}].{\rm S}_{3}.2&2&7&2\\ \hline\cr 3^{6}{.}(2{\rm A}_{4}{\times}{\rm SL}_{3}(3))&[3^{8}]{.}(2{\rm A}_{4}{\times}2{\rm S}_{4}).2&3&13&3\\ 3^{6}{.}(2{\rm S}_{4}{\times}{\rm SL}_{3}(3))&[3^{8}]{.}(2{\rm S}_{4})^{2}.2&3&13&3\\ \hline\cr 3^{3}{.}{\rm SL}_{3}(3)&[3^{5}]{.}2{\rm S}_{4}.2&2&13&3\\ 3^{3}{.}(2{\times}{\rm SL}_{3}(3))&[3^{5}]{.}(2{\times}2{\rm S}_{4}).2&2&13&3\\ \hline\cr 3^{6}{:}{\rm SL}_{3}(3)&[3^{8}]{.}2{\rm S}_{4}.2&2&13&3\\ \end{array}$$

$$({\mathrm{F}}_4(8).2,9^4.(2_{+}^{1+4}:{\mathrm{S}}_3^2).2),({\mathrm{PSL}}_5(3).2,[3^8]:(2{\mathrm{S}}_4{)}^2.2),({\mathrm{PSL}}_4(3).2,3_{+}^{1+4}:(2\times 2{\mathrm{S}}_4)).$$

$$\begin{array}[]{l|l|c|l}\hline\cr T&N&|T_{\{u,v\}}{:}N|&\mbox{Remark}\\ \hline\cr{\rm Ree}(3^{a})&\mathbb{Z}_{3^{a}{\pm}3^{a{+}1\over 2}{+}1}&6&\mbox{odd }a\geq 3\\ &\mathbb{Z}_{2}{\times}\mathbb{Z}_{3^{a}{+}1\over 2}&6&\&\,X=T\\ \hline\cr{\rm G}_{2}(3^{a})&\mathbb{Z}_{3^{a}{\pm}1}^{2}&12&\mbox{odd }a\geq 2\\ &\mathbb{Z}_{3^{2a}{\pm}3^{a}{+}1}&6&\\ \hline\cr&\mathbb{Z}_{2^{a}{+}1}^{2}&48&\mbox{odd }a\geq 3\\ {}^{2}{\rm F}_{4}(2^{a})&\mathbb{Z}_{2^{a}{\pm}2^{a{+}1\over 2}{+}1}^{2}&96&\&\,X=T\\ &\mathbb{Z}_{2^{2a}{\pm}2^{3a{+}1\over 2}{+}2^{a}{\pm}2^{a{+}1\over 2}{+}1}&12&\&\,2^{a}{\pm}2^{a{+}1\over 2}{+}1>5\\ \hline\cr&\mathbb{Z}_{2^{2a}{\pm}2^{a}{+}1}^{2}&72&\\ {\rm F}_{4}(2^{a})&\mathbb{Z}_{2^{2a}{+}1}^{2}&96&a\geq 2\\ &\mathbb{Z}_{2^{4a}-2^{2a}{+}1}&12&\\ \hline\cr{\rm E}_{8}(q^{\prime})&\mathbb{Z}_{q^{\prime 4}-q^{\prime 2}{+}1}^{2}&288&X=T\\ &\mathbb{Z}_{q^{\prime 8}{\pm}q^{\prime 7}{\mp}q^{\prime 5}-q^{\prime 4}{\mp}q^{\prime 3}{\pm}q^{\prime}{+}1}&30&\\ \hline\cr\end{array}$$

$$\begin{array}[]{l|l|l|l|l|l|l|l}X&{\rm J}_{1}&{\rm J}_{2}&{\rm J}_{4}&{\rm Co}_{1}&{\rm O^{\prime}N}.2&{\rm He}&{\rm B}\\ \hline\cr|X_{\{u,v\}}|&2{\cdot}3{\cdot}7&2^{2}{\cdot}3{\cdot}5^{2}&2{\cdot}7{\cdot}43&2^{3}{\cdot}3^{2}{\cdot}7^{2}&2{\cdot}3{\cdot}5{\cdot}31&2^{4}{\cdot}3{\cdot}5^{2}&2^{5}{\cdot}3^{4}{\cdot}13\\ \hline\cr r&7&5&43&7&31&5&13\\ \hline\cr p^{f}&2^{3}&2^{4}&2^{14}&2^{3},\,3^{6}&2^{5}&2^{4}&3^{3},5^{4},2^{12}\\ \hline\cr{p^{f}-1}\,\big{|}\,|G_{uv}|&\checkmark&\checkmark&{\times}&\checkmark,\,{\times}&\checkmark&\checkmark&\checkmark,\checkmark,{\times}\\ \hline\cr\hline\cr X&{\rm B}&{\rm B}&{\rm M}&{\rm M}&{\rm M}&{\rm M}&{\rm M}\\ \hline\cr|X_{\{u,v\}}|&2^{2}{\cdot}3^{2}{\cdot}19&2{\cdot}11{\cdot}23&2^{3}{\cdot}3{\cdot}11{\cdot}23&2^{2}{\cdot}3{\cdot}7{\cdot}29&2{\cdot}3^{2}{\cdot}5{\cdot}31&2^{3}{\cdot}5{\cdot}41&2{\cdot}23{\cdot}47\\ \hline\cr r&19&23&23&29&31&41&47\\ \hline\cr p^{f}&2^{18}&2^{11},3^{11}&2^{11},3^{11}&2^{28}&2^{5},5^{3}&2^{20},3^{8}&2^{23}\\ \hline\cr{p^{f}-1}\,\big{|}\,|G_{uv}|&{\times}&{\times},\,{\times}&{\times},\,{\times}&{\times}&\checkmark,{\times}&{\times},\,{\times}&{\times}\\ \end{array}$$

Full text

Keywords**. Edge-primitive graph, $2$-arc-transitive graph, almost simple group, $2$-transitive group, soluble group.**

2010 Mathematics Subject Classification. 05C25, 20B25

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

1. Introduction

In this paper, all graphs are assumed to be finite and simple, and all groups are assumed to be finite.

A graph is a pair ${\it\Gamma}=(V,E)$ of a nonempty set $V$ and a set $E$ of $2$-subsets of $V$. The elements in $V$ and $E$ are called the vertices and edges of ${\it\Gamma}$, respectively. For $v\in V$, the set ${\it\Gamma}(v)=\{u\in V\mid\{u,v\}\in E\}$ is called the neighborhood of $v$ in ${\it\Gamma}$, while $|{\it\Gamma}(v)|$ is called the valency of $v$. We say that the graph ${\it\Gamma}$ has valency $d$ or ${\it\Gamma}$ is $d$-regular if its vertices have equal valency $d$. For an integer $s\geq 1$, an $s$-arc in ${\it\Gamma}$ is an $(s{+}1)$-tuple $(v_{0},v_{1},\ldots,v_{s})$ of vertices with $\{v_{i},v_{i+1}\}\in E$ and $v_{i}\neq v_{i+2}$ for all possible $i$. A $1$-arc is also called an arc.

Let ${\it\Gamma}=(V,E)$ be a graph. A permutation $g$ on $V$ is called an automorphism of ${\it\Gamma}$ if $\{u^{g},v^{g}\}\in E$ for all $\{u,v\}\in E$. All automorphisms of ${\it\Gamma}$ form a subgroup of the symmetric group ${\rm Sym}(V)$, denoted by ${\sf Aut}{\it\Gamma}$, which is called the automorphism group of ${\it\Gamma}$. The group ${\sf Aut}{\it\Gamma}$ has a natural action on $E$, namely, $\{u,v\}^{g}=\{u^{g},v^{g}\}$ for $\{u,v\}\in E$ and $g\in{\sf Aut}{\it\Gamma}$. If this action is transitive, that is, for each pair of edges there exists some $g\in{\sf Aut}{\it\Gamma}$ mapping one edge to the other one, then ${\it\Gamma}$ is called edge-transitive. Similarly, we may define the vertex-transitivity, arc-transitivity and $s$-arc-transitivity** of ${\it\Gamma}$. The graph ${\it\Gamma}$ is called edge-primitive if ${\sf Aut}{\it\Gamma}$ acts primitively on $E$, that is, ${\it\Gamma}$ is edge-transitive and the stabilizer $({\sf Aut}{\it\Gamma})_{\{u,v\}}$ of some (and hence every) edge $\{u,v\}$ in ${\sf Aut}{\it\Gamma}$ is a maximal subgroup.**

Let ${\it\Gamma}=(V,E)$ be an edge-primitive graph of valency no less than $3$. Then, as observed in ****[9]****, ${\it\Gamma}$ is also arc-transitive. If ${\it\Gamma}$ is $2$-arc-transitive then Praeger’s reduction theorems ****[24, 25]**** will be effective tools for us to investigate the group-theoretic and graph-theoretic properties of ${\it\Gamma}$. However, ${\it\Gamma}$ is not necessarily $2$-arc-transitive; for example, by the Atlas ****[3]****, the sporadic Rudvalis group ${\rm Ru}$ is the automorphism group of a rank $3$ graph, which is edge-primitive and of valency $2304$ but not $2$-arc-transitive. Using O’Nan-Scott Theorem for (quasi)primitive groups ****[24]****, Giudici and Li ****[9]**** gave a reduction theorem on the automorphism group of ${\it\Gamma}$. They proved that, as a primitive group on $E$, only four of the eight O’Nan-Scott types for primitive groups may occur for ${\sf Aut}{\it\Gamma}$, say SD, CD, PA and AS. They also considered the possible O’Nan-Scott types for ${\sf Aut}{\it\Gamma}$ acting on $V$, and presented constructions or examples to verify the existence of corresponding graphs. Then what will happen if we assume further that ${\it\Gamma}$ is $2$-arc-transitive? The third author of this paper showed that either ${\sf Aut}{\it\Gamma}$ is almost simple or ${\it\Gamma}$ is a complete bipartite graph if ${\it\Gamma}$ is $2$-arc-transitive, see ****[20]****. This stimulate our interest in classifying those edge-primitive graphs which are $2$-arc-transitive.

In this paper, we present a classification result stated as follows.

Theorem 1.1**.**

Let ${\it\Gamma}=(V,E)$ be a graph of valency $d\geq 6$, and let $G\leq{\sf Aut}{\it\Gamma}$ such that $G$ acts primitively on the edge set and transitively on the $2$-arc set of ${\it\Gamma}$. Assume further that $G$ is almost simple and, for $\{u,v\}\in E$, the edge-stabilizer $G_{\{u,v\}}$ is soluble. Then either ${\it\Gamma}$ is $(G,4)$-arc-transitive, or $G$, $G_{\{u,v\}}$, $G_{v}$ and $d$ are listed as in Table 1.

Remark. If ${\it\Gamma}$ is edge-primitive and either $4$-arc-transitive or of valency less than $6$, then the edge-stabilizers must be soluble. The reader may find a complete list of such graphs in ****[10, 11, 17, 29**]****. For each triple $(G,G_{v},G_{\{u,v\}})$ listed in Table 1, the coset graph ${\sf Cos}(G,G_{v},G_{\{u,v\}})$, see Section 2 for the definition, is both $(G,2)$-arc-transitive and $G$-edge-primitive. **$\square$

2. Preliminaries

Let $G$ be a finite group and $H,K\leq G$ with $|K:(H\cap K)|=2$ and $\cap_{g\in G}H^{g}=1$. Let $[G:H]=\{Hx\mid x\in G\}$, and define a graph ${\sf Cos}(G,H,K)$ on $[G:H]$ such that $\{Hx,Hy\}$ is an edge if and only if $yx^{-1}\in HKH\setminus H$. The group $G$ can be viewed as a subgroup of ${\sf Aut}{\sf Cos}(G,H,K)$, where $G$ acts on $[G:H]$ by right multiplication. Then ${\sf Cos}(G,H,K)$ is $G$-arc-transitive and, for $x\in K\setminus H$, the edge $\{H,Hx\}$ has stabilizer $K$ in $G$. Thus ${\sf Cos}(G,H,K)$ is $G$-edge-primitive if and only if $K$ is maximal in $G$.

Assume that ${\it\Gamma}=(V,E)$ is a $G$-edge-primitive graph of valency $d\geq 3$. Then ${\it\Gamma}$ is $G$-arc-transitive by ****[9, Lemma 3.4]****. Take an edge $\{u,v\}\in E$, let $H=G_{v}$ and $K=G_{\{u,v\}}$. Then $K$ is maximal in $G$, and $H\cap K=G_{uv}$, which has index $2$ in $K$. Noting that $\cap_{g\in G}H^{g}$ fixes $V$ pointwise, $\cap_{g\in G}H^{g}=1$. Further, $\,v^{g}\mapsto G_{v}g,\,\forall g\in G$ gives an isomorphism from ${\it\Gamma}$ to ${\sf Cos}(G,H,K)$. Then, by ****[5, Theorem 2.1]****, the following lemma holds.

Lemma 2.1**.**

Let ${\it\Gamma}=(V,E)$ be a connected graph of valency $d\geq 3$, and $G\leq{\sf Aut}{\it\Gamma}$. Then ${\it\Gamma}$ is both $(G,2)$-arc-transitive and $G$-edge-primitive if and only if ${\it\Gamma}\cong{\sf Cos}(G,H,K)$ for some subgroups $H$ and $K$ of $G$ satisfying

• (1)

$|K:(H\cap K)|=2$, $\cap_{g\in G}H^{g}=1$ and $K$ is maximal in $G$;

• (2)

$H$* acts $2$-transitively on $[H:(H\cap K)]$ by right multiplication.*

Let ${\it\Gamma}=(V,E)$ be a connected graph of valency no less than $3$, $\{u,v\}\in E$ and $G\leq{\sf Aut}{\it\Gamma}$. Assume that ${\it\Gamma}$ is $(G,s)$-arc-transitive for some $s\geq 1$, that is, $G$ acts transitively on the $s$-arc set of ${\it\Gamma}$. Then $G_{v}$ acts transitively on the neighborhood ${\it\Gamma}(v)$ of $v$ in ${\it\Gamma}$. Let ${\rm G}_{v}^{{\it\Gamma}(v)}$ be the transitive permutation group induced by $G_{v}$ on ${\it\Gamma}(v)$, and let $G_{v}^{[1]}$ be the kernel of $G_{v}$ acting on ${\it\Gamma}(v)$. Then $G_{v}^{{\it\Gamma}(v)}\cong G_{v}/G_{v}^{[1]}$. Set $G_{uv}^{[1]}=G_{u}^{[1]}\cap G_{v}^{[1]}$. Then $G_{v}^{[1]}$ induces a normal subgroup of $(G_{u}^{{\it\Gamma}(u)})_{v}$ with the kernel $G_{uv}^{[1]}$. Since $G$ is transitive on the arcs of ${\it\Gamma}$, there is some element in $G$ interchanging $u$ and $v$. This implies that

[TABLE]

Writing $G_{v}^{[1]}$ and $G_{v}$ in group extensions, the next lemma follows.

Lemma 2.2**.**

1. (1)

$G_{v}^{[1]}=G_{uv}^{[1]}.(G_{v}^{[1]})^{{\it\Gamma}(u)}$, $(G_{v}^{[1]})^{{\it\Gamma}(u)}\unlhd(G_{u}^{{\it\Gamma}(u)})_{v}\cong(G_{v}^{{\it\Gamma}(v)})_{u}$. 2. (2)

$G_{\{u,v\}}=G_{uv}.2$, $G_{uv}=(G_{uv}^{[1]}.(G_{v}^{[1]})^{{\it\Gamma}(u)}).(G_{v}^{{\it\Gamma}(v)})_{u}$, $G_{v}=(G_{uv}^{[1]}.(G_{v}^{[1]})^{{\it\Gamma}(u)}).G_{v}^{{\it\Gamma}(v)}$. 3. (3)

If $G_{uv}^{[1]}=1$ then $G_{uv}\lesssim(G_{v}^{{\it\Gamma}(v)})_{u}\times(G_{u}^{{\it\Gamma}(u)})_{v}$.

By ****[32]****, $s\leq 7$ , and if $s\geq 2$ then $G_{uv}^{[1]}$ is a $p$-group for some prime $p$, refer to ****[7]****. Thus Lemma 2.2 yields a fact as follows.

Corollary 2.3**.**

Let ${\it\Gamma}=(V,E)$ be a connected $(G,2)$-arc-transitive graph, and $\{u,v\}\in E$. Then $G_{\{u,v\}}$ is soluble if and only if $(G_{v}^{{\it\Gamma}(v)})_{u}$ is soluble, and $G_{v}$ is soluble if and only if $G_{v}^{{\it\Gamma}(v)}$ is soluble.

Choose $s$ maximal as possible, that is, ${\it\Gamma}$ is $(G,s)$-arc-transitive but not $(G,s{+}1)$-arc-transitive. In this case, ${\it\Gamma}$ is said to be $(G,s)$-transitive. If further $G_{uv}^{[1]}\neq 1$, then one can read out the vertex-stabilizer $G_{v}$ from ****[8, 31]**** for $s\geq 4$ and from ****[28]**** for $2\leq s\leq 3$. In particular, we have the following result.

Theorem 2.4**.**

Let ${\it\Gamma}=(V,E)$ be a connected $(G,s)$-transitive graph of valency on less $3$, and $\{u,v\}\in E$. Assume that $s\geq 2$.

1. (1)

If $G_{uv}^{[1]}=1$ then $s=2$ or $3$. 2. (2)

If $G_{uv}^{[1]}\neq 1$ then $G_{uv}^{[1]}$ is a $p$-group for some prime $p$, ${\rm PSL}_{n}(q)\unlhd G_{v}^{{\it\Gamma}(v)}$, $|{\it\Gamma}(v)|={q^{n}-1\over q-1}$ and $6\neq s\leq 7$, where $n\geq 2$ and $q=p^{f}$ for some integer $f\geq 1$; moreover, either

1. (2.1)

$n=2$* and $s\geq 4$; or* 2. (2.2)

$n\geq 3$, $s\leq 3$ and ${\bf O}_{p}(G_{v})$ is given as in Table 2, where ${\bf O}_{p}(G_{v})$ is the maximal normal $p$-subgroup of $G_{v}$.

