#  The connected generalized Cayley graphs††thanks: This research was
supported by NSFC (No. 12071484), Hunan Provincial Natural Science Foundation
(2020JJ4675). E-mail addresses: liaoqianfen@163.com(Q. Liao),
wjliu6210@126.com(W. Liu, corresponding author).

Qianfen Liao1, Weijun Liu1,2†
1 School of Mathematics and Statistics, Central South University
New Campus, Changsha, Hunan, 410083, P.R. China.
2 College of General Education, Guangdong University of Science and
Technology,
Dongguan, Guangdong, 523083, P.R. China

###### Abstract

In this paper, we provide the sufficient and necessary conditions for
generalized Cayley graphs to be connected and bipartite, respectively. As a
consequence, we determine the groups whose all generalized Cayley cubic graphs
are connected and integral.

Key words: generalized Cayley graph; connected graph; bipartite graph.

##  1 Introduction

The earliest result about generalized Cayley graphs comes from D. Marušič et
al. [9], who introduced the definition of generalized Cayley graphs as below
and provided some examples of non-Cayley generalized Cayley graphs.

######  Definition 1.1.

Given a group $G$, for $S\subseteq G$ and $\alpha\in{\rm\textsf{Aut}}(G)$, if
they satisfy several conditions below:

  1. (a)

$\alpha^{2}={\rm\textsf{id}}$, where id is the identity of
${\rm\textsf{Aut}}(G)$;

  2. (b)

for any $g\in G$, $\alpha(g^{-1})g\notin S$;

  3. (c)

for $g,h\in G$, if $\alpha(h^{-1})g\in S$, then $\alpha(g^{-1})h\in S$.

Then the graph with vertices $G$ and edges $\\{\\{g,h\\}\mid\alpha(g^{-1})h\in
S\\}$ is denoted by $GC(G,S,\alpha)$. We call $S$ a generalized Cayley subset
of $G$ induced by $\alpha$ and $GC(G,S,\alpha)$ a generalized Cayley graph
with respect to the ordered pair $(S,\alpha)$. Especially, if
$\alpha={\rm\textsf{id}}$, then $GC(G,S,\alpha)$ is the Cayley graph.

After proposing the definition, research on generalized Cayley graphs focused
on finding examples of non-Cayley generalized Cayley graphs [5, 6], and
studying isomorphism problem of generalized Cayley graphs [8, 10, 11, 12].
Recently, Zhu et al. [13] characterize the spectra of generalized Cayley
graphs on finite abelian groups using the connection between Cayley graphs and
the double cover of generalized Cayley graphs. These are all the results about
generalized Cayley graphs of groups that we currently know.

Observe that the Cayley sum graph on abelian group is a special case of
generalize Cayley graph under the inverse automorphism. Inspired the results
of Cayley sum graphs in references [1, 2, 4], we are concerned with the
sufficient and necessary conditions for generalized Cayley graphs to be
connected and bipartite, respectively. In order to differ from Cayley graph,
all the automorphisms in this paper that induce generalized Cayley graphs are
assumed to be some involutions.

An outline of this paper is as follows. We provide the sufficient and
necessary conditions for generalized Cayley graphs of abelian groups to be
bipartite in Section $2$. In Section $3$, the sufficient and necessary
conditions for generalized Cayley graphs to be connected are studied. Based on
the results in Section $3$, we determined the groups whose all generalized
Cayley cubic graphs are connected and integral in Section $4$.

Before we get into the discussion, some useful notations are listed here. Let
$G$ be a finite group and $S$ a generalized Cayley subset of $G$ induced by
the involutory automorphism $\alpha$. With respect to $\alpha$, define subsets
$\omega_{\alpha}(G)=\\{\alpha(g^{-1})g\mid g\in G\\}$,
$\Omega_{\alpha}(G)=\\{g\mid g\in
G\rm\setminus\omega_{\alpha}~{}\text{and}~{}\alpha(g)=g^{-1}\\}$ and
$\mho_{\alpha}(G)=\\{g\mid\alpha(g)\neq g^{-1}\\}$. Clearly, the subsets
$\omega_{\alpha}(G)$, $\Omega_{\alpha}(G)$ and $\mho_{\alpha}(G)$ form a
partition of $G$. In addition, the conditions $(b)$ and $(c)$ in Definition
1.1 are equivalent to $S\cap\omega_{\alpha}(G)=\emptyset$ and
$\alpha(S^{-1})=S$ which indicates that, if $s\in S\cap\mho_{\alpha}(G)$, then
$\alpha(s^{-1})\in S$. Moreover, if the size of $S$ is odd, then
$S\cap\Omega_{\alpha}(G)\neq\emptyset$. For subsets $S_{1}$ and $S_{2}$ of
$G$, let $S_{1}S_{2}=\\{s_{1}s_{2}\mid s_{1}\in S_{1},s_{2}\in S_{2}\\}$. For
graphs $X$ and $Y$, the direct product $X\times Y$ of $X$ and $Y$ is the graph
with vertex set $V(X\times Y)=\\{(x,y)\mid x\in V(X),y\in V(Y)\\}$, and two
vertices $(x_{1},y_{1})$ and $(x_{2},y_{2})$ are adjacent in $X\times Y$ if
and only if $x_{1}$ is adjacent to $x_{2}$ in $X$ and $y_{1}$ is adjacent to
$y_{2}$ in $Y$. For convenience, if $G$ is abelian group and it is not
elementary abelian $2$-group, let $\iota$ be the involutory automorphism
$g\mapsto g^{-1}$ for all $g\in G$.

##  2 Generalized Cayley graphs of abelian groups

In this section, we investigate the conditions for generalized Cayley graphs
of abelian groups to be bipartite. Some basic but important lemmas are
presented.

######  Lemma 2.1.

Let $\alpha$ be an involutory automorphism of abelian group $G$. If
$g_{1}g_{2}\in\omega_{\alpha}(G)$ and $g_{1}\in\omega_{\alpha}(G)$, then
$g_{2}\in\omega_{\alpha}(G)$. Moreover, if $h_{1},h_{2}\in\omega_{\alpha}(G)$,
then $h_{1}^{n}h_{2}^{m}\in\omega_{\alpha}(G)$ for all integers $n$ and $m$.

###### Proof.

Assume $g_{1}g_{2}=\alpha(a_{1}^{-1})a_{1}$ and
$g_{1}=\alpha(a_{2}^{-1})a_{2}$ for some $a_{1},a_{2}\in G$. Then
$g_{2}=\alpha(a_{1}^{-1}a_{2})a_{1}a_{2}^{-1}\in\omega_{\alpha}(G)$. If
$h_{1}=\alpha(b_{1}^{-1})b_{1}$ and $h_{2}=\alpha(b_{2}^{-1})b_{2}$ for some
$b_{1},b_{2}\in G$, then for any integers $n$ and $m$,
$h_{1}^{n}h_{2}^{m}=\alpha(b_{1}^{-n}b_{2}^{-m})b_{1}^{n}b_{2}^{m}\in\omega_{\alpha}(G)$.
∎

######  Lemma 2.2.

If the graph $H$ contains a closed walk of odd length, then $H$ contains an
odd cycle.

###### Proof.

Let $W=v_{0}e_{1}v_{1}e_{2}\cdots v_{k-1}e_{k}v_{0}$ be a shortest odd-length
closed walk of $H$, where $k$ is odd. If there exist vertices $v_{i}$ and
$v_{j}$ such that $i\neq j$ and $v_{i}=v_{j}$. Then we obtain closed walks
$W_{1}=v_{i}e_{i+1}\cdots e_{j}v_{j}$ and $W_{2}=v_{0}e_{1}\cdots
v_{i-1}e_{i}v_{i}e_{j+1}v_{j+1}\cdots v_{k-1}e_{k}v_{0}$. Note that one of the
walks $W_{1}$ and $W_{2}$ has odd length, which contradicts the choice of $W$.
Thus, the vertices in $W$ are all different, which implies $W$ is an odd
cycle. ∎

Using Lemmas 2.1 and 2.2, we obtain the following results. During the proof, a
fact will be frequently used which is for any $s\in S\cap\Omega_{\alpha}(G)$,
we have $s^{2}=\alpha(s^{-1})s\in\omega_{\alpha}(G)$ and further
$s^{2k}\in\omega_{\alpha}(G)$ for any integer $k$.

######  Theorem 2.3.

Let $G$ be a finite abelian group and $S$ a generalized Cayley subset of $G$
induced by the involutory automorphism $\alpha$. If
$S\subseteq\Omega_{\alpha}$, then $GC(G,S,\alpha)$ is not bipartite if and
only if for each $s\in S$, there exists non-negative integer $k_{s}$ such that
$\prod_{s\in S}s^{k_{s}}\in\omega_{\alpha}(G)$ and $\Sigma_{s\in S}k_{s}$ is
odd.

###### Proof.

Assume $GC(G,S,\alpha)$ is not bipartite, then it contains an odd cycle
$C_{2k+1}=(x_{0},x_{1},\cdots,x_{2k},x_{2k+1}=x_{0})$. For each $s\in S$, let
$k_{s}$ denote the number of indices $j\in\\{0,1,\cdots,2k\\}$ such that
$\alpha(x_{j}^{-1})x_{j+1}=s$. Then we have

$\prod_{s\in
S}s^{k_{s}}=\prod_{j=0}^{2k}\alpha(x_{j}^{-1})x_{j+1}=\alpha\left(\left(\prod_{j=0}^{2k}x_{i}\right)^{-1}\right)\left(\prod_{j=0}^{2k}x_{i}\right)\in\omega_{\alpha}(G).$

Moreover, $\Sigma_{s\in S}k_{s}=2k+1$ is odd.

