# On $2$-integral Cayley graphs

Alireza Abdollahi aaaaE-mail<EMAIL_ADDRESS>Majid Arezoomand
bbbbEmail<EMAIL_ADDRESS>(corresponding author) Tao Feng ccccEmail:
<EMAIL_ADDRESS>and Shixin Wang cdddEmail<EMAIL_ADDRESS>
a Department of Pure Mathematics, Faculty of Mathematics and Statistics,
University of Isfahan, Isfahan 81746-73441, Iran
b University of Larestan, Larestan 74317-16137, Iran
b Department of Mathematics, Faculty of Science, Shahid Rajaee Teacher
Training University, Tehran, 16785-163, I. R. Iran
cSchool of Mathematics and Statistics, Beijing Jiaotong University, Beijing
100044, P. R. China

###### Abstract

In this paper, we introduce the concept of $k$-integral graphs. A graph
$\Gamma$ is called $k$-integral if the extension degree of the splitting field
of the characteristic polynomial of $\Gamma$ over rational field $\mathbb{Q}$
is equal to $k$. We prove that for any integer $k$, the set of all finite
connected $l$-integral graphs with $l\leqslant k$, is finite. We study
$2$-integral Cayley graphs over finite groups $G$ with respect to Cayley sets
which are a union of conjugacy classes of $G$. Among other general results, we
completely characterize all finite abelian groups having a connected
$2$-integral Cayley graph with valency $2,3,4$ and $5$.

Keywords: Cayley graph, algebraic degree, characters of groups, integral
eigenvalue.

Mathematics Subject Classifications: 05C25, 05C50.

##  1 Introduction and results

A graph is called integral if the eigenvalues of its adjacency matrix are all
integers. This concept was first introduced by Harary and Schwenk [10] and
they raised the question: which graphs are integral? After this,
classifications and construction of the integral graph has become an active
topic. The general way to study the integral graphs is to study some special
classes of graphs, such as Cayley graphs. A Cayley digraph over a group $G$
with respect to a subset $S$ of $G$, denoted by $\mathrm{Cay}(G,S)$, is a
digraph with vertex set $G$ such that $(g,h)$ is an arc if and only if
$hg^{-1}\in S$. If $S=\emptyset$ then $\mathrm{Cay}(G,S)$ is the empty graph,
and if $1_{G}\in S$ then the corresponding Cayley digraph has a loop at each
vertex. If $S$ is inverse-closed, i.e $S=S^{-1}=\\{s^{-1}\mid s\in S\\}$, then
$\mathrm{Cay}(G,S)$ is an undirected graph. Furthermore, $\mathrm{Cay}(G,S)$
is a connected digraph if and only if $G=\langle S\rangle$. The Cayley graph
over a cyclic group is also called circulant. A subset $S$ of a group $G$ is
called normal if for any $g\in G$, we have $g^{-1}Sg=S$. Clearly, a subset of
any abelian group is a normal set. The Cayley graph over a group $G$ with
respect to a normal subset $S$ is also called normal Cayley graph. A large
number of results on the eigenvalues of Cayley graphs have been produced over
the past more than four decades, for a survey on this topic see [14]. All
Cayley digraphs in this paper are loop-free and non-empty.

Mönius, Steuding and Stumpfm [17] introduced the concept of the splitting
fileds and the algebraic degrees of graphs to investigate which graph
properties prevent integral eigenvalues. For a graph $\Gamma$, its splitting
field $\mathbb{SF}(\Gamma)$ is the smallest field extension of the rational
number field $\mathbb{Q}$ which contains all the eigenvalues of $\Gamma$. The
extension degree $[\mathbb{SF}(\Gamma):\mathbb{Q}]$ is called the algebraic
degree of $\Gamma$, denoted by $\mathrm{deg}(\Gamma)$. Since eigenvalues of
graphs are all algebraic integers, a graph $\Gamma$ is integral if and only if
$\mathrm{deg}(\Gamma)=1$. Some works have been done, very recently, on
determining algebraic degrees of Cayley graphs and their generalization [11,
15, 20, 21, 22]. Note that the question raised by Harary and Schwenk can be
translated as: which graphs have algebraic degree 1? It is naturally to ask
which graphs have algebraic degree $k$ for a given positive integer $k$? For
studying this question we define $k$-integral graphs as follows.