Lemma 2.5**.**

Let ${\it\Gamma}=(V,E)$ be a connected $(G,2)$-arc-transitive graph, and $\{u,v\}\in E$. If $r$ is a prime divisor of $|{\it\Gamma}(v)|$ then ${\bf O}_{r}(G_{v}^{[1]})=1={\bf O}_{r}(G_{uv})$, and either ${\bf O}_{r}(G_{v})=1$, or $|{\it\Gamma}(v)|=r^{e}$ and ${\bf O}_{r}(G_{v})\cong\mathbb{Z}_{r}^{e}\cong{\rm soc}(G_{v}^{{\it\Gamma}(v)})$ for some integer $e\geq 1$.

*Proof.  *Since ${\it\Gamma}$ is $(G,2)$-arc-transitive, $G_{v}^{{\it\Gamma}(v)}$ is a $2$-transitive group, and thus $G_{uv}$ is transitive on ${\it\Gamma}(v)\setminus\{u\}$. Since ${\bf O}_{r}(G_{uv})\unlhd G_{uv}$, all ${\bf O}_{r}(G_{uv})$-orbits on ${\it\Gamma}(v)\setminus\{u\}$ have the same size. Noting that $r$ is coprime to $|{\it\Gamma}(v)\setminus\{u\}|$, it follows that ${\bf O}_{r}(G_{uv})\leq G_{v}^{[1]}$. Since $G_{v}^{[1]}\unlhd G_{uv}$, we have ${\bf O}_{r}(G_{v}^{[1]})\leq{\bf O}_{r}(G_{uv})$, and so ${\bf O}_{r}(G_{v}^{[1]})={\bf O}_{r}(G_{uv})$. Similarly, considering the action of $G_{uv}$ on ${\it\Gamma}(u)\setminus\{v\}$, we get ${\bf O}_{r}(G_{u}^{[1]})={\bf O}_{r}(G_{uv})$. Then ${\bf O}_{r}(G_{u}^{[1]})={\bf O}_{r}(G_{uv})={\bf O}_{r}(G_{v}^{[1]})\leq G_{uv}^{[1]}$. By Theorem 2.4, either $G_{uv}^{[1]}=1$ or $G_{uv}^{[1]}$ is a $p$-group, where $p$ is a prime divisor of $|{\it\Gamma}(v)|-1$. It follows that ${\bf O}_{r}(G_{u}^{[1]})={\bf O}_{r}(G_{uv})={\bf O}_{r}(G_{v}^{[1]})=1$.

**Note that ${\bf O}_{r}(G_{v})G_{v}^{[1]}/G_{v}^{[1]}\cong{\bf O}_{r}(G_{v})/({\bf O}_{r}(G_{v})\cap G_{v}^{[1]})$. Since ${\bf O}_{r}(G_{v})\cap G_{v}^{[1]}\leq{\bf O}_{r}(G_{v}^{[1]})=1$, we have ${\bf O}_{r}(G_{v})\cong{\bf O}_{r}(G_{v})G_{v}^{[1]}/G_{v}^{[1]}\unlhd G_{v}/G_{v}^{[1]}\cong{\rm G}_{v}^{{\it\Gamma}(v)}$. Thus ${\bf O}_{r}(G_{v})$ is isomorphic to a normal $r$-subgroup of ${\rm G}_{v}^{{\it\Gamma}(v)}$. This implies that either ${\bf O}_{r}(G_{v})=1$, or ${\rm G}_{v}^{{\it\Gamma}(v)}$ is an affine $2$-transitive group of degree $r^{e}$ for some $e$. Thus the lemma follows. **$\square$

Let $a\geq 2$ and $f\geq 1$ be integers. A prime divisor $r$ of $a^{f}-1$ is primitive if $r$ is not a divisor of $a^{e}-1$ for all $1\leq e<f$. By Zsigmondy’s theorem ****[35]****, if $f>1$ and $a^{f}-1$ has no primitive prime divisor then $a^{f}=2^{6}$, or $f=2$ and $a=2^{t}-1$ for some prime $t$. Assume that $a^{f}-1$ has a primitive prime divisor $r$. Then $a$ has order $f$ modulo $r$. Thus $f$ is a divisor of $r-1$, and if $r$ is a divisor of $a^{f^{\prime}}-1$ for some $f^{\prime}\geq 1$ then $f$ is a divisor of $f^{\prime}$. Thus we have the following lemma.

Lemma 2.6**.**

Let $a\geq 2$, $f\geq 1$ and $f^{\prime}\geq 1$ be integers. If $a^{f}-1$ has a primitive prime divisor $r$ then $f$ is a divisor of $r-1$, and $r$ is a divisor of $a^{f^{\prime}}-1$ if and only if $f$ is a divisor of $f^{\prime}$. If $f\geq 3$ then $a^{f}-1$ has a prime divisor no less than $5$.

We end this section with a fact on finite primitive groups.

Lemma 2.7**.**

Assume that $G$ is a finite primitive group with a point-stabilizer $H$. If $H$ has a normal Sylow subgroup $P\neq 1$, then $P$ is also a Sylow subgroup of $G$.

*Proof.  ***Assume that $P\neq 1$ is a normal Sylow subgroup of $H$. Clearly, $P$ is not normal in $G$. Take a Sylow subgroup $Q$ of $G$ with $P\leq Q$. Then $H\leq\langle{\bf N}_{Q}(P),H\rangle\leq{\bf N}_{G}(P)\neq G$. Since $H$ is maximal in $G$, we have $H=\langle{\bf N}_{Q}(P),H\rangle$ and so ${\bf N}_{Q}(P)\leq H$. It follows that ${\bf N}_{Q}(P)=P$, and hence $P=Q$. Then the lemma follows. **$\square$

3. Some restrictions on stabilizers

In Sections 4 and 5, we shall prove Theorem 1.1 using the result given in ****[17]**** which classifies finite primitive groups with soluble point-stabilizers. Let ${\it\Gamma}=(V,E)$ be a graph of valency $d\geq 6$, $\{u,v\}\in E$ and $G\leq{\sf Aut}{\it\Gamma}$. Assume that $G$ is almost simple, $G_{\{u,v\}}$ is soluble, ${\it\Gamma}$ is $G$-edge-primitive and $(G,2)$-arc-transitive. Clearly, each nontrivial normal subgroup of $G$ acts transitively on the edge set $E$. Choose a minimal $X$ among the normal subgroups of $G$ which act primitively on $E$. By the choice of $X$, we have ${\rm soc}(X)={\rm soc}(G)$, $X_{\{u,v\}}=X\cap G_{\{u,v\}}$, $G=XG_{\{u,v\}}$ and $G/X=XG_{\{u,v\}}/X\cong G_{\{u,v\}}/X_{\{u,v\}}$. Then, considering the restrictions on both $X_{\{u,v\}}$ and $X_{v}$ caused by the $2$-arc-transitivity of ${\it\Gamma}$, we may work out the pair $(X,X_{\{u,v\}})$ from ****[17, Theorem 1.1]****, and then determine the group $G$ and the graph ${\it\Gamma}$. Thus we make the following assumptions.

Hypothesis 3.1**.**

Let ${\it\Gamma}=(V,E)$ be a $G$-edge-primitive graph of valency $d\geq 6$, and $\{u,v\}\in E$, where $G$ is an almost simple group with socle $T$. Assume that

• (i)

${\it\Gamma}$* is $(G,2)$-arc-transitive, and the edge-stabilizer $G_{\{u,v\}}$ is soluble;*

• (ii)

$G$* has a normal subgroup $X$ such that ${\rm soc}(X)=T$, $X_{\{u,v\}}$ is maximal in $X$, and $(X,X_{\{u,v\}})$ is one of the pairs $(G_{0},H_{0})$ listed in [17, Tables 14-20].*

For the group $X$ in Hypothesis 3.1, we have $1\neq X_{v}^{{\it\Gamma}(v)}\unlhd G_{v}^{{\it\Gamma}(v)}$. Note that $G_{v}^{{\it\Gamma}(v)}$ is $2$-transitive (on ${\it\Gamma}(v)$). Then $G_{v}^{{\it\Gamma}(v)}$ is affine or almost simple, see ****[4, Theorem 4.1B]**** for example. It follows that ${\rm soc}(G_{v}^{{\it\Gamma}(v)})={\rm soc}(X_{v}^{{\it\Gamma}(v)})$.

3.1.

Assume that $G_{v}$ is insoluble. Then $G_{v}^{{\it\Gamma}(v)}$ is an almost simple $2$-transitive group (on ${\it\Gamma}(v)$). Recall that ${\rm soc}(G_{v}^{{\it\Gamma}(v)})={\rm soc}(X_{v}^{{\it\Gamma}(v)})$. Checking the point-stabilizers of almost simple $2$-transitive groups, since $(G_{v}^{{\it\Gamma}(v)})_{u}$ is soluble, we conclude that either $X_{v}^{{\it\Gamma}(v)}$ is $2$-transitive, or $G_{v}^{{\it\Gamma}(v)}\cong{\rm PSL}_{2}(8){.}3$ and $d=28$. For a complete list of finite $2$-transitive groups, the reader may refer to ****[2, Tables 7.3 and 7.4]****.

Lemma 3.2**.**

Suppose that Hypothesis 3.1 holds. If $d=28$ then $G_{v}^{{\it\Gamma}(v)}\not\cong{\rm PSL}_{2}(8){.}3$.

*Proof.  *Suppose that $G_{v}^{{\it\Gamma}(v)}\cong{\rm PSL}_{2}(8){.}3$ and $d=28$. Note that $X_{uv}^{[1]}\leq G_{uv}^{[1]}=1$, see Theorem 2.4. Thus $X_{uv}\lesssim(X_{v}^{{\it\Gamma}(v)})_{u}{\times}(X_{u}^{{\it\Gamma}(u)})_{v}$ by Lemma 2.2.

Assume that $X_{v}^{{\it\Gamma}(v)}\cong{\rm PSL}_{2}(8)$. Then $(X_{v}^{{\it\Gamma}(v)})_{u}\cong{\rm D}_{18}$, and $X_{uv}\cong{\rm D}_{18}$, $(\mathbb{Z}_{3}{\times}\mathbb{Z}_{9}){:}\mathbb{Z}_{2}$, $(\mathbb{Z}_{9}{\times}\mathbb{Z}_{9}){:}\mathbb{Z}_{2}$ or ${\rm D}_{18}{\times}{\rm D}_{18}$. In particular, the unique Sylow $3$-subgroup of $X_{\{u,v\}}=X_{uv}.2$ is isomorphic to $\mathbb{Z}_{m}{\times}\mathbb{Z}_{9}$, where $m=1$, $3$ or $9$. Checking the primitive groups listed in ****[17, Tables 14-20]****, we know that only the pairs $({\rm PSL}_{2}(q),{\rm D}_{2{q{\pm}1\over(2,q-1)}})$ possibly meet our requirements on $X_{\{u,v\}}$, yielding $X_{\{u,v\}}\cong{\rm D}_{2{q{\pm}1\over(2,q-1)}}$. Then ${\rm D}_{36}\cong X_{\{u,v\}}\cong{\rm D}_{2{q{\pm}1\over(2,q-1)}}$. Calculation shows that $q=37$; however, ${\rm PSL}_{2}(37)$ has no subgroup which has a quotient ${\rm PSL}_{2}(8)$, a contradiction.

Now let $X_{v}^{{\it\Gamma}(v)}=G_{v}^{{\it\Gamma}(v)}\cong{\rm PSL}_{2}(8){.}3$. Then $(X_{v}^{{\it\Gamma}(v)})_{u}\cong(X_{u}^{{\it\Gamma}(u)})_{v}\cong\mathbb{Z}_{9}{:}\mathbb{Z}_{6}$ and $X_{uv}\lesssim\mathbb{Z}_{9}{:}\mathbb{Z}_{6}{\times}\mathbb{Z}_{9}{:}\mathbb{Z}_{6}$. In particular, a Sylow $2$-subgroup of $X_{\{u,v\}}=X_{uv}.2$ is not a cyclic group of order $8$, and the unique Sylow $3$-subgroup of $X_{\{u,v\}}$ is nonabelian and contains elements of order $9$. Since $X_{\{u,v\}}=X_{uv}.2=X_{v}^{[1]}.(X_{v}^{{\it\Gamma}(v)})_{u}.2$ and $X_{v}^{[1]}\cong(X_{v}^{[1]})^{{\it\Gamma}(u)}\unlhd(X_{u}^{{\it\Gamma}(u)})_{v}$, we have $|X_{\{u,v\}}|=2^{2}3^{3}$, $2^{2}3^{4}$, $2^{2}3^{5}$, $2^{2}3^{6}$, $2^{3}3^{5}$ or $2^{3}3^{6}$. Checking the Tables 14-20 given in ****[17]****, we conclude that $X={\rm G}_{2}(3).2$, and $X_{\{u,v\}}\cong[3^{6}]{:}{\rm D}_{8}$. In this case, $X_{v}^{[1]}\cong\mathbb{Z}_{9}{:}\mathbb{Z}_{6}$ and $X_{v}\cong\mathbb{Z}_{9}{:}\mathbb{Z}_{6}.{\rm PSL}_{2}(8){.}3$; however, $X$ has no such subgroup by the Atlas ****[3**]****, a contradiction. This completes the proof. **$\square$

By Lemma 3.2, combining with Theorem 2.4, the next lemma follows from checking the point-stabilizers of finite almost simple $2$-transitive groups.

Lemma 3.3**.**

Suppose that Hypothesis 3.1 holds and $G_{v}^{{\it\Gamma}(v)}$ is almost simple. Then one of the following holds:

1. (s1)

$G_{v}^{{\it\Gamma}(v)}=X_{v}^{{\it\Gamma}(v)}={\rm PSL}_{3}(2)$* or ${\rm PSL}_{3}(3)$, and $d=7$ or $13$, respectively;* 2. (s2)

${\rm soc}(X_{v}^{{\it\Gamma}(v)})={\rm PSL}_{2}(q)$* with $q>4$, and $d=q{+}1$;* 3. (s3)

$G_{uv}^{[1]}=1$, ${\rm soc}(X_{v}^{{\it\Gamma}(v)})={\rm PSU}_{3}(q)$ with $q>2$, and $d=q^{3}{+}1$; 4. (s4)

$G_{uv}^{[1]}=1$, ${\rm soc}(X_{v}^{{\it\Gamma}(v)})={\rm Sz}(q)$ with $q=2^{2n{+}1}>2$, and $d=q^{2}{+}1$; 5. (s5)

$G_{uv}^{[1]}=1$, ${\rm soc}(X_{v}^{{\it\Gamma}(v)})={\rm Ree}(q)$ with $q=3^{2n{+}1}>3$, and $d=q^{3}{+}1$.

In particular, ${\it\Gamma}$ is $(X,2)$-arc-transitive.

Recall that the Fitting subgroup ${\rm Fit}(H)$ of a finite group $H$ is the direct product of ${\bf O}_{r}(H)$, where $r$ runs over the set of prime divisors of $|H|$.

Lemma 3.4**.**

Suppose that Hypothesis 3.1 holds and one of Lemma 3.3 (s2)-(s5)* occurs. Let $q=p^{f}$ for some prime $p$. Assume that $X_{u,v}^{[1]}=1$. Then ${\rm Fit}(X_{uv})={\bf O}_{p}(X_{uv})$, and either ${\rm Fit}(X_{uv})={\rm Fit}(X_{\{u,v\}})$ or ${\rm Fit}(X_{\{u,v\}})={\rm Fit}(X_{uv}).2$; in particular, $|{\rm Fit}(X_{\{u,v\}}):{\bf O}_{p}(X_{\{u,v\}})|\leq 2$.*

*Proof.  *Let $r$ be a prime divisor of $|X_{uv}|$. Then ${\bf O}_{r}(X_{uv})$ is normal in $X_{uv}$. Since ${\it\Gamma}$ is $(X,2)$-arc-transitive, $X_{uv}$ acts transitively on ${\it\Gamma}(v)\setminus\{u\}$. Thus all ${\bf O}_{r}(X_{uv})$-orbits (on ${\it\Gamma}(v)\setminus\{u\}$) have equal size, which is a power of $r$ and a divisor of $|{\it\Gamma}(v)\setminus\{u\}|$. Note that $|{\it\Gamma}(v)\setminus\{u\}|=d-1$, which is a power of $p$. It follows that either $r=p$ or ${\bf O}_{r}(X_{uv})=1$. Then ${\rm Fit}(X_{uv})={\bf O}_{p}(X_{uv})$.

Note that $X_{uv}$ is normal in $X_{\{u,v\}}$ as $|X_{\{u,v\}}:X_{uv}|=2$. Since ${\bf O}_{p}(X_{uv})$ is a characteristic subgroup of $X_{uv}$, it follows that ${\bf O}_{p}(X_{uv})$ is normal in $X_{\{u,v\}}$, and so ${\bf O}_{p}(X_{uv})\leq{\bf O}_{p}(X_{\{u,v\}})\leq{\rm Fit}(X_{\{u,v\}})$. For each odd prime divisor $r$ of $|X_{\{u,v\}}|$, since $|X_{\{u,v\}}:X_{uv}|=2$, we have ${\bf O}_{r}(X_{\{u,v\}})\leq X_{uv}$, and so ${\bf O}_{r}(X_{\{u,v\}})={\bf O}_{r}(X_{uv})$. It follows that ${\rm Fit}(X_{\{u,v\}})={\rm Fit}(X_{uv}){\bf O}_{2}(X_{\{u,v\}})={\bf O}_{p}(X_{uv}){\bf O}_{2}(X_{\{u,v\}})$. In particular, ${\bf O}_{p}(X_{uv})={\bf O}_{p}(X_{\{u,v\}})$ if $p\neq 2$.

