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\title{\bf Arc-transitive pentavalent graphs of order $4pq$\thanks{This work was partially supported by the National Natural
Science Foundation of China (11071210,11171292), and the Fundamental
Research Funds of Yunnan University(2012CG015).}}

% input author, affilliation, address and support information as follows;
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% the street address


\author{Jiangmin Pan{\thanks{Corresponding author. E-malis:jmpan@ynu.edu.cn(J.Pan), bglou@ynu.edu.cn(B.Lou).}
\quad Bengong Lou \quad Cuifeng Liu}\\
\small School of Mathematics and Statistics \\[-0.8ex]
\small Yunnan University\\[-0.8ex]
\small Kunming, Yunnan, 650031, P.R. China\\}



%\date{{submission:Jan 1, 2012},\quad {acceptence:Feb 2, 2013}\\
%\small Mathematics Subject Classifications: 05C25}



%\author{Dagwood Remifa\thanks{Supported by NASA grant ABC123.}\\
%\small Department of Inconsequential Studies\\[-0.8ex]
%\small Solatido College\\[-0.8ex]
%\small North Kentucky, U.S.A.\\
%\small\tt remifa@dis.solatido.edu\\
%\and
%Forgotten Second Author \qquad  Forgotten Third Author\\
%\small School of Hard Knocks\\[-0.8ex]
%\small University of Western Nowhere\\[-0.8ex]
%\small Nowhere, Australasiaopia\\
%\small\tt \{fsa,fta\}@uwn.edu.ao}

 %\date{\dateline{submission date}{acceptance date}\\
 %\small Mathematics Subject Classifications: comma separated list of
% MSC codes available from http://www.ams.org/mathscinet/freeTools.html}

\date{\dateline{May 22, 2012}{Feb 6, 2013}\\
\small Mathematics Subject Classifications: 05C25, 20B25}

\begin{document}

\maketitle

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% principal new results that are to be found in the paper. The abstract
% should be informative, clear, and as complete as possible. Phrases
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\begin{abstract}
 This paper determines all arc-transitive pentavalent graphs of order
$4pq$, where $q>p\ge 5$ are primes. The cases $p=1,2,3$ and $p=q$ is
a prime have been treated previously by Hua et al. [Pentavalent
symmetric graphs of order $2pq$, {\it Discrete Math.}
{\bf{311}}(2011), 2259-2267], Hua and Feng [Pentavalent symmetric
graphs of order $8p$, {\it J. Beijing Jiaotong University}
{\bf{35}}(2011), 132-135], Guo et al. [Pentavalent symmetric graphs
of order $12p$, {\it Electronic J. Combin.} {\bf{18}}(2011), \#P233]
and Huang et al. [Pentavalent symmetric graphs of order four time a
prime power, submitted for publication], respectively.

\bigskip\noindent \textbf{Keywords:}  arc-transitive graph; normal quotient; automorphism
group.

\end{abstract}

\section{Introduction}

 For a simple, connected and undirected graph $\Ga$, denoted by
$V\Ga$ and $A\Ga$ the vertex set and arc set of $\Ga$, respectively.
Let $G$ be a subgroup of the full automorphism group $\Aut\Ga$ of
$\Ga$. Then $\Ga$ is called {\it $G$-vertex-transitive} and {\it
$G$-arc-transitive} if $G$ is transitive on $V\Ga$ and $A\Ga$,
respectively. An arc-transitive graph is also called {\it
symmetric}. It is well known that $\Ga$ is $G$-arc-transitive if and
only if $G$ is transitive on $V\Ga$ and the stabilizer
$G_{\a}:=\{g\in G\mid \a^g=\a\}$ for some $\a\in V\Ga$ is transitive
on the neighbor set $\Ga(\a)$ of $\a$ in $\Ga$.

The cubic and tetravalent graphs have been studied extensively in
the literature. It would be a natural next step toward a
characterization of pentavalent graphs. In recent years, a series of
results regarding this topic have been obtained. For example, a
classification of arc-transitive pentavalent abelian Cayley graphs
is given in \cite{ATP}, a classification of $1$-regular pentavalent
graph (that is, the full automorphism group acts regularly on its
arc set) of square-free order is presented in \cite{Li-Feng}, and
all the possibilities of vertex stabilizers of pentavalent
arc-transitive graphs are determined in \cite{Guo-Feng,Zhou-Feng}.
Also, for distinct primes $p$ and $q$, classifications of
arc-transitive pentavalent graphs of order $8p, 12p, 2pq, 2p^2$ and
$4p^n$ are presented in \cite{Hua-2,Guo,Hua-1,Pan-1,Pan-2},
respectively. In the present paper, we shall classify arc-transitive
pentavalent graphs of order $4pq$ with $q>p\ge 5$ primes. By
using the {\it Fitting subgroup} (that is, the largest
nilpotent normal subgroup) and the {\it soluble radical} (that is,
the largest soluble normal subgroup), the method used in this paper
is more simple than some relative papers.


