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\def\div{\,\big|\,} \def\mod{{\sf mod~}}
\def\ZZ{\mathbb Z} \def\BB{{\mathcal B}} \def\CC{{\mathcal C}}
\def\Aut{{\sf Aut}}  \def\Out{{\sf Out}}
\def\soc{{\sf soc}} \def\Cos{{\sf Cos}}  \def\Cay{{\sf Cay}}
\def\K{{\sf K}} \def\D{{\rm D}} \def\S{{\rm S}} \def\G{{\rm G}} \def\A{{\rm A}} 
\def\J{{\rm J}} \def\M{{\rm M}} \def\E{{\rm E}} \def\F{{\rm F}} 
\def\C{{\bf C}}\def\N{{\bf N}}
\def\PG{{\rm PG}} \def\GF{{\rm GF}}
\def\PSL{{\rm PSL}}\def\PGL{{\rm PGL}}
\def\GL{{\rm GL}} \def\AGL{{\rm AGL}} \def\PSU{{\rm PSU}} \def\Sz{{\rm Sz}} 
\def\ETSQF{{\sf ETSQF}}



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\title{\bf On  edge-transitive graphs of square-free order}

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

\author{Cai Heng Li\thanks{Supported by ARC Grant DP1096525.}\\
\small School of Mathematics and Statistics\\[-0.8ex]
\small The University of Western Australia\\[-0.8ex]
\small Crawley, WA 6009, Australia\\
\small\tt cai.heng.li@uwa.edu.au\\
\and
Zai Ping Lu\thanks{Supported by National Natural Science Foundation of China (11271267, 11371204).}\\
\small  Center for Combinatorics\\[-0.8ex]
\small LPMC-TJKLC, Nankai University\\[-0.8ex]
\small Tianjin 300071, P. R. China\\
\small\tt lu@nankai.edu.cn\\
\and
Gai Xia Wang\thanks{Supported by Anhui Provincial Natural Science Foundation(1408085MA04).}\\
\small Department of Applied Mathematics\\[-0.8ex]
\small  Anhui University of Technology\\[-0.8ex]
\small  Maanshan 243002, P. R. China\\
\small\tt wgx075@163.com\\
}

% \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{Aug 4, 2014}{?? ??, 2015}\\
\small Mathematics Subject Classifications: 05C25, 20B25}

\begin{document}

\maketitle

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\begin{abstract}
  We study the class of  edge-transitive graphs   of
square-free order and valency at most $k$. It is shown that, except for a
few special families of graphs, only finitely many
members in this class are {\em basic} (namely, not a normal multicover of
another member). Using this result, we determine the automorphism groups
of locally primitive arc-transitive graphs with square-free order.

  % keywords are optional
  \bigskip\noindent \textbf{Keywords:} edge-transitive graph; arc-transitive graph; stabilizer; quasiprimitive permutation group; almost simple group
\end{abstract}

\section{Introduction}

For a graph $\Gamma=(V,E)$, the number of vertices $|V|$ is called the
{\em order} of $\Gamma$. A graph $\Gamma=(V,E)$ is called {\em edge-transitive} if its automorphism group $\Aut\Gamma$ acts transitively on the edge set $E$.
For convenience, denote by $\ETSQF(k)$ the class of connected
edge-transitive graphs with square-free order and valency at most
$k$.



The study of special subclasses of $\ETSQF(k)$ has a long history,
see for example \cite{AX,Chao,CO-2p,PWX,PX,Wang,WX,ZF} for those graphs of
order being a prime or a product of two primes. Recently, several classification results about the class $\ETSQF(k)$ were given.
Feng and Li \cite{Feng-Li} gave a classification of one-regular graphs of square-free order and prime valency.
By Li et al. \cite{Edge-Cay-val-4-sqf,sqf-v4}, one may obtain   a   classification of vertex-transitive and edge-transitive  tetravalent graphs of square-free order.
By Li et al. \cite{sqf-cubic} and Liu and Lu \cite{LiuLu}, one may deduce an explicitly  classification of $\ETSQF(3)$.
In this paper, we give a   characterization about
the class $\ETSQF(k)$.



A typical method for analyzing edge-transitive graphs is to take
 {\em normal quotient}. Let $\Gamma=(V,E)$ be a connected graph such that
 a subgroup $G\le \Aut\Gamma$ acts transitively on $E$.
Let $N$ be  a normal subgroup of $G$, denoted by $N\lhd G$. Then either $N$ is transitive on $V$, or each $N$-orbit is an independent set of $\Gamma$. Let
$V_N$ be the set of all $N$-orbits on $V$. The {\em normal quotient} $\Gamma_N$  (with respect to $G$ and $N$)
is defined as the graph with vertex set $V_N$ such that distinct
vertices $B,B'\in V_N$ are adjacent in $\Gamma_N$ if and only if some $\alpha\in B$
and some $\alpha'\in B'$  are adjacent in $\Gamma$.
We call $\Gamma_N$  non-trivial if $N\ne 1$ and $|V_N|\ge 3$.
It is well-known and  easily shown that $\Gamma_N$ is an edge-transitive graph.
Moreover, if all $N$-orbits have the same length (which is obvious if $G$ is transitive
on $V$), then $\Gamma_N$ is a regular graph of valency a divisor of the valency of $\Gamma$; in this case, $\Gamma$ is called a {\em normal multicover} of $\Gamma_N$.



A member   in $\ETSQF(k)$ is
called {\em basic} if it has no non-trivial normal quotients.
Then every member in $\ETSQF(k)$   is a  multicover
of some basic member, or has a  non-regular normal quotient  (which might occur for
vertex-intransitive graphs). Thus, to a great extent, basic members play an important role in characterizing   the graphs in $\ETSQF(k)$. The first result of this paper shows that, except for a
few special families of graphs, there are only finitely many basic
members in $\ETSQF(k)$.

\begin{theorem}\label{ET-reduction}
Let $\Gamma=(V,E)$ be a connected graph of square-free order and
valency $k\ge 3$. Assume that   $G\le \Aut\Gamma$ acts transitively
on $E$ and that each non-trivial normal subgroup of $G$ has at most $2$
orbits on $V$. Then one of the following  holds:
\begin{itemize}
\item[{\rm (1)}] $\Gamma$ is a complete bipartite graph, and $G$ is described in
(1) and (5) of Lemma~\ref{reduction};
\item[{\rm (2)}] $G$ is   one of the Frobenius groups $\ZZ_p{:}\ZZ_k$ and $\ZZ_p{:}\ZZ_{2k}$,
where $p$ is a prime;

\item[{\rm (3)}] $\soc(G)=\M_{11},\M_{12},\M_{22},\M_{23},\M_{24}$ or $\J_1$;

\item[{\rm (4)}] $G=\A_n$ or $\S_n$ with $n<3k$;

\item[{\rm (5)}] $G=\PSL(2,p)$ or $\PGL(2,p)$;

\item[{\rm (6)}] $\soc(G)=\PSL(2,p^f)$ with $f\ge 2$ and $p^f>9$, and either $k$ is divisible by $p^{f-1}$
or $f=2$ and $k$ is divisible by $p+1$;

\item[{\rm (7)}] $\soc(G)=\Sz(2^f)$
and $k$ is divisible by $2^{2f-1}$;


\item[{\rm (8)}]  $G$ is of Lie type defined over $\GF(p^f)$ with $p\le k$, and
either
\begin{itemize}
\item[{\rm (i)}] $[\frac{d}{2}]f<k$, and $G$ is a $d$-dimensional classical
group with $d\ge 3$; or

\item[{\rm (ii)}] $2f<k$, and $\soc(G)=\G_2(p^f)$, ${}^3\D_4(p^f)$, $\F_4(p^f)$, ${}^2\E_6(p^f)$, or $\E_7(p^f)$.
\end{itemize}
\end{itemize}
\end{theorem}

%\noindent
%{\bf Remarks on Theorem~$\ref{ET-reduction}$:}
\begin{remark}[Remarks on Theorem~\ref{ET-reduction}]
For a finite group $G$, the {\em socle} $\soc(G)$ of
$G$ is the subgroup generated by all minimal normal subgroups of
$G$. A finite group is called {\em almost simple} if $\soc(G)$ is a
non-abelian simple group. 
\begin{itemize}
\item[(a)] The groups $G$ in case (1) are known except for $G$ being almost simple.
\item[(b)] The vertex-transitive graphs in case (5) are characterized in Theorem~\ref{PSL(2,p)-graphs}.
\item[(c)] Some properties about the graphs in cases (6)-(7) are given in Lemmas~\ref{PSL(2,q)} and \ref{Sz}, respectively.
\end{itemize}
\end{remark}

It would be interest to give further characterization for some special cases.

%\noindent{\bf Problems}.
\begin{problem}
\begin{itemize}
\item[(i)] Characterize  edge-transitive graphs of square-free order which admits a group with socle $\PSL(2,q)$, $\Sz(q)$,
$\A_n$ or a sporadic simple group.
\item[(ii)] Classify edge-transitive graphs  of square-free order of small valencies.
\end{itemize}
\end{problem}

%\vskip0.1in

For a graph $\Gamma=(V,E)$ and $G\le \Aut\Gamma$, the graph $\Gamma$ is called {\em $G$-locally
primitive} if, for each $\alpha\in V$, the stabilizer  of $\alpha$ in
$G$ induces a primitive permutation group on the   neighbors of
$\alpha$ in $\Gamma$. The second result of this paper determines, on  the basis of Theorem \ref{ET-reduction},
the automorphism groups
of locally primitive  arc-transitive graphs of square-free order.




\begin{theorem}\label{th-locally-prim}
Let $\Gamma=(V,E)$ be a connected  $G$-locally primitive
graph of  square-free order and valency $k\ge3$. Assume that $G$ is transitive on $V$ and that $\Gamma$ is
not a complete bipartite graph. Then one of the following statements
is true.
\begin{itemize}
\item[{\rm (1)}] $G=\D_{2n}{:}\ZZ_k$,
$2nk$ is square-free, $k$ is the smallest prime divisor of
$nk$, and $\Gamma$ is a bipartite Cayley graph of the dihedral group
$\D_{2n}$;



\item[{\rm (2)}] $G=M{:}X$, where $M$ is of square-free order, $X$ is almost simple with socle $T$ descried as
in (3)-(6) and (8) of Theorem~\ref{ET-reduction} such that $MT=M\times T$, $T$ has at
most two orbits on $V$ and $\Gamma$ is $T$-edge-transitive; in
particular, if $T=\PSL(2,p)$, then $M$, $T_\alpha$ and $k$   are listed in Table~\ref{tab=PSL(2,p)-graph},
where $\alpha\in V$.
\end{itemize}
\end{theorem}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Preliminaries}\label{lemmas}
Let $\Gamma=(V,E)$ be a graph without isolated vertices, and let $G\le \Aut\Gamma$.
The graph $\Gamma$ is said to be  {\em
$G$-vertex-transitive} or  {\em $G$-edge-transitive}  if $G$ acts transitively on $V$ or $E$,  respectively.
Recall that an {\em arc} in $\Gamma$ is an ordered pair of adjacent vertices. The graph $\Gamma$ is called
{\em $G$-arc-transitive} if $G$ acts transitively on the  set of arcs of $\Gamma$.
For a vertex $\alpha\in V$, we denote by $\Gamma(\alpha)$
the set of neighbors of $\alpha$ in $\Gamma$, and by $G_\alpha$ the stabilizer of $\alpha$ in $G$. Then it is easily shown that
$\Gamma$ is $G$-arc-transitive if and only if $\Gamma$ is $G$-vertex-transitive and, for $\alpha\in V$, the vertex-stabilizer
$G_\alpha$ acts transitively on $\Gamma(\alpha)$.


