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\title{ Digraph representations of 2-closed permutation groups with a normal regular cyclic subgroup}

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\author{Jing Xu\thanks{This work
  was supported by NSFC (project number 10901110, 11371259).}\\
\small Department of Mathematics\\[-0.8ex]
\small Capital Normal University\\[-0.8ex]
\small Beijing 100048,  China\\
\small\tt xujing@cnu.edu.cn
}

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\date{\dateline{March 30, 2015}{ }\\
\small Mathematics Subject Classifications: 05C25, 20B25}

\begin{document}

\maketitle

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\begin{abstract}
  In this paper, we classify
2-closed (in Wielandt's sense) permutation groups which contain  a
normal  regular cyclic subgroup and prove that for each such  group
$G$, there exists a   circulant $\G$ such that $\Aut (\G)=G$.

\end{abstract}

\section{Introduction}

In 1969, Wielandt  \cite{we1} introduced the concept of the
2-closure of a   permutation group. Let $G$ be a finite permutation
group on a set $\O$, the \emph{$2$-closure} $G^{(2)}$ of $G$ on $\O$
is the largest subgroup of $\Sym(\O)$ containing $G$ that has the
same orbits as $G$ in the induced action on $\O\times \O$, and we
say $G$ is \emph{2-closed} if $G=G^{(2)}$. It seems impossible
  to classify all 2-closed transitive
permutation groups. However, certain classes of 2-closed transitive
groups have been determined. For example,  in \cite{xj, xjmeta}  the
author determined all  2-closed odd-order transitive permutation
groups of degree $pq$ where $p, q$
 are distinct odd primes.
 In this paper, one of our main purposes is to classify all  2-closed permutation groups
 with a normal regular cyclic subgroup, see Theorem \ref{mainthm}. Recall that a   permutation group  is
  \emph{regular} if  it is
transitive   and the only element that fixes a point is the
identity. And for more information about the 2-closures of permutation groups containing a cyclic regular subgroup, see also \cite{EP}.


Another research topic of this paper is  the study of  the
automorphism groups of (di)graphs. The full automorphism group of a
(di)graph $\G$ must be 2-closed since any permutation of the vertex
set that preserves the orbits of $\Aut(\Gamma)$ on ordered pairs
preserves adjacency.   However, not every 2-closed permutation group
is the full automorphism group of some (di)graph. Therefore, the
concept of   2-closed groups  is more general  than the concept of
the full automorphism groups of (di)graphs, and the classification
of 2-closed groups is closely related to the study of the full
automorphism groups of the corresponding digraphs. In this paper, in
order to determine  2-closed groups that contain a normal regular
cyclic subgroup, we also study circulant digraphs, that is Cayley digraphs
of cyclic groups. See Section \ref{preliminary} for a more detailed
explanation.






Furthermore, we   discuss the  following representation problem. A
digraph $\G$ with vertex set $\O$ is said to \emph{represent} a
permutation group $G\le \Sym(\O)$ if $\Aut(\G)=G$. In this case, we
also say that the permutation group $G$ has a digraph
\emph{representation} $\G$.

\bigskip
\noindent\textbf{Digraph representation problem:} given a 2-closed
group $G$, is there a digraph $\G$ that represents $G$?

Suppose the digraph $\G$  represents a 2-closed group $G\le
\Sym(\O)$. Then for any $g\in \Sym(\O)$, to determine whether $g$
lies in $G$ we only need to test   if $g$ preserves the single
2-relation given by the arc set of $\G$, instead of checking all
$G$-invariant 2-relations.
 We say
a digraph $\G$ is \emph{arc-transitive} if $\Aut(\G)$ is transitive
on the arc set of $\G$. This means, the arc set of $\G$ is actually
a minimal $\Aut(\G)$-invariant 2-relation. Suppose further that the
2-closed group $G$ can be represented by an arc-transitive digraph
$\G$. Then a permutation $g$ lies in $G$ if and only if $g$ leaves
invariant the minimal $G$-invariant 2-relation given by the arc set
of $\G$. We will show that there are arc-transitive digraph
representations for most 2-closed groups
 that contain   a normal regular cyclic subgroup, see the remark after Lemma \ref{arctran}.

Replacing digraph with graph, we obtain the graph representation
problem which asks for an undirected graph to  represent a 2-closed
group. These two questions have previously appeared in the
literature, see for example \cite{al, babai1995}. Clearly, the graph
version problem is much more complicated than the digraph one. Since
we are interested in   understanding   the concept of 2-closed
groups, we concentrate on the digraph representation problem in this
paper.



 A regular permutation group is 2-closed, and in 1980,
 Babai \cite{babai}   proved that with five
exceptions, every finite regular permutation group  occurs as the
automorphism group of a digraph. This is the famous DRR (digraphical regular representations) problem
\cite{babai}.  It is proved in \cite{PX} that for any prime power
$q$, the semilinear group $\GaL(1,q)$ can be represented by an
arc-transitive circulant digraph. Moreover, it is shown in \cite{xj,xjmeta}
that every 2-closed odd-order transitive permutation group   of
degree $pq$ has a tournament digraph representation.
As for graphical representation problem, see for example \cite{bg,DSV,godsil78,godsil80,godsil, MSV}.


In
this paper, we will    prove that   every 2-closed permutation group
$G$ with a normal   regular cyclic subgroup is the full automorphism
group of a circulant digraph.
We may suppose that $G=Z_n\rtimes G_0$ acting on $Z_n$ naturally
where $G_0\le \Aut(Z_n)$. We first describe the necessary and sufficient condition  for $G_0$ such  that $G$ is  2-closed.
 For the
detailed explanation of notation, see Section \ref{preliminary} and
Section \ref{notation}.







\begin{condition}\label{condition}
Let  $n=2^{d_1}p_2^{d_2}\cdots p_t^{d_t}, \ \ d_1\ge 0,\
d_2,...,d_t\ge 1, \ t\ge 1$    where $p_2,...,p_t$ are distinct odd
primes (also write $p_1=2$). And let $\Aut(Z_n)=
\Aut(Z_{2^{d_1}})\times\cdots\times\Aut(Z_{p_t^{d_t}})=D_1D_2\cdots
D_t,$ where $D_i$ is the direct factor subgroup of $\Aut(Z_n)$ that
fixes each component of the elements of $Z_n$ except for the $i$-th
component. So $D_i\cong\Aut(Z_{p_i^{d_i}})$ for each $i$. In fact $D_i$ induces a faithful action on the subgroup $Z_{p_i^{d_i}}$.
Note that
 the induced action $D_1$ on the subgroup $Z_{2^{d_1}}$ is permutation isomorphic to $\la (-1)^*\ra\times \la 5^*\ra (d_1\ge 3)$, the multiplicative group
 of units of the ring  $\Z_{2^{d_1}}$ acting on the additive group $\Z_{2^{d_1}}$,
   let $ \phi:   \la(-1)^*\ra\times \la (5)^*\ra\to D_1$  be the corresponding group
isomorphism.

Let   $G_0\le
\Aut(Z_n)$.
\begin{enumerate}
\item[\rm{(i)}] if $i\ge 2$, $d_i=1$ and $p_i\ge 5$,
 then $D_i  \nleqslant    G_0$.
\item[\rm{(ii)}] if $i\ge 2$ and $d_i\ge 2$,
then $ D_i\cap G_0\le Z_{p_i-1}$.

\item[\rm{(iii)}] if $d_1=3,$  then $D_1  \nleqslant
G_0$.

\item[\rm{(iv)}] if $d_1\ge 4$, then either $|D_1\cap G_0|\le 2$ or $|D_1\cap G_0|=4 $ and $D_1\cap G_0 \nleqslant  \la \phi(5 ^*)\ra$.