We now give some necessary preliminary results. The first one is a
property of the Fitting subgroup, see \cite[P. 30,
Corollary]{Suzuki}.

\begin{lemma}\label{Fitting-sg}
Let $F$ be the Fitting subgroup of a group $G$. If $G$ is soluble,
then $F\ne 1$ and the centralizer $\C_G(F)\le F$.
\end{lemma}

The maximal subgroups of $\PSL(2,q)$ are known, see \cite[Section
239]{Dickson}.

\begin{lemma}\label{Subg-PSL(2,q)}
Let $T=\PSL(2,q)$, where $q=p^n\ge 5$ with $p$ a prime. Then a
maximal subgroup of $T$ is isomorphic to one of the following
groups, where $d=(2,q-1)$.

\begin{enumerate}
\item[(1)] $\D_{2(q-1)/d}$, where $q\ne 5,7,9,11$;
\item[(2)] $\D_{2(q+1)/d}$, where $q\ne 7,9$;
\item[(3)] $\ZZ_{q}{:}\ZZ_{{(q-1)/d}}$;
\item[(4)] $\A_4$, where $q=p=5$ or $q=p\equiv 3,13,27,37~(mod~40)$;
\item[(5)] $\S_4$, where $q=p\equiv \pm 1~(mod~8)$
\item[(6)] $\A_5$, where $q=p\equiv\pm 1~(mod~5)$, or $q=p^2\equiv -1~(mod~5)$
with $p$ an odd prime;
\item[(7)] $\PSL(2,p^m)$ with $n/m$ an odd integer;
\item[(8)] $\PGL(2,p^{n/2})$ with $n$ an even integer.
\end{enumerate}
\end{lemma}



For a graph $\Ga$ and a positive integer $s$, an $s$-{\it arc} of
$\Ga$ is a sequence $\a_0,\a_1,\dots,\a_s$ of vertices such that
$\a_{i-1},\a_i$ are adjacent for $1\le i\le s$ and
$\a_{i-1}\not=\a_{i+1}$ for $1\le i\le s-1$. In particular, a
$1$-arc is just an arc. Then $\Ga$ is called {\it
$(G,s)$-arc-transitive} with $G\le\Aut\Ga$ if $G$ is transitive on
the set of $s$-arcs of $\Ga$. A $(G,s)$-arc-transitive graph is
called {\it $(G,s)$-transitive} if it is not
$(G,s+1)$-arc-transitive. In particular, a graph $\Ga$ is simply
called {\it $s$-transitive} if it is $(\Aut\Ga,s)$-transitive.

The following lemma determines the stabilizers of arc-transitive
pentavalent graphs, refer to \cite{Guo-Feng,Zhou-Feng}.

\begin{lemma}\label{weiss}
Let $\Ga$ be a pentavalent $(G,s)$-transitive graph, where
$G\le\Aut\Ga$ and $s\ge 1$. Let $\a\in V\Ga$. Then one of the
following holds, where $\D_{10}$, $\D_{20}$ and $\F_{20}$ denote the
dihedral groups of order $10$ and $20$,  and the Frobenius group of
order $20$, respectively.
\begin{itemize}
\item[(a)] If $G_{\a}$ is soluble, then $s\le 3$ and $|G_{\a}|\mid 80$.
Further, the couple $(s,G_{\a})$ lies in the following table.

\[\begin{array}{|l|l|l|l|} \hline
s & 1 & 2 & 3 \\ \hline G_{\a} &  \ZZ_5,~\D_{10},~\D_{20}&
\F_{20},~\F_{20}\times\ZZ_2 & \F_{20}\times\ZZ_4  \\ \hline
\end{array}\]
\vskip0.2in

\item[(b)] If $G_{\a}$ is insoluble, then $2\le s\le 5$, and $|G_{\a}|\mid 2^{9}\cdot 3^2\cdot 5$.
Further, the couple $(s,G_{\a})$ lies in the following table.

\[\begin{array}{|l|l|l|l|l|} \hline
s & 2 & 3 & 4 & 5 \\ \hline
G_{\a} &  \A_5,\S_5 &  \A_4\times\A_5,(\A_4\times\A_5){:}\ZZ_2,& \ASL(2,4),\AGL(2,4),& \ZZ_2^6{:}\GammaL(2,4)  \\
       & &  \S_4\times\S_5 & \ASigmaL(2,4),\AGammaL(2,4)&  \\ \hline
\end{array}\]
\end{itemize}
\end{lemma}
\vskip0.2in


The next result may easily follow from \cite[Proposition 2.3]{Hua-1}
and its proof.

\begin{lemma}\label{5-fac-sg}
Let $q> p\ge 5$ be primes, and let $T$ be a nonabelian simple group
of order $2^{i}\cdot3^{j}\cdot 5\cdot p \cdot q$, where $1\leq i
\leq 11$ and $0 \leq j \leq 2$. Then $T$ lies in the following Table
$1$.
\end{lemma}