Conversely, suppose that for each $s\in S$, there exists non-negative integer
$k_{s}$ such that $\prod_{s\in S}s^{k_{s}}\in\omega_{\alpha}(G)$ and
$\Sigma_{s\in S}k_{s}$ is odd. Let $\ell_{s}=k_{s}(mod~{}2)$, then
$\ell_{s}\in\\{0,1\\}$ and $\Sigma_{s\in S}\ell_{s}=t$ is odd. Let $A=\\{a\in
S:\ell_{a}\neq 0\\}=\\{a_{0},a_{1},\cdots,a_{t-1}\\}$. Since $k_{s}$ and
$k_{a}-1$ are even for any $s\in S\setminus A$ and $a\in A$, we have
$\prod_{s\in S\setminus A}s^{k_{s}}\prod_{a\in
A}a^{k_{a}-1}\in\omega_{\alpha}(G)$. Note that

$\prod_{s\in S}s^{k_{s}}=\left(\prod_{s\in S\setminus
A}s^{k_{s}}\right)\left(\prod_{a\in A}a^{k_{a}-1}\right)\left(\prod_{a\in
A}a\right),$

by Lemma 2.1, it follows that $\prod_{a\in
A}a=\prod_{i=0}^{t-1}a_{i}\in\omega_{\alpha}(G)$. Assume
$\prod_{i=0}^{t-1}a_{i}=\alpha(g^{-1})g$ for some $g\in G$. For $0\leq j\leq
t-1$, let $x_{j}=g\left(\prod_{i=1}^{\frac{t-1}{2}}a_{2i+j-1}\right)^{-1}$,
where the indices are taken module $t$. Then

$\alpha(x_{j}^{-1})x_{j+1}=\left(\prod_{i=0}^{t-1}a_{i}\right)\left(\prod_{i=1}^{\frac{t-1}{2}}a_{2i+j-1}\right)^{-1}\left(\prod_{i=1}^{\frac{t-1}{2}}a_{2i+j}\right)^{-1}=a_{j}\in
S,$

which implies that $(x_{0},x_{1},\cdots,x_{t-1},x_{0})$ is a closed walk of
odd length. By Lemma 2.2, $GC(G,S,\alpha)$ contains an odd cycle and then it
is not bipartite graph. ∎

Using Theorem 2.3, we characterize the generalized Cayley bipartite graphs of
small valency.

######  Proposition 2.4.

Let $G$ be a finite abelian group and $S$ a generalized Cayley subset of size
$2$ induced by the involutory automorphism $\alpha$. Then $GC(G,S,\alpha)$ is
bipartite if and only if one of the following conditions holds:

  1. (i)

$S\subseteq\Omega_{\alpha}(G)$;

  2. (ii)

$S=\\{s,\alpha(s^{-1})\\}\subseteq\mho_{\alpha}(G)$, where
$s^{2k+1}\notin\omega_{\alpha}$ for all integers $1\leq k<o(\alpha(s)s)$.

###### Proof.

The proof is divided into two cases according to the classification of $S$.

Case 1: $S=\\{s_{1},s_{2}\\}\subseteq\Omega_{\alpha}(G)$.

Assume there exist positive integers $\ell_{1}$ and $\ell_{2}$ such that
$\ell_{1}+\ell_{2}$ is odd and
$s_{1}^{\ell_{1}}s_{2}^{\ell_{2}}\in\omega_{\alpha}(G)$. Without loss of
generality, we assume $\ell_{1}$ is odd and $\ell_{2}$ is even. Since
$\alpha(s_{i}^{-1})=s_{i}$ for $i=1$ and $2$, we have

$s_{1}^{\ell_{1}-1}s_{2}^{\ell_{2}}=\alpha(s_{1}^{-\frac{\ell_{1}-1}{2}}s_{2}^{-\frac{\ell_{2}}{2}})(s_{1}^{\frac{\ell_{1}-1}{2}}s_{2}^{\frac{\ell_{2}}{2}})\in\omega_{\alpha}(G).$
(1)

By Lemma 2.1, $s_{1}\in\omega_{\alpha}(G)$, which is contrary to
$S\cap\omega_{\alpha}(G)=\emptyset$. Thus, for any integers $\ell_{1}$ and
$\ell_{2}$ with $\ell_{1}+\ell_{2}$ odd,
$s_{1}^{\ell_{1}}s_{2}^{\ell_{2}}\notin\omega_{\alpha}(G)$. According to
Theorem 2.3, in this case, $GC(G,S,\alpha)$ must be bipartite.

Case 2: $S=\\{s,\alpha(s^{-1})\\}\subseteq\mho_{\alpha}(G)$.

Assume that $GC(G,S,\alpha)$ is bipartite. If there exists integer $1\leq
k<o(\alpha(s)s)$ such that $s^{2k+1}\in\omega_{\alpha}(G)$, we define $t$ to
be the smallest integer satisfying this condition. Since
$\alpha(s^{t})s^{-t}\in\omega_{\alpha}(G)$, we obtain
$\alpha(s^{t})s^{t+1}=(\alpha(s^{t})s^{-t})s^{2t+1}\in\omega_{\alpha}(G)$. Let
$\alpha(s^{t})s^{t+1}=\alpha(g^{-1})g$ for some $g\in G$. Note that the form
of the cycle containing $g$ is

$g\rightarrow\alpha(g)s\rightarrow
g\alpha(s)s\rightarrow\alpha(g)\alpha(s)s^{2}\rightarrow
g(\alpha(s)s)^{2}\rightarrow\cdots g$

Clearly, $g\neq g(\alpha(s)s)^{\ell}$ for all $1\leq\ell\leq t$, as $t$ is
smaller than the order of $\alpha(s)s$. Moreover, if there exists integer
$1\leq m\leq t-1$ such that $g=\alpha(g)\alpha(s^{m})s^{m+1}$, then
$\alpha(g^{-1})g=\alpha(s^{m})s^{m+1}$. Consequently,

$\displaystyle
s^{2m+1}=\alpha(g^{-1})g(\alpha(s^{-1})s)^{m}=\alpha((s^{m}g)^{-1})s^{m}g\in\omega_{\alpha}(G),$

which is contrary to the choice of $t$. Thus, for all integers $1\leq m\leq
t-1$, $g\neq\alpha(g)\alpha(s^{m})s^{m+1}$. According to
$g=\alpha(g)\alpha(s^{t})s^{t+1}$ and the above analysis, we find the cycle
containing $g$ is $C_{2t+1}$, which contradicts the assumption that
$GC(G,S,\alpha)$ is bipartite. Therefore, for all integers $1\leq
k<o(\alpha(s)s)$, $s^{2k+1}\notin\omega_{\alpha}(G)$.

Conversely, assume that for all integers $1\leq k<o(\alpha(s)s)$,
$s^{2k+1}\notin\omega_{\alpha}$. If $GC(G,S,\alpha)$ is not bipartite, then it
contains an odd cycle $C_{2\ell+1}$. Let $g$ be a vertex in this cycle. Based
on the form of the cycle containing $g$, we have
$g=\alpha(g)\alpha(s^{\ell})s^{\ell+1}$. Since $g\neq g(\alpha(s)s)^{n}$ for
all $n\leq\ell$, which implies $\ell<o(\alpha(s)s)$. However,
$s^{2\ell+1}=\alpha(g^{-1}s^{-\ell})gs^{\ell}\in\omega_{\alpha}(G)$, which is
a contradiction. Therefore, we conclude that $GC(G,S,\alpha)$ is bipartite. ∎

######  Proposition 2.5.

Let $G$ be a finite abelian group and $S$ a generalized Cayley subset of $G$
induced by the involutory automorphism $\alpha$. Then $GC(G,S,\alpha)$
contains a triangle if and only if there exist elements $s_{1},s_{2},s_{3}\in
S$ such that $s_{1}s_{2}s_{3}\in\omega_{\alpha}(G)$.

###### Proof.

The necessity is obvious. If $x$, $y$ and $z$ form a triangle in
$GC(G,S,\alpha)$, then there exist elements $s_{1},s_{2},s_{3}\in S$ such that
$\alpha(x^{-1})y=s_{1}$, $\alpha(y^{-1})z=s_{2}$ and $\alpha(z^{-1})x=s_{3}$.
Thus, we have $s_{1}s_{2}s_{3}=\alpha((xyz)^{-1})xyz\in\omega_{\alpha}(G)$.

Sufficiency. Assume $s_{1}s_{2}s_{3}=\alpha(g^{-1})g$ for some $g\in G$. Let
$x=gs_{2}^{-1}$, $y=\alpha(gs_{2}^{-1})s_{1}$ and $z=\alpha(s_{1})g$. It is
easy to verify that $x$, $y$ and $z$ are distinct elements. If $x=y$, then
$s_{1}=\alpha((gs_{2}^{-1})^{-1})gs_{2}^{-1}\in\omega_{\alpha}(G)$, which is
contrary to the fact that $S\cap\omega_{\alpha}(G)=\emptyset$. Similarly, the
cases of $y=z$ and $z=x$ give that either $s_{2}$ or $s_{3}$ belongs to
$\omega_{\alpha}(G)$, which leads to the same contradiction. Note that
$\alpha(x^{-1})y=s_{1}$, $\alpha(y^{-1})z=s_{2}$ and $\alpha(z^{-1})x=s_{3}$,
all of which are contained in $S$. Thus, the elements $x$, $y$ and $z$ indeed
form a triangle. ∎

Now we provide the sufficient and necessary condition for $3$-valent
generalized Cayley graphs to be bipartite.