**It is easily shown that $X_{uv}\cap{\bf O}_{2}(X_{\{u,v\}})={\bf O}_{2}(X_{uv})$. If $X_{uv}\geq{\bf O}_{2}(X_{\{u,v\}})$ then $p=2$, ${\rm Fit}(X_{\{u,v\}})={\bf O}_{2}(X_{\{u,v\}})={\rm Fit}(X_{uv})$, and the lemma holds. Assume that ${\bf O}_{2}(X_{\{u,v\}})\not\leq X_{uv}$. Since $|X_{\{u,v\}}:X_{uv}|=2$, we have $X_{\{u,v\}}=X_{uv}{\bf O}_{2}(X_{\{u,v\}})$. Then $2|X_{uv}|=|X_{\{u,v\}}|=|X_{uv}||{\bf O}_{2}(X_{\{u,v\}}):(X_{uv}\cap{\bf O}_{2}(X_{\{u,v\}}))|=|X_{uv}||{\bf O}_{2}(X_{\{u,v\}}):{\bf O}_{2}(X_{uv})|$, yielding $|{\bf O}_{2}(X_{\{u,v\}}):{\bf O}_{2}(X_{uv})|=2$. If $p=2$ then ${\rm Fit}(X_{\{u,v\}})={\bf O}_{2}(X_{\{u,v\}})$ and ${\rm Fit}(X_{uv})={\bf O}_{2}(X_{uv})$, the lemma follows. If $p\neq 2$ then ${\bf O}_{2}(X_{uv})=1$, $|{\bf O}_{2}(X_{\{u,v\}})|=2$, and so ${\rm Fit}(X_{\{u,v\}})={\bf O}_{p}(X_{uv})\times\mathbb{Z}_{2}$. This completes the proof. **$\square$

3.2.

Assume that Hypothesis 3.1 holds and $G_{v}$ is soluble. Then $G_{v}^{{\it\Gamma}(v)}$ is an affine $2$-transitive group. Let ${\rm soc}(G_{v}^{{\it\Gamma}(v)})=\mathbb{Z}_{p}^{f}$. Then $d=p^{f}$. Recalling that $d\geq 6$, we have $G_{uv}^{[1]}=1$ by Theorem 2.4, and so $G_{uv}\lesssim(G_{v}^{{\it\Gamma}(v)})_{u}{\times}(G_{u}^{{\it\Gamma}(u)})_{v}$. If $G_{uv}$ is abelian then ${\it\Gamma}$ is known by ****[20]****. Thus we assume further that $G_{uv}$ is not abelian. Then $(G_{v}^{{\it\Gamma}(v)})_{u}$ is nonabelian, and so $(G_{v}^{{\it\Gamma}(v)})_{u}\not\leq{\rm GL}_{1}(p^{f})$; in particular, $f>1$. Since $(G_{v}^{{\it\Gamma}(v)})_{u}$ is soluble, by ****[2, Table 7.3]****, we have the following lemma.

Lemma 3.5**.**

Suppose that Hypothesis 3.1 holds, $G_{v}$ is soluble and $G_{uv}$ is not abelian. Let ${\rm soc}(G_{v}^{{\it\Gamma}(v)})=\mathbb{Z}_{p}^{f}$, where $p$ is a prime. Then $f>1$, and one of the following holds.

1. (a1)

$f=2$, and either ${\rm SL}_{2}(3)\unlhd(G_{v}^{{\it\Gamma}(v)})_{u}\leq{\rm GL}_{2}(p)$ and $p\in\{3,5,7,11,23\}$, or $p=3$ and $(G_{v}^{{\it\Gamma}(v)})_{u}={\rm Q}_{8}$; 2. (a2)

$2_{+}^{1{+}4}{:}\mathbb{Z}_{5}\leq(G_{v}^{{\it\Gamma}(v)})_{u}\leq 2_{+}^{1{+}4}.(\mathbb{Z}_{5}{:}\mathbb{Z}_{4})<2_{+}^{1{+}4}.{\rm S}_{5}$, and $p^{f}=3^{4}$; 3. (a3)

$(G_{v}^{{\it\Gamma}(v)})_{u}\not\leq{\rm GL}_{1}(p^{f})$, $(G_{v}^{{\it\Gamma}(v)})_{u}\leq{\rm\Gamma L}_{1}(p^{f})$ and $|(G_{v}^{{\it\Gamma}(v)})_{u}|$ is divisible by $p^{f}-1$.

Consider the case (a3) in Lemma 3.5. Write

[TABLE]

Let $\langle\tau\rangle\cap(G_{v}^{{\it\Gamma}(v)})_{u}=\langle\tau^{m}\rangle$, where m\,\big{|}\,(p^{f}-1). Then

[TABLE]

Set $(G_{v}^{{\it\Gamma}(v)})_{u}/\langle\tau^{m}\rangle\cong\langle\sigma^{e}\rangle$ for some divisor $e$ of $f$. Then

[TABLE]

Choose $\tau^{l}\sigma^{k}\in(G_{v}^{{\it\Gamma}(v)})_{u}$ such that $(G_{v}^{{\it\Gamma}(v)})_{u}=\langle\tau^{m}\rangle\langle\tau^{l}\sigma^{k}\rangle$. Then $(\tau^{l}\sigma^{k})^{f\over e}\in\langle\tau^{m}\rangle$ but $(\tau^{l}\sigma^{k})^{j}\not\in\langle\tau^{m}\rangle$ for $1\leq j<{f\over e}$. It follows that $\sigma^{k}$ has order $f\over e$. Then $\sigma^{k}=\sigma^{ie}$ for some $i$ with $(i,{f\over e})=1$, and then $(\sigma^{k})^{i^{\prime}}=\sigma^{e}$ for some $i^{\prime}$. Thus, replacing $\tau^{l}\sigma^{k}$ by a power of it if necessary, we may let $k=e$. Then

[TABLE]

Further, $(G_{v}^{{\it\Gamma}(v)})_{u}=\langle\tau^{m}\rangle\langle(\tau^{m})^{i}\tau^{l}\sigma^{e}\rangle$ for an arbitrary integer $i$, thus we may assume further $0\leq l<m$. By ****[6, Proposition 15.3]****, letting ${\rm\pi}(n)$ be the set of prime divisors of a positive integer $n$, we have

1. ($\divideontimes$)

${\rm\pi}(m)\subseteq{\rm\pi}(p^{e}-1)$, me\,\big{|}\,f and $(m,l)=1$; in particular, $m=1$ if $l=0$.

Suppose that $X_{uv}$ is nonabelian. (The case where $X_{uv}$ is abelian is left in Section 5.) Since $X_{uv}^{[1]}\leq{\rm G}_{uv}^{[1]}=1$, we have

[TABLE]

This yields that $(X_{v}^{{\it\Gamma}(v)})_{u}$ is nonabelian. Then a limitation on $\pi(|X_{uv}|)$ is given as follows.

Lemma 3.6**.**

Assume that Lemma 3.5 (a3)* holds and $X_{uv}$ is nonabelian. Then $(X_{v}^{{\it\Gamma}(v)})_{u}\cong\mathbb{Z}_{m^{\prime}}.\mathbb{Z}_{f\over e^{\prime}}$, where $m^{\prime}$ and $e^{\prime}$ satisfy*

1. (i)

$\mathbb{Z}_{m^{\prime}}\cong(X_{v}^{{\it\Gamma}(v)})_{u}\cap\langle\tau^{m}\rangle$, mm^{\prime}\,\big{|}\,p^{f}-1, e\,\big{|}\,e^{\prime}\,\big{|}\,f; and 2. (ii)

$m^{\prime}>1$, $e^{\prime}<f$, ${\rm\pi}(p^{f}-1)\setminus{\rm\pi}(p^{e^{\prime}}-1)\subseteq{\rm\pi}(m^{\prime})\subseteq\pi(|X_{uv}|)$.

*Proof.  *Recall that $(X_{v}^{{\it\Gamma}(v)})_{u}\unlhd(G_{v}^{{\it\Gamma}(v)})_{u}=\langle\tau^{m}\rangle\langle\tau^{l}\sigma^{e}\rangle\cong\mathbb{Z}_{p^{f}-1\over m}.\mathbb{Z}_{f\over e}$. Then

[TABLE]

yielding $(X_{v}^{{\it\Gamma}(v)})_{u}\cong\mathbb{Z}_{m^{\prime}}.\mathbb{Z}_{f\over e^{\prime}}$ with $m^{\prime}$ and $e^{\prime}$ satisfying (i). Since $X_{uv}$ is nonabelian, $(X_{v}^{{\it\Gamma}(v)})_{u}$ is nonabelian, and so $m^{\prime}>1$ and $e^{\prime}<f$.

**By the above $(\divideontimes)$, each $r\in{\rm\pi}(p^{f}-1)\setminus{\rm\pi}(p^{e^{\prime}}-1)$ is a divisor of $|\langle\tau^{m}\rangle|={p^{f}-1\over m}$. Let $R$ be the unique subgroup of order $r$ of $\langle\tau^{m}\rangle$. Then, noting that $R$ is normal in $(G_{v}^{{\it\Gamma}(v)})_{u}$, either $R\leq(X_{v}^{{\it\Gamma}(v)})_{u}$ or $R(X_{v}^{{\it\Gamma}(v)})_{u}=R{\times}(X_{v}^{{\it\Gamma}(v)})_{u}$. Suppose that the latter case occurs. Since $e^{\prime}<f$, we may let $\tau^{n}\sigma^{e^{\prime}}\in(X_{v}^{{\it\Gamma}(v)})_{u}\setminus\langle\tau^{m}\rangle$. Then $\sigma^{e^{\prime}}$ centralizes $R$. Thus $x^{p^{e^{\prime}}}=x$ for $x\in R$, yielding r\,\big{|}\,(p^{e^{\prime}}-1), a contradiction. Then $R\leq(X_{v}^{{\it\Gamma}(v)})_{u}\cap\langle\tau^{m}\rangle\cong\mathbb{Z}_{m^{\prime}}$. Noting that $m^{\prime}$ is a divisor of $|X_{uv}|$, the result follows. **$\square$

4. Graphs with insoluble vertex-stabilizers

In this and next sections, we prove Theorem 1.1. Thus, we let $G$, $T$, $X$ and ${\it\Gamma}=(V,E)$ be as in Hypothesis 3.1. Our task is to determine which pair $(G_{0},H_{0})$ listed in ****[17, Tables 14-20]**** is a possible candidate for $(X,X_{\{u,v\}})$, and determine whether or not the resulting triple $(G,G_{v},G_{\{u,v\}})$ meets the conditions (1) and (2) in Lemma 2.1.

In this section, we deal with the case where $G_{v}$ is insoluble, that is, $X_{v}$ is described as in Lemma 3.3. First, by the following lemma, Lemma 3.3 (s4) and (s5) are excluded.

Lemma 4.1**.**

Lemma 3.3 (s4) and (s5) do not occur.

*Proof.  *Suppose that Lemma 3.3 (s4) or (s5) holds. Then $X_{uv}^{[1]}=1$ by Theorem 2.4. Thus $X_{v}=X_{v}^{[1]}.X_{v}^{{\it\Gamma}(v)}$, $X_{v}^{[1]}\cong(X_{v}^{[1]})^{{\it\Gamma}(u)}\unlhd(X_{u}^{{\it\Gamma}(u)})_{v}\cong(X_{v}^{{\it\Gamma}(v)})_{u}$, and $X_{uv}\lesssim(X_{v}^{{\it\Gamma}(v)})_{u}{\times}(X_{u}^{{\it\Gamma}(u)})_{v}$. Set $q=p^{f}$ with $p$ a prime. Then the pair $(X_{v}^{{\it\Gamma}(v)},(X_{v}^{{\it\Gamma}(v)})_{u})$ is given as follows:

[TABLE]

In particular, ${\bf O}_{p}(X_{\{u,v\}})$ is not abelian.

We next show that none of the pairs $(G_{0},H_{0})$ in ****[17, Tables 14-20]**** gives a desired pair $(X,X_{\{u,v\}})$. Since ${\bf O}_{p}(X_{\{u,v\}})$ is nonabelian, those pairs $(G_{0},H_{0})$ with ${\bf O}_{p}(H_{0})$ abelian are not in our consideration. In particular, ${\rm soc}(X)$ is not isomorphic to an alternating group. Also, noting that $X_{\{u,v\}}$ has a subgroup of index $2$, those $H_{0}$ having no subgroup of index $2$ are excluded.

Case 1. Suppose that ${\rm soc}(X_{v}^{{\it\Gamma}(v)})={\rm Ree}(q)$. Then $p=3$, ${\bf O}_{3}(X_{\{u,v\}})$ is nonabelian and of order $3^{3f}$, $3^{4f}$, $3^{5f}$ or $3^{6f}$, $|X_{\{u,v\}}|$ is a divisor of $2(q-1)^{2}f^{2}$ and divisible by $2(q-1)$. Checking the orders of those $H_{0}$ given in ****[17, Tables 15]****, we conclude that ${\rm soc}(X)$ is not a sporadic simple group.

Suppose that ${\rm soc}(X)$ is a simple exceptional group of Lie type. By ****[17, Table 20]****, we conclude that $(X,X_{\{u,v\}})$ is one of $({\rm G}_{2}(3^{t}).\mathbb{Z}_{2^{l+1}},[3^{6t}]{:}\mathbb{Z}_{3^{t}-1}^{2}.\mathbb{Z}_{2^{l+1}})$ and $({\rm Ree}(3^{t}),[3^{3t}]{:}\mathbb{Z}_{3^{t}-1})$, where $2^{l}$ is the $2$-part of $t$. If $f>t$ then, by Zsigmondy’s theorem, $q-1$ has a prime divisor which is not a divisor of $3^{t}-1$, which contradicts that $|X_{\{u,v\}}|$ is divisible by $2(q-1)$. Thus $f=t$, and then $X={\rm G}_{2}(q).\mathbb{Z}_{2^{l+1}}$ and $X_{\{u,v\}}\cong[q^{6}]{:}\mathbb{Z}_{q-1}^{2}.\mathbb{Z}_{2^{l+1}}$. This implies that $X_{v}^{[1]}\neq 1$, in fact, $|{\bf O}_{3}(X_{v}^{[1]})|=q^{3}$. Thus ${\bf O}_{3}(X_{v})\neq 1$ and $X_{v}$ has a quotient ${\rm Ree}(q).e$. Checking the maximal subgroups of ${\rm G}_{2}(q).\mathbb{Z}_{2^{l+1}}$, refer to ****[15, Theorems A and B]****, we conclude that ${\rm G}_{2}(q).\mathbb{Z}_{2^{l+1}}$ has no maximal subgroup containing such $X_{v}$ as a subgroup, a contradiction.

Suppose that ${\rm soc}(X)$ is a simple classical group over a finite field of order $r^{t}$, where $r$ is a prime. Since $f>1$ is odd, $3^{f}-1$ has an odd prime divisor, and so $X_{\{u,v\}}$ is not a $\{2,3\}$-group as $|X_{\{u,v\}}|$ is divisible by $3^{f}-1$. Recall that ${\bf O}_{3}(X_{\{u,v\}})$ is nonabelian and of order $3^{3f}$, $3^{4f}$, $3^{5f}$ or $3^{6f}$. Checking the groups $H_{0}$ given in ****[17, Table 16-19]****, we conclude that ${\rm soc}(X)={\rm PSL}_{n}(r^{t})$ or ${\rm PSU}_{n}(r^{t})$, where $n\in\{3,4\}$. Take a maximal subgroup $M$ of $X$ such that $X_{v}\leq M$. Then $M$ has a simple section (i.e., a quotient of some subgroup) ${\rm Ree}(q)$. Recall that $q>3$. Checking Tables 8.3-8.6 and 8.8-8.11 given in ****[1]****, we conclude that none of ${\rm PSL}_{3}(r^{t})$, ${\rm PSL}_{4}(r^{t})$, ${\rm PSU}_{3}(r^{t})$ and ${\rm PSU}_{4}(r^{t})$ has such maximal subgroups, a contradiction.

Case 2. Suppose that ${\rm soc}(X_{v}^{{\it\Gamma}(v)})={\rm Sz}(q)$. Then $q=2^{f}$, $|{\bf O}_{2}(X_{\{u,v\}})|=2^{2f}a$, $2^{3f}a$ or $2^{4f}a$, where $f>1$ is odd, and $a=1$ or $2$. Noting that $X_{\{u,v\}}$ has order divisible by $2^{f}-1$, by Lemma 2.6, we conclude that $X_{\{u,v\}}$ is not a $\{2,3\}$-group. Since $X_{\{u,v\}}$ is nonabelian, it follows from ****[17, Table 15-20]**** that either $(X,X_{\{u,v\}})$ is one of $({}^{2}{\rm F}_{4}(2)^{\prime},[2^{9}]{:}5{:}4)$, $({\rm PSp}_{4}(2^{t}).\mathbb{Z}_{2^{l+1}},[2^{4t}]{:}\mathbb{Z}_{2^{t}-1}^{2}.\mathbb{Z}_{2^{l+1}})$ and $({\rm Sz}(2^{t}),[2^{2t}]{:}\mathbb{Z}_{2^{t}-1})$, or ${\rm soc}(X)$ is one of ${\rm PSL}_{n}(r^{t})$ and ${\rm PSU}_{n}(r^{t})$, where $n\in\{3,4\}$, $2^{l}$ is the $2$-part of $t$, and $r$ is odd if $n=4$. The first pair leads to $q=2^{3}$, and so $|X_{\{u,v\}}|$ is divisible by $7$, a contradiction. Checking the maximal subgroups of ${\rm soc}(X)$ (refer to ****[1, Tables 8.3-8.6, 8.8-8.14]****), the groups ${\rm PSL}_{3}(r^{t})$, ${\rm PSU}_{3}(r^{t})$, ${\rm PSL}_{4}(r^{t})$ and ${\rm PSU}_{4}(r^{t})$ are excluded as they have no maximal subgroup with a simple section ${\rm Sz}(q)$. Thus $(X,X_{\{u,v\}})=({\rm PSp}_{4}(2^{t}).\mathbb{Z}_{2^{l+1}},[2^{4t}]{:}\mathbb{Z}_{2^{t}-1}^{2}.\mathbb{Z}_{2^{l+1}})$ or $({\rm Sz}(2^{t}),[2^{2t}]{:}\mathbb{Z}_{2^{t}-1})$.