\vskip0.1in \nobreak
\[\begin{array}{|l|l|l|l|} \hline
 $4$-\PD & \Order & $5$-\PD & \Order\\ \hline

\PSL(2,5^2) & 2^3 \cdot 3 \cdot 5^2 \cdot 13 & \M_{22} & 2^7 \cdot
3^2 \cdot 5 \cdot 7 \cdot
11 \\

 \PSU(3,4) & 2^6 \cdot 3 \cdot 5^2 \cdot 13
 & \PSL(5,2) & 2^{10} \cdot 3^2 \cdot 5 \cdot 7 \cdot 31
\\

 \PSp(4,4) & 2^8 \cdot 3^2 \cdot 5^2 \cdot 17  & \PSL(2,2^6) & 2^6 \cdot 3^2 \cdot 5 \cdot 7
\cdot 13 \\

 &  & \PSL(2,2^8) & 2^8 \cdot 3 \cdot 5 \cdot 17
\cdot 257 \\

&  & \PSL(2,q) & q~ an ~odd~ prime \\
\hline
\end{array}\]
\vskip0.1in \centerline{{\bf Table 1. }} \vskip0.1in


A typical method for studying vertex-transitive graphs is taking
normal quotients. Let $\Ga$ be a $G$-vertex-transitive graph, where
$G\le\Aut\Ga$. Suppose that $G$ has a normal subgroup $N$ which is
intransitive on $V\Ga$. Let $V\Ga_N$ be the set of $N$-orbits on
$V\Ga$. The {\it normal quotient graph} $\Ga_N$ of $\Ga$ induced by
$N$ is defined as the graph with vertex set $V\Ga_N$, and $B$ is
adjacent to $C$ in $\Ga_N$ if and only if there exist vertices
$\b\in B$ and $\g\in C$ such that $\b$ is adjacent to $\g$ in $\Ga$.
In particular, if $\val(\Ga)=\val(\Ga_N)$, then $\Ga$ is called a
{\it normal cover } of $\Ga_N$.

A graph $\Ga$ is called {\it $G$-locally primitive} if, for each
$\a\in V\Ga$, the stabilizer $G_{\a}$ acts primitively on $\Ga(\a)$.
Obviously, an arc-transitive pentavalent graph is locally primitive.
The following theorem gives a basic method for studying
vertex-transitive locally primitive graphs, see \cite[Theorem
4.1]{Praeger} and \cite[Lemma 2.5]{L-Pan}.

\begin{theorem}\label{praeger}
Let $\Ga$ be a $G$-vertex-transitive locally primitive graph, where
$G\le\Aut\Ga$, and let $N\lhd G$ have at least three orbits on
$V\Ga$. Then the following statements hold.
\begin{itemize}
\item[(i)] $N$ is semi-regular on $V\Ga$, $G/N\le\Aut\Ga_N$,
and $\Ga$ is a normal cover of $\Ga_N$;
\item[(ii)] $G_{\a}\cong (G/N)_{\g}$, where $\a\in V\Ga$ and $\g\in V\Ga_N$;
\item[(iii)] $\Ga$ is $(G,s)$-transitive if and only if $\Ga_N$
is $(G/N,s)$-transitive, where $1\le s\le 5$ or $s=7$.
\end{itemize}
\end{theorem}


For reduction, we need some information of arc-transitive
pentavalent graphs of order $2pq$, stated in the following
proposition, see \cite[Theorem 4.2]{Hua-1}, where $\CC_n$, following
the notation in \cite{Hua-1}, denotes the corresponding graph of
order $n$. Noting that a $G$-arc-transitive graph is bipartite if
and only if  $G$ has a normal subgroup with index $2$ which has
exactly two orbits on the vertex set.

\begin{proposition}\label{2pq-gps}
Let $\Ga$ be an arc-transitive pentavalent graph of order $2pq$,
where $q> p\ge 5$ are primes. Then either $\Aut\Ga$ is soluble, or
the couple $(\Aut\Ga,(\Aut\Ga)_{\a})$ lies in the following Table
$2$, where $\a\in V\Ga$.

\vskip0.1in

\[\begin{array}{|lllllll|} \hline
\Row & \Ga & (p,q) & \Aut\Ga & (\Aut\Ga)_{\a} & \Transitivity &
\Remar  \\ \hline
1 & \CC_{574} & (7,41) & \PSL(2,41) & \A_5 & 2-\transitive & \no~\bipartite\\
2  & \CC_{406} & (7,29)& \PGL(2,29) & \A_5 & 2-\transitive & \bipartite\\
3 & \CC_{3422} & (29,59) & \PGL(2,59) & \A_5 & 2-\transitive & \bipartite\\
4 & \CC_{3782} & (31,61) & \PGL(2,61) & \A_5 &  2-\transitive & \bipartite\\
5 & \CC_{170} & (5,17) & \PSp(4,4).\ZZ_4 & \ZZ_2^6{:}\GammaL(2,4) &
5-\transitive & \bipartite \\ \hline
\end{array}\]
\end{proposition}
\centerline{{\bf Table 2. }} \vskip0.1in

\section{Examples}

In this section, we give two examples of arc-transitive pentavalent
graphs of order $4pq$ with $p,q\ge 5$ distinct primes.