######  Proposition 2.6.

Let $G$ be a finite abelian group and $S=\\{s_{1},s_{2},s_{3}\\}$ a
generalized Cayley subset of size $3$ induced by the involutory automorphism
$\alpha$ with $S\subseteq\Omega_{\alpha}(G)$. Then $GC(G,S,\alpha)$ is
bipartite if and only if $s_{1}s_{2}s_{3}\notin\omega_{\alpha}(G)$.

###### Proof.

The necessity follows directly from Theorem 2.3. We prove the sufficiency by
negation. If $GC(G,S,\alpha)$ is not bipartite, by Theorem 2.3, there exist
integers $\ell_{i}$ for $1\leq i\leq 3$ such that
$s_{1}^{\ell_{1}}s_{2}^{\ell_{2}}s_{3}^{\ell_{3}}\in\omega_{\alpha}(G)$ and
$\ell_{1}+\ell_{2}+\ell_{3}$ is odd. If exactly one of elements $\ell_{1}$,
$\ell_{2}$ and $\ell_{3}$ is odd, without loss of generality, assume
$\ell_{1}$ is odd. Then from
$s_{1}^{\ell_{1}}s_{2}^{\ell_{2}}s_{3}^{\ell_{3}}\in\omega_{\alpha}(G)$ and
$s_{1}^{\ell_{1}-1}s_{2}^{\ell_{2}}s_{3}^{\ell_{3}}\in\omega_{\alpha}(G)$, we
have $s_{1}\in\omega_{\alpha}(G)$, which is a contradiction. Thus, all
elements $\ell_{i}$ are odd, for $1\leq i\leq 3$. Since
$s_{i}^{\ell_{i}-1}\in\omega_{\alpha}(G)$ for $1\leq i\leq 3$, we have
$s_{1}^{\ell_{1}-1}s_{2}^{\ell_{2}-1}s_{3}^{\ell_{3}-1}\in\omega_{\alpha}(G)$.
Combining
$s_{1}^{\ell_{1}}s_{2}^{\ell_{2}}s_{3}^{\ell_{3}}\in\omega_{\alpha}(G)$, it
follows that $s_{1}s_{2}s_{3}\in\omega_{\alpha}(G)$. ∎

######  Proposition 2.7.

Let $G$ be a finite abelian group and $S=\\{s_{1},s_{2},\alpha(s_{2}^{-1})\\}$
a generalized Cayley subset of size $3$ induced by the involutory automorphism
$\alpha$, where $s_{1}\in S\cap\Omega_{\alpha}(G)$ and $s_{2}\in
S\cap\mho_{\alpha}(G)$. Then $GC(G,S,\alpha)$ is bipartite if and only if the
following conditions hold:

  1. (i)

for every odd integer $k$, $s_{2}^{k}\notin\omega_{\alpha}(G)$;

  2. (ii)

for every even integer $\ell$, $s_{1}s_{2}^{\ell}\notin\omega_{\alpha}(G)$.

###### Proof.

Suppose that $GC(G,S,\alpha)$ is bipartite. If there exists odd integer $k$
such that $s_{2}^{k}\in\omega_{\alpha}(G)$, let $t$ be the smallest odd
integer that satisfies this condition. Since
$s_{1}^{-(t-1)}=\alpha(s_{1}^{\frac{t-1}{2}})s_{1}^{-\frac{t-1}{2}}\in\omega_{\alpha}(G)$,
we have $s_{1}^{-(t-1)}s_{2}^{t}\in\omega_{\alpha}(G)$. Assume
$s_{1}^{-(t-1)}s_{2}^{t}=\alpha(g^{-1})g$ for some $g\in G$. Let
$x_{2i+1}=\alpha(g)s_{2}^{i+1}\alpha(s_{1}^{i})$ and
$x_{2i}=g\alpha(s_{2}^{i})s_{1}^{i}$ for $0\leq i\leq t-1$. Then
$(x_{0}=g,x_{1},\cdots,x_{2t-2},x_{2t-1}=x_{0})$ is a closed walk of odd
length in $GC(G,S,\alpha)$. Similarly, if there exists even integer $\ell$
such that $s_{1}s_{2}^{\ell}\in\omega_{\alpha}(G)$, let $m$ be the smallest
even integer that satisfies the condition. Then
$s_{1}^{-m+1}s_{2}^{m}=\alpha(s_{1}^{\frac{m}{2}})s_{1}^{-\frac{m}{2}}s_{1}s_{2}^{m}\in\omega_{\alpha}(G)$.
Assume $s_{1}^{-m+1}s_{2}^{m}=\alpha(h^{-1})h$ for some $h\in G$. Let
$y_{2j+1}=\alpha(h)s_{2}^{j+1}\alpha(s_{1}^{j})$ and
$y_{2j}=h\alpha(s_{2}^{j})s_{1}^{j}$ for $0\leq j\leq m-1$. Then
$(y_{0}=h,y_{1},\cdots,y_{2m-2},y_{2m-1}=y_{0})$ is a closed walk of odd
length in $GC(G,S,\alpha)$. For each case, by Lemma 2.2, there exists odd
cycle in $GC(G,S,\alpha)$, which is contrary to the assumption.

We prove the sufficiency by negation. Suppose $GC(G,S,\alpha)$ is not
bipartite and let $(z_{0},z_{1},\cdots,z_{2t+1}=z_{0})$ be an odd cycle in
$GC(G,S,\alpha)$. Since $\alpha(z_{j}^{-1})z_{j+1}\in S$ for $0\leq j\leq 2t$
and
$\prod_{j=0}^{2t}\alpha(z_{j}^{-1})z_{j+1}=\alpha(\prod_{j=0}^{2t}z_{j})^{-1}\prod_{j=0}^{2t}z_{j}\in\omega_{\alpha}(G)$,
it follows that there exist integers $\ell_{1}$, $\ell_{2}$ and $\ell_{3}$
such that
$s_{1}^{\ell_{1}}s_{2}^{\ell_{2}}\alpha(s_{2}^{-\ell_{3}})\in\omega_{\alpha}(G)$
and $\ell_{1}+\ell_{2}+\ell_{3}$ is odd. If $\ell_{1}$ is even, then the
parity of $\ell_{2}$ and $\ell_{3}$ are different. Since
$s_{1}^{\ell_{1}}\in\omega_{\alpha}(G)$ and
$s_{2}\alpha(s_{2}^{-1})\in\omega_{\alpha}(G)$, by Lemma 2.1, we have
$s_{1}^{\ell_{1}}\left(s_{2}\alpha(s_{2}^{-1})\right)^{\ell_{3}}\in\omega_{\alpha}(G)$.
As a result, using Lemma 2.1 again, from

$\displaystyle
s_{1}^{\ell_{1}}s_{2}^{\ell_{2}}\alpha(s_{2}^{-\ell_{3}})=s_{1}^{\ell_{1}}\left(s_{2}\alpha(s_{2}^{-1})\right)^{\ell_{3}}s_{2}^{\ell_{2}-\ell_{3}}\in\omega_{\alpha}(G),$

we have $s_{2}^{\ell_{2}-\ell_{3}}\in\omega_{\alpha}(G)$, which is contrary to
condition $\rm(i)$. If $\ell_{1}$ is odd, then the parity of $\ell_{2}$ and
$\ell_{3}$ are the same. From
$s_{1}^{\ell_{1}}s_{2}^{\ell_{2}}\alpha(s_{2}^{-\ell_{3}})\in\omega_{\alpha}(G)$
and
$s_{1}^{\ell-1}\left(s_{2}\alpha(s_{2}^{-1})\right)^{\ell_{3}}\in\omega_{\alpha}(G)$,
we obtain $s_{1}s_{2}^{\ell_{2}-\ell_{3}}\in\omega_{\alpha}(G)$, which is
contrary to condition $\rm(ii)$. Therefore, we complete the proof. ∎

An equivalent statement of Proposition 2.7 is

######  Proposition 2.8.

Let $G$ be a finite abelian group and $S=\\{s_{1},s_{2},\alpha(s_{2}^{-1})\\}$
a generalized Cayley subset of size $3$ induced by the involutory automorphism
$\alpha$, where $s_{1}\in S\cap\Omega_{\alpha}(G)$ and $s_{2}\in
S\cap\mho_{\alpha}(G)$. Then $GC(G,S,\alpha)$ is bipartite if and only if for
all integers $\ell_{1}$, $\ell_{2}$ and $\ell_{3}$ with
$\sum_{i=1}^{3}\ell_{i}$ is odd,
$s_{1}^{\ell_{1}}s_{2}^{\ell_{2}}\alpha(s_{2}^{-\ell_{3}})\notin\omega_{\alpha}(G)$
holds.

###### Proof.

According to Proposition 2.7, the sufficiency is obvious.

