$A\cong\Omega_{2m-1}(q).{(4,q-1)\over 2}$ (odd $q$) or $\mathrm{Sp}_{2m-2}(q)$
(even $q$), and $B\cong{q+1\over(4,q+1)}{.}\mathrm{PGU}_{m}(q)$.

For case (1), we have $|G:B|=3^{2}\cdot 5\cdot 11$ and $|G:A|=2^{11}\cdot
3\cdot 17$. Check the insolvable subgroups of $A$ and $B$ with order divisible
by $|G:B|$ and $|G:A|$, respectively. Recalling that $X$ is not an alternating
group, it follows that $X\cong\mathrm{M}_{11}$ or $\mathrm{M}_{12}$, and
$Y\gtrsim\Omega^{-}_{8}(2)$. Then the only possibility is that
$T=Y\cong\Omega^{-}_{8}(2)$ and $G_{\alpha}\cong\mathrm{M}_{11}$. Thus
$|T|_{2}|G_{\alpha}|_{2}=2^{16}<|G|_{2}$, a contradiction.

Suppose that (2) or (3) occurs. It is easy to check that $|G:A|$ is divisible
by $q^{m-1}{q^{m}+1\over(4,q^{m}+1)}$, and then $|Y|$ is divisible by
$q^{m-1}\Phi^{*}_{m(2f)}(p)$. Since $m\geq 4$ and $m$ is odd,
$\Phi^{*}_{m(2f)}(p)\neq 1$. By [2, Theorem 3.1], one of the following holds:
$Y\succcurlyeq\mathrm{U}_{m_{0}}(q^{m\over m_{0}})$ for some divisor
$m_{0}\geq 3$ of $m$; or $m=5$, $q^{2}=4$ and
$Y\succcurlyeq\mathrm{L}_{2}(11)$; or $m=9$, $q^{2}=4$ and
$Y\succcurlyeq\mathrm{J}_{3}$ or $\mathrm{L}_{2}(19)$. For the last two cases,
we have $|Y|_{p}<q^{m-1}$, a contradiction. Thus we have
$T=Y\cong\mathrm{U}_{m_{0}}(q^{m\over m_{0}})$, and then $G_{\alpha}=X\leq A$.

######  Theorem 8.3.

Assume that Hypothesis 6.1 holds, $G=\mathrm{P\Omega}_{2m}^{-}(q)$ for $m\geq
4$. Then $m=5$, $q>2$, $T\cong\mathrm{U}_{5}(q)$,
$\mathrm{SL}_{2}(q^{4})\unlhd G_{\alpha}^{\Gamma(\alpha)}$,
$|\mathbf{O}_{p}(G_{\alpha})|=q^{8}$ and $f$ is a power of $2$.

###### Proof.

By the argument ahead of this lemma, $m$ is odd, $\mathrm{U}_{m_{0}}(q^{m\over
m_{0}})\cong T\leq B$, $G_{\alpha}\leq A$, where $m_{0}$ is a divisor of $m$.
Note that $|G:B|=q^{m(m-1)\over 2}\prod_{i=1}^{m-1}(q^{i}-(-1)^{i-1})$. Then
$|G_{\alpha}|$ is divisible by $q^{m(m-1)\over
2}\Phi_{2(m-1)}^{*}(q)\Phi_{2(m-3)}^{*}(q)$.

Assume that $A\cong[q^{2m-2}]{:}[{q-1\over
a}]{.}\mathrm{P\Omega}^{-}_{2m-2}(q){.}a$. Let $\widehat{A}$ and
$\widehat{G_{\alpha}}$ be the preimages of
$A^{\prime}\mathbf{rad}(A)/\mathbf{rad}(A)$ and $(A^{\prime}\cap
G_{\alpha})\mathbf{rad}(A)/\mathbf{rad}(A)$ in $\mathrm{GO}^{-}_{2(m-1)}(q)$,
respectively. Then $S\preccurlyeq\widehat{G_{\alpha}}$, and
$|\widehat{G_{\alpha}}|$ is divisible by
$\Phi_{2(m-1)f}^{*}(p)\Phi_{2(m-3)}^{*}(q)$. By Lemma 7.1, either
$S=\mathrm{L}_{d}(q^{2(m-1)\over d})$ or $(S,2(m-1),q)$ is one of
$(\mathrm{A}_{13},12,2)$ and $(\mathrm{L}_{3}(3),12,2)$. For the latter case,
by Lemma 4.2, we conclude that $|G_{\alpha}|_{2}\leq 2^{11}$, which
contradicts that $|G_{\alpha}|$ is divisible by $q^{m(m-1)\over 2}$. Suppose
that $S=\mathrm{L}_{d}(q^{2(m-1)\over d})$. Then
$r={(q^{2(m-1)}-1)\over(q^{2(m-1)\over d}-1)}$. By Lemma 2.1, $d=2$, $q$ is
even and $m-1$ is a power of $2$. By the choice of $\widehat{G_{\alpha}}$, we
conclude that
$|G_{\alpha}|_{p}\leq|\mathbf{rad}(A)|_{p}|\widehat{G_{\alpha}}|_{p}$. Then,
by Lemma 7.1, we have
$|G_{\alpha}|_{p}\leq|\mathbf{rad}(A)|_{p}2(m-1)|S|_{p}=2(m-1)q^{3(m-1)}$.
Thus $q^{m(m-1)\over 2}\leq|G_{\alpha}|_{p}\leq 2(m-1)q^{3(m-1)}$. This
implies that $m\leq 7$, yielding $m=5$ as $m\geq 5$ and $m-1$ is a power of
$2$. Then $T\cong\mathrm{U}_{5}(q)$. Noting that
$8q^{12}\geq|G_{\alpha}|_{p}\geq q^{10}$, by Lemma 4.2, we have
$|\mathbf{O}_{p}(G_{\alpha})|=q^{8}$. To finish the proof, we next show that
$A\not\cong\Omega_{2m-1}(q).{(4,q-1)\over 2}$ or $\mathrm{Sp}_{2m-2}(q)$.

Suppose that $A\cong\Omega_{2m-1}(q).{(4,q-1)\over 2}$ for odd $q$. Recall
that $|G_{\alpha}|$ is divisible by $q^{m(m-1)\over
2}\Phi_{2(m-1)}^{*}(q)\Phi_{2(m-3)}^{*}(q)$, and $2(m-1)\geq 8$. By Lemma 7.2,
the only possibility is that $S\cong\mathrm{L}_{d}(q^{2(m-1)\over d})$, $d$ is
a prime and $2(m-1)f$ is a power of $d$. This implies that $d=2$, and so
$r=q^{m-1}+1$. Then $q$ is even by Lemma 2.1, a contradiction.

Suppose that $A\cong\mathrm{Sp}_{2m-2}(q)$ for even $q$. Clearly,
$G_{\alpha}\neq A$. By Lemma 7.1, we conclude that either
$S\cong\mathrm{L}_{2}(q^{m-1})$ or $(S,2(m-1),q)$ is one of
$(\mathrm{A}_{13},12,2)$ and $(\mathrm{L}_{3}(3),12,2)$. For the latter case,
$|G_{\alpha}|_{2}\leq 2^{11}$ by Lemma 4.2, and so $2^{21}=q^{m(m-1)\over
2}\leq|G_{\alpha}|_{2}\leq 2^{11}$, a contradiction. Then
$S\cong\mathrm{L}_{2}(q^{m-1})$ and, by Lemma 7.1, $|G_{\alpha}|_{p}\leq
2(m-1)|S|_{p}$. Thus $q^{m(m-1)\over 2}\leq 2(m-1)q^{m-1}\leq q^{2(m-1)}$,
yielding $m\leq 4$, contradiction. ∎

###  8.4. Orthogonal case III

Let $G=\mathrm{P\Omega}^{+}_{8}(q)$, $q=p^{f}$ for some prime $p$. Then $A$
and $B$ are as in [31, Table 4]. Recall that $\\{X,Y\\}=\\{T,G_{\alpha}\\}$
with $X\leq A$ and $Y\leq B$.

######  Lemma 8.5.

Either $q>3$, or one of the following holds.

  * (1)

$q=2$, $A\cong\mathrm{Sp}_{6}(2)\cong B$, $T\cong\mathrm{Sp}_{6}(2)$ and
$G_{\alpha}\cong\mathrm{S}_{5}$.

  * (2)

$q=3$, $T=A\cong\Omega^{+}_{8}(2)$,
$G_{\alpha}\cong[3^{6}]{:}\mathrm{SL}_{3}(3)$ and $r=13$.

###### Proof.

Let $q=2$ or $3$. Then, up to isomorphism, $A$ and $B$ are listed as follows:

$\tiny\begin{array}[]{l|l|l||l|l|l}X&A&|G:A|&|G:B|&B&Y\\\
\hline\cr&\mathrm{A}_{9}&2^{6}\cdot 3\cdot 5&2^{3}\cdot 3\cdot
5&\mathrm{Sp}_{6}(2)&\\\ &&&3^{3}\cdot 5&2^{6}{:}\mathrm{A}_{8}&\\\
&&&2^{5}\cdot 5\cdot 7&(3{\times}\mathrm{U}_{4}(2)){:}2\\\
\hline\cr&\mathrm{Sp}_{6}(2)&2^{3}\cdot 3\cdot 5&2^{6}\cdot 3^{3}\cdot
7&(\mathrm{A}_{5}\times\mathrm{A}_{5}){:}2^{4}&\\\ &&&2^{3}\cdot 3\cdot
5&\mathrm{Sp}_{6}(2)&\\\ &&&3^{3}\cdot 5&2^{6}{:}\mathrm{A}_{8}&\\\
&&&2^{5}\cdot 5\cdot 7&(3{\times}\mathrm{U}_{4}(2)){:}2&\\\
\hline\cr&(3{\times}\mathrm{U}_{4}(2)){:}2&2^{5}\cdot 5\cdot 7&3^{3}\cdot
5&2^{6}{:}\mathrm{A}_{8}&\\\ \hline\cr\hline\cr&\Omega_{8}^{+}(2)&3^{7}\cdot
13&2^{3}\cdot 3^{3}\cdot 5&\Omega_{7}(3)&\\\ &&&2^{5}\cdot 5\cdot
7&3^{6}{:}\mathrm{L}_{4}(3)&\\\ \hline\cr&\Omega_{7}(3)&2^{3}\cdot 3^{3}\cdot
5&&\Omega_{7}(3)&\\\ &&&2^{5}\cdot 5\cdot 7&3^{6}{:}\mathrm{L}_{4}(3)&\\\
&&&2^{2}\cdot 3^{6}\cdot 5\cdot 13&2^{.}\mathrm{U}_{4}(3){.}2^{2}&\\\
&&&2^{3}\cdot 3^{7}\cdot 5\cdot 7\cdot
13&(\mathrm{A}_{4}\times\mathrm{U}_{4}(2)){.}2&\\\
\hline\cr&2^{.}\mathrm{U}_{4}(3){.}2^{2}&2^{2}\cdot 3^{6}\cdot 5\cdot
13&2^{5}\cdot 5\cdot 7&3^{6}{:}\mathrm{L}_{4}(3)&\end{array}$

Case 1. Let $q=2$. With the help of GAP, we search the subgroups of $A$ with
order divisible by $|G:B|$. Up to isomorphism, either $A\cong\mathrm{A}_{9}$
and
$G_{\alpha}=X\cong\mathrm{S}_{5},\,2^{2}{\times}\mathrm{A}_{5},\,\mathrm{A}_{4}{\times}\mathrm{A}_{5},\,(2^{2}{\times}\mathrm{A}_{5}){.}2,\,(\mathrm{A}_{4}{\times}\mathrm{A}_{5}){.}2$
or $\mathrm{S}_{7}$, or $X$ and $Y$ are listed as follows:

$\tiny\begin{array}[]{l|l|l|l}A&X&B&Y\\\
\hline\cr\mathrm{Sp}_{6}(2)&\mathrm{Sp}_{6}(2)&(\mathrm{A}_{5}\times\mathrm{A}_{5}){:}2^{4}&\mathrm{S}_{5},2^{2}{\times}\mathrm{A}_{5},(2^{2}{\times}\mathrm{A}_{5}){.}2,\mathrm{A}_{4}{\times}\mathrm{A}_{5},(\mathrm{A}_{4}{\times}\mathrm{A}_{5}){.}\\\
&\mathrm{S}_{5},2^{4}{:}\mathrm{A}_{5},2^{4}{:}\mathrm{S}_{5},\mathrm{S}_{7}&\mathrm{Sp}_{6}(2)&\Omega_{5}(3),\mathrm{Sp}_{6}(2)\\\
&\Omega_{5}(3),\mathrm{Sp}_{6}(2)&\mathrm{Sp}_{6}(2)&\mathrm{S}_{5},2^{4}{:}\mathrm{A}_{5},2^{4}{:}\mathrm{S}_{5},\mathrm{S}_{7}\\\
&\Omega_{5}(3),\mathrm{Sp}_{6}(2)&2^{6}{:}\mathrm{A}_{8}&\mathrm{S}_{5},2^{2}{\times}\mathrm{A}_{5},(2^{2}{\times}\mathrm{A}_{5}){.}2,\mathrm{A}_{4}{\times}\mathrm{A}_{5},(\mathrm{A}_{4}{\times}\mathrm{A}_{5}){.}2\\\
&&&2^{4}{:}\mathrm{A}_{5},2^{4}{:}\mathrm{S}_{5},\mathrm{A}_{4}\times
2^{4}{:}\mathrm{A}_{5},(\mathrm{A}_{4}\times 2^{4}{:}\mathrm{A}_{5}){.}2\\\
&\mathrm{Sp}_{6}(2)&(3{\times}\mathrm{U}_{4}(2)){:}2&\mathrm{S}_{5},2^{4}{:}\mathrm{A}_{5},2^{4}{:}\mathrm{S}_{5}\\\
\hline\cr(3{\times}\mathrm{U}_{4}(2)){:}2&\Omega_{5}(3)&2^{6}{:}\mathrm{A}_{8}&\mbox{No}\end{array}$

Then the vertex-stabilizer $G_{\alpha}$ is easily determined. By calculation,
we have a desired graph only if $A\cong\mathrm{Sp}_{6}(2)\cong B$,
$T\cong\mathrm{Sp}_{6}(2)$ and $G_{\alpha}\cong\mathrm{S}_{5}$.

Case 2. Let $q=3$ and $B\not\cong\Omega_{7}(3)$. Set
$\overline{Y}=Y\mathbf{rad}(B)\mathbf{rad}(B)$ and
$\overline{B}=B/\mathbf{rad}(B)$.

Suppose that $B\cong 3^{6}{:}\mathrm{L}_{4}(3)$. Then
$A\cong\Omega_{8}^{+}(2)$, $\Omega_{7}(3)$ or
$2^{.}\mathrm{U}_{4}(3){.}2^{2}$, and $|\overline{Y}|$ has a divisor $5\cdot
13$. By [32, Table 10.3], we have
$\overline{Y}=\overline{B}\cong\mathrm{L}_{4}(3)$. Thus
$Y\succcurlyeq\mathrm{L}_{4}(3)$. Noting that $3^{4}-1\over 3-1$ is not a
prime, we have $S\not\cong\mathrm{L}_{4}(3)$, and then
$B>T\cong\mathrm{L}_{4}(3)$. Thus $|G:T|=|G:B||B:T|=2^{5}\cdot 5\cdot 7\cdot
3^{6}$, and so $G_{\alpha}$ is a subgroup of $A$ with order divisible by
$2^{5}\cdot 3^{6}\cdot 5\cdot 7$. By [32, Table 10.3], we conclude that
$A\not\cong\Omega_{8}^{+}(2)$. Then $A\cong\Omega_{7}(3)$ or
$2^{.}\mathrm{U}_{4}(3){.}2^{2}$. If $A\cong\Omega_{7}(3)$ then, checking the
maximal subgroup of $A$ with order divisible by $|G:T|$, we conclude that
$G_{\alpha}$ is contained in a maximal subgroup
$2^{.}\mathrm{U}_{4}(3){.}2^{2}$ of $A$. Thus we may let $G_{\alpha}\leq
A_{0}$, where $2^{.}\mathrm{U}_{4}(3){.}2^{2}\leq\mathrm{A}_{0}\leq A$.
Considering $G_{\alpha}\mathbf{rad}(A_{0})/\mathbf{rad}(A_{0})$ and
$A_{0}/\mathbf{rad}(A_{0})$, by [32, Table 10.4], we know that
$G_{\alpha}\succcurlyeq\mathrm{L}_{3}(4)$ or $\mathrm{A}_{7}$. Then
$S\cong\mathrm{L}_{3}(4)$ or $\mathrm{A}_{7}$. Noting $4^{3}-1\over 4-1$ is
not a prime, $S\cong\mathrm{A}_{7}$. By Lemma 4.2, $|G_{\alpha}|_{3}=3^{3}$, a
contradiction.