Let $\Gamma=(V,E)$ be a connected  $G$-edge-transitive graph. Note that each  edge of $\Gamma$ gives two arcs.
Then either $\Gamma$ is $G$-arc-transitive or $G$ has exactly two orbits (of the same size $|E|$) on the arc set of $\Gamma$.
If $\Gamma$ is not $G$-vertex-transitive  then $\Gamma$ is a bipartite graph and, for $\alpha\in V$, the stabilizer
$G_\alpha$ acts transitively on $\Gamma(\alpha)$.
 If $\Gamma$ is $G$-arc-transitive, then there exists $g\in G\setminus
G_\alpha$ such that $(\alpha,\beta)^g=(\beta,\alpha)$ and, since $\Gamma$ is connected, $\langle g,G_\alpha\rangle=G$; obviously, this $g$ can be
chosen as a 2-element in $\N_G(G_{\alpha\beta})$ with $g^2\in G_{\alpha\beta}$, where  $G_{\alpha\beta}=G_\alpha\cap G_\beta$.
Suppose  that $\Gamma$ is $G$-vertex-transitive but not $G$-arc-transitive.
Then the arc set of $\Gamma$ is partitioned into two $G$-orbits  $\Delta$ and $\Delta^*$, where $\Delta^*=\{(\alpha,\beta)\mid (\beta,\alpha)\in \Delta\}$.
Thus, for $\alpha\in V$, the set $\Gamma(\alpha)$ is  partitioned into two $G_\alpha$-orbits $\Delta(\alpha)=\{\beta\mid (\alpha,\beta)\in \Delta\}$ and $\Delta^*(\alpha)=\{\beta\mid (\beta,\alpha)\in \Delta\}$, which have   equal size. Then we have the next lemma.
\begin{lemma}\label{edge-trans-gh}
 Let $\Gamma=(V,E)$ be a connected  $G$-edge-transitive graph, and
$\{\alpha,\beta\}\in E$.  Then one of the following holds.
\begin{itemize}
\item[{\rm (1)}] The  stabilizer $G_\alpha$ is transitive on $\Gamma(\alpha)$,  $|\Gamma(\alpha)|=|G_\alpha:G_{\alpha\beta}|$, and either
 \begin{itemize}
\item[{\rm (i)}] $G$ is intransitive on $V$; or
\item[{\rm (ii)}] $G=\langle g,G_\alpha\rangle$ for a $2$-element $g\in \N_G(G_{\alpha\beta})\setminus G_\alpha$ with $(\alpha,\beta)^g=(\beta,\alpha)$  and $g^2\in G_{\alpha\beta}$.
\end{itemize}
\item[{\rm (2)}] $\Gamma$ is $G$-vertex-transitive,   $G_\alpha$ has exactly two orbits on
$\Gamma(\alpha)$ of the same size $|G_\alpha:G_{\alpha\beta}|$; in particular,
$|\Gamma(\alpha)|=2|G_\alpha:G_{\alpha\beta}|$.
\end{itemize}
\end{lemma}



%\vskip 5pt


Let $\Gamma=(V,E)$ be a  regular graph and $G\le \Aut\Gamma$. For $\alpha\in V$, the stabilizer $G_\alpha$ induces a permutation group
$G_\alpha^{\Gamma(\alpha)}$ (on $\Gamma(\alpha)$). Let $G_\alpha^{[1]}$ be the kernel of this action.
Then $G_\alpha^{\Gamma(\alpha)}\cong G_\alpha/G_\alpha^{[1]}$.
Considering the actions of   Sylow subgroups of $G_\alpha^{[1]}$ on $V$, it is easily shown that the next lemma holds, see \cite{CLP} for example.
%, and the next lemma follows.
\begin{lemma}\label{p<k} Let $\Gamma=(V,E)$ be a connected regular  graph, $G\le \Aut\Gamma$ and $\alpha\in V$.
Assume that $G_\alpha\ne 1$.
Let $p$ be a prime divisor of $|G_\alpha|$. Then $p\le |\Gamma(\alpha)|$. If further $\Gamma$ is $G$-vertex-transitive, then
$p$ divides $|G_\alpha^{\Gamma(\alpha)}|$ and,  for $\beta\in \Gamma(\alpha)$,
each prime divisor of $|G_{\alpha\beta}|$ is less than $|\Gamma(\alpha)|$.
\end{lemma}

A permutation group $G$ on a set $\Omega$ is {\em semiregular} if $G_\alpha=1$ for each $\alpha\in \Omega$.
A transitive permutation group is {\em regular} if further it is semiregular.


\begin{lemma}\label{abel-reg-1}
Let $\Gamma$ be a connected $G$-vertex-transitive graph,  $N\lhd G\le \Aut\Gamma$ and $\alpha\in V$. Assume
that $N_\alpha^{\Gamma(\alpha)}$ is semiregular on $\Gamma(\alpha)$. Then $N_\alpha^{[1]}=1$.
\end{lemma}
\proof
Let $\beta \in \Gamma(\alpha)$. Then $\beta=\alpha^x$ for some $x\in G$, and
hence $N_\beta=N_{\alpha^x}=N\cap G_{\alpha^x}=(N\cap G_\alpha)^x=(N_\alpha)^x$. It follows that $N_\beta^{\Gamma(\beta)}$ and
$N_\alpha^{\Gamma(\alpha)}$ are permutation isomorphic; in particular,
$N_\beta^{\Gamma(\beta)}$ is semiregular on $\Gamma(\beta)$. Thus $N_\alpha^{[1]}$ acts
trivially on $\Gamma(\beta)$, and so $N_\alpha^{[1]}=N_\beta^{[1]}$. Since $\Gamma$
is connected, $N_\alpha^{[1]}$ fixes each vertex of $\Gamma$, hence
$N_\alpha^{[1]}=1$. \qed


\begin{lemma}\label{bi-semireg}
Let $\Gamma=(V,E)$ be a connected graph,  $N\lhd G\le\Aut\Gamma$ and $\alpha\in V$.
Assume that either $N$ is regular on $V$, or $\Gamma$ is a bipartite graph such that $N$
is regular on both the bipartition subsets of $\Gamma$. Then $G_\alpha^{[1]}=1$.
\end{lemma}
\begin{proof}
Set $X=NG_\alpha^{[1]}$. Then $X_\alpha=G_\alpha^{[1]}$ and   $X_\alpha^{[1]}=G_\alpha^{[1]}$, and hence $X_\alpha^{\Gamma(\alpha)}=1$.

Assume first that $N$ is regular on $V$. Then $G=NG_\alpha$. It follows that $X$ is normal in $G$.
Thus our results follows from Lemma \ref{abel-reg-1}.

Now assume that  $\Gamma$ is a bipartite graph with bipartition subsets $U$ and $W$, and  that $N$
is regular on both $U$ and $W$. Without loss of generality, we assume that $\alpha\in U$. Then
$\Gamma(\alpha)\subseteq W$, and $X_\alpha=X_\beta$ for $\beta\in \Gamma(\alpha)$.
Let $\gamma\in \Gamma(\beta)$. Then $\gamma\in U$. Set
$E_0=\{\{\gamma,\beta\}^x\mid x\in X\}$. Then $\Sigma=(V,E_0)$ is a
spanning subgraph of $\Gamma$, and $X$ acts transitively on $E_0$. Thus
$\Sigma$ is a regular graph, and $X_\alpha$ is transitive on $\Sigma(\alpha)$.
Noting $\Sigma(\alpha)\subseteq \Gamma(\alpha)$, it follows that $|\Sigma(\alpha)|=1$,
and hence $\Sigma$ is a matching. In particular, $X_\beta=X_\gamma$. It follows that
$G_\alpha^{[1]}=X_\alpha=X_\beta=X_\gamma$. Since all vertices in $U$ are equivalent
under $X$, we have $X_\gamma$ acts trivially on $\Gamma(\gamma)$. Then a
similar argument as above leads to $G_\alpha^{[1]}=X_\gamma=X_\delta=X_\theta$ for
any $\delta\in \Gamma(\gamma)$ and $\theta\in \Gamma(\delta)$. Then, by the
connectedness, we conclude that $G_\alpha^{[1]}$ fixes each vertex of $\Gamma$.
Thus $G_\alpha^{[1]}=1$.
\end{proof}








We end this section by quoting a known result.

\begin{lemma}[\cite{Edge-Cay-val-4-sqf}]\label{normal-sub}
Let $\Gamma=(V, E)$ be a connected   $G$-edge-transitive graph,  $N\lhd G\le \Aut\Gamma$ and $\alpha\in V$. Then
all $N_\alpha$-orbits on $\Gamma(\alpha)$ have the same length.
\end{lemma}

%\vskip 10pt


\section{Complete bipartite graphs}\label{complete}

We first list a well-known result in number theory. For integers
$a>0$ and $n>0$, a prime divisor of $a^n-1$ is called {\em
primitive} if it does not divide $a^i-1$ for any $0<i<n$.

\begin{theorem}[Zsigmondy]\label{Zsigmondy}
For integers $a,n\ge2$, if $a^n-1$ does not have primitive prime
divisors, then either $(a,n)=(2,6)$, or $n=2$ and $a+1$ is a power
of $2$.
\end{theorem}

Let $G$ be a   permutation group on  $V$, and let $x$ be a permutation on $V$ which centralizes $G$.
If   $x$ fixes some point  $\alpha\in V$, then
$x$ fixes $\alpha^g$ for each $g\in G$. Thus the next simple result follows.


\begin{lemma}\label{centralizer}
Let $G$ be a   permutation group on  $V$. Assume that $N$ is a normal transitive subgroup  of $G$. Then the centralizer
$\C_G(N)$ is semiregular on $V$, and $\C_G(N)=N$ if further $N$ is
abelian.
\end{lemma}



Recall that a transitive permutation group $G$ is  {\em
quasiprimitive} if each non-trivial normal subgroup of $G$ is
transitive.  Let $G$ be a quasiprimitive  permutation group on  $V$, and
let $\BB$ be a $G$-invariant partition on $V$. Then $G$ induces a
 permutation group $G^\BB$  on $\BB$.  Assume that $|\BB|\ge 2$. Since $G$ is quasiprimitive, $G$ acts faithfully on $\BB$.
 Then $G^\BB\cong G$, and so $\soc(G^\BB)\cong \soc(G)$.

\begin{lemma}\label{quasi-sqf} Let $G$ be a quasiprimitive permutation group of
square-free degree. Then $\soc(G)$ is simple, so either $G$ is almost
simple or $G\le \AGL(1,p)$ for a prime $p$.
\end{lemma}
\begin{proof}
Let $G$ be a quasiprimitive permutation group on $V$ of
square-free degree.
Let $\BB$ be a $G$-invariant partition on $V$ such that $|\BB|\ge 2$ and
$G^\BB$ is primitive.
Noting that $|\BB|$ is square-free,
by \cite{L-S}, $\soc(G^\BB)$ is simple. Thus $\soc(G)\cong \soc(G^\BB)$
is simple, and the result follows. 
\end{proof}

Let $G$ be a permutation group on $V$. For a subset $U\subseteq V$, denote by
$G_U$ and $G_{(U)}$ the subgroups of $G$ fixing $U$ set-wise and point-wise, respectively.
For $X\le G$ and an $X$-invariant subset $U$ of $V$, denote by $X^U$ the restriction of $X$ on $U$.
Then $X^U\cong X/X_{(U)}$.


We now  prove a reduction lemma for Theorem~\ref{ET-reduction}.

\begin{lemma}\label{reduction}
Let $\Gamma=(V,E)$ be a connected $G$-edge-transitive graph of
square-free order and valency $k\ge 3$, where $G\le \Aut\Gamma$.
Assume that each minimal normal subgroup of $G$ has at most two
orbits on $V$. Then one of the following holds:
\begin{itemize}

\item[{\rm (1)}] $\Gamma\cong \K_{k,k}$,  $k$ is an odd prime,   $G\cong (\ZZ_k^2{:}\ZZ_l){.}\ZZ_2$ and $\Gamma$ is $G$-vertex-transitive, where
 $l$ is a divisor of $k-1$;
\item[{\rm (2)}]
$|V|=p$ with $p\ge3$ prime, $k$ is even,  $G\cong\ZZ_p{:}\ZZ_k$ and $\Gamma$ is $G$-vertex-transitive;

\item[{\rm (3)}] $|V|=2p$ with $p\ge3$ prime,   and   $G$ is isomorphic to one
of $\ZZ_p{:}\ZZ_k$ and $\ZZ_p{:}\ZZ_{2k}$;

\item[{\rm (4)}] $G$ is almost simple;

\item[{\rm (5)}] $\Gamma\cong\K_{k,k}$, $\Gamma$ is $G$-vertex-transitive, $\soc(G)$ is the unique minimal normal subgroup of $G$, $\soc(G)\cong T^2$ for a nonabelian simple group $T$  and, for $\alpha\in V$, either
\begin{itemize}
\item[{\rm (i)}]
$\soc(G)_\alpha\cong H\times T$ for a subgroup $H$ of $T$ with $k=|T:H|$; or

\item[{\rm (ii)}] $k=105$, $T\cong \A_7$ and  $\soc(G)_\alpha\cong \A_6\times\PSL(3,2)$.
\end{itemize}
\end{itemize}
\end{lemma}
\begin{proof}
Let $N$ be a minimal normal subgroup of $G$. Then $N$ is a
directed product of isomorphic simple groups. Since $\Gamma$ has valency $k\ge 3$,
we know that $|V|>3$.
Since $|V|$ is square-free and $N$ has at most
two orbits on $V$, we conclude that $N$ is not an elementary abelian
$2$-group. In particular, $N$ has no a subgroup of index $2$.

\vskip 5pt

{\bf Case 1}. Assume first that $G$ has two distinct minimal normal subgroups $N$ and $M$.
Then $N\cap M=1$, and hence $NM=N\times M$.

Suppose that both $N$ and $M$ are transitive on $V$.
By Lemma~\ref{centralizer}, $N$ and $M$ are regular on $V$; in particular, $|N|=|M|=|V|$.
Thus $N$ and $M$ are soluble, it implies that $N\cong M\cong \ZZ_p$ for an odd prime $p$.
Again by Lemma~\ref{centralizer}, $N=M$, a contradiction.

Without loss of generality, we assume that $N$ is intransitive on $V$. Then $\Gamma$ is a bipartite graph,
whose bipartition subsets are $N$-orbits, say $U$ and $V\setminus U$.
A similar argument as
above paragraph yields that $M$ has no  subgroups of index $2$. It
follows that $M$ fixes both $U$ and $V\setminus U$ set-wise, and hence $U$ and $V\setminus U$ are two $M$-orbits on $V$.