The standard double cover is a method to construct arc-transitive
graphs from small arc-transitive graphs. Let $\Ga$ be a graph. Its
{\it standard double cover}, denoted by $\Ga^{(2)}$, is defined as a
graph with vertex set $V\Ga\times\{1,2\}$ (Cartesian product) such
that vertices $(\a,i)$ and $(\b,j)$ are adjacent if and only if
$i\ne j$ and $\a$ is adjacent to $\b$ in $\Ga$. The following facts
are well known: $\val(\Ga)=\val(\Ga^{(2)})$, $\Ga^{(2)}$ is
connected $s$-transitive if and only if $\Ga$ is connected
$s$-transitive and is not a bipartite graph.

Thus, by Proposition~\ref{2pq-gps}, the standard double cover
$\CC^{(2)}_{574}$ is a connected $2$-transitive pentavalent graph of
order $1148=4\cdot 7\cdot 41$.

 For the proof of Theorem~\ref{Thm-1}
in Section 3, we need a necessary and sufficient condition for a
graph to be the standard double cover of its normal quotient graph,
which can be easily derived from \cite[Proposition 2.6]{Guo}.

\begin{lemma}\label{SDCover}
Let $\Ga$ be a $G$-arc-transitive graph with $G\le\Aut\Ga$. Suppose
that $N\lhd G$ acts semi-regularly on $V\Ga$. Then $\Ga$ is the
standard double cover of the normal quotient graph $\Ga_N$ if and
only if $N\cong\ZZ_2$, and there is $H\lhd X$ such that $G=N\times
H$ and $H$ has exactly two orbits on $V\Ga$.
\end{lemma}



Another useful tool for constructing and studying arc-transitive
graphs is the coset graph. For a group $G$, a {\it core-free
subgroup} $H$ of $G$ (that is, $H$ contains no nontrivial normal
subgroup of $G$), and an element $g\in G\setminus H$, the {\it coset
graph} $\Cos(G,H,HgH)$ is defined as the graph with vertex set
$[G{:}H]{:}=\{Hx\mid x\in G\}$ and  $Hx$ is adjacent to $Hy$ if and
only if $yx^{-1}\in HgH$. The following lemma is well known, see
\cite{Sabidussi}.

\begin{lemma}\label{coset-gph}
Using notation as above. Then the coset graph $\Ga:=\Cos(G,H,HgH)$
is $G$-arc-transitive and $\val(\Ga)=|H{:}H\cap H^g|$. Moreover,
$\Ga$ is undirected if and only if $g^2\in H$, and $\Ga$ is
connected if and only if $\l H,g\r=G$.

Conversely, each $G$-arc-transitive graph $\Sig$ with $G\le\Aut\Sig$
is isomorphic to the coset graph $\Cos(G,G_{\a},G_{\a}gG_{\a})$,
where $\a\in V\Sig$, and $g\in \N_G(G_{\a\b})$ with $\b\in\Ga(\a)$
is a $2$-element.
\end{lemma}

\begin{example}\label{exam(4108)}
Let $T=\PSL(2,79)$. Then $T$ has two maximal subgroups $H\cong\A_5$
and $K\cong\S_4$ such that $H\cap K\cong\A_4$. Take an involution
$g\in K\setminus H$ and define the coset graph
$\CC_{4108}=\Cos(T,H,HgH)$. Then $\CC_{4108}$ is a connected
arc-transitive pentavalent graph of order $4108$ and
$\Aut(\CC_{4108})=T$. Further, any connected arc-transitive
pentavalent graph of order $4108$ admitting $T$ as an arc-transitive
automorphism group is isomorphic to $\CC_{4108}$.
\end{example}

\proof By Lemma~\ref{Subg-PSL(2,q)}, $T$ has a maximal subgroup
$H\cong\A_5$. Let $L\cong\A_4$ be a subgroup of $H$. Then
$K{:}=\N_T(L)\cong\S_4$ is a maximal subgroup of $T$ and $H\cap
K=L$. Let $g\in K\setminus H$ be an involution and define the coset
graph $\CC_{4108}=\Cos(T,H,HgH)$. Since $\l H,g\r=T$ and $|H{:}H\cap
H^g|=5$, $\CC_{4108}$ is a connected arc-transitive pentavalent
graph of order $4108$.