Suppose that $B\cong 2^{.}\mathrm{U}_{4}(3){.}2^{2}$. Then
$A\cong\Omega_{7}(3)$. It is easily shown that $\overline{Y}$ and
$\overline{Y}\cap\overline{B}^{\infty}$ have same insolvable composition
factors, and $|\overline{Y}\cap\overline{B}^{\infty}|$ is divisible by
$3^{3}\cdot 5$. With the help of GAP, searching the subgroups of
$\mathrm{U}_{4}(3)$ with order divisible by $3^{3}\cdot 5$, we get
$\overline{Y}\cap\overline{B}^{\infty}\cong 3^{4}{:}\mathrm{A}_{5}$,
$\Omega_{5}(3)$, $3^{4}{:}\mathrm{A}_{6}$ or $\mathrm{U}_{4}(3)$. Noting that
$\overline{Y}=Y\mathbf{rad}(B)\mathbf{rad}(B)\cong Y/(Y\cap\mathbf{rad}(B))$,
it follows that $Y$ has a normal subgroup $R{.}(3^{4}{:}\mathrm{A}_{5})$,
$R{.}\Omega_{5}(3)$, $R{.}(3^{4}{:}\mathrm{A}_{6})$ or
$R{.}\mathrm{U}_{4}(3)$, where $R=Y\cap\mathbf{rad}(B)$. Considering the
limitations on $Y$, we conclude that $T=Y\cong\Omega_{5}(3)$, and
$G_{\alpha}=X\leq A$. Noting that $|G:T|=2^{6}\cdot 3^{8}\cdot\cdot 5\cdot
7\cdot 13$, this leads to a similar contradiction as in the above paragraph.

Suppose that $B\cong(\mathrm{A}_{4}\times\mathrm{U}_{4}(2)){.}2$. Then
$A\cong\Omega_{7}(3)$, and $|\overline{Y}\cap\overline{B}^{\infty}|$ is
divisible by $3^{2}\cdot 5$. Searching the subgroups of $\mathrm{U}_{4}(2)$
with order divisible by $3^{2}\cdot 5$, we conclude that
$\overline{Y}\cap\overline{B}^{\infty}\cong\mathrm{A}_{6},\,\mathrm{S}_{6}$ or
$\Omega_{5}(3)$. It follows that $T=Y\cong\Omega_{5}(3)$, and
$G_{\alpha}=X\leq A$, which leads to a contradiction.

Case 3. Let $B\cong\Omega_{7}(3)$ and $A\cong\Omega^{+}_{8}(2)$ or
$\Omega_{7}(3)$. Suppose that $G_{\alpha}\lesssim\Omega^{+}_{8}(2)$. Then, by
Lemma 7.4, $S\cong\mathrm{A}_{5}$ or $\mathrm{A}_{7}$, which leads to a
contradiction by calculating $|G_{\alpha}|_{3}$. Thus $T\leq
A\cong\Omega^{+}_{8}(2)$, and $G_{\alpha}\leq B$. In particular,
$|G:A|=3^{7}\cdot 13$. Then $G_{\alpha}$ is a subgroup of $B$ of order
divisible by $3^{7}\cdot 13$. Checking the orders of maximal subgroups of
$\Omega_{7}(3)$, we conclude that $\mathrm{L}_{3}(3)\lesssim
G_{\alpha}\lesssim[3^{6}]{:}\mathrm{L}_{3}(3)\lesssim B$. Then
$\mathbf{soc}(G_{\alpha}^{\Gamma(\alpha)}\cong\mathrm{L}_{3}(3)$, $r=13$ and
$|\mathbf{O}_{3}(G_{\alpha})|=3^{6}$ by Lemma 4.2. Thus
$G_{\alpha}\cong[3^{6}]{:}\mathrm{SL}_{3}(3)$, and
$|G:G_{\alpha}|=|G:B||B:G_{\alpha}|=2^{8}\cdot 3^{3}\cdot 5^{2}\cdot 7$. Since
$|T|$ is divisible by $|G:G_{\alpha}|$, by [32, Table 10.4], we have
$T=A\cong\Omega^{+}_{8}(2)$, and part (2) of this lemma follows. ∎

For $q>3$, we list $A$ and $B$ as follows:

$\tiny\begin{array}[]{l|l|l|l}A&|G:A|&B&|G:B|\\\
\hline\cr\Omega_{7}(q)&{1\over e}q^{3}(q^{4}-1)&\Omega_{7}(q)&{1\over
e}q^{3}(q^{4}-1)\\\ \Omega_{7}(q)&{1\over
e}q^{3}(q^{4}-1)&\mathsf{P}_{1},\mathsf{P}_{3}\mbox{ or
}\mathsf{P}_{4}&(q^{3}+1)(q^{2}+1)(q+1)\\\ \Omega_{7}(q)&{1\over
e}q^{3}(q^{4}-1)&{}^{\bf\hat{}}({q+1\over
e}{\times}\Omega^{-}_{6}(q)){.}2^{e}&{1\over 2}q^{6}(q^{3}-1)(q^{2}+1)(q-1)\\\
\Omega_{7}(q)&{1\over e}q^{3}(q^{4}-1)&{}^{\bf\hat{}}({q-1\over
e}{\times}\Omega^{+}_{6}(q)){.}2^{e}&{1\over 2}q^{6}(q^{3}+1)(q^{2}+1)(q+1)\\\
\Omega_{7}(q),q\mbox{ ood}&{1\over
2}q^{3}(q^{4}-1)&(\mathrm{PSp}_{2}(q)\otimes\mathrm{PSp}_{4}(q)){.}2&{1\over
2}q^{7}(q^{6}-1)(q^{2}+1)\\\ \Omega_{7}(q)&{1\over
e}q^{3}(q^{4}-1)&\Omega^{-}_{8}(\sqrt{q})&{1\over
e}q^{6}(q^{4}-1)(q^{3}+1)(q+1)\\\ \hline\cr{}^{\bf\hat{}}({q+1\over
e}{\times}\Omega^{-}_{6}(q)){.}2^{e}&{1\over
2}q^{6}(q^{3}-1)(q^{2}+1)(q-1)&\mathsf{P}_{1},\mathsf{P}_{3}\mbox{ or
}\mathsf{P}_{4}&(q^{3}+1)(q^{2}+1)(q+1)\end{array}$

where $e=(2,q-1)$. Note that $\Omega_{m}(q)=\mathrm{Sp}_{m-1}(q)$ for odd $m$
and even $q$.

######  Lemma 8.6.

Assume that $q>3$. Then $A\cong\Omega_{7}(q)$.

###### Proof.

Suppose that $A$ is of type ${}^{\bf\hat{}}({q+1\over
e}{\times}\Omega^{-}_{6}(q)){.}2^{e}$. Let
$\overline{A}=A^{\infty}\mathbf{rad}(A)/\mathbf{rad}(A)$ and
$\overline{X}=(X\cap A^{\infty})\mathbf{rad}(A)/\mathbf{rad}(A)$. Then
$\overline{A}\cong\mathrm{U}_{4}(q)$ and $|\overline{X}|$ is divisible by
$\Phi^{*}_{6}(q)\Phi^{*}_{4}(q)$. By [32, Table 10.3], either
$\overline{X}=\overline{A}\cong\mathrm{U}_{4}(q)$, or $p\geq 5$ and
$|\overline{X}|$ is indivisible by $p$. Assume that
$\overline{X}\neq\overline{A}$. Checking the orders of maximal subgroups of
$\mathrm{U}_{4}(q)$ (refer to [3, Tables 8.10 and 8.11]), we conclude that
$\overline{X}\cong\mathrm{A}_{7}$. Considering the order of $\mathrm{A}_{7}$,
it follows from Lemma 2.2 that $\Phi^{*}_{6}(q)=7$ and $\Phi^{*}_{4}(q)=5$.
Then $q=3$ by [20, Theorem 3.9], a contradiction. Thus
$\overline{X}=\overline{A}\cong\mathrm{U}_{4}(q)$, and then
$T=X\cong\mathrm{U}_{4}(q)$, yielding $(2,q-1)=1$ and
$|G:T|=q^{6}(q^{4}-1)(q^{3}-1)$. Moreover, $G_{\alpha}=Y\leq B$.

Let $\overline{B}=B^{\infty}\mathbf{rad}(B)/\mathbf{rad}(B)$ and
$\overline{Y}=(Y\cap B^{\infty})\mathbf{rad}(B)/\mathbf{rad}(B)$. Then
$\overline{B}\cong\mathrm{L}_{4}(q)$, and $|\overline{Y}|$ has a divisor
$\Phi^{*}_{3}(q)\Phi^{*}_{4}(q)$. By [32, Table 10.3], either
$\overline{Y}=\overline{B}\cong\mathrm{L}_{4}(q)$, or $p\geq 5$ and
$|\overline{Y}|$ is indivisible by $p$. For the former, we conclude that
$G_{\alpha}=Y\succcurlyeq\mathrm{L}_{4}(q)$, a contradiction. Then
$|\overline{Y}|$ is indivisible by $p$. Checking the orders of maximal
subgroups of $\mathrm{L}_{4}(q)$ (refer to [3, Tables 8.10 and 8.11]), we get
$\overline{Y}\lesssim 2^{.}\mathrm{L}_{2}(7)$ or $2^{.}\mathrm{A}_{7}$, and
$q=p$. Recalling $(2,q-1)=1$, we have $q=p=2$, a contradiction. ∎

######  Lemma 8.7.

Assume that $q>3$. Then $T\lesssim\Omega_{7}(q)$.

###### Proof.

Suppose that $T\not\lesssim\Omega_{7}(q)$. Then, by Lemma 8.6, $T\leq
B\not\cong\Omega_{7}(q)$, yielding $G_{\alpha}\leq A\cong\Omega_{7}(q)$. Then
$|G_{\alpha}|$ is divisible by $|G:B|$. It follows that either
$T\lesssim\mathrm{U}_{4}(q)$ or $|G_{\alpha}|$ is divisible by
$\Phi^{*}_{6}(q)\Phi^{*}_{4}(q)$. Assume that the latter case occurs. Noting
$6$ is not a prime power, it follows from Lemmas 7.1 and 7.2 that $q=5$ and
$S\cong\mathrm{A}_{7}$ or $\mathrm{L}_{3}(2)$. Then
$\Phi^{*}_{4}(q)=13\not\in\pi(S)$, which contradicts that
$\pi(G_{\alpha})=\pi(S)$, see Lemma 4.2. Thus we have
$B\cong{}^{\bf\hat{}}({q+1\over e}{\times}\Omega^{-}_{6}(q)){.}2^{e}$, and
$|G:B|={1\over 2}q^{6}(q^{3}-1)(q^{2}+1)(q-1)$, where $e=(2,q-1)$. Note that
$|G_{\alpha}|$ is divisible by $|G:B|$. Take a maximal subgroup $A_{0}$ of $A$
with $G_{\alpha}\leq A_{0}$. Then both $|A_{0}|$ and $|G_{\alpha}|$ are
divisible by ${1\over 2}q^{6}\Phi^{*}_{3}(q)\Phi^{*}_{4}(q)$. Checking the
orders of maximal subgroups of $\Omega_{7}(q)$ and $\mathrm{Sp}_{6}(q)$, refer
to [3, Tables 8.28, 8.29, 8.39 and 8.40], we conclude that
$A_{0}\cong\Omega^{+}_{6}(q){.}2\cong\mathrm{L}_{4}(q){.}2$. It follows from
[32, Table 10.3] that $\mathrm{L}_{4}(q)\preccurlyeq G_{\alpha}$, which is
impossible by Lemma 4.2. ∎

######  Theorem 8.4.

Assume that Hypothesis 6.1 holds and $G=\mathrm{P\Omega}^{+}_{8}(q)$, where
$q=p^{f}$ for some prime $p$. Then one of the following holds.

  * (1)

$p=2$, $f$ is a power of $2$, $T\cong\mathrm{Sp}_{6}(q)$,
$\mathrm{SL}_{2}(q^{2})\unlhd G_{\alpha}^{\Gamma(\alpha)}$ and $r=q^{2}+1$; in
this case, $G_{\alpha}\cong\mathrm{S}_{5}$ if $f=1$, and
$\mathbf{O}_{2}(G_{\alpha})\neq 1$ if $f>2$.

  * (2)

$q=3$, $T\cong\Omega^{+}_{8}(2)$,
$G_{\alpha}\cong[3^{6}]{:}\mathrm{SL}_{3}(3)$ and $r=13$.

###### Proof.

If $q\leq 3$ then our result holds by Lemma 8.5. Assume that $q>3$. By Lemmas
8.6 and 8.7, $T\leq A\cong\Omega_{7}(q)$ and $G_{\alpha}\leq B$. Then $S$ is
given as in Lemma 7.4. Note that $|G_{\alpha}|$ is divisible by
$|G:A|={1\over(2,q-1)}q^{3}(q^{4}-1)$. If $(S,q)=(\mathrm{A}_{7},7)$ then
$|G_{\alpha}|_{7}=7<|G:A|_{7}$, a contradiction. Thus
$S\cong\mathrm{L}_{2}(q^{2})$, $r=q^{2}+1$, $p=2$ and $f$ is a power of $2$.
If $\mathbf{O}_{p}(G_{\alpha})=1$ then, by Lemma 4.2, $|G_{\alpha}|_{2}$ is a
divisor of $2fq^{2}$, and so $q^{3}\leq 2fq^{2}$, yielding $f\leq 2$. Since
$q$ is even, $A\cong\mathrm{Sp}_{6}(q)$.

By Lemma 4.2, $|G_{\alpha}|$ is a divisor of
$4f^{2}q^{2}(q^{4}-1)(q^{2}-1)|\mathbf{O}_{2}(G_{\alpha})|$. It follows that
$|G:G_{\alpha}|$ is divisible by $(q^{4}-1)(q^{6}-1)$. Since $q>3$, by [32,
Table 10.1], we have $T=A\cong\mathrm{Sp}_{6}(q)$, and the result follows. ∎

###  8.5. Orthogonal case IV

Let $G=\mathrm{P\Omega}^{+}_{2m}(q)$, $m\geq 5$ and $q=p^{f}$ for some prime
$p$.

######  Lemma 8.8.

$A$ and $B$ are not described as in [31, Table 3].

###### Proof.

Suppose that $G\cong\Omega^{+}_{24}(2)$, $A\cong\mathrm{Co}_{1}$ and
$B\cong\mathrm{Sp}_{22}(2)$. Then $|G:B|=2^{11}(2^{12}-1)$ and $|G:A|$ is
divisible by $\Phi^{*}_{22}(2)\Phi^{*}_{20}(2)\Phi^{*}_{18}(2)=683\cdot
41\cdot 19$. Since $|Y|$ is divisible by $|G:A|$, by [32, Table 10.1], either
$Y=B$ or $Y$ has a normal subgroup $\Omega^{-}_{22}(2)$, yielding
$T=Y\cong\mathrm{Sp}_{22}(2)$ or $\Omega^{-}_{22}(2)$. Then $G_{\alpha}\leq
A$. Checking the insolvable subgroups of $\mathrm{Co}_{1}$ with order
divisible by $2^{11}(2^{12}-1)$, we conclude that
$G_{\alpha}\succcurlyeq\mathrm{G}_{2}(4)$ or $\mathrm{Suz}$, which contradicts
Lemma 4.2. ∎

######  Lemma 8.9.

$A$ and $B$ are not described as in [31, Table 2].

###### Proof.

Suppose that $G=\mathrm{P\Omega}^{+}_{16}(q)$, $A\cong\Omega_{9}(q){.}a$ and
$B=\mathsf{N}_{1}$, where $a\leq 2$. Then $|G:B|$ and $|G:A|$ are divisible by
$q^{7}{q^{8}-1\over(4,q^{8}-1)}$ and
$q^{40}{\prod_{i\in\\{8,10,12,14\\}}(q^{i}-1)\over a(4,q^{8}-1)}$.

Assume that $q$ is even. Then $B\cong\mathrm{Sp}_{14}(q)$, and $|Y|$ is
divisible by $\Phi^{*}_{14}(q)\Phi^{*}_{12}(q)\Phi^{*}_{7}(q)$. By [32, Table
10.1], we have $Y=B\cong\mathrm{Sp}_{14}(q)$, and so $G_{\alpha}=X\leq A$.
Assume that $q$ is odd. Then $B\cong\Omega_{15}(q){.}b$. By Lemmas 2.6 and
2.7, either $Y\succcurlyeq\Omega_{15}(q)$, or there exists a core-free maximal
factorization $B_{0}=HK$ such that $K$ contains a normal subgroup $N$ of $Y$
of index at most $2$. The former cases yields that $T=Y=\Omega_{15}(q)$.
Suppose that the latter case occurs. It is easily shown that $|N|$ is
divisible by $\Phi^{*}_{14}(q)\Phi^{*}_{12}(q)\Phi^{*}_{10}(q)$. Then, by [31,
Theorem A], we conclude that $K$ is the stabilizer of some $1$-dimensional
nonsingular subspace, and so $N\leq K\cong\Omega_{14}^{-}(q){.}2$. By [32,
Table 10.1], we have $N\succcurlyeq\mathrm{P\Omega}_{14}^{-}(q)$. Then
$Y\succcurlyeq\mathrm{P\Omega}_{14}^{-}(q)$, yielding
$T=Y\cong\mathrm{P\Omega}_{14}^{-}(q)$.

All in all, $T\leq B=\mathsf{N}_{1}$. By Lemma 7.6,
$S\cong\mathrm{L}_{2}(q^{4})$, $p=2$ and $f$ is power of $2$. Since
$G_{\alpha}\leq A\cong\mathrm{Sp}_{8}(q){.}a$, we have $S\preccurlyeq
G_{\alpha}\cap A^{\prime}$ and $|G_{\alpha}:(G_{\alpha}\cap A^{\prime})|\leq
a$. By Lemma 7.1, $|G_{\alpha}\cap A^{\prime}|_{2}\leq 4|S|_{2}=4q^{4}$, and
so $|G_{\alpha}|_{2}\leq 2^{3}q^{4}$. Recalling that $|G_{\alpha}|$ is
divisible by $|G:B|$, we have $q^{7}\leq|G_{\alpha}|_{2}\leq 2^{3}q^{4}$,
yielding $q=2$ and $|G_{\alpha}|_{2}=2^{7}$. Then
$\mathbf{O}_{2}(G_{\alpha})=1$, otherwise, $|G_{\alpha}|_{2}\geq q^{8}$ by
Lemma 4.2, a contradiction. Thus $G_{\alpha}\lesssim\mathrm{P\Gamma
L}_{2}(2^{4})$, and then $|G_{\alpha}|_{2}\leq 2^{6}$, a contradiction. ∎

We next deal with the case that $A$ and $B$ are described as in [31, Table 1].
In particular, either $m$ is even, or $|G:B|=\prod_{i=1}^{m-1}(q^{i}+1)$.

