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% If needed, include a line break (\\) at an appropriate place in the title.
\title{Semiregular automorphisms in vertex-transitive graphs of order $3p^2$}

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\author{Dragan Maru\v si\v c\thanks{This work is supported in part by the Slovenian Research Agency (I0-0035, research program P1-0285 and research projects N1-0032, N1-0038, N1-0062,  and J1-6720), and in part by H2020 Teaming InnoRenew CoE
(grant no. 739574).}\\
\small University of Primorska\\[-0.8ex]
\small UP FAMNIT $\&$ UP IAM\\[-0.8ex] 
\small Glagolja\v ska 8, Koper, Slovenia\\
\small IMFM\\[-0.8ex]
\small Jadranska 19, Slovenia\\
\small\tt dragan.marusic@upr.si\\
}

\begin{document}

\maketitle

% E-JC papers must include an abstract. The abstract should consist of a
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\begin{abstract}
It has been conjectured that automorphism groups of vertex-transitive (di)graphs, and more generally $2$-closures of transitive permutation groups, must necessarily possess a fixed-point-free element of prime order, and thus a non-identity element with all orbits of the same length, in other words, a {\em semiregular} element. It is the purpose of this paper to prove that vertex-transitive graphs of order $3p^2$, where $p$ is a prime, contain semiregular automorphisms.
\end{abstract}

It is known that every finite transitive permutation group
contains a fixed-point-free element of prime power order~(see \cite[Theorem 1]{FKS81}), 
but not necessarily a fixed-point-free
element of prime order (which is equivalent to existence of a semiregular element)   \cite{seven,FKS81}.
In 1981 it was asked if every vertex-transitive digraph admits a
semiregular automorphism (see \cite[Problem 2.4]{DM81}). 
The existence of such automorphisms plays an important role in solutions to many 
important open problems in algebraic graph theory, 
such as, for example,  in the classifications of graphs satisfying 
certain prescribed symmetry conditions (see 
\cite{KM09,KMSWX13,RRS17,S08,Z09}). Semiregular automorphisms have
also proved useful  in a long standing
hamiltonicity problem for connected vertex-transitive 
graphs and in a recently explored dichotomy of even/odd automorphisms
(see \cite{BA89,HKM,LL70}).

In 1997 Klin generalized the  semiregularity problem conjecturing that every transitive 
$2$-closed permutation group contains a semiregular element (see \cite{bcc15}) --
the term {\em polycirculant conjecture} is sometimes used for the semiregularity problem
in this wider context. 
(Recall that for   a finite permutation group $G$  on a set $V$ the $2$-{\em closure} $G^{(2)}$
of $G$ is the largest subgroup of the symmetric group $Sym(V)$ containing $G$ and having
the same orbits as $G$ in the induced action on $V\times V$.)
%Both terms will be used throughout the paper, this should cause no confusion. 
The problem
has spurred a lot  of interest in the mathematical community 
producing several partial results. In particular,
Giudici~\cite{MG03} settled the question for  quasiprimitive group actions,
leaving as one of the main open cases graphs admitting solvable group actions (see \cite{M17}).
Furthermore, there have also been a number of papers dealing with semiregularity
problem for vertex-transitive graphs satisfying certain valency
and order restrictions (see, for instance, \cite{seven,CSS06,FKS81, DMMN07,DMMN07a,DM11,MG03,MG07,GX07,KM08AMC,MS98,S14,V15,X08}). 
For example, it is known that 
every $2$-closed group of square-free degree admits semiregular elements (see \cite{DMMN07a}).
As for composite non-square orders the only positive result is, if we disregard prime power orders,  that every 
vertex-transitive graph of order $2p^2$, $p$ a prime, admits semiregular automorphisms
(see \cite{MS98}). It is the object of this paper to prove the existence of semiregular
automorphisms in vertex-transitive graphs of order $3p^2$, where $p$ is a prime.
We hope that this will motivate further  research, leading 
eventually to the solution of the semiregularity  problem in
the case of vertex-transitive (di)graphs of cube-free order.