Now, let $\Ga$ be a connected arc-transitive pentavalent graph of
order $4108$ admitting $T$ as an arc-transitive automorphism group.
Then, for $\a\in V\Ga$,  $|T_{\a}|=|T|/4108=60$ and so
$T_{\a}\cong\A_5$ by Lemma~\ref{Subg-PSL(2,q)}. Noting that $T$ has
two conjugate classes of subgroups isomorphic to $\A_5$, and let
$H_1=H$ and $H_2$ be representatives of the two classes. Then up to
isomorphism of the graphs, we may assume that $T_{\a}=H_1$ or $H_2$.

Suppose $T_{\a}=H_1$. By Lemma~\ref{coset-gph},
$\Ga\cong\Cos(T,H,HfH)$ for some $f\in T\setminus H$ such that
$H\cap H^f\cong\A_4$. Since $H\cong\A_5$ has unique conjugate class
of subgroups isomorphic to $\A_4$, $L=(H\cap H^f)^h$ for some $h\in
H$. Then, as $H\cap H^{fh}=(H\cap H^f)^h=L$, $HfH=HfhH$ and
$\Cos(T,H,HfH)=\Cos(T,H,HfhH)$, without lose of generality, we may
assume that $H\cap H^f=L$ and $f\in \N_T(L)\setminus L$. Now, since
$\N_T(L)/L\cong\S_4/\A_4\cong\ZZ_2$, we have $\N_T(L)=L\cup Lg$, so
$f\in Lg$. It follows that $HfH=HgH$, and hence
$\Ga\cong\Cos(T,H,HfH)=\Cos(T,H,HgH)=\CC_{4108}$. Moreover, by
\cite{Magma}, $|\Aut\Ga|=246480=|T|$, we have $\Aut\Ga=\PSL(2,79)$.
Since $(\Aut\Ga)_{\a}\cong\A_5$, $\Ga\cong\CC_{4108}$ is
$2$-transitive.

Suppose next $T_{\a}=H_2$. Arguing similarly as above, there also
exists unique $T$-arc-transitive pentavalent graph. Further, by
\cite{Magma}, this graph and $\CC_{4108}$ are isomorphic, thus
completes the proof.\qed




\section{Classification}

For a given group $G$, the {\it socle} of $G$, denoted by $\soc(G)$,
is the product of all minimal normal subgroups of $G$. Obviously,
$\soc(G)$ is a characteristic subgroup of $G$. Now, we prove the
main result of this paper.

\begin{theorem}\label{Thm-1}
Let $\Ga$ be an arc-transitive pentavalent graph of order $4pq$,
where $q>p\ge 5$ are primes. Then $\Ga$ lies in the following Table
$3$, where $\a\in V\Ga$.
\end{theorem}

\[\begin{array}{|lllll|} \hline
\Ga & (p,q) &  \Aut\Ga & (\Aut\Ga)_{\a} & \Transitivity  \\ \hline
\CC^{(2)}_{574} & (7,41) & \PSL(2,41)\times\ZZ_2 & \A_5 & 2-\transitive \\
\CC_{4108} & (13,79) & \PSL(2,79) & \A_5 & 2-\transitive \\ \hline
\end{array}\]
\centerline{{\bf Table 3. }} \vskip0.1in

\proof Set $\A=\Aut\Ga$. By Lemma~\ref{weiss}, $|\A_{\a}|\mid
2^{9}\cdot 3^{2}\cdot 5$, and hence $|\A|\mid 2^{11}\cdot 3^{2}\cdot
5\cdot p\cdot q$. We divide our discussion into the following two
cases. \vskip0.1in

\noindent\textbf{Case 1. Assume $\A$ has a soluble normal subgroup.}

Let $R$ be the soluble radical of $\A$ and let $F$ be the Fitting
subgroup of $\A$. Then $R\ne 1$ and $F$ is also the Fitting subgroup
of $R$. By Lemma~\ref{Fitting-sg}, $F\ne 1$ and $\C_R(F)\le F$. As
$|V\Ga|=4pq$, $\A$ has no nontrivial normal Sylow $s$-subgroup where
$s\ne 2,p$ or $q$. So $F=\O_2(\A)\times\O_p(\A)\times\O_q(\A)$,
where $\O_2(\A),\O_p(\A)$ and $\O_q(\A)$ denote the largest normal
Sylow $2$-, $p$- and $q$-subgroups of $\A$, respectively.

For each $r\in\{2,p,q\}$, since $q>p\ge 5$, $\O_r(\A)$ has at least
4 orbits on $V\Ga$, by Proposition~\ref{praeger}, $\O_r(\A)$ is
semi-regular on $V\Ga$. Therefore, $|\O_2(\A)|\le 4$, $|\O_p(\A)|\le
p$ and $|\O_q(\A)|\le q$.