######  Lemma 8.10.

Assume that $A$ and $B$ are described as in [31, Table 1], and $(m,q)=(6,2)$.
Then one of the following holds.

  * (1)

$T=\Omega^{-}_{10}(2)$ or $\mathrm{Sp}_{10}(2)$, and $\mathrm{SL}_{5}(2)\leq
G_{\alpha}\leq[2^{20}]{:}\mathrm{SL}_{5}(2)$;

  * (2)

$T=\mathrm{Sp}_{10}(2)$, $G_{\alpha}^{\Gamma(\alpha)}\cong\mathrm{SL}_{3}(2)$
and $\mathbf{O}_{2}(G_{\alpha})\neq 1$.

###### Proof.

Let $(m,q)=(6,2)$. Then $A$ and $B$ are listed as follows:

$\tiny\begin{array}[]{l|l|l|l|l|l}X^{\infty}&&A&|G:A|&B&|G:B|\\\
\hline\cr\Omega^{-}_{10}(2)&\mathrm{SO}^{-}_{10}(2)&\mathrm{Sp}_{10}(2)&2^{5}\cdot
3^{2}\cdot 7&2^{15}{:}\mathrm{GL}_{6}(2)&3^{4}\cdot 5\cdot 11\cdot 17\\\
\Omega^{+}_{10}(2)&\mathrm{SO}^{+}_{10}(2)&&&3{.}\mathrm{U}_{6}(2){.}[6]&2^{14}\cdot
5\cdot 7\cdot 17\cdot 31\\\
&\mbox{No}&&&\mathrm{Sp}_{2}(2)\times\mathrm{Sp}_{6}(2)&2^{20}\cdot 3^{3}\cdot
5\cdot 7\cdot 11\cdot 17\cdot 31\\\
\Omega^{-}_{10}(2)&\mathrm{SO}^{-}_{10}(2)&&&\mathrm{SL}_{6}(2){.}2&2^{14}\cdot
3^{4}\cdot 5\cdot 11\cdot 17\\\
&\mbox{No}&&&\Omega^{+}_{6}(4){.}2^{2}&2^{16}\cdot 3^{4}\cdot 7\cdot 11\cdot
31\\\ \hline\cr&\mbox{No}&(3\times\Omega^{-}_{10}(2)){.}2&2^{9}\cdot 3\cdot
7\cdot 31&\mathrm{SL}_{6}(2){.}2&2^{14}\cdot 3^{4}\cdot 5\cdot 11\cdot 17\\\
&\mbox{No}&&&2^{15}{:}\mathrm{GL}_{6}(2)&3^{4}\cdot 5\cdot 11\cdot 17\\\
\hline\cr&\mbox{No}&2^{10}{:}\Omega^{+}_{10}(2)&3^{3}\cdot 7\cdot
11&3{.}\mathrm{U}_{6}(2){.}[6]&2^{14}\cdot 5\cdot 7\cdot 17\cdot
31\end{array}$

If $A^{\infty}\leq\mathbf{rad}(A)X$ then we have $X\cong\mathrm{PSp}_{10}(2)$,
$\Omega^{-}_{10}(2)$ or $\Omega^{+}_{10}(2)$. Suppose that
$A^{\infty}\not\leq\mathbf{rad}(A)X$. Then $\mathbf{rad}(A)X\cap A^{\infty}$
is properly contained in $A^{\infty}$, and $X^{\infty}<A^{\infty}$. It is
easily shown that every odd prime divisor is also a divisor of
$|\mathbf{rad}(A)X\cap A^{\infty}|$. Check the maximal subgroups of
$A^{\infty}$ with order divisible by these primes, refer to [3, Tables
8.64-8.69]. Combining [32, Table 10.4], we list $X^{\infty}$ and those maximal
subgroups of $A$ which possibly contain $X^{\infty}$ in the first and second
columns of the above table, respectively. Then $X\cong\Omega^{-}_{10}(2)$ or
$\Omega^{+}_{10}(2)$. Therefore, $X\cong\mathrm{PSp}_{10}(2)$ or
$\Omega^{\pm}_{10}(2)$, and so $T=X<A$ and $G_{\alpha}\leq B$.

Assume that $T\cong\Omega^{-}_{10}(2)$. Then $|G:T|=2^{10}\cdot 3^{2}\cdot
7\cdot 31$, which is a divisor of $|G_{\alpha}|$ and $|B|$. It follows that
$B\cong 2^{15}{:}\mathrm{GL}_{6}(2)$ or $\mathrm{SL}_{6}(2){.}2$. Since
$2^{6}-1$ is not a prime, by Lemma 4.2, $\mathrm{SL}_{6}(2)\not\preccurlyeq
G_{\alpha}$. Thus $G_{\alpha}\mathbf{rad}(B)/\mathbf{rad}(B)$ is a proper
subgroup of $B/\mathbf{rad}(B)$. It is easy to see that
$|G_{\alpha}\mathbf{rad}(B)/\mathbf{rad}(B)|$ is divisible by $3^{2}\cdot
7\cdot 31$. By [32, Table 10.4],
$G_{\alpha}\mathbf{rad}(B)/\mathbf{rad}(B)\lesssim\mathrm{SL}_{5}(2)$ or
$2^{5}{:}\mathrm{SL}_{5}(2)$. By the Atlas [5], $\mathrm{SL}_{5}(2)$ has no
proper insolvable subgroup with order divisible by $7\cdot 31$. It follows
that $\mathrm{SL}_{5}(2)\lesssim G_{\alpha}\mathbf{rad}(B)/\mathbf{rad}(B)$.
Thus we have $\mathrm{SL}_{5}(2)\lesssim
G_{\alpha}\lesssim[2^{20}]{:}\mathrm{SL}_{5}(2)$.

Assume next that $T\cong\mathrm{Sp}_{10}(2)$ or $\Omega^{+}_{10}(2)$. Then
$|G:T|=2^{5}\cdot 3^{2}\cdot 7$ or $2^{10}\cdot 3^{3}\cdot 7\cdot 11$,
respectively. Suppose that $11\in\pi(G_{\alpha})$. Then
$7,\,11\in\pi(G_{\alpha})$. Checking those $B$ with order divisible by $77$,
we have $B\cong 3{.}\mathrm{U}_{6}(2){.}[6]$. By [32, Table 10.4],
$G_{\alpha}\mathbf{rad}(B)/\mathbf{rad}(B)$ has a normal subgroup
$\mathrm{U}_{6}(2)$ or $\mathrm{M}_{22}$, which is impossible by Lemma 4.2.
Thus $11\not\in\pi(G_{\alpha})$. Recalling that $|G_{\alpha}|$ is divisible by
$|G:T|$, we have $T\cong\mathrm{Sp}_{10}(2)$. If $31\in\pi(G_{\alpha})$ then
$\mathrm{SL}_{5}(2)\lesssim G_{\alpha}\lesssim[2^{20}]{:}\mathrm{SL}_{5}(2)$
by a similar argument as in above paragraph. Thus we assume that neither $11$
nor $31$ is a divisor of $|G_{\alpha}|$. Moreover, for
$B\cong\Omega^{+}_{6}(4){.}2^{2}\cong\mathrm{SL}_{4}(4){.}2^{2}$, by [32,
Table 10.3], $B$ has no proper subgroup of order divisible by $7\cdot 17$, and
thus $17\not\in\pi(G_{\alpha})$. Then the prime divisors of $|G_{\alpha}|$ are
$2$, $3$, $7$ or and $5$. By Lemma 4.2, we conclude that
$(G_{\alpha}\mathbf{rad}(B)/\mathbf{rad}(B))^{\infty}$ is a proper subgroup of
$(B/\mathbf{rad}(B))^{\infty}$. Checking the maximal subgroups of
$\mathrm{SL}_{6}(2)$, $\mathrm{U}_{6}(2)$ and $\mathrm{SL}_{4}(4)$ (refer to
[5] and [3, Tables 8.3 and 8.4]), we conclude that
$(G_{\alpha}\mathbf{rad}(B)/\mathbf{rad}(B))^{\infty}\succcurlyeq\mathrm{SL}_{3}(2)$
or $\mathrm{A}_{7}$. For $\mathrm{A}_{7}$, we have
$G_{\alpha}\lesssim\mathrm{S}_{7}$ by Lemma 4.2, and thus $|G_{\alpha}|$ is
indivisible by $2^{5}$, a contradiction. Thus
$G_{\alpha}\succcurlyeq\mathrm{SL}_{3}(2)$. If $\mathbf{O}_{2}(G_{\alpha})=1$
then $|G_{\alpha}|_{2}\leq 2^{3}$ by Lemma 4.2, and so
$|G_{\alpha}|_{2}|T|_{2}\leq 2^{28}<|G|_{2}$, a contradiction. Thus part (2)
of this lemma follows. ∎

######  Lemma 8.11.

Let $(m,q)\neq(6,2)$ and $A=\mathsf{N}_{1}$. Then $T\leq A$ and
$G_{\alpha}\leq B$.

###### Proof.

By the assumption, $|G:A|={1\over(2,q-1)}q^{m-1}(q^{m}-1)$, and $|G:B|$ is
listed as follows:

$\tiny\begin{array}[]{l|l|l}B&|G:B|&\\\ \hline\cr\mathsf{P}_{m}\mbox{ or
}\mathsf{P}_{m-1}&\prod_{i=1}^{m-1}(q^{i}+1)&\\\
\hline\cr{}^{\bf\hat{}}\mathrm{GU}_{m}(q){.}2&{1\over 2}q^{m(m-1)\over
2}\prod_{i=1}^{m-1}(q^{i}-(-1)^{i-1})&m\mbox{ even}\\\
\hline\cr(\mathrm{PSp}_{2}(q){\times}\mathrm{PSp}_{m}(q)).c&{1\over
c(q^{2}-1)}q^{(3m+2)(m-2)\over 4}\prod_{i={m\over
2}+1}^{m-1}(q^{2i}-1)&m\mbox{ even},c\leq 2,c=2\mbox{ for odd $q$ and
}4\,\,\big{|}\,\,m\\\ \hline\cr{}^{\bf\hat{}}\mathrm{GL}_{m}(q){.}2&{1\over
2}q^{m(m-1)\over 2}\prod_{i=1}^{m-1}(q^{i}+1)&m\mbox{ even}\\\
\hline\cr\Omega^{+}_{m}(4).2^{2}&q^{(m+2)(m-2)\over 2}\prod_{i=1}^{m\over
2}(q^{2(2i-1)}-1)&q=2,m\mbox{ even}\end{array}$

Clearly, $|G:B|$ is divisible by $\Phi^{*}_{2(m-1)}(q)$ or
$\Phi^{*}_{2(m-2)}(q)$. Moreover,
$A^{\prime}=\mathbf{soc}(A)\cong\Omega_{2m-1}$, $|A:A^{\prime}|\leq 2$, and
$|A:A^{\prime}|=2$ if and only if both $q$ and ${m(q-1)\over 2}$ are odd.

Suppose that $T\leq B$. Then $|G:T|=|G:B||B:T|$, and thus $|G:T|$ is divisible
by ${1\over(2,p)}q^{m(m-1)\over 2}$. Since $G=TG_{\alpha}$, by Lemma 2.4,
$|G_{\alpha}|_{p}\geq{1\over(2,p)}q^{m(m-1)\over 2}$, and $|G_{\alpha}|$ is
divisible by $\Phi^{*}_{2(m-1)}(q)$ or $\Phi^{*}_{2(m-2)}(q)$. Noting
$G=G_{\alpha}B$ and $A=G_{\alpha}(A\cap B)$ is core-free, next we deduce the
contradiction in two cases.

Case 1. Let $q$ be odd. By Lemma 2.7, choose a core-free maximal factorization
$A_{0}=A_{1}B_{1}$ such that $A_{0}=X_{1}B_{1}$ with $A_{0}=A$ or
$A^{\prime}$, $G_{\alpha}^{\infty}\leq X_{1}=A_{1}\cap G_{\alpha}$ and
$|G_{\alpha}:X_{1}|$ is divisor of $|A:A_{0}|$. Noting that $2m-1\geq 9$, by
[31, Theorem A], $A_{1}$ and $B_{1}$ are given in [31, Tables 1 and 2]. On the
other hand, it is easy to see $|X_{1}|$ is divisible by $\Phi^{*}_{2(m-1)}(q)$
or $\Phi^{*}_{2(m-2)}(q)$. Checking the orders of $A_{1}$ and $B_{1}$, we
conclude that $A_{1}\cong\Omega^{-}_{2(m-1)}(q){.}[2|A_{0}:A^{\prime}|]$. Note
that $\Omega^{-}_{2(m-1)}(q)\not\cong S\preccurlyeq X_{1}$. By [32, Table
10.1], $|X_{1}|$ and so $|G_{\alpha}|$ is indivisible by
$\Phi^{*}_{2(m-1)}(q)\Phi^{*}_{2(m-2)}(q)\Phi^{*}_{2(m-3)}(q)$. Then $|G:B|$
is indivisible by $(q^{m-1}+1)(q^{m-2}+1)(q^{m-3}+1)$. It follows that
$B\cong{}^{\bf\hat{}}\mathrm{GU}_{m}(q){.}2$ for some even $m$, and then
$|G_{\alpha}|$ divisible by $q^{m(m-1)\over 2}\Phi^{*}_{2(m-2)f}(p)$. Since
$m$ is even, we have $A=A^{\prime}=A_{0}$, and so $G_{\alpha}=X_{1}\leq
A_{1}\cong\Omega^{-}_{2(m-1)}(q){.}2$.

It is easily shown that $S\preccurlyeq G_{\alpha}\cap A_{1}^{\prime}$, and
$|G_{\alpha}\cap A^{\prime}_{1}|$ is divisible by $q^{m(m-1)\over
2}\Phi^{*}_{2(m-2)f}(p)$. Combining Lemma 4.2, it follows from [2, Theorem
3.1] that $G_{\alpha}\cap
A^{\prime}_{1}\lesssim\mathrm{GO}^{+}_{2}(q)\times\mathrm{GO}^{-}_{2m-4}(q)$,
$[q^{2m-4}].(q-1\times\mathrm{GO}^{-}_{2m-4}(q))$, $\mathrm{GO}_{2m-3}(q)$,
$\mathrm{GO}^{-}_{2{m-1\over d}}(q^{d}){.}[d]$, $\mathrm{GU}_{m-1}(q){.}2$ or
$\mathrm{GO}_{m-1}(q^{2}){.}2$, where $d>1$ is a divisor of $m-1$. The last
three groups are excluded as their orders are indivisible by $q^{m(m-1)\over
2}$. If $G_{\alpha}\cap A^{\prime}_{1}\lesssim\mathrm{GO}_{2m-3}(q)$ then, by
Lemma 7.2, we have $(S,2m-3,q)=(\mathrm{A}_{7},7,3)$ or
$(\mathrm{A}_{7},7,5)$, which yields that $|G_{\alpha}|$ is indivisible by
$q^{m(m-1)\over 2}$, a contradiction. Therefore, $G_{\alpha}\cap
A^{\prime}_{1}\lesssim\mathrm{GO}^{+}_{2}(q)\times\mathrm{GO}^{-}_{2m-4}(q)$
or $[q^{2m-4}].(q-1\times\mathrm{GO}^{-}_{2m-4}(q))$. It follows from Lemma
7.1 that $S\cong\mathrm{L}_{2}(q^{m-2})$, and so $r=q^{m-2}+1$. By Lemma 2.1,
$q$ is even, a contradiction.

Case 2. Let $q$ be even. Then $A\cong\mathrm{Sp}_{2m-2}(q)$. Since $m\geq 5$,
we have $\Omega^{-}_{2(m-1)}(q)\not\preccurlyeq G_{\alpha}$. Suppose that
$(m,q)=(5,2)$. Then $|G:A|=2^{4}\cdot 31$, and $T\leq B\cong
2^{10}{:}\mathrm{GL}_{5}(2)$. Recalling $G=TA$, by Lemma 2.4, $|T|$ is
divisible $31$, yielding $T=\mathrm{GL}_{5}(2)$. Thus $|G:T|=2^{10}\cdot
3^{3}\cdot 5\cdot 17$, which is a divisor of $|G_{\alpha}|$. Checking the
maximal subgroups of $\mathrm{Sp}_{8}(2)$, we conclude that
$G_{\alpha}\lesssim\mathrm{SO}^{-}_{8}(2)$. Checking the maximal subgroups of
$\mathrm{SO}^{-}_{8}(2)$, we get $\Omega^{-}_{8}(2)\unlhd G_{\alpha}$, which
is impossible by Lemma 4.2. Then $m\geq 7$ if $q=2$.