 
\begin{theorem}
\label{the:main}
A vertex-transitive graph of order $3p^2$, where $p$ is a prime,
admits a semiregular automorphism.
\end{theorem}

Theorem~\ref{the:main} is proved in Section~\ref{sec:main} after
a series of propositions each of which considers vertex-transitive graphs in question
with particular (im)primitivity actions
of their automorphism groups.
A comment is in order. There are two reasons for the restriction to  vertex-transitive graphs
in the main theorem.
First, in the proof of Theorem~\ref{the:main} we use  certain 
results from \cite{M17},  proved  within a restricted setting of vertex-transitive graphs.
The second reason is somewhat more  philosophical  and reflects 
author's personal bias.
If one's  goal is a complete solution of the semiregularity problem, then
rather than worrying over the distinction between the original question 
and its generalization to $2$-closed groups, 
one should primarily aim at advancements for groups which are not quasiprimitive
- say be it solvable or of particular degrees - even if only in the context of vertex-transitive (di)graphs.



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\section{Vertex-transitive graphs of order $3p^2$}
\label{sec:main}
\noindent

 
Let us first recall the concept imprimitive groups.
Given a transitive permutation group $G$ on a set $V$,
we say that a partition $\B$ of $V$ is a $G$-{\em invariant}
if the elements of $G$ permute the parts, that is, {\em blocks} of $\B$, setwise.
If the {\em trivial} partitions $\{V\}$ and $\{\{v\}\mid v \in V\}$ are the only
$G$-invariant partitions of $V$, then $G$ is said to be {\em primitive},
and is said to be {\em imprimitive} otherwise.
In the latter case we shall refer to a corresponding
$G$-invariant partition as to a  {\em complete imprimitivity block system}, in short 
an {\em imprimitivity block system}, of $G$. 
A transitive permutation group is {\em quasiprimitive} if each of its
non-identity normal subgroups is transitive, and is said to be {\em genuinely imprimitive} otherwise. Note that in the latter case there exists an imprimitivity block system of $G$
arising from orbits of an intransitive normal subgroup of $G$.
A vertex-transitive graph is {\em primitive} if its automorphism group is primitive.
Otherwise it is called an {\em imprimitive graph}.

The following proposition, proved by Giudici in \cite{MG03}, implies the existence
of semiregular automophisms in a vertex-transitive graph  in case its automorphism group  
is quasiprimitive.  (A finite transitive permutation group is said to be {\em elusive} 
if it has no semiregular element.)

\begin{proposition}
\label{pro:quasi}
{\rm \cite{MG03}} A 2-closed quasiprimitive group is not elusive.
\end{proposition}

Proposition~\ref{pro:quasi} implies that only those graphs with genuinely imprimitive automorphism groups 
need to be considered. In particular, let $X$ be a vertex-transitive graph of order $3p^2$, 
where $p$ is a prime. We may assume that there exists
an intransitive normal subgroup $N$ of the automorphism group $\Aut(X)$ of $X$. In fact, we may,
without loss of generality, assume that $N$ is a minimal normal subgroup of $\Aut(X)$.
The size of the blocks arising from the orbits of $N$ divides the order of $X$,
and is therefore   $3$, $p$, $p^2$ or $3p$.
The proposition below was recently proved in \cite{M17}, 
where semiregularity in vertex-transitive graphs with a
solvable automorphism group is  considered.
Note, however, that this particular  result  does not require the permutation group to be solvable.
The proposition implies the existence of semiregular automorphisms in $X$ in case the blocks
arising from the orbits of $N$ are of prime size.
 

\begin{proposition}
\label{pro:amc}
{\rm \cite[Corollary~2.2]{M17}} 
Let $G$ be a permutation group acting transitively on a set $V$
and let $M$ be  a minimal normal subgroup  of $G$ having orbits of prime length $q$ on $V$.
Then $G^{(2)}$ contains  a semiregular element of order $q$.
\end{proposition}