Definition. For a given positive integer $k$, a graph $\Gamma$ is called
$k$-integral if $\mathrm{deg}(\Gamma)=k$.

To study $k$-integral graphs is crucial to know that the number of such
connected graphs are finite. In section 3, we will give a positive answer to
this question. We prove that for any integer $k$, the set of all finite
connected graphs with algebraic degree at most $k$ is finite (see Theorem
3.2). Our result further improves [7, Theorem 2] which state that all regular,
connected, integral graphs of a fixed degree is finite.

Let $G=\langle a\rangle\cong\mathbb{Z}_{n}$ be a cyclic group of order $n$ and
$S$ be an inverse-closed subset of $G$. In 2005, W. So proved that
$\mathrm{Cay}(G,S)$ is integral if and only if $S$ is a union of $G_{n}(d)$s,
where $d\neq n$ is a divisor of $n$ and $G_{n}(d)=\\{a^{k}\mid(k,n)=d\\}$ [19,
Theorem 7.1]. Since then some authors tried to generalize So’s result to
finite abelian groups and non-abelian groups [4, 5, 13]. An interesting
question is how to construct $k$-integral Cayley graphs over cyclic groups.
For $G=\langle a\rangle\cong\mathbb{Z}_{n}$ and any divisor $d\neq n$ of $n$,
if we put $[a^{d}]=\\{a^{k}\mid(k,n)=d\\}$, then by So’s result,
$\mathrm{Cay}(G,[a^{d}])$ is integral. In section 4, we have found that using
some special subsets of $[a^{d}]$, one can construct $k$-integral Cayley
graphs over cyclic groups (see Proposition 4.3). As a corollary, we also give
a method to construct a $p$-integral Cayley graph over cyclic group
$\mathbb{Z}_{n}$, where $n\geqslant 5$ and $p$ is a prime divisor of $n$ (see
Corollary 4.4).

The problem of classifying all finite groups having a connected integral
Cayley graph of given valency is started by Abdollahi and Vatandoost [2]. They
classified all such groups for valency $2$ and $3$ in [2] and partially for
valency $4$ in [3]. Then, the problem completely solved, using computer, for
valency $4$ in [16] by Minchenko and Wanless. We also are interest to classify
all finite abelian groups having a connected $2$-integral Cayley graph with
small valency. To this purpose, we define the set $\mathcal{G}_{k}$ to be the
set of all finite groups having a connected undirected $2$-integral Cayley
graph, and using Corollary 4.6 and some other general results, we completely
characterize all finite abelian groups belong to $\mathcal{G}_{k}$ for
$2\leqslant k\leqslant 5$ (see Theorem 5.1, 5.3, 5.5, and 5.10). As a
corollary, we also give the classification of $2$-integral abelian Cayley
graphs with valency $2,3,4$ and $5$.

##  2 Preliminaries and notations

In this paper, (di)graphs are finite, loop-free and without multiple edges.
Also the groups are finite. For a graph $\Gamma$, we use $V(\Gamma)$ and
$E(\Gamma)$ to denote the set of vertices and the set of edges of $\Gamma$,
respectively. The notation $i\sim j$ means that there is an edge between the
vertices $i$ and $j$ of $\Gamma$. Our notations are standard and mainly taken
from [9] and [12], but for the reader’s convenience we recall some of them as
follows:

  * •

$\langle g\rangle$ : the cyclic group generated by $g$.

  * •

$\mathbb{Z}_{n}$ : the additive group of integers module $n$.

  * •

$\mathbb{Z}_{n}^{*}$ : the multiplicative group of $\mathbb{Z}_{n}$.

  * •

$\mathrm{Aut}(G)$ : the automorphism group of the group $G$.

  * •

$\tau$ : the automorphism of an abelian group which maps every element to its
inverse.

  * •

$G_{1}\times G_{2}$ : the direct product of groups $G_{1}$ and $G_{2}$.

  * •

$[g]$ : the set $\\{h\in G\mid\langle h\rangle=\langle g\rangle\\}$, where
$g\in G$.

  * •

$Z(G)$ : the center of $G$.