Necessity. Assume that $GC(G,S,\alpha)$ is bipartite. For any integers
$\ell_{1}$, $\ell_{2}$ and $\ell_{3}$ with $\sum_{i=1}^{3}\ell_{i}$ is odd, if
$\ell_{1}$ is odd, then $s_{1}^{\ell_{1}-1}\in\omega_{\alpha}(G)$, which gives
that
$s_{1}^{\ell_{1}-1}s_{2}^{\ell_{3}}\alpha(s_{2}^{-\ell_{3}})\in\omega_{\alpha}(G)$.
By the condition (ii) of Proposition 2.7,
$s_{1}s_{2}^{\ell_{2}-\ell_{3}}\notin\omega_{\alpha}(G)$, thus we have

$s_{1}^{\ell_{1}}s_{2}^{\ell_{2}}\alpha(s_{2}^{-\ell_{3}})=s_{1}s_{2}^{\ell_{2}-\ell_{3}}s_{1}^{\ell_{1}-1}s_{2}^{\ell_{3}}\alpha(s_{2}^{-\ell_{3}})\notin\omega_{\alpha}(G).$

Otherwise, $\ell_{1}$ is even, which implies
$s_{1}^{\ell_{1}}\in\omega_{\alpha}(G)$ and then
$s_{1}^{\ell_{1}}s_{2}^{\ell_{3}}\alpha(s_{2}^{-\ell_{3}})\in\omega_{\alpha}(G)$.
By the condition (i) of Proposition 2.7,
$s_{2}^{\ell_{2}-\ell_{3}}\notin\omega_{\alpha}(G)$, thus

$s_{1}^{\ell_{1}}s_{2}^{\ell_{2}}\alpha(s_{2}^{-\ell_{3}})=s_{1}^{\ell_{1}}s_{2}^{\ell_{3}}\alpha(s_{2}^{-\ell_{3}})s_{2}^{\ell_{2}-\ell_{3}}\notin\omega_{\alpha}(G).$

∎

######  Example 2.9.

For the cyclic group $Z_{14}=\langle g\rangle$, $S_{1}=\\{g,g^{3},g^{5}\\}$ is
a generalized Cayley subset of $Z_{14}$ induced by $\iota$ and contained in
$\Omega_{\iota}(Z_{14})$. Since $\omega_{\iota}(Z_{14})=\langle g^{2}\rangle$
and $gg^{3}g^{5}=g^{9}\notin\omega_{\iota}(Z_{14})$, $GC(Z_{14},S_{1},\iota)$
is bipartite. For the group $Z_{2}^{2}\times Z_{6}$,
$S_{2}=\\{(1,0,0),(0,0,2),(0,0,4)\\}$ is a generalized Cayley subset of
$Z_{2}^{2}\times Z_{6}$ induced by the involutory automorphism
$\alpha:(1,0,0)\mapsto(1,0,0),(0,1,0)\mapsto(0,1,0),(0,0,1)\mapsto(0,1,1)$,
where $(1,0,0)\in\Omega_{\alpha}(Z_{2}^{2}\times Z_{6})$ and
$\\{(0,0,2),(0,0,4)\\}\subseteq\mho_{\alpha}(Z_{2}^{2}\times Z_{6})$. Since
$(0,0,2)^{3}=(0,0,0)\in\omega_{\alpha}(Z_{2}^{2}\times Z_{6})$,
$GC(Z_{2}^{2}\times Z_{6},S_{2},\alpha)$ is not bipartite.

At the end of this section, we propose the following question:

######  Question 2.10.

Investigate the conditions of generalized Cayley graph $GC(G,S,\alpha)$ of
abelian group $G$ to be bipartite, where $|S|>3$ and
$S\cap\mho_{\alpha}(G)\neq\emptyset$.

##  3 Connected generalized Cayley graphs

In this section, we provide the sufficient and necessary conditions for the
generalized Cayley graphs to be connected. Firstly, a simple observation
should be pointed out.

######  Proposition 3.1.

Let $G$ be a finite group and $S$ a generalized Cayley subset of $G$ induced
by the involutory automorphism $\alpha$. Then $\\{x,y\\}\in E(Cay(G,S^{-1}S))$
if and only if there exists a path of length $2$ from $x$ to $y$ in
$GC(G,S,\alpha)$.

###### Proof.

If $\\{x,y\\}\in E(Cay(G,S^{-1}S))$, then there exist $a,b\in S$ such that
$y=x(a^{-1}b)$, which implies
$\alpha(y)\alpha(b^{-1})=\alpha(x)\alpha(a^{-1})$. Note that
$\alpha(S^{-1})=S$, hence both $\alpha(a^{-1})$ and $\alpha(b^{-1})$ belong to
$S$. Therefore, $(y,\alpha(y)\alpha(b^{-1}),x)$ is a path of length $2$ from
$x$ to $y$ in $GC(G,S,\alpha)$.

Conversely, if there exists a path $(x,\alpha(x)s_{1},y)$ from $x$ to $y$,
where $s_{1}\in S$, then $y=\alpha(\alpha(x)s_{1})s_{2}=x(\alpha(s_{1})s_{2})$
for some $s_{2}\in S$. Recall $\alpha(S^{-1})=S$, we have
$\alpha(s_{1})s_{2}\in S^{-1}S$. Thus, $\\{x,y\\}\in E(Cay(G,S^{-1}S))$. ∎

For any generalized Cayley graph $GC(G,S,\alpha)$, we define an equivalence
relation on $G$ by declaring that $x$ is equivalent to $y$ if there exists a
walk of even length from $x$ to $y$ in $GC(G,S,\alpha)$. Let $\Theta$ denote
the equivalence class containing the identity element $e$. Using this
equivalence relation, we obtain the following result.

######  Lemma 3.2.

Let $G$ be a finite group and $S$ a generalized Cayley subset of $G$ induced
by the involutory automorphism $\alpha$. If $GC(G,S,\alpha)$ is connected,
then $\Theta=\langle S^{-1}S\rangle$ and $|\langle
S^{-1}S\rangle|\geq\frac{1}{2}|G|$.

###### Proof.

We first show that $\Theta\subseteq\langle S^{-1}S\rangle$ and
$|\Theta|\geq\frac{1}{2}|G|$. For any $y\in\Theta$, there exists a walk of
even length from $e$ to $y$. Then by Proposition 3.1, there exists a walk
$(e,y_{1},y_{2},\cdots,y_{m}=y)$ from $e$ to $y$ in $Cay(G,S^{-1}S)$. Observe
that $y_{1}\in S^{-1}S$, then $y_{2}=y_{1}x_{1}\in\langle S^{-1}S\rangle$ for
some $x_{1}\in S^{-1}S$. Similarly, we obtain $y_{i}\in\langle S^{-1}S\rangle$
for each integer $2\leq i\leq m$. Thus, $\Theta\subseteq\langle
S^{-1}S\rangle$. For each element $z\in G\setminus\Theta$, let
$(e,z_{1},z_{2},\cdots,z_{2t+1}=z)$ be a walk of odd length from $e$ to $z$.
Let $s\in S$. Then $(e,z_{1},z_{2},\cdots,z_{2t+1},\alpha(z)s)$ is a walk of
even length from $e$ to $\alpha(z)s$, which implies $\alpha(z)s\in\Theta$. It
follows that each element $z$ in $G\setminus\Theta$ corresponds to an element
$\alpha(z)s$ in $\Theta$. Hence, we have $|\Theta|\geq\frac{1}{2}|G|$. As a
result, one of conditions $\Theta=\langle S^{-1}S\rangle$ and,
$\Theta\subsetneq\langle S^{-1}S\rangle$ and $\langle S^{-1}S\rangle=G$ holds.
However, the latter case is impossible. In fact, if $\langle
S^{-1}S\rangle=G$, then $Cay(G,S^{-1}S)$ is connected. Thus, for any vertex
$x\in G$, there is a path from $x\in G$ to $e$ in $Cay(G,S^{-1}S)$. By
Proposition 3.1, there is an even walk from $x$ to $e$ in $GC(G,S,\alpha)$,
which yields that $\Theta=G=\langle S^{-1}S\rangle$. Therefore, if
$GC(G,S,\alpha)$ is connected, then $\Theta=\langle S^{-1}S\rangle$ always
hold. ∎

To prove the main theorem, the next lemmas are also needed.

######  Lemma 3.3.

Let $G$ be a finite group and $S$ a subset of $G$. If there exists an
automorphism $\alpha$ of $G$ such that $\alpha(S)=S^{-1}$, then
$\alpha(\langle S^{-1}S\rangle)=\langle SS^{-1}\rangle$.

###### Proof.

Since $(S^{-1}S)^{-1}=S^{-1}S$, any $g\in\langle S^{-1}S\rangle$ can be
expressed as $g=\prod_{i=1}^{m}s_{i}^{-1}t_{i}$, where $s_{i},t_{i}\in S$.
Then $\alpha(g)=\prod_{i=1}^{m}\alpha(s_{i}^{-1})\alpha(t_{i})\in\langle
SS^{-1}\rangle$, which implies that $\alpha(\langle
S^{-1}S\rangle)\subseteq\langle SS^{-1}\rangle$.

Conversely, for any $h\in\langle SS^{-1}\rangle$, we have
$h=\prod_{j=1}^{n}s_{j}t_{j}^{-1}$ for some $s_{j},t_{j}\in S$. According to
$\alpha(S^{-1})=S$, there exist $u_{j}$ and $v_{j}$ in $S$ such that
$\alpha(u_{j}^{-1})=s_{j}$ and $\alpha(v_{j})=t_{j}^{-1}$ for $1\leq j\leq n$.
Thus, we obtain
$h=\prod_{j=1}^{n}\alpha(u_{j}^{-1})\alpha(v_{j})=\alpha(\prod_{j=1}^{n}u_{j}^{-1}v_{j})\in\alpha(\langle
S^{-1}S\rangle)$, which implies $\langle SS^{-1}\rangle\subseteq\alpha(\langle
S^{-1}S\rangle)$. Therefore, we conclude that $\alpha(\langle
S^{-1}S\rangle)=\langle SS^{-1}\rangle$. ∎

######  Lemma 3.4.

Let $G$ be a finite group and $S$ a generalized Cayley subset of $G$ induced
by the involutory automorphism $\alpha$. Let $C(e)$ denote the set of vertices
of the connected component containing $e$ in $GC(G,S,\alpha)$. Then
$C(e)=\langle S^{-1}S\rangle\cup\langle SS^{-1}\rangle s$, where $s$ is any
element in $S$.