The remaining two cases, that is graphs whose automorphism groups
admit  an intransitive normal subgroup giving rise to imprimitivity block system
consisting of blocks of size $p^2$ and $3p$, are considered in Propositions~\ref{pro:p2} and~\ref{pro:3p}.
For the sake of completeness we first state the following classical result which will be
used in the proofs. It implies that in 
a vertex-transitive graph of order $3p^2$, where $p$ is a prime, the orbits of 
a Sylow $p$-subgroup of the automorphism group are of length $p^2$.

\begin{proposition}{\rm \cite[Theorem~3.4]{W64}}
\label{pro:wielandt}
Let $p$ be a prime and let $P$ be a Sylow $p$-subgroup of 
a permutation group $G$ acting on a set $\Omega$.
Let $\omega \in \Omega$. If $p^m$ divides the length of the $G$-orbit containing $\omega$,
then $p^m$ also divides the length of the $P$-orbit containing $\omega$. 
\end{proposition}

Before considering the remaining two cases let us recall a recent result about 
existence of semiregular automorphisms in a vertex-transitive graph with solvable
automorphism group of order $mp^2$, where $m$ satisfies certain conditions.
This result implies that only vertex-transitive graphs of order $3p^2$
with non-solvable automorphism groups need to be considered.

\begin{proposition}
\label{pro:amc2}
{\rm \cite[Theorem~2.4]{M17}} 
Let $X$ be a connected vertex-transitive graph of order $p^2q$, where $p$ and $q$ are primes, 
and either $q\le p$ or $p^2 < q$.
Then either 
\begin{enumerate}[(i)]
\itemsep=0pt
\item $X$ admits a semiregular automorphism, or 
\item $2< q < p$ and $\Aut(X)$ is nonsolvable with
an intransitive non-abelian minimal normal subgroup whose orbits are either of length $p^2$ or of length $pq$.
\end{enumerate}
\end{proposition}


In the next proposition we consider the case
where the blocks of imprimitivity are of size $p^2$.

%
% from here on changes made, Dragan, October 31, 2017
%

\begin{proposition}
\label{pro:p2}
Let $p$ be a prime and let $X$ be a vertex-transitive graph of order $3p^2$, where $p>3$ is a prime, admitting 
an imprimitivity block system consisting of three blocks of size $p^2$ arising from orbits of an intransitive normal 
subgroup $N$ of $\Aut(X)$. Then $X$ admits a semiregular automorphism.
\end{proposition}

\begin{proof}
We may assume that $X$ is connected as otherwise
a semiregular automorphism in $X$ can be easily constructed via semiregular
automorphisms in the connected components.
Namely, if $X$ is disconnected then its connected components
are vertex-transitive graphs of order $3$, $p$, $p^2$ or $3p$, and 
it is well known that such graphs admit semiregular automorphisms.

%In view of Proposition~\ref{pro:amc2} we may also assume that $\Aut(X)$ is not solvable
%and thus there exists $k\ge 1$ such that $N\cong U^k$ where $U$ is a non-abelian simple group.

Let $\B=\{A,B,C\}$ be the imprimitivity block system arising 
from the orbits of $N$, each of length $p^2$.
Clearly, in view of Proposition~\ref{pro:wielandt}, the orbits $A$, $B$ and $C$ coincide with the orbits of 
a Sylow $p$-subgroup $P$ of $\Aut(X)$.
Observe also that there must exist  an automorphism $\pi\in\Aut(X)$ 
which cyclically permutes the three blocks in $\B$. We may, without
loss of generality, assume that $\pi|_\B = (A\, B\, C)$.

The center $Z(P)$ of the Sylow $p$-subgroup $P$ is non-trivial and thus
there exists a central element $\alpha\in Z(P)$ of order $p$. Clearly, 
for each $Y\in\{A,B,C\}$ either $\alpha^Y$ is trivial or $\alpha^Y$  
is semiregular of order $p$.  If $\alpha$ is not semiregular then there are essentially 
only two possibilities that need to  be considered, depending on the number of orbits 
$Y \in \{ A,B,C\}$ for which the restriction $\alpha^Y$ is trivial.

\begin{case}\end{case} $\alpha^A$ is semiregular and $\alpha^B=\alpha^C=1$.