If $|\O_2(\A)|=4$, by Proposition~\ref{praeger}, the normal quotient
graph $\Ga_{\O_2(\A)}$ is an $\A/\O_2(\A)$-arc-transitive
pentavalent graph of odd order $pq$, not possible.

If $|\O_p(\A)|=p$, then $\Ga_{\O_p(\A)}$ is an arc-transitive
pentavalent graph of order $4q$, by \cite[Theorem 4.1]{Hua-2}, we
have $q=3$, which is not the case. Similarly, we may exclude the
case where $|\O_q(\A)|=q$.

Thus, $F\cong\ZZ_2$. Then $\C_R(F)=F$ and
$R/F=R/\C_R(F)\le\Aut(F)=1$, it follows that $R=F\cong\ZZ_2$. In
particular, $\A\ne R$, that is, $\A$ is insoluble.

Now, $\Ga_R$ is an $\A/R$-arc-transitive pentavalent graph of order
$2pq$. Since $\Aut(\Ga_R)\ge \A/R$ is insoluble, by
Proposition~\ref{2pq-gps}, $\Ga_R$ and $\Aut(\Ga_R)$ lie in Table 2,
and as $\A/R$ is transitive on $A\Ga_R$, we have $10pq\mid |\A/R|$,
then checking the subgroups of $\Aut(\Ga_R)$ in the Atlas
\cite{Atlas}, we easily conclude that $\soc(\A/R)=\soc(\Aut\Ga_R)$.
Set $T=\soc(\A/R)$. We consider all the possibilities of $T$ lying
in Table 2 one by one.

For row 1, $(p,q)=(7,41)$ and
$\A=R.\PSL(2,41)\cong\ZZ_2.\PSL(2,41)$. It follows that either
$\A=\ZZ_2\times H{:}=\ZZ_2\times\PSL(2,41)$ or $\A=\SL(2,41)$. For
the former, by Theorem~\ref{praeger}, $H$ has at most two orbits on
$V\Ga$. Further, if $H$ is transitive on $V\Ga$, then
$|H_{\a}|={|H|\over 4\cdot 7\cdot 41}=30$, which is not possible as
$H=\PSL(2,41)$ has no subgroup of order $30$. Thus, $H$ has exactly
two orbits on $V\Ga$, it then follows from Lemma~\ref{SDCover} that
$\Ga=\CC^{(2)}_{574}$, as in row 1 of Table 3. For the latter, by
Theorem~\ref{praeger}(ii), $\A=\SL(2,41)\ge\A_{\a}\cong\A_5$.
However, by \cite[Lemma 2.7]{Du-M}, the group $\GL(2,a)$ for each
prime $a\ge 5$ contains no nonabelian simple subgroup, which is a
contradiction.

For row 2, $(p,q)=(7,29)$, and $T=\PSL(2,29)$ has exactly two orbits
on $V\Ga_R$. Since $\Out(\PSL(2,29))=\ZZ_2$, we have
$\A/R=\Aut(\Ga_R)=\PGL(2,29)$, and by Theorem~\ref{praeger}~(ii),
$\A_{\a}\cong\A_5$. Since $\A\cong\ZZ_2.\PSL(2,29).\ZZ_2$, we have
that either $\A=(\ZZ_2\times\PSL(2,29)).\ZZ_2$ or $\SL(2,29).\ZZ_2$.
For the former case, $\A$ has a normal subgroup $M_1\cong
\PSL(2,29)$. By Theorem\ref{praeger}~(i), $M_1$ has at most two
orbits on $V\Ga$. If $M_1$ is transitive on $V\Ga$, then
$T\cong(R\times M_1)/R$ is transitive on $V\Ga$, a contradiction.
Therefore, $M_1$ has exactly two orbits on $V\Ga$, and hence
$|(M_1)_{\a}|={|M_1|\over 2\cdot 7\cdot 29}=30$. By \cite{Atlas},
$(M_1)_{\a}\cong \D_{30}$, which is not possible as $(M_1)_{\a}\le
\A_{\a}\cong\A_5$. For the latter case, $\A$ has a normal subgroup
$M_2\cong\SL(2,29)$ which has exactly two orbits on $V\Ga$. It
follows that $|(M_2)_{\a}|={|M_2|\over 2\cdot 7\cdot 29}=60$. Then
as $(M_2)_{\a}\le \A_{\a}\cong\A_5$, we obtain that
$(M_2)_{\a}\cong\A_5$. which is not possible as $\SL(2,29)$ has no
nonabelian simple subgroup by \cite[Lemma 2.7]{Du-M}.

Similarly, we may exclude the cases where $T=\PSL(2,59)$ and
$\PSL(2,61)$, lying in rows 3 and 4 of Table 2.

