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\title[Tetravalent non-normal Cayley graphs of order 4p]{Tetravalent non-normal Cayley graphs of order  $5p^2$  }
\date{}
\subjclass[2010]{20D60, 05B25}
\author{S. Khazaei and H.~Sharifi }

\begin{document}
\maketitle

\begin{center}
Department of Mathematics, Faculty of Science,

Shahed University, Tehran, Iran.
\footnote{Correspondence: hsharifi@shahed.ac.ir}
\end{center}



\begin{abstract}
In this paper, we determine all the connected  Cayley graphs on
non-abelian groups of order $5p^2(p$ prime greater than 5),
with cyclic Sylow $p$-subgroup
with respect to tetravalent sets including elements with different orders.
Moreover, we prove that such graphs are normal but not normal edge transitive, nor edge transitive, nor arc-transitive.

\\
\par

\textbf{ Key words:} Cayley graph, normal edge-transitive, arc-transitive, graph automorphism, group automorphism
\let\thefootnote\relax\footnotetext{ 2010 AMS Mathematics Subject Classification: 20D60, 05B25.}


\end{abstract}
\section{Introduction and Preliminary}


Suppose that  $G$ is a group and $S$ is a  subset  of $G$ which does not include $1$.
The Cayley graph associated with $(G, S)$, which is denoted by $Cay(G, S)$, is the directed graph with the vertex set $G$ and the edge set consisting of $(u,v)\in G\times G$ when $uv^{-1}\in S$.
We notice that the Cayley graph $Cay(G,S)$, may depend on the choice of $S$, and is connected if and only if $S$ generates $G$.
Also, we care that the edge set can be identified with set of ordered pairs $\{(g, sg)| g\in G, s\in S\}$.
The $Cay(G,S)$ can be considered as an undirected graph when $S$ is closed under taking inverse i.e., $S=S^{-1}$.
It is easy to see that the degree of each vertex is $|S|$.

Following \cite{pra}, the graph $\Gamma$ is called vertex-transitive or edge-transitive
if the automorphism group $\mathbb{A}ut(\Gamma)$, acts
transitively on vertex-set or edge-set of $\Gamma$, respectively.

Assume that $\Gamma=Cay(G,S)$.
Let $\rho(G)=\{\rho_{g}\mid g\in G \}$, in which,
for $g\in G$, the map $\rho_{g}: G\rightarrow G$ is given by
$\rho_{g}(x)=xg$.
Obviously, $\rho_{g}\in\mathbb{A}ut(\Gamma)$.
The set  $\rho(G)$
forms a subgroup (isomorphic to $G$) of $\mathbb{A}ut(\Gamma)$.
Therefore, since $\rho(G)\leqslant\mathbb{A}ut(\Gamma)$,
 acting right regularly on the vertices of $\Gamma$, $\Gamma$ is vertex-transitive. However $\Gamma$ is not edge-transitive in general.



Some notations used here are as follows.
The distance between two vertices $u$ and $v$, denoted by $d(u,v)$ in a graph $\Gamma$ is the number of edge in a shortest path connecting them.
Let us introduce the $D_{i}(u)=\{v\in V(\Gamma)\mid d(u,v)=i \}$.
For Semi-direct product of $K$ by~$H$ in which H act on K, we write $K \rtimes H$.
$\mathbb{Z}_{n}$ denotes a cyclic group of order $n$ as well as the ring of integers modulo $n$, by $\mathbb{Z}_n^{\times$} we mean the multiplicative group  consisting of the elements in $\mathbb{Z}_n$, which are coprime to $n$.
By  $\mathbb{A}ut(G, S)=\{\sigma\in\mathbb{A}ut(G) | \sigma(S)=S\}$.
It is easy to see that $\mathbb{A}ut(G, S)$
is a subgroup of the automorphisms group of $Cay(G, S)$. For two subsets $S$ and $T$ of $G$, if there is an  $f\in\mathbb{A}ut(G)$ such that $f(S)=T$, then $S$ and $T$ are said to be equivalent.  

We review  some fundamental facts of normal edge-transitive Cayley graphs.


A Cayley graph $\Gamma = Cay(G, S)$ is called normal if $\rho(G)$
is a normal subgroup of $\mathbb{A}ut(\Gamma)$, i.e,
$N_{\mathbb{A}ut(\Gamma)}(\rho(G))=\mathbb{A}ut({\Gamma})$; and $\Gamma$ is called normal edge-transitive or normal vertex-transitive if $N_{\mathbb{A}ut(\Gamma)}(\rho(G))$ is
transitive on the edges or vertices of $\Gamma$, respectively.

 \begin{lemma}(\cite[Lemma 2.1]{god1} or \cite{pra})\label{2.1}
For a Cayley graph $\Gamma=Cay(G, S)$, we have
$N_{\mathbb{A}ut(\Gamma)}(\rho(G))=\rho(G)\rtimes\mathbb{A}ut(G, S)$.
\end{lemma}

Therefore, $\Gamma$ is normal edge-transitive when
$\rho(G)\rtimes\mathbb{A}ut(G, S)$ is transitive on
the edge-set of $\Gamma$.


\begin{lemma}(\cite[Proposition 1(c)]{pra})\label{2.2}
Consider the Cayley graph $\Gamma=Cay(G, S)$.
Then the following are equivalent:

\begin{itemize}
\item[(i)] $\Gamma$ is normal edge-transitive;
\item[(ii)] $S = T\cup T^{-1}$, where $T$ is an $\mathbb{A}ut(G,S)$-orbit in $G$;
\end{itemize}
moreover, $\rho(G)\rtimes\mathbb{A}ut(G,S)$ is transitive on the arcs
of $\Gamma$ if and only if $\mathbb{A}ut(G,S)$ is transitive on $S$.
\end{lemma}

\begin{lemma}(\cite[Proposition 1.5]{xu})\label{2.2.1}
Let $A= \mathbb{A}ut(Cay(G,S))$. The following are equivalent:
\begin{itemize}