\medskip

\noindent
Then $(\pi\alpha\pi^{-1})^B$ and $(\pi^2\alpha\pi^{-2})^C$ are semiregular, and 
$$
(\pi\alpha\pi^{-1})^A=(\pi\alpha\pi^{-1})^C=(\pi^2\alpha\pi^{-2})^A=(\pi^2\alpha\pi^{-2})^B=1,
$$
implying that $\alpha\cdot \pi\alpha\pi^{-1}\cdot \pi^2\alpha\pi^{-2}=(\alpha\pi)^3\pi^{-3}$
is the desired semiregular automorphism.

\begin{case}\end{case} $\alpha^A=1$, and $\alpha^B$ and $\alpha^C$ are semiregular of order $p$.

\medskip

\noindent
Recall that $\B=\{A,B,C\}$ consists of orbits of  the normal subgroup 
$N$. %$\cong U^k$, where $U$ is a non-abelian simple group.
For each $Y \in \B$ let $K_{(Y)} = Ker(N \rightarrow N^Y)$.
Then $\alpha \in K_{(A)}$, and there exists $\beta \in K_{(B)}$ (without loss of generality
we may assume that $\beta=\alpha^\pi$) such that 
$\beta^A$ and $\beta^C$ are semiregular of order $p$.
Now consider $\alpha\beta$.
Clearly, $(\alpha\beta)^A$ and $(\alpha\beta)^B$ are semiregular of order $p$.
We need to consider the action of $\alpha\beta$ on $C$.
Observe that either the orbits of $\la\alpha\ra$ on $C$ are blocks of imprimitivity for $N^C$ -- as $K_{(A)}$
is normal in $N$ and so $K_{(A)}^C$ is normal in $N^C$ --  or $K_{(A)}$ is transitive on $C$.
In the latter case the bipartite subgraphs $X[A,B]$, $X[B,C]$ and $X[A,C]$
are all isomorphic to  the complete bipartite graph $K_{p^2,p^2}$, and $X$ clearly
admits a semiregular automorphism. 
(For disjoint subsets $U, W$ of the vertex set $V(X)$
the subgraph of $X$ induced by the set $U$ is denoted by $X[ U ]$, and
similarly,  the bipartite subgraph
of $X$ induced by the edges having one endvertex in $U$
and the other endvertex in $W$ is denoted by $X[U,W]$.)
Hence, we may assume that the orbits of $\la\alpha\ra$ 
on $C$ are blocks of imprimitivity for $N^C$. Note that
$(\alpha\beta)^C$
either fixes the orbits of $\alpha$ or cyclically permutes them.
We deal with these two cases in the two subcases below.

\begin{subcase}\end{subcase} 
$(\alpha\beta)^C$ fixes the orbits of $\alpha$.  

\medskip

\noindent
It follows that the orbits of $\alpha$ and $\beta$ on $C$ coincide. 
Denote these orbits by $C_i$, $i\in\ZZ_p$. 
If all of the restrictions $(\alpha\beta)^{C_i}$, $i\in\ZZ_p$, 
are of order $p$ then $\alpha\beta$ is a semiregular automorphism of $X$.
If not, then there exists $ r \in\ZZ_p$ such that the restrictions  $(\alpha\beta)^{C_j}$ are of order $p$
for $j\in\{r+1,\ldots, p-1\}$ and are not of order $p$ for $j\in\{0,1,\ldots, r\}$.
We now define a semiregular automorphism $\sigma$ of $X$ in the following way:
$$
\sigma(u)=\left\{\begin{array}{cl}
\alpha\beta(u)  , & u\in A\cup B\cup C_{r+1}\cup C_{r+2}\cup \ldots \cup C_{p-1}\\
\alpha(u)  , & u\in C_0\cup C_1 \cup \ldots \cup C_{r}
\end{array}\right..
$$
To show that $\sigma$ is indeed an automorphism of $X$ observe first that
the bipartite graph $X[C_i,C_j]$, where $i\in\{0,1,\ldots, r\}$ and $j\in\{r+1,\ldots,p-1\}$,
is either isomorphic to the complete bipartite graph $K_{p,p}$ or is totally disconnected.
Combining this with the fact that $(\alpha\beta)^B=\alpha^B$
we obtain that
$\sigma ^{B\cup C}$ is an automorphism of
the subgraph of $X$ induced on $B\cup C$.
To complete the proof we need to check the edges of the induced bipartite graph $X[A,C]$.
Since for each $j\in\{0,1,\ldots,r\}$ there exists $k_j$ coprime with $p$
such that $((\alpha\beta)^{k_j})^{C_j}=1$ if follows that each of 
the bipartite graphs $X[A_i,C_j]$, where 
$A=\{A_i\mid i\in\ZZ_p\}$ is a partition of $A$ into the orbits of $\beta$
and $j\in\{0,1,\ldots, r\}$,
is either isomorphic to the complete bipartite graph $K_{p,p}$ or is totally disconnected.
It follows that $\sigma$ preserves the edges of $X[A,C]$, and consequently $\sigma$
is an automorphism of $X$.