###### Proof.

Before starting the proof, observe that for any $s\in S$, we have
$t=ts^{-1}s\in\langle SS^{-1}\rangle s$ for every $t\in S$. It follows that
$S\subseteq\langle SS^{-1}\rangle s$ and $\langle SS^{-1}\rangle t=\langle
SS^{-1}\rangle s$.

For any $x\in C(e)$, there exists a path $P=(e,x_{1},x_{2},\cdots,x_{n}=x)$
from $e$ to $x$. We claim that either $x_{i}\in\langle S^{-1}S\rangle$ or
$x_{i}\in\langle SS^{-1}\rangle s$ for $1\leq i\leq n$. It is easy to see that
$x_{1}=\alpha(e)s_{1}=s_{1}\in\langle SS^{-1}\rangle s$ and
$x_{2}=\alpha(x_{1})s_{2}\in\alpha(\langle SS^{-1}\rangle)\alpha(s)s_{2}$, for
some $s_{1},s_{2}\in S$. By Lemma 3.3 and $\alpha(S)=S^{-1}$, we obtain
$x_{2}\in\langle S^{-1}S\rangle$. By induction, suppose that for every integer
$t$ smaller than $k$, where $k\leq n$, $x_{t}\in\langle SS^{-1}\rangle s$ if
$t$ is odd and $x_{t}\in\langle S^{-1}S\rangle$ if $t$ is even. Then from
$x_{k-1}\in\langle SS^{-1}\rangle s\cup\langle S^{-1}S\rangle$ and
$x_{k}=\alpha(x_{k-1})s_{k}$ for some $s_{k}\in S$, it follows that
$x_{k}\in\langle S^{-1}S\rangle$ or $\langle SS^{-1}\rangle s$. Thus,
$C(e)\subseteq\langle S^{-1}S\rangle\cup\langle SS^{-1}\rangle s$.

On the contrary, we assert that $\langle S^{-1}S\rangle\cup\langle
SS^{-1}\rangle s\subseteq C(e)$. For any $y\in\langle S^{-1}S\rangle$, assume
that $y=\prod_{i=1}^{m}s_{i}^{-1}t_{i}$, where $s_{i},t_{i}\in S$ for $1\leq
i\leq m$. Since $\alpha(S)=S^{-1}$, there exists $u_{i}\in S$ such that
$\alpha(u_{i})=s_{i}^{-1}$ for each integer $1\leq i\leq m$. Then

$W_{1}=\left(e,u_{1},s_{1}^{-1}t_{1},u_{1}\alpha(t_{1})u_{2},\prod_{i=1}^{2}s_{i}^{-1}t_{i},\cdots,\left(\prod_{i=1}^{m-1}u_{i}\alpha(t_{i})\right)u_{m},\prod_{i=1}^{m}s_{i}^{-1}t_{i}=y\right)$

is a walk from $e$ to $y$. For any $y\in\langle SS^{-1}\rangle s$, we assume
$y=(\prod_{j=1}^{n}s_{j}t_{j}^{-1})s$, where $s_{j},t_{j}\in S$ for $1\leq
j\leq n$. Let $v_{j}=\alpha(t_{j}^{-1})$ for each integer $1\leq j\leq n$.
Then

$W_{2}=\left(e,s_{1},\alpha(s_{1})v_{1},s_{1}t_{1}^{-1}s_{2},\prod_{j=1}^{2}\alpha(s_{j})v_{j},\cdots,\prod_{j=1}^{n}\alpha(s_{j})v_{j},\left(\prod_{j=1}^{n}s_{j}t_{j}^{-1}\right)s=y\right)$

is a walk from $e$ to $y$.

Therefore, we arrive at the conclusion that $C(e)=\langle
S^{-1}S\rangle\cup\langle SS^{-1}\rangle s$. ∎

Based on the above lemmas, the main results of this section follows.

######  Theorem 3.5.

Let $G$ be a finite group and $S$ a generalized Cayley subset of $G$ induced
by the involutory automorphism $\alpha$. Then $GC(G,S,\alpha)$ is connected if
and only if one of the following conditions holds:

  1. (i)

$S$ is not contained in a right coset of any proper subgroup of $G$;

  2. (ii)

$S$ is contained in a nontrivial right coset of proper subgroup $H$ of $G$,
where $H=\langle SS^{-1}\rangle$ and it satisfies $|G:H|=2$ and $\alpha(H)=H$.

###### Proof.

Suppose $GC(G,S,\alpha)$ is connected and let $H$ be the smallest subgroup
such that $S$ is contained in some right coset of $H$. Assume $S\subseteq Hg$
for some $g\in G$. Then $S^{-1}\subseteq g^{-1}H$, which yields that
$SS^{-1}\subseteq Hgg^{-1}H=H$. Thus, $\langle SS^{-1}\rangle\leq H$. Since
for any $s\in S$, we have $S\subseteq\langle SS^{-1}\rangle s$, which implies
that $S$ is also contained in a right coset of $\langle SS^{-1}\rangle$. By
the choice of $H$, $\langle SS^{-1}\rangle=H$ follows. From Lemma 3.2, either
$H=G$ or the order of $H$ is $\frac{1}{2}|G|$. For the latter case, $H$ is a
normal subgroup of $G$. It follows that $S^{-1}S\subseteq g^{-1}HHg=H$ and
then $\langle S^{-1}S\rangle\leq H$. By Lemma 3.3, we obtain $\alpha(H)\leq
H$. Since the order of $\alpha(H)$ and $H$ are the equal, we have
$\alpha(H)=H$. In addition, we assert that $g\notin H$. Otherwise, assume
$g\in H$, then $S\subseteq H$. From Lemma 3.4, we obtain $C(e)=H$, which
contradicts that $GC(G,S,\alpha)$ is connected. Moreover, if $S$ is contained
in a right coset of other proper subgroup $K$, with similar discussion above,
we have $H=\langle SS^{-1}\rangle\leq K$. Recall that $|H|=\frac{1}{2}|G|$,
which yields that $H=K$.

Conversely, we show that both the conditions (i) and (ii) imply
$GC(G,S,\alpha)$ is connected. If $S$ is not contained in any right coset of
proper subgroup of $G$, then we have $G=\langle SS^{-1}\rangle$ as
$S\subseteq\langle SS^{-1}\rangle s$. Thus, by Lemma 3.4, $C(e)=G$, which
means that $GC(G,S,\alpha)$ is connected. Let $H=\langle SS^{-1}\rangle$. If
$S$ is contained in $Hg$ for some $g\in G\setminus H$, where $\alpha(H)=H$ and
$|G:H|=2$, then $C(e)=H\cup\alpha(H)g=G$. Thus, $GC(G,S,\alpha)$ is connected.
∎

By Theorem 3.5, we obtain another equivalent condition for generalized Cayley
graphs to be connected. Usually, this one is more convenient to use.

######  Theorem 3.6.

Let $G$ be a finite group and $S$ a generalized Cayley subset of $G$ induced
by the involutory automorphism $\alpha$. Then $GC(G,S,\alpha)$ is connected if
and only if $G=\langle S\rangle$, $|G:\langle SS^{-1}\rangle|\leq 2$ and
$\alpha(\langle SS^{-1}\rangle)=\langle SS^{-1}\rangle$.

###### Proof.

Necessity. Assume $GC(G,S,\alpha)$ is connected. Since $C(e)=\langle
S^{-1}S\rangle\cup\langle SS^{-1}\rangle s\subseteq\langle S\rangle$ and
$C(e)=G$, we have $G=\langle S\rangle$. Note that for any $s\in S$,
$S\subseteq\langle SS^{-1}\rangle s$. By Theorem 3.5, $\langle
SS^{-1}\rangle=G$ or $\alpha(\langle SS^{-1}\rangle)=\langle SS^{-1}\rangle$
and $|G:\langle SS^{-1}\rangle|=2$.

Sufficiency. We divide the proof into two cases:

Case 1: $G=\langle S\rangle=\langle SS^{-1}\rangle=\langle S^{-1}S\rangle$.

In this case, we claim that there exists no proper subgroup of $G$ such that
$S$ is contained in right coset of this subgroup. If the claim is not true,
let $H<G$ and $S\subseteq Hg$ for some $g\in G$. Then $SS^{-1}\subseteq
Hgg^{-1}H=H$. Further, $G=\langle SS^{-1}\rangle\leq H$, which is a
contradiction. Thus, by Theorem 3.5, the result follows.

Case 2: $G=\langle S\rangle$, $|G:\langle SS^{-1}\rangle|=2$ and
$\alpha(\langle SS^{-1}\rangle)=\langle SS^{-1}\rangle$.

If there exists a proper subgroup $H$ of $G$ such that $S\subseteq Hg$ for
some $g\in G$, then we obtain $\langle SS^{-1}\rangle\leq H$. Since
$|G:\langle SS^{-1}\rangle|=2$, it follows that $H=\langle SS^{-1}\rangle$.
Clearly, $g\notin H$. Otherwise, $S\subseteq H$ and then $\langle
S\rangle=G\leq H$, which is a contradiction. Using Theorem 3.5 again, we
conclude that $GC(G,S,\alpha)$ is connected. ∎

According to Theorem 3.6, we give the equivalent conditions of the connected
generalized Cayley graphs to be bipartite.