  * •

$g^{H}$ : $\\{g^{\sigma}\mid\sigma\in H\\}$, where $g\in G$ and
$H\leqslant\mathrm{Aut}(G)$.

  * •

$\mathrm{Irr}(G)$ : the set of all inequivalent $\mathbb{C}$-irreducible
characters of group $G$.

  * •

$K_{n}$ : the complete graph with $n$ vertices.

  * •

$K_{n,m}$ : the complete bipartite graph with parts having $m$ and $n$
vertices.

  * •

$\Gamma_{1}\vee\Gamma_{2}$ : the join of graphs $\Gamma_{1}$ and $\Gamma_{2}$.

  * •

$\Gamma_{1}\square\Gamma_{2}$ : the Cartesian product of graphs $\Gamma_{1}$
and $\Gamma_{2}$.

  * •

$\Gamma_{1}\otimes\Gamma_{2}$ : the direct product of $\Gamma_{1}$ and
$\Gamma_{2}$.

  * •

$\Gamma_{1}\boxtimes\Gamma_{2}$ : the strong product of $\Gamma_{1}$ and
$\Gamma_{2}$.

  * •

$\Gamma_{1}[\Gamma_{2}]$ : the lexicographic product of $\Gamma_{1}$ and
$\Gamma_{2}$.

  * •

$\mathbb{SF}(\Gamma)$ : the splitting field of a graph $\Gamma$.

Let $\Gamma$ be a graph and $\mathbb{SF}(\Gamma)$ be the splitting field of
$\Gamma$. By the definition of $k$-integral graph, $\Gamma$ is $2$-integral
means that $[\mathbb{SF}(\Gamma):\mathbb{Q}]=2$. It is known that
$[\mathbb{SF}(\Gamma):\mathbb{Q}]=2$ implies that
$\mathbb{SF}(\Gamma)=\mathbb{Q}(\alpha)$, where
$\alpha\in\mathbb{SF}(\Gamma)\setminus\mathbb{Q}$ and
$\alpha^{2}\in\mathbb{Z}$. Next we will show that there exists a $2$-integral
graph with $n$ vertices for any integer $n\geqslant 3$.

Before the proof, we recall that the eigenvalues of the complete bipartite
graph $K_{m,n}$ are $0$ with multiplicity $m+n-2$ and $\pm\sqrt{mn}$ with
multiplicity $1$. Hence if $mn$ is not a square integer,
$\mathbb{SF}(K_{m,n})=\mathbb{Q}(\sqrt{mn})$ and so $K_{m,n}$ is a
$2$-integral graph.

######  Lemma 2.1.

For any integer $n\geqslant 3$, there exists a $2$-integral graph with $n$
vertices.

###### Proof.

The eigenvalues of $P_{3}$ are $0$ and $\pm\sqrt{2}$, and the eigenvalues of
$P_{4}$ are $\frac{\pm 1\pm\sqrt{5}}{2}$. Thus $P_{3}$ and $P_{4}$ are both
$2$-integral. So we may assume that $n\geqslant 5$. If $n$ is odd, then there
exist distinct integers $n_{1}$ and $n_{2}$ such that $n=n_{1}+n_{2}$ and
$\sqrt{n_{1}n_{2}}$ is not an integer. Now $K_{n_{1},n_{2}}$ is $2$-integral
with $n$ vertices. Let $n=2m$ is even and $m\geqslant 3$. In this case, all
distinct eigenvalues of graph $K_{m-2,m+2}$ are $0$ and $\pm\sqrt{m^{2}-4}$.
Clearly, $\pm\sqrt{m^{2}-4}\notin\mathbb{Z}$ and $K_{m-2,m+2}$ is
$2$-integral. This completes the proof. ∎

Next we introduce a special subset of a group, which will be useful to study
$k$-integral Cayley graphs. Let $G$ be a finite group, and for any $g\in G$,
define $[g]:=\\{h\in G\mid\langle g\rangle=\langle h\rangle\\}$. If $g$ and
$h$ be two elements of $G$, then $[g]=[h]$ or $[g]\cap[h]=\emptyset$. Hence
the set $\Omega(G):=\\{[g]\mid g\in G\\}$ is a partition of $G$. In the case
that $G=\langle a\rangle\cong\mathbb{Z}_{n}$ is a cyclic group of order $n$
generated by $a$, for a divisor $d\neq n$ of $n$, we have
$[a^{d}]=\\{a^{k}\mid 1\leqslant k\leqslant n-1,~{}(k,n)=d\\}$ which is
denoted by $G_{n}(d)$ in the literature, which is important for a see [19].