Assume that $|G_{\alpha}|$ is divisible by $\Phi^{*}_{2m-2}(q)$. Recall that
$|G_{\alpha}|_{2}\geq{1\over 2}q^{m(m-1)\over 2}$. Then $|G_{\alpha}|$ is
divisible by ${1\over 2}q^{m(m-1)\over 2}\Phi^{*}_{2m-2}(q)$. Applying Lemma
7.1 to $A$ and $G_{\alpha}$, either $S\cong\mathrm{L}_{2}(q^{m-1})$ and
$|G_{\alpha}|_{2}\leq(m-1)|S|_{2}$ with $(m-1)f$ a power of $2$, or
$(S,2m-2,q)$ is one of $(\mathrm{A}_{13},12,2)$, $(\mathrm{A}_{19},18,2)$ and
$(\mathrm{L}_{3}(3),12,2)$. The former case implies that ${1\over
2}q^{m(m-1)\over 2}\leq(m-1)q^{m-1}$, which is impossible as $m\geq 5$. For
the latter case, we have $|G_{\alpha}|_{2}<2^{m-2}$ by Lemma 4.2, again a
contradiction.

Assume that $|G_{\alpha}|$ is indivisible by $\Phi^{*}_{2m-2}(q)$. Then
$B=^{\bf\hat{}}\mathrm{GU}_{m}(q){.}2$, $m$ is even and $|G_{\alpha}|$ is
divisible by ${1\over 2}q^{m(m-1)\over 2}\Phi^{*}_{2m-4}(q)$. Applying [2,
Theorem 3.1] to $A$ and $G_{\alpha}$, we have $G_{\alpha}\leq
A_{1}\cong[q^{2m-3}]{:}((q-1){\times}\mathrm{Sp}_{2m-4}(q))$ or
$\mathrm{Sp}_{2}(q)\times\mathrm{Sp}_{2m-4}(q)$. If
$S\lesssim\mathrm{Sp}_{2}(q)$ then, by Lemma 4.2, $|G_{\alpha}|_{2}$ is a
divisor of $q^{4}f^{2}$, and so ${1\over 2}q^{m(m-1)\over
2}\leq(m-1)q^{m-1}\leq|G_{\alpha}|_{2}\leq q^{5}$, yielding $m\leq 4$, a
contradiction. Let $N\unlhd A_{1}$ with $A_{1}/N\cong\mathrm{Sp}_{2m-4}(q))$.
Then $S\preccurlyeq G_{\alpha}N/N$, and $|G_{\alpha}N/N|$ is divisible by
$\Phi^{*}_{2m-4}(q)$. By Lemma 7.1, either $S\cong\mathrm{L}_{2}(q^{m-2})$ and
$|G_{\alpha}N/N|_{2}\leq(m-2)|S|_{2}$ with $(m-2)f$ a power of $2$, or
$(S,2m-4,q)$ is one of $(\mathrm{A}_{13},12,2)$ and
$(\mathrm{L}_{3}(3),12,2)$. For the latter case, $m=8$ and
$|G_{\alpha}|_{2}\geq 2^{27}$; however, by Lemma 4.2, $|G_{\alpha}|_{2}\leq
2^{10}$, a contradiction. The former case implies that ${1\over
2}q^{m(m-1)\over
2}\leq|G_{\alpha}|_{2}\leq|G_{\alpha}N/N|_{2}|N|_{2}\leq(m-2)q^{3m-5}$,
yielding $(m,q)=(6,2)$, again a contradiction. ∎

######  Lemma 8.12.

Let $(m,q)\neq(6,2)$ and $A=\mathsf{N}_{1}$. Then one of the following holds.

  * (1)

$T\cong\mathrm{Sp}_{2m-2}(q)$, $\mathrm{SL}_{2}(q^{m\over 2})\unlhd
G_{\alpha}^{\Gamma(\alpha)}$, $\mathbf{O}_{2}(G_{\alpha})\neq 1$, $m\geq 8$,
$p=2$ and $mf$ is a power of $2$.

  * (2)

$T\cong\mathrm{Sp}_{2m-2}(q)$ or $\Omega^{-}_{2m-2}(q)$,
$\mathrm{L}_{d}(q^{m\over d})\unlhd G_{\alpha}^{\Gamma(\alpha)}$, $p=2$, $d$
is an odd prime and $mf$ is a power of $d$.

  * (3)

$T\cong\Omega_{2m-1}(q)$, $\mathrm{L}_{d}(q^{m\over d})\unlhd
G_{\alpha}^{\Gamma(\alpha)}$, $p$ and $d$ are odd, and $mf$ is a power of $d$.

###### Proof.

By Lemma 8.11, $T\leq A$ and $G_{\alpha}\leq B$, and then Lemma 7.6 works.
Note that $|G:A|={1\over(2,q-1)}q^{m-1}(q^{m}-1)$. In particular,
$|G_{\alpha}|_{p}\geq q^{m-1}$. For those triples in Lemma 7.6 (2), by Lemma
4.2, we have $|G_{\alpha}|_{p}\leq q^{m-2}$, and thus they are excluded. Thus
$S\cong\mathrm{L}_{d}(q^{m\over d})$, $d$ is a prime and $mf$ is a power of
$d$. In particular, $m$ is even if and only if $d=p=2$.

Assume that $d=2$. Then $p=2$ and $m$ is a power of $2$, and so
$A\cong\mathrm{Sp}_{2m-2}(q)$ and $m\geq 8$. Noting that $|G_{\alpha}|$ is a
divisor of $|\mathbf{O}_{p}(G_{\alpha})|\mathrm{GL}_{2}(q^{m\over
2})|{m^{2}f^{2}\over 4}$, it follows that
$(|G_{\alpha}|,\Phi^{*}_{2m-2}(q)\Phi^{*}_{2m-4}(q)\Phi^{*}_{m-1}(q))=1$.
Since $G=TG_{\alpha}=TB$, by Lemma 2.4, $|T|$ is divisible by
$\Phi^{*}_{2m-2}(q)\Phi^{*}_{2m-4}(q)\Phi^{*}_{m-1}(q)$. By [32, Table 10.1],
we have $T=A\cong\mathrm{Sp}_{2m-2}(q)$. Suppose that
$\mathbf{O}_{p}(G_{\alpha})=1$. Then $|G_{\alpha}|_{p}$ is a divisor of
${mf\over 2}q^{m\over 2}$. It follows that
$q^{m-1}\leq|G_{\alpha}|_{p}\leq{mf\over 2}q^{m\over 2}<q^{{3m\over 4}+1}$,
yielding $m<8$, a contradiction. Thus $\mathbf{O}_{p}(G_{\alpha})\neq 1$. Then
part (1) of this lemma follows.

Assume $d$ is odd. Then $m$ is odd, and $B=\mathsf{P}_{m}$ or
$\mathsf{P}_{m-1}$. Suppose that $q$ is even. Then
$A\cong\mathrm{Sp}_{2m-2}(q)$. Noting that $|T|$ is divisible by
$\Phi^{*}_{2m-2}(q)\Phi^{*}_{2m-4}(q)\Phi^{*}_{2m-6}(q)$, by [32, Table 10.1],
we have $T=\mathrm{Sp}_{2m-2}(q)$ or $\Omega^{-}_{2m-2}(q)$, and part (2) of
this lemma follows. Suppose that $q$ is odd. Then
$A\cong\Omega_{2m-1}(q){.}{4\over(4,m(q-1))}$. Noting that $|G_{\alpha}|$ is a
divisor of $|\mathbf{O}_{p}(G_{\alpha})|\mathrm{GL}_{d}(q^{m\over
d})|{m^{2}f^{2}\over d^{2}}$, it follows that $|B:G_{\alpha}|$ is divisible by
$\prod_{m>i\geq 2,(i,d)=1}\Phi^{*}_{i}(q)$. In particular, $|B:G_{\alpha}|$ is
divisible by $\Phi^{*}_{m-1}(q)$. Noting that
$|G:G_{\alpha}|=|G:B||B:G_{\alpha}|$, it follows that $|G:G_{\alpha}|$ is
divisible by $\Phi^{*}_{m-1}(q)\prod_{i=1}^{m-1}(q^{i}+1)$. Then $|T|$ is
divisible by $\Phi^{*}_{m-1}(q)\prod_{i=1}^{m-1}(q^{i}+1)$, and thus $|T|$ has
a divisor $(\Phi^{*}_{m-1}(q))^{2}\Phi^{*}_{2m-2}(q)\Phi^{*}_{2m-4}(q)$. Since
$m\geq 5$ and $q$ is odd, $\Phi^{*}_{m-1}(q)\neq 1$. By [32, Table 10.2],
$T\cong\Omega_{2m-1}$, desired as in part (3) of this lemma. ∎

######  Lemma 8.13.

Let $(m,q)\neq(6,2)$ and $A=\mathsf{N}_{2}^{-}$. Then
$T\cong\mathrm{P\Omega}^{-}_{2m-2}(q)$, $\mathrm{L}_{m}(q)\unlhd
G_{\alpha}^{\Gamma(\alpha)}$, $m$ is an odd prime and $f$ is a power of $m$.

###### Proof.

In this case, $|G:A|=q^{2m-2}{(q^{m}-1)(q^{m-1}-1)\over 2(q+1)}$, and $B$ is
$\mathsf{P}_{m}$, $\mathsf{P}_{m-1}$ or ${}^{\bf\hat{}}\mathrm{GL}_{m}(2){.}2$
(with $q=2$ and $m$ even). Then $|G:B|$ is divisible by
$\prod_{i=1}^{m-1}(q^{i}+1)$. Let $\overline{A}=A/\mathbf{rad}(A)$ and
$\overline{X}=X\mathbf{rad}(A)/\mathbf{rad}(A)$. It is easily shown that
$\mathbf{soc}(\overline{A})\cong\mathrm{P\Omega}^{-}_{2m-2}(q)$ and
$|\overline{X}|$ is divisible by
$\Phi^{*}_{2m-2}(q)\Phi^{*}_{2m-4}(q)\Phi^{*}_{2m-6}(q)$. Noting that
$(m,q)\neq(6,2)$, it follows from [32, Table 10.1] that either
$\mathrm{P\Omega}^{-}_{2m-2}(q)\lesssim\overline{X}$, or $(m-1,q)=(4,2)$ and
$\Omega^{-}_{8}(2)\not\lesssim\overline{X}$. Suppose that the latter case
occurs. Checking the maximal subgroups of $\Omega^{-}_{8}(2)$, we get
$\overline{X}\lesssim\mathrm{L}_{2}(16){.}2$. Since $\mathrm{L}_{2}(16){.}2$
has no proper subgroup of order divisible by $\Phi^{*}_{8}(2)\Phi^{*}_{4}(2)$,
we have $\mathrm{L}_{2}(16)\lesssim\overline{X}$. If
$T=X\cong\mathrm{L}_{2}(16)$ then $|T|$ is indivisible by $|G:B|$, a
contradiction. Thus $X=G_{\alpha}$, and $T\leq B$, yielding
$T\lesssim\mathrm{SL}_{5}(2)$. By Lemma 2.10, it is easy to show that
$|G_{\alpha}|_{3}\leq 9$, and so $|T|_{3}|G_{\alpha}|_{3}\leq
3^{4}<3^{5}=|G|_{3}$, a contradiction. Therefore, we have
$\mathrm{P\Omega}^{-}_{2m-2}(q)\lesssim\overline{X}$. This implies that $A\geq
T\cong\mathrm{P\Omega}^{-}_{2m-2}(q)$. Then our result follows from Lemma 7.5.
∎

######  Theorem 8.5.

Assume that Hypothesis 6.1 holds and $G=\mathrm{P\Omega}_{2m}^{+}(q)$, where
$m\geq 5$ and $q=p^{f}$. Then one of the following holds.

  * (1)

$(m,q)=(6,2)$, $T=\Omega^{-}_{10}(2)$ or $\mathrm{Sp}_{10}(2)$, and
$\mathrm{SL}_{5}(2)\leq G_{\alpha}\leq[2^{20}]{:}\mathrm{SL}_{5}(2)$;

  * (2)

$(m,q)=(6,2)$, $T=\mathrm{Sp}_{10}(2)$,
$G_{\alpha}^{\Gamma(\alpha)}\cong\mathrm{SL}_{3}(2)$ and
$\mathbf{O}_{2}(G_{\alpha})\neq 1$.

  * (3)

$T\cong\mathrm{Sp}_{2m-2}(q)$, $\mathrm{SL}_{2}(q^{m\over 2})\unlhd
G_{\alpha}^{\Gamma(\alpha)}$, $\mathbf{O}_{2}(G_{\alpha})\neq 1$, $m\geq 8$,
$p=2$ and $mf$ is a power of $2$.

  * (4)

$T\cong\mathrm{Sp}_{2m-2}(q)$ or $\Omega^{-}_{2m-2}(q)$,
$\mathrm{L}_{d}(q^{m\over d})\unlhd G_{\alpha}^{\Gamma(\alpha)}$, $p=2$, $d$
is and odd prime and $mf$ is a power of $d$.

  * (5)

$T\cong\Omega_{2m-1}(q)$, $\mathrm{L}_{d}(q^{m\over d})\unlhd
G_{\alpha}^{\Gamma(\alpha)}$, $p$ and $d$ are odd, and $mf$ is a power of $d$.

  * (6)

$T=\mathrm{P\Omega}^{-}_{2m-2}(q)$, $\mathrm{L}_{m}(q)\unlhd
G_{\alpha}^{\Gamma(\alpha)}$, $m$ is an odd prime and $f$ is a power of $m$.

###### Proof.

By Lemmas 8.10, 8.12 and 8.13, we assume that $A=\mathsf{P}_{1}$ and
$(m,q)\neq(6,2)$. In this case, $m$ is even,
$|G:A|={(q^{m}-1)(q^{m-1}+1)\over(q-1)}$,
$B={}^{\bf\hat{}}\mathrm{GU}_{m}(q){.}2$ and $|G:B|={1\over 2}q^{m(m-1)\over
2}\prod_{i=1}^{m-1}(q^{i}+(-1)^{i})$. We next deduce a contradiction.

Suppose that $X=G_{\alpha}$. Let $\overline{A}=A/\mathbf{rad}(A)$ and
$\overline{X}=X\mathbf{rad}(A)/\mathbf{rad}(A)$. Then
$|\mathbf{rad}(A)|_{p}=q^{2m-2}$,
$\mathbf{soc}(\overline{A})\cong\mathrm{P\Omega}^{+}_{2m-2}(q)$,
$|\overline{A}:\mathbf{soc}(\overline{A})|\leq 2$ and $|\overline{X}|$ is
divisible by ${1\over 2}q^{(m-4)(m-1)\over
2}\prod_{i=2}^{m-1}(q^{i}+(-1)^{i})$. By Lemma 2.6,
$\overline{A}=\overline{A}\,\overline{A\cap B}$ is core-free, where
$\overline{A\cap B}=(A\cap B)\mathbf{rad}(A)/\mathbf{rad}(A)$. By Lemma 2.7,
we have a core-free maximal factorization $G_{0}=A_{0}B_{0}$ such that
$\overline{A}\geq G_{0}=X_{0}B_{0}$, $X_{0}=A_{0}\cap\overline{X}$,
$\mathbf{soc}(G_{0})\cong\mathrm{P\Omega}^{+}_{2m-2}(q)$,
$\overline{X}^{\infty}\leq X_{0}$ and $|X_{0}|$ has a divisor ${1\over
o}|\overline{X}|$, where $o\leq 2$. In particular, $|X_{0}|$ is divisible by
${1\over(4,p^{2})}q^{(m-4)(m-1)\over
2}\Phi^{*}_{2m-4}(q)\Phi^{*}_{m-1}(q)\Phi^{*}_{m-3}(q)$. Noting that $m-1$ is
odd and $|A_{0}|$ is divisible by $\Phi^{*}_{2m-4}(q)$, by [31, Theorem A and
Proposition 2.5], $A_{0}$ has type $\mathsf{N}_{1}$ or $\mathsf{N}^{-}_{2}$.
However, in this case, $|A_{0}|$ is indivisible by $\Phi^{*}_{m-1}(q)$, a
contradiction.

Now, we have $T\leq A$ and $G_{\alpha}\leq
B\cong{}^{\bf\hat{}}\mathrm{GU}_{m}(q){.}2$. Then $|G_{\alpha}|$ is divisible
by $|G:A|$; in particular, $|G_{\alpha}|$ is divisible by
$\Phi^{*}_{m}(q)\Phi^{*}_{2m-2}(q)$. Applying Lemma 7.2 to the preimages of
$G_{\alpha}\cap B^{\prime}$ and $B^{\prime}$ in $\mathrm{GL}_{m}(q^{2})$, we
conclude that $(m,q^{2})=(4,9),\,(4,25)$ or $(6,4)$. Then $m<5$ or
$(m,q)=(6,2)$, a contradiction. ∎

###  8.6. Symplectic case

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

######  Lemma 8.14.

$A$ and $B$ are not in [31, Table 3].