######  Theorem 3.7.

Let $G$ be a finite group and $S$ a generalized Cayley subset of $G$ induced
by the involutory automorphism $\alpha$. If $GC(G,S,\alpha)$ is connected,
then the following statements are equivalent:

  1. (i)

$GC(G,S,\alpha)$ is bipartite;

  2. (ii)

$|\langle SS^{-1}\rangle|=\frac{1}{2}|G|$;

  3. (iii)

$S\cap\langle SS^{-1}\rangle=\emptyset$.

###### Proof.

As $GC(G,S,\alpha)$ is connected, by Theorem 3.6, we have $\langle
S\rangle=G$, $|G:\langle SS^{-1}\rangle|\leq 2$ and $\alpha(\langle
SS^{-1}\rangle)=\langle S^{-1}S\rangle=\langle SS^{-1}\rangle$. Next, we show
that the conditions $\rm(i)$, $\rm(ii)$ and $\rm(iii)$ are equivalent.

$\rm(i)\Rightarrow(ii):$ Since $GC(G,S,\alpha)$ is bipartite, by Lemma 3.2,
$\langle SS^{-1}\rangle$ is contained in one of the partition set of
$GC(G,S,\alpha)$. Thus, $\langle SS^{-1}\rangle\neq G$.

$\rm(ii)\Rightarrow(iii):$ Suppose that $|\langle
SS^{-1}\rangle|=\frac{1}{2}|G|$ but $S\cap\langle
SS^{-1}\rangle\neq\emptyset$. Let $a\in S\cap\langle SS^{-1}\rangle$. Then,
for any $b\in S$, $b=(ba^{-1})a\in\langle SS^{-1}\rangle$, which yields that
$S\subseteq\langle SS^{-1}\rangle$. As a result, $\langle S\rangle\neq G$,
which leads to a contradiction.

$\rm(iii)\Rightarrow(i):$ If $S\cap\langle SS^{-1}\rangle=\emptyset$, then
$\langle SS^{-1}\rangle\neq G$, which implies $|\langle
SS^{-1}\rangle|=\frac{1}{2}|G|$. For any edge $\\{x,y\\}$ of $GC(G,S,\alpha)$,
we have $y=\alpha(x)s$ for some $s\in S$. Observe that $s\not\in\langle
SS^{-1}\rangle$, thus $G=\langle SS^{-1}\rangle\cup\langle SS^{-1}\rangle s$.
Without loss of generality, assume that $x\in\langle SS^{-1}\rangle$. Then
$y=\alpha(x)s\in\alpha(\langle SS^{-1}\rangle)s=\langle SS^{-1}\rangle s$.
Thus, $GC(G,S,\alpha)$ is bipartite with bipartition sets $\langle
SS^{-1}\rangle$ and $\langle SS^{-1}\rangle s$. ∎

A simple observation is, if $G$ is abelian, then $\alpha(\langle
SS^{-1}\rangle)=\langle SS^{-1}\rangle$ always hold. Thus, the generalized
Cayley graph of abelian group $G$ with respect to the pair $(S,\alpha)$ is
connected if and only if $G=\langle S\rangle$ and $|G:\langle
SS^{-1}\rangle|\leq 2$.

##  4 Connected and integral generalized Cayley graphs

A graph is called integral if all eigenvalues of its adjacency matrix are
integers. With the help of results about generalized Cayley graphs in section
$3$, we determine the groups whose all $3$-valent generalized Cayley graphs
are connected and integral. Firstly, the conception of Cayley sum graph are
given.

######  Definition 4.1.

Let $G$ be an abelian group and $S$ a subset of $G$. The Cayley sum graph
$Cay^{+}(G,S)$ is the graph having the vertex set $G$ and the edge set
$\\{\\{g,h\\}\mid g,h\in G,gh\in S\\}$.

If $S$ is a square free subset of $G$, then $Cay^{+}(G,S)$ is simple graph.
Clearly, the Cayley sum graph is special generalized Cayley graph of abelian
group under the involutory automorphism $\iota$.

The following are some lemmas will be used in this section.

######  Lemma 4.2.

[11] Let $G$ be a finite group and $S$ a generalized Cayley subset of $G$
induced by $\alpha$. Then $GC(G,S,\alpha)\cong GC(G,S^{\beta},\alpha^{\beta})$
for all $\beta\in{\rm\textsf{Aut}}(G)$, where
$\alpha^{\beta}=\beta\alpha\beta^{-1}$.

From Lemma 4.2, to study all generalized Cayley graphs of $G$, it suffices to
study the generalized Cayley graphs induced by involutory automorphisms in
different conjugate classes of ${\rm\textsf{Aut}}(G)$.

######  Lemma 4.3.

[3] A connected cubic integral graph has $n$ vertices, where
$n\in\\{4,6,8,10,12,20,24,30\\}$. Moreover, all connected, integral, cubic
graphs are shown in Figure 1.

Figure 1: All connected, integral, cubic graphs

######  Lemma 4.4.

[2] Let $G$ be an abelian group. For every square free subset $S$ of $G$ of
size $3$, $Cay^{+}(G,S)$ is connected and integral if and only if $G$ is
isomorphic to one of the groups $Z_{2}^{2}$, $Z_{2}^{3}$, $Z_{6}$ and $Z_{8}$.

######  Lemma 4.5.

[7] If the direct product $X\times Y$ of graphs $X$ and $Y$ is connected, then
both $X$ and $Y$ are connected.

######  Lemma 4.6.

[6] Let $X=GC(G_{1},S_{1},\alpha_{1})$, $Y=GC(G_{2},S_{2},\alpha_{2})$, and
let $\alpha_{1}\times\alpha_{2}$ be the automorphism of $G_{1}\times G_{2}$
defined by
$(\alpha_{1}\times\alpha_{2})(g_{1},g_{2})=(\alpha_{1}(g_{1}),\alpha_{2}(g_{2}))$
for all $g_{1}\in G_{1}$ and $g_{2}\in G_{2}$. Then $X\times Y=GC(G_{1}\times
G_{2},S_{1}\times S_{2},\alpha_{1}\times\alpha_{2})$.

According to Lemma 4.4, it is straightforward to obtain the following theorem.

######  Theorem 4.7.

Let $G$ be a finite abelian group admitting $3$-valent generalized Cayley
graph. Then each $3$-valent generalized Cayley graph of $G$ is connected and
integral if and only if $G\cong Z_{6}$, $Z_{2}^{3}$ or $Z_{8}$.

###### Proof.

By Lemma 4.4, to determined the abelian groups whose all generalized Cayley
cubic graphs are connected and integral, it suffices to consider the groups
$Z_{2}^{2}$, $Z_{2}^{3}$, $Z_{6}$ and $Z_{8}$.

  1. $\bullet$

For $Z_{2}^{2}$, its all involutory automorphisms are in the same conjugate
class. Let $\delta:(0,1)\mapsto(1,0),(1,0)\mapsto(0,1)$ be an involutory
automorphism of $Z_{2}^{2}$ and we obtain subsets
$\omega_{\delta}(Z_{2}^{2})=\\{(0,0),(1,1)\\}$,
$\Omega_{\delta}(Z_{2}^{2})=\emptyset$ and
$\mho_{\delta}(Z_{2}^{2})=\\{(0,1),(1,0)\\}$. Thus, $Z_{2}^{2}$ does not have
$3$-valent generalized Cayley graph.

  2. $\bullet$

For $Z_{6}$, it has unique involutory automorphism $\iota$, which implies the
generalized Cayley graphs of $Z_{6}$ are Cayley sum graphs.

  3. $\bullet$

For $Z_{2}^{3}$, ${\rm\textsf{Aut}}(Z_{2}^{3})\cong GL(3,2)$ and all
involutory automorphisms of $Z_{2}^{3}$ are in the same conjugate class. By
Lemma 4.2, it is enough to discuss the generalized Cayley graphs induced by
$\alpha:(1,0,0)\mapsto(1,0,0),(0,1,0)\mapsto(0,1,0),(0,0,1)\mapsto(0,1,1)$.
With respect to $\alpha$, we obtain subsets
$\omega_{\alpha}(Z_{2}^{3})=\\{(0,0,0),(0,1,0)\\}$,
$\Omega_{\alpha}(Z_{2}^{3})=\\{(1,0,0),(1,1,0)\\}$ and
$\mho_{\alpha}(Z_{2}^{3})=\\{(0,0,1),(0,1,1),(1,0,1),(1,1,1)\\}$. Thus, the
generalized Cayley subsets of size $3$ induced by $\alpha$ are
$S_{1}=\\{(1,0,0),(0,0,1),(0,1,1)\\}$, $S_{2}=\\{(1,0,0),(1,0,1),(1,1,1)\\}$,
$S_{3}=\\{(1,1,0),(0,0,1),(0,1,1)\\}$ and
$S_{4}=\\{(1,1,0),(1,0,1),(1,1,1)\\}$. It is easy to verify that for $1\leq
i\leq 4$, $\langle S_{i}\rangle=Z_{2}^{3}$ and $|\langle
S_{i}S_{i}^{-1}\rangle|\geq\frac{1}{2}|Z_{2}^{3}|$. By Theorem 3.6, all
$3$-valent generalized Cayley graphs are connected. Moreover, for each $1\leq
i\leq 4$, $GC(Z_{2}^{3},S_{i},\alpha)\cong G_{4}$ which is shown in Figure 1.

  4. $\bullet$

For $Z_{8}$, the involutory automorphisms of which are $\iota$,
$\alpha_{1}:g\mapsto g^{3}$ and $\alpha_{2}:g\mapsto g^{5}$ for all $g\in
Z_{8}$. Since
$\Omega_{\alpha_{1}}(Z_{8})=\Omega_{\alpha_{2}}(Z_{8})=\emptyset$, there is no
$3$-valent generalized Cayley graph induced by $\alpha_{1}$ and $\alpha_{2}$.
Thus, all generalized Cayley graphs of $Z_{8}$ are Cayley sum graphs.