Finally, we treat the case $T=\PSp(4,4)$, as in row 5 of Table 2.
Then $(p,q)=(5,17)$ and $\PSp(4,4).\ZZ_2\le \A/R\le \PSp(4,4).\ZZ_4$
as $T=\PSp(4,4)$ is not transitive on $V\Ga_R$. Since the Schur
Multiplier of $\PSp(4,4)$ is trivial, we conclude that
$\A=(\ZZ_2\times\PSp(4,4)).\ZZ_2$ or $(\ZZ_2\times\PSp(4,4)).\ZZ_4$.
Thus $\A$ always has a normal subgroup $M$ such that
$M\cong\PSp(4,4)$. By Theorem~\ref{praeger}, $M$ has at most two
orbits on $V\Ga$. However, by \cite{Atlas}, $\PSp(4,4)$ has no
subgroup with index $170$ or $340$, which is a contradiction as
$|V\Ga|=340$. \vskip0.1in

\noindent\textbf{Case 2. Assume $\A$ has no soluble normal
subgroup.}

Let $N$ be a minimal normal subgroup of $\A$. Then $N=S^d$, where
$S$ is a nonabelian simple group and $d\geq 1$.

If $N$ is semi-regular on $V\Ga$, then $|N|$ divides $4pq$, we
conclude that $|S|=4pq$ because $S$ is insoluble. Noting that
$q>p\ge 5$, by \cite[P. 12-14]{Gorenstein}, no such simple group
exists, a contradiction.

Hence, $N$ is not semi-regular on $V\Ga$. Then by
Theorem~\ref{praeger}, $N$ has at most two orbits on $V\Ga$, so
$2pq$ divides $|\a^N|$. Moreover, since $\Ga$ is connected and $1\ne
N\lhd \A$, we have $1\ne N_{\a}^{\Ga(\a)}\lhd \A_{\a}^{\Ga(\a)}$, it
follows that $5\mid |N_{\a}|$, we thus have $10pq\mid |N|$. Since
$q>5$, $q\mid |N|$ and $q^2$ does not divide $|N|$ as $|\A|\mid
2^{11}\cdot 3^{2}\cdot 5\cdot p\cdot q$, we conclude that $d=1$ and
$N=S$ is a nonabelian simple group. Let $C=\C_\A(S)$. Then $C\lhd
\A$, $C\cap S=1$ and $\l C,S\r=C\times S$. Because $|C\times S|$
divides $2^{11}\cdot 3^{2}\cdot 5\cdot p\cdot q$ and $10pq\mid |S|$,
$C$ is a $\{2,3\}$-group, and hence soluble. So $C=1$ as $R=1$. This
implies $\A=\A/C\le\Aut(S)$, that is, $\A$ is almost simple with
socle $S$.

Thus, $\soc(\A)=S$ is a nonabelian simple group and satisfies the
following condition. \vskip0.1in

\textbf{Condition $(*)$:} $|S|$ lies in Table 1 such that $10pq\mid
|S|$, and $|S{:}S_{\a}|=2pq$ or $4pq$. \vskip0.1in

Suppose first that $S$ has exactly four prime factors. Then
$S=\PSL(2,5^2), \PSU(3,4)$ or $\PSp(4,4)$. By Condition $(*)$ and
\cite{Atlas}, the only possibility is
$(S,S_{\a})=(\PSL(2,25),\A_5)$. Now, $(p,q)=(5,13)$,
$|S{:}S_{\a}|=130$ and $S$ has two orbits on $V\Ga$. Since
$\Out(\PSL(2,25))=\ZZ_2^2$, $\A\le\PSL(2,5^2).\ZZ_2^2$, hence either
$\A=\PGammaL(2,25)$ and $\A_{\a}=\S_5$, or $\A=\PSL(2,25){:}\ZZ_2\le
\PGammaL(2,25)$ and $\A_{\a}\cong\A_5$. For the former, $\A$ has
three conjugate classes of subgroups isomorphic to $\S_5$. By
\cite{Magma}, for each case, $\A$ has no suborbit of length $5$,
that is, there is no pentavalent graph admitting $\A$ as an
arc-transitive automorphism group, no example appears. For the
latter, $\PGammaL(2,25)$ has three subgroups which are semi-products
$\PSL(2,25){:}\ZZ_2$, and $\A$ is isomorphic to one of the three.
Then by using \cite{Magma}, a direct computation shows that $\A$
also has no suborbit of length $5$ for each of the three cases, and
thus can not give rise example.