\end{enumerate}
\end{condition}
The main result of this paper is the following theorem.

\begin{theorem}\label{mainthm} Suppose $G=Z_n\rtimes G_0$ acting on $Z_n$ naturally
where $G_0\le \Aut(Z_n)$. Then $G$ is 2-closed if and only if $G_0$
satisfies  Conditions  \ref{condition}. Moreover, if $G$ is 2-closed
then $G$ can be represented by a circulant digraph.
\end{theorem}



\section{Preliminary results and notation}\label{preliminary}
First we introduce   some concepts and notation concerning
 Cayley digraphs.  Given a
finite group $H$, and a subset $S\subset H\setminus \{1\}$, the
Cayley digraph $\G=\Cay(H,S)$ with respect to $S$ is
 defined as the   directed graph with vertex set $H$ and arc set $A\G=\{(g,sg)\mid
 g\in H, s\in S\}$. Moreover, a Cayley digraph of a cyclic group is called a
\emph{circulant}. It is easy to check that the right regular
representation $\hat H$ is contained
 in $\Aut(\G)$.
  In fact, a  digraph is a Cayley digraph if and
  only if its automorphism group contains a regular subgroup.
   Moreover let $\Aut(H,S)=\{\s\in\Aut(H)\mid S^{\s}=S\}$, then each element in
$\Aut(H,S)$ induces an automorphism of the Cayley digraph
$\G=\Cay(H,S)$. It is proved in \cite{godsil} that  the normalizer of   $\hat H$ in
$\Aut(\G)$ is $\hat H\rtimes \Aut(H,S)$. We say a Cayley digraph
$\G=\Cay(H,S)$ is \emph{normal} if $\hat H$ is normal in $\Aut(\G)$,
that is, $\Aut(\G)=\hat H\rtimes \Aut(H,S)$, see \cite{godsil,
xumy1}. So the automorphism group of a normal circulant must be a
2-closed group that contains a normal regular cyclic group.
Conversely, we will show that each such 2-closed    group is the
automorphism group of some normal circulant.




Throughout the rest of this paper, let $Z_n$ be an abstract cyclic
group of order $n$ and let $G\le \Sym(Z_n)$ be  a transitive
permutation group which contains a normal regular cyclic group $\hat
Z_{n}$ where
\begin{equation}\label{S}\hat Z_{n}=\{ \hat g: x\to xg  \ \forall x\in Z_n |\ g\in Z_n\}.\end{equation}
Therefore $G$ is a semidirect product $\hat Z_n\rtimes G_0$ for some
subgroup $G_0\leq\Aut(Z_n)$ acting naturally on $Z_n$. Since $\hat
Z_n\cong Z_n$, we may also write $G=Z_n\rtimes G_0$ directly. Our
goal is to determine all such 2-closed   groups.






The mail tool used in this paper is the   Kov\'acs-Li classification
of arc-transitive circulants \cite{kovacs, li arc circ}.
 Praeger and the author \cite{PX} refined   the Kov\'acs-Li
classification  and obtained the following theorem.



\begin{theorem}\label{PXthm}{\rm \cite[Theorem 1.1]{PX}}
Let $G=Z_n\rtimes G_0\le Z_n\rtimes \Aut(Z_n)$    acting naturally
on $Z_n$. Then, up to isomorphism, there is a unique connected
$Z_n$-circulant $\G$ on which $G$ acts arc-transitively. Moreover
either $\Aut(\G)=G$ or one of the following holds.
\begin{enumerate}
 \item[$(a)$] $n=p\geq5$ is prime, $\G=K_p$, and $G={\rm AGL}(1,p)$;
\item[$(b)$] $n=bm>4$, where $b\geq2$, $p$ divides $m$ for each
prime $p$ dividing $b$, $\G=\si[\ov K_b]$; \item[$(c)$] $n=pm$,
where $p$ is prime, $5\leq p<n$, and $\gcd(m,p)=1$, $\G=\si[\ov
K_p]-p.\si$, $G_0=\Aut(Z_p)\times H\leq \Aut(Z_p)\times\Aut(Z_m)$,
and $\si$ is a connected $(Z_m\rtimes H)$-arc-transitive
$Z_m$-circulant.
\end{enumerate}
\end{theorem}

  We   point out
that up to isomorphism, in the above theorem $\G$ can be defined as
$\Cay(Z_n, z^{G_0})$ where $z$ is a generator of $Z_n$ and $z^{G_0}$
is the orbit of $z$ under $G_0$. Moreover, if case (b) happens, then
the group $Z_n$ has a subgroup $Y$ of order $b$, and
$\G=\Cay(Z_n,S)$ where $S$ is a union of $Y$-cosets each consisting
of generators for $Z$.

As a simple application of Theorem \ref{PXthm}, we determine the
2-closed transitive permutation groups of degree $p$ where $p$ is a
prime.

\begin{cor}\label{2closeddegreep} Let $p$ be a  prime.
Let $G\le \Sym(\O)$ be a 2-closed transitive permutation group of
degree $p$. Then there exists a digraph  representing $G$.
Moreover, $G$ is one of the following.
\begin{enumerate}
\item The symmetric group $S_p\ (p\ge 2)$ which is 2-transitive on $\O$.

\item An affine subgroup $Z_p\rtimes Z_k$ where  $p\ge 3$, $1\le k< (p-1)$ and
$k|(p-1)$.
\end{enumerate}
Conversely, each group of the above two types is 2-closed.
\end{cor}
\begin{proof}  Suppose $G$ is a 2-closed transitive permutation group of degree
$p$. By a classical result of Burnside, $G$ is either 2-transitive
or is affine. If $G$ is 2-transitive, then $G=G^{(2)}=S_p$ and $p\ge
2$. If $G$ is not 2-transitive, then $G=Z_p\rtimes Z_k$ where $p\ge
3$, $1\le k<(p-1)$ and $k|(p-1)$.

For the converse,   note that $S_p$ is the full automorphism group
of the complete graph $K_p$ and so $S_p$ is indeed 2-closed. Next,
let $G=Z_p\rtimes Z_k$ where $p\ge 3$, $1\le k< (p-1)$ and $k|(p-1)$.
By Theorem \ref{PXthm}, there is a connected arc-transitive
circulant $\G$ of order $p$ such that $\Aut(\G)=G$, and so $G$ is
2-closed.
\end{proof}

\noindent\textbf{Remark:}    If $p=2,3$ then $S_p=Z_p\rtimes
\Aut(Z_p)$ is 2-closed; and if $p\ge 5$ then $Z_p\rtimes \Aut(Z_p)$
is not 2-closed.

We also need the following  theorem.

\begin{theorem}{\rm \cite[Theorem 5.1]{sevenauthor}}
\label{product action} Let $G_1\le \Sym(\O_1)$ and $G_2\le
\Sym(\O_2)$ be transitive permutation groups. Consider the natural
product action of $G_1\times G_2$
  on $\O_1\times \O_2$. Then $(G_1\times
G_2)^{(2)}=G_1^{(2)}\times G_2^{(2)}$.
\end{theorem}


Finally, we fix the following notation. Let $A\le \Sym(\O)$. Suppose
that $A_B$ is the setwise stabilizer of $B\subseteq \O$ and $g\in
A_B$, we denote $A_B^B$ to be the induced permutation group on $B$
by $A_B$ and denote $g^B$ to be the induced permutation   on $B$ by
$g$.