##  3 The number of $k$-integral graphs

One of the main questions to study $k$-integral graphs is to know that the
number of such connected graphs are finite. In this section, we give a
positive answer to this question.

######  Proposition 3.1.

Let $\Gamma$ be a graph with algebraic degree $k$. Then there exists a
function $f:\mathbb{N}\times\mathbb{N}\rightarrow\mathbb{N}$ such that the
number of distinct eigenvalues of $\Gamma$ is at most $f(k,\Delta(\Gamma))$,
where $\Delta(\Gamma)$ is the maximum degree of vertices of $\Gamma$.

###### Proof.

Let $p(x)$ be the characteristic polynomial of $\Gamma$. Then we have
$p(x)=p_{1}(x)^{t_{1}}\cdots p_{l}(x)^{t_{l}}$, where $p_{i}(x)$s are monic
polynomials on $\mathbb{Q}[x]$, irreducible and pairwise relatively prime and
$t_{i}$’s are positive integers. Since $p_{i}(x)$s are irreducible in
$\mathbb{Q}$ and so in $\mathbb{C}$. Then the roots of $p_{i}(x)$s are
pairwise distinct and $p_{i}(x)$s has no repeated roots.

Let $p_{i}(x)=x^{k_{i}}+a_{1}x^{k_{i}-1}+\cdots+a_{k_{i}-1}x+a_{k_{i}}$, where
$a_{i}$s are integers. Since the roots of $p_{i}(x)$ are also the roots of
$p(x)$, $k_{i}$ divides $k$. Furthermore, since the roots of $p_{i}(x)$ are
eigenvalues of $\Gamma$, the absolute value of any root of $p_{i}(x)$ is less
than or equal to $\Delta(\Gamma)$, the maximum degree of vertices of $\Gamma$.
Now by Vieta’s formula and triangle inequality, for each $j$, we have

$|a_{j}|\leqslant\binom{k_{i}}{j}\Delta(\Gamma)^{k_{i}-j}.$

This means that the number of such polynomials and so the number of distinct
eigenvalues of $\Gamma$ is less than or equal to a function of $k$ and
$\Delta(\Gamma)$, as desired. ∎

######  Theorem 3.2.

For any integer $k$, the set of all finite connected graphs with algebraic
degree at most $k$ is finite.

###### Proof.

Since for any connected graph with $s$ distinct eigenvalues and diameter $D$,
we have $D\leqslant s-1$, the result directly follows from Proposition 3.1 and
boundlessness of the size of a connected graph with given diameter and maximum
degree. ∎

##  4 $k$-integral Cayley graphs

In this section, we are going to study the algebraic degree of normal Cayley
(di)graphs. Let $G$ be a finite group with exponent $m$, that is the smallest
positive integer such that $g^{m}=1$ for all $g\in G$, and
$\Gamma=\mathrm{Cay}(G,S)$ for some normal subset $S$ of $G$. Then eigenvalues
of $\Gamma$ are of the form $\frac{\chi(S)}{\chi(1)}$, where $\chi$ is an
irreducible character of $G$ and $\chi(S)=\sum_{s\in S}\chi(s)$ [8, 23]. Also
by [21, Corollary 3.10], $\deg(\Gamma)=\frac{\varphi(m)}{|H^{\prime}|}$, where
$H^{\prime}=\\{k\in\mathbb{Z}_{m}^{*}\mid S^{k}=S\\}$, where
$S^{k}=\\{s^{k}\mid s\in S\\}$. Next let us start with some basic results.

######  Lemma 4.1.

Let $G$ be a finite group, $\Omega(G)=\\{[g_{1}],\ldots,[g_{n}]\\}$,
$\emptyset\neq S$ be a normal subset of $G$, and $\Gamma=\mathrm{Cay}(G,S)$.
Then $S=\bigcup_{i=1}^{n}S_{i}$, where $S_{i}=[g_{i}]\cap S$ for each
$1\leqslant i\leqslant n$, and for $S_{i}\neq\emptyset$, $\deg(\Gamma_{i})$ is
a divisor of $\deg(\Gamma)$, where $\Gamma_{i}=\mathrm{Cay}(G,S_{i})$.