Let $X=NM$ and $\Delta=U$ or $V\setminus U$.
By Lemma~\ref{centralizer}, both $N^\Delta$ and $M^\Delta$ are regular subgroups of $X^\Delta$.
Set $N\cong T^i$, where $T$  is a simple group. Then $N_{(\Delta)}\cong T^j$ for some $j<i$, and so $N^\Delta\cong N/N_{(\Delta)}\cong T^{i-j}$.
It follows that $|\Delta|=|N^\Delta|=|T|^{i-j}$. Since $T$ is simple and $|\Delta|$ is square-free, $i-j=1$ and $N^\Delta\cong T\cong \ZZ_p$, where $p=|\Delta|$
is an odd prime. Similarly, $M^\Delta\cong  \ZZ_p$, and so $M$ is abelian.
In particular, $X=N\times M$ is abelian and $|X|$ is a power of $p$.
It implies that $X^\Delta\cong \ZZ_p$. Then, by Lemma~\ref{centralizer}, $N^\Delta=M^\Delta=X^\Delta$.
Thus $N\times M=X\le  X^\Delta\times X^{V\setminus \Delta}\cong \ZZ_p^2$.
Then $X\cong \ZZ_p^2$, and hence $N\cong M\cong \ZZ_p$. Moreover, $X_{(\Delta)}\cong \ZZ_p$.


Let $\alpha\in \Delta$. Then $G_\alpha\ge X_{(\Delta)}$. By Lemma~\ref{p<k}, $k=|\Gamma(\alpha)|\ge p$, and so $\Gamma\cong \K_{p,p}$.
Noting that $N$ is regular on $\Delta$ and $V\setminus \Delta$, by Lemma~\ref{bi-semireg}, $G_\alpha$ acts faithfully on $\Gamma(\alpha)$, and so
$G_\alpha$ is isomorphic to a subgroup of the symmetric group $\S_p$.
Noting that $G_\alpha$ has a normal subgroup $X_{(\Delta)}\cong \ZZ_p$,
it follows that
$G_\alpha$ is isomorphic to a subgroup of  the Frobenius group $\ZZ_p{:}\ZZ_{p-1}$. Write $G_\alpha\cong \ZZ_p{:}\ZZ_l$, where $l$ is a divisor of $p-1$.
Then $G_\Delta=NG_\alpha\cong \ZZ_p^2{:}\ZZ_l$.

Clearly,  $X_{(\Delta)}$ has at least $p+1$ orbits on $V$. Then, by the assumptions of this lemma,  $X_{(\Delta)}$ is not normal in $G$. On the other hand, $(X_{(\Delta)})^g=(X^g)_{(\Delta^g)}=X_{(\Delta)}$
for each $g\in G_\Delta$, yielding $X_{(\Delta)}\lhd G_\Delta$. It follows that $G\ne G_\Delta$, and hence $G$ is transitive on $V$.
Note that $|G:G_\Delta|\le 2$. Then part (1) of this lemma follows.


\vskip 5pt

{\bf Case 2}. Assume that $N:=\soc(G)$ is the unique minimal normal subgroup of $G$.

Assume  that $N$ is simple. If $N$ is nonabelian then (4) occurs. Assume that $N\cong \ZZ_p$ for some odd prime $p$.
Then $N$ is regular on each $N$-orbit on $V$. Thus $G_\alpha$ is faithful on $\Gamma(\alpha)$ by Lemma~\ref{bi-semireg}, where $\alpha\in V$.
Noting that $\C_G(N)$ is normal in $G$, we conclude   that $\C_G(N)=N$. Thus $G/N=\N_G(N)/\C_G(N)\lesssim \Aut(N)\cong \ZZ_{p-1}$,
and so $G\lesssim\AGL(1,p)$. Set $G\cong \ZZ_p{:}\ZZ_m$, where $m$ is a divisor of $p-1$.
Let $\alpha\in U$. Then    $G_\alpha\cong NG_\alpha/N\le G/N\cong \ZZ_m$; in particular,
$G_\alpha$ is cyclic. Recalling that $G_\alpha$ is faithful on $\Gamma(\alpha)$, it implies that $G_\alpha\cong \ZZ_k$.
Thus one of (2) and (3) occurs by noting that $|G:(NG_\alpha)|\le 2$.


In the following we assume that  $N\cong T^l$ for an integer $l\ge 2$ and a simple group $T$.
If $N$ is transitive on $V$ then $G$ is quasiprimitive on $V$, and hence $\soc(G)=N$ is simple  by Lemma~\ref{quasi-sqf}, a contradiction.
If $G$ is intransitive on $V$, then $G$ is faithful on each of its orbits, and
then $N$ is simple by Lemma~\ref{quasi-sqf}, again  a contradiction.
 Thus, in the following,
we assume further that  $\Gamma$ is $G$-vertex-transitive   and $N$ has two orbits $U$ and $W$ on $V$.
Note that $|U|=|W|=\frac{|V|}{2}$ is odd and square-free.


Since $\Gamma$ is $G$-vertex-transitive, $|G:G_U|=2$.
Let $x\in G\setminus G_U$.
Then $G=G_U\langle x\rangle$, $x^2\in G_U$, $U^x=W$ and $W^x=U$.
Let $\BB$ be a $G_U$-invariant partition of $U$ such
that $(G_U)^\BB$ is primitive. Set $\CC=\{B^x\mid B\in \BB\}$. Then
$(G_U)^\CC$ is also primitive. By \cite{L-S}, both $\soc((G_U)^\BB)$ and
$\soc((G_U)^\CC)$ are simple. Then $\soc((G_U)^\BB)\cong \soc((G_U)^\CC)\cong
T$. Let $K$ be the kernel of $G_U$ acting on $\BB$. Then $K^x$ is the
kernel of $G_U$ acting on $\CC$, and $K^{x^2}=K$. Since $K,K^x\lhd G_U$,
we have $K\cap K^x\lhd G_U$. Noting that $(K\cap K^x)^x=K\cap K^x$, it
follows that $K\cap K^x\lhd G$. Since $K\cap K^x$ has at least
$2|\BB|>2$ orbits on $V$, we have $K\cap K^x=1$. Then $G_U\lesssim
G_U/K\times G_U/K^x\cong (G_U)^\BB\times (G_U)^\CC$, yielding $N\cong T^2$.


We claim that $T$ is a nonabelian simple group.
Suppose that $T\cong \ZZ_p$ for some (odd) prime $p$. Then $(G_U)^\BB\cong (G_U)^\CC\lesssim \ZZ_p{:}\ZZ_{p-1}$,
and so $G=G_U{.}\ZZ_2\lesssim ((\ZZ_p{:}\ZZ_{p-1})\times (\ZZ_p{:}\ZZ_{p-1})){.}\ZZ_2$.
Let $H$ be a $p'$-Hall subgroup of $G$ with $x\in H$. Then $G=N{:}H$, $H\lesssim (\ZZ_{p-1}\times\ZZ_{p-1}){.}\ZZ_2$.
Moreover, $H_U$ is $p'$-Hall subgroup of $G_U$, $H=H_U\langle x\rangle$ and $H_U\lesssim \ZZ_{p-1}\times\ZZ_{p-1}$.
Note that $N$ is the unique minimal normal subgroup of $G$. Then  $H$ is maximal in $G$, and thus $G$ can be viewed as a primitive subgroup
of the affine group $\AGL(2,p)$. Since $H_U$ is an abelian normal subgroup of $H$, by \cite[2.5.10]{256},
$H_U$ is cyclic. It follows that $H_U\lesssim\ZZ_{p-1}$. Since $H_U$ has index $2$ in $H$, by \cite[2.5.7]{256},
$H_U$ is an irreducible subgroup of $\GL(2,p)$. Then, by \cite[2.3.2]{256}, $|H_U|$ is not a divisor of $p-1$, a contradiction.
Therefore, $T$ is a nonabelian simple group.


Set $N=T_1\times T_2$, where $T_1\cong T_2\cong T$. Since $T_1$ and $T_2$ are isomorphic nonabelian simple groups,
$T_1$ and $T_2$ are the only non-trivial normal subgroups of $N$. Thus
$N_{(U)}\in \{1, T_1,T_2\}$. For $g\in G_U$, we have $(N_{(U)})^g=(N^g)_{(U^g)}=N_{(U)}$. Thus
$N_{(U)}\lhd G_U$. Let $x\in G\setminus G_U$. Then $U^x=W$ and $W^x=U$, yielding $(N_{(U)})^x=N_{(W)}$
and $(N_{(W)})^x=N_{(U)}$. It follows that either $\{N_{(U)},N_{(W)}\}=\{T_1,T_2\}$ or $N$ is faithful on both $U$ and $W$.
The former case yields that $N_{(U)}$   acts transitively on  $W$, and so (i) of part~(5) follows.

Assume   that $N$  is faithful on both $U$ and $W$. Then neither  $T_1$ nor $T_2$
is transitive on $U$.
Let $\mathcal{O}$ be the set of $T_1$-orbits on $U$, and let
$O\in\mathcal{O}$. Then $T_2$ is transitive on $\mathcal{O}$.
Thus $T$ has two transitive permutation representations of degrees $|O|$ and $|\mathcal{O}|$, respectively.
Then $T$ has two  primitive permutation representations of  degrees $n_1$ and $n_2$, where
$n_1>1$ is a divisor of $|O|$ and $n_2>1$ is a divisor of $|\mathcal{O}|$. Since $|V|=2|U|=2|O||\mathcal{O}|$ is   square-free, $n_1$ and $n_2$ are odd, square-free and coprime.
Inspecting
\cite[Tables~1-4]{L-S}, we conclude that $T$ is either an
alternating group or a classical group of Lie type.



Suppose that $T=\PSL(d,q)$ with $d\ge3$. By the Atlas \cite{atlas}, neither
$\PSL(3,2)$ nor $\PSL(4,2)$ has maximal subgroups of coprime indices.
Thus  we assume that $(d,q)\not=(3,2)$ or $(4,2)$.
Then, by \cite[Table~3]{L-S}, \[\{n_1,n_2\}\subseteq\left\{\frac{\prod_{j=0}^{i-1}(q^{m-j}-1)}{\prod_{j=1}^i(q^j-1)}\mid 1\le i< d\right\}\cup\left\{\frac{\prod_{j=0}^{2i-1}(q^{m-j}-1)}{(\prod_{j=1}^i(q^j-1))^2}\mid 1\le i< \frac{d}{2}\right\}.\]
If $q^d-1$ has a primitive prime divisor $r$,
then both $n_1$ and $n_2$ are divisible by $r$, which is not possible. Thus  $q^d-1$
has no primitive prime divisor,  and so $(q,d)=(2,6)$ by Theorem~\ref{Zsigmondy}. Computation of $n_1$ and $n_2$
shows that this is not the case.

Similarly, we exclude other classes of classical groups of Lie type
except for $\PSL(2,p^f)$, where $p$ is a prime. By the Atlas \cite{atlas},
we exclude $\PSL(2,p^f)$ while  $p^f\le31$. Suppose that $T=\PSL(2,p^f)$ with $p^f\ge32$.
By \cite[Table~3]{L-S}, one of $n_1$ and $n_2$   is $p^f+1$ and the
other one is divisible by   $p$. This is
not possible since one of $p^f+1$ and $p$ is even.

Now let  $T=\A_c$ for some $c\ge 5$. By the above argument, we may assume that $\A_c$ is not
isomorphic to a classical simple group of Lie type. Then $c\ne 5$,
$6$ or $8$. Note that for $c\ge 5$ and $a<b<\frac{c}{2}$, the
binomial coefficient $(^c_b)=(^c_a)(^{c-a}_{b-a})/(^b_{b-a})$. It is
easily shown that $(^c_a)>(^b_{b-a})=(^b_a)$; in particular,
$(^c_a)$ is not a divisor of $(^b_{b-a})$. Thus $(^c_a)$ and
$(^c_b)$ are not comprime, and so at most one of $n_1$ and $n_2$
equals to a binomial coefficient. Checking the actions listed in
\cite[Table 1]{L-S} implies that either $c=7$, or $c=2a$ for $a\in
\{6,9,10,12,36\}$. Suppose the later case occurs. Then one of $n_1$ and $n_2$ is $\frac{1}{2}(^{2a}_a)$ and the other one is a
binomial coefficient, say $(^{2a}_b)$. But computation shows that
such two integers are not coprime, a contradiction.
Therefore, $T=\A_7$.