Therefore, our result follows. ∎

Using the fact that each Cayley sum graph is generalized Cayley graph, we
quickly determined the abelian groups whose all generalized Cayley cubic
graphs are connected and integral as above discussion. Next, we aim to
determine the non-abelian groups whose all $3$-valent generalized Cayley
graphs are connected and integral.

######  Lemma 4.8.

For any integer $n\geq 2$, the dicyclic group $T_{4n}$ has $3$-valent
generalized Cayley graph which is not connected.

###### Proof.

Let $T_{4n}=\langle a,b\mid
a^{2n}=b^{4}=e,a^{n}=b^{2},b^{-1}ab=a^{-1}\rangle$. Note that $\alpha:a\mapsto
a,b\mapsto a^{n}b$ is an involutory automorphism of $T_{4n}$ and
$S=\\{b,a^{2}b,a^{4}b\\}$ is a generalized Cayley subset of $T_{4n}$ induced
by $\alpha$. From $S^{-1}=\\{a^{n}b,a^{n+2}b,a^{n+4}b\\}$, we have $\langle
SS^{-1}\rangle=\langle a^{2}\rangle$ and $|\langle
SS^{-1}\rangle|=n<\frac{1}{2}|T_{4n}|$. By Theorem 3.6, $GC(T_{4n},S,\alpha)$
is not connected. ∎

######  Lemma 4.9.

For the dihedral group $D_{4n}(n\geq 2)$, all $3$-valent generalized Cayley
graphs of $D_{4n}$ are connected and integral if and only of $n=2$.

###### Proof.

Let $D_{4n}=\langle a,b\mid a^{2n}=b^{2}=e,bab=a^{-1}\rangle$ and
$\alpha:a\mapsto a^{-1},b\mapsto a^{2}b$ an involutory automorphism of
$D_{4n}$. If $n\geq 3$, then $S=\\{a^{3}b,a^{-1}b,ab\\}$ is a generalized
Cayley subset of $D_{4n}$ induced by $\alpha$. As $|\langle
SS^{-1}\rangle|=|\langle a^{2}\rangle|=n<\frac{1}{2}|D_{4n}|$, by Theorem 3.6,
$GC(D_{4n},S,\alpha)$ is not connected. If $n=2$, then the involutory
automorphisms of $D_{8}$ and the corresponding partition of $D_{8}$ are shown
in Table 1 where we omit $\mho=D_{8}\setminus(\omega\cup\Omega)$.

Table 1:  $~{}\text{Involutions of}~{}{\rm\textsf{Aut}}(D_{8})$ |  $\omega$ |  $\Omega$
---|---|---
$\alpha:a\mapsto a,b\mapsto a^{2}b$ |  $\\{e,a^{2}\\}$ |  $\emptyset$
$\beta_{1}:a\mapsto a^{-1},b\mapsto b$ |  $\\{e,a^{2}\\}$ |  $\\{a,a^{3},b,a^{2}b\\}$
$\beta_{2}:a\mapsto a^{-1},b\mapsto ab$ |  $\langle a\rangle$ |  $\emptyset$
$\beta_{3}:a\mapsto a^{-1},b\mapsto a^{2}b$ |  $\\{e,a^{2}\\}$ |  $\\{a,a^{3},ab,a^{3}b\\}$
$\beta_{4}:a\mapsto a^{-1},b\mapsto a^{3}b$ |  $\langle a\rangle$ |  $\emptyset$

From the information in Table 1, we have there is no $3$-valent generalized
Cayley graph induced by $\alpha$, $\beta_{2}$ and $\beta_{4}$. In addition, it
can be verified that all $3$-valent generalized Cayley graphs induced by
$\beta_{1}$ and $\beta_{3}$ are isomorphic to $G_{4}$ in Figure 1. Thus, all
$3$-valent generalized Cayley graphs of $D_{8}$ are connected and integral. ∎

An immediate consequence of above lemmas is

######  Corollary 4.10.

Each of groups $D_{12}\times Z_{2}$, $T_{12}\times Z_{2}$ and $D_{6}\times
Z_{4}$ has $3$-valent generalized Cayley graph which is not connected.

###### Proof.

By Lemmas 4.8 and 4.9, both $D_{12}$ and $T_{12}$ have $3$-valent generalized
Cayley graphs which are not connected. Let $GC(G,S,\alpha)$ be a disconnected
generalized Cayley graph of valency $3$, where $G=D_{12}$ or $T_{12}$. Then
from Lemmas 4.5 and 4.6, $GC(G\times
Z_{2},S\times\\{1\\},\alpha\times{\rm\textsf{id}})$ is disconnected $3$-valent
generalized Cayley graph of $G\times Z_{2}$.

Note that $GC(Z_{4},\\{1\\},\iota)$ is a disconnected generalized Cayley graph
of $Z_{4}$ and $GC(D_{6},\\{b,ab,a^{2}b\\},\beta)$ is a generalized Cayley
graph of $D_{6}$ induced by $\beta:a\mapsto a^{-1},b\mapsto a^{2}b$. Then
$GC(D_{6}\times Z_{4},\\{1\\}\times\\{b,ab,a^{2}b\\},\iota\times\beta)$ is a
disconnected generalized Cayley graph of $D_{6}\times Z_{4}$. ∎

######  Theorem 4.11.

Let $G$ be a finite non-abelian group admitting $3$-valent generalized Cayley
graph. Then all $3$-valent generalized Cayley graphs of $G$ are connected and
integral if and only if $G$ is isomorphic to the dihedral groups $D_{6}$,
$D_{8}$ or the quaternion group $Q_{8}$.

###### Proof.

Since $G$ is non-abelian, its order cannot be equal to $4$. By Lemma 4.3, the
order of $G$ belongs to$\\{6,8,10,12,20,24,30\\}$. We divide the argument into
following conditions:

Case 1: Let $|G|=6$. Then $G\cong D_{6}$. Observe that the generalized Cayley
graphs of $D_{6}$ are $GC(D_{6},\\{a,ab,a^{2}b\\},\alpha_{j})$, where
$\alpha_{j}:a\mapsto a^{-1},b\mapsto a^{j}b$. These generalized Cayley graphs
are isomorphic to $G_{2}$ in Figure 1.

Case 2: Let $|G|=8$. Then $G\cong D_{8}$ or $Q_{8}=\langle a,b\mid
a^{4}=e,a^{2}=b^{2},b^{-1}ab=a^{-1}\rangle$.

Since ${\rm\textsf{Aut}}(Q_{8})\cong S_{4}$, the involutory automorphisms of
$Q_{8}$ are divided into two classes. Let $\alpha:a\mapsto a^{-1},b\mapsto ab$
and $\beta:a\mapsto a^{-1},b\mapsto b$ be involutory automorphisms in
different conjugate class of ${\rm\textsf{Aut}}(Q_{8})$. By Lemma 4.2, it
suffices to discuss the generalized Cayley graphs induced by $\alpha$ and
$\beta$. As $\omega_{\alpha}=\langle a\rangle$ and
$\Omega_{\alpha}=\emptyset$, there is no $3$-valent generalized Cayley graph
induced by $\alpha$. With respect to $\beta$, we obtain subsets
$\omega_{\beta}=\\{e,a^{2}\\}$, $\Omega_{\beta}=\\{a,a^{3},ab,a^{3}b\\}$ and
$\mho_{\beta}=\\{b,a^{2}b\\}$. Thus, there are several $3$-valent generalized
Cayley graphs. It is easy to verify that all $3$-valent generalized Cayley
graphs of $Q_{8}$ induced by $\beta$ are isomorphic to $G_{4}$ in Figure 1.
Therefore, all $3$-valent generalized Cayley graphs of $Q_{8}$ are connected
and integral.

In addition, by Lemma 4.9, all $3$-valent generalized Cayley graphs of $D_{8}$
are connected and integral.

Case 3: Let $|G|=10$. Then $D\cong D_{10}$. Let $S=\\{b,ab,a^{4}b\\}$ be a
generalized Cayley subset of $D_{10}$ induced by $\alpha:a\mapsto
a^{-1},b\mapsto b$ and the corresponding graph is shown in Figure 2. It can be
observed that $GC(D_{10},S,\alpha)$ is a connected graph but it is not
isomorphic to any graph in Figure 1.

Figure 2:

Case 4: Let $|G|=12$. Then $G\cong D_{12}$, $T_{12}$ or $A_{4}$. By Lemmas 4.8
and 4.9, there exists $3$-valent generalized Cayley graph of $D_{12}$ and
$T_{12}$ which is not connected, respectively. For $A_{4}$, let
$S=\\{(123),(124),(12)(34)\\}$, which is a generalized Cayley subset induced
by $\alpha:x\mapsto(12)(34)x(12)(34)$ for all $x\in A_{4}$. As the order of
$\langle SS^{-1}\rangle=\langle(243)\rangle$ is smaller than
$\frac{1}{2}|A_{4}|$, by Theorem 3.6, $GC(G,S,\alpha)$ is disconnected.