Since $f>1$ is odd, $2^{f}-1$ has a primitive prime divisor, say $s$. Recalling that $|X_{\{u,v\}}|$ is divisible by $2^{f}-1$, it follows that $s$ is a divisor of $2^{t}-1$, and so $t\geq f$. Then $t=f$. It follows that $X={\rm PSp}_{4}(q).\mathbb{Z}_{2^{l+1}}$, and $X_{v}^{[1]}\cong[q^{2}]{:}\mathbb{Z}_{q-1}$. However, by ****[1**, Table 8.14]****, ${\rm PSp}_{4}(q).\mathbb{Z}_{2^{l+1}}$ has no maximal subgroup containing $[q^{2}]{:}\mathbb{Z}_{q-1}.{\rm Sz}(q)$, a contradiction. **$\square$

Lemma 4.2**.**

Assume that (s1) of Lemma 3.3 occurs. Then $G$, $X$, $X_{\{u,v\}}$ and $X_{v}$ are listed as in Table 3.

*Proof.  *Assume first that $X_{uv}^{[1]}=1$. Then, $X_{v}=X_{v}^{[1]}.X_{v}^{{\it\Gamma}(v)}$, $X_{v}^{[1]}\cong(X_{v}^{[1]})^{{\it\Gamma}(u)}\unlhd(X_{u}^{{\it\Gamma}(u)})_{v}\cong(X_{v}^{{\it\Gamma}(v)})_{u}$, and $X_{uv}\lesssim(X_{v}^{{\it\Gamma}(v)})_{u}{\times}(X_{u}^{{\it\Gamma}(u)})_{v}$.

Suppose that $X_{v}^{{\it\Gamma}(v)}={\rm PSL}_{3}(2)$. Then $(X_{v}^{{\it\Gamma}(v)})_{u}\cong{\rm S}_{4}$, and thus $X_{v}^{[1]}$ and $X_{\{u,v\}}$ are given as follows:

[TABLE]

In particular, $2^{2}\leq|{\bf O}_{2}(X_{\{u,v\}})|\leq 2^{5}$. Check all possible pairs $(X,X_{\{u,v\}})$ in ****[17, Tables 14-20]****. Noting that ${\rm A}_{8}\cong{\rm PSL}_{4}(2)$ and ${\rm PSU}_{4}(2)\cong{\rm PSp}_{4}(3)$, we conclude that $X\cong{\rm A}_{8}$, $X_{\{u,v\}}\cong 2^{4}{:}{\rm S}_{3}^{2}$ and $X_{v}^{[1]}\cong{\rm A}_{4}$; or $X={\rm M}_{12}$ with $X_{\{u,v\}}\cong 2_{+}^{1{+}4}{:}{\rm S}_{3}$; or $X\cong{\rm PSU}_{4}(2)$ with $X_{\{u,v\}}\cong 2{\rm A}_{4}^{2}.2$. The group ${\rm A}_{8}$ is excluded as it has no subgroup of the form of $X_{v}^{[1]}.{\rm PSL}_{3}(2)$. The groups ${\rm M}_{12}$ and ${\rm PSU}_{4}(2)$ are excluded as their orders are not divisible by $d=7$.

Suppose that $X_{v}^{{\it\Gamma}(v)}={\rm PSL}_{3}(3)$. Then $(X_{v}^{{\it\Gamma}(v)})_{u}\cong 3^{2}{:}2{\rm S}_{4}$. Thus $X_{v}^{[1]}$ and $X_{\{u,v\}}$ are given as follows:

[TABLE]

Note that ${\bf O}_{3}(X_{\{u,v\}})\cong 3^{2}$ or $3^{4}$. Checking the possible pairs $(X,X_{\{u,v\}})$, we have $X_{\{u,v\}}\cong 3^{4}{:}2^{3}{.}{\rm S}_{4}$ and $X={\rm A}_{12}$ or ${\rm P\Omega}^{+}_{8}(2)$; in this case, $d=13$ is not a divisor of $|X|$, a contradiction.

Now let $X_{uv}^{[1]}$ be a nontrivial $p$-group. Then, by Theorem 2.4, $X_{v}$ and $X_{\{u,v\}}$ are given as follows:

[TABLE]

Suppose that $p=2$. Then $|X_{\{u,v\}}|$ is divisible by $9$ if and only if $|{\bf O}_{2}(X_{\{u,v\}})|\geq 8$, and ${\bf O}_{2}(X_{\{u,v\}})$ contains no elements of order $8$ unless $|{\bf O}_{2}(X_{\{u,v\}})|\geq 2^{22}$. Check the pairs $(G_{0},H_{0})$ given in ****[17, Tables 14-20]**** by estimating $|H_{0}|$ and $|{\bf O}_{2}(H_{0})|$. We conclude that one of the following holds:

1. (i)

$X={\rm PSL}_{4}(2).2\cong{\rm S}_{8}$** and $X_{\{u,v\}}=2^{4}{:}{\rm S}_{4}$;** 2. (ii)

$X={\rm PSL}_{5}(2).2$** and $X_{\{u,v\}}=[2^{8}]{.}{\rm S}_{3}^{2}.2$;** 3. (iii)

$X={\rm F}_{4}(2).2$** and $X_{\{u,v\}}=[2^{22}]{.}{\rm S}_{3}^{2}.2$;** 4. (iv)

${\rm soc}(X)={\rm PSL}_{3}(4)$** and $|{\bf O}_{2}(X_{\{u,v\}})|=2^{6}$;** 5. (v)

$X={\rm PSU}_{4}(3).2_{3}$** and $|{\bf O}_{2}(X_{\{u,v\}})|=2^{7}$;** 6. (vi)

$X={\rm He}.2$** and $X_{\{u,v\}}=[2^{8}]{:}{\rm S}_{3}^{2}.2$.**

Case (iv) yields that $X_{v}\cong 2^{3}{:}{\rm SL}_{3}(2)$ or $2^{4}{:}{\rm SL}_{3}(2)$; however, $X$ has no such subgroup by the Atlas ****[3]****. Similarly, cases (v) and (vi) are excluded. For (i), $G=X$ and ${\it\Gamma}$ is (isomorphic to) the point-plane incidence graph of the projective geometry ${\rm PG}(3,2)$. For (ii), $G=X$ and ${\it\Gamma}$ is (isomorphic to) the line-plane incidence graph of the projective geometry ${\rm PG}(4,2)$. If (iii) holds then $G=X$ and ${\it\Gamma}$ is the line-plane incidence graph of the metasymplectic space associated with ${\rm F}_{4}(2)$, see ****[30]****.

Now let $p=3$. Then $|{\bf O}_{3}(X_{\{u,v\}})|=3^{5}$ or $3^{8}$, and $X_{\{u,v\}}$ has no normal Sylow subgroups. Checking all possible pairs $(X,X_{\{u,v\}})$ in ****[17, Tables 14-20]****, we know that $(X,X_{\{u,v\}})$ is one of the following pairs:

[TABLE]

**Note that ${\bf O}_{3}(X_{v})\leq{\bf O}_{3}(X_{\{u,v\}})$. Then, for the first pair, ${\bf O}_{3}(X_{\{u,v\}})\cong\mathbb{Z}_{9}^{4}$ has no subgroup isomorphic to $\mathbb{Z}_{3}^{6}$, which is impossible. For the second pair, $G=X$ and ${\it\Gamma}$ is (isomorphic to) the line-plane incidence graph of the projective geometry ${\rm PG}(4,3)$. The last pair implies that $X\not\cong{\rm PGL}_{4}(3)$, $G=X$ or $X.2$, and ${\it\Gamma}$ is (isomorphic to) the line-plane incidence graph of the projective geometry ${\rm PG}(3,3)$. **$\square$

Lemma 4.3**.**

Assume that Lemma 3.3 (s2)* holds. Then $d=q+1$, and either ${\it\Gamma}$ is $(X,4)$-arc-transitive, or $G$, $X$, $X_{\{u,v\}}$ and $X_{v}$ are listed in Table 4.*

*Proof.  *Let $X_{v}^{{\it\Gamma}(v)}={\rm PSL}_{2}(q).[o]$, and $q=p^{f}>4$, where $p$ is a prime and o\,\big{|}\,(2,q-1)f. Note that ${\it\Gamma}$ is $(X,2)$-arc-transitive, see Lemma 3.3. By Theorem 2.4, if $X_{uv}^{[1]}\neq 1$ then ${\it\Gamma}$ is $(X,4)$-arc-transitive. Thus we assume next that $X_{uv}^{[1]}=1$, and then Lemma 3.4 works.

Note that $X_{v}=X_{v}^{[1]}.X_{v}^{{\it\Gamma}(v)}$, $X_{v}^{[1]}\cong(X_{v}^{[1]})^{{\it\Gamma}(u)}\unlhd(X_{u}^{{\it\Gamma}(u)})_{v}\cong(X_{v}^{{\it\Gamma}(v)})_{u}=p^{f}{:}{q-1\over(2,q-1)}{.}[o]$, and $X_{uv}\lesssim(X_{v}^{{\it\Gamma}(v)})_{u}{\times}(X_{u}^{{\it\Gamma}(u)})_{v}$. We have ${\bf O}_{p}(X_{\{u,v\}})=\mathbb{Z}_{p}^{if}.a$, where $i\in\{1,2\}$ and $a$ is a divisor of $(2,p)$. It is easily shown that $i=2$ if and only if ${\bf O}_{p}(X_{v}^{[1]})=\mathbb{Z}_{p}^{f}$. Combining with Lemma 3.4, we need only consider those pairs $(G_{0},H_{0})$ in ****[17, Tables 14-20]**** which satisfy

1. (a)

${\bf O}_{p}(H_{0})=\mathbb{Z}_{p}^{if}.a$, where $i\in\{1,2\}$ and $a$ is a divisor of $(2,p)$; $|{\rm Fit}(H_{0}):{\bf O}_{p}(H_{0})|\leq 2$; $G_{0}$ has a subgroup, say $M_{0}$, such that $|M_{0}:(M_{0}\cap H_{0})|=q+1$, $|H_{0}:(M_{0}\cap H_{0})|=2$, and $M_{0}$ has a simple section ${\rm PSL}_{2}(q)$; 2. (b)

$|H_{0}{:}{\bf O}_{p}(H_{0})|$** is a divisor of $2(q-1)^{2}f^{2}$ and divisible by $q-1$; if $i=1$ then $|H_{0}{:}{\bf O}_{p}(H_{0})|$ is a divisor of $2(q-1)f$.**

Case 1. Assume that ${\rm soc}(X)$ is an alternating group. Using ****[17, Table 14]****, we have $G=X={\rm S}_{p}$ and $X_{\{u,v\}}\cong\mathbb{Z}_{p}{:}\mathbb{Z}_{p-1}$, where $p\in\{7,11,17,23\}$. Then $X_{v}={\rm PSL}_{2}(p)$ and $d=p{+}1$. For $p=17$ or $23$, the group ${\rm PSL}_{2}(p)$ has no transitive permutation representation of degree $p$, and thus it cannot occur as a subgroup of ${\rm S}_{p}$. Therefore, $p=7$ or $11$, and $G$, $X$ and $X_{\{u,v\}}$ are listed in Table 4. In fact, $X_{uv}$ and $X_{\{u,v\}}$ are the normalizers of some Syolw $p$-subgroup in ${\rm PSL}_{2}(p)$ and ${\rm S}_{p}$, respectively. (Note that, for $p=7$, the resulting graph is the point-plane non-incidence graph of ${\rm PG}(3,2)$.)

Case 2. Assume that ${\rm soc}(X)$ is a simple sporadic group. By ****[17, Table 15]****, with the restrictions (a) and (b), the only pairs $(G_{0},H_{0})$ are listed as follows: $({\rm M}_{11},3^{2}{:}{\rm Q}_{8}.2)$, $({\rm J}_{1},\mathbb{Z}_{11}{:}\mathbb{Z}_{10})$, $({\rm J}_{1},\mathbb{Z}_{7}{:}\mathbb{Z}_{6})$, $({\rm J}_{3}.2,\mathbb{Z}_{19}{:}\mathbb{Z}_{18})$, $({\rm J}_{4},\mathbb{Z}_{29}{:}\mathbb{Z}_{28})$, $({\rm O^{\prime}N}.2,\mathbb{Z}_{31}{:}\mathbb{Z}_{30})$, $({\rm B},\mathbb{Z}_{19}{:}\mathbb{Z}_{18}{\times}\mathbb{Z}_{2})$, $({\rm B},\mathbb{Z}_{23}{:}\mathbb{Z}_{11}{\times}\mathbb{Z}_{2})$, $({\rm M},\mathbb{Z}_{41}{:}\mathbb{Z}_{40})$ and $({\rm M},\mathbb{Z}_{47}{:}\mathbb{Z}_{23}{\times}\mathbb{Z}_{2})$. In particular, ${\bf O}_{p}(H_{0})$ is a Sylow $p$-subgroup of $G_{0}$. This yields that $X_{v}^{[1]}=1$, and so ${\rm soc}(X_{v})={\rm PSL}_{2}(p^{f})$.

Suppose that $(X,X_{\{u,v\}})$ is one of $({\rm J}_{1},\mathbb{Z}_{7}{:}\mathbb{Z}_{6})$, $({\rm J}_{4},\mathbb{Z}_{29}{:}\mathbb{Z}_{28})$ and $({\rm M},\mathbb{Z}_{47}{:}\mathbb{Z}_{23}{\times}\mathbb{Z}_{2})$. Then $X_{v}={\rm PSL}_{2}(p)$ for $p=7$, $29$ and $47$, respectively; however, by the Altas ****[3]**** and ****[34, Tables 5.6 and 5.11]****, $X$ has no subgroup ${\rm PSL}_{2}(p)$, a contradiction. Thus $G$, $X$ and $X_{\{u,v\}}$ are listed in Table 4. (Note that the Monster ${\rm M}$ has a maximal subgroup ${\rm PSL}_{2}(41)$ by ****[22]****.)

Case 3. Assume that ${\rm soc}(X)$ is a simple group of Lie type over a finite field of order $r^{t}$, where $r$ is a prime. We first show $r\neq p$.

Suppose that $r=p$. Then, by (a), either ${\bf O}_{p}(H_{0})$ is abelian or $r=p=2$. For $r=p>2$, noting that $|H_{0}|$ has a divisor $q-1$, there does not exist $H_{0}$ in ****[17, Tables 16-20]**** such that ${\bf O}_{p}(H_{0})$ is abelian. Thus we have $r=p=2$. Recalling that $p^{f}>4$ and $|H_{0}/{\bf O}_{p}(H_{0})|$ is divisible by $2^{f}-1$, it follows from Lemma 2.6 that $H_{0}/{\bf O}_{p}(H_{0})$ is not a $\{2,3\}$-group. Checking those $H_{0}$ given in ****[17, Tables 16-20]****, we conclude that $(G_{0},H_{0})$ is one of the following pairs:

• $({\rm PSL}_{2}(2^{t}),\mathbb{Z}_{2}^{t}{:}\mathbb{Z}_{2^{t}-1})$, $({\rm PSL}_{3}(2^{t}),[2^{3t}]{:}[{(2^{t}-1)^{2}\over(3,2^{t}-1)}].2)$; $({\rm PSU}_{3}(2^{t}),[2^{3t}]{:}\mathbb{Z}_{2^{2t}-1\over(3,2^{t}+1)})$;

$({\rm PSp}_{4}(2^{t}).\mathbb{Z}_{2^{l+1}},[2^{4t}]{:}\mathbb{Z}_{2^{t}-1}^{2}.\mathbb{Z}_{2^{l+1}})$, where $2^{l}$ is the $2$-part of $t$;

$({\rm Sz}(2^{t}),[2^{2t}]{:}\mathbb{Z}_{2^{t}-1})$; $({}^{3}{\rm D}_{4}(2),[2^{11}]{:}(\mathbb{Z}_{7}\times{\rm S}_{3}))$; $({}^{2}{\rm F}_{4}(2)^{\prime},[2^{9}]{:}5{:}4)$.

First, the pair $({\rm Sz}(2^{t}),[2^{2t}]{:}\mathbb{Z}_{2^{t}-1})$ is excluded as ${\rm Sz}(2^{t})$ has no subgroup with a section ${\rm PSL}_{2}(2^{f})$. For the last two pairs, we have $f=5$ and $4$ respectively, which yields that $2^{f}-1$ is not a divisor of $|H_{0}|$, a contradiction. For the three pairs after the first one, we have $t<f$, thus $G_{0}$ has no maximal subgroup with a section ${\rm PSL}_{2}(2^{f})$, a contradiction. Suppose finally that $(X,X_{\{u,v\}})=({\rm PSL}_{2}(2^{t}),\mathbb{Z}_{2}^{t}{:}\mathbb{Z}_{2^{t}-1})$. Then $3\leq f<t\leq 2f+1$. Noting that $2^{f}-1$ is a divisor of $2^{t}-1$, it follows that $f$ is a divisor of $t$, and so $t=2f$. Then ${\bf O}_{2}(X_{\{u,v\}})=2^{2f}$, yielding $|{\bf O}_{2}(X_{v}^{[1]})|=2^{f}$. Thus ${\bf O}_{2}(X_{v})\neq 1$ and $X_{v}$ has a simple section ${\rm PSL}_{2}(2^{f})$. Checking the subgroups of ${\rm PSL}_{2}(2^{2f})$, refer to ****[12, II.8.27]****, we conclude that ${\rm PSL}_{2}(2^{2f})$ has no subgroup isomorphic to $X_{v}$, a contradiction.

Assume that $r\neq p$ in the following.