\begin{subcase}\end{subcase} 
$(\alpha\beta)^C$ cyclically permutes the orbits of $\alpha$.  

\medskip

\noindent
Then either $(\alpha\beta)^C$ is of order $p$ and clearly semiregular, in which case
$\alpha\beta$ is a semiregular automorphism of $X$. Alternatively, $(\alpha\beta)^C$ is of order $p^2$ in which case
$(\alpha\beta)^p$ is trivial on $A\cup B$ and semiregular of order $p$ on $C$.
In this case take $(\alpha\beta)^p\alpha^{\pi^2}$ to get the desired semiregular
automorphism.
\end{proof}

\setcounter{case}{0}

In the next proposition we deal with blocks of size $3p$.

\begin{proposition}
\label{pro:3p}
Let $X$ be a vertex-transitive graph of order $3p^2$, where $p >3$ is a prime,
admitting an imprimitivity block system  consisting of $p$ blocks of size $3p$
arising from orbits of 
an intransitive normal subgroup $N$ of $\Aut(X)$. 
Then $X$ admits a semiregular automorphism.
\end{proposition}

\begin{proof}
We may again assume that $X$ is connected.
Let $\B$ be the imprimitivity block system arising from orbits of $N$.
Let $P$ be a Sylow $p$-subgroup of $\Aut(X)$ with orbits $A$, $B$ and $C$.
Observe that each block in $\B$ intersects each of $A$, $B$ and $C$
in exactly $p$ vertices. 
We have 
$$
\begin{array}{ccc}
A&=&A_0 \cup A_1 \cup \ldots \cup A_{p-1},\\
B&=&B_0 \cup B_1 \cup \ldots \cup B_{p-1},\\
C&=&C_0 \cup C_1 \cup \ldots \cup C_{p-1},
\end{array}
$$
and $\B=\{Y_i \mid i\in\ZZ_p\}$ where $Y_i=A_i\cup B_i \cup C_i$. 

The center $Z(P)$ of $P$ is non-trivial and thus
there exists a central element $\alpha\in Z(P)$ of order $p$
 such that for each $Y\in\{A,B,C\}$ either $\alpha^Y$ is trivial or $\alpha^Y$  
is semiregular of order $p$.  
If $\alpha$ is not semiregular then there are essentially only two possibilities
 depending on the number of orbits $Y \in \{ A,B,C\}$ for which the restriction
$\alpha^Y$ is trivial.

\begin{case}\end{case} $\alpha^A$ is semiregular of order $p$,  and $\alpha^B=\alpha^C=1$ .

\medskip

\noindent
First observe that, since $\B$ is an imprimitivity block system, 
the set of orbits of $\alpha^A$ is equal to
the set $\{A_i\mid i\in\ZZ_p\}$. 
By \cite[Proposition~3.2]{DM85}, every transitive group of degree $p^2$ contains
a regular (abelian) subgroup, and so there 
exists $Q\le P$ such that $Q^B$ is either a cyclic or an elementary abelian
subgroup acting regularly on $B$. Thus we distinguish two subcases. 

\begin{subcase}\end{subcase} $Q^B\cong \ZZ_{p^2}$.