\item[(i)] $\rho(G)\unlhd A.$
\item[(ii)] $\mathbb{A}ut(\Gamma)=\rho(G)\rtimes\mathbb{A}ut(G,S).$
\item[(iii)]$A_1\leqslant\mathbb{A}ut(G,S)$, where $A_1$ is the stabilizer of the identity $1$
of $G$ in $A$.
\end{itemize}
\end{lemma}

In \cite{sha}, all the normal edge-transitive Cayley graphs of modular groups of order  $8n$, where $n$ is a natural number, are determind and in \cite{dar, li, pak} all the tetravalent edge-transitive Cayley graphs
on non-Abelian  groups of order $p^2$, $3p^2$, $4p^2$ are determined. In  \cite{yu} all connected cubic non-normal Cayley graphs of order $2p^2$ are studied.

In this paper, motivated by \cite{dar, kha, pak, zh},
we determine the structure of Cayley graphs on non-Abelian
groups of orders $5p^2$ with cyclic Sylow $p$-subgroup, 
with respect to tetravalent sets such that all the elements of sets don't have same orders, 
where $p$ is a prime number greater than 5. And, we show that in this case Cayley graphs are non-normal edge transitive, 
non-edge transitive and  non-arc-transitive.


Let $G$ be a finite group of order $5p^2$ with cyclic Sylow p-subgroups of order $p^2$, where $p$ is a prime number greater than 5.
It is not difficult to see, by the Sylow theorems, if the Sylow $p$-subgroup of $G$ is cyclic then $G$ is isomorphic to $\langle x,y | x^5=y^{p^2}=1, x^{-1}yx=y^k  \rangle  $
where $1<k<p^2$, $p\nmid k$, $xy{x^{ - 1}} = {y^{{k^4}}}$ , $k^5\equiv 1$ (mod$p^2$) and $o(x^iy^j)=5$, for $1\leq i \leq 4$, $0\leq j <p^2$.

\begin{lemma}(\cite[Lemmas 2.4 and 2.5]{kha})\label{2.3} Suppose $G$ is a finite group of order $5p^2$. The following then hold.
\begin{itemize}
\item[(1)]$\mid\mathbb{A}ut(G)\mid\leqslant p^3(p-1).$
\item[(2)]$\mathbb{A}ut(G)\cong\mathbb{Z}_{p^2}\rtimes\mathbb{Z}^{\times}_{p^2}$, where $\mathbb{Z}^{\times}_{p^2}$ denotes
the group of the units (invertible elements) of the ring $\mathbb{Z}_{p^2}$.
\end{itemize}
\end{lemma}


%Since the calculations are very large and we have page limitations,
We are interested in the Cayley graph $\Gamma=Cay(G, S)$ with
$|S|=4$, $S=S^{-1}$, $G=\langle S \rangle$ and  elements of $S$ are of order 5 and ${p^2}$.



Elements of $S$ are of the form $x^iy^j$, with $0\leq i \leq 4$ and $0\leq j <p^2$.
Since $G=\langle S \rangle$ and the inverses of $x^3y^j$ and  $x^4y^j$ are $x^2y^j'$ and $xy^j'$, respectively, we conclude that $S=S_i$ with $i\in\{1,2 \}$, where
if $x^iy^j\in S$ then $(x^iy^j)^{-1}\in S$.

$$S_1=\{xy^j, (xy^j)^{-1}, y^j', y^{-j'}  \},\ \  (j', p)=1$$



$$S_2=\{x^2y^j, (x^2y^j)^{-1}, y^j', y^{-j'} \}, \ \  (j', p)=1$$


\
Our main result  is the following.
\begin{main theorem}
Let $G$ be a finite group of order $5p^2$ with cyclic Sylow subgroups and let $S$ be a subset of $G$ satisfying the following conditions $(1)$ $|S|=4$, $(2)$ $S=S^{-1}$, $(3)$ $S$ contains elements of order 5 and elements of order $p^2$.
Then one of the following holds.
\begin{itemize}

\item[$(1)$]$S_1=\{x, y, x^{-1}, y^{-1} \}$, where $x$ and $y$ are of order $5$ and $p^2$, respectively.
$\Gamma_1=Cay(G, S_1)$ is normal but it is not normal edge transitive, nor edge transitive, nor arc-transitive, nor normal arc-transitive. $\mathbb{A}ut(G, S_1)\cong\mathbb{Z}_2$ and $\mathbb{A}ut(\Gamma_1)\cong \rho(G)\rtimes \mathbb{Z}_2$.

\item[$(2)$]$S_2=\{x^2, y, x^{-2}, y^{-1}\}$, where $x^2$ and $y$ are of order $5$ and $p^2$, respectively. $\Gamma_2=Cay(G, S_2)$ is normal but it is not normal edge transitive, nor edge transitive, nor arc-transitive, nor normal arc-transitive. $\mathbb{A}ut(G, S_2)\cong\mathbb{Z}_2$ and $\mathbb{A}ut(\Gamma_2)\cong \rho(G)\rtimes \mathbb{Z}_2$.
\end{itemize}
\end{main theorem}