Subcase 3.1**. We first deal with those pairs $(G_{0},H_{0})$ such that $H_{0}$ is included in some infinite families in [17, Table 16-20]. Note that $r\neq p$, and we consider only those $H_{0}$ having subgroups of index $2$. It follows that either $H_{0}/{\rm Fit}(H_{0})$ is a $\{2,3\}$-group or $(G_{0},H_{0})=({\rm E}_{8}(q^{\prime}),\mathbb{Z}_{q^{\prime 8}{\pm}q^{\prime 7}{\mp}q^{\prime 5}-q^{\prime 4}{\mp}q^{\prime 3}{\pm}q^{\prime}{+}1}.\mathbb{Z}_{30})$, where $q^{\prime}=r^{t}$. Suppose that $(G_{0},H_{0})=({\rm E}_{8}(q^{\prime}),\mathbb{Z}_{q^{\prime 8}{\pm}q^{\prime 7}{\mp}q^{\prime 5}-q^{\prime 4}{\mp}q^{\prime 3}{\pm}q^{\prime}{+}1}.\mathbb{Z}_{30})$. Then $q^{\prime 8}{\pm}q^{\prime 7}{\mp}q^{\prime 5}-q^{\prime 4}{\mp}q^{\prime 3}{\pm}q^{\prime}{+}1$ is divisible by some primitive prime divisor $s$ of $q^{\prime 15}-1$ or of $q^{\prime 30}-1$. Noting that $s\geq 17$, we know that $H_{0}$ has normal cyclic Sylow $s$-subgroup. It follows from (a) that $17\leq p=s=q^{\prime 8}{\pm}q^{\prime 7}{\mp}q^{\prime 5}-q^{\prime 4}{\mp}q^{\prime 3}{\pm}q^{\prime}{+}1$. In particular, ${\bf O}_{p}(H_{0})=\mathbb{Z}_{p}$ and $f=1$. By (b), $|H_{0}|$ is divisible by $p-1$, and then $30$ is divisible by $p-1$. This implies that $30=p-1=q^{\prime 8}{\pm}q^{\prime 7}{\mp}q^{\prime 5}-q^{\prime 4}{\mp}q^{\prime 3}{\pm}q^{\prime}$, which is impossible. Therefore, $H_{0}/{\rm Fit}(H_{0})$ is a $\{2,3\}$-group.**

By (a), ${\rm Fit}(H_{0})$ a $\{2,p\}$-group. Then $|H_{0}|$ has no prime divisor other than $2$, $3$ and $p$. Since $p^{f}-1$ is a divisor of $|H_{0}|$, by Lemma 2.6, we have $f<3$. Recall that $(X_{u}^{{\it\Gamma}(u)})_{v}\cong(X_{v}^{{\it\Gamma}(v)})_{u}=p^{f}{:}{q-1\over(2,q-1)}{.}[o]$, and $X_{uv}\lesssim(X_{v}^{{\it\Gamma}(v)})_{u}{\times}(X_{u}^{{\it\Gamma}(u)})_{v}$, where $o$ is a divisor of $(2,q-1)f$. It follows that $X_{uv}/{\bf O}_{p}(X_{uv})$ has an abelian Hall $2^{\prime}$-subgroup. Note that $X_{uv}{\bf O}_{p}(X_{\{u,v\}})/{\bf O}_{p}(X_{\{u,v\}})\cong X_{uv}/({\bf O}_{p}(X_{\{u,v\}}\cap X_{uv})=X_{uv}/{\bf O}_{p}(X_{uv})$, and $|X_{\{u,v\}}:X_{uv}{\bf O}_{p}(X_{\{u,v\}})|\leq 2$. It follows that $X_{\{u,v\}}/{\bf O}_{p}(X_{\{u,v\}})$ has an abelian Hall $2^{\prime}$-subgroup. Thus, as a possible candidate for $X_{\{u,v\}}$, the quotient of $H_{0}$ over ${\bf O}_{p}(H_{0})$ has abelian Hall $2^{\prime}$-subgroups. In particular, $H_{0}/{\bf O}_{p}(H_{0})$ has no section $A_{4}$.

Considering the restrictions on $H_{0}$, $r$ and $f$, we conclude that $(G_{0},H_{0})$ can only be one of the following pairs:

• $({\rm PSL}_{2}(r^{t}),\mathbb{Z}_{r^{t}\pm 1\over(2,r^{t}-1)}{:}\mathbb{Z}_{2})$, $({\rm PSL}_{3}(r^{t}),[{(r^{t}-1)^{2}\over(3,r^{t}-1)}].{\rm S}_{3})$, $({\rm PSU}_{3}(r^{t}),[{(r^{t}+1)^{2}\over(3,r^{t}+1)}].{\rm S}_{3})$;

$({\rm PSp}_{4}(2^{t}).\mathbb{Z}_{2^{l+1}},\mathbb{Z}_{2^{t}\pm 1}^{2}.[2^{l+4}])$, $({\rm PSp}_{4}(2^{t}).\mathbb{Z}_{2^{l+1}},\mathbb{Z}_{2^{2t}+1}.[2^{l+3}])$, where $t\geq 3$;

$({\rm Sz}(2^{t}),\mathbb{Z}_{2^{t}-1}{:}\mathbb{Z}_{2})$**, $({\rm Sz}(2^{t}),\mathbb{Z}_{2^{t}\pm\sqrt{2^{t+1}}+1}{:}\mathbb{Z}_{4})$, where $t\geq 3$;; **

$({\rm Ree}(3^{t}),\mathbb{Z}_{3^{t}\pm\sqrt{3^{t+1}}+1}{.}\mathbb{Z}_{6})$**, $({\rm Ree}(3^{t}),\mathbb{Z}_{3^{t}+1}{.}\mathbb{Z}_{6})$, where $t\geq 3$;; **

$({\rm G}_{2}(3^{t}).\mathbb{Z}_{2^{l+1}},\mathbb{Z}_{3^{t}\pm 1}^{2}{.}[3\cdot 2^{l+3}])$, $({\rm G}_{2}(3^{t}).\mathbb{Z}_{2^{l+1}},\mathbb{Z}_{3^{2t}\pm 3^{t}+1}{.}[3\cdot 2^{l+2}])$, where $t\geq 2$;

$({}^{3}{\rm D}_{4}(r^{t}),\mathbb{Z}_{r^{4t}-r^{2t}+1}{:}\mathbb{Z}_{4})$**; $({}^{2}{\rm F}_{4}(2^{t}),\mathbb{Z}_{2^{2t}\pm\sqrt{2^{3t+1}}+2^{t}\pm\sqrt{2^{t+1}}+1}{.}\mathbb{Z}_{12})$, where $t\geq 3$; **

$({\rm F}_{4}(2^{t}).\mathbb{Z}_{2^{l+1}},\mathbb{Z}_{2^{4t}-2^{2t}+1}{.}[3\cdot 2^{l+3}])$, where $t\geq 2$;

where the power $2^{l}$ appeared means the $2$-part of $t$. Recall that $|{\rm Fit}(H_{0}):{\bf O}_{p}(H_{0})|\leq 2$ and $|H_{0}:{\bf O}_{p}(H_{0})|$ is divisible by $p^{f}-1$. This allows us determine the values of $p^{f}$ and $r^{t}$. As an example, we only deal with the second pair. Suppose that $(G_{0},H_{0})=({\rm PSL}_{3}(r^{t}),[{(r^{t}-1)^{2}\over(3,r^{t}-1)}].{\rm S}_{3})$. Considering the structures of ${\rm Fit}(H_{0})$ and ${\bf O}_{p}(H_{0})$, either $(3,r^{t}-1)=1$, $p=r^{t}-1$ and $f\in\{1,2\}$, or $f=1$ and $p=r^{t}-1=3$. The latter implies that ${\rm PSL}_{2}(q)$ is soluble, which is not the case. Assume that the former case holds. Then $|{\rm S}_{3}|$ is divisible by $r^{t}-1-1$ or $(r^{t}-1)^{2}-1$. Then the only possibility is that $(p^{f},r^{t})=(7,8)$. The other pairs can be fixed out in a similar way, the details is omitted here. Eventually, we conclude that $(G_{0},H_{0},p,f)$ is one of $({\rm PSL}_{2}(19),{\rm D}_{20},5,1)$, $({\rm PSL}_{3}(8),7^{2}{:}{\rm S}_{3},7,1)$ and $({\rm Sz}(8),\mathbb{Z}_{5}{:}\mathbb{Z}_{4},5,1)$. By the Atlas ****[3]****, neither ${\rm PSL}_{3}(8)$ nor ${\rm Sz}(8)$ has subgroup with a section ${\rm PSL}_{2}(p)$. Thus, in this case, $G$, $X$ and $X_{\{u,v\}}$ are given as in Table 4.

Subcase 3.2**. For the pairs $(G_{0},H_{0})$ not appearing in Subcase 3.1, we check the finite number of $H_{0}$ one by one. We observed that either $p=2$, or $H_{0}/{\bf O}_{p}(H_{0})$ is a $\{2,3\}$-group. Recall that $r\neq p$.**

Suppose that $p=2$. Recalling that $q=2^{f}>4$, we have $f\geq 3$. In particular, since $|H_{0}|$ is divisible by $2^{f}-1$, $H_{0}$ is not a $\{2,3\}$-group by Lemma 2.6. Then the only possibility is that $G_{0}={}^{2}{\rm F}_{4}(2)^{\prime}$ and $H_{0}=[2^{9}]{:}5{:}4$. Thus $|{\bf O}_{2}(H_{0})|=2^{9}$, it follows from (a) that $f=4$ or $9$, and then $G_{0}$ has a section ${\rm PSL}_{2}(2^{4})$ or ${\rm PSL}_{2}(2^{9})$, which is impossible by checking the (maximal) subgroups of ${}^{2}{\rm F}_{4}(2)^{\prime}$. Thus $p>2$, and $H_{0}/{\bf O}_{p}(H_{0})$ is a $\{2,3\}$-group; in particular, by (a), ${\bf O}_{p}(H_{0})=\mathbb{Z}_{p}^{if}$ for some $i\in\{1,2\}$.

Suppose that $H_{0}$ has a section ${\rm A}_{4}$. Then $H_{0}$ has no normal Sylow $3$-subgroup. Further, $H_{0}$ has no quotient ${\rm A}_{4}$ as $H_{0}$ has a subgroup of index $2$. If $(3,(q-1)f)=1$ then, by (b), we conclude that $p=3$ and ${\bf O}_{p}(H_{0})$ is the unique Sylow $3$-subgroup of $H_{0}$, a contradiction. Thus $3$ is a divisor of $(q-1)f$. Check those $H_{0}$ in ****[17, Table 16-20]**** which have a section ${\rm A}_{4}$ and do not appear in Subcase 3.1. Recalling $r\neq p>2$ and ${\bf O}_{p}(H_{0})=\mathbb{Z}_{p}^{if}$, it follows that either ${\bf O}_{p}(H_{0})=\mathbb{Z}_{3}^{2}$ or $(G_{0},H_{0})=({\rm F}_{4}(2).4,\mathbb{Z}_{7}^{2}{:}(3\times{\rm SL}_{2}(3)).4)$. Since $3$ is a divisor of $(q-1)f$, we get $G_{0}={\rm F}_{4}(2).4$ and $q=p^{f}=7$ or $7^{2}$. By (b), for $q=7$ or $7^{2}$, the order of $H_{0}$ should be a divisor of $72$ or $192$ respectively, which is impossible.

The above argument allows us ignore many cases without further inspection. Inspecting carefully the remaining pairs, the possible candidates for $(X,X_{\{u,v\}})$ are as follows:

• $({\rm PGL}_{2}(9),{\rm D}_{20})$, $({\rm M}_{10},\mathbb{Z}_{5}{:}\mathbb{Z}_{4})$, $({\rm PGL}_{2}(11),{\rm D}_{20})$;
• $({\rm PSL}_{3}(r),3^{2}{:}{\rm Q}_{8})$, where $r\equiv 4,7\,({\sf mod~{}}9)$;
• $({\rm PSp}_{4}(4).4,\mathbb{Z}_{17}{:}\mathbb{Z}_{16})$, $({\rm PSp}_{4}(4).4,5^{2}{:}[2^{5}])$;
• $({\rm PSU}_{3}(r),3^{2}{:}{\rm Q}_{8})$, where $5<r\equiv 2,5\,({\sf mod~{}}9)$;
• $({\rm PSU}_{3}(2^{t}),3^{2}{:}{\rm Q}_{8})$, where $t$ is a prime no less than $5$;
• $({}^{2}{\rm F}_{4}(2),\mathbb{Z}_{13}{:}\mathbb{Z}_{12})$.

For the first three pairs, $G$, $X$ and $X_{\{u,v\}}$ are easily determined and given as in Table 4. The pair $({\rm PSp}_{4}(4).4,\mathbb{Z}_{17}{:}\mathbb{Z}_{16})$ is excluded as ${\rm PSp}_{4}(4).4$ has no subgroup ${\rm PSL}_{2}(17)$, and the pair $({}^{2}{\rm F}_{4}(2),\mathbb{Z}_{13}{:}\mathbb{Z}_{12})$ is excluded as ${}^{2}{\rm F}_{4}(2)$ has no subgroup ${\rm PSL}_{2}(13)$. Suppose that $X\cong{\rm PSU}_{3}(2^{t})$ and $X_{\{u,v\}}\cong 3^{2}{:}{\rm Q}_{8}$. Then we have $X_{v}\cong{\rm PSL}_{2}(9)$; however, by ****[1, Tables 8.3, 8.4]****, ${\rm PSU}_{3}(2^{t})$ has no subgroup ${\rm PSL}_{2}(9)$, a contradiction. Suppose that $(X,X_{\{u,v\}})=({\rm PSp}_{4}(4).4,5^{2}{:}[2^{5}])$. Then $X_{v}$ contains a Sylow $5$-subgroup $P$ of $X$ and has a section ${\rm PSL}_{2}(5)$ or ${\rm PSL}_{2}(25)$. By the information for ${\rm PSp}_{4}(4).4$ given in the Atlas ****[3]****, we conclude that $X_{v}\leq M\cong({\rm A}_{5}\times{\rm A}_{5}){:}2^{2}<{\rm PSp}_{4}(4).2<{\rm PSp}_{4}(4).4$. Note that $X_{uv}=5^{2}{:}[2^{4}]$, which should be the normalizer of $P$ in $X_{v}$. Using GAP ****[27]****, computation shows that $|{\bf N}_{L}(P)|\leq 200$ for any maximal subgroup $L$ of $M$ with $P\leq L$. It follows that $X_{v}=M\cong({\rm A}_{5}\times{\rm A}_{5}){:}2^{2}$, yielding $d=|X_{v}:X_{uv}|=36\neq q+1$, a contradiction.

Let $(X,X_{\{u,v\}})=({\rm PSL}_{3}(r),3^{2}{:}{\rm Q}_{8})$. Then $X_{uv}\cong 3^{2}{:}4$. It is easily shown that $p=3$ and $X_{v}\cong{\rm PSL}_{2}(9)$. Since $r\equiv 4,7\,({\sf mod~{}}9)$, we know that ${\rm PSL}_{3}(r)$ has a Sylow $3$-subgroup $\mathbb{Z}_{3}^{2}$. By ****[1, Tables 8.3, 8.4]****, ${\rm PSL}_{3}(r)$ has a subgroup ${\rm PSL}_{2}(9)$ if and only if $r\equiv 1,4\,({\sf mod~{}}15)$. Thus, in this case, we have $r\equiv 11,14,29,41\,({\sf mod~{}}45)$. For a subgroup ${\rm PSL}_{2}(9)$ of ${\rm PSL}_{3}(r)$, taking a Sylow $3$-subgroup $Q$ of ${\rm PSL}_{2}(9)$, the normalizers of $Q$ in ${\rm PSL}_{2}(9)$ and ${\rm PSL}_{3}(r)$ are (isomorphic to) $3^{2}{:}4$ and $3^{2}{:}{\rm Q}_{8}$, respectively. Then these two normalizers of $Q$ can serve as the roles of $X_{uv}$ and $X_{\{u,v\}}$, respectively. Thus $X$ and $X_{\{u,v\}}$ are given as in Table 4. Noting that $G=XG_{\{u,v\}}$, we have $G_{\{u,v\}}/X_{\{u,v\}}\cong G/X\lesssim{\sf Out}({\rm PSL}_{3}(r))\cong{\rm S}_{3}$, and so $G=X.[m]$ and $G_{\{u,v\}}=X_{\{u,v\}}.[m]$, where $m$ is a divisor of $6$. Thus $|G_{uv}{:}X_{uv}|=m$, since $|G_{v}{:}G_{uv}|=10=|X_{v}{:}X_{uv}|$, we have $|G_{v}{:}X_{v}|=m$. By ****[1**, Table 8.4]****, ${\bf N}_{{\sf Aut}{\rm PSL}_{3}(r)}(X_{v})=X_{v}.2$. Since $X_{v}\unlhd G_{v}$, it follows that $m\leq 2$. Thus $G=X$ or $X.2$, and if $G=X.2$ then $G_{v}=X_{v}.2\cong{\rm PGL}_{2}(9)$ and $G_{\{u,v\}}\cong 3^{2}{:}{\rm Q}_{8}.2$. The pair $({\rm PSU}_{3}(r),3^{2}{:}{\rm Q}_{8})$ is similarly dealt with. **$\square$

Lemma 4.4**.**

If Lemma 3.3 (s3)* holds then $G$, $X$, $X_{\{u,v\}}$ and $X_{v}$ are listed in Table 5.*

Proof.  Let $X_{v}^{{\it\Gamma}(v)}={\rm PSU}_{3}(q).[o]$ and $q=p^{f}>2$, where $p$ is a prime and o\,\big{|}\,2(3,q{+}1)f. Then $(X_{v}^{{\it\Gamma}(v)})_{u}=p^{f{+}2f}{:}{q^{2}-1\over(3,q{+}1)}.[o]$, and $X_{uv}^{[1]}=1$ by Theorem 2.4. Thus $|{\bf O}_{p}(X_{\{u,v\}})|=p^{3f}.a$, $p^{4f}.a$, $p^{5f}.a$ or $p^{6f}.a$, where $a$ is a divisor of $(2,p)$. Moreover, ${\bf O}_{p}(X_{\{u,v\}})$ is nonabelian, and $X_{\{u,v\}}/{\bf O}_{p}(X_{\{u,v\}})$ has a subgroup $\mathbb{Z}_{(q^{2}-1)\over(3,q{+}1)}$. We next determine which pair $(G_{0},H_{0})$ in ****[17**, Tables 14-20]** is a possible candidate for $(X,X_{\{u,v\}})$. Note that we may ignore those $H_{0}$ which either has no subgroup of index $2$ or has abelian maximal normal $p$-subgroup. In particular, ${\rm soc}(X)$ is not an alternating group.**

Case 1. Let $(G_{0},H_{0})$ be a pair with $H_{0}$ included in some infinite families given in ****[17, Table 16-20]****. Since ${\bf O}_{p}(X_{\{u,v\}})$ is nonabelian, we conclude that $(X,{\bf O}_{p}(X_{\{u,v\}}))$ is one of the following pairs:

• $({\rm PSL}_{3}(p^{t}).2,[p^{3t}])$, $({\rm PGL}_{3}(p^{t}).2,[p^{3t}])$ (with $p=2$),

$({\rm PSU}_{3}(p^{t}),[p^{3t}])$**, $({\rm PSp}_{4}(p^{t}).\mathbb{Z}_{2^{l+1}},[p^{4t}])$ (with $p=2$), **

$({\rm Sz}(p^{t}),[p^{2t}])$, $({\rm Ree}(p^{t}),[p^{3t}])$ and $({\rm G}_{2}(p^{t}).\mathbb{Z}_{2^{l+1}},[p^{6t}])$,

where $2^{l}$ is the $2$-part of $t$. Check the maximal subgroups of ${\rm PSp}_{4}(p^{t}).\mathbb{Z}_{2^{l+1}}$, ${\rm Sz}(p^{t})$ and ${\rm Ree}(p^{t})$, refer to ****[1, Table 8.14]****, ****[26, Theorem 9]**** and ****[15, Theorem C]****, respectively. We conclude that none of ${\rm PSp}_{4}(t^{f}).\mathbb{Z}_{2^{l+1}}$, ${\rm Sz}(p^{t})$ and ${\rm Ree}(p^{t})$ has maximal subgroups with a simple section ${\rm PSU}_{3}(q)$, and they are excluded. For the first three and the last pairs, $|X/{\bf O}_{p}(X_{\{u,v\}})|$ is a divisor of $2(p^{t}-1)^{2}$, and ${\bf O}_{p}(X_{\{u,v\}})=[p^{3t}]$ or $[p^{6t}]$. Clearly, $t\leq 2f$.