\medskip

\noindent
There exists $\rho\in Q$ of order $p^2$ such that $\rho^B$ 
is also of order $p^2$ and maps 
according to the rule 
$$
\rho\colon B_i\to B_{i+1}, \,\, i\in\ZZ_p.
$$ 
Further, since $A_i\cup B_i \cup C_i$  
are blocks of imprimitivity we have that
$\rho(A_i)=A_{i+1}$  and $\rho(C_i)=C_{i+1}$ for every $i\in\ZZ_p$.
%
Let $e$, $f$, and $g$ denote the respective orders of
$\rho^A$, $\rho^B$, and $\rho^C$.
Then $(e,f,g)$  is one of the following ordered triples:
\begin{center}
$(p^2,p^2,p^2)$, $(p,p^2,p^2)$,  $(p^2,p^2,p)$ or $(p,p^2,p)$.
\end{center}
In the first case $\rho$ is semiregular.
In the second case $\alpha\rho^p$ is semiregular.
In the third and the fourth case, the existence of automorphisms $\alpha$
and $\sigma^p$ implies that each of the bipartite subgraphs $X[A_i,B_j]$, $X[B_i,C_j]$, and $X[A_i,C_j]$  ($i ,j \in \ZZ_p $) is either isomorphic to the  complete bipartite graph $K_{p,p}$ or is totally disconnected.
Consequently, any permutation $\omega$ fixing each of $A_i$, $B_i$ and $C_i$, $i\in \ZZ_p$, set-wise and satisfying 
the property that $\omega^A$, $\omega^B$, and 
$\omega^C$ is, respectively, an automorphism of $X[A]$, $X[B]$, and $X[C]$, 
is in fact an automorphism of $X$.
As in the case of the orbit $B$ and the subgroup $Q \leq P$ there exists a subgroup $R \leq P$ such that $R^C$ is a regular abelian group. Since $C_i$, $ i \in \ZZ_p$,  are the intersections of the blocks $Y_i$, $ i \in \ZZ_p$, with the orbit $C$ there must exist an automorphism $\tau \in R$ fixing these blocks and such that $\tau^C$ is semiregular (and of order $p$ on each $C_i$). We now define $\omega$ as follows
$$
\omega(u)=\left\{\begin{array}{cc}
\alpha(u)  , & u\in A\\
\sigma^p(u)  , & u\in B\\
\tau(u)  , & u\in C
\end{array}\right..
$$
Clearly, $\omega$ is the desired semiregular automorphism of $X$.

\begin{subcase}\end{subcase} $Q^B \cong\ZZ_{p}^2$.

\medskip

\noindent
There exist $\rho,\sigma\in Q$ such that both
$\rho^B$ and $\sigma^B$ are of order $p$. 
Furthermore, we may assume that the orders of
$\rho^C$ and $\sigma^C$ are either $p$ or $1$,
for otherwise an argument analogous to the one used in Subcase 1.~1,
with $B$ replaced by $C$, applies. As for the orders of
$\rho^A$ and $\sigma^A$ they can be $1$, $p$ or $p^2$.

%
% do tu
%

We may assume that $\rho^B$ permutes the sets $B_i$ and that $\sigma^B$ 
fixes the sets $B_i$. Consequently, $\rho^C$ permutes the sets $C_i$,
and so $\rho$ is semiregular on both
$B$ and $C$. If $\rho^A$ is of order $p$ then 
it permutes the sets $A_i$, and so $\rho^A$ is semiregular, and thus
$\rho$ is semiregular.
Hence we may assume that $\rho^A$ is of order $p^2$.
Consider now $\sigma$. Clearly, $\sigma^B$ is semiregular.
If $\sigma^C$ is of order $p$ and semiregular then we are done because
we can construct the desired automorphism $\omega$ as follows:
$$
\omega(u)=\left\{\begin{array}{cc}
\alpha(u)  , & u\in A\\
\sigma(u)  , & u\in B\\
\sigma(u)  , & u\in C
\end{array}\right..
$$
The mapping $\omega$ is an automorphism of $X$
since each of the bipartite subgraphs $X[A_i,B_j]$ and $X[A_i,C_j]$ 
is either isomorphic to  the complete bipartite graph $K_{p,p}$ or is totally disconnected.