\section{$\Gamma_1=Cay(G, S_1)$ }

In this section Cayley graph $Cay(G, S_1)$ is denoted by $\Gamma_1$.
\begin{lemma}\label{6.1}
$S_1$ is equivalent to $\{x, y, x^{-1}, y^{-1}\}$.
\end{lemma}
\begin{proof}
Let's remind that $S_1=\{xy^{j}, y^{j'}, (xy^{j})^{-1}, y^{-j'} \}$,
where $0\leqslant j < p$ and $(j', p)=1$.
It is sufficient to consider $f\in\mathbb{A}ut(G)$ such that $f(x)=xy^j$ and $f(y)=y^{j'}$.
Since $(j', p^2)=1$, there exists such an automorphism $f$.
\end{proof}

\begin{theorem}\label{6.2}
$\mathbb{A}ut(G, S_1)\cong\mathbb{Z}_2.$
\end{theorem}
\begin{proof}
Suppose that $f \in Aut(G,{S_1})$. By order of elements of $S_1$ and proof of Lemma \ref{2.3}(1), clearly $f(x)=x$ and for $f(y)$ we have two cases;
$f(y)=y$ or $f(y)=y^{-1}$.
In the first case, $f=id$ and in the second case $o(f)=2$.
\end{proof}

\begin{lemma}\label{6.3}
If $\varphi\in\mathbb{A}ut(\Gamma_1)_g$ and $\varphi(xg)=xg$, then $\varphi(x^{i}g)=x^{i}g$ for $1 \leqslant i \leqslant 4$.
\end{lemma}
\begin{proof}
 Because ${x^4}g \in {D_1}\left( g \right)$ and $\varphi $ preserves distance, so: $$\varphi \left( {{x^4}g} \right) \in \varphi \left( {{D_1}\left( g \right)} \right) = {D_1}\left( {\varphi \left( g \right)} \right) = {D_1}\left( g \right) = \left\{ {xg,{x^4}g,yg,{{\left {y} \right}^{ - 1}}g} \right\}$$$$ = \left\{ {\varphi \left( {xg} \right),{x^4}g,yg,{{\left {y} \right}^{ - 1}}g} \right\}.$$
Since $\varphi $ is one-to-one, so $\varphi \left( {{x^4}g} \right) \in \left\{ {{x^4}g,yg,{{\left {y} \right}^{ - 1}}g} \right\}$.
On the other hand $\left( {{x^2}g,{x^3}g} \right)$ is an edge and ${x^2}g \in {D_1}\left( {xg} \right)$, ${x^3}g \in {D_1}\left( {{x^4}g} \right)$, so there exists a member of $\varphi \left( {{D_1}\left( {xg} \right)} \right)$ that is adjacent with a member of $\varphi \left( {{D_1}\left( {{x^4}g} \right)} \right)$.
But if $\varphi \left( {{x^4}g} \right) = yg$ or ${\left {y} \right^{ - 1}}g$, these members don't exist.  Thus $\varphi \left( {{x^4}g} \right) = {x^4}g$.
${x^2}g$ and ${x^3}g$ are only vertices ${D_1}\left( {xg} \right)$ and  ${D_1}\left( {{x^4}g} \right)$ that are adjacent, so $\varphi $ will fix them.
\end{proof}

Similarly, if $\varphi(xg)=x^{-1}g$, then $\varphi(x^{i}g)=x^{-i}g$ for $1 \leqslant i \leqslant 4$ and if $\varphi(x^{-1}g)=xg$, then $\varphi(x^{-i}g)=x^{i}g$ for $1 \leqslant i \leqslant 4$.

\begin{lemma}\label{6.4}
If $\varphi\in\mathbb{A}ut(\Gamma_1)_g$, then $\varphi \left( {xg} \right)\notin\{yg, y^{-1}g\}$.
\end{lemma}
\begin{proof}
If $\varphi \left( {xg} \right) = yg$, then according to the above description, $\varphi \left( {{x^{ - 1}}g} \right) \ne xg$, thus $\varphi \left( {{x^{ - 1}}g} \right) = {y^{ - 1}}g$. $\left( {{x^2}g,{x^3}g} \right)$ is an edge and ${x^2}g \in {D_1}\left( {xg} \right)$, ${x^3}g \in {D_1}\left( {{x^4}g} \right)$, so there exists a member of $\varphi \left( {{D_1}\left( {xg} \right)} \right) = {D_1}\left( {yg} \right)$ that is adjacent with a member of $\varphi \left( {{D_1}\left( {{x^4}g} \right)} \right) = {D_1}\left( {{y^{ - 1}}g} \right)$, but these members do not exist.
Similarly $\varphi \left( {xg} \right) \ne {y^{ - 1}}g$.
\end{proof}