\section{2-closed groups containing a normal regular cyclic group}
  In this
section we classify   2-closed groups $G$ that contain a normal
regular cyclic group $Z_n$. With notation in Section
\ref{preliminary}, we may suppose that $ G=  Z_n\rtimes G_0\le
 Z_n\rtimes \Aut(Z_n)$  acting naturally on $Z_n$. We first
handle the special case  that $n$ is a prime power in Subsection
\ref{p^dsection} and Subsection \ref{section 2^d}. The notation
needed for the statement of Theorem \ref{mainthm} is given in
Subsection \ref{notation} and  the proof is given in Subsection
\ref{proof}.
\subsection{The case   $n=p^d$ with $p$  an odd
prime}\label{p^dsection} Let $n=p^d$ where $p$ is an odd prime and
$d\ge 2$ is an integer. Then $\Aut(Z_n)=Z_{(p-1)}\times Z_{p^{d-1}}$
is a cyclic group. We take  $\a\in \Aut(Z_n)$ such that $o(\a)=p$, then there exists $\g\in \Aut(Z_n)$ with order $p^{d-1}$
such that $\a=\g^{p^{d-2}}$.
We first look at the action of $\a$ on $Z_n$.

Let $H=Z_{p^{d-1}}$ be the unique subgroup of $Z_n$ of order
$p^{d-1}$. Let $N=Z_{n}\rtimes \Aut(Z_n)$. Then  the cosets of $H$
form a block system $\mathcal B$ of $N$ on $Z_n$. Denote $\mathcal
B=\{B_1=H, B_2,...,B_p\}$. Since the elements in $B_2,...,B_p$ are
of order $p^d$, $\g$ fixes each block setwise and $\g^{B_i}$ is a
$p^{d-1}-$cycle for each $i\ge 2$. However, $\g$ fixes the point
$1\in H=B_1$, so the order of $\g^{B_1}$  is strictly less than
$p^{d-1}$. It then follows that $\a$ fixes $B_1$ pointwise and is
fixed point free on each $B_i$ for $i\ge 2$.


On the other hand,   let $N_{B_i}^{B_i}$ be the induced permutation
group of the setwise stabilizer $N_{B_i}$ on $B_i$. Then
$N_{B_i}^{B_i}= \hat Z_{p^{d-1}}\rtimes K_i$ and
$K_i\cong\Aut(Z_{p^{d-1}})$, ($\hat Z_{p^{d-1}}$ is
defined  in equation $(\ref{S})$). For each $i\ge 2$, since $\g^{B_i}$
is fixed point free, we have that
$\g^{B_i}=\hat y_i^{B_i}\tau$ where $1\ne y_i\in H\le Z_n$ and
$\tau\in K_i$. Since $\tau$ normalizes $\hat Z_{p^{d-1}}$, $(\g^{B_i})^2=\hat y_i^{B_i}(\tau\hat y_i^{B_i} \tau^{-1})\tau\tau=a_{i2}\tau^2$ where $a_{i2}$ is some element in $\hat Z_{p^{d-1}}$.  By induction, we have that for each $k\ge 1$, $(\g^{B_i})^k= a_{ik}\tau^k$ where $a_{ik}$ is some element in $\hat Z_{p^{d-1}}$. Since $\g^{B_i}$
is of order $p^{d-1}$ and $\hat Z_{p^{d-1}}\cap K_i=\{1\}$, we have that $\t^{p^{d-1}}=1$. Since
$\t\in\Aut(Z_{p^{d-1}})=Z_{p-1}\times Z_{p^{d-2}}$, $\t^{p^{d-2}}=1$. Recall that $\a=\g^{p^{d-2}}$, it then follows
  that $\a^{B_i}$ is $\hat x_i^{B_i}$ for some
$x_i\in Z_n$ with order $p$. Note that $x_i$ may not equal $x_j$ for
$2\le i<j\le p$, but they are all of order $p$. We have proved the
following lemma.

\begin{lemma}\label{lemma4.1} Let $\a\in \Aut(Z_{p^{d}})$ with order $p$. Let $\mathcal
B=\{B_1=H, B_2,...,B_p\}$ be the cosets of the subgroup $H$ where
$H< Z_{p^d}$ is of order $p^{d-1}$. Then $\a$ fixes $B_1=H$
pointwise and for each $i\ge 2$, $\a^{B_i}$ is $\hat x_i^{B_i}$ for
some $x_i\in Z_n$ with order $p$.
\end{lemma}

\begin{cor}\label{zZ_p} Let $n=p^d$ and  $Z_n=\la z\ra$. Let
 $Z_p\le Z_n$ be the
subgroup of order $p$. Suppose that $G= Z_n\rtimes G_0$ where
$G_0\le \Aut(Z_n)$. Then the coset $zZ_p\subseteq z^{G_0}$ if and
only if  $p||G_0|$.
\end{cor}
\noindent\textbf{Remark:} Let $S=z^{G_0}$ and $\G=\Cay(Z_n, S)$.  If
case (b) of Theorem \ref{PXthm} occurs for $\G$, then $zZ_p\subseteq
z^{G_0}$. That is why we consider this corollary.
\begin{proof} Let $\Aut(Z_{p^d})=\la \mu \ra\times \la
\g\ra=Z_{p-1}\times Z_{p^{d-1}}$  and $\a=\g^{p^{d-2}}$. Then $p||G_0|$ if and only if $\a\in G_0$.

Let
$\mathcal B=\{B_1=H, B_2,...,B_p\}$ be the cosets of the subgroup
$H$ where $H< Z_{p^d}$ is of order $p^{d-1}$. Then it is easy to
show that $\mu$ fixes $B_1$ setwise, and permutes $B_2,...,B_{p}$ as
a $(p-1)$-cycle.

By Lemma \ref{lemma4.1}, if $\a\in G_0$ then $zZ_p\subseteq
z^{G_0}$. Conversely, suppose that $zZ_p\subseteq z^{G_0}$.  Note
that the generator $z\in B_k$ for some $k\ge 2$ and $zZ_p\subseteq
B_k$. By the action of $\mu$ and $\g$, we conclude that $\a\in G_0$.
\end{proof}

\begin{prop}\label{p^d} Let $n=p^d$ where $p$ is an odd prime and $d\ge 2$. Let $G=Z_n\rtimes G_0\le Z_n\rtimes \Aut(Z_n)$
acting naturally on $Z_n$. Then $G$ is 2-closed if and only if
$ G_0\le Z_{p-1}$. Moreover, if $G$ is 2-closed then $G$ can be
represented by  an arc-transitive circulant.
\end{prop}

\begin{proof}
As defined at the beginning of Subsection \ref{p^dsection}, let $\a\in \Aut(Z_{p^{d}})$ be an element of order $p$. Let $\mathcal B=\{B_1=H,
B_2,...,B_p\}$ be the cosets of the subgroup $H$ where $H< Z_{p^d}$
is of order $p^{d-1}$.

Suppose first that $G_0\not\le Z_{p-1}$, that is $p||G_0|$,  then $\a\in G_0$. By Lemma
\ref{lemma4.1},  $\a$ fixes $B_1=H$ pointwise and for each $i\ge 2$,
$\a^{B_i}$ is $\hat x_i^{B_i}$ for some $x_i\in Z_n$ with order $p$.