Finally, suppose that $\sigma^C$ is not semiregular. 
In this case apart from the bipartite subgraphs $X[A_i,B_j]$ and $X[A_i,C_j]$
also any of the induced bipartite subgraphs $X[B_i,C_j]$
is either isomorphic to  the complete bipartite graph $K_{p,p}$ or is totally disconnected.
Recall that $\sigma^B$ is semiregular with orbits $B_i$.
Analogously, we may assume that
there exits $\tau\in P$ such that $\tau^C$ is semiregular
with orbits $C_i$. Hence the permutation defined by the rule
$$
\omega(u)=\left\{\begin{array}{cc}
\alpha(u)  , & u\in A\\
\sigma(u)  , & u\in B\\
\tau(u)  , & u\in C
\end{array}\right.
$$
is a semiregular automorphism of $X$.

\begin{case}\end{case} $\alpha^A=1$, and $\alpha^B$ and $\alpha^C$ are semiregular of order $p$.

\medskip

\noindent
There exists $Q\le P$ such that $Q^A$ is abelian and regular, and so either cyclic or elementary abelian.
Observe also that the set of orbits of $\alpha^B$ is equal to
the set $\{B_i\mid i\in\ZZ_p\}$ and that the set of orbits of $\alpha^C$ coincides with 
the set $\{C_i\mid i\in\ZZ_p\}$. 

\begin{subcase}\end{subcase} $Q^A\cong \ZZ_{p^2}$.

\medskip

\noindent
Note that non-identity elements of $P$ are all of order $p$ or $p^2$.
There exists $\rho\in Q$ of order $p^2$ such that $\rho^A$ is also of order $p^2$.
Let $\sigma=\rho^p$. Hence $\sigma^A$ is semiregular of order $p$.
We now analyze possibilities for $\sigma^B$ and $\sigma^C$:
they are either trivial or semiregular of order $p$.
%These two are either trivial or semiregular of order $p$.
If $\sigma^B$ and $\sigma^C$ are both semiregular then $\sigma$ is the desired automorphism. 
If $\sigma^B=\sigma^C$ are both trivial then $\sigma\alpha$ is the desired automorphism.
Finally, without loss of generality, assume that $\sigma^B$ is semiregular 
and $\sigma^C=1$.
Then $\la\rho\ra^B\cong \ZZ_{p^2}$ and 
being contained in $\la\alpha,\rho\ra^B$ which is abelian
(since $\alpha \in Z(P)$), it follows that $\la\rho\ra^B$=$\la\alpha,\rho\ra^B$. 
Therefore $\alpha^B\in\la\rho\ra^B$. In particular
$\alpha^B=(\rho^{pj})^B$, for some $j\in\ZZ_{p}^*$.
It follows that $\alpha\rho^{pj}$ is the desired automorphism.

\begin{subcase}\end{subcase} $Q^A\cong \ZZ_{p}^2$.