Checking the subgroups of $\A_7$, we conclude that
$\{n_1,n_2\}=\{|O|,|\mathcal{O}|\}=\{7,15\}$. Take $\alpha\in O$.
Recall that $\Gamma$ is $G$-vertex-transitive. Then there is an element
$x\in G\setminus G_U$ such that $\{\alpha,\alpha^x\}\in E$, $U^x=W$ and $W^x=U$.
Since $N=T_1\times T_2$ is the unique minimal normal subgroup of $G$, we know that
$T_1^x=T_2$ and $T_2^x=T_1$. It follows $O^x$ is a $T_2$-orbit on $W$, and so $\mathcal{O}^x:=\{O^{hx}\mid h\in G_U\}$
is the set of $T_2$-orbits on $W$. Moreover, $T_1$ acts transitively on $\mathcal{O}^x$.
Note that $|O|=|O^x|$ and $|\mathcal{O}|=|\mathcal{O}^x|$. Thus, without loss of generality, we may assume that $|O|=7$ and
$|\mathcal{O}|=15$. Then $(T_2)_O\cong \PSL(3,2)$ and $(T_1)_\alpha\cong \A_6$, where $\alpha\in O$.
Recall that $T_2$ is intransitive on $V$.  Since $T_2\lhd N$ and $N$ is transitive on $U$,
we conclude that each $T_2$-orbit on $U$ has size $15$. It follows that $(T_2)_O=(T_2)_\alpha$.
Then $N_\alpha\ge (T_1)_\alpha\times (T_2)_\alpha$, and so $N_\alpha=(T_1)_\alpha\times (T_2)_\alpha\cong \A_6\times\PSL(3,2)$ as
$|N:N_\alpha|=|U|=|O||\mathcal{O}|=105$.
Note that $N_{\alpha^x}=(N_\alpha)^x=((T_1)_\alpha\times (T_2)_\alpha)^x=(T_2)_{\alpha^x}\times (T_1)_{\alpha^x}$.
Then it is easily shown that $N_\alpha\cap N_{\alpha^x}=((T_1)_\alpha \cap (T_1)_{\alpha^x})\times ((T_2)_\alpha\cap (T_2)_{\alpha^x})\cong \S_4\times \S_4$.
By the choice of $x$, we conclude that  $|\Gamma(\alpha)|\geq |N_\alpha:(N_\alpha\cap N_{\alpha^x})|\geq 105$.
Thus $\Gamma=\K_{105,105}$, and hence (ii) of part (5) occurs.
\end{proof}
%\vskip 10pt


\section{Graphs associated with $\PSL(2,p^f)$ and $\Sz(2^f)$}

Let $\Gamma=(V,E)$ be a connected   graph of square-free order and valency $k$. Assume
that $G\le \Aut\Gamma$ is almost simple with socle $T$.
Assume further that $G$ is transitive on $E$ and that $T$
has at most two orbits on $V$. Let $\{\alpha,\beta\}\in E$. Then $|T_\alpha|=|T_\beta|$ as $\Gamma$ is a regular graph.   Then
$|T_\beta:T_{\alpha\beta}|=|T_\alpha:T_{\alpha\beta}|$ and,
by Lemma~\ref{normal-sub}, $|T_\alpha:T_{\alpha\beta}|$
is a divisor of $k=|\Gamma(\alpha)|$. Moreover, since $|V|$ is square-free, it is easily shown that $T_\alpha\ne T_\beta$.



\begin{lemma}\label{PSL(2,q)}
Let $\Gamma=(V,E)$ be a connected  $G$-edge-transitive graph of square-free order and valency $k$. Assume
that $\soc(G)=\PSL(2,p^f)$ with $f\ge 2$ and $p^f>9$, and that $\soc(G)$
has at most two orbits on $V$.  Then one of
the following statements  holds:
\begin{itemize}
\item[{\rm (i)}] $f=2$, $T_\alpha=\PGL(2,p)$ or $\PSL(2,p)$, and  $k$ is divisible by $p$  or $p+1$;

\item[{\rm (ii)}] $T_\alpha=\ZZ_p^{f-1}{:}\ZZ_l$ for  a divisor $l$ of $p-1$,
and $k$ is divisible by $p^{f-1}$;
further, if $\Gamma$ is $G$-locally primitive  then
$k=p^{f-1}$;

\item[{\rm (iii)}] $T_\alpha=\ZZ_p^f{:}\ZZ_l$ for  a divisor $l$ of $p^f-1$, and
$k$ is divisible by $p^f$;
further, if $\Gamma$ is $G$-locally primitive  then $k=p^f$.
\end{itemize}
\end{lemma}
\begin{proof}
Let $T=\soc(G)$. Take $\alpha\in V$ and a maximal subgroup $M$ of $T$ with $T_\alpha\le M$. Then both $|T:M|$ and $|M:T_\alpha|$ are square-free as $|T:T_\alpha|$ is square-free. By \cite{L-S}, either
$M=\ZZ_p^f{:}\ZZ_{\frac{p^f-1}{(2,p-1)}}$  and $|T:M|=p^f+1$, or
$f=2$, $M=\PGL(2,p)$ and $|T:M|=\frac{p(p^2+1)}{2}$.

Assume that $T_\alpha$ is insoluble. Then
$f=2$ and $T_\alpha=\PGL(2,p)$ or $\PSL(2,p)$. Let $\beta\in \Gamma(\alpha)$. Recall that $T_\alpha\ne T_\beta$ and  $|T_\beta:T_{\alpha\beta}|=|T_\alpha:T_{\alpha\beta}|$
is a divisor of $k$. If $T_\alpha=\PSL(2,p)$ then, by  \cite[II.8.27]{Huppert},  $|T_\alpha:T_{\alpha\beta}|$ is divisible by $p$ or   $p+1$.
Suppose that $T_\alpha=\PGL(2,p)$. Then $T_\alpha$ is maximal in $T$, and so $T=\langle T_\alpha,T_\beta\rangle$. Thus
$|T_\beta:T_{\alpha\beta}|>2$ as $T$ is simple; in particular, $\PSL(2,p)\ne T_{\alpha\beta}$.
Checking  the subgroups of $T_\alpha$ which do not contain $\PSL(2,p)$ (refer to \cite{COT}), we conclude that $|T_\alpha:T_{\alpha\beta}|$ is divisible by $p$ or   $p+1$.
Thus part (i) occurs.


In the following, we  assume that $T_\alpha$ is soluble. Since $p^2$ is not a divisor of $|T:T_\alpha|$, each Sylow $p$-subgroup of $T_\alpha$
has $p^f$ or $p^{f-1}$. Then, inspecting the subgroups of $T$, we conclude that $T_\alpha\cong T_\beta$ for $\beta\in \Gamma(\alpha)$, and that
$T_\alpha$ has a unique Sylow $p$-subgroup.

Let $Q$ be a Sylow $p$-subgroup of $T_{\alpha\beta}$.
Then $Q$ is normal
in  $T_{\alpha\beta}$. Suppose that $Q\ne 1$. Let $P_1$ and $P_2$
be the Sylow
$p$-subgroups of $T_\alpha$ and $T_\beta$, respectively. Then $P_1\cap P_2=Q\ne 1$. By \cite[II.8.5]{Huppert}, any two distinct Sylow $p$-subgroups of $T$ intersect trivially. It follows $P_1$ and $P_2$ are contained the same Sylow $p$-subgroup, say $P$ of $T$. In particular,
$P_1=P_\alpha$ and $P_2=P_\beta$. For $\gamma\in \Gamma(\beta)$, since $\Gamma$ is $G$-edge-transitive, we have
$|T_{\alpha\beta}|=|T_{\beta\gamma}|$.
A similar argument implies that $P_\gamma$ is
the Sylow $p$-subgroup of $T_\gamma$. It follows from the connectedness
of $\Gamma$ that $P_\delta$ is  the Sylow $p$-subgroup of $T_\delta$ for any
$\delta\in V$. Thus $P$ contains a normal subgroup $\langle P_\delta\mid \delta\in
V\rangle\ne 1$ of $G$, a contradiction. Thus, $T_{\alpha\beta}$ is of order
coprime to $p$, and so $|T_\alpha:T_{\alpha\beta}|$ is divisible by $|P_1|=p^{f-1}$ or $p^f$.
Thus, by Lemma~\ref{normal-sub},  $k$ is divisible by $p^{f-1}$ or $p^f$, respectively.

If $M=\PGL(2,p)$ then, inspecting the subgroups of $M$, we conclude that
$T_\alpha=\ZZ_p{:}\ZZ_l$, where $l$ is a divisor of $p-1$ and divisible by $4$.
Assume that $M=\ZZ_p^f{:}\ZZ_{\frac{p^f-1}{(2,p-1)}}$. Then $T_\alpha=\ZZ_p^f{:}\ZZ_l$  or $\ZZ_p^{f-1}{:}\ZZ_l$ with $l$ dividing $\frac{p^f-1}{(2,p-1)}$.
Suppose that $T_\alpha=\ZZ_p^{f-1}{:}\ZZ_l$.
Noting that $M$ is a Frobenius
group, $T_\alpha$ is also a Frobenius
group. It follows that  $l$ is a divisor of  $p^{f-1}-1$, and so $l$ divides $p-1$.




Assume further that   $\Gamma$ is $G$-locally primitive.
Then $T_\alpha^{\Gamma(\alpha)}$ is  a  normal transitive soluble subgroup of the primitive permutation group $G_\alpha^{\Gamma(\alpha)}$ of degree $k$.
Since $k$ is divisible by $|P_1|$, we have $\soc(G_\alpha^{\Gamma(\alpha)})\cong \ZZ_p^t$ for some integer $t\ge 1$ such that $k=p^t\ge |P_1|$.
It follows $T_\alpha^{\Gamma(\alpha)}\cong \ZZ_p^t{:}\ZZ_{l'}$, where $l'$ is a divisor of $l$. Since $P_1$ is the Sylow $p$-subgroup of
$T_\alpha$, we have $p^t\le |P_1|$. Then $k=|P_1|=p^{f-1}$ or $p^f$. Thus one of (ii) and (iii) follows.
 \end{proof}


The following lemma gives a characterization of graphs admitting
Suzuki groups.

\begin{lemma}\label{Sz}
Let $\Gamma=(V,E)$ be a connected  $G$-edge-transitive graph of square-free order and valency $k$. Assume
that $\soc(G)=\Sz(2^f)$ with odd $f\ge 3$, and that $\soc(G)$
has at most two orbits on $V$.
Then
$k$ is divisible by $2^{2f-1}$
and $\Gamma$ is not $G$-locally primitive.
\end{lemma}
\begin{proof}
Let $\alpha\in V$ and $\beta\in \Gamma(\alpha)$. Since $|T:T_\alpha|$ is square-free, 4 does not divide $|T:T_\alpha|$, and
hence $2^{2f-1}$ divides $|T_\alpha|$. Then, inspecting the subgroups of $T$ (see \cite{Suz}), we get
$T_\alpha=[2^n]{:}\ZZ_l$, where $n=2f$ or $2f-1$, and $l$ is a divisor of $2^f-1$. So $T_\alpha$ has a unique Sylow $2$-subgroup.
By \cite{Suz}, for a Sylow $2$-subgroup $Q$ of $T$, all involutions of $Q$ are contained in the center of $Q$.
Noting that any two distinct conjugations of $Q$ generate $T$, it follows any two distinct Sylow $2$-subgroups of $T$ intersect
trivially. Thus, by a similar argument as in the above lemma, we know that $T_{\alpha\beta}$ has odd order. Thus $k=|\Gamma(\alpha)|$ is
divisible by $n=2^{2f}$ or $2^{2f-1}$.





Finally, suppose that $G_\alpha^{\Gamma(\alpha)}$ is a primitive group. Let $Q_1$ be the Sylow
$2$-subgroup   of $T_\alpha$, and $Q$ be a  Sylow
$2$-subgroup   of   $T=\Sz(2^f)$ with $Q\ge Q_1$. Then $Q=Q_1$ or $Q_1{.}\ZZ_2$.
By a similar argument as in the above lemma, we conclude that  $Q_1$  is isomorphic to
$\soc(G_\alpha^{\Gamma(\alpha)})$. It follows that $Q_1$ is an elementary abelian $2$-group. By \cite{Suz}, $Q_1$ lies in the center of $Q$, and so
$Q$ is abelian,  which is
impossible. Then this lemma follows. 
\end{proof}






\section{Proof of Theorem~\ref{ET-reduction}}\label{proof}

Let $\Gamma=(V,E)$ be a connected graph of square-free order and
valency $k$. Assume that a subgroup $G\le \Aut\Gamma$ acts transitively
on $E$ and that each non-trivial normal subgroup of $G$ has at most $2$
orbits on $V$.
By Lemma~\ref{reduction},
to complete the proof of the theorem, we may assume that $G$ is
almost simple. Let $T=\soc(G)$ and $\alpha\in V$. Then $T$ is transitive
or has exactly two orbits on $V$, and every prime divisor of $|T_\alpha|$
is at most $k$.

Let $U$ be a $T$-orbit, and let $\BB$ be a $T$-invariant partition on $U$ such that $|\BB|\ge 2$ and
$T^\BB$ is primitive.
Noting that $|\BB|$ is square-free, $T$ is listed in
\cite[Tables 1-4]{L-S}. In particular, if $T$ is one of sporadic simple groups then
part~(3) of
Theorem~\ref{ET-reduction} follows.

Assume that $T=\A_n$, where $n\ge5$. Suppose that $n\ge 3k$. By
\cite{L-S}, there exists a prime $p$ such that $k<p<3k/2$, and thus
$p^2$ divides $|T|$, and $p$ divides $|T_\alpha|$. So $p\le k$, which is
a contradiction. Therefore, $n<3k$, as in   part~(4) of
Theorem~\ref{ET-reduction}.

We next deal with the classical groups and the exceptional groups of Lie type.
If $T=\PSL(2,p^f)$ or $\Sz(2^f)$ then, by Lemmas~\ref{PSL(2,q)} and \ref{Sz}, one of parts
(5), (6) and (7) of Theorem~\ref{ET-reduction} follows. Thus the following two lemmas
will fulfill the proof of Theorem~\ref{ET-reduction}.


%\vskip 5pt


\begin{lemma}\label{classical gps}
Let $T$ be a $d$-dimensional classical simple group of Lie type over
$\GF(p^f)$, where  $p$ is a prime. Then either $T=\PSL(2,p)$, or $p\le k$ and one of the following holds:
\begin{itemize}

\item[{\rm (i)}] $T=\PSL(2,p^f)$ with $f\ge2$;

\item[{\rm (ii)}] $[\frac{d}{2}]f<k$; if further $T=\PSU(d,p^f)$ then $2[\frac{d}{2}]f< k$ and $[\frac{d}{2}]$ is odd.