Let $1\ne\b\in \Sym(Z_{n})$ such that $\b$ fixes every element of
$B_1,...,B_{p-1}$ and $\b^{B_p}=\a^{B_p}$. That means $\b^{B_p}=\hat
x_p^{B_p}$, (recall that $\hat x: z\mapsto zx $ for any $z\in Z_n$). We claim that $\b\in (Z_{p^d}\rtimes \la
\a\ra)^{(2)}$ and so $\b\in G^{(2)}$. Take any pair $
(y_1,y_2)\in Z_n\times Z_n$. If both $y_1$ and $y_2$ belong to
$B_p$,   then $(y_1, y_2)^{\b}=(y_1x_p, y_2x_p)$ is in the orbital
$(y_1, y_2)^G$. Suppose next that exactly one of $\{y_1, y_2\}$ lies
in $B_p$, say $y_2\in B_p$. Since the stabilizer $G_{y_1}$ is the
conjugate of $G_0$ in $G$ by an element in $\hat Z_n$, a conjugate
of $\a$, say $\r$, is in $G_{y_1}$.  Therefore $\b^{B_p}$ equals
  $(\r^j)^{B_p}$ for some $j\in \{1,...,p-1\}$, and so
$(y_1,y_2)^{\b}\in (y_1,y_2)^G$. It then follows    that $\b\in
(Z_{p^d}\rtimes \la
\a\ra)^{(2)}\le G^{(2)}$. However, since $\b$ fixes $B_1$ and $B_2$ pointwise, $\b\notin Z_{p^d}\rtimes \Aut(Z_{p^d})$, and so
$\b\notin G$ and   $G$ is not 2-closed.


Suppose next that $G_0\le Z_{p-1}$. Let $S=z^{G_0}$ where $z\in
Z_{p^d}$ is an element of order $p^d$ and let $\G=\Cay(Z_n, S)$.
 Since $(p, |G_0|)=1$, $p\nmid |S|$ and so $S$ is not a union of cosets of
  any subgroup of $Z_n$. By Theorem \ref{PXthm},  $\Aut(\G)=G$ and so $G$
is 2-closed. This completes the proof.
\end{proof}


\noindent\textbf{Remark:} In above proof, note that $\b$ is   in $(Z_{p^d}\rtimes \la
\a\ra)^{(2)}$. Hence we actually proved that $(Z_{p^d}\rtimes \la \a\ra)^{(2)}\nleq
Z_{p^d}\rtimes \Aut(Z_{p^d})$ where $\a\in \Aut(Z_{p^{d}})$ is of order $p$.



\subsection{The case   $n=2^d$ for $d\ge 2$}\label{section 2^d}


\textbf{Notation:} For convenience, in this subsection we write
$Z_n$ additively as the group $\mathbb{Z}_n$ of integers modulo $n$,
so in this case
$$\hat Z_n=\hat \Z_n=\{\hat x: g\to g+x\ |\ x\in Z_n\}.$$  Moreover $\Aut(Z_n)$ is the
multiplicative group $\mathbb{Z}_n^*$ so that $i^*\in \Aut(Z_n)$
denotes the map $j\mapsto ij$.


\subsubsection{ $d=2$:}
  In this case, $\Aut(Z_4)=\la (-1)^*\ra\cong Z_2$. We have
the following result.

\begin{lemma}\label{2^2} Suppose that  $\hat Z_4\le G\le \hat Z_4\rtimes \la (-1)^*\ra\cong D_8$. Then
$G$ is 2-closed and is the full automorphism group of an
arc-transitive circulant.
\end{lemma}
\begin{proof} Either $G\cong Z_4$ is regular or $G\cong D_8$. Note
that $\Aut(\Cay(\Z_4,\{ 1\}))=Z_4$ and $\Aut(\Cay(\Z_4,
\{1,-1\}))=D_8=Z_4\rtimes Z_2$, this proves the lemma.
\end{proof}

\noindent\textbf{Remark:} By \cite[Lemma 2.3]{PX}, a connected
arc-transitive
 circulant $\G$ is both normal and   of lexicographic
product form if and only if $\G=\Cay(\Z_4, \{1,-1\})$ and
$\Aut(\G)=Z_4\rtimes \Aut(Z_4)$. In this case the orbit
$1^{\Aut(\Z_4)}=\{1,3\}=1+Z_2$ is a coset of $Z_2$.

\subsubsection{$d\ge 3$:}
In this case, $\Aut(Z_n)=\la (-1)^*\ra\times \la 5^*\ra\cong
Z_2\times Z_{2^{d-2}}.$ Denote $N=\hat \Z_n\rtimes \Z_n^*$. Let $H$
be the unique subgroup of $\Z_n$ with order $2^{d-2}$. Let $B_0=H,
B_1=1+H, B_2=2+H, B_3=3+H$ be  the cosets of $H$, then $\mathcal
B=\{B_0, B_1, B_2, B_3\}$ forms a complete block system of $N$ on
$\Z_n$.


We   first study the action of $5^*$. By computation $5^*$ preserves each
block $B_i$,  we determine the induced permutation $(5^*)^{B_i}$
next. Since $B_1\cup B_3$ consists of all elements of   order $2^d$,
$(5^*)^{B_1}$ and $(5^*)^{B_3}$ are
 $2^{d-2}$-cycles. As $B_0=\la 4\ra= Z_{2^{d-2}}$ and $B_0\cup
B_2=\la 2\ra= Z_{2^{d-1}}$, it is easy to deduce that $(5^*)^{B_2}$
is a product of two $2^{d-3}$-cycles (if $d=3$, then $(5^*)^{B_2}$
is trivial). Therefore the orders of $(5^*)^{B_1}$ and $(5^*)^{B_3}$
are $2^{d-2}$, the order of $(5^*)^{B_2}$ is $2^{d-3}$, and the
order of $(5^*)^{B_0}$ is $2^{d-4}$ (if $d=3$, then the order is 1).

\bigskip
\noindent\textbf{Case 1: $d=3$}


In this case, $n=8$ and  $\Aut(Z_8)=\la (-1)^*\ra\times \la
5^*\ra\cong Z_2\times Z_2$. By computation, $5^*$ fixes $B_0$ and $B_2$
pointwise,  and  the induced action $(5^*)^{B_1}=\hat 4 ^{B_1}$ and
$(5^*)^{B_3}=\hat 4 ^{B_3}$.   The element $(-1)^*$ fixes $B_0$
pointwise and $((-1)^*)^{B_2}=\hat 4 ^{B_2}$.

\begin{lemma}\label{coset 8} Let   $\Z_8=\la z\ra$. Suppose that $G=
\Z_8\rtimes G_0$ where $G_0\le \Aut(\Z_8)=\la (-1)^*\ra\times \la
5^*\ra$.     Then the coset $z+Z_2\subseteq z^{G_0}$ if and only if
$5^{*}\in G_0$ where $Z_2=\la 4\ra$ is the subgroup of order 2.
 \end{lemma}
 \begin{proof} Note that both $z$ and $z+Z_2$  are contained in $  B_1$ or $B_3$ and $(-1)^*$
 interchanges two blocks $B_1$ and $B_3$. The result follows from
 the analysis of the actions of $(-1)^*$ and $5^*$ easily.
 \end{proof}
\begin{prop}\label{n=8} With above notation, let $G=Z_8\rtimes G_0$ where
$G_0\le \Aut(Z_8)=\la (-1)^*\ra\times \la 5^*\ra$. Then

1.  if $G_0=\Aut(Z_8)$ then $G$ is not 2-closed.

2.  if $G_0 \lvertneqq \Aut(Z_8)$ and  $G_0\ne \la 5^*\ra$, then $G$ is 2-closed and can be represented by an arc-transitive
circulant.

3. if $G_0=\la 5^*\ra$, then $G$ is 2-closed and can be represented by a circulant.