###### Proof.

Suppose that $A$ and $B$ are described as in [31, Table 3]. Since $A$ and $B$
are insolvable, they are listed as follows:

$\tiny\begin{array}[]{l|l|l|l|l}G&A&|G:A|&B&|G:B|\\\
\hline\cr\mathrm{PSp}_{6}(3)&\mathrm{L}_{2}(13)&2^{7}\cdot 3^{8}\cdot
5&[3^{5}]{:}2\mathrm{U}_{4}(2)&2^{2}\cdot 7\cdot 13\\\
\hline\cr\mathrm{PSp}_{8}(2)&\Omega^{-}_{8}(2){.}2&2^{3}\cdot 3\cdot
5&\mathrm{S}_{10}&2^{8}\cdot 3\cdot 17\\\ &\mathrm{L}_{2}(17)&2^{12}\cdot
3^{3}\cdot 5^{2}\cdot 7&\Omega^{+}_{8}(2){.}2&2^{3}\cdot 17\end{array}$

Assume that $G=\mathrm{PSp}_{6}(3)$. Then $X=A\cong\mathrm{L}_{2}(13)$. Let
$\overline{B}=B/\mathbf{rad}(B)$ and
$\overline{Y}=Y\mathbf{rad}(B)/\mathbf{rad}(B)$. Then
$\overline{B}\cong\mathrm{U}_{4}(2)$ and $|\overline{Y}|$ has a divisor
$2^{6}\cdot 3^{3}\cdot 5$. Searching the subgroups of $\mathrm{U}_{4}(2)$, we
conclude that $\overline{Y}\succcurlyeq\mathrm{U}_{4}(2)$ or $\Omega_{5}(3)$.
By Lemma 4.2, neither $X$ nor $Y$ plays the role of $G_{\alpha}$, a
contradiction. Thus $G=\mathrm{PSp}_{8}(2)$.

Suppose that $A\cong\mathrm{L}_{2}(17)$ and $B\cong\Omega^{+}_{8}(2){.}2$. It
is easily shown that $X=A\cong\mathrm{L}_{2}(17)$, and $|Y\cap B^{\prime}|$
has a divisor $5^{2}\cdot 7$. By [32, Table 10.4], $Y\cap
B^{\prime}\cong\Omega^{+}_{8}(2)$. Then $X\neq G_{\alpha}$ and $Y\neq
G_{\alpha}$, a contradiction. Suppose that $A\cong\Omega^{-}_{8}(2){.}2$. Then
$X\cong\Omega^{-}_{8}(2)$ due to searching the subgroups of
$\Omega^{-}_{8}(2)$ with order divisible by $2^{7}\cdot 3\cdot 17$. Thus
$X=T$, and $G_{\alpha}=Y\leq B$, which contradicts Lemma 7.7. ∎

######  Lemma 8.15.

If $A$ and $B$ are as in [31, Table 2], then $p=2$, $f>2$ is a power of $2$,
$G=\mathrm{Sp}_{6}(q)$, $T\cong\mathrm{G}_{2}(q)$,
$\mathrm{SL}_{2}(q^{2})\unlhd G_{\alpha}^{\Gamma(\alpha)}$ and
$\mathbf{O}_{2}(G_{\alpha})\neq 1$.

###### Proof.

Suppose that $A$ and $B$ are described as in [31, Table 2]. Then $q$ is even,
and $(G,A)$ is one of $(\mathrm{Sp}_{4}(q),\mathrm{Sz}(q))$ and
$(\mathrm{Sp}_{6}(q),\mathrm{G}_{2}(q))$.

Assume that $G=\mathrm{Sp}_{4}(q)$ and $A\cong\mathrm{Sz}(q)$. Then $f>1$ is
odd. Noting that $A=X(A\cap B)$, by [21], we have $A=X$. Thus
$T=X\cong\mathrm{Sz}(q)$, and $G_{\alpha}=Y\leq B\cong\mathrm{GO}^{+}_{4}(q)$.
Since $G=TG_{\alpha}$ and $|G:T|=q^{2}(q^{2}-1)(q+1)$, by Lemma 2.4,
$|G_{\alpha}|$ divisible by $q^{2}(q^{2}-1)(q+1)$. Then $|B:G_{\alpha}|$ is a
divisor of $2(q-1)$. This implies that $G_{\alpha}$ has a normal subgroup
$\mathrm{L}_{2}(q)\times\mathrm{L}_{2}(q)$, which is impossible by Lemma 4.2.

Assume that $G=\mathrm{Sp}_{6}(q)$ and $A\cong\mathrm{G}_{2}(q)$. Then
$|G:A|=q^{3}(q^{4}-1)$ and
$B\in\\{\mathrm{GO}^{\pm}_{6}(q),\mathsf{P}_{1},\mathsf{N}_{2}\\}$. Suppose
that $X\neq A$. Noting that $A=X(A\cap B)$, by [21], $q=4$ and
$X\cong\mathrm{J}_{2}$, $\mathrm{SU}_{3}(4)$ or $\mathrm{SU}_{3}(4){.}2$. Then
$T=X\cong\mathrm{J}_{2}$ or $\mathrm{U}_{3}(4)$, and $G_{\alpha}=Y\leq B$.
Noting that $|T|$ is divisible by $|G:B|$, the factorization $G=TB$ is listed
as follows:

$\tiny\begin{array}[]{l|l|l|l}T&|G:T|&B&|G:B|\\\
\hline\cr\mathrm{J}_{2}&2^{11}\cdot 3\cdot 5\cdot 13\cdot
17&\mathrm{U}_{4}(4){.}2&2^{5}\cdot 3^{3}\cdot 7\\\
\hline\cr\mathrm{U}_{3}(4)&2^{12}\cdot 3^{3}\cdot 5\cdot 7\cdot
17&\mathrm{L}_{4}(4){.}2&2^{5}\cdot 5\cdot 13\\\ \end{array}$

Clearly, $S\preccurlyeq Y\cap B^{\prime}$, and then $\pi(Y)=\pi(S)=\pi(Y\cap
B^{\prime})$. Then, by [32, Table 10.3], we get $Y\cap
B^{\prime}\cong\mathrm{U}_{4}(4)$ or $\mathrm{L}_{4}(4)$. Thus
$G_{\alpha}=Y\succcurlyeq\mathrm{U}_{4}(4)$ or $\mathrm{L}_{4}(4)$, a
contradiction. Therefore, $X=A\cong\mathrm{G}_{2}(q)$, and so
$T\cong\mathrm{G}_{2}(q)$ for $q>2$.

Let $\overline{B}=B^{\infty}\mathbf{rad}(B)/\mathbf{rad}(B)$ and
$\overline{Y}=(Y\cap B^{\infty})\mathbf{rad}(B)/\mathbf{rad}(B)$. Then
$|\overline{Y}|$ has a divisor $\Phi^{*}_{4}(q)$, and
$S\preccurlyeq\overline{Y}$. If $\overline{B}\cong\mathrm{L}_{4}(q)$,
$\mathrm{PSp}_{4}(q)$ or $\mathrm{PSp}_{2}(q)\times\mathrm{PSp}_{4}(q)$ then,
by Lemma 7.1, we conclude that $S\cong\mathrm{L}_{2}(q^{2})$. Assume that
$\overline{B}\cong\mathrm{U}_{4}(q)$. Then $|\overline{Y}|$ has a divisor
${1\over 2}q^{3}\Phi^{*}_{4}(q)$. Checking the maximal subgroups
$\mathrm{U}_{4}(q)$, we have $\overline{Y}\lesssim
q^{4}{:}([a]{.}\mathrm{L}_{2}(q^{2}){.}[b])$, where
$ab={(q-1)(2,q^{2}-1)\over(4,q+1)}$. This implies that
$S\cong\mathrm{L}_{2}(q^{2})$. All in all, $r=q^{2}+1$, and so $p=2$ and $f$
is a power of $2$ by Lemma 2.1. If $\mathbf{O}_{2}(G_{\alpha})=1$ then
$q^{3}\leq|G_{\alpha}|_{2}\leq q^{2}2f$, yielding $q=4$. For $q=4$, confirmed
by GAP, there exists no a desired graph. Then the lemma follows. ∎

Assume that $A$ and $B$ are as in [31, Table 1]. For $(m,q)=(3,2)$ or $(2,4)$,
$A$ and $B$ are listed as follows:

$\tiny\begin{array}[]{l|l|l|l|l}(m,q)&A&|G:A|&B&|G:B|\\\
\hline\cr(3,2)&\mathrm{Sp}_{2}(8){.}3&2^{6}\cdot 3\cdot
5&2^{5}{:}\mathrm{S}_{6}&3^{2}\cdot 7\\\ &&&\mathrm{S}_{8}&2^{2}\cdot 3^{2}\\\
&&&\Omega^{-}_{6}(2){.}2&2^{2}\cdot 7\\\ &\Omega^{-}_{6}(2){.}2&2^{2}\cdot
7&2^{6}{:}\mathrm{L}_{3}(2)&3^{3}\cdot 5\\\ &&&\mathrm{S}_{8}&2^{2}\cdot
3^{2}\\\ \hline\cr(2,4)&\mathrm{Sp}_{2}(16){.}2&2^{3}\cdot 3\cdot
5&2^{6}{:}(3\times\mathrm{Sp}_{2}(4))&5\cdot 17\\\
&&&\Omega^{+}_{4}(4){.}2&2^{3}\cdot 17\\\ &&&\Omega^{-}_{4}(4){.}2&2^{3}\cdot
3\cdot 5\\\ &\Omega^{-}_{4}(4){.}2&2^{3}\cdot 3\cdot
5&2^{6}{:}\mathrm{GL}_{2}(4)&5\cdot 17\\\
&&&(\mathrm{Sp}_{2}(4)\times\mathrm{Sp}_{2}(4)){:}2&2^{3}\cdot 17\end{array}$

We first search the insolvable subgroups of $A$ and $B$ with orders divisible
by $|G:B|$ and $|G:A|$, respectively. It follows that $X^{\prime}=A^{\prime}$,
and either $Y^{\prime}=B^{\prime}$ or $G_{\alpha}=Y$, and then all possible
pairs $(T,G_{\alpha})$ are easily determined. Calculation shows that no
desired graph arises for $(m,q)=(3,2)$ or $(4,4)$.

######  Lemma 8.16.

Let $A$ and $B$ be as in [31, Table 1]. Then $(m,q)\neq(3,2),\,(2,4)$.

For convenience, we list $A$, $B$ and their indices in $G$ as follows:

$\tiny\begin{array}[]{l|l|l|l|l}A&|G:A|&B&|G:B|&\\\
\hline\cr\mathrm{PSp}_{2a}(q^{b}){.}b&{1\over
b}q^{m(m-a)}\prod_{(i,b)=1}(q^{2i}-1)&\mathsf{P}_{1}&{q^{2m}-1\over q-1}&\\\
&&\mathrm{GO}^{\pm}_{2m}(q)&q^{m}{q^{m}\pm 1\over 2}&p=2\\\
&&\mathsf{N}_{2}&q^{2(m-1)}{q^{2m}-1\over q^{2}-1}&b=q=2,2\,\,\big{|}\,\,m\\\
\hline\cr\mathrm{GO}^{-}_{2m}(q)&q^{m}{q^{m}-1\over
2}&\mathsf{P}_{m}&\prod_{i=1}^{m}(q^{i}+1)&p=2\\\
&&\mathrm{Sp}_{m}(q)\wr\mathrm{S}_{2}&{1\over 2}q^{m^{2}\over
2}\prod_{i=1}^{m\over 2}{q^{m+2i}-1\over q^{2i}-1}&2\,\,\big{|}\,\,m,p=2\\\
&&\mathrm{GO}^{+}_{2m}(2)&q^{m}{q^{m}+1\over 2}&q=2\par\end{array}$

where $b$ is a prime such that $m=ab$.

######  Lemma 8.17.

Let $A$ and $B$ be as in [31, Table 1], and $m=2$. Then $p=2$, $f>2$ is a
power of $2$, $T\cong\mathrm{L}_{2}(q^{2})$, and either
$G_{\alpha}\cong\mathrm{L}_{2}(q^{2})$ or $\mathrm{SL}_{2}(q)\unlhd
G_{\alpha}^{\Gamma(\alpha)}$ and $\mathbf{O}_{2}(G_{\alpha})\neq 1$.

###### Proof.

Note that $A^{\prime}\cong\mathrm{L}_{2}(q^{2})$ and $q>2$. Assume that
$B\cong\mathrm{GO}^{-}_{4}(q)$. Then $A\cong\mathrm{Sp}_{2}(q^{2}){.}2$,
$B^{\prime}\cong\mathrm{L}_{2}(q^{2})$ for even $q$, and both $|X\cap
A^{\prime}|$ and $|Y\cap B^{\prime}|$ are divisible by ${1\over
4}q^{2}(q^{2}-1)$. Checking the insolvable subgroups of
$\mathrm{L}_{2}(q^{2})$, it follows that either $X\cap A^{\prime}\cong Y\cap
B^{\prime}\cong\mathrm{L}_{2}(q^{2})$, or one of $X\cap A^{\prime}$ and $Y\cap
B^{\prime}$ is isomorphic to $\mathrm{A}_{5}$ if $q=4$. By Lemma 8.16, $q\neq
4$. Thus $X\cap A^{\prime}\cong Y\cap B^{\prime}\cong\mathrm{L}_{2}(q^{2})$.
It follows that $T\cong\mathrm{L}_{2}(q^{2})$,
$G_{\alpha}\cong\mathrm{L}_{2}(q^{2}){.}o$ and $r=q^{2}+1$, where $o\leq 2$.
Then, by Lemma 2.1, $f$ is a power of $2$. In particular, $f\geq 4$. Suppose
that $G_{\alpha}\cong\mathrm{L}_{2}(q^{2}){.}2$. Then $|T:(T\cap
G_{\alpha})|=|G:G_{\alpha}|={1\over 2}q^{2}(q^{2}-1)$, and thus
$T_{\alpha}=T\cap G_{\alpha}\cong\mathrm{D}_{2r}$, a maximal subgroup of $T$.
Thus both $T$ and $G$ are primitive on $V$, and $T_{\alpha}$ and $G_{\alpha}$
have a common orbit on $V$ of size $r$, which is impossible by [33, Theorem 1
and Table 7.1]. Thus $G_{\alpha}\cong\mathrm{L}_{2}(q^{2})$.

Assume next that $B\not\cong\mathrm{GO}^{-}_{4}(q)$. Then $|X\cap A^{\prime}|$
is divisible by $\Phi^{*}_{2}(q^{2})$. Note that $X\cap A^{\prime}$ is
insoluble. Checking the subgroups of $\mathrm{L}_{2}(q^{2})$, it follows that
either $X\cap A^{\prime}=A^{\prime}$, or $q=p\equiv\pm
3\,({\mathrm{mod}~{}}10)$ and $X\cap A^{\prime}\cong\mathrm{A}_{5}$. For the
latter, we have $\Phi^{*}_{2}(q^{2})=5$ or $15$, which is impossible by [20,
Theorem 3.9]. Thus $X=A^{\prime}$ or $A$.

Let $\overline{B}=B^{\infty}\mathbf{rad}(B)/\mathbf{rad}(B)$ and
$\overline{Y}=(Y\cap B^{\infty})\mathbf{rad}(B)/\mathbf{rad}(B)$. Then
$\overline{B}\cong\mathrm{L}_{2}(q)$ or
$\mathrm{L}_{2}(q)\times\mathrm{L}_{2}(q)$, $\overline{Y}$ is insolvable and
has order divisible by $\Phi^{*}_{2}(q)$. Clearly, if $q\neq p$ then
$\Phi^{*}_{2}(q)\neq 1$. Checking the insolvable subgroups of
$\mathrm{L}_{2}(q)$, we conclude that either
$\mathrm{L}_{2}(q)\preccurlyeq\overline{Y}$, or
$\mathrm{A}_{5}\preccurlyeq\overline{Y}$ and $q=p\equiv\pm
1\,({\mathrm{mod}~{}}10)$. Then $Y\succcurlyeq\mathrm{L}_{2}(q)$ or
$\mathrm{A}_{5}$.

If $T=Y$, recalling $T\not\cong\mathrm{A}_{5}$, then
$q^{4}=|G|_{p}\leq|T|_{p}|G_{\alpha}|=|T|_{p}|X|\leq q(2,p)q^{2}$, which is
impossible as $q>2$. Thus we have $Y=G_{\alpha}$, $X=T$ and
$S=\mathbf{soc}(G_{\alpha}^{\Gamma(\alpha)})\cong\mathrm{L}_{2}(q)$ or
$\mathrm{A}_{5}$. If $S\cong\mathrm{A}_{5}$, recalling $q=p\equiv\pm
1\,({\mathrm{mod}~{}}10)$, then $r=5$ and $|G_{\alpha}|_{p}=1$ by Lemma 4.2,
and so $q^{4}=|G|_{p}\leq|T|_{p}|G_{\alpha}|_{p}=q^{2}$, a contradiction. If
$q=11$ then $r=11$ and $|G_{\alpha}|_{p}=q$, so
$q^{4}=|G|_{p}\leq|T|_{p}|G_{\alpha}|_{p}=q^{3}$, a contradiction. Therefore,
$r=q+1$, $p=2$ and $f$ is power of $2$. If $\mathbf{O}_{2}(G_{\alpha})=1$ then
$q^{4}=|G|_{2}\leq|G_{\alpha}|_{2}|T|_{2}\leq|\mathrm{\Gamma
L}_{2}(q)|_{2}|T|_{2}=q^{3}f$, yielding $q\leq f$, a contradiction. Thus
$\mathbf{O}_{2}(G_{\alpha})\neq 1$, and the lemma follows. ∎

######  Lemma 8.18.

Let $A$ and $B$ be as in [31, Table 1], and $m\geq 3$. Assume that
$A\cong\mathrm{GO}^{-}_{2m}(q)$ for even $q$. Then one of the following holds.

  * (1)

$T\cong\Omega^{-}_{2m}(q)$ and $S\cong\mathrm{L}_{d}(q^{m\over d})$, $d$ is an
odd prime and $mf$ is a power of $d$.

