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

Note that $\textbf {O}_r(G_v)G_v^{[1]}/G_v^{[1]}\cong \textbf {O}_r(G_v)/(\textbf {O}_r(G_v)\cap G_v^{[1]})$ . Clearly, $\textbf {O}_r(G_v)\cap G_v^{[1]}\le \textbf {O}_r(G_v^{[1]})$ and we have $\textbf {O}_r(G_v)\cap G_v^{[1]}=1$ . It follows that $\textbf {O}_r(G_v)\cong \textbf {O}_r(G_v)G_v^{[1]}/G_v^{[1]}\unlhd G_v/G_v^{[1]}\cong {\mathrm{G}}_v^{{{\Gamma}}(v)}$ . Thus, $\textbf {O}_r(G_v)$ is isomorphic to a normal r-subgroup of ${\mathrm{G}}_v^{{{\Gamma}}(v)}$ . This implies that either $\textbf {O}_r(G_v)=1$ , or ${\mathrm{G}}_v^{{{\Gamma}}(v)}$ is an affine $2$ -transitive group of degree $r^e$ for some e. Thus, the lemma follows.

Proof. Assume that $P\ne 1$ is a normal Sylow subgroup of H. Clearly, P is not normal in G. Take a Sylow subgroup Q of G with $P\le Q$ . Then $H\le \langle \textbf {N}_Q(P),H\rangle \le \textbf {N}_G(P)\ne G$ . Since H is maximal in G, we have $H=\langle \textbf {N}_Q(P),H\rangle $ and so $\textbf {N}_Q(P)\le H$ . It follows that $\textbf {N}_Q(P)=P$ and hence $P=Q$ . Then the lemma follows.

Hypothesis 3.1. Let ${{\Gamma}}=(V,E)$ be a G-edge-primitive graph of valency $d\ge 6$ and $\{u,v\}\in E$ , where G is an almost simple group with socle T. Assume that:

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

2. (2) G has a normal subgroup X such that ${\mathrm{soc}}(X)=T$ , $X_{\{u,v\}}$ is maximal in X and $(X,X_{\{u,v\}})$ is one of the pairs $(G_0,H_0)$ listed in [Reference Li and Zhang17, Tables 14–20].

Proof. Suppose that Lemma 3.3(4) or (5) holds. By Theorem 2.4, $X_{uv}^{[1]}=1$ . Then $X_v=X_v^{[1]}.X_v^{{{\Gamma}}(v)}$ , $X_v^{[1]}\cong (X_v^{[1]})^{{{\Gamma}}(u)}\unlhd (X_u^{{{\Gamma}}(u)})_v\cong (X_v^{{{\Gamma}}(v)})_u$ and $X_{uv}\lesssim (X_v^{{{\Gamma}}(v)})_u{\times } (X_u^{{{\Gamma}}(u)})_v$ . Set $q=p^f$ with p a prime. Then the pair $(X_v^{{{\Gamma}}(v)}, (X_v^{{{\Gamma}}(v)})_u)$ is given as follows:

$$ \begin{align*} \begin{array}{lll} X_v^{{{\Gamma}}(v)} & (X_v^{{{\Gamma}}(v)})_u & \\ {\mathrm{Sz}}(q).e & p^{f{+}f}{:}(q-1).e& e\mbox{ a divisor of } f,\,p=2,\, \mbox{odd } f>1 \\ {\mathrm{Ree}}(q).e & p^{f{+}2f}{:}(q-1).e &e\mbox{ a divisor of } f,\,p=3,\, \mbox{odd } f>1. \end{array} \end{align*} $$

In particular, $\textbf {O}_p(X_{\{u,v\}})$ is not abelian.

We next show that none of the pairs $(G_0,H_0)$ in [Reference Li and Zhang17, Tables 14–20] gives a desired pair $(X,X_{\{u,v\}})$ . Since $\textbf {O}_p(X_{\{u,v\}})$ is nonabelian, those pairs $(G_0,H_0)$ with $\textbf {O}_p(H_0)$ abelian are not in our consideration. In particular, ${\mathrm{soc}}(X)$ 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 ${\mathrm{soc}}(X_v^{{{\Gamma}}(v)})={\mathrm{Ree}}(q)$ . Then $p=3$ , $\textbf {O}_3(X_{\{u,v\}})$ is nonabelian and of order $3^{3f}$ , $3^{4f}$ , $3^{5f}$ or $3^{6f}$ and $|X_{\{u,v\}}|$ is a divisor of $2\cdot 3^{6f}\cdot (q-1)^2f^2$ and divisible by $2(q-1)$ . Checking the orders of those $H_0$ given in [Reference Li and Zhang17, Table 15], we conclude that ${\mathrm{soc}}(X)$ is not a sporadic simple group.

Suppose that ${\mathrm{soc}}(X)$ is a simple exceptional group of Lie type. Checking [Reference Li and Zhang17, Table 20], we conclude that $(X,X_{\{u,v\}})$ is one of the pairs $({\mathrm{Ree}}(3^t),[3^{3t}]{:}\mathbb Z_{3^t-1})$ and $({\mathrm{G}}_2(3^t).\mathbb Z_{2^{l+1}}, [3^{6t}]{:}\mathbb Z_{3^t-1}^2. \mathbb Z_{2^{l+1}})$ , where $2^l$ is the $2$ -part of t. Recall that $|X_{\{u,v\}}|$ is a divisor of $2\cdot 3^{6f}\cdot (q-1)^2f^2$ and divisible by $2(q-1)$ . It follows that $f=t$ , $X={\mathrm{G}}_2(q).\mathbb Z_{2^{l+1}}$ and $X_{\{u,v\}}\cong [q^6]{:}\mathbb Z_{q-1}^2.\mathbb Z_{2^{l+1}}$ . This implies that $X_v^{[1]}\ne 1$ ; in fact, $|\textbf {O}_3(X_v^{[1]})|=q^3$ . Thus, $\textbf {O}_3(X_v)\ne 1$ and $X_v$ has a quotient ${\mathrm{Ree}}(q).e$ . Checking the maximal subgroups of ${\mathrm{G}}_2(q).\mathbb Z_{2^{l+1}}$ (refer to [Reference Kleidman15, Theorems A and B]) we conclude that ${\mathrm{G}}_2(q).\mathbb Z_{2^{l+1}}$ has no maximal subgroup containing such $X_v$ as a subgroup, which is a contradiction.

Suppose that ${\mathrm{soc}}(X)$ 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 $\textbf {O}_3(X_{\{u,v\}})$ is nonabelian and of order $3^{3f}$ , $3^{4f}$ , $3^{5f}$ or $3^{6f}$ . Checking the groups $H_0$ given in [Reference Li and Zhang17, Tables 16–19], we conclude that ${\mathrm{soc}}(X)={\mathrm{PSL}}_n(r^t)$ or ${\mathrm{PSU}}_n(r^t)$ , where $n\in \{3,4\}$ . Take a maximal subgroup M of X such that $X_v\le M$ . Then M has a simple section (that is, a quotient of some subgroup) ${\mathrm{Ree}}(q)$ . Recall that $q>3$ . Checking Tables 8.3–8.6 and 8.8–8.11 given in [Reference Bray, Holt and Roney-Dougal1], we conclude that none of ${\mathrm{PSL}}_3(r^t)$ , ${\mathrm{PSL}}_4(r^t)$ , ${\mathrm{PSU}}_3(r^t)$ and ${\mathrm{PSU}}_4(r^t)$ has such maximal subgroups, which is a contradiction.