\end{prop}
\begin{proof} (1) Suppose first that $G_0=\Aut(Z_8)$. Let $\b\in S_8$
such that $\b$ fixes $B_0, B_1$ and $B_3$ pointwise and
$\b^{B_2}=\hat 4 ^{B_2}$. Take any pair $ (y_1,y_2)\in Z_8\times
Z_8$. If both $y_1$ and $y_2$ belong to $B_2$,   then $(y_1,
y_2)^{\b}=(y_1, y_2)^{\hat 4}$ is in the orbital $(y_1, y_2)^G$.
Suppose next that exactly one of $\{y_1, y_2\}$ belongs to $B_2$,
say $y_2\in B_2$. It is straightforward to check that
$(y_1,y_2)^{\b}=(y_1,y_2)^{(-1)^*}$ if $y_1\in B_0$. Let $G_1$   be
the point stabilizer of point $1$, then $G_1$ is the conjugate  of
$G_0$ by  $\hat 1\in \hat \Z_n$. Let $\a_1$  be the corresponding
conjugate  of $5^{*}$ in $G_1$. It follows that
$(y_1,y_2)^{\b}=(y_1,y_2)^{\a_1}$ if $y_1\in B_1\cup B_3$. Hence
$\b\in G^{(2)}$. However since $\b$ fixes 0 and 1, $\b\notin G$ and
so $G$ is not 2-closed.

(2) In this case, $5^*\notin G_0$. Let $S=1^{G_0}$ and let
$\G=\Cay(\Z_8, S)$. It follows from Lemma \ref{coset 8} and Theorem
\ref{PXthm} that $G=\Aut(\G)$ and  is 2-closed.

(3) Finally we show that $\Z_8\rtimes \la 5^*\ra$ is 2-closed. Let
$S_1=1^{\la 5^*\ra}=\{1,5\}$ and $S_2=2^{\la 5^*\ra}=\{2\}$. Let
$\G=\Cay(\Z_8, S_1\cup S_2)$. By \cite[Theorem 1.3]{li arc circ}, it
is easy to deduce that $\G$ is not arc-transitive. Suppose $g\in
\Aut(\G)$ such that $g$ fixes 0 and 1, it is straightforward to
check that $g=1$. We conclude that $\Aut(\G)=\Z_8\rtimes \la 5^*\ra$
as required.
\end{proof}


 \noindent\textbf{Case 2: $d\ge 4$}

 Let $\a=(5^*)^{2^{d-4}}$ be an element of order 4 in $\la 5^*\ra$.
By the analysis of action of $5^*$, we deduce that  $\a$ fixes $B_0$
pointwise and $o(\a^{B_2})=2$, $o(\a^{B_1})=o(\a^{B_3})=4$.

Suppose first that $d=4$, then $\a=5^*$. By direct computation,   $\a^{B_2}=
\hat {8} ^{B_2}$, $\a^{B_1}=\hat 4^{B_1}$   and $\a^{B_3}= \widehat {-4} ^{B_3}$.

Next suppose $d\ge 5$. Denote $N=\hat \Z_n\rtimes \Z_n^*$. Note that $N_{B_i}^{B_i}\cong
\hat Z_{2^{d-2}}\rtimes K_i$ where $K_i\cong\Aut(Z_{2^{d-2}})$ for
each $i\in \{1,2,3\}$. Since $(5^*)^{B_i}$ is fixed point free on $B_i$ for $i=1,2,3$,  $(5^*)^{B_i}=\hat
y_i^{B_i}\t_i$ where $0\ne y_i\in \Z_n$ and $\t_i\in K_i$.  Since $\tau_i$ normalizes $\hat Z_{2^{d-2}}$, $((5^*)^{B_i})^2=\hat
y_i^{B_i}(\t_i\hat y_i^{B_i} \tau_i^{-1})\tau_i\tau_i=a_{i2}\tau_i^2$ where $a_{i2}$ is some element in $\hat Z_{2^{d-2}}$.  By induction, we have that for each $k\ge 1$, $((5^*)^{B_i})^k= a_{ik}\tau_i^k$ where $a_{ik}$ is some element in $\hat Z_{2^{d-2}}$. Since
$\t_i\in\Aut(Z_{2^{d-2}})$ and $d\ge 5$, $\t_i^{2^{d-4}}=1$.  By the order  of $\a^{B_i}$,  we have that
$\a^{B_i}=\hat x_i^{B_i}$,    where $x_1, x_3\in Z_n$ are of  order 4 and
$x_2=2^{d-1}$  is the unique involution in $Z_n$. In addition,
$2x_1=2x_3=2^{d-1}$. Therefore we have proved the following lemma.

\begin{lemma}\label{a-action} Suppose $d\ge 4$. With above notation, let $\a=(5^*)^{2^{d-4}}$ be an element of order 4 in $\la
5^*\ra$. Then  $\a$ fixes $B_0$ pointwise, $\a^{B_2}=
(\widehat{2^{d-1}})^{B_2}$, $\a^{B_1}=\hat x_1^{B_1}$ for some
$x_1\in Z_n$ with order 4 and $\a^{B_3}=\hat x_3^{B_3}$ for some
$x_3\in Z_n$ with order 4.
\end{lemma}

\begin{cor}\label{coset 2^d} Let $n=2^d$ for $d\ge 4$ and let $Z_n=\la z\ra$. Suppose that $G=
Z_n\rtimes G_0$ where $G_0\le \Aut(Z_n)=\la (-1)^*\ra\times \la
5^*\ra$. Let $\a\in \la 5^*\ra$ be of order $4$.   Then

\begin{enumerate}\item the coset $z+Z_4\subseteq z^{G_0}$ if and only if $\a\in G_0$
where $Z_4\le Z_n$ is the subgroup of order 4.
\item the coset $z+Z_2\subseteq z^{G_0}$ if and only if $\a^2\in G_0$
where $Z_2\le Z_n$ is the subgroup of order 2.
\end{enumerate}
\end{cor}

\begin{proof}
By Lemma \ref{a-action}, we have that $z+Z_4\subseteq z^{G_0}$ if
$\a\in G_0$ and $z+Z_2\subseteq z^{G_0}$ if $\a^2\in G_0$.

With the notation in Lemma \ref{a-action}, suppose that
$z+Z_4\subseteq z^{G_0}$. Note that $z\in B_1$ or $B_3$ and
$z+Z_4\subseteq B_1$ or $B_3$ respectively.
  Since $(-1)^*$
interchanges $B_1$ and $B_3$, it is easy to deduce that $\a\in G_0$.
Similarly, if $z+Z_2\subseteq z^{G_0}$ then  $\a^2\in G_0$.
\end{proof}
\begin{prop}\label{2^dnot} With above notation, let $G=Z_n\rtimes G_0\le Z_n\rtimes
 \Aut(Z_n)$ where $n=2^{d}$ for $d\ge 4$. If
$\a=(5^*)^{2^{d-4}}\in G_0$, then  $(Z_{n}\rtimes \la\a\ra)^{(2)}\nleq Z_n\rtimes
 \Aut(Z_n)$. In particular, $G$ is not 2-closed on $Z_n$.
\end{prop}

\begin{proof}Let $1\ne \b\in
Sym(Z_{2^d})$ such that $\b$ fixes $B_0, B_2, B_3$ pointwise and
$\b^{B_1}=\widehat{(2^{d-1})}^{B_1}$ is of order 2. Therefore
$\b^{B_1}=(\a^2)^{B_1}$.
  We will show next that  $\b\in (Z_{2^d}\rtimes \la\a\ra)^{(2)}\le G^{(2)}$.