  * (2)

$T\cong\Omega^{-}_{2m}(q)$ and $S\cong\mathrm{L}_{2}(q^{m\over 2})$,
$\mathbf{O}_{2}(G_{\alpha})\neq 1$ and $mf$ is a power of $2$.

  * (3)

$q=2$, $T\cong\Omega^{+}_{2m}(2)$, $S\cong\mathrm{L}_{2}(2^{m})$ and
$\mathbf{O}_{2}(G_{\alpha})=1$, where $m$ is a power of $2$.

  * (4)

$G=\mathrm{Sp}_{12}(2)$, $T\cong\Omega^{-}_{12}(2)$, either
$S\cong\mathrm{L}_{3}(2)$ and $\mathbf{O}_{2}(G_{\alpha})\neq 1$ or
$S\cong\mathrm{L}_{5}(2)$.

###### Proof.

Recall that $\\{X,Y\\}=\\{T,G_{\alpha}\\}$ with $X\leq A$ and $Y\leq B$. We
discuss in three cases according the structure of $B$.

Case 1. Let $B=\mathsf{P}_{m}$. Then $|G:B|$ has a divisor
$\Phi^{*}_{2m}(q)\Phi^{*}_{2m-2}(q)\Phi^{*}_{2m-4}(q)$. Since $|X|$ is
divisible by $|G:B|$, it is easily shown that $|X\cap A^{\prime}|$ is
divisible by $\Phi^{*}_{2m}(q)\Phi^{*}_{2m-2}(q)\Phi^{*}_{2m-4}(q)$. By [32,
Tables 10.1 and 10.3], either $q=2$ and $m\in\\{4,5\\}$ or $X\cap
A^{\prime}=A^{\prime}\cong\Omega^{-}_{2m}(q)$. For the former,
$|G:B|=3^{3}\cdot 5\cdot 17$ or $3^{4}\cdot 5\cdot 11\cdot 17$, by the Atlas
[5] and [32, Table 10.4], also $X\cap
A^{\prime}=A^{\prime}\cong\Omega^{-}_{2m}(q)$. Since $m\geq 3$, it follows
that $T=X\cong\Omega^{-}_{2m}(q)$, and $G_{\alpha}=Y\leq B$. Then $S$ is
described as in Lemma 7.7.

If $S\cong\mathrm{L}_{d}(q^{m\over d})$ for some prime $d$ then part (1) or
(2) of this lemma occurs. For the triples given in Lemma 7.7 (2) and (3), by
Lemma 4.2 and calculating $|G_{\alpha}|_{2}$, we conclude that $(S,m,q)$ is
one of $(\mathrm{A}_{7},4,2)$, $(\mathrm{L}_{3}(2),6,2)$ and
$(\mathrm{L}_{5}(2),6,2)$. Suppose that $(S,m,q)=(\mathrm{A}_{7},4,2)$. By
Lemma 4.2, noting $|G|_{2}\leq|T|_{2}|G_{\alpha}|_{2}$, we have
$G_{\alpha}\cong\mathrm{S}_{7}$. Then $\mathrm{S}_{7}\cong
G_{\alpha}\mathbf{O}_{2}(B)/\mathbf{O}_{2}(B)\leq\overline{B}\cong\mathrm{A}_{8}$,
which is impossible. Thus $T$ and $S$ are desired as in part (4) of this
lemma.

Case 2. Let $B\cong\mathrm{Sp}_{m}(q)\wr\mathrm{S}_{2}$. Suppose that $T=Y\leq
B$. Then $G_{\alpha}=X\leq A$, $T\lesssim\mathrm{Sp}_{m}(q)$, and $|G:T|$ is
divisible by $q^{3m^{2}\over 4}\prod_{i={m\over 2}+1}^{m}(q^{2i}-1)$. Since
$|G_{\alpha}|$ is divisible by $|G:T|$, it is easily shown that
$|G_{\alpha}\cap A^{\prime}|$ is divisible ${1\over 2}q^{3m^{2}\over
4}\prod_{i={m\over 2}+1}^{m}(q^{2i}-1)$. Checking [32, Tables 10.1, 10.3 and
10.4], we conclude that $G_{\alpha}\cap
A^{\prime}=A^{\prime}\cong\Omega^{-}_{2m}(q)$, which is impossible by Lemma
4.2. Then $T=X\lesssim\Omega^{-}_{2m}(q)$ and $G_{\alpha}=Y\leq B$. By a
similar argument as in Case 1, $T$ and $S$ are described as in part (2) or (4)
of this lemma.

Case 3. Let $B\cong\mathrm{GO}^{+}_{2m}(2)$, $m\geq 4$ and $q=2$. If
$G_{\alpha}=Y\leq B$ then Lemma 7.7 works, and $T$ and $S$ are described as in
part (2) or (4) of this lemma.

Suppose next that $G_{\alpha}=X\leq A$. Then $T\leq
B^{\prime}\cong\Omega^{+}_{2m}(2)$, and $|G:T|$ is divisible by
$2^{m}(2^{m}+1)$. Considering $G_{\alpha}\cap A^{\prime}$, by Lemma 7.1,
either $S\cong\mathrm{L}_{2}(2^{m})$ with $m$ a power of $2$, or $(S,2m,q)$ is
one of $(\mathrm{A}_{11},10,2)$, $(\mathrm{A}_{13},12,2)$,
$(\mathrm{A}_{19},18,2)$, $(\mathrm{M}_{11},10,2)$,
$(\mathrm{L}_{2}(11),10,2)$ and $(\mathrm{L}_{3}(3),12,2)$. The triples
$(\mathrm{M}_{11},10,2)$ and $(\mathrm{L}_{2}(11),10,2)$ are excluded by
calculation of $|G_{\alpha}|_{2}$. If $(S,2m,q)=(\mathrm{L}_{3}(3),12,2)$ then
$5\not\in\pi(G_{\alpha})$ by Lemma 4.2, which contradicts that $|G_{\alpha}|$
is divisible by $2^{m}+1$. Assume that $(S,2m,q)=(\mathrm{A}_{11},10,2)$,
$(\mathrm{A}_{13},12,2)$ or $(\mathrm{A}_{19},18,2)$. By [31, pp. 52,
3.2.4(e)], $A\cap B\cong\mathrm{Sp}_{2m-2}(2)\times 2$, which is maximal in
both $A$ and $B$. Since $A=G_{\alpha}(A\cap B)$, we have a core-free maximal
factorization of $A$ with a factor $\mathrm{Sp}_{2m-2}(2)\times 2$ and the
other factor containing $\mathrm{A}_{2m+1}$, which leads to a contradiction by
checking the maximal factorizations of $\mathrm{GO}^{-}_{2m}(2)$.

Now $S\cong\mathrm{L}_{2}(2^{m})$, where $m$ is a power of $2$. By Lemma 7.1,
$|G_{\alpha}\cap A^{\prime}|_{2}$ is a divisor of $m|S|_{2}$. Then
$|G_{\alpha}|$ is a divisor of $2m2^{m}$. If $\mathbf{O}_{2}(G_{\alpha})\neq
1$ then $|G_{\alpha}|_{2}\geq 2^{2m}$, and so $2^{m}\leq 2m$, yielding $m\leq
2$, a contradiction. Thus $\mathbf{O}_{2}(G_{\alpha})=1$. It is easily shown
that $|G:G_{\alpha}|$ has a divisor ${2^{m(m-1)}\over
4m(2^{m}-1)}\prod_{i=1}^{m-1}(2^{2i}-1)$, and then $|T|$ is divisible by
${2^{m(m-1)}\over 4m(2^{m}-1)}\prod_{i=1}^{m-1}(2^{2i}-1)$. If $m>4$ then, by
[32, Table 10.1], $T\cong\mathrm{Sp}_{2m-2}(2)$ or $\Omega^{+}_{2m}(2)$.
Assume that $m=4$. Then $|T|$ is divisible by $2^{8}\cdot 3^{3}\cdot 7$.
Checking the subgroups of $\Omega^{+}_{8}(2)$, we have
$T\cong\Omega^{+}_{8}(2)$ or $\mathrm{Sp}_{6}(2)$. Suppose that
$T\cong\mathrm{Sp}_{2m-2}(2)$. Then $|G:T|$ is divisible by $2^{2m-1}$, which
is a divisor of $|G_{\alpha}|$. Since $\mathbf{O}_{2}(G_{\alpha})=1$, we have
$|G_{\alpha}|_{2}\leq 2^{m}mf$. It follows that $2^{2m-1}\leq 2^{m}m$, and so
$2^{m-1}\leq m$, which is impossible as $m>2$. Thus
$T\cong\Omega^{+}_{2m}(2)$, and part (3) of this lemma occurs. ∎

######  Lemma 8.19.

Let $m\geq 3$ and $A\cong\mathrm{PSp}_{2a}(q^{b}){.}b$, where $b$ is a prime
and $m=ab$. Assume that $B\not\cong\mathrm{GO}^{-}_{2m}(q)$. Then $p=2$ and
one of the following holds.

  * (1)

$T\cong\Omega^{+}_{2m}(q)$, $S\cong\mathrm{L}_{2}(q^{m})$,
$\mathbf{O}_{2}(G_{\alpha})=1$ and $mf$ is a power of $2$.

  * (2)

$m=3$, $q>2$, $T\cong\mathrm{L}_{2}(q^{3})$, $S\cong\mathrm{SL}_{2}(q^{2})$,
$\mathbf{O}_{2}(G_{\alpha})\neq 1$ and $f$ is a power of $2$.

###### Proof.

Assume that $X=G_{\alpha}$. Then $T\leq B$, and so
$T\lesssim\mathrm{PSp}_{2m-2}(q)$ or $\Omega^{+}_{2m}(q)$. In particular,
$|G:T|$ has a divisor $q^{m}(q^{m}+1)$. Then $|X\cap A^{\prime}|$ is divisible
by ${1\over(b,p)}q^{m}\Phi^{*}_{2m}(q)$. Since $(m,q)\neq(3,2)$, we have
$\Phi^{*}_{2m}(q)\neq 1$. Clearly, $S\preccurlyeq X\cap A^{\prime}$. Applying
Lemma 7.1 to $X\cap A^{\prime}$ and $A^{\prime}$, we conclude
$S\cong\mathrm{L}_{d}(q^{b{2a\over d}})$, where $d$ is a prime divisor of $2a$
and $mf$ is power of $d$. By Lemma 2.1, we have $d=b=p=2$. Again by Lemma 7.1,
$|X\cap A^{\prime}|_{2}$ is a divisor of $q^{m}a$, and so $|G_{\alpha}|_{2}$
is a divisor of $q^{m}m$. Thus $q^{m}m\geq|G:T|_{2}$ as $|G_{\alpha}|$ is
divisible by $|G:T|$. It follows that $T\leq B\cong\mathrm{GO}^{+}_{2m}(q)$.
It is easily shown that $|G:G_{\alpha}|$ is divisible by ${1\over
m}q^{m^{2}-m}\Phi^{*}_{2m-2}(q)\Phi^{*}_{2m-4}(q)\Phi^{*}_{m-1}(q)$, which is
a divisor of $|T|$. By [32, Table 10.1], we conclude that either
$T=B^{\prime}\cong\Omega^{+}_{2m}(q)$, or $(m,q)=(4,2)$. For the latter case,
noting that $|T|$ is divisible by $2^{10}\cdot 5\cdot 7$, we have
$T\cong\Omega^{+}_{8}(2)$ by [32, Table 10.4]. Then part (1) of this lemma
occurs.

We assume next that $X=T$. Then $T\lesssim\mathrm{PSp}_{2a}(q^{b})$,
$G_{\alpha}=Y\leq B$, and $|G_{\alpha}|$ is divisible by
$q^{m(m-a)}\prod_{(i,b)=1}(q^{2i}-1)$. Let
$\overline{B}=B^{\prime}\mathbf{rad}(B)/\mathbf{rad}(B)$ and
$\overline{Y}=(Y\cap B^{\prime})\mathbf{rad}(B)/\mathbf{rad}(B)$. Then
$S\preccurlyeq\overline{Y}$ and $|\overline{Y}|$ is divisible by
$\prod_{(i,b)=1,i>1}(q^{2i}-1)$. In particular, $|\overline{Y}|$ is divisible
by $\prod_{(i,b)=1,i>1}\Phi^{*}_{2i}(q)$.

Suppose that $B=\mathsf{P}_{1}$ or $\mathsf{N}_{2}$. Then
$\overline{B}\cong\mathrm{PSp}_{2m-2}(q)$. By Lemma 7.1, either
$S\cong\mathrm{L}_{2}(q^{m-1})$, $p=2$ and $(m-1)f$ is a power of $2$, or
$(S,2(m-1),q)$ is one of the triples $(\mathrm{A}_{5},4,3)$,
$(\mathrm{A}_{7},4,7)$, $(\mathrm{A}_{7},6,3)$, $(\mathrm{A}_{7},6,5)$,
$(\mathrm{A}_{11},10,2)$, $(\mathrm{A}_{13},12,2)$, $(\mathrm{A}_{19},18,2)$,
$(\mathrm{M}_{11},10,2)$, $(\mathrm{L}_{3}(2),6,3)$,
$(\mathrm{L}_{3}(2),6,5)$, $(\mathrm{L}_{2}(11),10,2)$,
$(\mathrm{L}_{3}(3),12,2)$. Recalling that $|G_{\alpha}|$ is divisible by
$q^{m(m-a)}\prod_{(i,b)=1}(q^{2i}-1)$, the latter case is excluded by Lemma
4.2 and calculating $|G_{\alpha}|_{p}$ or $\pi(G_{\alpha})$. Then
$S\cong\mathrm{L}_{2}(q^{m-1})$. Assume $m>3$. Then $|G:G_{\alpha}|$ has a
divisor $\prod_{i={m+1\over 2}}^{m-2}\Phi^{*}_{2i}(q)$, and so $|T|$ is
divisible by $\prod_{i={m+1\over 2}}^{m-2}\Phi^{*}_{2i}(q)$. Considering
$\pi(\mathrm{PSp}_{2a}(q^{b}))$, we have $b\geq{m+1\over 2}$, yielding $b=m$
and $T\lesssim\mathrm{PSp}_{2}(q^{m})$. Since $\Phi^{*}_{2m-4}(q)$ is a
divisor of $|T|$, it implies that $(m,q)=(5,2)$ and
$S\cong\mathrm{L}_{2}(16)$. In this case,
$q^{m(m-a)}\prod_{(i,b)=1}(q^{2i}-1)$ has a divisor $7$, and so
$7\in\pi(G_{\alpha})=\pi(S)=\\{2,3,5,17\\}$, a contradiction. Thus $m=3$,
$T\lesssim\mathrm{L}_{2}(q^{3})$. Then $T\cong\mathrm{L}_{2}(q^{3})$ by
checking the subgroups of $\mathrm{L}_{2}(q^{3})$. If
$\mathbf{O}_{2}(G_{\alpha})=1$ then
$q^{6}\leq|G_{\alpha}|_{2}\leq|\mathrm{P\Gamma L}_{2}(q^{2})|_{2}$, yielding
$q^{4}\leq 2f$, a contradiction. Therefore, part (2) of this lemma occurs.

Now let $B\cong\mathrm{GO}^{+}_{2m}(q)$. Recall that $|G_{\alpha}|$ is
divisible by $q^{m(m-a)}\prod_{(i,b)=1}(q^{2i}-1)$. Assume that $(m,q)=(4,2)$.
Then $|G_{\alpha}|$ is divisible by $2^{8}\cdot 3^{3}\cdot 7$. Checking the
maximal subgroups of $\mathrm{GO}^{+}_{8}(2)$, we have
$G_{\alpha}\lesssim\mathrm{Sp}_{6}(2)\times 2$. Searching subgroups of
$\mathrm{Sp}_{6}(2)$ with order divisible by $3^{3}\cdot 7$, we conclude that
$\mathrm{U}_{3}(3)\preccurlyeq G_{\alpha}$, a contradiction. Then
$(m,q)\neq(4,2)$, and $\Phi^{*}_{(2m-2)f}(q)\neq 1$. Recalling the limitations
on $G_{\alpha}$, by [2, Theorem 3.1], we conclude that
$G_{\alpha}\lesssim\mathrm{GO}^{-}_{2}(q)\times\mathrm{GO}^{-}_{2m-2}(q)$, or
$\mathrm{Sp}_{2m-2}(q)$. Then Lemma 7.1 works for $G_{\alpha}$ modulo a
suitable solvable group. Arguing similarly as above, we have part (2) of this
lemma. ∎

######  Theorem 8.6.

Assume that Hypothesis 6.1 holds and $G=\mathrm{PSp}_{2m}(q)$, where $m\geq 2$
and $q=p^{f}$. Then $q$ is even and one of the following holds.

  * (1)

$m=2$, $f>2$ is a power of $2$, $T\cong\mathrm{L}_{2}(q^{2})$, and either
$G_{\alpha}\cong\mathrm{L}_{2}(q^{2})$ or $\mathrm{SL}_{2}(q)\unlhd
G_{\alpha}^{\Gamma(\alpha)}$ and $\mathbf{O}_{2}(G_{\alpha})\neq 1$.

  * (2)

$m=3$, $q>2$, $f$ is a power of $2$, $T\cong\mathrm{G}_{2}(q)$ (with $f>2$) or
$\mathrm{L}_{2}(q^{3})$, $\mathrm{SL}_{2}(q^{2})\unlhd
G_{\alpha}^{\Gamma(\alpha)}$ and $\mathbf{O}_{2}(G_{\alpha})\neq 1$.