Case 2. Suppose that ${\mathrm{soc}}(X_v^{{{\Gamma}}(v)})={\mathrm{Sz}}(q)$ . Then $q=2^f$ , $|\textbf {O}_2(X_{\{u,v\}})|=2^{2f}a$ , $2^{3f}a$ or $2^{4f}a$ , where $f>1$ is odd and $a=1$ or $2$ . Noting that $|X_{\{u,v\}}|$ is divisible by $2(2^f-1)$ , 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 [Reference Li and Zhang17, Tables 15–20] that either $(X,X_{\{u,v\}})$ is one of $({}^2{\mathrm{F}}_4(2)', [2^9]{:}5{:}4)$ , $({\mathrm{Sz}}(2^t),[2^{2t}]{:}\mathbb Z_{2^t-1})$ and $({\mathrm{PSp}}_4(2^t).\mathbb Z_{2^{l+1}}, [2^{4t}]{:}\mathbb Z_{2^t-1}^2.\mathbb Z_{2^{l+1}})$ , or ${\mathrm{soc}}(X)$ is one of ${\mathrm{PSL}}_n(r^t)$ and ${\mathrm{PSU}}_n(r^t)$ , 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 $7$ , which is a contradiction. Checking the maximal subgroups of ${\mathrm{soc}}(X)$ (refer to [Reference Bray, Holt and Roney-Dougal1, Tables 8.3–8.6 and 8.8–8.14]), the groups ${\mathrm{PSL}}_3(r^t)$ , ${\mathrm{PSU}}_3(r^t)$ , ${\mathrm{PSL}}_4(r^t)$ and ${\mathrm{PSU}}_4(r^t)$ are excluded as they have no maximal subgroup with a simple section ${\mathrm{Sz}}(q)$ . Thus, $(X,X_{\{u,v\}})=({\mathrm{PSp}}_4(2^t).\mathbb Z_{2^{l+1}}, [2^{4t}]{:}\mathbb Z_{2^t-1}^2.\mathbb Z_{2^{l+1}})$ or $({\mathrm{Sz}}(2^t),[2^{2t}]{:}\mathbb Z_{2^t-1})$ . Note that $|X_{\{u,v\}}|$ is a divisor of $2\cdot 2^{4f}\cdot (q-1)^2f^2$ and is divisible by $2^{2f+1}(2^f-1)$ . It follows that $X={\mathrm{PSp}}_4(q).\mathbb Z_{2^{l+1}}$ and $X_v^{[1]}\cong [q^2]{:}\mathbb Z_{q-1}$ . However, by [Reference Bray, Holt and Roney-Dougal1, Table 8.14], ${\mathrm{PSp}}_4(q).\mathbb Z_{2^{l+1}}$ has no maximal subgroup containing $[q^2]{:}\mathbb Z_{q-1}.{\mathrm{Sz}}(q)$ , which is a contradiction. This completes the proof.

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

Suppose that $X_v^{{{\Gamma}}(v)}={\mathrm{PSL}}_3(2)$ . Then $(X_v^{{{\Gamma}}(v)})_u\cong {\mathrm{S}}_4$ and thus $X_v^{[1]}$ and $X_{\{u,v\}}$ are given as follows:

$$ \begin{align*} \begin{array}{lllll} X_v^{[1]}& 1 & 2^2 &{\mathrm{A}}_4&{\mathrm{S}}_4\\ X_{\{u,v\}}& 2^2{:}{\mathrm{S}}_3.2& 2^4.{\mathrm{S}}_3.2&2^4{:}3^2.[4]& 2^4{:}{\mathrm{S}}_3^2.2. \end{array} \end{align*} $$

In particular, $2^2\le |\textbf {O}_2(X_{\{u,v\}})|\le 2^5$ . Check all possible pairs $(X,X_{\{u,v\}})$ in [Reference Li and Zhang17, Tables 14–20]. Noting that ${\mathrm{A}}_8\cong {\mathrm{PSL}}_4(2)$ and ${\mathrm{PSU}}_4(2)\cong {\mathrm{PSp}}_4(3)$ , we conclude that $X\cong {\mathrm{A}}_8$ , $X_{\{u,v\}}\cong 2^4{:}{\mathrm{S}}_3^2$ and $X_v^{[1]}\cong {\mathrm{A}}_4$ ; or $X= {\mathrm{M}}_{12}$ with $X_{\{u,v\}}\cong 2_{+}^{1{+}4}{:}{\mathrm{S}}_3$ ; or $X\cong {\mathrm{PSU}}_4(2)$ where $X_{\{u,v\}}\cong 2{\mathrm{A}}_4^2.2$ . The group ${\mathrm{A}}_8$ is excluded as it has no subgroup of the form of $X_v^{[1]}.{\mathrm{PSL}}_3(2)$ . The groups ${\mathrm{M}}_{12}$ and ${\mathrm{PSU}}_4(2)$ are excluded as their orders are not divisible by $d=7$ .

Suppose that $X_v^{{{\Gamma}}(v)}={\mathrm{PSL}}_3(3)$ . Then $(X_v^{{{\Gamma}}(v)})_u\cong 3^2{:}2{\mathrm{S}}_4$ . Thus, $X_v^{[1]}$ and $X_{\{u,v\}}$ are given as follows:

$$ \begin{align*} \begin{array}{cllllll} X_v^{[1]}& 1 & 3^2 &3^2{:}2&3^2{\mathrm{Q}}_8& 3^2{:}2{\mathrm{A}}_4& 3^2{:}2{\mathrm{S}}_4\\ X_{\{u,v\}}& 3^2{:}2{\mathrm{S}}_4{.}2& 3^4{:}2{\mathrm{S}}_4{.}2& 3^4{:}([4].{\mathrm{S}}_4){.}2&3^4{:}{\mathrm{Q}}_8^2{.}{\mathrm{S}}_3{.}2& 3^4{:}(2{\mathrm{A}}_4)^2{.}[4] &3^4{:}(2{\mathrm{S}}_4)^2{.}2. \end{array} \end{align*} $$

Note that $\textbf {O}_3(X_{\{u,v\}})\cong 3^2$ or $3^4$ . Checking the possible pairs $(X,X_{\{u,v\}})$ , we have $X_{\{u,v\}}\cong 3^4{:}2^3{.}{\mathrm{S}}_4$ and $X= {\mathrm{A}}_{12}$ or ${\mathrm {P\Omega }}^{+}_8(2)$ ; in this case, $d=13$ is not a divisor of $|X|$ , which is 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 shown in Table 4.

Table 4 Edge-stabilizers with $d=7$ or 13.

Suppose that $p=2$ . Then $|X_{\{u,v\}}|$ is divisible by $9$ if and only if $|\textbf {O}_2(X_{\{u,v\}})|\ge 8$ , and $\textbf {O}_2(X_{\{u,v\}})$ contains no elements of order 8 unless $|\textbf {O}_2(X_{\{u,v\}})|\ge 2^{22}$ . Check the pairs $(G_0,H_0)$ given in [Reference Li and Zhang17, Tables 14–20] by estimating $|H_0|$ and $|\textbf {O}_2(H_0)|$ . We conclude that one of the following holds:

1. (i) $X={\mathrm{PSL}}_4(2).2\cong {\mathrm{S}}_8$ and $X_{\{u,v\}}= 2^4{:}{\mathrm{S}}_4$ ;

2. (ii) $X= {\mathrm{PSL}}_5(2).2$ and $X_{\{u,v\}}= [2^8]{.}{\mathrm{S}}_3^2.2$ ;

3. (iii) $X= {\mathrm{F}}_4(2).2$ and $X_{\{u,v\}}= [2^{22}]{.}{\mathrm{S}}_3^2.2$ ;

4. (iv) ${\mathrm{soc}}(X)={\mathrm{PSL}}_3(4)$ and $|\textbf {O}_2(X_{\{u,v\}})|=2^6$ ;

5. (v) $X= {\mathrm{PSU}}_4(3).2_3$ and $|\textbf {O}_2(X_{\{u,v\}})|=2^7$ ;

6. (vi) $X= {\mathrm{He}}.2$ and $X_{\{u,v\}}= [2^8]{:}{\mathrm{S}}_3^2.2$ .