\end{itemize}
\end{lemma}
\begin{proof}
Let $\alpha\in V$. Then $|T:T_\alpha|$ is square-free and, by
Lemma~\ref{p<k}, each prime divisor of $|T_\alpha|$ is at most $k$.
Assume that $T\not=\PSL(2,p)$. Let $P$ be a Sylow $p$-subgroup of
$T$. Then $p^2$ divides $|P|$. Since $|T:T_\alpha|$ is square-free, $p$
divides $|T_\alpha|$, and so $p\le k$.

Assume  that $d\ge3$. Let $d_0=[\frac{d}{2}]$, the largest integer no
more than $\frac{d}{2}$. Check the orders of classical simple groups of Lie type, see
\cite[Section 47]{Aschbacher} for example. We conclude that  either
\begin{itemize}
\item[(1)] $(p^{2d_0f}-1)(p^{d_0f}-1)$ divides $(d,p^f-1)|T|$; or

\item[(2)] $T=\PSU(d,p^f)$ with $d_0$ odd, and $(p^{2d_0f}-1)(p^{d_0f}+1)$ divides $(d,p^f+1)|T|$.
\end{itemize}

Consider part~(1) first. Suppose that $p^{d_0f}-1$ has a primitive
prime divisor $r$. Then $r>d_0f$, and hence either $r=d=3$ and
$f=1$, or $r^2$ divides $|T|$. For the former, $T=\PSL(3,p)$, and so
$[\frac{d}{2}]f=1<k$. For the latter, $r$ divides $|T_\alpha|$, and so
$d_0f<r\le k$. Suppose that $p^{d_0f}-1$ has no primitive prime
divisor. By Theorem~\ref{Zsigmondy}, either $d_0f=2$ and $p+1$ is a
power of $2$, or $(p,d_0f)=(2,6)$. For the former,
$[\frac{d}{2}]f=d_0f=2<k$. Assume that $(p,d_0f)=(2,6)$. Then $(d_0,f)=(1,6)$, $(2,3)$, $(3,2)$, or $(6,1)$. It follows that
$(d,f)=(3,6)$, $(4,3)$, $(5,3)$, $(6,2)$, $(7,2)$, $(12,1)$ or
$(13,1)$. Thus  $|T|$ is  divisible by $7^2$, and so $|T_\alpha|$ is divisible by $7$.
Then $[\frac{d}{2}]f=d_0f=6<7\le k$ by Lemma~\ref{p<k}.

Now assume that $T=\PSU(d,p^f)$ with $d_0=[\frac{d}{2}]$ odd. Then
$(p^{2d_0f}-1)(p^{d_0f}+1)$ divides $(d,p^f+1)|T|$. A similar
argument shows that either $p^{2d_0f}-1$ has no primitive prime
divisor, or $2d_0f<k$. Assume that $p^{2d_0f}-1$ has no primitive
prime divisor. Then either $2d_0f=2$, or $(p,{2d_0f})=(2,6)$. For
the former, $2d_0f=2<k$. Suppose that  $(p,{2d_0f})=(2,6)$. Then $d_0f=3$, and so $(d,p^f)=(3,2^3)$, $(6,2)$ or $(7,2)$. Thus
$|T|$ is  divisible by $7^2$,
and so $2d_0f=6<7\leq k$. 
\end{proof}




Finally we consider the exceptional simple groups of Lie type.

\begin{lemma}\label{exceptional gps}
Let $T$ be an exceptional simple group of Lie type  defined over
$\GF(p^f)$ with $p$ prime. Then $p\le k$, and one of the following
holds:
\begin{itemize}
\item[{\rm (i)}] $T=\Sz(2^f)$;

\item[{\rm (ii)}] $T=\G_2(p^f)$ or ${}^3\D_4(p^f)$,   $p^f\ne 2^3$ and $2f<k$;

\item[{\rm (iii)}] $T=\F_4(p^f)$, ${}^2\E_6(p^f)$ or $\E_7(p^f)$, $p^f\ne 2$ and $6f<k$.
\end{itemize}
\end{lemma}
\begin{proof}
Note that $T$ has order divisible by $p^2$. Then $p$ divides
$|T_\alpha|$, and so $p\le k$. By \cite[Table 4]{L-S}, $T$ is one of
$\Sz(2^f)$, $\G_2(p^f)$, ${}^3\D_4(p^f)$, $\F_4(p^f)$,
${}^2\E_6(p^f)$ and $\E_7(p^f)$.

For $T=\G_2(p^f)$ or ${}^3\D_4(p^f)$, the order $|T|$ is divisible
by $(p^f+1)^2$ and $|T:T_\alpha|$ is divisible by $p^f+1$. If $p^{2f}-1$
has a primitive prime divisor $r$, then $|T|$ is divisible by $r^2$, and
$|T_\alpha|$ is divisible by $r$,  hence $2f<r\le m$. Assume that $p^{2f}-1$ has
no primitive prime divisor. Then either $f=1$ and $2f=2<k$, or
$(p,2f)=(2,6)$. For the latter, $T=\G_2(8)$ or  ${}^3\D_4(8)$, and
so $9$ is a divisor  of $|T:T_\alpha|$, which contradicts that
$|T:T_\alpha|$ is square-free. Thus $T$ is described as in part~(ii) of this lemma.

Assume that $T$ is one of $\F_4(p^f)$, ${}^2\E_6(p^f)$ and 
$\E_7(p^f)$. Then $|T|$ is divisible by $(p^{6f}-1)^2$ and
$|T:T_\alpha|$ is divisible by $\frac{p^{6f}-1}{p^f-1}$. If $p^{6f}-1$ has
a primitive divisor $r$ say, then $r$ divides $|T_\alpha|$, and hence $6f<r\leq k$. If $p^{6f}-1$ has no primitive prime divisor, then
$p=2$ and $f=1$, and so $|T:T_\alpha|$ is not square-free as it is divisible by $9$, and hence  $T$ is described as in part(iii) of this lemma. 
\end{proof}


%\vskip 10pt

\section{Graphs associated with $\PSL(2,p)$}

In this section, we investigate vertex- and edge-transitive graphs associated with $\PSL(2,p)$, and then give a
characterization for such graphs.

\subsection{Examples}
It is well-known that vertex- and edge-transitive graphs  can be described as coset graphs.
Let $G$ be a finite group and $H$ be a core-free  subgroup of $G$, where core-free means that $\cap_{g\in G}H^g=1$. Let
$[G:H]=\{Hx\mid x\in G\}$, the set of right cosets of $H$ in $G$.
For an element $g\in G\setminus H$, define the {\em coset graph}
$\Gamma:=\Cos(G,H,H\{g,g^{-1}\}H)$ on $[G:H]$ such that $(Hx,Hy)$ is an arc of $\Gamma$
if and only if $yx^{-1}\in H\{g,g^{-1}\}H$.
Then $\Gamma$ is a well-defined regular graph, and $G$ induces a subgroup of $\Aut\Gamma$ acting on
$[G:H]$ by right multiplication. The next lemma collects several basic  facts on
coset graphs.

\begin{lemma}\label{coset-pty}
Let $G$ be a finite group and $H$   a core-free subgroup of  $G$. Take $g\in G\setminus H$ and set
$\Gamma=\Cos(G,H,H\{g,g^{-1}\}H)$. Then $\Gamma$
is $G$-vertex-transitive and $G$-edge-transitive. Moreover,
\begin{itemize}
\item[{\rm (1)}] $\Gamma$ is $G$-arc-transitive if and only if $H\{g,g^{-1}\}H=HxH$ for some $2$-element
$x\in \N_G(H\cap H^g)\setminus H$ with  $x^2\in H\cap H^g$;

\item[{\rm (2)}] $\Gamma$ is connected if and only if $\langle H,g\rangle=G$.
\end{itemize}
\end{lemma}

Now we construct several examples.


\begin{example}\label{PSL(2,p)-vp}
{\rm Let $T=\PSL(2,p)$, $\ZZ_p{:}\ZZ_l\cong H<T$ and $\ZZ_l\cong K<H$, where $l$ is an even divisor of
$\frac{p-1}{2}$ with $\frac{p-1}{2l}$ odd. Then $\N_T(K)\cong \D_{p{-}1}$. Set $\N_T(K)=\langle a\rangle {:}\langle b\rangle$. It is easily shown that $\langle b, H\rangle=T$.
Then $\Cos(T,H, H bH)$ is a connected $T$-arc-transitive graph of valency $p$.}
\end{example}

\begin{example}\label{PSL(2,p)-vr}
{\rm Let $T=\PSL(2,p)$ and $H$ a dihedral subgroup of $T$.
\begin{itemize}
\item[{\rm (1)}] Let $\ZZ_2\cong K<H\cong \D_{2r}$ for an odd prime $r$ such that $|T:H|$ is square-free. Let $\epsilon=\pm1$ such that $4$ divides $p+\epsilon$. Then
$\N_T(K)=K{\times}\langle a\rangle{:}\langle b\rangle\cong \ZZ_2{\times}\D_{\frac{p+\epsilon}{2}}\cong \D_{p+\epsilon}$, where $b$ is an involution and, if $r$ divides $p+\epsilon$, we may choose $b$ such that $b$ centralizes $H$.
Then, for $1\le i< \frac{p+\epsilon}{2}$, the
coset graph $\Cos(T,H,Ha^ibH)$ is a connected $T$-arc-transitive graph of valency $r$.

\item[{\rm (2)}] Let $G=T$ or $\PGL(2,p)$ and $\ZZ_2^2\cong K<H\cong \D_{4r}$ for an odd prime $r$ with $|G:H|$ square-free.
Suppose that $G$ contains a subgroup isomorphic to $\S_4$. Then $\N_G(K)=K{:}\langle y,z\rangle\cong \S_4$, where $z$ is an involution
with $y^z=y^{-1}$. Then $\Cos(G,T_\alpha,T_\alpha yzT_\alpha)$ is a $G$-arc-transitive graph of valency $r$.
\end{itemize}}
\end{example}


\begin{example}\label{A4-graphs}
{\rm Let $\A_4\cong H<T=\PSL(2,p)<G=\PGL(2,p)$ with $|T:H|$ square-free and $\ZZ_3\cong K<H$. Let $\epsilon=\pm1$ with
$3$ dividing $p+\epsilon$. Then $\N_T(K)\cong \D_{p+\epsilon}$ and $\N_{G}(K)\cong \D_{2(p+\epsilon)}$. Moreover,
\begin{enumerate}
\item[{\rm (1)}] $\Cos(T, H, HxH)$ is a connected $(T,2)$-arc-transitive graph of valency $4$, where $x\in \N_T(K)\setminus \N_T(H)$ is an involution;

\item[{\rm (2)}] $\Cos(G, H, HxH)$ is a connected $(G,2)$-arc-transitive graph of valency $4$, where $x\in \N_G(K)\setminus (T\cup\N_G(H))$ is an involution.
\end{enumerate}}
\end{example}

\begin{example}\label{S4-graphs}
{\rm Let $\S_4\cong H<T=\PSL(2,p)$ with $|T:H|$ square-free.
\begin{enumerate}
\item[(1)]
Let $\D_8\cong K<H$, and $X=T$ or $\PGL(2,p)$ such that $|X:H|$ is square-free and $X$ has a Sylow $2$-subgroup isomorphic to $\D_{16}$. Then
$\D_{16}\cong \N_X(K)=K{:}\langle z\rangle$ for an involution $z\in X\setminus H$, and $\Cos(X, H, HzH)$ is a connected $(X,2)$-arc-transitive graph of valency $3$.

\item[(2)] Let $\S_3\cong K<H$ and $G=\PGL(2,p)$, and $\epsilon=\pm1$ with $3$ dividing $p+\epsilon$.
Then $\N_G(K)=\langle o\rangle{\times}K$ for an involution $o$. Set $X=\langle o, H\rangle$. Then $X=T$ or $\PGL(2,p)$ depending on whether or not $12$ divides $p+\epsilon$. Thus $\Cos(X, H, HoH)$ is a connected $(X,2)$-arc-transitive graph of valency $4$.
\end{enumerate}}
\end{example}

\begin{example}\label{A5-graphs}
{\rm Let $\A_5\cong H<T=\PSL(2,p)<G=\PGL(2,p)$ and $K<H$ with $K\cong \A_4$, $\D_{10}$ or $\S_3$.
Then $\N_G(K)=K{:}\langle z\rangle\cong \S_4$, $\D_{20}$ or $\D_{12}$, respectively, where $z\in G\setminus H$ is an involution. Set $X=\langle z, H\rangle$. Then $X=T$ or $\PGL(2,p)$, and $\Cos(X, H, HzH)$ is either a connected $(X,2)$-arc-transitive graph of valency $5$ or $6$, or a connected $X$-locally primitive
graph of valency $10$.
}
\end{example}

\subsection{A characterization}



Let $\Gamma=(V,E)$ be a connected $G$-edge-transitive graph of square-free order and valency $k\ge 3$, where $G\le \Aut\Gamma$. Assume
that  $T:=\soc(G)=\PSL(2,p)$ for a prime $p\ge 5$, and that $G$ acts transitively  on  $V$.

Let $\alpha\in V$.
Then $|T:T_\alpha|$ is square-free; in particular, $T_\alpha$ has
even order. Since $|G:T|\le 2$, either $T$ is transitive on $V$, or
$T$ has two orbits on $V$ of the same length $\frac{|V|}{2}$.
Thus $|V|=|T:T_\alpha|$ or $2|T:T_\alpha|$.