  * (3)

$(m,q)=(6,2)$, $T\cong\Omega^{-}_{12}(2)$, either $S\cong\mathrm{L}_{3}(2)$
and $\mathbf{O}_{2}(G_{\alpha})\neq 1$ or $S\cong\mathrm{L}_{5}(2)$.

  * (4)

$m\geq 5$, $T\cong\Omega^{-}_{2m}(q)$, $S\cong\mathrm{L}_{d}(q^{m\over d})$,
$d$ is an odd prime and $mf$ is a power of $d$.

  * (5)

$m\geq 4$, $T\cong\Omega^{-}_{2m}(q)$ and $S\cong\mathrm{L}_{2}(q^{m\over
2})$, $\mathbf{O}_{2}(G_{\alpha})\neq 1$ and $mf$ is a power of $2$.

  * (6)

$T\cong\Omega^{+}_{2m}(q)$, $S\cong\mathrm{L}_{2}(q^{m})$,
$\mathbf{O}_{2}(G_{\alpha})=1$, $m\geq 4$ and $mf$ is a power of $2$.

###### Proof.

Based on the previous discussion, we assume that
$A\cong\mathrm{PSp}_{2a}(q){.}b$ and $B\cong\mathrm{GO}^{-}_{2m}(q)$, where
$q$ is even, $m\geq 3$, $(m,q)\neq(3,2)$, and $b$ is a prime divisor of $m$
with $m=ab$. If $G_{\alpha}\leq B$ then part (2) of this lemma occurs by a
similar argument as in the last paragraph of the proof of Lemma 8.19.

Now Let $G_{\alpha}\leq A$ and $T\leq B$. By Lemma 7.7 and Case 4 in its
proof, we conclude that $a\geq 3$, $S\cong\mathrm{L}_{d}(q^{m\over d})$, $d$
is a prime divisor of $a$ and $mf$ is a power of $d$. Assume that $m$ is even.
Then $d=b=2$, $a\geq 4$ and $S\cong\mathrm{L}_{2}(q^{m\over 2})$. It is easily
shown that $|G:G_{\alpha}|$ is divisible by $\prod_{i={m+2\over
2}}^{m}\Phi^{*}_{2i}(q)$. By [32, Table 10.1], $T\cong\Omega^{-}_{2m}(q)$, and
then $T$ and $S$ are desired as in part (5) of this lemma. Now let $m$ be odd;
in particular, $a,\,b$ and $d$ are odd. Then $A\cap
B\cong\mathrm{GO}_{2a}^{-}(q^{b}){.}b$, which is maximal in $B$. Suppose that
$T\not\cong\Omega_{2m}^{-}(q)$. Noting $B=X(A\cap B)$, by Lemma 2.6, we have a
core-free maximal factorization $G_{0}=A_{0}B_{0}$ with
$\mathbf{soc}(G_{0})\cong\Omega_{2m}^{-}(q)$ and
$\Omega^{-}_{2a}(q^{b})\lesssim B_{0}$. It follows from [31, Theorem A] that
$q\in\\{2,4\\}$ and $b=2$, a contradiction. Thus $T\cong\Omega_{2m}^{-}(q)$,
and $T$ and $S$ are desired as in part (4) of this lemma. ∎

###  8.7. Linear case

Let $G=\mathrm{L}_{n}(q)$, where $n\geq 2$ and $q=p^{f}$ for some prime $p$.
Since $A$ and $B$ are insolvable, by [31, Theorem A] and Corollary 5.1, $n$ is
not a prime, $A\cong{}^{\bf\hat{}}\mathrm{GL}_{a}(q^{b}){.}b$ or
$\mathrm{PSp}_{n}(q){.}[c]$, and
$B\cong[q^{n-1}]{:}({q-1\over(n,q-1)}{.}\mathrm{PGL}_{n-1}(q))$, where $a\geq
2$, $b$ is a prime with $n=ab$, and $c={(2,q-1)({n\over 2},q-1)\over(n,q-1)}$.
Recalling $G\not\cong\mathrm{A}_{8}$, we have $(m,q)\neq(4,2)$.

######  Lemma 8.20.

If $G_{\alpha}\leq A$ then $T\cong\mathrm{L}_{n-1}(q)$,
$S\cong\mathrm{L}_{d}(q^{n\over d})$, where $d$ is an odd prime and $nf$ is a
power of $d$.

###### Proof.

Let $G_{\alpha}\leq A$. Then $T\lesssim\mathrm{L}_{n-1}(q)$, and
$|G:T|={(n-1,q-1)\over(n,q-1)}q^{n-1}(q^{n}-1)$. Let $X_{0}=G_{\alpha}\cap
A^{\prime}$. Then $S\preccurlyeq X_{0}$ and $|X_{0}|$ is divisible by
${1\over(e,p)}q^{n-1}(q^{n}-1)$, where $e=b$ or $c$. Suppose that
$(n,q)=(6,2)$. Then $A^{\prime}\cong\mathrm{GL}_{3}(4)$ or
$\mathrm{Sp}_{6}(2)$, and $|G_{\alpha}|$ is divisible by $2^{4}\cdot
3^{2}\cdot 7$. Checking the subgroups of $A^{\prime}$, we conclude that
$X_{0}$ has a composition factor $\mathrm{L}_{3}(4)$, $\mathrm{U}_{3}(3)$,
$\mathrm{A}_{8}$ or $\mathrm{Sp}_{6}(2)$, which is impossible by Lemma 4.2.
Thus $(n,q)\neq(6,2)$, and $\Phi^{*}_{nf}(p)\neq 1$. Then Lemma 7.1 works
here. By calculation of $\pi(S)$ or $|G_{\alpha}|_{p}$, we conclude that
$A\cong{}^{\bf\hat{}}\mathrm{GL}_{a}(q^{b}){.}b$,
$S\cong\mathrm{L}_{d}(q^{n\over d})$, $d$ is an odd prime divisor of $a$ and
$nf$ is a power of $d$. In particular, $a\geq 3$ and $n$ is odd. Then $|G:A|$
has a divisor $(q^{n-1}-1)(q^{n-2}-1)$. By [32, Table 10.1], we have
$T\cong\mathrm{L}_{n-1}(q)$, and the lemma holds. ∎

######  Lemma 8.21.

If $G_{\alpha}\leq B$ then one of the following holds.

  * (1)

$(n,q)=(4,4)$, $T\cong\mathrm{Sp}_{4}(4)$, $S\cong\mathrm{L}_{3}(2)$ and
$\mathbf{O}_{2}(G_{\alpha})\neq 1$.

  * (2)

$S\cong\mathrm{L}_{n-1}(q)$, $n-1$ is an odd prime and $f$ is a power of
$n-1$.

###### Proof.

Let $G_{\alpha}\leq B$, set $\overline{B}=B/\mathbf{rad}(B)$ and
$\overline{Y}=Y\mathbf{rad}(B)/\mathbf{rad}(B)$. Then
$S\preccurlyeq\overline{Y}$. Note that $|G:A|$ has a divisor $q^{n-1}-1\over
q-1$, which is also a divisor of $|G_{\alpha}|$. Then $|\overline{Y}|$ is
divisible by $q^{n-1}-1\over q-1$, and hence divisible by
$\Phi^{*}_{(n-1)f}(p)$.

Assume that $(n,q)=(4,4)$. Then $\overline{B}\cong\mathrm{PGL}_{3}(4)$.
Checking the subgroups of $\mathrm{PGL}_{3}(4)$, we conclude that
$S\cong\overline{Y}\cong\mathrm{L}_{3}(2)$. Note that
$T\lesssim\mathrm{L}_{2}(16)$ or $\mathrm{Sp}_{4}(4)$. Checking the subgroups
of $\mathrm{L}_{2}(16)$ and $\mathrm{Sp}_{4}(4)$, we conclude that
$T\cong\mathrm{L}_{2}(16)$ or $\mathrm{Sp}_{4}(4)$. By Lemma 4.2,
$|G_{\alpha}|_{5}=1$. If $T\cong\mathrm{L}_{2}(16)$ then
$5^{2}=|G|_{5}\leq|T|_{5}|G_{\alpha}|_{5}=5$, a contradiction. Then
$T\cong\mathrm{Sp}_{4}(4)$, and $|G:T|=2^{4}\cdot 3^{2}\cdot 7$. By Lemma 4.2,
$\mathbf{O}_{2}(G_{\alpha})\neq 1$, and part (1) of this lemma follows.

Now $(n,q)\neq(4,4)$, and $\Phi^{*}_{(n-1)f}(p)\neq 1$. By Lemma 7.1, one of
the following holds:

  * (i)

$S\cong\mathrm{L}_{d}(q^{n-1\over d})$, $d$ is a prime and $(n-1)f$ is a power
of $d$;

  * (ii)

$(S,n-1,q)$ is one of $(\mathrm{A}_{7},3,25)$, $(\mathrm{L}_{3}(2),3,25)$,
$(\mathrm{L}_{3}(2),3,9)$ and $(\mathrm{L}_{2}(11),5,4)$.

The case (ii) is easily excluded by calculating $|G_{\alpha}|_{p}$ and
$|G:A|_{p}$. Thus (i) occurs.

Suppose that $A\cong\mathrm{PSp}_{n}(q){.}[c]$. Then
$T\lesssim\mathrm{PSp}_{n}(q)$, $n$ is even and
$|G:T|={(2,q-1)\over(n,q-1)}q^{n(n-2)\over 4}\prod_{i=2}^{n\over
2}(q^{2i-1}-1)$. In particular, $|G_{\alpha}|$ is divisible by
$\Phi^{*}_{n-3}(q)$. Clearly, $\Phi^{*}_{n-3}(q)\neq 1$. Since
$\pi(G_{\alpha})=\pi(S)$, we have $(\Phi^{*}_{n-3}(q),|S|)\neq 1$. By Lemma
2.2, $n-3$ is a divisor of $(n-1)i\over d$ for some $2\leq i\leq d$. It
follows that $d=n-1$, and so $S\cong\mathrm{L}_{n-1}(q)$.

Suppose that $A\cong{}^{\bf\hat{}}\mathrm{GL}_{a}(q^{b}){.}b$. If $b=2$ or
$b\geq 5$ then $|G_{\alpha}|$ is divisible by $\Phi^{*}_{n-3}(q)$, and so
$d=n-1$ by a similar argument as above. Now let $b=3$. Then $|G_{\alpha}|$ is
divisible by $\Phi^{*}_{n-2}(q)$. Clearly, $n\neq 8$ and $n-2>2$. Then
$\Phi^{*}_{n-2}(q)\neq 1$, this also yields $d=n-1$. Thus the lemma follows. ∎

######  Theorem 8.7.

Assume that Hypothesis 6.1 holds and $G=\mathrm{L}_{n}(q)$, where $n\geq 4$
and $q=p^{f}$. Then one of the following holds.

  * (1)

$T\cong\mathrm{L}_{n-1}(q)$, $S\cong\mathrm{L}_{d}(q^{n\over d})$, $d$ is an
odd prime and $nf$ is a power of $d$.

  * (2)

$(n,q)=(4,4)$, $T\cong\mathrm{Sp}_{4}(4)$, $S\cong\mathrm{L}_{3}(2)$ and
$\mathbf{O}_{2}(G_{\alpha})\neq 1$.

  * (3)

$S\cong\mathrm{L}_{n-1}(q)$, $n-1$ is an odd prime, $f$ is a power of $n-1$,
and

    * $T\cong\mathrm{L}_{m}(q^{n\over m})$, $m$ is a proper divisor of $n$; or

    * $T\cong\mathrm{PSp}_{m}(q^{n\over m})$, $m\geq 4$ is even, and $q$ is even if $m=n$; or

    * $T\cong\mathrm{P\Omega}^{-}_{m}(q^{n\over m})$, $4\,\,\big{|}\,\,m$, $m\geq 8$, and $q$ is even if $m=n$; or

    * $T\cong\mathrm{G}_{2}(q^{n\over 6})$, $6\,\,\big{|}\,\,n$, $q$ is even; or

    * $T\cong\mathrm{U}_{3}(3)$ and $(n,q)=(6,2)$.

###### Proof.

By Lemmas 8.20 and 8.21, we assume that $T\leq A$, $S\cong\mathrm{L}_{n-1}(q)$
and $f$ is a power of $n-1$. We shall show that $T$ is described as in part
(3) of lemma.

Since $G=TB$, by Lemma 2.4, $|T|$ is divisible by $q^{n}-1\over q-1$. Assume
that $(n,q)=(6,2)$. Then $T\lesssim\mathrm{L}_{2}(8)$, $\mathrm{L}_{3}(4)$ or
$\mathrm{Sp}_{6}(2)$. Checking the subgroups of $\mathrm{L}_{2}(8)$,
$\mathrm{L}_{3}(4)$ and $\mathrm{Sp}_{6}(2)$, it follows that
$T\cong\mathrm{L}_{2}(8),\,\mathrm{U}_{3}(3)$ or $\mathrm{Sp}_{6}(2)$, desired
as in part (3). Thus we assume next that $(n,q)\neq(6,2)$. Then
$\Phi^{*}_{nf}(p)\neq 1$.

Case 1. Assume that $A\cong{}^{\bf\hat{}}\mathrm{GL}_{a}(q^{b}){.}b$. Then
$T\lesssim\mathrm{L}_{a}(q^{b})$. Choose a minimal divisor $m$ of $n$ such
that $T\lesssim\mathrm{L}_{m}(q_{0})$, where $q_{0}=q^{n\over m}$. Then $m\leq
a$, $|T|$ is divisible by $q_{0}^{m}-1\over q-1$ and $\Phi^{*}_{m{nf\over
m}}(p)$. If $m=2$ then $T\lesssim\mathrm{L}_{2}(q^{n\over 2})$ and $q_{0}$ is
not a prime, and so $T\cong\mathrm{L}_{2}(q^{n\over 2})$ by checking the
subgroups of $\mathrm{L}_{2}(q^{n\over 2})$. Thus we let $m\geq 3$, and so
$a\geq 3$. Noting that $T$ is not an alternating group, by [2, Theorem 3.1],
one of the following holds:

  * (i)

$T\cong\mathrm{P\Omega}^{-}_{m}(q_{0})$, $\mathrm{PSp}_{m}(q_{0})$ or
$\mathrm{L}_{m}(q_{0})$;

  * (ii)

$T\cong\mathrm{U}_{m}(\sqrt{q_{0}})$, $q_{0}$ is a square and $m$ is odd;

  * (iii)

$(T,m,q_{0})$ is one of the following triples: $(\mathrm{J}_{3},9,4)$,
$(\mathrm{L}_{3}(2),3,9)$, $(\mathrm{L}_{3}(2),3,25)$,
$(\mathrm{L}_{2}(11),5,4)$, $(\mathrm{L}_{2}(13),6,4)$,
$(\mathrm{L}_{2}(19),9,4)$, $(\mathrm{Suz}(q_{0}),4,2^{nf\over m})$ and
$(\mathrm{G}_{2}(q_{0}),6,2^{nf\over m})$.

We only need deal with cases (ii) and (iii).

Suppose that (ii) occurs. Since $m\geq 3$ is odd, by Lemma 2.1,
$\Phi^{*}_{m}(\sqrt{q_{0}})\neq 1$. Since $m$ is odd, it is easy to see that
$|\mathrm{U}_{m}(\sqrt{q_{0}})|$ is indivisible by
$\Phi^{*}_{m}(\sqrt{q_{0}})$; however, $|T|$ is divisible by $q_{0}^{m}-1\over
q-1$, a contradiction. Suppose that (iii) occurs. For the first six triples,
we have $b={n\over m}=2$ and $q=p$, and then they are excluded by calculating
$|T|$ and ${q^{n}-1\over q-1}$. Similarly, the triple
$(\mathrm{Suz}(q_{0}),4,2^{nf\over m})$ is excluded as $|\mathrm{Suz}(q_{0})|$
is indivisible by $q_{0}+1$. Thus $T\cong\mathrm{G}_{2}(q_{0})$, desired as in
part (3).

Case 2. Assume that $A\cong\mathrm{PSp}_{n}(q){.}[c]$. Then
$T\lesssim\mathrm{PSp}_{n}(q)$ for even $n$. Noting that $T$ is not an
alternating group, $n-1$ is an odd prime and $f$ is a power of $n-1$, by [2,
Theorem 3.1], one of the following holds:

  * (iv)

$T\cong\Omega^{-}_{n}(q)$ (with $q$ even) or $\mathrm{PSp}_{n}(q)$;

  * (v)

$T\lesssim\mathrm{PSp}_{m}(q^{n\over m})$, where $m$ is a proper divisor of
$n$;

  * (vi)

$T\lesssim\mathrm{U}_{n\over 2}(q)$, where ${n\over 2}q$ is odd;

  * (vii)

$(T,n,q)$ is one of the following triples:
$(\mathrm{J}_{2},6,5)$, $(\mathrm{L}_{3}(2),6,3)$, $(\mathrm{L}_{3}(2),6,5)$,
$(\mathrm{L}_{2}(13),6,3)$, $(\mathrm{L}_{2}(13),6,4)$,
$(\mathrm{L}_{2}(17),8,2)$,
$(\mathrm{L}_{2}(25),12,2)$, $(\mathrm{L}_{2}(37),18,2)$,
$(\mathrm{L}_{2}(41),20,2)$, $(\mathrm{L}_{3}(3),12,2)$,
$(\mathrm{L}_{3}(4),6,3)$,
$(\mathrm{PSp}_{4}(5),12,2)$, $(\mathrm{U}_{3}(9),6,5)$,
$(\mathrm{Suz}(q),4,2^{f})$ and $(\mathrm{G}_{2}(q),6,2^{f})$.