Case (iv) yields that $X_v\cong 2^3{:}{\mathrm{SL}}_3(2)$ or $2^4{:}{\mathrm{SL}}_3(2)$ ; however, X has no such subgroup by the Atlas [Reference Conway, Norton, Parker and Wilson3]. Similarly, cases (v) and (vi) are excluded. For (i), $G=X$ and ${{\Gamma}}$ is (isomorphic to) the point–plane incidence graph of the projective geometry ${\mathrm{PG}}(3,2)$ . For (ii), $G=X$ and ${{\Gamma}}$ is (isomorphic to) the line–plane incidence graph of the projective geometry ${\mathrm{PG}}(4,2)$ . If (iii) holds, then $G=X$ and ${{\Gamma}}$ is the line–plane incidence graph of the metasymplectic space associated with ${\mathrm{F}}_4(2)$ ; see [Reference Weiss30].

Now let $p=3$ . Then $|\textbf {O}_3(X_{\{u,v\}})|=3^5$ or $3^8$ , and $X_{\{u,v\}}$ has no normal Sylow subgroup. Checking all possible pairs $(X,X_{\{u,v\}})$ in [Reference Li and Zhang17, Tables 14–20], we know that $(X,X_{\{u,v\}})$ is one of the following pairs:

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

Note that $\textbf {O}_3(X_v)\le \textbf {O}_3(X_{\{u,v\}})$ . Then, for the first pair, $\textbf {O}_3(X_{\{u,v\}})\cong \mathbb Z_9^4$ has no subgroup isomorphic to $\mathbb Z_3^6$ , which is impossible. For the second pair, $G=X$ and ${{\Gamma}}$ is (isomorphic to) the line–plane incidence graph of the projective geometry ${\mathrm{PG}}(4,3)$ . The last pair implies that $X\cong {\mathrm{PGL}}_4(3)$ , $G=X$ or $X.2$ , and ${{\Gamma}}$ is (isomorphic to) the line–plane incidence graph of the projective geometry ${\mathrm{PG}}(3,3)$ . This completes the proof.

Proof. Let $X_v^{{{\Gamma}}(v)}= {\mathrm{PSL}}_2(q).[o]$ and $q=p^f> 4$ , where p is a prime and o is a divisor of $(2,p-1)f$ . Note that ${{\Gamma}}$ is $(X,2)$ -arc-transitive; see Lemma 3.3. By Theorem 2.4, if $X_{uv}^{[1]}\ne 1$ , then ${{\Gamma}}$ is $(X,4)$ -arc-transitive. Thus, we assume next that $X_{uv}^{[1]}=1$ and then Lemma 3.4 works.

Note that $X_v=X_v^{[1]}.X_v^{{{\Gamma}}(v)}$ ,

$$\begin{align*}X_v^{[1]}\cong (X_v^{[1]})^{{{\Gamma}}(u)}\unlhd (X_u^{{{\Gamma}}(u)})_v\cong (X_v^{{{\Gamma}}(v)})_u= p^f{:}({q-1})/{(2,q-1)}{.}[o]\end{align*}$$

and $X_{uv}\lesssim (X_v^{{{\Gamma}}(v)})_u{\times } (X_u^{{{\Gamma}}(u)})_v$ . We have $\textbf {O}_p(X_{\{u,v\}})=\mathbb Z_p^{if}.a$ , where $i\in \{1,2\}$ and a is a divisor of $(2,p)$ . It is easily shown that $i=2$ if and only if $\textbf {O}_p(X_v^{[1]})=\mathbb Z_p^f$ . Combining with Lemma 3.4, we need only consider those pairs $(G_0,H_0)$ in [Reference Li and Zhang17, Tables 14–20] that satisfy:

1. (a) $\textbf {O}_p(H_0)=\mathbb Z_p^{if}.a$ , where $i\in \{1,2\}$ $a\mid (2,p)$ ; $|{\mathrm{Fit}}(H_0):\textbf {O}_p(H_0)|\le 2$ ; $G_0$ 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 ${\mathrm{PSL}}_2(q)$ ;

2. (b) $|H_0{:}\textbf {O}_p(H_0)|$ is a divisor of $2(q-1)^2f^2$ and divisible by $q-1$ ; if $i=1$ , then $|H_0{:}\textbf {O}_p(H_0)|$ is a divisor of $2(q-1)f$ .

Case 1. Assume that ${\mathrm{soc}}(X)$ is an alternating group. Using [Reference Li and Zhang17, Table 14], we have $G=X={\mathrm{S}}_p$ and $X_{\{u,v\}}\cong \mathbb Z_p{:}\mathbb Z_{p-1}$ , where $p\in \{7,11,17,23\}$ . Then $X_v={\mathrm{PSL}}_2(p)$ and $d=p{+}1$ . In particular, ${{\Gamma}}$ is a bipartite graph with two parts being the orbits of ${\mathrm{A}}_p$ on the vertex set V. For $p=17$ or $23$ , the group ${\mathrm{PSL}}_2(p)$ has no transitive permutation representation of degree p and thus it cannot occur as a subgroup of ${\mathrm{S}}_p$ . Therefore, $p=7$ or $11$ , and G, X and $X_{\{u,v\}}$ are as listed in Table 5. In fact, $X_{uv}$ and $X_{\{u,v\}}$ are the normalizers of some Sylow p-subgroup in ${\mathrm{PSL}}_2(p)$ and ${\mathrm{S}}_p$ , respectively. (Note that ${\mathrm{A}}_7$ can be embedded in ${\mathrm{PSL}}_4(2)$ acting on the projective points or the hyperplanes of the projective geometry ${\mathrm{PG}}(3,2)$ ; see [Reference Liebeck, Praeger and Saxl18, Table III], for example. Then, for $p=7$ , it is easily shown that the resulting graph is the point–plane nonincidence graph of ${\mathrm{PG}}(3,2)$ .)

Case 2. Assume that ${\mathrm{soc}}(X)$ is a simple sporadic group. By [Reference Li and Zhang17, Table 15], with the restrictions (a) and (b), the only pairs $(G_0,H_0)$ are listed as follows:

$$ \begin{align*}\begin{array}{l} ({\rm M}_{11}, 3^2{:}{\rm Q}_8.2),\, ({\rm J}_1,\mathbb Z_{11}{:}\mathbb Z_{10}),\, ({\rm J}_1,\mathbb Z_{7}{:}\mathbb Z_{6}),\, ({\rm J}_3.2,\mathbb Z_{19}{:}\mathbb Z_{18}),\, ({\rm J}_4,\mathbb Z_{29}{:}\mathbb Z_{28}),\\ ({\rm O'N}.2,\mathbb Z_{31}{:}\mathbb Z_{30}),\, ({\rm B},\mathbb Z_{19}{:}\mathbb Z_{18}{\times} \mathbb Z_2),\, ({\rm B},\mathbb Z_{23}{:}\mathbb Z_{11}{\times} \mathbb Z_2),\\ ({\rm M},\mathbb Z_{41}{:}\mathbb Z_{40}), ({\rm M},\mathbb Z_{47}{:}\mathbb Z_{23}{\times} \mathbb Z_2).\end{array}\end{align*} $$

In particular, $\textbf {O}_p(H_0)$ is a Sylow p-subgroup of $G_0$ . This yields that $X_v^{[1]}=1$ and so ${\mathrm{soc}}(X_v)={\mathrm{PSL}}_2(p^f)$ .