\medskip

\noindent
There are elements $\sigma, \rho\in P$ such that $\la \sigma,\rho\ra^A\cong \ZZ_p^2$.
Of course, both $\rho^A$ and $\sigma^A$ are semiregular. 
Moreover, since the sets $\{A_i\cup B_i\cup C_i\}$, $i\in\ZZ_p$ are blocks, 
we may assume that $\rho^A$  maps $A_i$ to $A_{i+1}$, and similarly 
$\rho^B$  maps $B_i$ to $B_{i+1}$ and $\rho^C$  maps $C_i$ to $C_{i+1}$,
whereas $\sigma^A$, $\sigma^B$ and $\sigma^C$ fix these sets. In particular,
this means that $\sigma^B$ and $\sigma^C$ fix the orbits of 
$\alpha$ on $B$ and $C$.
Consider now 
the action of the conjugates $\epsilon_k=\rho^{-k}\sigma\rho^k$ ($k\in\ZZ_p$)
on $B$ and $C$. Clearly, $\epsilon_k^A=\sigma^A$.
If $\sigma$ is semiregular on $B$ and $C$ then we are done.
If $\sigma$ is not semiregular on $B$ then there exists $B_i$ such that
$\sigma^{B_i}=1$. Consequently, every bipartite subgraph
$X[A_j,B_i]$, $i,j\in\ZZ_p$,
is either isomorphic to  the complete bipartite graph $K_{p,p}$ or is totally disconnected.
Applying the automorphisms $\epsilon_k$ we see that each of the bipartite subgraphs $X[A_j,B_k]$
is isomorphic to  the complete bipartite graph $K_{p,p}$ or is totally disconnected.
An analogous argument holds for the case when $\sigma$ is not semiregular on $C$, implying that
$X[A_j,C_k]$
is either isomorphic to  the complete bipartite graph $K_{p,p}$ or is totally disconnected.
Then the permutation 
$\omega$ mapping according to the rule:
$$
\omega(u)=\left\{\begin{array}{cc}
\sigma(u) , & u\in A\\
\alpha(u)  , & u\in B\\
\alpha(u)  , & u\in C
\end{array}\right.
$$
is a semiregular automorphism of $X$.
We are now left with the case where $\sigma$ is semiregular 
on one of the two orbits $B$ and $C$ and not semiregular on the other.
Without loss of generality we assume that $\sigma^B$ is semiregular
and $\sigma^C$ is not semiregular.
Then applying the same argument as above it follows that
each of the bipartite subgraphs $X[A_j,C_i]$ and $X[B_j,C_i]$, $j\in\ZZ_p$,
is either isomorphic to  the complete bipartite graph $K_{p,p}$ or is totally disconnected.
Applying then the automorphisms $\epsilon_k$
it follows that the same holds for all of the subgraphs 
$X[A_j,C_k]$ and $X[B_j,C_k]$, $j,k\in\ZZ_p$.
Then the permutation 
$\omega$ mapping according to the rule:
$$
\omega(u)=\left\{\begin{array}{cc}
\sigma(u)  , & u\in A\\
\sigma(u)  , & u\in B\\
\alpha(u)  , & u\in C
\end{array}\right.
$$
is the desired semiregular automorphism of $X$.
\end{proof}

We are now ready to prove Theorem~\ref{the:main}.

\bigskip

\noindent
\begin{proofT}
Let $X$ be a vertex-transitive graph of order $3p^2$, 
where $p$ is a prime, and let $\Aut(X)$ be its automorphism group.
If $p\in\{2,3\}$ then $X$ is of order $12$ or $3^3$, and 
the existence of semiregular automorphisms follows from the fact that $X$ is a Cayley
graph in both of these two cases (see \cite{DM85}). 
We may therefore assume that $p>3$.

If $\Aut(X)$ is quasiprimitive then Proposition~\ref{pro:quasi} implies 
the existence of semiregular automorphisms in $\Aut(X)$.
We may thus assume that $\Aut(X)$ is genuinely imprimitive.
Let $N$ be an intransitive minimal normal subgroup of $\Aut(X)$,
and let $\B$ be an $\Aut(X)$-invariant partition of $V(X)$ arising
from the orbits of $N$. Then the blocks in $\B$ are of size $3$, $p$, $p^2$ or $3p$.
If the blocks in $\B$ are of prime size then
the existence of semiregular automorphisms is assured by Proposition~\ref{pro:amc}.
If the blocks in $\B$ are of prime-squared size then
the existence of semiregular automorphisms follows from
Proposition~\ref{pro:p2}. If, however, the blocks in $\B$ are of 
size $3p$ then semiregular automorphisms in $\Aut(X)$ exist
by Proposition~\ref{pro:3p}. 
 \hfill $\square$
\end{proofT} 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{Acknowledgements}

The author wishes to thank  Raffaele Scapellato 
for   conversations about the material of this paper
and the  referee for helpful comments.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% You do not have to use the same format for your references, but 
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\end{document}