Recall that $|T|$ is divisible by $q^{n}-1\over q-1$. For
$T\lesssim\mathrm{U}_{n\over 2}(q)$, since ${n\over 2}q$ is odd,
$\Phi^{*}_{n\over 2}(q)\neq 1$; however, it is easy to see that
$|\mathrm{U}_{n\over 2}(q)|$ is indivisible by $\Phi^{*}_{n\over 2}(q)$, a
contradiction.

For case (vii), only the triples $(\mathrm{L}_{2}(13),6,3)$ and
$(\mathrm{G}_{2}(q),6,2^{f})$ are left. If $(n,q)=(6,3)$ then $r={3^{5}-1\over
3-1}$, which is not a prime, a contradiction. Thus $T\cong\mathrm{G}_{2}(q)$,
desired as in part (3) of this lemma. For case (iv), either $T$ is desired in
part (3) of this lemma, or $T\cong\Omega^{-}_{n}(q)$ with odd ${n\over 2}$;
the latter does occur as $|T|$ has a divisor $q^{n\over 2}-1\over q-1$.

The left case is that $T\lesssim\mathrm{PSp}_{m}(q^{n\over m})$. We choose $m$
minimal as possible. If $m=2$ then $T\lesssim\mathrm{PSp}_{2}(q^{n\over 2})$,
yielding $T\cong\mathrm{PSp}_{2}(q^{n\over 2})$, desired as in part (3) of
this lemma. Thus we may let $m\geq 4$. Clearly $q^{n\over m}$ is not a prime,
and $m-1$ is odd. Using [2, Theorem 3.1], by a similar argument as above, $T$
is described as in part (3). ∎

##  9\. The proof of Theorem 2

Let $\Gamma=(V,E)$ be a connected $X$-symmetric Cayley graph of a finite
simple group $T$ of prime valency $r\geq 7$, where $T\leq
X\leq\mathbf{Aut}(\Gamma)$. Assume that $T$ is not normal in $X$, and
$X_{\alpha}$ is insolvable, where $\alpha\in V$. By Theorem 1 (2), we choose a
characteristic subgroup $G$ of $X$ such that that
$G^{\infty}=G\gneq\mathbf{rad}(G)T$, $G/\mathbf{rad}(G)$ is simple,
$r\not\in\pi(\mathbf{rad}(G))$, $\mathbf{rad}(G)$ is intransitive on $V$ and
$\Gamma$ is $G$-symmetric. By Lemma 2.8, $\mathbf{rad}(G)$ is semiregular on
$V$, and so $|\mathbf{rad}(G)|$ is a proper divisor of $|V|=|T|$.

Set $\overline{G}=G/\mathbf{rad}(G)$ and
$\overline{T}=T\mathbf{rad}(G)/\mathbf{rad}(G)$, and identify $\overline{G}$
with a subgroup of $\mathbf{Aut}(\Gamma_{G/\mathbf{rad}(G)})$. Then
$\Gamma_{G/\mathbf{rad}(G)}$ is $\overline{G}$-symmetric, and
$\overline{G}=\overline{T}\,\overline{G}_{\overline{a}}$, where
$\overline{\alpha}=\alpha^{\mathbf{rad}(G)}$. Thus
$(\overline{G},\overline{T},\overline{G}_{\overline{a}})$ is one of the
triples satisfying Hypothesis 4.1. Since $G\unlhd X$, we have
$G_{\alpha}\unlhd X_{\alpha}$, and so $G_{\alpha}^{\Gamma(\alpha)}\unlhd
X_{\alpha}^{\Gamma(\alpha)}$. Since $X_{\alpha}$ is insolvable,
$X_{\alpha}^{\Gamma(\alpha)}$ is almost simple, and then
$G_{\alpha}^{\Gamma(\alpha)}$ is almost simple. In particular, $G_{\alpha}$ is
insolvable. Note that $\overline{G}_{\overline{a}}\cong
G_{\overline{a}}/\mathbf{rad}(G)=\mathbf{rad}(G)G_{\alpha}/\mathbf{rad}(G)\cong
G_{\alpha}$. Then $\overline{G}_{\overline{a}}$ is insolvable. Thus
$(\overline{G},\overline{T},\overline{G}_{\overline{a}})$ can be read out from
Theorems 6.1, 6.2 and 8.1-8.7.

Note that $T\cong\overline{T}$ and
${|T||G_{\alpha}|\over|\mathbf{rad}(G)|}=|G:\mathbf{rad}(G)|=|\overline{G}|=|\overline{T}||\overline{G}_{\overline{a}}:(\overline{G}_{\overline{a}}\cap\overline{T})|$.
Then
$|\mathbf{rad}(G)|=|\overline{G}_{\overline{a}}\cap\overline{T}|={|G_{\alpha}|\over|\overline{G}:\overline{T}|}$,
and so $|\overline{G}:\overline{T}|_{r}=r$. Thus we only need to deal with
those triples $(\overline{G},\overline{T},\overline{G}_{\overline{a}})$ with
$|\overline{G}:\overline{T}|_{r}=r$. For further argument, we need more
precise structural information about $G$. Note that $G$ is perfect, i.e.,
$G^{\prime}=G$.

######  Lemma 9.1.

Let $L=\overline{G}$ and $R_{s}=\min\\{m\mid
L\lesssim\mathrm{PGL}_{m}(q_{0}),q_{0}\mbox{ a power of }s\\}$, where $s$ is a
prime. Then either $G$ is a perfect central extension of $L$, or
$|\mathbf{rad}(G)|_{s}\geq q_{0}^{R_{s}}$ for some prime $s$ and its a power
$q_{0}$ with $L\lesssim\mathrm{PGL}_{m}(q_{0})$. For the latter case, if
$L\cong\mathrm{A}_{n}$ for $n\geq 9$ then $n$ is a power of $2$ and
$|\mathbf{rad}(G)|_{2}\geq 2^{n-2}$.

###### Proof.

Assume that $G$ is not a perfect central extension of $L$. Recalling
$G^{\infty}=G$, it is easily shown that $G/K$ is perfect for each normal
subgroup $K$ of $G$ with $K\leq\mathbf{rad}(G)$. Choose a minimal $K$ such
that $G/K$ is a perfect central extension of $L$. Then $K\neq 1$. Choose
$J\unlhd G$ with $J\leq K$ such that $K/J$ is a minimal normal subgroup of
$G/J$. Since $K$ is solvable, we set $K/J\cong\mathbb{Z}_{s}^{l}$ for some
prime $s$ and integer $l\geq 1$.

Since $G/J=(K/J){.}(G/K)$ and $G/J$ is perfect, by the choice of $K$,
$G/J=(K/J){.}(G/K)$ is not a central extension; otherwise, $G/J$ should be a
perfect central extension of $L$, refer to [1, pp. 167, (33.5)]. Thus $G/J$
acts nontrivially on $K/J$ induced by the conjugation of $G$ on $K$, and the
kernel is $\mathbf{C}_{G/J}(K/J)$. Set $C/J=\mathbf{C}_{G/J}(K/J)$. Then
$C\leq\mathbf{rad}(G)$, and
$G/C\cong(G/J)/(C/J)\lesssim\mathrm{GL}_{l_{0}}(q_{0})$, where
$s^{l}=q_{0}^{l_{0}}$. It follows that $l_{0}\geq R_{s}$, refer to [25,
pp.184, Proposition 5.3.2]. Then $|\mathbf{rad}(G)|_{s}\geq q_{0}^{l_{0}}\geq
q_{0}^{R_{s}}$.

Suppose that $L\cong\mathrm{A}_{n}$ for some $n\geq 9$. Then, by [25,
Propositions 5.3.2 and 5.3.7], $l\geq R_{s}\geq n-2$. Thus
$|\mathrm{A}_{n}|_{s}\geq s^{n-2}$, and then
$n-2+(2,s)\leq\sum_{i=1}^{\infty}\lfloor{n\over s^{i}}\rfloor$. The only
possibility is that $s=2$ and $n$ is a power of $2$. Then the lemma follows. ∎

We next finish the proof of Theorem 2 by checking one by one the possible
candidates for $(\overline{G},\overline{T},\overline{G}_{\overline{a}})$
described in Theorems 6.1, 6.2 and 8.1-8.7. Recall that
$|\mathbf{rad}(G)|=|\overline{G}_{\overline{a}}\cap\overline{T}|$ and
$|\overline{G}:\overline{T}|_{r}=r$.

Assume that $L\cong\mathrm{A}_{n}$. Then, by Theorem 6.1, either
$T\cong\mathrm{A}_{n-1}$ as at Line 1 of Table 1.1, or
$\overline{G}\cong\mathrm{A}_{r+1}$ for $r\geq 7$,
$\overline{G}_{\overline{a}}\cong\mathrm{A}_{r}$,
$\Gamma_{G/\mathbf{rad}(G)}\cong\mathsf{K}_{r+1}$ and $T\cong\overline{T}$ is
isomorphic to a transitive subgroup of $\mathrm{A}_{r+1}$. Since
$|\overline{G}:\overline{T}|_{r}=r$, we have $r\not\in\pi(T)$. Checking the
subgroups of $\mathrm{A}_{8}$, we conclude that $\mathrm{A}_{8}$ has no
insolvable transitive subgroup with order indivisible by $7$. Thus we have
$r>7$, and $r+1>9$. By Lemma 9.1 either $G$ is a perfect central extension of
$\mathrm{A}_{r+1}$, or $|\mathbf{rad}(G)|_{2}\geq 2^{r-1}$. Recalling
$|\mathbf{rad}(G)|$ is a divisor of $|\overline{G}_{\overline{a}}|$, if
$|\mathbf{rad}(G)|_{2}\geq 2^{r-1}$ then $|\mathrm{A}_{r}|_{2}\geq 2^{r-1}$,
which is impossible. Thus $G$ is a perfect central extension of
$\mathrm{A}_{r+1}$, and so $|\mathbf{rad}(G)|\leq 2$, refer to [1, pp. 170,
(33.15)]. If $|\mathbf{rad}(G)|=2$ then $\Gamma$ is bipartite by [10, Theorem
1.1], which is not the case. Then $\mathbf{rad}(G)=1$, and $T$ is desired at
Line $2$ of Table 1.1.

Assume that $L$ is a sporadic simple group. By Theorem 6.2,
$\overline{G}\cong\mathrm{M}_{n}$,
$\overline{G}_{\overline{a}}\cong\mathrm{M}_{n-1}$,
$\Gamma_{G/\mathbf{rad}(G)}\cong\mathsf{K}_{n}$, $n=12$ or $24$ and
$\overline{T}\cong\mathrm{A}_{5}$ or $\mathrm{L}_{3}(2)$. In this case,
$|\mathbf{rad}(G)|=|\overline{G}_{\overline{a}}\cap\overline{T}|=s\in\\{5,7\\}$.
It follows that $G\cong\mathbb{Z}_{s}\times\mathrm{M}_{n}$, which contradicts
the choice of $G$.

Assume that $L$ is a classical simple group. Then Theorems 8.1-8.7 work here,
and either the pair $(L,T)$ is desired as in Table 1.1 or
$(\overline{G},\overline{T},S,r)$ is listed as follows:

$\tiny\begin{array}[]{l|l|l|l|l}|\overline{G}:\overline{T}|&\overline{G}&\overline{T}&\overline{G}_{\overline{\alpha}}&r\\\
\hline\cr 3^{7}\cdot
13&\mathrm{P\Omega}^{+}_{8}(3)&\Omega^{+}_{8}(2)&\overline{G}_{\overline{a}}\cong[3^{6}]{:}\mathrm{SL}_{3}(3)&13\\\
\hline\cr 2^{10}\cdot 3^{2}\cdot 7\cdot
31&\Omega^{+}_{12}(2)&\Omega^{-}_{10}(2)&\mathrm{SL}_{5}(2)\preccurlyeq\overline{G}_{\overline{\alpha}}&31\\\
\hline\cr 2^{4}\cdot 3^{2}\cdot
7&\mathrm{L}_{4}(4)&\mathrm{Sp}_{4}(4)&\mathrm{SL}_{3}(2)\preccurlyeq\overline{G}_{\overline{\alpha}},\,\mathbf{O}_{p}(\overline{G}_{\overline{\alpha}})\neq
1&7\\\ \hline\cr 2^{10}\cdot 3\cdot 5\cdot 7\cdot
31&\mathrm{L}_{6}(2)&\mathrm{U}_{3}(3)&\mathrm{SL}_{5}(2)\preccurlyeq\overline{G}_{\overline{\alpha}}&31\\\
\hline\cr 2^{mf}(2^{mf}+1)&\mathrm{Sp}_{2m}(2^{f}),m\geq
4&\Omega^{+}_{2m}(2^{f})&\mathrm{SL}_{2}(2^{mf})\lesssim\overline{G}_{\overline{a}}\lesssim\mathrm{\Gamma
L}_{2}(2^{mf})&2^{mf}+1\\\ \end{array}$

Recall that
$|\mathbf{rad}(G)|=|\overline{G}_{\overline{a}}\cap\overline{T}|={|\overline{G}_{\overline{\alpha}}|\over|\overline{G}:\overline{T}|}$.

Suppose that $\overline{G}\cong\mathrm{Sp}_{2m}(2^{f})$ and
$\overline{T}\cong\Omega^{+}_{2m}(2^{f})$, where $m\geq 4$ and $mf$ is a power
of $2$. Then $|\mathbf{rad}(G)|$ is a divisor of $mf(2^{mf}-1)^{2}$ and
divisible by $2^{mf}-1$. If $G$ is a perfect central extension of
$\mathrm{Sp}_{2m}(2^{f})$ then $|\mathbf{rad}(G)|\leq(2,2^{f}-1)=1$ by [25,
pp. 173, Theorem 5.1.4], a contradiction. Thus, by Lemma 9.1,
$|\mathbf{rad}(G)|_{s}\geq q_{0}^{R_{s}}$, where $s\in\pi(\mathbf{rad}(G))$,
$R_{s}$ is defined as in Lemma 9.1, $q_{0}$ is a power of $s$ and
$q_{0}=2^{f}$ if $s=2$. Note that $R_{s}$ is given as in [25, Theorem 5.3.9
and Proposition 5.4.13]. If $s=2$ then $R_{s}=2m$, and so
$mf\geq|\mathbf{rad}(G)|_{s}\geq(2^{f})^{R_{s}}=2^{2mf}$, which is impossible.
Thus $s$ is odd, and so $R_{s}\geq{1\over 2}2^{(m-1)f}(2^{(m-1)f}-1)(2^{f}-1)$
by [25, Theorem 5.3.9]. It follows that
$\log_{s}(2^{mf}-1)^{2}\geq\log_{s}|\mathbf{rad}(G)|_{s}\geq R_{s}\geq{1\over
2}2^{(m-1)f}(2^{(m-1)f}-1)(2^{f}-1)$. Then $4\log_{s}(2^{mf}-1)\geq
2^{(m-1)f}(2^{(m-1)f}-1)(2^{f}-1)$, and so $\log_{s}(2^{mf}-1)>2^{(m-1)f}-1$
as $m\geq 4$. We have $mf>\log_{s}(2^{mf}-1)>2^{(m-1)f}-1$, yielding
$mf+1>2^{(m-1)f}>2(m-1)f$. This implies that $m\leq 2$, a contradiction.

Suppose that $\overline{G}\cong\Omega^{+}_{12}(2)$ and
$\overline{T}\cong\Omega^{-}_{10}(2)$. By Lemma 4.2, either
$|\mathbf{O}_{2}(\overline{G}_{\overline{\alpha}})|=2^{30}$ or
$|\mathbf{O}_{2}(\overline{G}_{\overline{\alpha}})|\leq 2^{10}$. Noting that
$2^{30}=|\Omega^{+}_{12}(2)|\geq|\mathbf{O}_{2}(\overline{G}_{\overline{\alpha}})||\mathrm{SL}_{5}(2)|_{2}$,
we have $|\mathbf{O}_{2}(\overline{G}_{\overline{\alpha}})|\leq 2^{20}$, and
so $|\mathbf{O}_{2}(\overline{G}_{\overline{\alpha}})|\leq 2^{10}$. Then
$|\overline{G}_{\overline{\alpha}}|$ is a divisor of
$2^{10}|\mathrm{SL}_{5}(2)|$ and divisible by $|\mathrm{SL}_{5}(2)|$. It
follows that
$|\mathbf{rad}(G)|={|\overline{G}_{\overline{\alpha}}|\over|\overline{G}:\overline{T}|}=2^{i}\cdot
3\cdot 5$ for some $i\leq 10$. By [25, Theorem 5.3.9 and Proposition 5.4.13],
we have $|\mathbf{rad}(G)|_{s}<s^{R_{s}}$ for $s\in\\{2,3,5\\}$, and then $G$
is a perfect central extension of $\mathrm{P\Omega}^{+}_{12}(2)$ by Lemma 9.1.
Thus $|\mathbf{rad}(G)|\leq(4,2^{6}-1)=1$ by [25, pp. 173, Theorem 5.1.4], a
contradiction.

Finally, if $(\overline{G},\overline{T})$ is one of
$(\mathrm{P\Omega}^{+}_{8}(3),\mathrm{P\Omega}^{+}_{8}(2))$,
$(\mathrm{L}_{4}(4),\mathrm{Sp}_{4}(4))$ and
$(\mathrm{L}_{6}(2),\mathrm{U}_{3}(3))$ then a similar contradiction arises as
above, we omit the details here. This completes the proof of Theorem 2.

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