If $(X,X_{\{u,v\}})$ is one of $({\mathrm{J}}_1,\mathbb Z_{7}{:}\mathbb Z_{6})$ , $({\mathrm{J}}_4,\mathbb Z_{29}{:}\mathbb Z_{28})$ and $({\mathrm{M}},\mathbb Z_{47}{:}\mathbb Z_{23}{\times } \mathbb Z_2)$ , then $X_v= {\mathrm{PSL}}_2(p)$ for $p=7$ , $29$ and $47$ , respectively; however, by the Atlas [Reference Conway, Norton, Parker and Wilson3] and [Reference Wilson36, Tables 5.6 and 5.11], X has no subgroup ${\mathrm{PSL}}_2(p)$ , which is a contradiction. Thus, G, X and $X_{\{u,v\}}$ are as listed in Table 5. (Note that the Monster ${\mathrm{M}}$ has a maximal subgroup ${\mathrm{PSL}}_2(41)$ by [Reference Norton and Wilson23].)

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

Suppose that $r=p$ . Then, by (a), either $\textbf {O}_p(H_0)$ 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 [Reference Li and Zhang17, Tables 16–20] such that $\textbf {O}_p(H_0)$ is abelian. Thus, we have $r=p=2$ . Recalling that $p^f>4$ and $|H_0/\textbf {O}_p(H_0)|$ is divisible by $2^f-1$ , it follows from Lemma 2.6 that $H_0/\textbf {O}_p(H_0)$ is not a $\{2,3\}$ -group. Checking those $H_0$ given in [Reference Li and Zhang17, Tables 16–20], we conclude that $(G_0,H_0)$ is one of the following pairs:

$$ \begin{align*}\begin{array}{l} ({\mathrm{PSL}}_2(2^t), \mathbb Z_2^t{:}\mathbb Z_{2^t-1}),\,({\mathrm{PSL}}_3(2^t), [2^{3t}]{:}[{(2^t-1)^2}/{(3,2^t-1)}].2),\\ ({\mathrm{PSU}}_3(2^t), [2^{3t}]{:}\mathbb Z_{({2^{2t}-1})/{(3,2^t+1)}}),\\ ({\mathrm{PSp}}_4(2^t).\mathbb Z_{2^{l+1}}, [2^{4t}]{:}\mathbb Z_{2^{t}-1}^2.\mathbb Z_{2^{l+1}}), \mbox{ where }2^l \mbox{ is the }2\mbox{-part of }t,\\ ({\mathrm{Sz}}(2^t), [2^{2t}]{:}\mathbb Z_{2^{t}-1}), ({}^3{\mathrm{D}}_4(2), [2^{11}]{:}(\mathbb Z_7\times{\mathrm{S}}_3)),\, ({}^2{\mathrm{F}}_4(2)', [2^9]{:}5{:}4). \end{array}\end{align*} $$

First, the pair $({\mathrm{Sz}}(2^t), [2^{2t}]{:}\mathbb Z_{2^{t}-1})$ is excluded as ${\mathrm{Sz}}(2^t)$ has no subgroup with a section ${\mathrm{PSL}}_2(2^f)$ . For the last two pairs, we have $f=5$ and $4$ , respectively, which yields that $2^f-1$ is not a divisor of $|H_0|$ , which is a contradiction. For the three pairs after the first one, we have $t<f$ and thus $G_0$ has no maximal subgroup with a section ${\mathrm{PSL}}_2(2^f)$ , which is a contradiction. Suppose finally that $(X,X_{\{u,v\}})=({\mathrm{PSL}}_2(2^t), \mathbb Z_2^t{:}\mathbb Z_{2^t-1})$ . Then $3\le f<t\le 2f+1$ . 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 $\textbf {O}_2(X_{\{u,v\}})=2^{2f}$ , yielding $|\textbf {O}_2(X_v^{[1]})|=2^f$ . Thus, $\textbf {O}_2(X_v)\ne 1$ and $X_v$ has a section ${\mathrm{PSL}}_2(2^f)$ . Check the subgroups of ${\mathrm{PSL}}_2(2^{2f})$ ; refer to [Reference Huppert12, Hauptsatz II.8.27]. We conclude that ${\mathrm{PSL}}_2(2^{2f})$ has no subgroup isomorphic to $X_v$ , which is a contradiction.

We assume that $r\ne p$ from now on.

Subcase 3.1. We first deal with those pairs $(G_0,H_0)$ such that $H_0$ is included in some infinite families in [Reference Li and Zhang17, Tables 16–20]. Note that $r\ne p$ and we consider only those $H_0$ having subgroups of index 2. It follows that either $H_0/{\mathrm{Fit}}(H_0)$ is a $\{2,3\}$ -group, or $G_0={\mathrm{E}}_8(q')$ and $|H_0|=30(q^{\prime 8}{\pm } q^{\prime 7}{\mp } q^{\prime 5}-q^{\prime 4}{\mp } q^{\prime 3}{\pm } q'{+}1)$ , where $q'=r^t$ . Suppose the latter case occurs. It is easily shown that $q^{\prime 8}{\pm } q^{\prime 7}{\mp } q^{\prime 5}-q^{\prime 4}{\mp } q^{\prime 3}{\pm } q'{+}1$ is divisible by some primitive prime divisor s of $q^{\prime 15}-1$ or of $q^{\prime 30}-1$ . Noting that $s\ge 17$ , we know that $H_0$ has a normal cyclic Sylow s-subgroup. It follows from (a) that $17\le p=s=q^{\prime 8}{\pm } q^{\prime 7}{\mp } q^{\prime 5}-q^{\prime 4}{\mp } q^{\prime 3}{\pm } q'{+}1$ . In particular, $\textbf {O}_p(H_0)=\mathbb Z_p$ and $f=1$ . By (b), $|H_0|$ is divisible by $p-1$ and then $30$ is divisible by $p-1$ . This implies that $30=p-1=q^{\prime 8}{\pm } q^{\prime 7}{\mp } q^{\prime 5}-q^{\prime 4}{\mp } q^{\prime 3}{\pm } q'$ , which is impossible. Therefore, $H_0/{\mathrm{Fit}}(H_0)$ is a $\{2,3\}$ -group.