Suppose that $t=2f$. Then ${\rm soc}(X)={\rm PSL}_{3}(q^{2})$ or ${\rm PSU}_{3}(q^{2})$, and ${\bf O}_{p}(X_{\{u,v\}})=[q^{6}]$. It follows that ${\bf O}_{p}(X_{v}^{[1]})=[q^{3}]$. Thus ${\bf O}_{p}(X_{v})\neq 1$ and $X_{v}$ has an almost simple quotient ${\rm PSU}_{3}(q).[o]$. Checking Tables 8.3 and 8.5 given in ****[1]****, we conclude that $X$ has no maximal subgroup containing $X_{v}$, a contradiction. If $t=f$ then we have $(X,{\bf O}_{p}(X_{\{u,v\}}))=({\rm G}_{2}(p^{t}).\mathbb{Z}_{2^{l+1}},[q^{6}])$, and we get a similar contradiction by checking the maximal subgroups of ${\rm G}_{2}(p^{t}).\mathbb{Z}_{2^{l+1}}$.

Suppose that $f\neq t<2f$. Then $f>1$. Recalling that $X_{\{u,v\}}/{\bf O}_{p}(X_{\{u,v\}})$ has a subgroup $\mathbb{Z}_{(q^{2}-1)\over(3,q{+}1)}$, we know that $p^{2f}-1$ is a divisor of $2(3,q+1)(p^{t}-1)^{2}$. If $p^{2f}-1$ has a primitive prime divisor say $s$, then $s\geq 2f+1\geq 5$, and $s$ is not a divisor of $2(3,q+1)(p^{t}-1)^{2}$, a contradiction. It follows from Zsigmondy’s theorem that $2f=6$ and $p=2$, and so $t=1$ or $2$. Then $7$ is a divisor of $p^{2f}-1$ but not a divisor of $2(3,q+1)(p^{t}-1)^{2}$, a contradiction.

Case 2. Let $(G_{0},H_{0})$ be one of the pairs in ****[17, Table 15-20]**** which is not considered in Case 1. Assume that $X_{\{u,v\}}/{\bf O}_{p}(X_{\{u,v\}})$ is a $\{2,3\}$-group. Then $p^{2f}-1$ has no prime divisor other than $2$ and $3$. It follows that $f=1$, and so $p=q>2$. Calculation shows that $p\in\{3,5,7\}$. For $q=p=3$, it is easily shown that $X_{\{u,v\}}/{\bf O}_{p}(X_{\{u,v\}})$ is a $2$-group. These observations yield that either $q=p=3$ and $X_{\{u,v\}}/{\bf O}_{p}(X_{\{u,v\}})$ is a $2$-group, or $X_{\{u,v\}}$ is not a $\{2,3\}$-group.

Recalling that $X_{\{u,v\}}/{\bf O}_{p}(X_{\{u,v\}})$ has a subgroup $\mathbb{Z}_{(q^{2}-1)\over(3,q{+}1)}$ and $|{\bf O}_{p}(X_{\{u,v\}})|=p^{if}.a$ for $3\leq i\leq 6$, it follows that $(X,X_{\{u,v\}})$ is one of the following pairs:

• $({\rm HS}.2,[5^{3}]{:}[2^{5}])$, $({\rm Ru},[5^{3}]{:}[2^{5}])$, $({\rm McL},[5^{3}]{:}3{:}8)$, $({\rm Co}_{2},[5^{3}]{:}4{\rm S}_{4})$,

$({\rm Th},[5^{3}]{:}4{\rm S}_{4})$, $({\rm J}_{4},[11^{3}]{:}(5\times 2{\rm S}_{4}))$.

Then $q=p\in\{5,11\}$ and $X_{v}^{[1]}=1$. In particular, ${\rm soc}(X_{v})={\rm PSU}_{3}(p)$, and $X_{\{u,v\}}$ is the normalizer ${\bf N}_{X}(P)$ of some Sylow $p$-subgroup $P$ of $X$. Thus $X_{uv}=X_{v}\cap X_{\{u,v\}}\leq{\bf N}_{X_{v}}(P)$. For the pairs $({\rm HS}.2,[5^{3}]{:}[2^{5}])$ and $({\rm Ru},[5^{3}]{:}[2^{5}])$, by the Atlas ****[3]****, $X_{\{u,v\}}$ is a normalizer of some Sylow $5$-subgroup, which intersects a maximal subgroup ${\rm PSU}_{3}(5){:}2$ of ${\rm soc}(X)$ at $[5^{3}]{:}8{:}2$, thus $G$, $X$ and $X_{\{u,v\}}$ are listed in Table 5. The other pairs are excluded as follows.

First, the group ${\rm Th}$ is excluded as it has no maximal subgroup with a simple section ${\rm PSU}_{3}(5)$, refer to ****[34, Table 5.8]****. For the pair $({\rm McL},[5^{3}]{:}3{:}8)$, by the Atlas ****[3]****, we have $X_{v}={\rm PSU}_{3}(5)$, and so $X_{uv}\leq{\bf N}_{{\rm PSU}_{3}(5)}(P)=[5^{3}]{:}8$, which contradicts that $|X_{\{u,v\}}:X_{uv}|=2$. For the pair $({\rm J}_{4},[11^{3}]{:}(5\times 2{\rm S}_{4}))$, by ****[34, Table 5.8]****, $X_{v}={\rm PSU}_{3}(11).2$, yielding $X_{uv}\leq{\bf N}_{X_{v}}(P)=[11^{3}]{:}(5\times 8{:}2)$, we get a similar contradiction. For the pair $(X,X_{\{u,v\}})=({\rm Co}_{2},[5^{3}]{:}4{\rm S}_{4})$, by the Atlas ****[3**]****, $X_{v}<{\rm HS}.2<{\rm Co}_{2}$. Checking the maximal subgroups of ${\rm HS}.2$, we have $X_{v}={\rm PSU}_{3}(5)$ or $X_{v}={\rm PSU}_{3}(5){:}2$. It follows that $X_{uv}\leq{\bf N}_{X_{v}}(P)=[5^{3}]{:}8$ or $[5^{3}]{:}[2^{5}]$, and then $|X_{\{u,v\}}:X_{uv}|\neq 2$, a contradiction. **$\square$

5. Graphs with soluble vertex-stabilizers

Let $G$, $T$, $X$ and ${\it\Gamma}=(V,E)$ be as in Hypothesis 3.1. We now deal with the case that $G_{v}$ is soluble. First, the following lemma says that ${\it\Gamma}$ is not a complete bipartite graph if $G_{v}^{{\it\Gamma}(v)}$ is soluble.

Lemma 5.1**.**

Assume that ${\it\Gamma}\cong{\sf K}_{d,d}$. Then $T\cong{\rm A}_{6}$, $d=6$, $T_{v}={\rm PSL}_{2}(5)$ and $T_{uv}\cong{\rm D}_{10}$. In particular, $X_{uv}$ is nonabelian.

Proof.  Let $G^{+}$ be the subgroup of $G$ fixing the bipartition of ${\it\Gamma}$. Then $G_{v}\leq G^{+}$, and $G_{v}$ is $2$-transitive on the partite set which does not contain $v$. Thus $G^{+}$ acts $2$-transitively on each partite set, and these two actions are not equivalent. Check the almost simple $2$-transitive groups, refer to ****[2**, Table 7.4]**. We conclude that $T\cong{\rm A}_{6}$ or ${\rm M}_{12}$, and $T_{v}\cong{\rm A}_{5}$ or ${\rm M}_{11}$, respectively. Since $T_{uv}$ is soluble, the lemma follows. **$\square$

By Lemma 5.1 and ****[20]****, we have the following result.

Lemma 5.2**.**

Assume that $X_{uv}$ is abelian. Then either $T\cong{\rm PSL}_{2}(q)$ and ${\it\Gamma}\cong{\sf K}_{q{+}1}$, or $T={\rm Sz}(2^{2m{+}1})$, $T_{\{u,v\}}\cong{\rm D}_{2(2^{2m{+}1}-1)}$, $T_{v}\cong\mathbb{Z}_{2}^{2m{+}1}{:}\mathbb{Z}_{2^{2m{+}1}-1}$, ${\sf Aut}{\it\Gamma}={\sf Aut}({\rm Sz}(q))$ and ${\it\Gamma}$ is $(T,2)$-arc-transitive.

Lemma 5.3**.**

Assume that (a1) or (a2) of Lemma 3.5 holds, and $X_{uv}$ is nonabelian. Then one of the following holds:

1. (1)

$G=X$* or $X.2$, $X={\rm M}_{10}$, $X_{\{u,v\}}\cong\mathbb{Z}_{8}{:}\mathbb{Z}_{2}$, $X_{v}\cong 3^{2}{:}{\rm Q}_{8}$ and ${\it\Gamma}\cong{\sf K}_{10}$.* 2. (2)

$G=X={\rm PSL}_{3}(3).2$, $X_{\{u,v\}}\cong{\rm GL}_{2}(3){:}2$, $X_{v}\cong 3^{2}{:}{\rm GL}_{2}(3)$, and ${\it\Gamma}$ is the point-line non-incidence graph of ${\rm PG}(2,3)$.

Proof.  Case 1. Assume that Lemma 3.5 (a1) holds. Suppose that $(X_{v}^{{\it\Gamma}(v)})_{u}={\rm Q}_{8}$. Then $X_{uv}\lesssim{\rm Q}_{8}{\times}{\rm Q}_{8}$. This implies that $|X_{\{u,v\}}|$ is a divisor of $2^{7}$ and divisible by $2^{4}$. Checking the Tables 14-20 in ****[17**], we have $X\cong{\rm PSL}_{2}(9).2={\rm M}_{10}$ and $X_{\{u,v\}}\cong\mathbb{Z}_{8}{:}\mathbb{Z}_{2}$; in this case, $X_{v}\cong 3^{2}{:}{\rm Q}_{8}$, and $d=9$. Then part (1) of this lemma follows.

Suppose that $(X_{v}^{{\it\Gamma}(v)})_{u}\neq{\rm Q}_{8}$. If $p=3$ and $(G_{v}^{{\it\Gamma}(v)})_{u}={\rm Q}_{8}$, then $(X_{v}^{{\it\Gamma}(v)})_{u}$ is abelian, it follows that $X_{uv}$ is abelian, a contradiction. Thus we have ${\rm SL}_{2}(3)\unlhd(G_{v}^{{\it\Gamma}(v)})_{u}\leq{\rm GL}_{2}(p)$, and $p\in\{3,5,7,11,23\}$. Then $(G_{v}^{{\it\Gamma}(v)})_{u}\leq{\bf N}_{{\rm GL}_{2}(p)}({\rm SL}_{2}(3))=\mathbb{Z}_{p-1}\circ{\rm GL}_{2}(3)$. Since $(X_{v}^{{\it\Gamma}(v)})_{u}$ is nonabelian and normal in $(G_{v}^{{\it\Gamma}(v)})_{u}$, we have ${\rm Q}_{8}\unlhd(X_{v}^{{\it\Gamma}(v)})_{u}$, and hence ${\rm SL}_{2}(3)\unlhd(X_{v}^{{\it\Gamma}(v)})_{u}$. Moreover, $|X_{\{u,v\}}|$ is a divisor of $2^{7}3^{2}(p-1)^{2}$ and divisible by $2^{4}$. Let $M$ be an arbitrary normal abelian subgroup of $X_{\{u,v\}}$. Then $M\cap X_{uv}$ has index at most $2$ in $M$, and $(M\cap X_{uv})X_{v}^{[1]}/X_{v}^{[1]}$ is isomorphic to a normal subgroup of $(X_{v}^{{\it\Gamma}(v)})_{u}$. Thus $(M\cap X_{uv})X_{v}^{[1]}/X_{v}^{[1]}\lesssim\mathbb{Z}_{p-1}$. Since $M\cap X_{v}^{[1]}\unlhd X_{v}^{[1]}$ and $X_{v}^{[1]}$ is isomorphic to a normal subgroup of $(X_{v}^{{\it\Gamma}(v)})_{u}$, we have $M\cap X_{v}^{[1]}\lesssim\mathbb{Z}_{p-1}$. Noting that $(M\cap X_{uv})X_{v}^{[1]}/X_{v}^{[1]}\cong M\cap X_{uv}/(M\cap X_{v}^{[1]})$, it follows that $|M\cap X_{uv}|$ is a divisor of $(p-1)^{2}$. Thus $|M|$ is a divisor of $2(p-1)^{2}$.

The above observations allow us to consider only the pairs $(G_{0},H_{0})$ in ****[17, Tables 14-20]**** which satisfy the following conditions:

1. (c1)

$|H_{0}|$** is a divisor of $2^{7}3^{2}(p-1)^{2}$ and divisible by $2^{4}$; $H_{0}$ has a factor (a quotient of some subnormal subgroup) ${\rm Q}_{8}$; and $H_{0}$ has no element of order $3^{2}$, $5^{2}$ or $11^{2}$.** 2. (c2)

If $M$ is a normal abelian subgroup of $H_{0}$ then $|M|$ is a divisor of $2(p-1)^{2}$; if $p\in\{7,11,23\}$, the order of ${\bf O}_{p-1\over 2}(H_{0})$ is a divisor of ${(p-1)^{2}\over 4}$.

Checking the those $H_{0}$ which satisfy conditions (c1) and (c2), we conclude that the possible pairs $(X,X_{\{u,v\}})$ are listed as follows:

1. $({\rm M}_{11},3^{2}{:}{\rm Q}_{8}.2)$**, $({\rm M}_{11},2{\rm S}_{4})$, $({\rm M}_{12},[2^{5}].{\rm S}_{3})$, $({\rm M}_{12},3^{2}{:}2{\rm S}_{4})$, $({\rm J}_{2},[2^{6}]{:}(3{\times}{\rm S}_{3}))$, **

$({\rm J}_{3},[2^{6}]{:}(3{\times}{\rm S}_{3}))$**, $({\rm Co}_{3},[2^{9}].3^{2}.{\rm S}_{3}))$, $({\rm He}.2,[2^{8}]{:}3^{2}.{\rm D}_{8}))$, $({\rm McL}.2,[2^{6}]{:}{\rm S}_{3}^{2})$, **

$({\rm PSL}_{3}(3),3^{2}{:}2{\rm S}_{4})$**, $({\rm PSL}_{3}(3).2,2{\rm S}_{4}{:}2)$, $({\rm PSL}_{3}(4).2,2^{2{+}4}.3.2)$, **

$({\rm PGL}_{3}(4){.}2,[2^{6}]{.}3{.}{\rm S}_{3})$**, $({\rm PSL}_{4}(3){.}2,2{.}{\rm S}_{4}^{2}{.}2)$, $({\rm PSL}_{5}(2){.}2,[2^{8}]{.}{\rm S}_{3}^{2}{.}2)$, **

$({\rm PSp}_{4}(4){.}4,[2^{8}]{:}3{.}12)$**, $({\rm PSp}_{4}(4).4,5^{2}{:}[2^{5}])$, $({\rm PSp}_{6}(2),[2^{7}]{:}{\rm S}_{3}^{2})$, **

$({\rm PSp}_{6}(3),[2^{8}]{:}3^{3}.S_{3})$, $({\rm PSU}_{3}(3),4.{\rm S}_{4})$, $({\rm PSU}_{4}(2),2.{\rm A}_{4}^{2}.2)$, $({\rm PSU}_{4}(3),2.{\rm A}_{4}^{2}.4)$,

$({\rm PSU}_{4}(3).2,[2^{5}].{\rm S}_{4})$**, $({\rm P\Omega}^{+}_{8}(3){.}{\rm A}_{4},10^{2}{:}4{\rm A}_{4})$, **

$({\rm G}_{2}(2)^{\prime},4.{\rm S}_{4})$, $({\rm G}_{2}(3),{\rm SL}_{2}(3)\circ{\rm SL}_{2}(3){:}2)$, $({}^{2}{\rm F}_{4}(2)^{\prime},5^{2}{:}4{\rm A}_{4})$.

Note these groups $X$ are included in the Atlas ****[3]****. Inspecting the subgroups of $X$, only the pair $({\rm PSL}_{3}(3).2,2{\rm S}_{4}{:}2)$ gives a desired $X_{v}\cong 3^{2}{:}{\rm GL}_{2}(3)$. Then part (2) of this lemma follows.