\begin{lemma}\label{6.5}
If $\varphi  \in {\Bbb A}ut{({\Gamma _1})_{{y^{k - 2}}g}}$ and $\varphi  \in {\Bbb A}ut{({\Gamma _1})_{{y^{k - 1}}g}}$, then $\varphi  \in {\Bbb A}ut{({\Gamma _1})_{{y^{k }}g}}$.
\end{lemma}
\begin{proof}
Since $\varphi $ is one-to-one, $\varphi \left( {{D_1}\left( {y^{k-1}g} \right)} \right) = {D_1}\left( {y^{k-1}g} \right)$ and the fact that
$${D_1}\left( {y^{k-1}g} \right) = \left\{ {{y^{k-2}g,x^{k-1}yg,{x^4}y^{k-1}g,{y^{k}}g} \right\} $$$$= \left\{ {\varphi \left({y^{k-2}g \right),x^{k-1}yg,{x^4}y^{k-1}g,{y^{k}}g} \right\},$$
we have
$$\varphi \left( {x^{k-1}yg} \right),\varphi \left( {{x^4}y^{k-1}g} \right),\varphi \left( {{y^k}g} \right) \in \left\{ {x^{k-1}yg,{x^4}y^{k-1}g,{y^k}g} \right\}.$$
We know, ${x^3}y^{k-1}g \in {D_1}\left( {{x^4}y^{k-1}g} \right)$, ${x^2}y^{k-1}g \in {D_1}\left( {x^{k-1}yg} \right)$ and $\left( {{x^3}y^{k-1}g,{x^2}y^{k-1}g} \right)$
is an edge, therefore we have
$$\varphi \left( {{x^3}y^{k-1}g} \right) \in \varphi \left( {{D_1}\left( {{x^4}y^{k-1}g} \right)} \right) = {D_1}(\varphi \left( {{x^4}y^{k-1}g} \right)),$$
$$\varphi \left( {{x^2}y^{k-1}g} \right) \in \varphi \left( {{D_1}\left( {xy^{k-1}g} \right)} \right) = {D_1}(\varphi \left( {xy^{k-1}g} \right))$$
and
$\left( {\varphi \left( {{x^3}y^{k-1}g} \right),\varphi \left( {{x^2}y^{k-1}g} \right)} \right)$
is an edge.
Since there is no element of
${D_1}\left( {{y^k}g} \right)$
adjacent to an element of ${D_1}\left( {xy^{k-1}g} \right)$
or an element of  ${D_1}\left( {{x^4}y^{k-1}g} \right).$
Therefore, we have
 $$\varphi \left( {xy^{k-1}g} \right),\varphi \left( {{x^4}y^{k-1}g} \right) \in \left\{ {xy^{k-1}g,{x^4}y^{k-1}g} \right\}.$$
Thus
$\varphi \left( {{y^k}g} \right) = {y^k}g$
and the lemma is proved.
\end{proof}

\begin{result}\label{6.6}
If $\varphi\in\mathbb{A}ut(\Gamma_1)_g$ and $\varphi(yg)=yg$, then $\varphi(y^{i}g)=y^{i}g$ for
$1 \leqslant i < {p^2}$.
\end{result}
\begin{proof}
It follows immediately from Lemma \ref{6.5}.
\end{proof}

\begin{lemma}\label{6.7}
If $\varphi  \in {\Bbb A}ut{({\Gamma _1})_{g}}$ and $\varphi(yg)=y^{-1}g$, then $\varphi(y^{2}g)=y^{-2}g$.
\end{lemma}
\begin{proof}
Since $\varphi(yg)=y^{-1}g$, $\varphi \left( {{D_1}\left( {yg} \right)} \right)= {D_1}(\varphi \left( {yg})) = {D_1}\left( {y^{-1}g} \right)$, so we have

$$ \left\{ {\varphi \left(g \right),\varphi \left({xy}g \right),\varphi \left({x^{4}yg \right),\varphi \left({y^{2}g \right)} \right\} $$$$= \left\{ {\varphi \left(g \right),xy^{-1}g,{x^4}y^{-1}g,{y^{-2}}g} \right\}.$$
On the other hand  ${x^2}yg \in {D_1}\left( {xyg} \right)$, ${x^3}yg \in {D_1}\left( {x^{4}yg} \right)$ and $\left( {{x^2}yg,{x^3}yg} \right)$
is an edge, so we have
$$\varphi \left( {{x^2}yg} \right) \in \varphi \left( {{D_1}\left( {xyg} \right)} \right) = {D_1}(\varphi \left( {xyg} \right)),$$
$$\varphi \left( {{x^3}yg} \right) \in \varphi \left( {{D_1}\left( {x^{4}yg} \right)} \right) = {D_1}(\varphi \left( {x^{4}yg} \right))$$
and
$\left( {\varphi \left( {{x^2}yg} \right),\varphi \left( {{x^3}yg} \right)} \right)$
is an edge.
Since there is no element of
${D_1}\left( {{y^{-2}}g} \right)$
adjacent to an element of ${D_1}\left( {xy^{-1}g} \right)$
or an element of  ${D_1}\left( {x^{4}y^{-1}g} \right).$
So, we have
 $$\varphi \left( {xyg} \right),\varphi \left( {{x^4}yg} \right) \in \left\{ {xy^{-1}g,{x^4}y^{-1}g} \right\}.$$
Thus $\varphi \left( {{y^2}g} \right) = {y^{-2}}g$.
\end{proof}

\begin{result}\label{6.8}
If $\varphi\in\mathbb{A}ut(\Gamma_1)_g$ and $\varphi(yg)=y^{-1}g$, then $\varphi(y^{i}g)=y^{-i}g$ for
$1 \leqslant i < {p^2}$.
\end{result}
\begin{proof}
Similar to proof of Lemma \ref{6.7}.
\end{proof}


\begin{lemma}\label{6.9}
If $\varphi\in\mathbb{A}ut(\Gamma_1)_g$, then $\varphi(xg)=xg$.
\end{lemma}
\begin{proof}
We know that $xg \in {D_1}\left( g \right)$, so we have
$$\varphi \left( {xg} \right) \in \varphi \left( {{D_1}\left( g \right)} \right) = {D_1}\left( {\varphi \left( g \right)} \right) = {D_1}\left( g \right) = \left\{ {xg,x^4g,yg,y^{  - 1} g} \right\}.$$
On the contrary  assume that $\varphi (xg) \ne xg$. Thus by Lemma \ref{6.4} we have
\\ $\varphi(xg)=x^{-1}g$, so we encounter with two cases for $\varphi(yg)$:


\begin{center}
\begin{tikzpicture}
	\begin{pgfonlayer}{nodelayer}
	\node [style=newstyle] (0) at (-2.5, 1) {$x^{2}g$};
		\node [style=newstyle] (1) at (-2.5, -0.75) {$x^{3}g$};
		\node [style=newstyle] (2) at (-0.75, 1) {$xg$};
		\node [style=newstyle] (3) at (-0.75, -0.75) {$x^{4}g$};
		\node [style=newstyle] (4) at (0.25, -0) {$g$};
		\node [style=newstyle] (5) at (1.5, 1) {$yg$};
		\node [style=newstyle] (6) at (1.75, -0.75) {$y^{-1}g$};
		\node [style=newstyle] (7) at (3.5, 1) {$y^{2}g...$};
		\node [style=newstyle] (8) at (3.75, -0.75) {$y^{-2}g...$};
				\end{pgfonlayer}
	\begin{pgfonlayer}{edgelayer}
		\draw (0) to (2);
		\draw (1) to (3);
		\draw (0) to (1);
		\draw (2) to (4);
		\draw (4) to (3);
		\draw (5) to (4);
		\draw (4) to (6);
		\draw (5) to (7);
		\draw (6) to (8);
	\end{pgfonlayer}
\end{tikzpicture}
\end{center}


\begin{itemize}

\item[Case 1.]
$\varphi(yg)=yg$;
since  $\left( {xg,x{y^k}g} \right)$ and $\left( {y^kg,x{y^k}g} \right)$ are two edges of graph and $\varphi$ preserves distance, so we have:
$$2=d(y^kg, xg)=d(\varphi(y^kg), \varphi(xg))=d(y^kg, x^{-1}g)> 2$$
which is a contradiction.

\item[Case 2.]
$\varphi(yg)=y^{-1}g$;
by Corollary \ref{6.8}, we have:
$$2=d(y^kg, xg)=d(\varphi(y^kg), \varphi(xg))=d(y^{-k}g, x^{-1}g)> 2$$
which is a contradiction.
Therefore this case does not happen.
\end{itemize}
\end{proof}


Using Lemma \ref{6.9}, if $\varphi\in \mathbb{A}ut(\Gamma_1)_1$, then one of the following two cases  occurs:
$\varphi(x)=x, \varphi(y)=y$ or $\varphi(x)=x, \varphi(y)=y^{-1}$.

\begin{lemma}\label{6.10}
If $\varphi\in A_g$, and $\varphi $ fixes all the elements of ${D_1}(g)$, then it fix all the elements of ${D_2}(g)$.
\end{lemma}
\begin{proof}
We know that
$$D_{2}(g)=\{x^2g, x^3g, xy^kg, xy^{-k}g, x^4yg, xyg, y^2g, y^{-2}g,$$

$$x^4y^{-1}g, xy^{-1}g, x^4y^{-k^4}g, x^4y^{k^4}g\}.$$

In fact by Lemma \ref{6.9} and Corollary \ref{6.6}, we conclude that $\varphi$ keeps fixed them.
\end{proof}

Immediately, the following result is obtained.

\begin{result}\label{6.11}
If $\varphi\in \mathbb{A}ut(\Gamma_1)_1$ and $\varphi$ fixes all the elements of $S_1$, then $\varphi= id$.
\end{result}
\begin{proof}
Since graph is connected, it suffices to show that for every integer $i \geqslant 2$, the statement
$$g' \in {D_i}\left( 1 \right) \Rightarrow \varphi \left( g' \right) = g'$$
holds.
By Lemma \ref{6.10} the statement is true for $i = 2$. Now assume that the statement is true for
$1 \leqslant i \leqslant n$,
and we will show that the statement holds for $n + 1$. Let $g' \in {D_{n + 1}}\left( 1 \right)$.
Hence, there is a sequence of adjacent vertices
$$1 = {{g'}_0},{{g'}_1},...,{{g'}_{n - 1}},{\kern 1pt} {{g'}_n},{{g'}_{n + 1}}.$$
Clearly ${g'_{n - 1}} \in {D_{n - 1}}(1)$ and ${g'_n} \in {D_n}(g)$.
Therefore by hypothesis, $\varphi \left( {{g'_{n - 1}}} \right) = {g'_{n - 1}}$ and $\varphi \left( {{g'_n}} \right) = {g'_n}$.
By applying Lemma \ref{6.10} for $g: = {g'_{n}}$ and $\varphi \left( {{g'_{n - 1}}} \right) = {g'_{n - 1}}$, we conclude that $\varphi \left( {{g'_{n + 1}}} \right) = {g'_{n + 1}}$, or equivalently $\varphi \left( g' \right) = g'$.
\end{proof}

\begin{lemma}\label{6.12}
If $\varphi\in \mathbb{A}ut(\Gamma_1)_1$ such that $\varphi(x)=x$, $\varphi \left( y \right) = {y^{ - 1}}$ and $f\in\mathbb{A}ut(G, S_1)$ is non trivial, then $\varphi=f$
\end{lemma}
\begin{proof}
Since $\varphi  \circ f(1) = 1$, $\varphi  \circ f(x) = x$ and $\varphi  \circ f(y) = y$, by Corollary \ref{6.11}, the statement is obtained.
\end{proof}


\begin{theorem}\label{6.13}
$\mathbb{A}ut(\Gamma_1)_1\cong\mathbb{Z}_{2}$.
\end{theorem}
\begin{proof}
By Lemmas \ref{6.9} and \ref{6.12} and Corollary \ref{6.11}, the proof is straightforward.
\end{proof}

Therefore, the first part of the main theorem is a consequence of Theorems \ref{6.2} and \ref{6.13} and Lemmas \ref {2.2} and \ref {2.2.1} .