By (a), ${\mathrm{Fit}}(H_0)$ 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 $(X_u^{{{\Gamma}}(u)})_v\cong (X_v^{{{\Gamma}}(v)})_u= p^f{:}({q-1})/{(2,q-1)}{.}[o]$ and $X_{uv}\lesssim (X_v^{{{\Gamma}}(v)})_u{\times } (X_u^{{{\Gamma}}(u)})_v$ , where o is a divisor of $(2,q-1)f$ . Then $X_{uv}/\textbf {O}_p(X_{uv})$ has an abelian Hall $2'$ -subgroup. Note that $X_{uv}\textbf {O}_p(X_{\{u,v\}})/\textbf {O}_p(X_{\{u,v\}})\cong X_{uv}/(\textbf {O}_p(X_{\{u,v\}}\cap X_{uv})=X_{uv}/\textbf {O}_p(X_{uv})$ and also $|X_{\{u,v\}}:X_{uv}\textbf {O}_p(X_{\{u,v\}})|\le 2$ . It follows that $X_{\{u,v\}}/\textbf {O}_p(X_{\{u,v\}})$ has an abelian Hall $2'$ -subgroup. Thus, as a possible candidate for $X_{\{u,v\}}$ , the quotient of $H_0$ over $\textbf {O}_p(H_0)$ has abelian Hall $2'$ -subgroups. In particular, $H_0/\textbf {O}_p(H_0)$ 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:

$$ \begin{align*}\def\arraystretch{1.18}\begin{array}{l} ({\mathrm{PSL}}_2(r^t), \mathbb Z_{({r^t\pm 1})/{(2,r^t-1)}}{:}\mathbb Z_2), \,({\mathrm{PSL}}_3(r^t), [{(r^t-1)^2}/{(3,r^t-1)}].{\mathrm{S}}_3),\\ ({\mathrm{PSU}}_3(r^t), [{(r^t+1)^2}/{(3,r^t+1)}].{\mathrm{S}}_3);\\ ({\mathrm{PSp}}_4(2^t).\mathbb Z_{2^{l+1}}, \mathbb Z_{2^t\pm 1}^2.[2^{l+4}]), \,({\mathrm{PSp}}_4(2^t).\mathbb Z_{2^{l+1}}, \mathbb Z_{2^{2t}+1}.[2^{l+3}]), \,t\ge 3;\\ ({\mathrm{Sz}}(2^t), \mathbb Z_{2^t-1}{:}\mathbb Z_2), \,({\mathrm{Sz}}(2^t), \mathbb Z_{2^t\pm \sqrt{2^{t+1}}+1}{:}\mathbb Z_4), \,t\ge 3; \\ ({\mathrm{Ree}}(3^t), \mathbb Z_{3^t\pm \sqrt{3^{t+1}}+1}{.}\mathbb Z_6), \,({\mathrm{Ree}}(3^t), \mathbb Z_{3^t+1}{.}\mathbb Z_6), \,t\ge 3;\\ ({\mathrm{G}}_2(3^t).\mathbb Z_{2^{l+1}}, \mathbb Z_{3^t\pm 1}^2{.}[3\cdot2^{l+3}]), \,({\mathrm{G}}_2(3^t).\mathbb Z_{2^{l+1}}, \mathbb Z_{3^{2t}\pm3^t+1}{.}[3\cdot2^{l+2}]), \,t\ge 2;\\ ({}^3{\mathrm{D}}_4(r^t), \mathbb Z_{r^{4t}-r^{2t}+1}{:}\mathbb Z_4), \,({}^2{\mathrm{F}}_4(2^t), \mathbb Z_{2^{2t}\pm \sqrt{2^{3t+1}}+2^{t}\pm \sqrt{2^{t+1}}+1}{.}\mathbb Z_{12}), \,t\ge 3; \\ ({\mathrm{F}}_4(2^t).\mathbb Z_{2^{l+1}}, \mathbb Z_{2^{4t}-2^{2t}+1}{.}[3\cdot2^{l+3}]), \,t\ge 2, \end{array}\end{align*} $$

where the power $2^l$ appearing means the $2$ -part of t. Recall that $|{\mathrm{Fit}}(H_0):\textbf {O}_p(H_0)|\le 2$ and $|H_0:\textbf {O}_p(H_0)|$ 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 $(G_0,H_0)=({\mathrm{PSL}}_3(r^t), [{(r^t-1)^2}/{(3,r^t-1)}].{\mathrm{S}}_3)$ . Considering the structures of ${\mathrm{Fit}}(H_0)$ and $\textbf {O}_p(H_0)$ , 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 ${\mathrm{PSL}}_2(q)$ is soluble, which is not the case. Assume that the former case holds. Then $|{\mathrm{S}}_3|$ 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 determined in a similar way; the details are omitted here. Eventually, we conclude that $(G_0,H_0, p,f)$ is one of $({\mathrm{PSL}}_2(19), {\mathrm{D}}_{20}, 5,1)$ , $({\mathrm{PSL}}_3(8), 7^2{:}{\mathrm{S}}_3,7,1)$ and $({\mathrm{Sz}}(8),\mathbb Z_5{:}\mathbb Z_4,5,1)$ . By the Atlas [Reference Conway, Norton, Parker and Wilson3], neither ${\mathrm{PSL}}_3(8)$ nor ${\mathrm{Sz}}(8)$ has subgroup with a section ${\mathrm{PSL}}_2(p)$ . Thus, in this case, G, X and $X_{\{u,v\}}$ are as given in Table 5.

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 observe that either $p=2$ or $H_0/\textbf {O}_p(H_0)$ is a $\{2,3\}$ -group. Recall that $r\ne p$ .

Suppose that $p=2$ . Recalling that $q=2^f>4$ , we have $f\ge 3$ . 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 $G_0={}^2{\mathrm{F}}_4(2)'$ and $H_0=[2^9]{:}5{:}4$ . Thus, $|\textbf {O}_2(H_0)|=2^9$ ; it follows from (a) that $f=4$ or $9$ and then $G_0$ has a section ${\mathrm{PSL}}_2(2^4)$ or ${\mathrm{PSL}}_2(2^9)$ , which is impossible by checking the (maximal) subgroups of ${}^2{\mathrm{F}}_4(2)'$ . Thus, $p>2$ and $H_0/\textbf {O}_p(H_0)$ is a $\{2,3\}$ -group; in particular, by (a), $\textbf {O}_p(H_0)=\mathbb Z_p^{if}$ for some $i\in \{1,2\}$ .

Suppose that $H_0$ has a section ${\mathrm{A}}_4$ . Then $H_0$ has no normal Sylow $3$ -subgroup. Further, $H_0$ has no quotient ${\mathrm{A}}_4$ as $H_0$ has a subgroup of index two. If $(3,(q-1)f)=1$ , then, by (b), we conclude that $p=3$ and $\textbf {O}_p(H_0)$ is the unique Sylow $3$ -subgroup of $H_0$ , which is a contradiction. Thus, $3$ is a divisor of $(q-1)f$ . Check those $H_0$ in [Reference Li and Zhang17, Tables 16–20] which have a section ${\mathrm{A}}_4$ and do not appear in Subcase 3.1. Recalling that $r\ne p>2$ and $\textbf {O}_p(H_0)=\mathbb Z_p^{if}$ , it follows that either $\textbf {O}_p(H_0)=\mathbb Z_3^2$ or $(G_0,H_0)=({\mathrm{F}}_4(2).4, \mathbb Z_7^2{:}(3\times {\mathrm{SL}}_2(3)).4)$ . Since $3$ is a divisor of $(q-1)f$ , we get $G_0={\mathrm{F}}_4(2).4$ 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 to ignore many cases without further inspection. Inspecting carefully the remaining pairs, the possible candidates for $(X,X_{\{u,v\}})$ are as follows:

$$ \begin{align*}\begin{array}{l} ({\mathrm{PGL}}_2(9), {\mathrm{D}}_{20}), \,({\mathrm{M}}_{10}, \mathbb Z_5{:}\mathbb Z_4), \,({\mathrm{PGL}}_2(11), {\mathrm{D}}_{20});\\ ({\mathrm{PSL}}_3(r), 3^2{:}{\mathrm{Q}}_8), \quad\mbox{where }r\equiv 4,7 \ \mod 9;\\ ({\mathrm{PSp}}_4(4).4, \mathbb Z_{17}{:}\mathbb Z_{16}), \,({\mathrm{PSp}}_4(4).4, 5^2{:}[2^5]);\\ ({\mathrm{PSU}}_3(r), 3^2{:}{\mathrm{Q}}_8), \quad\mbox{where }5<r\equiv 2,5 \ \mod 9;\\ ({\mathrm{PSU}}_3(2^t), 3^2{:}{\mathrm{Q}}_8), \quad\mbox{where }t \mbox{ is a prime no less than }5;\\ ({}^2{\mathrm{F}}_4(2), \mathbb Z_{13}{:}\mathbb Z_{12}). \end{array} \end{align*} $$