Case 2. Let $2_{+}^{1{+}4}{:}\mathbb{Z}_{5}\leq(G_{v}^{{\it\Gamma}(v)})_{u}\leq 2_{+}^{1{+}4}.(\mathbb{Z}_{5}{:}\mathbb{Z}_{4})$. Then $2_{+}^{1{+}4}\unlhd(X_{v}^{{\it\Gamma}(v)})_{u}$, and so, $|X_{\{u,v\}}|$ is a divisor of $2^{15}5^{2}$ and divisible by $2^{6}$. Further, if $M$ is a normal abelian subgroup of $X_{\{u,v\}}$ then a similar argument as in Case 1 yields that $|M|$ is a divisor of $2^{5}$. It is easily shown that ${\bf O}_{2}(X_{uv})\neq 1$, and hence ${\bf O}_{2}(X_{\{u,v\}})\neq 1$. Checking the pairs $(G_{0},H_{0})$ in ****[17**, Tables 14-20]****, either ${\bf O}_{2}(H_{0})=1$ or $|H_{0}|$ has an odd prime divisor other than $5$. Thus, in this case, no desired pair $(X,X_{\{u,v\}})$ exists. **$\square$

We assume next that Lemma 3.5 (a3) occurs. Thus $(G_{v}^{{\it\Gamma}(v)})_{u}\not\leq{\rm GL}_{1}(p^{f})$ and $(G_{v}^{{\it\Gamma}(v)})_{u}\leq{\rm\Gamma L}_{1}(p^{f})$. Then $f>1$ and $(G_{v}^{{\it\Gamma}(v)})_{u}\lesssim\mathbb{Z}_{p^{f}-1}{:}\mathbb{Z}_{f}$. Recalling $X_{uv}\lesssim(X_{u}^{{\it\Gamma}(u)})_{v}{\times}(X_{v}^{{\it\Gamma}(v)})_{u}\leq(G_{u}^{{\it\Gamma}(u)})_{v}{\times}(G_{v}^{{\it\Gamma}(v)})_{u}$, we have the following simple fact.

Lemma 5.4**.**

If (a3) of Lemma 3.5 occurs then $X_{\{u,v\}}$ has no section $\mathbb{Z}_{t}^{3}$, $\mathbb{Z}_{r}^{5}$ or $\mathbb{Z}_{2}^{6}$, where $t$ is a primitive prime divisor of $p^{f}-1$ and $r$ is an arbitrary odd prime.

Lemma 5.5**.**

Assume that $X_{uv}$ is nonabelian and (a3) of Lemma 3.5 occurs. Then $p^{f}\neq 2^{6}$.

Proof.  Suppose that $p^{f}=2^{6}$. Then $X$ has order divisible by $2^{6}$, $X_{uv}\lesssim\mathbb{Z}_{63}{:}\mathbb{Z}_{6}{\times}\mathbb{Z}_{63}{:}\mathbb{Z}_{6}$, and thus $X_{\{u,v\}}$ has a normal Hall $2^{\prime}$-subgroup and $|X_{\{u,v\}}|$ is indivisible by $2^{4}$. Checking Tables 14-20 given in ****[17**], $(X,X_{\{u,v\}})$ is one of the following pairs:

1. $({\rm S}_{7},\mathbb{Z}_{7}{:}\mathbb{Z}_{6})$**, $({\rm M}_{12}.2,3_{+}^{1{+}2}{:}{\rm D}_{8})$, $({\rm PSL}_{2}(2^{6}),{\rm D}_{126})$, $({\rm PSL}_{2}(5^{3}),{\rm D}_{126})$, **

$({\rm PSL}_{2}(7937),{\rm D}_{7938})$, $({\rm PSL}_{3}(8),7^{2}{:}{\rm S}_{3})$, $({\rm Sz}(8),{\rm D}_{14})$, $({\rm G}_{2}(3).2,[3^{6}]{:}{\rm D}_{8})$.

**The pair $({\rm PSL}_{2}(2^{6}),{\rm D}_{126})$ yields that $X_{v}\cong 2^{6}{:}\mathbb{Z}_{63}$, and thus $X_{uv}$ is abelian, this is not the case. The other pairs are easily excluded as none of them gives a desired $X_{v}$. **$\square$

Lemma 5.6**.**

Assume that $X_{uv}$ is nonabelian and (a3) of Lemma 3.5 occurs. Suppose that $X_{uv}$ has a normal abelian Hall $2^{\prime}$-subgroup. Then $G=X$ or $X.2$, $X={\rm M}_{10}$, $X_{\{u,v\}}\cong\mathbb{Z}_{8}{:}\mathbb{Z}_{2}$, $X_{v}\cong 3^{2}{:}{\rm Q}_{8}$ and ${\it\Gamma}\cong{\sf K}_{10}$.

Proof.  Note that $X_{\{u,v\}}=X_{uv}.2$. The unique Hall $2^{\prime}$-subgroup of $X_{uv}$ is also the Hall $2^{\prime}$-subgroup of $X_{\{u,v\}}$. Checking Tables 14-20 given in ****[17**], we know that $(X,X_{\{u,v\}})$ is one of the following pairs:

1. (i)

$({\rm PGL}_{2}(7),{\rm D}_{16})$, $({\rm PSL}_{3}(2).2,{\rm D}_{16})$, $({\rm PGL}_{2}(9),{\rm D}_{16})$, $({\rm M}_{10},\mathbb{Z}_{8}{:}\mathbb{Z}_{2})$; $({\rm A}_{5},{\rm D}_{10})$,

$({\rm A}_{6},3^{2}{:}\mathbb{Z}_{4})$**, $({\rm M}_{11},3^{2}{:}{\rm Q}_{8}.2)$, $({\rm J}_{1},{\rm D}_{6}{\times}{\rm D}_{10})$, $({\rm PGL}_{2}(7),{\rm D}_{12})$, $({\rm PGL}_{2}(9),{\rm D}_{20})$, **

$({\rm M}_{10},\mathbb{Z}_{5}{:}\mathbb{Z}_{4})$, $({\rm PGL}_{2}(11),{\rm D}_{20})$, $({\rm PSL}_{2}(t^{a}),{\rm D}_{2(t^{a}{\pm}1)\over(2,t-1)})$; $({\rm PSp}_{4}(4).4,\mathbb{Z}_{17}{:}\mathbb{Z}_{16})$; 2. (ii)

$({\rm PSL}_{2}(t^{a}),\mathbb{Z}_{t}^{a}{:}\mathbb{Z}_{t^{a}-1\over 2})$, $t$ is a prime $a\leq 4$ and $t^{a}-1$ a power of $2$;

$({\rm PSL}_{3}(t),\mathbb{Z}_{3}^{2}{:}{\rm Q}_{8})$**, $t$ is a prime with $t\equiv 4,7\,({\sf mod~{}}9)$; **

$({\rm PSU}_{3}(t),\mathbb{Z}_{3}^{2}{:}{\rm Q}_{8})$**, $t$ is a prime with $t\equiv 2,5\,({\sf mod~{}}9)$; **

$({\rm PSU}_{3}(2^{a}),\mathbb{Z}_{3}^{2}{:}{\rm Q}_{8})$** with prime $a>3$;**

$({\rm PSp}_{4}(2^{a}).\mathbb{Z}_{2^{b{+}1}},{\rm D}_{2(q{\pm}1)}^{2}{:}2.\mathbb{Z}_{2^{b{+}1}})$, $({\rm PSp}_{4}(2^{a}).\mathbb{Z}_{2^{b{+}1}},\mathbb{Z}_{2^{2a}{+}1}.4.\mathbb{Z}_{2^{b{+}1}})$, where $2^{b}\parallel a$;

$({\rm Sz}(2^{2a{+}1}),{\rm D}_{2(2^{2a{+}1}-1)})$, $({\rm Sz}(2^{2a{+}1}),\mathbb{Z}_{2^{2a{+}1}{\pm}2^{a{+}1}{+}1}{:}\mathbb{Z}_{4})$;

$({}^{3}{\rm D}_{4}(t^{a}),\mathbb{Z}_{t^{4a}-t^{2a}{+}1}{:}\mathbb{Z}_{4})$, $t$ is a prime.

The pair $({\rm M}_{10},\mathbb{Z}_{8}{:}\mathbb{Z}_{2})$ yields that $X_{v}\cong 3^{2}{:}{\rm Q}_{8}$ and $d=9$. The third pair in (i) implies that $X_{v}\cong\mathbb{Z}_{3}^{2}{:}\mathbb{Z}_{8}$; however, $X_{uv}$ is abelian, which is not the case. For $({\rm PSL}_{2}(t^{a}),{\rm D}_{2(t^{a}{\pm}1)\over(2,t^{a}-1)})$, checking the subgroups of ${\rm PSL}_{2}(t^{a})$, we have $t^{a}=p^{f}$ and $X_{v}\cong\mathbb{Z}_{p}^{f}{:}\mathbb{Z}_{p^{f}-1\over(2,p-1)}$, and then $X_{uv}$ is abelian, a contradction. The other pairs in (i) are also excluded as $|X|$ is indivisible by $p^{f}$. (Note that $f>1$.)

Now we deal with the pairs in (ii). Note that, for an odd prime $r$, the edge-stabilizer $X_{\{u,v\}}$ has a unique Sylow $r$-subgroup ${\bf O}_{r}(X_{\{u,v\}})$. Then ${\bf O}_{r}(X_{\{u,v\}})$ is a Sylow subgroup of $X$ by Lemma 2.7. This implies that the unique Hall $2^{\prime}$-subgroup of $X_{\{u,v\}}$, say $K$, is a Hall subgroup of $X$. Since $X_{\{u,v\}}=X_{uv}.2$, we have $K\leq X_{uv}$. Note that $|X_{v}{:}X_{uv}|=d=p^{f}$ and $X_{v}$ is contained in a maximal subgroup of $X$. We now check the maximal subgroups of $X$ which contain $K$, refer to ****[12, II.8.27]****, ****[1, Tables 8.3-8.6, 8.14, 8.15]**** and ****[14, 26]****. Then one of the following occurs:

1. (iii)

$X={\rm Sz}(2^{2a{+}1})$** and $X_{v}\cong\mathbb{Z}_{2}^{2a{+}1}{:}\mathbb{Z}_{2^{2a{+}1}-1}$;** 2. (iv)

$X={\rm PSp}_{4}(2^{a}).\mathbb{Z}_{2^{b{+}1}}$** and $X_{v}\lesssim{\rm Sp}_{2}(2^{2a}){:}2.\mathbb{Z}_{2^{b}}$;** 3. (v)

$X={\rm PSp}_{4}(2^{a}).\mathbb{Z}_{2^{b{+}1}}$** and $X_{v}\lesssim{\rm Sp}_{2}(2^{a})\wr{\rm S}_{2}.\mathbb{Z}_{2^{b}}$.**

**Item (iii) yields that $X_{uv}$ is abelian, which is not the case. Item (iv) gives $X_{uv}=X_{v}$, a contradiction. Suppose that (v) occurs, we have $X_{v}\cong(\mathbb{Z}_{2}^{a}{:}\mathbb{Z}_{2^{a}-1})^{2}{:}2.\mathbb{Z}_{2^{b}}$. Then $1\neq{\bf O}_{2}(X_{v})\leq{\bf O}_{2}(G_{v})$, and hence $d=|{\bf O}_{2}(G_{v})|$ by Lemma 2.5. Since $X_{v}$ is transitive on ${\it\Gamma}(v)$, it follows that $p^{f}=d=2^{2a}$. Thus $|X_{uv}|=(2^{a}-1)^{2}2^{b{+}1}$, and so $|X_{\{u,v\}}{:}X_{uv}|=8>2$, a contradiction. **$\square$

Corollary 5.7**.**

Assume that $X_{uv}$ is nonabelian and (a3) of Lemma 3.5 occurs. If $f=2$ then $G=X$ or $X.2$, $X={\rm M}_{10}$, $X_{\{u,v\}}\cong\mathbb{Z}_{8}{:}\mathbb{Z}_{2}$, $X_{v}\cong 3^{2}{:}{\rm Q}_{8}$ and ${\it\Gamma}\cong{\sf K}_{10}$.

*Proof.  ***Let $f=2$. Then $(X_{v}^{{\it\Gamma}(v)})_{u}\lesssim\mathbb{Z}_{p^{2}-1}.\mathbb{Z}_{2}$. Note that $X_{\{u,v\}}=X_{uv}.2$ and $X_{uv}\lesssim\mathbb{Z}_{p^{2}-1}.\mathbb{Z}_{2}{\times}\mathbb{Z}_{p^{2}-1}.\mathbb{Z}_{2}$. Then Lemma 5.6 is applicable, and the result follows. **$\square$

Let ${\rm\pi}_{0}(p^{f}-1)$ be the set of primitive primes of $p^{f}-1$. By Zsigmondy’s theorem, if ${\rm\pi}_{0}(p^{f}-1)=\emptyset$ and $f>1$ then $p^{f}=2^{6}$, or $f=2$ and $p=2^{t}-1$, where $t$ is a prime. Thus, in view of Lemma 5.5 and Corollary 5.7, we assume next that ${\rm\pi}_{0}(p^{f}-1)\neq\emptyset$.

Lemma 5.8**.**

Assume that $\pi{:}={\rm\pi}_{0}(p^{f}-1)\neq\emptyset$, $X_{uv}$ is nonabelian and (a3) of Lemma 3.5 occurs. Then $f\geq 3$, and

• (1)

$\pi\neq\pi(|X_{\{u,v\}}|)\setminus\{2\}$, $\min(\pi)\geq\max\{5,f{+}1\}$;

• (2)

$p\not\equiv{\pm}1\,({\sf mod~{}}r)$* and ${\bf O}_{r}(X_{\{u,v\}})\neq 1$ for each $r\in\pi$;*

• (3)

$X_{\{u,v\}}$* has a unique *(nontrivial) Hall $\pi$-subgroup, which is either cyclic or a direct product of two cyclic subgroups.

*Proof.  ***By the assumption and Lemma 3.6, we have $(X_{v}^{{\it\Gamma}(v)})_{u}\cong\mathbb{Z}_{m^{\prime}}.\mathbb{Z}_{f\over e^{\prime}}$, and $\emptyset\neq\pi={\rm\pi}_{0}(p^{f}-1)\subseteq{\rm\pi}(m^{\prime})$. For $r\in{\rm\pi}$, since $p^{r-1}\equiv 1\,({\sf mod~{}}r)$, we have $f\leq r-1$, and so $r\geq f{+}1$. In particular, $r\geq 5$ and $p\not\equiv{\pm}1\,({\sf mod~{}}r)$. Recall that $X_{\{u,v\}}=X_{uv}.2$ and $X_{uv}\lesssim\mathbb{Z}_{m^{\prime}}.\mathbb{Z}_{f\over e^{\prime}}{\times}\mathbb{Z}_{m^{\prime}}.\mathbb{Z}_{f\over e^{\prime}}$. It follows that ${\bf O}_{r}(X_{\{u,v\}})\neq 1$, and ${\bf O}_{r}(X_{\{u,v\}})$ is the unique Sylow $r$-subgroup of $X_{\{u,v\}}$. Clearly, ${\bf O}_{r}(X_{\{u,v\}})$ is either cyclic or a direct product of two cyclic subgroups. Then $X_{\{u,v\}}$ has a unique Hall $\pi$-subgroup $F$, which is either cyclic or a direct product of two cyclic subgroups. Clearly, $F\neq 1$ and, by Lemma 5.6, $X_{\{u,v\}}$ has no normal abelian Hall $2^{\prime}$-subgroup. Then $\pi\neq\pi(|X_{\{u,v\}}|)\setminus\{2\}$, the lemma follows. **$\square$

Recall that $X_{\{u,v\}}$ has no section $\mathbb{Z}_{2}^{6}$ or $\mathbb{Z}_{3}^{5}$, see Lemma 5.4. Combining with Lemma 5.8, we next check the pairs $(G_{0},H_{0})$ listed in ****[17, Tables 14-20]****.

Lemma 5.9**.**

Assume that ${\rm\pi}_{0}(p^{f}-1)\neq\emptyset$, $X_{uv}$ is nonabelian and (a3) of Lemma 3.5 occurs. Then $T={\rm soc}(X)$ is not a simple group of Lie type.

*Proof.  *Suppose that $T$ is a simple group of Lie type over a finite field of order $q^{\prime}=t^{a}$, where $t$ is a prime. Since $T\unlhd G$, we know that $T$ is transitive on the edge set of ${\it\Gamma}$. Then $T_{v}^{{\it\Gamma}(v)}\neq 1$. Noting that $T_{v}^{{\it\Gamma}(v)}\unlhd G_{v}^{{\it\Gamma}(v)}$, we have ${\rm soc}(G_{v}^{{\it\Gamma}(v)})\leq T_{v}^{{\it\Gamma}(v)}$. In particular, $T_{v}$ is transitive on ${\it\Gamma}(v)$, and so $|T_{v}|=p^{f}|T_{uv}|$. In view of this, noting that $T_{v}=T\cap X_{v}=T\cap G_{v}$ and $T_{\{u,v\}}=T\cap X_{\{u,v\}}=T\cap G_{\{u,v\}}$, we sometimes work on the triple $(T,T_{v},T_{\{u,v\}})$ instead of $(X,X_{v},X_{\{u,v\}})$.

By Lemmas 5.6 and 5.8, $X_{\{u,v\}}$ is not a $\{2,3\}$-group and has no normal abelian Hall $2^{\prime}$-subgroup. Assume that $t\in\pi_{0}(p^{f}-1)$. By Lemmas 5.4 and 5.8, $t\geq 5$, $X_{\{u,v\}}$ has no section $\mathbb{Z}_{t}^{3}$ and ${\bf O}_{t}(X_{\{u,v\}})\neq 1$ is abelian. Checking the pairs $(G_{0},H_{0})$ listed in ****[17, Tables 16-20]****, we have $X={\rm PSL}_{2}(t^{2})$ and $X_{\{u,v\}}\cong\mathbb{Z}_{t}^{2}{:}\mathbb{Z}_{t^{2}-1\over 2}$. For this case, checking the subgroups of ${\rm PSL}_{2}(t^{2})$, no desired $X_{v}$ arises, a contradiction. Therefore, $t\not\in\pi_{0}(p^{f}-1)$.

By Lemma 5.8, ${\bf O}_{r}(X_{\{u,v\}})\neq 1$ for each $r\in\pi_{0}(p^{f}-1)$. Recall that $X_{\{u,v\}}$ is not a $\{2,3\}$-group and has a subgroup of index $2$. Checking the pairs $(G_{0},H_{0})$ listed in ****[17, Tables 16-20]****, we conclude that ${\bf O}_{t}(X_{\{u,v\}})=1$. Further, we observe that a desired $X_{\{u,v\}}$ if exists has the form of $N.K$, where $N$ is an abelian subgroup of $T$ and either $K$ is a $\{2,3\}$-group or $(X,K)=({\rm E}_{8}(q^{\prime}),\mathbb{Z}_{30})$. For the case where $K\not\cong\mathbb{Z}_{30}$, by Lemma 3.6, $\pi_{0}(p^{f}-1)\subseteq\pi(|N|)$, and thus, by Lemma 5.4, $N$ has no subgroup $\mathbb{Z}_{r}^{3}$ for $r\in\pi_{0}(p^{f}-1)$. With these restrictions, only one of the following (i)-(iv) occurs.