\section{$\Gamma_2=Cay(G, S_2)$}

We remind that $S_2=\{x^2y^{j}, (x^2y^j)^{-1}, y^{j'}, y^{-j'} \}$ where $(j', p)=1$. In this section Cayley graph $Cay(G, S_2)$ is denoted by $\Gamma_2$.

\begin{lemma}\label{7.1}
$S_2$ is equivalent to $\{x^2, x^{-2}, y, y^{-1} \}$.
\end{lemma}
\begin{proof}
It is sufficient, consider $f\in\mathbb{A}ut{(G)}$ such that $f(x)=xy^{i}$ and $f(y)=y^{\alpha}$, where $i$ have two following cases
$$ \left\{
   \begin{array}{ll}
     i=\frac{j}{2}(k^4-k^3+k^2-k+1), & \hbox{if $j$ is even;} \\
     i=\frac{p^2+j}{2}(k^4-k^3+k^2-k+1), & \hbox{if $j$ is odd.}
   \end{array}
 \right.
$$
We also set $\alpha=j'$. Therefore, $f(x^2)=x^2y^j$ and $f(y)=y^{j'}$.
\end{proof}

From now on, we use the above mentioned equivalent for $S_2$.

\begin{theorem}\label{7.2}
$\mathbb{A}ut(G, S_2)\cong\mathbb{Z}_2.$
\end{theorem}
\begin{proof}
If we consider $f\in\mathbb{A}ut(G, S_2)$, then by attention to the orders of elements of $S_2$ and proof of Lemma \ref{2.3}(1) we have,  $f(x^2)=x^2$ and $f(y)$ equals to $y$ or $y^{-1}$.

Assume that  $f(y)=y$.  Then $x^2=f(x^2)=f(x)^2=xy^ixy^i=x^2y^{i(k+1)}$.
Hence, $p^2\mid i(k+1)$.
We know, $p\nmid k+1$, i.e., $(p, k+1)=1$,
because otherwise, for some integer
$r$, $k=rp-1$,
we will have
and so,
$p^2 \mid k^5 -1$,
thus $1\equiv k^5 (mod \ p^2)\equiv 5rp-1$;
hence, $p^2\mid 5rp-2$, which is a contradiction. Thus $p^2\mid i$, then $i=0$.
Therefore, $f(x)=x$ and $f(y)=y$, i.e., $f=id$.

Now assume that $f(y)=y^{-1}$.
Since $f(x^2)=x^2$, similar to the previous case $i=0$.
Thus, $f(x)=x$.
It means that $f$ is an element of order $2$.
\end{proof}



\begin{lemma}\label{7.3}
If $\varphi\in\mathbb{A}ut(\Gamma_2)_g$ and $\varphi(x^2g)=x^2g$, then $\varphi(x^{i}g)=x^{i}g$ for $1 \leqslant i \leqslant 4$.
\end{lemma}
\begin{proof}
 Since ${x^3}g \in {D_1}\left( g \right)$ and $\varphi $ preserves distance, so we have
 $$\varphi \left( {{x^3}g} \right) \in \varphi \left( {{D_1}\left( g \right)} \right) = {D_1}\left( {\varphi \left( g \right)} \right) = {D_1}\left( g \right) = \left\{ {x^2g,{x^3}g,yg,{{\left {y} \right}^{ - 1}}g} \right\}$$
 $$ = \left\{ {\varphi \left( {x^2g} \right),{x^3}g,yg,{{\left {y} \right}^{ - 1}}g} \right\}.$$
Since $\varphi $ is one-to-one, $\varphi \left( {{x^3}g} \right) \in \left\{ {{x^3}g,yg,{{\left {y} \right}^{ - 1}}g} \right\}$.
On the other hand
$\left( {{x^4}g,{x}g} \right)$
is an edge and
${x^4}g \in {D_1}\left( {x^2g} \right)$, ${x}g \in {D_1}\left( {{x^3}g} \right)$,
so there exists a member of
$\varphi \left( {{D_1}\left( {x^2g} \right)} \right)$
that is adjacent with a member of $\varphi \left( {{D_1}\left( {{x^3}g} \right)} \right)$.
But if
$\varphi \left( {{x^3}g} \right) = yg$ or ${\left {y} \right^{ - 1}}g$,
such members do not exist.
Thus $\varphi \left( {{x^3}g} \right) = {x^3}g$.
Also ${x^4}g$ and ${x}g$ are the only vertices in
${D_1}\left( {x^2g} \right)$ and  ${D_1}\left( {{x^3}g} \right)$ respectively, which adjacent, so $\varphi $ fixes them.
\end{proof}
 Similarly, if $\varphi(x^{2}g)=x^3g$, then $\varphi(x^{-i}g)=x^{i}g$ for $1 \leqslant i \leqslant 4$ and if $\varphi(x^{3}g)=x^2g$, then $\varphi(x^{-i}g)=x^{i}g$ for $1 \leqslant i \leqslant 4$.

\begin{lemma}\label{7.4}
 If $\varphi\in\mathbb{A}ut(\Gamma_1)_g$, then $\varphi \left( {x^2g} \right) \notin\{yg, y^{-1}g\}$.
\end{lemma}
\begin{proof}
If $\varphi \left( {x^2g} \right) = yg$, then according to above description, $\varphi \left( {{x^{3}}g} \right) \ne x^2g$, thus $\varphi \left( {{x^{3}}g} \right) = {y^{ - 1}}g$.
Also $\left( {{x^4}g,{x}g} \right)$ is an edge and ${x^4}g \in {D_1}\left( {x^2g} \right)$ and ${x}g \in {D_1}\left( {{x^3}g} \right)$, so there exists a member of $\varphi \left( {{D_1}\left( {x^2g} \right)} \right) = {D_1}\left( {yg} \right)$ that is adjacent with a member of $\varphi \left( {{D_1}\left( {{x^3}g} \right)} \right) = {D_1}\left( {{y^{ - 1}}g} \right)$, but there are no such elements.
\\Similarly $\varphi \left( {x^2g} \right) \ne {y^{ - 1}}g$ and the result now follows.
\end{proof}