For the first three pairs, G, X and $X_{\{u,v\}}$ are easily determined and as given in Table 5. The pair $({\mathrm{PSp}}_4(4).4, \mathbb Z_{17}{:}\mathbb Z_{16})$ is excluded as ${\mathrm{PSp}}_4(4).4$ has no subgroup ${\mathrm{PSL}}_2(17)$ and the pair $({}^2{\mathrm{F}}_4(2), \mathbb Z_{13}{:}\mathbb Z_{12})$ is excluded as ${}^2{\mathrm{F}}_4(2)$ has no subgroup ${\mathrm{PSL}}_2(13)$ . Suppose that $X\cong {\mathrm{PSU}}_3(2^t)$ and $X_{\{u,v\}}\cong 3^2{:}{\mathrm{Q}}_8$ . Then we have $X_v\cong {\mathrm{PSL}}_2(9)$ ; however, by [Reference Bray, Holt and Roney-Dougal1, Tables 8.3 and 8.4], ${\mathrm{PSU}}_3(2^t)$ has no subgroup ${\mathrm{PSL}}_2(9)$ , which is a contradiction. Suppose that $(X,X_{\{u,v\}})=({\mathrm{PSp}}_4(4).4, 5^2{:}[2^5])$ . Then $X_v$ contains a Sylow $5$ -subgroup P of X and has a section ${\mathrm{PSL}}_2(5)$ or ${\mathrm{PSL}}_2(25)$ . By the information for ${\mathrm{PSp}}_4(4).4$ given in the Atlas [Reference Conway, Norton, Parker and Wilson3], we conclude that $X_v\le M\cong ({\mathrm{A}}_5\times {\mathrm{A}}_5){:}2^2<{\mathrm{PSp}}_4(4).2<{\mathrm{PSp}}_4(4).4$ . Note that $X_{uv}=5^2{:}[2^4]$ , which should be the normalizer of P in $X_v$ . Using GAP [28], computation shows that $|\textbf {N}_L(P)|\le 200$ for any maximal subgroup L of M with $P\le L$ . It follows that $X_v=M\cong ({\mathrm{A}}_5\times {\mathrm{A}}_5){:}2^2$ , yielding $d=|X_v:X_{uv}|=36\ne q+1$ , which is a contradiction.

Let $(X,X_{\{u,v\}})=({\mathrm{PSL}}_3(r), 3^2{:}{\mathrm{Q}}_8)$ . Then $X_{uv}\cong 3^2{:}4$ . It is easily shown that $p=3$ and $X_v\cong {\mathrm{PSL}}_2(9)$ . Since $r\equiv 4,7 \ \mod 9$ , we know that ${\mathrm{PSL}}_3(r)$ has a Sylow $3$ -subgroup $\mathbb Z_3^2$ . By [Reference Bray, Holt and Roney-Dougal1, Tables 8.3 and 8.4], ${\mathrm{PSL}}_3(r)$ has a subgroup ${\mathrm{PSL}}_2(9)$ if and only if $r\equiv 1,4 \ \mod 15$ . Thus, in this case, we have $r\equiv 11,14,29,41 \ \mod 45$ . For a subgroup ${\mathrm{PSL}}_2(9)$ of ${\mathrm{PSL}}_3(r)$ , taking a Sylow $3$ -subgroup Q of ${\mathrm{PSL}}_2(9)$ , the normalizers of Q in ${\mathrm{PSL}}_2(9)$ and ${\mathrm{PSL}}_3(r)$ are (isomorphic to) $3^2{:}4$ and $3^2{:}{\mathrm{Q}}_8$ , 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 as given in Table 5. Noting that $G=XG_{\{u,v\}}$ , we have $G_{\{u,v\}}/X_{\{u,v\}}\cong G/X\lesssim {\sf Out}({\mathrm{PSL}}_3(r))\cong {\mathrm{S}}_3$ 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 [Reference Bray, Holt and Roney-Dougal1, Table 8.4], $\textbf {N}_{{\sf Aut}({\mathrm{PSL}}_3(r))}(X_v)=X_v.2$ . Since $X_v\unlhd G_v$ , it follows that $m\le 2$ . Thus, $G=X$ or $X.2$ and, if $G=X.2$ , then $G_v=X_v.2\cong {\mathrm{PGL}}_2(9)$ and $G_{\{u,v\}}\cong 3^2{:}{\mathrm{Q}}_8.2$ . The pair $({\mathrm{PSU}}_3(r), 3^2{:}{\mathrm{Q}}_8)$ is similarly dealt with; the details are omitted. This completes the proof.

Proof. Let $X_v^{{{\Gamma}}(v)}= {\mathrm{PSU}}_3(q).[o]$ and $q=p^f>2$ , where p is a prime and $o\mid 2(3,q{+}1)f$ . Then $(X_v^{{{\Gamma}}(v)})_u=p^{f{+}2f}{:}(({q^2-1})/{(3,q{+}1)}).[o]$ and $X_{uv}^{[1]}=1$ , by Theorem 2.4. Thus, $|\textbf {O}_p(X_{\{u,v\}})|=p^{3f}.a$ , $p^{4f}.a$ , $p^{5f}.a$ or $p^{6f}.a$ , where a is a divisor of $(2,p)$ . Moreover, $\textbf {O}_p(X_{\{u,v\}})$ is nonabelian and $X_{\{u,v\}}/\textbf {O}_p(X_{\{u,v\}})$ has a subgroup $\mathbb Z_{({q^2-1})/{(3,q{+}1)}}$ . We next determine which pair $(G_0,H_0)$ in [Reference Li and Zhang17, Tables 14–20] is a possible candidate for $(X,X_{\{u,v\}})$ . Note that we may ignore those $H_0$ that either have no subgroup of index two or have an abelian maximal normal p-subgroup. In particular, ${\mathrm{soc}}(X)$ 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 [Reference Li and Zhang17, Tables 16–20]. Since $\textbf {O}_p(X_{\{u,v\}})$ is nonabelian, we conclude that $(X,\textbf {O}_p(X_{\{u,v\}}))$ is one of the following pairs:

$$ \begin{align*}\begin{array}{l} ({\mathrm{PSL}}_3(p^t).2, [p^{3t}]), \,({\mathrm{PGL}}_3(p^t).2, [p^{3t}]) \mbox{ (with } p=2),\\ ({\mathrm{PSU}}_3(p^t), [p^{3t}]), \,({\mathrm{PSp}}_4(p^t).\mathbb Z_{2^{l+1}}, [p^{4t}]) \mbox{ (with }p=2), \\ ({\mathrm{Sz}}(p^t), [p^{2t}]), \,({\mathrm{Ree}}(p^t), [p^{3t}])\mbox{ and }({\mathrm{G}}_2(p^t).\mathbb Z_{2^{l+1}}, [p^{6t}]), \end{array}\end{align*} $$

where $2^{l}$ is the $2$ -part of t. Check the maximal subgroups of ${\mathrm{PSp}}_4(p^t).\mathbb Z_{2^{l+1}}$ , ${\mathrm{Sz}}(p^t)$ and ${\mathrm{Ree}}(p^t)$ ; refer to [Reference Bray, Holt and Roney-Dougal1, Table 8.14], [Reference Suzuki27, Theorem 9] and [Reference Kleidman15, Theorem C], respectively. We conclude that none of ${\mathrm{PSp}}_4(t^f).\mathbb Z_{2^{l+1}}$ , ${\mathrm{Sz}}(p^t)$ and ${\mathrm{Ree}}(p^t)$ has maximal subgroups with a simple section ${\mathrm{PSU}}_3(q)$ and they are excluded. For the first three and the last pairs, $|X/\textbf {O}_p(X_{\{u,v\}})|$ is a divisor of $2(p^t-1)^2$ and $\textbf {O}_p(X_{\{u,v\}})=[p^{3t}]$ or $[p^{6t}]$ . Clearly, $t\le 2f$ .