Note that the subgroups of $T$ are known, refer to
\cite[II.8.27]{Huppert}. We next analyze one by one the possible
candidates for $T_\alpha$.


\begin{lemma}\label{cyc-stab}
Assume that $T_\alpha$ is cyclic. Then $T_\alpha\cong\ZZ_m$ for an even divisor $m$
of $\frac{p\pm1}{2}$, $T$ is transitive on $V$,
$\Gamma$ is not $G$-locally-primitive, and one of the following holds:
\begin{itemize}
\item[{\rm (i)}] $\Gamma$ is $T$-edge-transitive, and $k=m$  or  $2m$;

\item[{\rm (ii)}] $G=\PGL(2,p)$, $G_\alpha\cong \ZZ_{2m}$ or $\D_{2m}$, and  $k=2m$ or $4m$.
\end{itemize}
\end{lemma}
\begin{proof}
Note $T_\alpha$ is a cyclic group of even order. By
Lemma \ref{abel-reg-1}, $T_\alpha$ is faithful and semiregular on
$\Gamma(\alpha)$. It is easy to check that no primitive group contains a
normal semiregular cyclic subgroup of even order.  Thus $\Gamma$ is not
$G$-locally-primitive. By \cite[II.8.5]{Huppert}, $T_\alpha$ is contained
in a subgroup conjugate to $\ZZ_{\frac{p\pm 1}{2}}$ in $T$. Thus
$T_\alpha\cong\ZZ_m$ for an even divisor $m$ of $\frac{p\pm1}{2}$.
Then $p(p\mp 1)$ is a divisor of
$|T:T_\alpha|$, and so $|T:T_\alpha|$ is even. It follows that
 $T$ is transitive on $V$.  Note that
$|G_\alpha|=m$ or $2m$. It follows that $\Gamma$ has valency $m$, $2m$ or
$4m$. Then (i) or (ii) is associated with the case that $T$ is transitive or intransitive on $E$,
respectively. 
\end{proof}



\begin{lemma}\label{val=p}
 Assume that $|T_\alpha|$ is divisible by $p$.
Then $T_\alpha\cong \ZZ_p{:}\ZZ_l$,  $T$ is transitive on $V$ and $\Gamma$ has valency
divisible by $p$, where $l$ is an even divisor of
$\frac{p-1}{2}$ with $\frac{p-1}{2l}$ odd. If $\Gamma$ is
$G$-locally primitive, then  $\Gamma$ is isomorphic to the graph in Example ~$\ref{PSL(2,p)-vp}$.
\end{lemma}
\begin{proof}
By \cite[II.8.27]{Huppert}, recalling that $T_\alpha$ has even order, $T_\alpha\cong \ZZ_p{:}\ZZ_l$ for an even divisor $l$ of
$\frac{p-1}{2}$. Since $|T:T_\alpha|=\frac{p^2{-1}}{2l}=(p{+}1)\frac{p{-1}}{2l}$ is even and square-free, $\frac{p-1}{2l}$ is odd and $T$ is transitive on $V$.
By Lemma \ref{abel-reg-1}, noting that $T_\alpha$ is a Frobenius group, $T_\alpha$ acts faithfully on $\Gamma(\alpha)$.
In particular, each $T_\alpha$-orbit on $\Gamma(\alpha)$ has size divisible by $p$.


Assume that $\Gamma$ is $G$-locally primitive. Then $T_\alpha$ is transitive on $\Gamma(\alpha)$ as $T_\alpha\lhd G_\alpha$. It implies that
$\Gamma$ has valency $p$ and $\Gamma$ is $T$-arc-transitive. Then $\Gamma\cong \Cos(T,T_\alpha,T_\alpha xT_\alpha)$ for some $x\in \N_T(T_{\alpha\beta})$ with
$x^2\in T_{\alpha\beta}$ and $\langle x, T_\alpha\rangle=T$, where $\beta\in \Gamma(\alpha)$.
Note that $\N_T(T_{\alpha\beta})\cong \D_{p{-}1}$. We write $\N_T(T_{\alpha\beta})=\langle a\rangle {:}\langle b\rangle$.
Let $M$ be a maximal subgroup of $T$ with $T_\alpha\le M\cong \ZZ_p{:}\ZZ_{\frac{p{-1}}{2}}$. Then $\ZZ_{\frac{p{-1}}{2}}\cong
\N_M(T_{\alpha\beta})\le \N_T(T_{\alpha\beta})$. Thus $a\in M$.
Write $\frac{p{-1}}{2}=ij$, where $i$ is odd and $j$ is a power of $2$. Then $\langle a\rangle=\langle a^i\rangle{\times}\langle a^j\rangle$. Since $T_{\alpha\beta}\cong \ZZ_l$ and $\frac{p-1}{2l}$ is odd, we have $a^i\in T_{\alpha\beta}\le T_\alpha$. Since $l$ is even, $j\ne 1$. It follows from $\langle x, T_\alpha\rangle=T$
that $x=a^{si}a^{tj}b$ for some $s$ and $t$. Then $T_\alpha xT_\alpha=T_\alpha a^{tj}bT_\alpha=(T_\alpha bT_\alpha)^{a^{-\frac{tj}{2}}}$. Noting that
$a^{-\frac{tj}{2}}$ normalizes $T_\alpha$, we have
$\Gamma\cong \Cos(T,T_\alpha,T_\alpha xT_\alpha)\cong \Cos(T,T_\alpha, T_\alpha bT_\alpha)$ as constructed in Example $\ref{PSL(2,p)-vp}$.
\end{proof}





\begin{lemma}\label{di-stab}
Assume that  $T_\alpha\cong \D_{2m}$ with $m>1$ coprime to $p$. Then $m$
is a divisor of $\frac{p\pm1}{2}$,  and $\Gamma$ has valency divisible by
$\frac{m}{2}$ or $m$. If $\Gamma$ is $G$-locally-primitive, then $\Gamma$
has odd prime valency $r$,
$T_\alpha\cong\D_{2r}$ or $\D_{4r}$, and
$\Gamma$ is isomorphic to
one of the graphs given in Example~$\ref{PSL(2,p)-vr}$.
\end{lemma}
\begin{proof}
The first part follows from that $|T_\alpha|$ is a divisor of
$|T|=\frac{p(p^2-1)}{2}$.

Let  $\{\alpha,\beta\}$ be an edge of $\Gamma$. Suppose that $T_{\alpha\beta}$ contains a
cyclic subgroup $C$ of order no less than $3$. Then $C$ is the unique
subgroup of order $|C|$ in both $T_\alpha$ and $T_\beta$. For an arbitrary
edge $\{\gamma,\delta\}$, since $\Gamma$ is $G$-edge-transitive,
$\{\gamma,\delta\}=\{\alpha,\beta\}^x$ for $x\in G$, so
$T_{\gamma\delta}=T_{\alpha^x\beta^x}=T\cap G_{\alpha^x\beta^x}=T\cap
(G_{\alpha\beta})^x=(T_{\alpha\beta})^x$. Then $C^x$ is the unique subgroup of
order $|C|$ in both $T_\gamma$ and $T_\delta$. So $C\le T_\gamma$ for $\gamma\in
\Gamma(\alpha)\cup \Gamma(\beta)$. Since $\Gamma$ is connected, $C$ fixes each
vertex of $\Gamma$, and so $C=1$ as $C\le \Aut\Gamma$, a contradiction.
Thus $|T_{\alpha\beta}|$ is a divisor of $4$, and hence $\Gamma$ has
valency divisible by $\frac{m}{2}$ or $m$.


Assume that $\Gamma$ is $G$-locally primitive. Then
$T_\alpha^{\Gamma(\alpha)}$ contains a  transitive normal cyclic subgroup. Thus $|\Gamma(\alpha)|=r$ is an odd prime,
and $T_\alpha^{\Gamma(\alpha)}\cong T_\alpha/T_\alpha^{[1]}\cong (T_\alpha
G_\alpha^{[1]})/G_\alpha^{[1]}\cong \D_{2r}$.
Note that $T_\alpha^{[1]}$ is a normal cyclic subgroup of $T_\alpha$.
By the argument in above paragraph,
$|T_\alpha^{[1]}|\le 2$.
It follows that $T_\alpha\cong \D_{2r}$ or $\D_{4r}$.

Let $T_\alpha\cong \D_{2r}$. Then $|T:T_\alpha|$ is even, so $T$ is transitive on $V$, and hence
$\Gamma$ is $T$-arc-transitive. Then $\Gamma\cong \Cos(T,T_\alpha,T_\alpha xT_\alpha)$ for some $x\in \N_T(T_{\alpha\beta})$ with
$x^2\in T_{\alpha\beta}$ and $\langle x, T_\alpha\rangle=T$. Let $\epsilon=\pm1$ such that $4$ divides $p+\epsilon$. Then
$\N_T(T_{\alpha\beta})=T_{\alpha\beta}{\times}\langle a\rangle{:}\langle b\rangle\cong \ZZ_2{\times}\D_{\frac{p+\epsilon}{2}}\cong \D_{p+\epsilon}$. It implies that
$x$ is an involution.
If $r$ does not  divides $p+\epsilon$, then $x=a^ib$ for some $1\le i\le \frac{p+\epsilon}{2}$. Assume that $r$ is a divisor of $p+\epsilon$. Then $T_\alpha$ is contained
in a maximal subgroup $M\cong \D_{p+\epsilon}$ of $T$, and  $\N_M(T_{\alpha\beta})\cong \ZZ_2^2$ contains the center of $M$. Without loss of
generality, we choose $b$ in the center of $M$, and so $x=a^ib$ for $1\le i< \frac{p+\epsilon}{2}$. Thus
$\Gamma$ is isomorphic to a graph given in Example~\ref{PSL(2,p)-vr}~(1).

Now let $T_\alpha\cong \D_{4r}$. Then $T_{\alpha\beta}\cong \ZZ_2^2$. If $T$ is not transitive on $V\Gamma$, then
$G=\PGL(2,p)$,  $\Gamma$ is a bipartite graph, and $T_\alpha=G_\alpha$. Thus we set $X=\PSL(2,p)$ or $\PGL(2,p)$ depending respectively on whether or not $T$ is
is transitive on $V\Gamma$.
Then $\Gamma\cong \Cos(X,T_\alpha,T_\alpha xT_\alpha)$ for some $x\in \N_X(T_{\alpha\beta})\setminus T_{\alpha\beta}$ with
$x^2\in T_{\alpha\beta}$; in particular, $\N_X(T_{\alpha\beta})/T_{\alpha\beta}$ is of even order
It implies that $\N_T(T_{\alpha\beta})\cong \S_4$.
Let $M$ be the maximal subgroup of $X$ with $T_\alpha\le M$. Then $8$ divides $|M|$, and $\N_M(T_{\alpha\beta})\cong \D_8$. Take
$\D_{8r}\cong M_1\ge T_\alpha$. Then $\N_M(T_{\alpha\beta})=\N_{M_1}(T_{\alpha\beta})$. We write $\N_X(T_{\alpha\beta})=T_{\alpha\beta}{:}(\langle y\rangle{:}\langle z\rangle)$, where $z\in \N_M(T_{\alpha\beta})$ and $\langle y\rangle{:}\langle z\rangle\cong \S_3$. Noting that $x\not\in \N_M(T_{\alpha\beta})$ and $x$ is of even order, we have $x=x_1y^iz$ for some $x_1\in T_{\alpha\beta}$ and $i=1$ or $2$. Noting that $z$ normalizes $T_\alpha$ and $y^z=y^{-1}$, we have $\Cos(X,T_\alpha,T_\alpha xT_\alpha)=\Cos(X,T_\alpha,T_\alpha y^izT_\alpha)\cong \Cos(X,T_\alpha,T_\alpha yzT_\alpha)$. Thus $\Gamma$ is isomorphic to the graph given in Example~\ref{PSL(2,p)-vr}~(2).
\end{proof}


\begin{theorem}\label{PSL(2,p)-graphs}
Let $\Gamma=(V,E)$ be a connected $G$-edge-transitive graph of square-free order and
valency $k\ge 3$, where $G\le \Aut\Gamma$. Assume that $\soc(G)=\PSL(2,p)$ for a prime $p\ge 5$, and that $G$ is transitive on $V$.
Then, for $\alpha\in V$, the pair $(\soc(G)_\alpha,k)$ lies in
Table~$\ref{PSL(2,p)-graphs-1}$. Further, if $\Gamma$ is $G$-locally
primitive, then $(\soc(G)_\alpha,k)$ lies in
Table~$\ref{PSL(2,p)-graphs-2}$.
\end{theorem}
\begin{table}
\[\begin{array}{|c|c|c|} \hline
\soc(G)_\alpha & k & \mbox{remark} \\ \hline

\ZZ_m & m, 2m, 4m & \mbox{$m$ is an even divisor of $\frac{p\pm1}{2}$} \\

\ZZ_p{:}\ZZ_l & pm, 2pm, 4pm & \mbox{$\frac{(p-1)}{2l}$ is odd, $m\div l$} \\



\D_{2m} & \frac{m}{2}, m, 2m,4m & \mbox{$m$ divides $\frac{p\pm1}{2}$} \\

\A_4 & l,2l & l\in\{4,6, 12\}, 32\not\div p^2-1, T^E \mbox{ is transitive} \\

 & 2l, 4l &    p\equiv\pm 3 (\mod 8), G=\PGL(2,p) \\


\S_4 & l,2l & l\ge 3, l\div 24, p\equiv\pm1 (\mod 8), G_\alpha=T_\alpha\\

\A_5 & l,2l & l\ge 5,  l\div60, p\equiv\pm1 (\mod 10) , G_\alpha=T_\alpha\\
\hline