\begin{lemma}\label{7.5}
If $\varphi  \in {\Bbb A}ut{({\Gamma _2})_{{y^{k - 2}}g}}$ and $\varphi  \in {\Bbb A}ut{({\Gamma _2})_{{y^{k - 1}}g}}$, then $\varphi  \in {\Bbb A}ut{({\Gamma _2})_{{y^{k}}g}}$ for $k \geqslant 2$.
\end{lemma}

\begin{proof}
Since $\varphi $ is one-to-one,
$\varphi \left( {{D_1}\left( {y^{k-1}g} \right)} \right) = {D_1}\left( {y^{k-1}g} \right)$
and
$${D_1}\left( {y^{k-1}g} \right) = \left\{ {y^{k-2}g,x^2y^{k-1}g,{x^3}y^{k-1}g,{y^k}g} \right\} =$$

$$\left\{ {\varphi \left( y^{k-2}g \right),x^2y^{k-1}g,{x^3}y^{k-1}g,{y^k}g} \right\}.$$
So

$$\varphi\left( {x^2y^{k-1}g} \right),\varphi \left( {{x^3}y^{k-1}g} \right),\varphi \left( {{y^k}g} \right) \in \left\{ {x^2y^{k-1}g,{x^3}y^{k-1}g,{y^k}g} \right\}.$$


We know

$${x}y^{k-1}g \in {D_1}\left( {{x^3}y^{k-1}g} \right),{x^4}y^{k-1}g \in {D_1}\left( {x^2y^{k-1}g} \right)$$
and
$\left( {{x}y^{k-1}g,{x^4}y^{k-1}g} \right)$
is an edge, therefore we have
$$\varphi \left( {{x}y^{k-1}g} \right) \in \varphi \left( {{D_1}\left( {{x^3}y^{k-1}g} \right)} \right) = {D_1}(\varphi \left( {{x^3}y^{k-1}g} \right)),$$
$$\varphi \left( {{x^4}y^{k-1}g} \right) \in \varphi \left( {{D_1}\left( {x^2y^{k-1}g} \right)} \right) = {D_1}(\varphi \left( {x^2y^{k-1}g} \right)),$$
and
$\left( {\varphi \left( {{x}y^{k-1}g} \right),\varphi \left( {{x^4}y^{k-1}g} \right)} \right)$
is an edge. Because there is no element of
${D_1}\left( {{y^k}g} \right)$
adjacent to an element of
${D_1}\left( {x^2y^{k-1}g} \right)$
or an element of
${D_1}\left( {{x^3}y^{k-1}g} \right)$. So we have
$$\varphi \left( {x^2y^{k-1}g} \right),\varphi \left( {{x^3}y^{k-1}g} \right) \in \left\{ {x^2y^{k-1}g,{x^3}y^{k-1}g} \right\}.$$ Thus $\varphi \left( {{y^k}g} \right) = {y^k}g$
and the lemma is proved.
\end{proof}

\begin{result}\label{7.6}
If $\varphi\in\mathbb{A}ut(\Gamma_2)_g$ and $\varphi(yg)=yg$, then $\varphi(y^{i}g)=y^{i}g$ for $1 \leqslant i < {p^2}$.
\end{lemma}
\begin{proof}
It follows immediately from Lemma \ref{7.5}.
\end{proof}

\begin{lemma}\label{7.7}
If $\varphi  \in {\Bbb A}ut{({\Gamma _2})_{g}}$ and $\varphi(yg)=y^{-1}g$, then $\varphi(y^{2}g)=y^{-2}g$.
\end{lemma}
\begin{proof}
Since $\varphi(yg)=y^{-1}g$, $\varphi \left( {{D_1}\left( {yg} \right)} \right)= {D_1}(\varphi \left( {yg})) = {D_1}\left( {y^{-1}g} \right)$, so we have

$$ \left\{ {\varphi \left(g \right),\varphi \left({x^{2}y}g \right),\varphi \left({x^{3}yg \right),\varphi \left({y^{2}g \right)} \right\} $$$$= \left\{ {\varphi \left(g \right),x^{2}y^{-1}g,{x^3}y^{-1}g,{y^{-2}}g} \right\},$$
On the other hand   ${x^4}yg \in {D_1}\left( {x^{2}yg} \right)$, $xyg \in {D_1}\left( {x^{3}yg} \right)$ and $\left( {{x^4}yg,{x^3}yg} \right)$
is an edge, so we have
$$\varphi \left( {{x^4}yg} \right) \in \varphi \left( {{D_1}\left( {x^{2}yg} \right)} \right) = {D_1}(\varphi \left( {x^{2}yg} \right)),$$
$$\varphi \left( {{x}yg} \right) \in \varphi \left( {{D_1}\left( {x^{3}yg} \right)} \right) = {D_1}(\varphi \left( {x^{3}yg} \right))$$
and
$\left( {\varphi \left( {{x^4}yg} \right),\varphi \left( {{x}yg} \right)} \right)$
is an edge.
Since there is no element of
${D_1}\left( {{y^{-2}}g} \right)$
adjacent to an element of ${D_1}\left( {x^{2}y^{-1}g} \right)$
or ${D_1}\left( {x^{3}y^{-1}g} \right).$
So, we have
 $$\varphi \left( {x^{2}yg} \right),\varphi \left( {{x^3}yg} \right) \in \left\{ {x^{2}y^{-1}g,{x^3}y^{-1}g} \right\}.$$
Thus $\varphi \left( {{y^2}g} \right) = {y^{-2}}g$.
\end{proof}

\begin{result}\label{7.8}
If $\varphi\in\mathbb{A}ut(\Gamma_1)_g$ and $\varphi(yg)=y^{-1}g$, then $\varphi(y^{i}g)=y^{-i}g$ for
$1 \leqslant i < {p^2}$.
\end{result}
\begin{proof}
Similar to proof of Lemma \ref{7.7}.
\end{proof}