(i) Either $X={\rm PSL}_{3}(q^{\prime})$ and $X_{\{u,v\}}\cong{1\over(3,q^{\prime}-1)}\mathbb{Z}_{q^{\prime}-1}^{2}.{\rm S}_{3}$ with $q^{\prime}\neq 2,\,4$, or $X={\rm PSU}_{3}(q^{\prime})$ and $X_{\{u,v\}}\cong{1\over(3,q^{\prime}{+}1)}\mathbb{Z}_{q^{\prime}{+}1}^{2}.{\rm S}_{3}$. Then $|X_{v}|={3\over(3,q^{\prime}{\mp}1)}p^{f}(q^{\prime}{\mp}1)^{2}$. Checking Tables 8.3-8.6 given in ****[1]****, we have $X={\rm PSL}_{3}(q^{\prime})$ and $X_{v}\lesssim[q^{\prime 3}]{:}{1\over(3,q^{\prime}-1)}\mathbb{Z}_{q^{\prime}-1}^{2}$. It follows that $p=t=3$, and $|{\bf O}_{3}(X_{v})|=3^{f{+}1}=3d$, which contradicts Lemma 2.5.

(ii) $T={\rm soc}(X)={\rm P\Omega}_{8}^{+}(q^{\prime})$ and $T_{\{u,v\}}\cong{\rm D}_{2(q^{\prime 2}{+}1)\over(2,q^{\prime}-1)}^{2}.[2^{2}]$. In this case, noting that $|T_{\{u,v\}}{:}T_{uv}|\leq 2$, we have $|T_{v}|=2^{4}p^{f}{(q^{\prime 2}{+}1)^{2}\over(2,q^{\prime}-1)^{2}}$ or $2^{3}p^{f}{(q^{\prime 2}{+}1)^{2}\over(2,q^{\prime}-1)^{2}}$. Let $M$ be a maximal subgroup of $T$ with $T_{v}\leq M$. By ****[14]****, since $|M|$ is divisible by $(q^{\prime 2}{+}1)^{2}$, we have $M\cong{\rm PSL}_{2}(q^{\prime 2})^{2}.2^{2}$. It is easily shown that ${\rm PSL}_{2}(q^{\prime 2})^{2}.2^{2}$ does not have subgroups of order $2^{4}p^{f}{(q^{\prime 2}{+}1)^{2}\over(2,q^{\prime}-1)^{2}}$ or $2^{3}p^{f}{(q^{\prime 2}{+}1)^{2}\over(2,q^{\prime}-1)^{2}}$, a contradiction.

(iii) $(X,X_{\{u,v\}})$ is one of $({}^{2}{\rm F}_{4}(2)^{\prime},5^{2}{:}4{\rm A}_{4})$ and $({}^{2}{\rm F}_{4}(2),13{:}12)$. For the first pair, we have $\pi_{0}(p^{f}-1)=\{5\}$ and, since $p^{f}$ is a divisor of $|{}^{2}{\rm F}_{4}(2)^{\prime}|$, we conclude that $p^{f}=2^{4}$ or $3^{4}$. The second pair implies that $\pi_{0}(p^{f}-1)=\{13\}$, and then $p^{f}=2^{12}$ or $3^{3}$. By the Atlas ****[3]****, $X$ has no maximal subgroup containing $X_{uv}$ as a subgroup of index divisible by $p^{f}$, a contradiction.

(iv) $T_{\{u,v\}}$ has a normal abelian subgroup $N$ listed as follows:

[TABLE]

Let $M$ be a maximal subgroup of $T$ with $T_{v}\leq M$. Then $|M|$ is divisible by $p^{f}|N|$. Check the maximal subgroups of $T$ of order divisible by $|N|$, refer to ****[15, 19, 21]****. Then we may deduce a contradiction. First, by ****[15, Theorem C]****, we conclude that ${\rm Ree}(3^{a})$ has no maximal subgroups of order divisible by $p^{f}|N|$. Similarly, by ****[21]****, the group ${}^{2}{\rm F}_{4}(2^{a})$ is excluded. We next deal with the remaining cases.

(1) Let $T={\rm G}_{2}(3^{a})$. Suppose that $|N|=3^{2a}{\pm}3^{a}{+}1$. By ****[15, Theorems A and B]****, since $|M|$ is divisible by $3^{2a}{\pm}3^{a}{+}1$, we have $M\cong{\rm SL}_{3}(3^{a}){:}2$ or ${\rm SU}_{3}(3^{a}){:}2$. By ****[1, Tables 8.3-8.6]****, we conclude that $T_{v}\lesssim\mathbb{Z}_{3^{2a}{\pm}3^{a}{+}1}{:}[6]$, which is impossible.

Similarly, for $|N|=(3^{a}{\pm}1)^{2}$, we have $T_{v}\lesssim({\rm SL}_{2}(3^{a})\circ{\rm SL}_{2}(3^{a})).2$, ${\rm SL}_{3}(3^{a}){:}2$ or ${\rm SU}_{3}(3^{a}){:}2$. Since $|T_{v}|$ is divisible by ${1\over 2}|T_{\{u,v\}}|p^{f}=6p^{f}(3^{a}{\pm}1)^{2}$, checking the maximal subgroups of ${\rm SL}_{2}(3^{a})$, ${\rm SL}_{3}(3^{a})$ and ${\rm SU}_{3}(3^{a})$, we have $p=3$ and $T_{v}\lesssim[3^{ba}]{:}\mathbb{Z}_{3^{a}-1}^{2}.2$ for $b=2$ or $3$. Since $T_{uv}$ has order divisible by $3$, it follows that ${\bf O}_{3}(T_{uv})\neq 1$, which contradicts Lemma 2.5.

(3) Let $T={\rm F}_{4}(2^{a})$. By ****[18, 19]****, noting that $|M|$ is divisible by $p^{f}|N|$, we conclude that $M\cong{\rm Sp}_{8}(2^{a})$ or ${\rm P\Omega}_{8}^{+}(2^{a}).{\rm S}_{3}$ with $|N|=(2^{2a}{+}1)^{2}$, or $M\cong c.{\rm PSL}_{3}(2^{a})^{2}.c.2$ or $c.{\rm PSU}_{3}(2^{a})^{2}.c.2$ with $|N|=(2^{2a}{\pm}2^{a}{+}1)^{2}$, where $c=(3,2^{a}\pm 1)$. Then a contradiction follows from checking the maximal subgroups of ${\rm Sp}_{8}(2^{a})$, ${\rm P\Omega}_{8}^{+}(2^{a})$, ${\rm PSL}_{3}(2^{a})$ and ${\rm PSU}_{3}(2^{a})$, refer to ****[1, Tables 8.3-8.6, 8.48-8.50]****.

**(4) Let $T={\rm E}_{8}(q^{\prime})$. Then $|N|=(q^{\prime 4}-q^{\prime 2}{+}1)^{2}$ and $M\cong{\rm PSU}_{3}(q^{\prime 2})^{2}.8$. For this case, checking the maximal subgroups ${\rm PSU}_{3}(q^{\prime 2})$, we get a contradiction. **$\square$

Lemma 5.10**.**

Assume that ${\rm\pi}_{0}(p^{f}-1)\neq\emptyset$, $X_{uv}$ is nonabelian and (a3) of Lemma 3.5 occurs. Then $G=X={\rm J}_{1}$, $X_{\{u,v\}}\cong\mathbb{Z}_{7}{:}\mathbb{Z}_{6}$, $X_{v}\cong\mathbb{Z}_{2}^{3}{:}\mathbb{Z}_{7}{:}\mathbb{Z}_{3}$ and $d=8$.

*Proof.  *By Lemma 5.9, $T={\rm soc}(X)$ is either an alternating group or a sporadic simple group. Note that $X_{\{u,v\}}$ is not a $\{2,3\}$-group and has no normal abelian Hall $2^{\prime}$-subgroup.

Assume that $T$ is an alternating group. Then, by ****[17, Table 14]****, either $X={\rm A}_{r}$ and $X_{\{u,v\}}\cong\mathbb{Z}_{r}{:}\mathbb{Z}_{r-1\over 2}$ for $r\not\in\{7,11,17,23\}$, or $X={\rm S}_{r}$ and $X_{\{u,v\}}\cong\mathbb{Z}_{r}{:}\mathbb{Z}_{r-1}$ for $r\in\{7,11,17,23\}$. For these two cases, $X_{v}$ is a transitive subgroup of ${\rm S}_{r}$ in the natural action of ${\rm S}_{r}$. Then either $X_{v}$ is almost simple or $X_{v}\lesssim\mathbb{Z}_{r}{:}\mathbb{Z}_{r-1}$ (refer to ****[4, page 99, Corollary 3.5B]****), a contradiction.

Assume that $T$ is a sporadic simple group, and let $r\in\pi_{0}(p^{f}-1)$. Then $(X,X_{\{u,v\}},r)$ is one of the following triples:

1. $({\rm J}_{1},\mathbb{Z}_{7}{:}\mathbb{Z}_{6},7)$**, $({\rm J}_{1},\mathbb{Z}_{11}{:}\mathbb{Z}_{10},11)$, $({\rm J}_{1},\mathbb{Z}_{19}{:}\mathbb{Z}_{6},19)$, $({\rm J}_{2},\mathbb{Z}_{5}^{2}{:}{\rm D}_{12},5)$, **

$({\rm J}_{3}.2,\mathbb{Z}_{19}{:}\mathbb{Z}_{18},19)$**, $({\rm J}_{4},\mathbb{Z}_{29}{:}\mathbb{Z}_{28},29)$, $({\rm J}_{4},\mathbb{Z}_{37}{:}\mathbb{Z}_{12},37)$, $({\rm J}_{4},\mathbb{Z}_{43}{:}\mathbb{Z}_{14},43)$, **

$({\rm O^{\prime}N}.2,\mathbb{Z}_{31}{:}\mathbb{Z}_{30},31)$, $({\rm He},\mathbb{Z}_{5}^{2}{:}4{\rm A}_{4},5)$, $({\rm Co}_{1},\mathbb{Z}_{7}^{2}{:}(3{\times}2{\rm A}_{4}),7)$,

$({\rm Ly},\mathbb{Z}_{37}{:}\mathbb{Z}_{18},37)$**, $({\rm Ly},\mathbb{Z}_{67}{:}\mathbb{Z}_{22},67)$, $({\rm Fi}_{24}^{\prime},\mathbb{Z}_{29}{:}\mathbb{Z}_{14},29)$, **

$({\rm B},\mathbb{Z}_{13}{:}\mathbb{Z}_{12}{\times}{\rm S}_{4},13)$**, $({\rm B},\mathbb{Z}_{19}{:}\mathbb{Z}_{18}{\times}\mathbb{Z}_{2},19)$, $({\rm B},\mathbb{Z}_{23}{:}\mathbb{Z}_{11}{\times}2,23)$, **

$({\rm M},\mathbb{Z}_{23}{:}\mathbb{Z}_{11}{\times}{\rm S}_{4},23)$**, $({\rm M},(\mathbb{Z}_{29}{:}\mathbb{Z}_{14}{\times}3).2,29)$, $({\rm M},\mathbb{Z}_{31}{:}\mathbb{Z}_{15}{\times}{\rm S}_{3},31)$, **

$({\rm M},\mathbb{Z}_{41}{:}\mathbb{Z}_{40},41)$, $({\rm M},\mathbb{Z}_{47}{:}\mathbb{Z}_{23}{\times}2,47)$.

Recall that $p^{f}$ is a divisor of $|X|$ and $r$ is a primitive prime divisor of $p^{f}-1$. Searching all possible pairs $(p^{f},r)$, we get the following table:

[TABLE]

Recalling that $G_{\{u,v\}}=X_{\{u,v\}}.(G/X)$, we have $2|G_{uv}|=|G_{\{u,v\}}|=|X_{\{u,v\}}||G{:}X|=2|X_{uv}||G{:}X|$, and so $|G_{uv}|=|X_{uv}||G{:}X|$. Since $G_{v}$ is $2$-transitive on ${\it\Gamma}(v)$, we know that $(p^{f}-1)$ is a divisor of $|G_{uv}|=|X_{uv}||G{:}X|$. It follows that $(X,X_{\{u,v\}},r,p^{f})$ is one of

• $({\rm J}_{1},\mathbb{Z}_{7}{:}\mathbb{Z}_{6},7,2^{3})$, $({\rm J}_{2},\mathbb{Z}_{5}^{2}{:}{\rm D}_{12},5,2^{4})$, $({\rm Co}_{1},\mathbb{Z}_{7}^{2}{:}(3{\times}2{\rm A}_{4}),7,2^{3})$,
• $({\rm O^{\prime}N}.2,\mathbb{Z}_{31}{:}\mathbb{Z}_{30},31,2^{5})$, $({\rm He},\mathbb{Z}_{5}^{2}{:}4{\rm A}_{4},5,2^{4})$, $({\rm B},\mathbb{Z}_{13}{:}\mathbb{Z}_{12}{\times}{\rm S}_{4},13,3^{3})$,
• $({\rm B},\mathbb{Z}_{13}{:}\mathbb{Z}_{12}{\times}{\rm S}_{4},13,5^{4})$, $({\rm M},\mathbb{Z}_{31}{:}\mathbb{Z}_{15}{\times}{\rm S}_{3},31,2^{5})$.

For $({\rm Co}_{1},\mathbb{Z}_{7}^{2}{:}(3{\times}2{\rm A}_{4}),7,2^{3})$, we have $X_{uv}\lesssim{\rm\Gamma L}_{1}(2^{3}){\times}{\rm\Gamma L}_{1}(2^{3})$, yielding that $X_{uv}$ has odd order, a contradiction. Similarly, for $({\rm B},\mathbb{Z}_{13}{:}\mathbb{Z}_{12}{\times}{\rm S}_{4},13,3^{3})$, the order of $X_{uv}$ is indivisible by $2^{4}$, a contradiction; for $({\rm M},\mathbb{Z}_{31}{:}\mathbb{Z}_{15}{\times}{\rm S}_{3},31,2^{5})$, the order of $X_{uv}$ is indivisible by $3$, also a contradiction. For $({\rm He},\mathbb{Z}_{5}^{2}{:}4{\rm A}_{4},5,2^{4})$, the order of $X_{uv}$ is divisible by $2^{3}\cdot 3\cdot 5^{2}$ and, since $p^{f}=2^{4}$, the order of $X_{u}$ is divisible by $2^{7}\cdot 3\cdot 5^{2}$; however, ${\rm He}$ has no soluble subgroups of order divisible by $2^{7}\cdot 3\cdot 5^{2}$, a contradiction. Similarly, the quadruple $({\rm O^{\prime}N}.2,\mathbb{Z}_{31}{:}\mathbb{Z}_{30},31,2^{5})$ is excluded as ${\rm O^{\prime}N}.2$ has no soluble subgroups with order divisible by $2^{5}\cdot 31$. (Note that $G_{v}$ is soluble.) By the Altas ****[3]****, ${\rm J}_{2}$ has no subgroups with order divisible by $2^{4}\cdot 5^{2}$, and then $({\rm J}_{2},\mathbb{Z}_{5}^{2}{:}{\rm D}_{12},5,2^{4})$ is excluded. By the Altas ****[3]**** and ****[33**, Theorem 2.1]****, ${\rm B}$ has no subgroups with order divisible by $3^{2}\cdot 5^{4}\cdot 13$, and then $({\rm B},\mathbb{Z}_{13}{:}\mathbb{Z}_{12}{\times}{\rm S}_{4},13,5^{4})$ is excluded. Then $({\rm J}_{1},\mathbb{Z}_{7}{:}\mathbb{Z}_{6},7,2^{3})$ is left, which gives $X_{v}\cong\mathbb{Z}_{2}^{3}{:}\mathbb{Z}_{7}{:}\mathbb{Z}_{3}$, $d=p^{f}=8$ and $G=X$. **$\square$

Finally, we summarize the argument for proving Theorem 1.1 as follows.

Proof of Theorem 1.1. Clearly, each $(G,G_{v},G_{\{u,v\}})$ in Table 1 gives a $G$-edge-primitive graph ${\sf Cos}(G,G_{v},G_{\{u,v\}})$. It is not difficult to check the $2$-arc-transitivity of $G$ on ${\sf Cos}(G,G_{v},G_{\{u,v\}})$, we omit the details.

**Now let $G$ and ${\it\Gamma}=(V,E)$ satisfy the assumptions in Theorem 1.1. If $G_{v}^{{\it\Gamma}(v)}$ is an almost simple $2$-transitive group then, by Lemmas 3.3, 4.1-4.3, the triple $(G,G_{v},G_{\{u,v\}})$ is listed in Table 1. Assume that $G_{v}^{{\it\Gamma}(v)}$ is a soluble $2$-transitive group of degree $d=p^{f}$, where $p$ is a prime. Then either $G_{v}^{{\it\Gamma}(v)}\leq{\rm\Gamma L}_{1}(p^{f})$, or $G_{v}^{{\it\Gamma}(v)}$ has a normal subgroup ${\rm SL}_{2}(3)$ or $2_{+}^{1+4}$. For the latter case, the triple $(G,G_{v},G_{\{u,v\}})$ is known by Lemmas 5.2 and 5.3. Let $G_{v}^{{\it\Gamma}(v)}\leq{\rm\Gamma L}_{1}(p^{f})$ and consider the primitive prime divisors of $p^{f}-1$. If $p^{f}-1$ has no primitive prime divisor then, by Lemmas 5.2, 5.5 and Corollary 5.7, $(G,G_{v},G_{\{u,v\}})$ is listed in Table 1. If $p^{f}-1$ has primitive prime divisors, then $(G,G_{v},G_{\{u,v\}})$ is known by Lemmas 5.2 and 5.10. This completes the proof. **$\square$