Suppose now that $S$ has five prime divisors, as in column 3 of
Table 1. Assume $S\ne\PSL(2,q)$. If $S=\PSL(5,2)$, then by
\cite{Atlas}, we have $S_{\a}=\ZZ_2^4{:}\S_6$. As
$\Out(\PSL(5,2))=\ZZ_2$, $\A_{\a}=\ZZ_2^4{:}\S_6$ or
$\ZZ_2^4{:}(\S_6\times\ZZ_2)$, both have no permutation
representation of degree $5$, not possible. If $S=\M_{22}$, then
$(p,q)=(7,11)$. By \cite{Atlas}, $\M_{22}$ has no subgroup with
index $154$ or $308$, that is, $S$ does not satisfy the Condition
$(*)$, not the case. If $S=\PSL(2,2^6)$, then $(p,q)=(7,13)$, and
either $|S_{\a}|=|S|/2pq=1440$ or $|S_{\a}|=|S|/4pq=720$. However,
by Lemma~\ref{Subg-PSL(2,q)}, it is easy to verify that
$\PSL(2,2^6)$ has no subgroup with order $720$ or $1440$, a
contradiction. For $S=\PSL(2,2^8)$, then $(p,q)=(17,257)$, and
either $|S_{\a}|=|S|/2pq=1920$ or $|S_{\a}|=|S|/4pq=960$. By
Lemma~\ref{Subg-PSL(2,q)}, the only possibility is that
$S_{\a}\le\ZZ_{2^8}{:}\ZZ_{2^8-1}$ is soluble, then as
$\Out(S)=\ZZ_8$, $\A_{\a}\le S_{\a}.\ZZ_8$ is also soluble. However,
as $|\A_{\a}|\ge 960$, Lemma~\ref{weiss} implies that $\A_{\a}$ is
insoluble, which is a contradiction.

Now, assume that $S=\PSL(2,q)$ has five prime divisors. Then $q>p>5$
and $S$ is a $\{ 2,3,5,p,q\}$-group. Since $|S{:}S_{\a}|=2pq$ or
$4pq$, we obtain $3\mid |S_{\a}|$, so $|\A_{\a}|{\not |} 80$. hence
$\A_{\a}$ is insoluble by Lemma~\ref{weiss}. Since
$\Out(\PSL(2,q))=\ZZ_2$, $\A_{\a}\le S_{\a}.\ZZ_2$, we have $S_{\a}$
is insoluble, that is, $S_{\a}$ is an insoluble subgroup of
$\PSL(2,q)$. It then follows from Lemma~\ref{Subg-PSL(2,q)} that
$S_{\a}=\A_{\a}=\A_5$ as $\Aut(\PSL(2,q))=\PGL(2,q)$ has no subgroup
isomorphic to $\A_5.\ZZ_2$. Since $S$ has at most two orbits on
$V\Ga$, $|S|=60\cdot 2pq$ or $60\cdot 4pq$. So $|S|$ has exactly one
3-divisor, one 5-divisor, and three or four 2-divisors. Moreover, as
$8$ divides $|S|=|\PSL(2,q)|={q(q-1)(q+1)\over 2}$, $16\mid
(q^2-1)$, it implies $q\equiv \pm 1~(mod~8)$. Now, since $({q-1\over
2},{q+1\over 2})=1$, if $p\mid {(q-1)}$, then ${q+1}=2^i3^j5^k$,
where $1\leq i\leq 4, 0\leq j,k\leq 1$, we easily conclude that
$q\in \{7,23,47,79,239\}$ as $q\equiv \pm 1~(mod~8)$. Similarly, if
$p\mid (q+1)$, then $q\in \{7,17,31,41,241\}$. Further, as
$|S|=|\PSL(2,q)|$ has exactly five prime divisors, a simple
computation shows that $q\ne 7,17,23,31,47,239$ or $241$, we finally
conclude $q=41$ or $79$.


Now, by \cite{Atlas}, the only possibilities are as in the following
table.

\nobreak
\[\begin{array}{|l|ll|} \hline
 S   & \PSL(2,41) & \PSL(2,79)     \\ \hline
S_{\a} & \A_5 & \A_5 \\ \hline (p,q) &  (13,41)  & (13,79)     \\
\hline

\end{array}\]
\vskip0.1in



If $S=\PSL(2,41)$, then $S$ has two orbits on $V\Ga$, so
$\A=\PGL(2,41)$ and $\A_{\a}=S_{\a}=\A_5$. Let $\b\in\Ga{(\a)}$. By
Lemma~\ref{coset-gph}, we may suppose
$\Ga=\Cos(\A,\A_{\a},\A_{\a}g\A_{\a})$ for some $g\in
\N_{\A}(\A_{\a\b})$. Since $\val(\Ga)=5$, $\A_{\a\b}\cong\A_4$, it
follows that $\N_{\A}(\A_{\a\b})=\N_S(\A_{\a\b})\cong\S_4$. Hence
$\l \A_{\a},g\r\subseteq S\subset \A$, which contradicts the
connectivity of $\Ga$.

Finally, for $S=\PSL(2,79)$, $S$ is transitive on $V\Ga$, and so
$\A=\PSL(2,79)$ and $\A_{\a}=\A_5$. By Example~\ref{exam(4108)},
$\Ga\cong\CC_{4108}$ is a $2$-transitive graph. This completes the
proof.\qed

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{Acknowledgements}
The authors are very grateful to the referee for the valuable
comments.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% \bibliographystyle{plain}
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\end{document}