\end{array}\]
{\caption{}\label{PSL(2,p)-graphs-1}}
\end{table}
\begin{table}
\[\begin{array}{|c|c|c|c|} \hline
\soc(G)_\alpha & k & \Gamma & \mbox{remark} \\ \hline

\ZZ_p{:}\ZZ_l & p&    {\mbox Example~\ref{PSL(2,p)-vp}}   & \mbox{$(p-1)/2l$ is odd} \\

\D_{4r} & r &     {\mbox Example~\ref{PSL(2,p)-vr}~(1)}   &\mbox{prime } r\ne p,\ 32\not\div(p^2-1) \\

\D_{2r} & r &        {\mbox Example~\ref{PSL(2,p)-vr}~(2)}    &    \mbox{prime } r\ne p,\ 16\not\div(p^2-1) \\

\A_4 & 4 &       {\mbox Example~\ref{A4-graphs}}      &32\not\div (p^2-1) \\

\S_4 & 3,\ 4 &     {\mbox Example~\ref{S4-graphs}}        &p\equiv\pm1 (\mod 8)\\

\A_5 & 5,\ 6,\ 10 &        {\mbox Example~\ref{A5-graphs}}     &p\equiv\pm1 (\mod 10) \\ \hline

\end{array}\]
{\caption{}\label{PSL(2,p)-graphs-2} }
\end{table}

\begin{table}[ht]
\[\begin{array}{|c|c|c|c|c|} \hline
M&T_\alpha & k &\mbox{$T$-orbits} & \mbox{remark} \\ \hline

\ZZ_m& \ZZ_p{:}\ZZ_l & p & 1&\mbox{$m$ and $(p-1)/2ml$ are odd} \\
\hline

1&\D_{4r} & r & 1,2&\mbox{prime } r\ne p,\ 32\not\div(p^2-1) \\
\hline

1&\D_{2r} & r &1,2 &\mbox{prime } r\ne p,\ 16\not\div(p^2-1)
\\ \hline

1&\A_4 & 4 & 1,2&32\not\div (p^2-1) \\
\ZZ_3&\ZZ_2^2 & 4 &1,2 &32\not\div (p^2-1) \\
\ZZ_6,\S_3&\ZZ_2^2 & 4& 2 & 16\not\div (p^2-1) \\
\hline
1&\S_4 & 3,\ 4 &1,2 &p\equiv\pm1 (\mod 8)\\
\ZZ_2&\A_4 &  4 &1 &32\not\div (p^2-1)\\
\S_3&\ZZ_2^2 &  4 &1 &32\not\div (p^2-1)\\
\ZZ_2&\S_4 &  4 & 2&32\not\div (p^2-1)\\

\hline

1&\A_5 & 5,\ 6,\ 10 & 1&p\equiv\pm1 (\mod 10) \\

\ZZ_2&\A_5 & 6,\ 10 & 2&p\equiv\pm1 (\mod 10), 16\not\div (p^2-1)
\\ \hline

\end{array}\]
{\caption{}\label{tab=PSL(2,p)-graph}}
\end{table}
\begin{proof}
Let $\Gamma=(V,E)$ be a connected $G$-edge-transitive graph of square-free order and valency $k\ge 3$, where $G\le \Aut\Gamma$. Assume
that  $T:=\soc(G)=\PSL(2,p)$ for a prime $p\ge 5$, and that $G$ acts transitively  on  $V$.
Let $\{\alpha,\beta\}\in E$.

Noting that $|G:G_\alpha|=|T:T_\alpha|$ or $2|T:T_\alpha|$, we have
$|G_\alpha:T_\alpha|=1$ or $2$. Then $T_\alpha$ has at most two orbits on each
$G_\alpha$-orbits on $\Gamma(\alpha)$. By Lemma~\ref{normal-sub}, we have
$k=|\Gamma(\alpha)|=l$, $2l$ or $4l$, where $l=|T_\alpha:T_{\alpha\beta}|$. By Lemmas~\ref{cyc-stab}, \ref{val=p} and \ref{di-stab},
we need only consider the remaining case: $T_\alpha\cong\A_4$, $\S_4$ or $\A_5$.


Let $T_\alpha\cong\S_4$ or $\A_5$. Checking  the maximal subgroups of
$\PGL(2,p)$ (see \cite{COT}, for example), we know that $\PGL(2,p)$
has no   subgroups of order $2|T_\alpha|$. It follows that $G_\alpha=T_\alpha$.
Then $k=l$ or $2l$ depending whether or not
$T_\alpha^{\Gamma(\alpha)}$ is transitive. If $T_\alpha\cong\S_4$, then
$T_\alpha^{\Gamma(\alpha)}\cong\S_3$ or $\S_4$, which implies that $l\ge 3$ and $l$
divides $24$. If $T_\alpha\cong\A_5$, Then $l\ge 5$ is a divisor of $60$.


Let $T_\alpha\cong\A_4$. Assume that $T$ is transitive on $E$. Then
$k=l$ or $2l$, where $l=|T_\alpha:T_{\alpha\beta}|$ for $\alpha\in
\Gamma(\alpha)$. By Lemma~\ref{abel-reg-1}, $l\ne 3$. Thus $l\in
\{4,6,12\}$. Assume that $T$ is intransitive on $E$.  Then $G=\PGL(2,p)$
and $G_\alpha\cong\S_4$, and hence $p\equiv\pm 3 (\mod 8)$ by checking the
maximal subgroups of $G$.  By Lemma~\ref{abel-reg-1}, we conclude
that $T_\alpha^{\Gamma(\alpha)}\cong \A_4$ and $G_\alpha^{\Gamma(\alpha)}\cong \S_4$. It
follows that $k=2l$ or $4l$ for $l\in \{4,6,12\}$.

Further, if $\Gamma$ is $G$-locally primitive, then  $k=4$ for $T_\alpha\cong \A_4$, $k=3$ or $4$ for
$T_\alpha\cong\S_4$, and $k=5,6$ or $10$ for $T_\alpha\cong\A_5$. Next we determine the $G$-locally primitive
graphs.

Let $T_\alpha\cong \A_4$. Then $T_{\alpha\beta}\cong \ZZ_3$, and $\Gamma$ is $(G,2)$-arc-transitive and of valency $4$. Let $X=T$ or $\PGL(2,p)$ depending $T$ is transitive
or intransitive on $V$. Then $\N_X(T_{\alpha\beta})\cong \D_{t(p+\epsilon)}$, where $t=|X:T|$ and $\epsilon=\pm1$ such that
$3$ divides $p+\epsilon$. Let $x\in \N_X(T_{\alpha\beta})$ with $x^2\in T_{\alpha\beta}$ and $\langle x, T_\alpha\rangle=X$. Then $x$ is either an involution or of order $6$,
and $xy$ is an involution for some $y\in T_{\alpha\beta}$. Note that $T_\alpha xT_\alpha=T_\alpha xyT_\alpha$. Thus, writing $\Gamma$ as a coset graph,
$\Gamma$ is isomorphic to one of the graphs in Example~\ref{A4-graphs}.

Let $T_\alpha\cong \S_4$. Then $G_\alpha=T_\alpha$. If $\Gamma$ has valency $3$, then $\Gamma$ is isomorphic the graph given in Example~\ref{S4-graphs}~(1).
If $\Gamma$ has valency $4$, then $G_{\alpha\beta}\cong\S_3$ and $\N_G(G_{\alpha\beta})=\ZZ_2{\times}\S_3$, it follows that $\Gamma$ is isomorphic the graph given in Example~\ref{S4-graphs}~(2).

Finally, if $T_\alpha=\A_5$ then  $G_\alpha=T_\alpha$ and $G_{\alpha\beta}\cong \A_4$, $\D_{10}$ or $\S_3$, and thus
$\Gamma$ is isomorphic one of the graphs given in Example~\ref{A5-graphs}.
\end{proof}


%\vskip 10pt

\section{Locally primitive arc-transitive graphs}

In this section we give a proof of Theorem~\ref{th-locally-prim}.
We first prove a technical lemma.





\begin{lemma}\label{split}
Let $G$ be a transitive permutation group on $V$ of square-free
degree and let $M$ be a normal subgroup of $G$. Assume that $M$ is
semiregular on $V$ and $G/M$ acts faithfully on the $M$-orbits. Then
there is $X\le G$ such that $G=M{:}X$.
\end{lemma}
\begin{proof}
The result is trivial if $M=1$. Thus we assume that $M\ne 1$.
Note that $M$ has square-free order. Let $p$ be the largest prime
divisor of $|M|$ and $P$ be the Sylow $p$-subgroup of $M$. Then $P$
is cyclic and is normal in $G$.
Let $\alpha\in V$ and  $B$ be the $P$-orbit with $\alpha\in B$. Let $V_P$
be the set of $P$-orbits. Then $|B|=p$ is coprime to $|V_P|$.
Then $G_{B}=P{:}G_\alpha$ contains a Sylow $p$-subgroup $P\times Q$
of $G$, where $Q$ is a Sylow $p$-subgroup of $G_\alpha$. It follows from
\cite[10.4]{Aschbacher} that the extension
$G=P{.}(G/P)$ splits over $P$. Thus $G=P{:}H$ for some $H<G$ with $H\cap
P=1$. If $M=P$, then the result follows. We assume $M\ne P$ in the
following.

Let $K$ be the kernel of $G$ acting on $V_P$. Noting that each
$M$-orbit on $V$ consists of several $P$-orbits, we know that $K$
fixes each $M$-orbits set-wise. It follows from the assumptions that
$K\le M$. Then, considering the action of $M$ on its an orbit, we
conclude that $K=P$. Thus $H$ is faithful and transitive on $V_P$.
Further, $M=M\cap PH=P(M\cap H)$ implies that $M\cap H$ is
semiregular on $V_P$. It is easily shown that $H/(M\cap H)$ acts
faithfully on the $(M\cap H)$-orbits on $V_P$. Noting that
$|V_P|<|V|$, we may assume by induction that $H=(M\cap H)X$ with
$X\cap (M\cap H)=1$. Then $G=P((M\cap H)X)=MX$, and $M\cap X\le
M\cap H$ yielding $M\cap X\le M\cap H\cap X=1$, hence our result
follows. 
\end{proof}

Let $\Gamma=(V,E)$ be a connected $G$-locally primitive graph.
Suppose that $G$ has  a normal subgroup $N$ which has at least three orbits on $V$.
Then either the quotient graph $\Gamma_N$ is a star, or   $\Gamma$ is a normal cover of $\Gamma_N$, refer
to \cite[Theorem~1.1]{GLP}.
Then following lemma is easily shown.
\begin{lemma}\label{irregular-N}
Let $\Gamma=(V,E)$ be a connected $G$-locally primitive and $G$-symmetric
graph. Let $N$ be a normal subgroup of $G$. If $N$ is not
semiregular on $V$, then $N$ is transitive on $E$ and has at most
two orbits on $V$.
\end{lemma}


\begin{theorem}\label{loc-reduction}
Let $\Gamma=(V,E)$ be a connected $G$-locally primitive graph
of square-free order and valency $k>2$.
Let $M\lhd G$ be maximal subject to that $M$ has at least three
orbits on $V$.   Assume further that  $\Gamma_M$ is not a star. Then
one of the following holds.
\begin{itemize}

\item[{\rm (1)}]  $M=1$, $\Gamma$ and $G$ are described as in  (1) or (5) of Lemma~\ref{reduction};



\item[{\rm (2)}] $\Gamma$ is a bipartite graph, $G\cong\D_{2n}{:}\ZZ_k$, $\ZZ_n{:}\ZZ_k$ or
$\ZZ_{\frac{n}{k}}{:}\ZZ_k^2$,  and $k$ is the smallest  prime divisor of $nk$;



\item[{\rm (3)}]$G=M{:}X$, $M\soc(X)=M\times \soc(X)$ and $\soc(X)$ is  a simple group descried in (3)-(6) and (8) of Theorem~\ref{ET-reduction}.
\end{itemize}
\end{theorem}
\begin{proof}
Since $\Gamma_M$ is not a star, $\Gamma$ is a normal cover of
$\Gamma_M$, hence $M$ is semiregular on $V$; in particular, $|M|$ is
coprime to $|V_M|$.  By the choice of $M$, we know that $G/M$ is
faithful on either $V_M$ or one of two $G/M$-orbits on $V_M$. Then,
by Lemma~\ref{split}, we have $G=M{:}X$. Note that $\Gamma_M$ is
$G/M$-locally primitive, and the pair  $G/M$ and $\Gamma_M$ satisfies
the assumptions in Theorem~\ref{ET-reduction}.  Let $Y=\soc(X)$.
Then, by Lemma~\ref{reduction},  $\Gamma_M\cong \K_{k,k}$ and $Y\cong T^2$ for a simple
group $T$, or  $Y$ is a
minimal normal subgroup of $X$.

Since $|M|$ is square-free, $M$ has soluble automorphism group $\Aut(M)$. Noting that
$G/\C_G(M)=\N_G(M)/\C_G(M)\lesssim \Aut(M)$, it follows that $G/\C_G(M)$ is soluble.
If $Y$ is a nonabelian simple
group then    $Y\le \C_G(M)$, and
hence $MY=M\times Y$, and so part~(3) of this theorem occurs.
We next complete the proof in two cases.