###### Proof.

For each $1\leqslant i\leqslant n$, put $S_{i}=[g_{i}]\cap S$. Since
$\Omega(G)$ is a partition of $G$, $\\{S_{i}\mid 1\leqslant i\leqslant n\\}$
is a partition of $S$. Let $A=\mathrm{Aut}(G)$, $m$ be the exponent of $G$. By
[21, Corollary 3.10], $\deg(\Gamma)=\frac{\varphi(m)}{|H|}$, where
$H=\\{k\in\mathbb{Z}_{m}^{*}\mid S^{k}=S\\}$, and if $S_{i}\neq\emptyset$,
$\deg(\Gamma_{i})=\frac{\varphi(m)}{|H_{i}|}$, where
$H_{i}=\\{k\in\mathbb{Z}_{m}^{*}\mid S_{i}^{k}=S_{i}\\}$. Next we will show
that $H\leqslant H_{i}$, and so by
$|\mathbb{Z}_{m}^{*}:H|=|\mathbb{Z}_{m}^{*}:H_{i}||H_{i}:H|$, we have
$\deg(\Gamma)=|H_{i}:H|\deg(\Gamma_{i})$ and then complete the proof.

Suppose that $k\in\mathbb{Z}_{m}^{*}$. Then for any $x\in G$, we have $\langle
x^{k}\rangle=\langle x\rangle$. If $S^{k}=S$, since $\\{S_{i}\mid 1\leqslant
i\leqslant n\\}$ is a partition of $S$, we conclude that $S_{i}^{k}=S_{i}$.
Then for every $1\leqslant i\leqslant n$ such that $S_{i}\neq\emptyset$, we
have $H\leqslant H_{i}$. ∎

######  Lemma 4.2.

Let $G$ be a finite group, $g\in G$ and $\emptyset\neq S\subseteq[g]$. Then
$\langle S\rangle=\langle g\rangle$ and the splitting field of
$\mathrm{Cay}(G,S)$ and $\mathrm{Cay}(\langle g\rangle,S)$ are the same.

###### Proof.

Since $x\in[g]$ if and only if $\langle x\rangle=\langle g\rangle$, the first
part is clear. On the other hand, $\mathrm{Cay}(G,S)$ is isomorphic to
$|G:\langle S\rangle|$ copies of $\mathrm{Cay}(\langle S\rangle,S)$, where
$\mathrm{Cay}(\langle S\rangle,S)$ is a connected component of
$\mathrm{Cay}(G,S)$. Hence the set of all distinct eigenvalues of
$\mathrm{Cay}(G,S)$ and $\mathrm{Cay}(\langle S\rangle,S)$ are the same and so
their splitting fields are the same, as desired. ∎

By the above, two lemmas, it seems that to study $k$-integral Cayley graphs,
the studying of $k$-integral $\mathrm{Cay}(\langle g\rangle,S)$, where
$\emptyset\neq S\subseteq[g]$ is important. Let $G=\langle
a\rangle\cong\mathbb{Z}_{n}$ and $\Gamma=\mathrm{Cay}(G,S)$. By [15, Theorem
1], we know that $\deg(\Gamma)$ is a divisor of $\varphi(n)$. In what follows,
we are going to give a way to find possible inverse-closed generating sets of
$G$ such that $\deg(\mathrm{Cay}(G,S))=d^{\prime}$ for some
$d^{\prime}\mid\varphi(n)$.