Suppose that $t=2f$ . Then ${\mathrm{soc}}(X)={\mathrm{PSL}}_3(q^2)$ or ${\mathrm{PSU}}_3(q^2)$ , and $\textbf {O}_p(X_{\{u,v\}})=[q^6]$ . It follows that $\textbf {O}_p(X_v^{[1]})=[q^3]$ . Thus, $\textbf {O}_p(X_v)\ne 1$ and $X_v$ has an almost simple quotient ${\mathrm{PSU}}_3(q).[o]$ . Checking Tables 8.3 and 8.5 given in [Reference Bray, Holt and Roney-Dougal1], we conclude that X has no maximal subgroup containing $X_v$ , which is a contradiction. If $t=f$ , then we have $(X,\textbf {O}_p(X_{\{u,v\}}))=({\mathrm{G}}_2(p^t).\mathbb Z_{2^{l+1}}, [q^6])$ and we get a similar contradiction by checking the maximal subgroups of ${\mathrm{G}}_2(p^t).\mathbb Z_{2^{l+1}}$ .

Suppose that $f\ne t<2f$ . Then $f>1$ . Recalling that $X_{\{u,v\}}/\textbf {O}_p(X_{\{u,v\}})$ has a subgroup $\mathbb Z_{({q^2-1})/{(3,q{+}1)}}$ , 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 $s\ge 2f+1\ge 5$ , and s is not a divisor of $2(3,q+1)(p^t-1)^2$ , which is 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 $2(3,q+1) (p^t-1)^2$ , which is a contradiction.

Case 2. Let $(G_0,H_0)$ be one of the pairs in [Reference Li and Zhang17, Tables 15–20] that is not considered in Case 1. Assume that $X_{\{u,v\}}/\textbf {O}_p(X_{\{u,v\}})$ 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 $X_{\{u,v\}}/\textbf {O}_p(X_{\{u,v\}})$ is a $2$ -group. These observations yield that either $q=p=3$ and $X_{\{u,v\}}/\textbf {O}_p(X_{\{u,v\}})$ is a $2$ -group, or $X_{\{u,v\}}$ is not a $\{2,3\}$ -group.

Recall that $X_{\{u,v\}}/\textbf {O}_p(X_{\{u,v\}})$ has a subgroup $\mathbb Z_{({q^2-1})/{(3,q{+}1)}}$ and $\textbf {O}_p(X_{\{u,v\}})$ has order $p^{if}.a$ , where $3\le i\le 6$ . It follows that $(X,X_{\{u,v\}})$ is one of the following pairs:

$$ \begin{align*}\begin{array}{l} ({\mathrm{HS}}.2, [5^3]{:}[2^5]), \,({\mathrm{Ru}}, [5^3]{:}[2^5]), \,({\mathrm{McL}}, [5^3]{:}3{:}8), \,({\mathrm{Co}}_2, [5^3]{:}4{\mathrm{S}}_4),\\ ({\mathrm{Th}}, [5^3]{:}4{\mathrm{S}}_4), \,({\mathrm{J}}_4, [11^3]{:}(5\times 2{\mathrm{S}}_4)). \end{array}\end{align*} $$

Then $q=p\in \{5,11\}$ and $X_v^{[1]}=1$ . In particular, ${\mathrm{soc}}(X_v)={\mathrm{PSU}}_3(p)$ , and $X_{\{u,v\}}$ is the normalizer $\textbf {N}_X(P)$ of some Sylow p-subgroup P of X. Thus, $X_{uv}=X_v\cap X_{\{u,v\}}\le \textbf {N}_{X_v}(P)$ . For the pairs $({\mathrm{HS}}.2, [5^3]{:}[2^5])$ and $({\mathrm{Ru}}, [5^3]{:}[2^5])$ , by the Atlas [Reference Conway, Norton, Parker and Wilson3], $X_{\{u,v\}}$ is a normalizer of some Sylow $5$ -subgroup that intersects a maximal subgroup ${\mathrm{PSU}}_3(5){:}2$ of ${\mathrm{soc}}(X)$ at $[5^3]{:}8{:}2$ ; thus G, X and $X_{\{u,v\}}$ are as listed in Table 6. The other pairs are excluded as follows.

First, the group ${\mathrm{Th}}$ is excluded as it has no maximal subgroup with a simple section ${\mathrm{PSU}}_3(5)$ ; refer to [Reference Wilson36, Table 5.8]. For the pair $({\mathrm{McL}}, [5^3]{:}3{:}8)$ , by the Atlas [Reference Conway, Norton, Parker and Wilson3], we have $X_v={\mathrm{PSU}}_3(5)$ and so $X_{uv}\le \textbf {N}_{{\mathrm{PSU}}_3(5)}(P)=[5^3]{:}8$ , which contradicts that $|X_{\{u,v\}}:X_{uv}|=2$ . For the pair $({\mathrm{J}}_4, [11^3]{:}(5\times 2{\mathrm{S}}_4))$ , by [Reference Wilson36, Table 5.8], $X_v={\mathrm{PSU}}_3(11).2$ , yielding $X_{uv}\le \textbf {N}_{X_v}(P)=[11^3]{:}(5\times 8{:}2)$ and we get a similar contradiction. For the pair $(X,X_{\{u,v\}})=({\mathrm{Co}}_2, [5^3]{:}4{\mathrm{S}}_4)$ , by the Atlas [Reference Conway, Norton, Parker and Wilson3], $X_v<{\mathrm{HS}}.2<{\mathrm{Co}}_2$ . Checking the maximal subgroups of ${\mathrm{HS}}.2$ , we have $X_v={\mathrm{PSU}}_3(5)$ or $X_v={\mathrm{PSU}}_3(5){:}2$ . It follows that $X_{uv}\le \textbf {N}_{X_v}(P)=[5^3]{:}8$ or $[5^3]{:}[2^5]$ and then $|X_{\{u,v\}}:X_{uv}|\ne 2$ , which is a contradiction. This completes the proof.

Proof. Let $G^{+}$ be the subgroup of G fixing the bipartition of ${{\Gamma}}$ . Then $G_v\le G^{+}$ and $G_v$ is $2$ -transitive on the partite set that 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 [Reference Cameron2, Table 7.4]. We conclude that $T\cong {\mathrm{A}}_6$ or ${\mathrm{M}}_{12}$ , $T_v\cong {\mathrm{A}}_5$ or ${\mathrm{M}}_{11}$ and $T_{uv}\cong {\mathrm{D}}_{10}$ or ${\mathrm{PSL}}_2(11)$ , respectively. Since $T_{uv}$ is soluble, the lemma follows.