\vskip 5pt

{\bf Case 1}. $\Gamma_M\cong \K_{k,k}$ and $Y\cong T^2$ for a simple
group $T$. In this case, by Lemma~\ref{reduction}, $X$ is transitive on $V_M$, and so
$\Gamma_M$ is $X$-arc-transitive. Then $\Gamma$ is $G$-arc-transitive.
Moreover, $Y$ has exactly two orbits on $V_M$ of size $k$. Thus $MY$ has
exactly two orbits $U$ and $W$ on $V$ of  length  $k|M|$.
Let $U_M$ and $W_M$ be the sets of $M$-orbits on $U$ and $W$, respectively.
Then $U_M$ and $W_M$ are $Y$-orbits on $V_M$.

Assume first that $T$ is a nonabelian simple group. Then part (5) of Lemma~\ref{reduction}
holds for the pair $(X, \Gamma_M)$. In particular, $Y$ is the unique minimal normal subgroup of $X$.
Let $\Delta$ be an $M$-orbit on $V$.
Suppose that  $T\cong \A_7$. Then  $k=105$ and $T_\Delta\cong \A_6\times \PSL(3,2)$.
It is easily shown that $\Gamma_M$ is not $X$-locally
primitive, which is not the case.  Thus   $Y$ is unfaithful on both $U_M$ and $W_M$. Let
$K$ be the kernel of $Y$ acting on $U_M$.  Then $K\cong T$ and, $Y=K\times K^x$ for $x\in X\setminus Y$.
It is easily shown that $K\cong T$ is transitive on $W_M$.
Recalling that $G/\C_G(M)$ is soluble,
it follows that $K\le \C_{MK}(M)$ and so $MK=M\times K$.
Considering the action of $MK$ on
$\Delta$, we conclude that $K$ acts trivially on $\Delta$. Then $K$ acts
trivially on $U$. Since $K$ is transitive on $W_M$, we conclude that
$\Gamma\cong \K_{k,k}$. It follows that $M=1$, and so (1) of this theorem occurs.


Now let $T\cong \ZZ_p$ for an odd prime $p$. Then  $k=p$ is
coprime to $|M|$, and so $|V|=2k|M|$.
Noting that $\Gamma_M$ has odd valency $k$, it implies that
$\Gamma_M$ has even order, and so $|M|$ is odd.
Moreover, by Lemma~\ref{reduction},
$X\cong G/M\cong (\ZZ_k^2{:}\ZZ_l){.}\ZZ_2$ is nonabelian, where
 $l$ is a divisor of $k-1$.
Since $|M|$ is square-free, $M$ is soluble, and so $G$ is soluble.
Let $F$ be the Fitting subgroup of $G$. Then $\C_G(F)\le F\ne 1$.
Suppose that $F$ has at least three orbits on $V$. Since $\Gamma$ is $G$-locally primitive and
$G$-vertex-transitive, $\Gamma$ is a normal cover of $\Gamma_F$; in particular,
$F$ has square-free order. Then $G/F$ is isomorphic to a subgroup of $\Aut\Gamma_F$  acting transitively on the arcs of $\Gamma_F$,
and so $G/F$ is not abelian.
On other hand, since $|F|$ is square-free, $F$ is cyclic, and hence $\C_G(F)=F$ and $\Aut(F)$ is abelian.
Since $G/F=\N_G(F)/\C_G(F)\lesssim \Aut(F)$, we know that $G/F$ is abelian, a contradiction.
Thus $F$ has one or two orbits on $V$.
Suppose that $|F|$ is even. Let $Q$ be the Sylow $2$-subgroup of $F$. Then $Q\lhd G$.
Consider the quotient $\Gamma_Q$. Since $|V|$ is square-free and $\Gamma$ is $G$-vertex-transitive, we get a graph of odd order $k|M|$ and odd valency $k$,
which is impossible. Then $F$ has odd order, and hence $F$ has exactly two orbits on $V$.


Assume $|F|$ is divisible by $k^2$. Let $P$ be the Sylow $k$-subgroup of $F$. Then $\ZZ_k^2\cong Y=\soc(X)=P\lhd G$.
By Lemma \ref{irregular-N}, we conclude that $\Gamma\cong \K_{k,k}$. This implies that $M=1$, and
$\Gamma$ and $G$ are described as in  {\rm (1)} of Lemma~{\rm \ref{reduction}}. Then (1) of this theorem occurs.

Assume that $|F|$ is not divisible by $k^2$.
Then $M\ne 1$; otherwise $\ZZ_k^2\cong Y\le F$, a contradiction.
Since   $F$ has exactly two orbits on $V$, we know that $|F|$ is divisible by $k|M|$.
Let $P$ be the Sylow $k$-subgroup of $F$. Then $\ZZ_k\cong P\lhd G$.
Let $q$ be the smallest prime divisor of $|M|$, and the let
$N$ be the $q'$-Hall subgroup of $M$. Then $NP$ is a normal subgroup of $G$.
It is easy to see that $NP$ is intransitive on both $U$ and $W$. Then the quotient graph $\Gamma_{NP}$ is bipartite and
of order $2q$ and valency $k$, and so $q>k$. Thus, since $l$ is a divisor of $k-1$, each possible
prime divisor of $l$ is less than $q$.
Note that $FM$ is a subgroup of $G$. Then $|G|=2lk^2|M|$ is divisible by $|FM|=\frac{|F||M|}{|F\cap M|}$.
Recalling that $|F|$ is divisible by $k|M|$, it follows that $M\le F$. Let $r$ be an arbitrary prime divisor of $|F|$, and let $R$ be the Sylow $r$-subgroup of $F$. Then $R\lhd G$
and $r$ is odd.
Since $G$ is transitive on $V$, all $R$-orbits on $V$ have the same length. It implies that $r$ is a divisor of $|V|$, and so
$r$ is a divisor of $k|M|$. The above argument yields that $|F|=k|M|$, and so
$|F|$ is square-free. Then $F$ is cyclic and semiregular on $V$, $\C_G(F)=F$ and $\Aut(F)$ is abelian.
Since $G_\alpha\cong G_\alpha F/F \le G/F=\N_G(F)/\C_G(F)\lesssim \Aut(F)$, we know that both $G_\alpha$ and $G/F$ are abelian.
By Lemma \ref {bi-semireg}, $G_\alpha\cong \ZZ_k$. Since $|G:(FG_\alpha)|=2$, we have $G=F{.}\ZZ_{2k}$.
Thus $G$ has  a normal regular subgroup $F{:}\ZZ_2$.
Then $\Gamma$ is isomorphic a Cayley
graph $\Cay(F{:}\ZZ_2,S)$, where $S=\{s^{\tau^i}\mid 0\le i\le k-1\}$ for an
involution $s\in F{:}\ZZ_2$ and   $\tau\in \Aut(F{:}\ZZ_2)$ of order $k$
such that $\langle S\rangle=F{:}\ZZ_2$. Noting that $|F{:}\ZZ_2|$ is square-free, it follows
that $F{:}\ZZ_2$ is a dihedral group, say $\D_{2n}$. Then part (2) of this theorem occurs.





\vskip 5pt

{\bf Case 2}. $\soc(G/M)\cong \soc(X)=Y\cong \ZZ_p$. Since $\Gamma_M$
is $X$-locally primitive, by Lemma~\ref{reduction},  either $X\cong \ZZ_p{:}\ZZ_k$, or $X\cong\ZZ_p{:}\ZZ_{2k}$
and $X$ is transitive on $V_M$.  Moreover, $|V_M|=2p$,  $(p,|M|)=1$,
$p>k$ and $k$ is an odd prime.
Let $L=MY$. Then $L$ is a semiregular normal
subgroup of $G$, and $L$ has exactly two orbits $U$ and $W$ on $V$.




Let $X\cong \ZZ_p{:}\ZZ_k$. Then $|G|=kp|M|=k|L|$.
Assume that $|L|$ has a prime divisor $q$ such that
either a Sylow $q$-subgroup of $L$ is not normal in $L$ or $q$ is
the smallest prime divisor of $|L|$. It is easily shown that $L$ has
a unique $q'$-Hall subgroup $N$; in particular, $N$ is normal in
$L$. Then $N$ is normal in $G$, and $N$ has $q$-orbits on each of
$U$ and $W$. Thus the quotient graph $\Gamma_N$ is bipartite and of  order $2q$ and  valency $k$.
In particular, $k\le q$.
Further, $G/N=\ZZ_q{:}\ZZ_k$ is not abelain unless $q=k$.
Since $|N|$ is square-free,
the outer automorphism group $\Out(N)$ of $N$ is
abelian, refer to \cite{Edge-Cay-val-4-sqf}.
Note that $G/(N\C_G(N))$ is isomorphic a quotient of a
subgroup of  $\Out(N)$. Then $G/(N\C_G(N))$ is abelian.
Thus either $q=k$, or $N\C_G(N)$ has order divisible by $q$.
Suppose that $q>k$. Then $q$ is not a divisor of $|N|$ as $N\le L$ and $|L|$ is square-free.
Note that $N\C_G(N)/N\cong  \C_G(N)/(N\cap \C_G(N))$. It follows that
$|\C_G(N)|$ is divisible by $q$. Let $Q$ be a Sylow $q$-subgroup of $\C_G(N)$.
Then $Q$ is also a Sylow $q$-subgroup of $G$, and hence $Q\le L$.
Moreover, $NQ/N\lhd G/N$, and so $NQ\lhd G$. Since $NQ=N\times Q$,
we know that $Q\lhd G$, which contradicts the choice of $q$.
Therefore, $q=k$.
This says that $k$ is the smallest prime divisor of $|G|$, and
either $L\cong \ZZ_n$ or $L\cong \ZZ_{\frac{n}{k}}{:}\ZZ_k$, where $n=|L|$.
Thus $G=\ZZ_n{:}\ZZ_k$ or $\ZZ_{\frac{n}{k}}{:}\ZZ_k^2$, and $k$ is the
smallest prime divisor of $nk$.


Now let
$X\cong\ZZ_p{:}\ZZ_{2k}$. Then $G$ has a normal regular
subgroup $R=L{:}\ZZ_2$,  and $\Gamma$ is isomorphic a Cayley
graph $\Cay(R,S)$, where $S=\{s^{\tau^i}\mid 0\le i\le k-1\}$ for an
involution $s\in R$ and an automorphism $\tau\in \Aut(R)$ of order $k$
such that $\langle S\rangle=R$. Noting that $|R|$ is square-free, it follows
that $R$ is a dihedral group, say $\D_{2n}$. Then
$G=\D_{2n}{:}\ZZ_k$.
Let $q$ be the smallest prime divisor of $n$. Then $G$ has a normal subgroup $N$
with $|G:N|=2qk$. It is easily shown that the quotient graph $\Gamma_N$ is bipartite and
of valency $k$ and order $2q$. Then $k\le q$, and so $k$ is the
smallest prime divisor of $nk$.
Thus part (2)   follows.
\end{proof}

Now we are ready to give a proof of Theorem~\ref{th-locally-prim}.

\begin{proof}[Proof of Theorem~\ref{th-locally-prim}]
Let $\Gamma=(V,E)$ be a $G$-locally primitive arc-transitive graph, and
let $M\lhd G$ be maximal subject to that $M$ has at least three
orbits on $V$. Then $\Gamma$ is a normal cover of $\Sigma:=\Gamma_M$. Note that $\Gamma$ and $\Sigma$ has even valency if $|M|$ is
even.

If $G$ is soluble then, by
Theorem~\ref{loc-reduction}, one of part  (1) of Theorem \ref{th-locally-prim} occurs.
Thus we assume that $G$ is insoluble. Then
$G=M{:}X$, where  $T:=\soc(X)$ is  a simple group descried in {\rm (3)-(6)} and {\rm (8)} of Theorem~{\rm \ref{ET-reduction}}.
By Lemma~\ref{irregular-N}, we conclude that either $\Gamma$ is $T$-arc-transitive, or
$\Gamma$ is $T$-edge-transitive and $T$ has exactly two orbits on $V$.
We next consider the case where
$T=\PSL(2,p)$ for a prime $p\ge 5$.



Let $\Delta$ be an $M$-orbit on $V$. Then either $T_\Delta$ is
transitive on $\Delta$; or $T_\Delta$ has exactly two orbits on
$\Delta$ and, in this case, $T$ is intransitive on $V$ and $M\times
T$ is transitive on $V$. We take a normal subgroup $N$ of $G$ such
that $N=M$ if the first case occurs, or $N$ is the $2'$-Hall
subgroup of $M$ if the second case occurs. Let $\Delta_1$ be an
$N$-orbit contained in $\Delta$. Then $T_\Delta=T_{\Delta_1}$ is
transitive on $\Delta_1$ and $N$ is regular on $\Delta_1$.
Considering the action of $N\times T_\Delta$, we conclude that
$N\cong T_\Delta/K$, where $K$ is the kernel of $T_\Delta$ on
$\Delta_1$. Note that $T_\Delta$ is known  by
Theorem~\ref{PSL(2,p)-graphs}, and that
$|V|=|T:T_\alpha|$ or $2|T:T_\alpha|$ is square-free. Then we get
Table~\ref{tab=PSL(2,p)-graph}  by
checking possible normal subgroups of $T_\Delta$ with square-free
index. 
\end{proof}





%\vskip 20pt


%\vskip 30pt




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