\begin{lemma} \label{7.9}
If $\varphi\in\mathbb{A}ut(\Gamma_2)_g$, then $\varphi(x^2g)=x^2g$.
\end{lemma}
\begin{proof}
We know $x^2g \in {D_1}\left( g \right)$, so we have
$$\varphi \left( {x^2g} \right) \in \varphi \left( {{D_1}\left( g \right)} \right) = {D_1}\left( {\varphi \left( g \right)} \right) = {D_1}\left( g \right) = \left\{ {x^2g,x^3g,yg,y^{  - 1} g} \right\}.$$
On the contrary  assume that $\varphi (x^2g) \ne x^2g$. Thus by Lemma \ref{7.4}, we have
\\$\varphi(x^2g)=x^{3}g$.

\begin{center}
\begin{tikzpicture}
	\begin{pgfonlayer}{nodelayer}
		\node [style=newstyle] (0) at (-2.5, 1) {$x^{4}g$};
		\node [style=newstyle] (1) at (-2.5, -0.75) {$xg$};
		\node [style=newstyle] (2) at (-0.75, 1) {$x^2g$};
		\node [style=newstyle] (3) at (-0.75, -0.75) {$x^{3}g$};
		\node [style=newstyle] (4) at (0.25, -0) {$g$};
		\node [style=newstyle] (5) at (1.5, 1) {$yg$};
		\node [style=newstyle] (6) at (1.75, -0.75) {$y^{-1}g$};
		\node [style=newstyle] (7) at (3.5, 1) {$y^{2}g...$};
		\node [style=newstyle] (8) at (3.75, -0.75) {$y^{-2}g...$};
					\end{pgfonlayer}
	\begin{pgfonlayer}{edgelayer}
		\draw (0) to (2);
		\draw (1) to (3);
		\draw (0) to (1);
		\draw (2) to (4);
		\draw (4) to (3);
		\draw (5) to (4);
		\draw (4) to (6);
		\draw (5) to (7);
		\draw (6) to (8);
	\end{pgfonlayer}
\end{tikzpicture}
\end{center}




 Then we have two cases for $\varphi(y)$:
\begin{itemize}

\item[Case 1.]
$\varphi(yg)=yg$;
since $\left( {x^2g,x^2{y^k^2}g} \right)$ and $\left( {y^k^2g,x^2{y^k^2}g} \right)$ are two edges of graph and $\varphi$ preserves distance, so we have
$$2=d(y^{k^2}g, x^2g)=d(\varphi(y^{k^2}g), \varphi(x^2g))=d(y^{k^2}g, x^{3}g)> 2,$$
which is a contradiction.

\item[Case 2.]
$\varphi(yg)=y^{-1}g$;
Similar to the previous case, we have
$$2=d(y^{k^2}g, x^2g)=d(\varphi(y^{k^2}g), \varphi(x^2g))=d(y^{-k^2}g, x^{3}g)> 2,$$
which is a contradiction,
therefore this case does not happen.
\end{itemize}
\end{proof}

\begin{result}\label{7.10}
If $\varphi\in\mathbb{A}ut(\Gamma_2)_g$, then $\varphi(xg)=xg$.
\end{result}
\begin{proof}
By applying Lemma \ref{7.9} for  $g: =x^2g$ and $g: =x^4g$, it follows.
\end{proof}

\begin{lemma}\label{7.11}
If $\varphi\in A_g$, and $\varphi $ fixes all the elements of ${D_1}(g)$, then it fixes all the elements of ${D_2}(g)$.
\end{lemma}
\begin{proof}
We know
$$D_{2}(g)=\{x^4g, xg, x^2y^{k^2}g,  x^2y^{-k^2}g, x^3y^{k^3}g,x^3y^{-k^3}g,  x^2y^{-1}g,$$

$$ x^3y^{-1}g, x^2yg, x^3yg, y^2g, y^{-2}g\}.$$

Already by Corollaries \ref{7.6} and \ref{7.10}, we know, $\varphi$ keeps fixed them.
\end{proof}

\begin{result}\label{7.12}
If $\varphi\in \mathbb{A}ut(\Gamma_2)_1$ and $\varphi$ fixes all the elements of ${D_1}(g)$, then $\varphi= id$.
\end{result}
\begin{proof}
It follows immediately from Lemma \ref{7.11} and graph connectivity.
\end{proof}


\begin{lemma}\label{7.13}
Let $\varphi\in \mathbb{A}ut(\Gamma_2)_1$. If $\varphi(y)=y$, then $\varphi=id$. If $\varphi(y)=y^{-1}$,
then $\varphi=f$, where $f$ is non-identity element of $\mathbb{A}ut(G, S_2)$.
\end{lemma}
\begin{proof}
The first case is obtained using Corollaries \ref{7.10}, \ref{7.12} and graph connectivity.
In the second case, since $\varphi\circ f(x)=x$, $\varphi\circ f(y)=y$ and $\varphi\circ f\in \mathbb{A}ut(\Gamma_2)_1$, by Corollary \ref{7.12}, $\varphi\circ f=id$ and because $f$ is of order $2$, so we have $\varphi=f$.
\end{proof}


\begin{theorem}\label{7.14}
$\mathbb{A}ut(\Gamma_2)_1\cong\mathbb{Z}_2$.
\end{theorem}
\begin{proof}
The statement is a consequence of Lemma \ref{7.13} and Corollary \ref{7.10}.
\end{proof}

Therefore, Lemmas \ref{2.2}, \ref{2.2.1} and Theorems \ref{7.2}, \ref{7.14} prove the second part of the Main Theorem.

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\end{document}