Case 5: Let $|G|=20$. Then $G\cong D_{20}$, $T_{20}$ or $F_{5,4}=\langle
a,b\mid a^{5}=b^{4}=e,b^{-1}ab=a^{2}\rangle$. Using Lemmas 4.8 and 4.9 again,
both $D_{20}$ and $T_{20}$ have disconnected generalized Cayley graphs. For
$F_{5,4}$, let $\alpha:a\mapsto a^{-1},b\mapsto b$ be its involutory
automorphism of it and $S=\\{b,b^{2},b^{3}\\}$ a generalized Cayley subset of
$F_{5,4}$ induced by $\alpha$. Since $\langle S\rangle\neq F_{5,4}$, by
Theorem 3.6, $GC(G,S,\alpha)$ is not connected.

Case 6: Let $|G|=24$. Then $G\cong D_{12}\times Z_{2}$, $T_{12}\times Z_{2}$,
$D_{6}\times Z_{4}$, $A_{4}\times Z_{2}$, $Q_{8}\times Z_{3}$, $D_{24}$,
$T_{24}$, $Sym(4)$, $D_{8}\times Z_{3}$, $U_{24}=\langle a,b\mid
a^{8}=b^{3}=e,a^{-1}ba=b^{-1}\rangle$, $V_{24}=\langle a,b\mid
a^{6}=b^{4}=e,ba=a^{-1}b^{-1},b^{-1}a=a^{-1}b\rangle$ or $SL(2,3)$.

The groups $D_{12}\times Z_{2}$, $T_{12}\times Z_{2}$ and $D_{6}\times Z_{4}$
are discussed in Corollary 4.10. As stated in Case $4$, $A_{4}$ has
disconnected $3$-valent generalized Cayley graph. With the same argument in
Corollary 4.10, it follows that $A_{4}\times Z_{2}$ also has disconnected
$3$-valent generalized Cayley graph.

For $Q_{8}\times Z_{3}$, let
$\alpha_{1}:(e,1)\mapsto(e,1),(a,0)\mapsto(a^{-1},0),(b,0)\mapsto(b,0)$ be an
involutory automorphism and $S_{1}=\\{(a,0),(a,1),(a,2)\\}$ a generalized
Cayley subset of $Q_{8}\times Z_{3}$ induced by $\alpha_{1}$. Since $\langle
S_{1}\rangle\neq Q_{8}\times Z_{3}$, $GC(Q_{8}\times Z_{3},S_{1},\alpha_{1})$
is disconnected.

For $D_{24}$, let $\alpha_{2}:a\mapsto a^{5},b\mapsto b$ be its involutory
automorphism and $S_{2}=\\{a^{2},a^{6},a^{10}\\}$ a generalized Cayley subset
induced by $\alpha_{2}$. Clearly, $\langle S_{2}\rangle\neq D_{24}$, hence
$GC(D_{24},S_{2},\alpha_{2})$ is disconnected.

For $T_{24}$, by Lemma 4.8, it has a disconnected $3$-valent generalized
Cayley graph.

For $Sym(4)$, let $\alpha_{3}:x\mapsto(12)x(12)$ be its involutory
automorphism and $S_{3}=\\{(12),(13),(23)\\}$ a generalized Cayley subset
induced by $\alpha_{3}$. Then $GC(Sym(4),S_{3},\alpha_{3})$ is disconnected as
$\langle S_{3}\rangle\neq Sym(4)$.

For $D_{8}\times Z_{3}$, let
$\alpha_{4}:(e,1)\mapsto(e,1),(a,0)\mapsto(a^{-1},0),(b,0)\mapsto(a^{2}b,0)$
be its involutory automorphism and $S_{4}=\\{(a,0),(a,1),(a,2)\\}$ a
generalized Cayley subset induced by $\alpha_{4}$. Then $GC(D_{8}\times
Z_{3},S_{4},\alpha_{4})$ is disconnected as $\langle S_{4}\rangle\neq
D_{8}\times Z_{3}$.

For $U_{24}$, let $\alpha_{5}:a\mapsto a,b\mapsto b^{-1}$ be its involutory
automorphism and $S_{5}=\\{a,a^{7},a^{4}\\}$ a generalized Cayley subset
induced by $\alpha_{5}$. Then $GC(U_{24},S_{5},\alpha_{5})$ is disconnected.

For $V_{24}$, let $\alpha_{6}:a\mapsto a,b\mapsto b^{-1}$ be its involutory
automorphism and $S_{6}=\\{a,a^{5},a^{3}\\}$ a generalized Cayley subset
induced by $\alpha_{6}$. Then $GC(V_{24},S_{6},\alpha_{6})$ is disconnected.

Finally, we show that there is no $3$-valent generalized Cayley graph of
$SL(2,3)=\langle A,B\rangle$, where

$A=\begin{bmatrix}0&1\\\ 2&0\\\ \end{bmatrix},B=\begin{bmatrix}1&1\\\ 0&1\\\
\end{bmatrix}.$

Observe that there are $9$ involutory automorphisms of $SL(2,3)$ which are
divided into two conjugate classes. Let

$\alpha_{7}:\begin{bmatrix}0&1\\\ 2&0\\\
\end{bmatrix}\mapsto\begin{bmatrix}0&2\\\ 1&0\\\
\end{bmatrix},\begin{bmatrix}1&1\\\ 0&1\\\
\end{bmatrix}\mapsto\begin{bmatrix}0&1\\\ 2&2\\\ \end{bmatrix}$

and

$\alpha_{8}:\begin{bmatrix}0&1\\\ 2&0\\\
\end{bmatrix}\mapsto\begin{bmatrix}2&1\\\ 1&1\\\
\end{bmatrix},\begin{bmatrix}1&1\\\ 0&1\\\
\end{bmatrix}\mapsto\begin{bmatrix}1&2\\\ 0&1\\\ \end{bmatrix}$

be involutory automorphisms in different conjugate classes. For $\alpha_{7}$,
we obtain subsets
$\omega_{\alpha_{7}}=\\{E,A,A^{2},A^{3},B^{2}A^{3}B,B^{2}AB\\}$ and
$\Omega_{\alpha_{7}}=\emptyset$. Similarly, with respect to $\alpha_{8}$, we
have
$\omega_{\alpha_{8}}=\\{E,B,B^{2},A^{2},AB^{2}A^{3}B,AB^{2}AB,B^{2}A^{2},AB,B^{2}A^{3},A^{3}B,BA^{2},B^{2}A\\}$
and $\Omega_{\alpha_{8}}=\emptyset$. It follows that there is no $3$-valent
generalized Cayley graph of $SL(2,3)$.

Case 7: Let $|G|=30$. Then $G\cong D_{30}$ or $U_{30}=\langle a,b\mid
a^{10}=b^{3}=e,a^{-1}ba=b^{-1}\rangle$. Let $\alpha:a\mapsto a^{4},b\mapsto b$
be an involutory automorphism of $D_{30}$ and $S=\\{b,a^{5}b,a^{10}b\\}$ a
generalized Cayley subset induced by $\alpha$. Then $GC(D_{30},S,\alpha)$ is
disconnected as $\langle S\rangle\neq D_{30}$. Similarly, let $\beta:a\mapsto
a,b\mapsto b^{-1}$ be an involutory automorphism of $U_{30}$ and
$T=\\{a,a^{-1},a^{5}\\}$ a generalized Cayley subset induced by $\beta$. Then
$GC(U_{30},T,\beta)$ is disconnected as $\langle T\rangle\neq U_{30}$.

In conclusion, among groups having $3$-valent generalized Cayley graph, only
the groups $D_{6}$, $D_{8}$ and $Q_{8}$ satisfy the property that all
$3$-valent generalized Cayley graphs are connected and integral. ∎

## References

  * [1] M. Amooshahi, B. Taeri, On Cayley sum graphs of non-abelian group, Graphs Combin. 32 (2016), 17–29.
  * [2] M. Amooshahi, B. Taeri, On integral Cayley sum graphs, Indian J. Pure Appl. Math. 47 (2016), 583–601.
  * [3] F.C. Bussemaker, D.M. Cvetković, There are exactly 13 connected, cubic, integral graphs, Univ. Beograd Publ. Elektrotehn Fak. Ser. Mat. Fiz. 544 (1976), 43–48.
  * [4] B. Cheyne, V. Gupta, C. Wheeler, Hamilton cycle in addition graphs, Rose-Hulman Undergrad. Math J. 4 (2003), 1–17.
  * [5] A. Hujdurović, K. Kutnar, D. Marušič, Vertex-transitive generalized Cayley graphs which are not Cayley graphs, European J. Combin. 46(2015), 45–50.
  * [6] A. Hujdurović, K. Kutnar, P. Petecki, A. Tanana, On automorphisms and structural properties of generalized Cayley graphs, Filomat 31 (2017), 4033–4040.
  * [7] W. Imrich, S. Klavzar, Product graphs : structure and recognition, Wiley, (2000).
  * [8] Qianfen Liao, Weijun Liu, GCI-property of some groups, Appl. Math. Comput. 438 (2023).
  * [9] D. Marušič, R. Scapellato, N. Zagaglia Salvi, Generalized Cayley graphs, Discrete Math. 102 (1992), 279–285.
  * [10] X. Yang, W.J. Liu, J. Chen, L.H. Feng, GCI-groups in the alternating groups, Appl. Math. Comput. 303 (2017), 42–47.
  * [11] X. Yang, W.J. Liu, L.H. Feng, Isomorphisms of generalized Cayley graphs, Ars. Math. Contemp. 15 (2018), 407–424.
  * [12] X.M. Zhu, W.J. Liu, X. Yang, The isomorphism of generalized Cayley graphs on finite non-abelian simple groups, Discrete Math. 346 (2023), 1–10.
  * [13] X.M. Zhu, X. Yang, J. Chen, Spectra of generalized Cayley graphs on finite abelian groups, Algebra Colloq. 30 (2023), 97–110.