Take any pair $(y_1,y_2)\in Z_n\times  Z_n$.   If both $y_1$ and
$y_2$ belong to $B_1$,   then  $(y_1, y_2)^{\b}=(y_1,
y_2)^{\widehat{2^{d-1}} }$ is in the orbital $(y_1, y_2)^G$. Suppose
next that exactly one of $\{y_1, y_2\}$ belongs to $B_1$, say
$y_2\in B_1$. By Lemma \ref{a-action},
$(y_1,y_2)^{\b}=(y_1,y_2)^{\a^2}$ if $y_1\in B_0$ or $B_2$. Let
$G_3$ be the point stabilizer of point
 $3$, then  $G_3$ is the conjugate of $G_0$
by $\hat 3\in \hat Z_n$. Let   $\a_3$ be the corresponding conjugate
of $\a$ in $G_3$, it follows from Lemma \ref{a-action} that
 $(y_1,y_2)^{\b}=(y_1,y_2)^{\a_3}$ if $y_1\in B_3$. Thus  $\b\in (Z_{2^d}\rtimes \la\a\ra)^{(2)}\le G^{(2)}$. However since $\b$ fixes $B_0$ and $B_3$
pointwise, $\b\notin Z_{2^d}\rtimes
\Aut(Z_{2^d})$ and so $(Z_{2^d}\rtimes \la\a\ra)^{(2)}\nleq Z_{2^d}\rtimes
\Aut(Z_{2^d})$.  In particular $G$ is not 2-closed.
\end{proof}

Next we will show that if $\a\notin G_0$ then $G$ is 2-closed.
 Note that $\a\notin G_0$ is equivalent to the condition that either $|G_0|\le 2$  or $|G_0|=4 $ and $ G_0 \nleqslant  \la  5 ^* \ra$.

We first discuss the case that
$\a^2\notin G_0$.
\begin{lemma}\label{normal2^d} With above notation, let $n=2^d$
for $d\ge 4$. Let $G=Z_n\rtimes G_0$. Suppose $\a^2\notin G_0$.  Then $G$ is the full automorphism group of an
arc-transitive circulant and so $G$ is 2-closed.
\end{lemma}
\begin{proof} Let $S=1^{G_0}$ be the orbit of $1$ under $G_0$, and  let
 $\G=\Cay(Z_n, S)$.
 Since $\a^2\notin G_0$, it follows from corollary
\ref{coset 2^d} that $S$ is not a union of cosets of any subgroup of
$Z_n$. By Theorem \ref{PXthm},
 $\Aut(\G)=G$ as required.
\end{proof}





It remains to show that if $G=Z_n\rtimes G_0$ where $\a^2\in G_0$
but $\a\notin G_0$ then $G$ is the full automorphism
group of some circulant. We will prove this in   Proposition
\ref{case2,3} when we handle the more general case.




\subsection{The general case.}
\subsubsection{The notation for the main theorem.}\label{notation}
We explain  Conditions \ref{condition} in more detail first.

Let $$n=2^{d_1}p_2^{d_2}\cdots p_t^{d_t}, \ \ d_1\ge 0,\
d_2,...,d_t\ge 1, \ t\ge 1$$   where $p_2,...,p_t$ are distinct odd
primes. For convenience, we also write $p_1=2$.  In addition, the
notion $p_i^{d_i}||n$ means $p_i^{d_i}|n$ but $p_i^{d_i+1}\nmid n$.

Let $G=\hat Z_n\rtimes G_0$ acting on $Z_n$ naturally where $G_0\le
\Aut(Z_n)$. In order to reduce the proof in the general case to the
prime power case, we choose the product action form to describe $G$.
Let   $Z_{m}$ be the unique subgroup of $Z_n$ of order $m$ for
$m|n$. Then we may write
$$Z_n=Z_{2^{d_1}}\times Z_{p_2^{d_2}}\times \cdots\times
Z_{p_t^{d_t}}=\{(z_1,...,z_t)=z_1z_2\cdots z_t | z_i\in
Z_{p_i^{d_i}}, \mbox{ where }p_1=2\}.$$ For any $g=(g_1,...,g_t)\in
Z_n$,   we have $ \hat g: (z_1,...,z_t)\mapsto (z_1g_1,...,z_tg_t)$.
Moreover,
$$\Aut(Z_n)=
\Aut(Z_{2^{d_1}})\times\cdots\times\Aut(Z_{p_t^{d_t}})=D_1D_2\cdots
D_t,$$where $D_i$ is the direct factor subgroup of $\Aut(Z_n)$ that
fixes each component of the elements of $Z_n$ except for the $i$-th
component. So $D_i\cong\Aut(Z_{p_i^{d_i}})$.

In fact $D_i$ induces a faithful action on the subgroup $Z_{p_i^{d_i}}$.
  With  notation in $\S$\ref{section 2^d},  if $d_1\ge 3$ then the induced action $D_1$ on the subgroup $Z_{2^{d_1}}$ is permutation isomorphic to $\la (-1)^*\ra\times \la 5^*\ra (d_1\ge 3)$, the multiplicative group
 of units of the ring  $\Z_{2^{d_1}}$ acting on the additive group $\Z_{2^{d_1}}$.
   Let $ \phi:   \la(-1)^*\ra\times \la (5)^*\ra\to D_1$  be the corresponding group
isomorphism.

The normalizer of $\hat Z_n$ in $\Sym(Z_n)$ is
$$N=\hat Z_n\rtimes \Aut(Z_n)=(\hat Z_{2^{d_1}}\rtimes
\Aut(Z_{2^{d_1}})) \times \cdots \times (\hat Z_{p_t^{d_t}}\rtimes
\Aut(Z_{p_t^{d_t}}))$$ acting on $Z_n$ by the natural product
action.
Therefore $G=\hat Z_n\rtimes G_0\le N$ has the natural product
action.

We need the following two easy observations in the proof below.

\noindent(1) Note that when $i\ge 2$, $\Aut(Z_{p_i^{d_i}})=Z_{p-1}\times Z_{p_i ^{d_i-1}}$. Conditions \ref{condition} [ii] is equivalent to $\a_i\notin G_0$
where $\a_i \in D_i\cong Z_{p_i-1} \times Z_{p_i^{d_i-1}}$ is of
order $p_i$.

\noindent(2) When $i=1$ and $d_1\ge 4$,  denote $\a_1= \phi((5^*)^{2^{d_1-4}})\in D_1$, then the order of $\a_1$ is 4. Conditions \ref{condition} [iv] is equivalent to
$ \a_1 \notin  G_0$.


\subsubsection{The proof of Theorem \ref{mainthm}.}\label{proof}



\begin{lemma}\label{not2-closed} With notation in Subsection \ref{notation},
  suppose $G=\hat Z_n\rtimes G_0$  where $G_0\le \Aut(Z_n)$.
  If $G_0$ fails to  satisfy one of
conditions \ref{condition}, then $G$ is not 2-closed.
\end{lemma}

\begin{proof} If
 condition (i)  does not hold, then there exists an odd prime
$p_i\ge 5$ where $i\ge 2$ such that  $p_i||n$ and $D_i\le G_0$.   In
this case we take $K=\hat Z_{p_i}\rtimes D_i$. By hypothesis, $K$
  is the subgroup of $G$ which fixes each component of elements of
$Z_n$ except for the $i$-th component.  Hence the action of $K$ on
$Z_n$ is  the product action of $\bar K \times \{1\}$ on $Z_n=
Z_{p_i}\times Z_{\frac{n}{p_i}}$ where $\bar K\cong K$ acts on
$Z_{p_i}$ naturally. It follows from Theorem \ref{product action}
that   $K^{(2)}=(\ov K)^{(2)}\times \{1\}$. By the remark after
Corollary \ref{2closeddegreep}, $(\ov K)^{(2)}\nsubseteq
Z_{p_i}\rtimes \Aut(Z_{p_i})$.
 Since $G^{(2)}\ge K^{(2)}$, we have that
$G$ is not 2-closed in this case.




If condition (ii) does not hold, then there exists an odd prime
$p_i$ where $i\ge 2$ such that $p_i^{d_i}||n$ and $d_i\ge2$. Since
$\a_i\in G_0$ in this case, we take  $K=\hat Z_{p_i^{d_i}}\rtimes
\la \a_i\ra\le G$. Hence the action of $K$ on $Z_n$ is  the product
action of $\bar K \times \{1\}$ on $Z_n= Z_{p_i^{d_i}}\times
Z_{\frac{n}{p_i^{d_i}}}$ where $\bar K\cong K$ acts on
$Z_{p_i^{d_i}}$ naturally.  By the remark after
  Proposition \ref{p^d}, $(\ov
K)^{(2)}\nsubseteq Z_{p_i^{d_i}}\rtimes \Aut(Z_{p_i^{d_i}})$. The
same argument as above proves that $G$ is not 2-closed in this case
either.