Proof. Case 1. Assume that Lemma 3.5(1) holds. Suppose first that $(X_v^{{{\Gamma}}(v)})_u={\mathrm{Q}}_8$ . Then $X_{uv}\lesssim {\mathrm{Q}}_8{\times }{\mathrm{Q}}_8$ . This implies that $|X_{\{u,v\}}|$ is a divisor of $2^7$ and divisible by $2^4$ . Checking Tables 14–20 in [Reference Li and Zhang17], we have $X\cong {\mathrm{PSL}}_2(9).2={\mathrm{M}}_{10}$ and $X_{\{u,v\}}\cong \mathbb Z_8{:}\mathbb Z_2$ . In this case, $X_v\cong 3^2{:}{\mathrm{Q}}_8$ and $d=9$ . Since ${{\Gamma}}$ has valency nine and order $|X:X_v|=10$ , we have ${{\Gamma}}\cong {\sf K}_{10}$ , desired as in part (1).

Suppose that $(X_v^{{{\Gamma}}(v)})_u\ne {\mathrm{Q}}_8$ . If $p=3$ and $(G_v^{{{\Gamma}}(v)})_u={\mathrm{Q}}_8$ , then $(X_v^{{{\Gamma}}(v)})_u$ is abelian; it follows that $X_{uv}$ is abelian, which is a contradiction. Thus, we have ${\mathrm{SL}}_2(3)\unlhd (G_v^{{{\Gamma}}(v)})_u\le {\mathrm{GL}}_2(p)$ and $p\in \{3,5,7,11,23\}$ . Then $(G_v^{{{\Gamma}}(v)})_u\le \textbf {N}_{{\mathrm{GL}}_2(p)}({\mathrm{SL}}_2(3))=\mathbb Z_{p-1}\circ {\mathrm{GL}}_2(3)$ . Since $(X_v^{{{\Gamma}}(v)})_u$ is nonabelian and normal in $(G_v^{{{\Gamma}}(v)})_u$ , we have ${\mathrm{Q}}_8\unlhd (X_v^{{{\Gamma}}(v)})_u$ and hence ${\mathrm{SL}}_2(3)\unlhd (X_v^{{{\Gamma}}(v)})_u$ . Moreover, $|X_{\{u,v\}}|$ is a divisor of $2^7\cdot 3^2\cdot (p-1)^2$ 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 $(X_v^{{{\Gamma}}(v)})_u$ . 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 $(X_v^{{{\Gamma}}(v)})_u$ , 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 [Reference Li and Zhang17, Tables 14–20] that satisfy the following conditions:

1. (c1) $|H_0|$ is a divisor of $2^7\cdot 3^2\cdot (p-1)^2$ and divisible by $2^4$ ; $H_0$ has a factor (a quotient of some subnormal subgroup) ${\mathrm{Q}}_8$ ; 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 $\textbf {O}_{({p-1})/{2}}(H_0)$ is a divisor of ${(p-1)^2}/{4}$ .

Checking those $H_0$ that satisfy conditions (c1) and (c2), we conclude that the possible pairs $(X,X_{\{u,v\}})$ are listed as follows:

$$ \begin{align*}\begin{array}{l} ({\rm M}_{11},3^2{:}{\rm Q}_8.2),\, ({\rm M}_{11},2{\rm S}_4),\, ({\rm M}_{12}, [2^5].{\rm S}_3),\, ({\rm M}_{12},3^2{:}2{\rm S}_4),\\ ({\rm J}_{2},[2^6]{:}(3{\times}{\rm S}_3)),\, ({\rm J}_{3},[2^6]{:}(3{\times}{\rm S}_3)),\, ({\rm Co}_3,[2^9].3^2.{\rm S}_3)),\\ ({\rm He}.2,[2^8]{:}3^2.{\rm D}_8)),\, ({\rm McL}.2, [2^6]{:}{\rm S}_3^2),\\ ({\rm PSL}_3(3), 3^2{:}2{\rm S}_4),\, ({\rm PSL}_3(3).2, 2{\rm S}_4{:}2),\, ({\rm PSL}_3(4).2, 2^{2{+}4}.3.2),\\ ({\rm PGL}_3(4){.}2, [2^6]{.}3{.}{\rm S}_3),\, ({\rm PSL}_4(3){.}2, 2{.}{\rm S}_4^2{.}2),\, ({\rm PSL}_5(2){.}2, [2^8]{.}{\rm S}_3^2{.}2),\\ ({\rm PSp}_4(4){.}4, [2^8]{:}3{.}12),\, ({\rm PSp}_4(4).4, 5^2{:}[2^5]),\, ({\rm PSp}_6(2), [2^7]{:}{\rm S}_3^2),\\ ({\rm PSp}_6(3), [2^8]{:}3^3.S_3),\, ({\rm PSU}_3(3), 4.{\rm S}_4),\, ({\rm PSU}_4(2), 2.{\rm A}_4^2.2),\\ ({\rm PSU}_4(3), 2.{\rm A}_4^2.4),\, ({\rm PSU}_4(3).2, [2^5].{\rm S}_4),\, ({\rm P\Omega}^{+}_8(3){.}{\rm A}_4, 10^2{:}4{\rm A}_4),\\ ({\rm G}_2(2)', 4.{\rm S}_4),\, ({\rm G}_2(3), {\rm SL}_2(3)\circ{\rm SL}_2(3){:}2),\, ({}^2{\rm F}_4(2)', 5^2{:}4{\rm A}_4).\end{array}\end{align*} $$

Note that these groups X are included in the Atlas [Reference Conway, Norton, Parker and Wilson3]. Inspecting the subgroups of X, only the pair $({\mathrm{PSL}}_3(3).2, 2{\mathrm{S}}_4{:}2)$ gives a desired $X_v\cong 3^2{:}{\mathrm{GL}}_2(3)$ and then the desired graph ${{\Gamma}}$ has valency $d=9$ . In this case, the socle ${\mathrm{PSL}}_3(3)$ of X has two orbits on the vertex set of ${{\Gamma}}$ ; each of them has size $13$ and can be viewed as the point set or the line set of the projective plane ${\mathrm{PG}}(2,3)$ . This forces that ${{\Gamma}}$ is (isomorphic to) one of the following graphs: ${\sf K}_{13,13}-13{\sf K}_2$ , the point–line incidence graph and the point–line nonincidence graph of ${\mathrm{PG}}(2,3)$ . Since ${{\Gamma}}$ has valency 9, the graph ${{\Gamma}}$ is the point–line nonincidence graph of ${\mathrm{PG}}(2,3)$ . Then part (2) of this lemma follows.

Case 2. Let $2_{+}^{1{+}4}{:}\mathbb Z_5\le (G_v^{{{\Gamma}}(v)})_u\le 2_{+}^{1{+}4}.(\mathbb Z_5{:}\mathbb Z_4)$ . Then $2_{+}^{1{+}4}\unlhd (X_v^{{{\Gamma}}(v)})_u$ and so $|X_{\{u,v\}}|$ is a divisor of $2^{15}\cdot 5^2$ 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 $\textbf {O}_2(X_{uv})\ne 1 $ and hence $\textbf {O}_2(X_{\{u,v\}})\ne 1$ . Checking the pairs $(G_0,H_0)$ in [Reference Li and Zhang17, Tables 14–20], either $\textbf {O}_2(H_0)=1$ or $|H_0|$ has an odd prime divisor other than $5$ . Thus, in this case, no desired pair $(X,X_{\{u,v\}})$ exists. This completes the proof.