 Suppose $2^{d_1}||n$ and $d_1\ge 3$, suppose also that either
condition (iii) or (iv)  fails. Take $K=\widehat{Z_{8}}\rtimes D_1$
if $d_1=3$ and take $K=\widehat{Z_{2^{d_1}}}\rtimes \la \a_1\ra$ if
$d_1\ge 4$. By the same argument as above, it follows from
Proposition \ref{n=8}(1) and Proposition \ref{2^dnot} that $G$ is
not 2-closed.
\end{proof}


\begin{lemma}\label{arctran} With notation in Subsection \ref{notation},
suppose $G=\hat Z_n\rtimes G_0$  where $G_0\le \Aut(Z_n)$ and $G_0$
satisfies   Conditions   \ref{condition}. Let $S=z^{G_0}$ where
$Z_n=\la z\ra$, and let  $\G=\Cay(Z_n, S)$.
   Then exactly one of the following holds.
\begin{enumerate}
\item $G$ is the full automorphism group of $\G$ and so $G$ is
2-closed and can be represented by an arc-transitive circulant.

\item $2^{d_1}|| n$, $d_1\ge 4$, and $\a_1^2\in G_0\cap D_1$.

\item $2^3|| n$, and $D_1\cap G_0=\la \phi(5^*)\ra\cong Z_2$.

\item $n=4m$ where $m>1$ is odd. $D_1\cap G_0=D_1\cong Z_2$, that is
 $G_0=\Aut(Z_4)\times
K$ where $K\le \Aut(Z_m)$.
\end{enumerate}
Moreover, in the latter three cases, $\G=\si[\ov K_2]$ is a
lexicographic product and the pointwise stabilizer of $\{1,z\}$ in
$\Aut(\G)$ preserves each coset of $Z_2$.
\end{lemma}

\begin{proof} Suppose that $G$ is not the full
automorphism group of $\G$.  By the condition (i), for any odd prime
$p_i\ge 5$ such that $p_i||n$,  we have $G_0\ne \Aut(Z_{p_i})\times
H$ for some $H\le \Aut(Z_{n/p_i})$. It then follows from Theorem
\ref{PXthm} that case (b) of Theorem \ref{PXthm} occurs for $\G$.
That is $n=bk>4$ where $b\ge 2$ and $\G=\si[\ov K_b]$. Moreover, the
group $Z_n$ has a subgroup $Y$ of order $b$ and $S$ is a union of
$Y$-cosets each consisting of generators for $Z_n$.

Recall that $n=2^{d_1}p_2^{d_2}\cdots p_t^{d_t}$. Suppose that
$p_j|b$ for some $j\in\{1,...,t\}$. Then $zZ_{p_j}\subseteq S$ where
$Z_{p_j}$ is the subgroup of order $p_j$ and $d_j\ge 2$ by Theorem
\ref{PXthm} (b). Let $z=(z_1,...,z_t)$ where $z_i$ is a generator of
$Z_{p_i^{d_i}}$ for each $i$. Thus $zZ_{p_j}\subseteq S=z^{G_0}$
implies that $z_jZ_{p_j}\subseteq z_j^{D_j\cap G_0}$ in the $j$-th
component. By Corollary \ref{zZ_p},   the condition (ii) implies
that $b=2^l$ is a power of 2. Similarly, by Corollary \ref{coset
2^d}, Lemma \ref{coset 8} and the action of  $\Aut(Z_4)$,  the
condition (iii) and (iv) imply that $b$ must be 2 and one of cases
2-4 happens.

Suppose next that one of cases 2-4 occurs. Thus $\G=\si[\ov K_2]$
where
 $\si=\Cay(Z_n/Z_2, \ov S)$ and $\ov S=\{sZ_2|s\in S\}$. Moreover, by
 \cite[Lemma 2.3]{PX}, the set $\{xZ_2|x\in Z_n\}$ forms a block
 system of $\Aut(\G)$, and so $\Aut(\G)=Z_2\wr \Aut(\si)$.

 Let $\ov G_0=G_0/\la \a_1^2\ra$ in case 2, and
 let $\ov G_0=G_0/(D_1\cap G_0)$ in case 3 or 4. Then
  $\ov G_0\le \Aut(Z_n/Z_2)$  and $\ov
S=(zZ_2)^{\ov G_0}$. Note that $G_0$ satisfies Conditions
\ref{condition}, it follows  that $\ov S$ is not the union of
cosets of any subgroup of $Z_n/Z_2$. By Theorem \ref{PXthm}, $\si$
is normal and $\Aut(\si)=(Z_n/Z_2)\rtimes \ov G_0$. Therefore the
pointwise stabilizer of $\{1,z\}$ in $\Aut(\G)$ preserves each coset
of $Z_2$.\end{proof}

 \noindent\textbf{Remark:} Suppose $G$ satisfies Conditions
  \ref{condition}. By the
 above lemma,
$G$ can be represented by an arc-transitive circulant if and only if
$G$ does not arise in any of the cases 2-4 of Lemma \ref{arctran}.

\bigskip

  Next we will show that if one of cases 2-4 occurs then there exists a circulant $\G$ which is not
arc-transitive such that $\Aut(\G)=G$. We discuss case 4   first.
\begin{lemma} \label{4m} Suppose $n=4m$ where   $m>1$ is odd
 and   $G=Z_{4m}\rtimes G_0$ where $G_0=\Aut(Z_4)\times
K$ and $K\le \Aut(Z_m)$. Suppose further that $G_0$ satisfies
Conditions    \ref{condition}. Then $G$ is 2-closed and can be
represented by a circulant.
\end{lemma}

\begin{proof} Let $z=z_1z_2\in Z_{4m}$ where $z_1$ is a generator of
$Z_{4}$ and $z_2$ is a generator of $Z_m$. Let $S_1=z^{G_0}$ and
$\G_1=\Cay(Z_{4m}, S_1)$. By Lemma \ref{arctran}, $S_1$ is the union
of some cosets of $Z_2=\la z_1^2\ra$. Let   $S_2=z_2^{G_0}\subseteq
Z_m$ and $\G_2=\Cay(Z_{4m},S_2)$. Thus $B_0=Z_m, B_1=zZ_m,
B_2=z^2Z_m$ and $B_3=z^3Z_m$ are the connected components of $\G_2$.

Let $S=S_1\cup S_2$ and $\G=\Cay(Z_{4m}, S)$.   Suppose first that
$\G$ is   arc-transitive. Note that $S_1$ consists of elements of
order $4m$ and $S_2$ contains elements of order $m$. We observe
that $S$ is not the union of cosets of any subgroup. By
\cite[Theorem 1.3]{li arc circ}, $\G=\si[\ov K_b]-b.\si$ where
$n=br$, $4\leq b<n$ and $\gcd(b,r)=1$.  Thus writing $Z_n=Y\times M$
with  $Y\cong Z_b$  and $M\cong Z_r$, we have that
$S=Y\setminus\{1\}\times T$ and $T\subseteq M\setminus\{1\}$.
Analyzing the orders of elements of $S$, we have that $b=p_i$ is
prime, $p_i\ge 5$ and $ p_i||m$ as $(b,r)=1$. As $z^{G_0}\subset
Y\setminus\{1\}\times T$, $D_i\cong \Aut(Z_{p_i})\subseteq G_0$,
  contradicting the
condition (i). Thus $\G$ is not arc-transitive.

Let $P$ be the point stabilizer  of $\Aut(\G)$ on vertex $1$. Since
$P\ge G_0$, $P$ has two orbits $S_1$ and $S_2$ and so
$\Aut(\G)=\Aut(\G_1)\cap \Aut(\G_2)$


Assume that $g\in \Aut(\G)$ fixing $1\in B_0$ and $z\in B_1$.
Consider $z^2\in B_2\cap zS_1$ which is adjacent to $z$. It follows
from Lemma \ref{arctran} that $g$ fixes each coset of $Z_2=\la
z_1^2\ra$. Hence $(z^2)^g\in \{ z^2, z^2z_1^2\}=z^2Z_2$ and $g$
fixes both $z\in B_1$ and $zz_1^2\in B_3$. Moreover, as $g\in
\Aut(\G_2)$, we conclude that $g$ must fix $B_0$, $B_1$, $B_2$ and
$B_3$ setwise. Therefore, $g$ fixes $z^2$. Continuing in this
fashion, we conclude that $g$ fixes $z^3, z^4,...$ and so on. Thus
$g=1$ and $P=G_0$. It follows that $\Aut(\G)=G$ as required.
\end{proof}

It remains to handle case 2 and case 3 in Lemma \ref{arctran}. By
Lemma \ref{arctran}, we may suppose that $8|n$ and $G=\hat
Z_n\rtimes G_0$  where $G_0\le \Aut(Z_n)$. Let $S_1=z^{G_0}$ where
$Z_n=\la z\ra$ and let $S_2=(z^2)^{G_0}\subseteq Z_{n/2}=\la
z^2\ra$. We construct $\G=\Cay(Z_n, S_1\cup S_2)$. We will show that
$\G$ can represent $G$ in both case 2 and case 3. In order for
proving this, let $\G_1=\Cay(Z_n,S_1)$ and $\G_2=\Cay(Z_n,S_2)$ we
need to study $\G_1$ and $\G_2$.  Note that $\G_1$   has been
studied in Lemma \ref{arctran}. We
  study $\G_2$  in the following lemma.


\begin{lemma}\label{G_2} Suppose that case 2 or 3 of Lemma \ref{arctran}
occurs. With above notation,   we have that $\G_2=2. \Cay(\la
z^2\ra, S_2)$. Let $A_3=\Aut(\Cay (\la z^2\ra, S_2))$ and
$A_2=\Aut(\G_2)$. Then $A_2=A_3\wr Z_2$. Moreover,
 $\Cay(\la z^2\ra, S_2)$ is a normal arc-transitive circulant and
$A_3=\la z^2\ra\rtimes G_0^{\la z^2\ra}$.
\end{lemma}
\begin{proof}
Let $\D_1=\la z^2\ra$ and $\D_2=z\la z^2\ra$. Then $\G_2=2. \Cay(\la
z^2\ra, S_2)$ such that $\D_1$ and $\D_2$ are  two connected
components of $\G_2$. Thus $A_2=A_3\wr Z_2$.

 Let $\ov
G_0=G_0/\la \a_1^2\ra$ in case 2, and let $\ov G_0=G_0/(D_1\cap
G_0)$ in case 3. Note that $G_0$ preserves $\D_1$,   it is easy to
check that the induced permutation group $G_0^{\D_1}\cong \ov G_0$
and $G_0^{\D_1}\le \Aut(\la z^2\ra)$. Also $S_2=(z^2)^{G_0^{\D_1}}$
is an orbit of $G_0^{\D_1}$. Since $G_0$ satisfies conditions in
Theorem \ref{mainthm}, $S_2$ is not the union of cosets of any
subgroup of $\la z^2\ra$. By Theorem \ref{PXthm} and Conditions
\ref{condition}, we conclude that $\Cay(\la z^2\ra, S_2)$ is normal
and $\Aut(\Cay (\la z^2\ra, S_2))=\la z^ 2\ra\rtimes G_0^{\D_1}$.
\end{proof}





\begin{prop}\label{case2,3} With   notation in Subsection
\ref{notation}, suppose $G=\hat Z_n\rtimes G_0$
 where $G_0\le \Aut(Z_n)$ and $G_0$ satisfies
Conditions     \ref{condition}. Suppose further that case 2 or 3 of
Lemma \ref{arctran} occurs. Let $S_1=z^{G_0}$ where $Z_n=\la z\ra$
and let $S_2=(z^2)^{G_0}$. Let $\G=\Cay(Z_n, S_1\cup S_2)$ and let
$P$ be the point stabilizer of vertex 1 in $\Aut(\G)$. Then
\begin{enumerate}
\item  $\G$ is not arc-transitive, and   $S_1,
S_2$ are two orbits of   $P$.

\item  For any $g\in \Aut(\G)$ such that $g$ fixes 1 and $z$, we
have that $g=1$.
\item $\Aut(\G)=G=Z_n\rtimes G_0$. So $\G$ is normal and $G$ is 2-closed.
\end{enumerate}
\end{prop}
\begin{proof}

(1) Suppose, to the contrary, that  $\G$ is arc-transitive. Note
that $S_1$ consists of elements of order $n$ and $S_2$ contains
elements of order $n/2\ne n$. Also observe that $S$ is not the union
of cosets of any subgroup. By \cite[Theorem 1.3]{li arc circ},
$\G=\si[\ov K_b]-b.\si$, where $n=br$, $4\leq b<n$ and
$\gcd(b,r)=1$.  Thus writing $Z_n=Y\times M$ with  $Y\cong Z_b$  and
$M\cong Z_r$, we have that $S=Y\setminus\{1\}\times T$ and
$T\subseteq M\setminus\{1\}$. Analyzing the orders of elements of $S$,  by conditions (i)(ii) we have that $b=4$.
 As $(b,r)=1$,   $ 4||n$,  contradicting the fact that $8|n$.
  Thus $\G$ is not arc-transitive.
As $P\ge G_0$, $S_1, S_2$ are two orbits of   $P$.





 (2) Let $\G_1=\Cay(Z_n,S_1)$, $\G_2=\Cay(Z_n,S_2)$ and
$A_1=\Aut(\G_1)$, $A_2=\Aut(\G_2)$. It follows from (1) that
$\Aut(\G)=A_1\cap A_2$.


  Let $g\in \Aut(\G)$ such that $g$ fixes 1 and $z$.
By Lemma \ref{arctran}, $g$ preserves each coset of $Z_2$ and so
$(z^2)^g\in\{ z^2, z^2z^{n/2}\}$.   Moreover, since $z^2\in S_2$ and
$g$ preserves $S_2$, we have $(z^2)^g\in S_2$. By the proof of Lemma
\ref{G_2}, we have that $z^2Z_2\nsubseteq S_2$ and so
$z^2z^{n/2}\notin S_2$. Thus $g$ fixes $z^2$. Let $\D_1=\la z^2\ra$
and $\D_2=z\la z^2\ra$ be two connected components of $\G_2$. By
Lemma \ref{G_2}, $g^{\D_1}\in \Aut(\la z^2\ra)$ fixes $\D_1$
pointwise. Now $g$ fixes $z$ and $z^2$ and consider $(z^3)^g$. Using
the same argument we deduce that $g$ fixes $\D_2$ pointwise and so
$g=1$.

(3)  It follows from (2) that
  $P=G_0$ and so $A=G=\hat Z_{n}\rtimes G_0$.
Therefore $\G$ is normal and $G$ is 2-closed on $Z_n$.
\end{proof}


Theorem \ref{mainthm} now follows from Lemma \ref{not2-closed},
Lemma \ref{arctran}, Lemma \ref{4m} and Proposition \ref{case2,3}.


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