######  Proposition 4.3.

Let $n\geqslant 3$, $d\neq n$, $d\mid n$, $G=\langle
a\rangle\cong\mathbb{Z}_{n}$ and $A=\mathrm{Aut}(G)$. Let $T_{d}=\\{\sigma\in
A\mid g^{\sigma}=g,~{}\forall g\in[a^{d}]\\}$ and $K_{d}$ be a subgroup of $A$
containing $T_{d}$. For a fixed integer $1\leqslant k\leqslant n-1$ with
$(k,n)=d$, put $S_{k,d}=\\{(a^{k})^{\sigma}\mid\sigma\in K_{d}\\}$ and
$\Gamma_{k,d}=\mathrm{Cay}(G,S_{k,d})$. Then

  * $(1)$

$S_{k,d}$ is inverse-closed if $K_{d}$ contains the element $\tau$ of $A$
which maps $a$ to $a^{-1}$;

  * $(2)$

$|S_{k,d}|=|K_{d}:T_{d}|\leqslant\varphi(\frac{n}{d})$;

  * $(3)$

$S_{k,d}$ is a generating set of $G$ if and only if $d=1$. In this case,
$T_{1}=\\{1_{A}\\}$.

  * $(4)$

$\deg(\Gamma_{k,d})$ is a divisor of $|A:K_{d}|$;

  * $(5)$

$\Gamma_{k,d}$ is integral if and only if $A=K_{d}$. In this case,
$S_{k,d}=[a^{d}]$.

In particular, if $|A:K_{d}|=p$ for some prime $p$, then $\Gamma_{k,d}$ is
$p$-integral.

###### Proof.

Part $(1)$ is obvious by the definition of $\tau$. Since $[a^{d}]$ is the set
of all elements of $G$ with order $\frac{n}{d}$, and each automorphism of $G$
preserves the order of elements of $G$, $S_{k,d}\subseteq[a^{d}]$ and $A$ acts
on $[a^{d}]$. Furthermore, $T_{d}$ is the kernel of this action. Hence
$A/T_{d}$ is a permutation group on $[a^{d}]$. Furthermore, $S_{k,d}$ is an
orbit of $K_{d}$ in this action. Since $A$ is abelian, $K_{d}/T_{d}$ is
abelian. This implies that $K_{d}/T_{d}$ is regular on $S_{k,d}$, which means
$|S_{k,d}|=|K_{d}:T_{d}|\leqslant|[a^{d}]|=\varphi(\frac{n}{d})$. This proves
$(2)$. Since $S_{k,d}\subseteq[a^{d}]$, $\langle
S_{k,d}\rangle\subseteq\langle a^{d}\rangle$. Hence $S_{k,d}$ generates $G$ if
and only if $d=1$. Clearly if $d=1$ then $T_{1}$ fixes $a$ and so
$T_{1}=\\{1_{A}\\}$. This proves $(3)$.

By [15, Theorem 1], $\deg(\Gamma_{k,d})=\frac{|A|}{|H|}$, where $H=\\{a\in
A\mid(S_{k,d})^{a}=S_{k,d}\\}$. Clearly, we have $T_{d}\leqslant H$. On the
other hand $K_{d}\leqslant H$. Hence we have $K_{d}/T_{d}\leqslant
H/T_{d}\leqslant A/T_{d}$ and so
$|A:K_{d}|=|A/T_{d}:K_{d}/T_{d}|=|A/T_{d}:H/T_{d}||H/T_{d}:K_{d}/T_{d}|=|A:H||H/T_{d}:K_{d}/T_{d}|$
which implies that $\deg(\Gamma_{k,d})$ divides $|A:K_{d}|$, which proves
$(4)$.

If $K_{d}=A$ then $\deg(\Gamma_{k,d})=1$, by $(4)$, which means $\Gamma_{k,d}$
is integral. Conversely, suppose that $\Gamma_{k,d}$ is integral. Since
$S_{k,d}$ is a subset of $[a^{d}]$, [19, Theorem 7.1] implies
$S_{k,d}=[a^{d}]$. Since $S_{k,d}$ is an orbit of $K_{d}$, we conclude that
$A$ acts transitively on $[a^{d}]$ and so $A/T_{d}$ is a regular permutation
group on $[a^{d}]$. Thus $|A/T_{d}|=|K_{d}/T_{d}|$, which means $A=K_{d}$.
This completes the proof. ∎

The above proposition enables us to construct a $p$-integral Cayley graph over
$\mathbb{Z}_{n}$ of valency $\frac{\varphi(n)}{p}$, where $n\geqslant 5$,
$n\neq 6$ and $p$ is a prime divisor of $\varphi(n)$.

######  Corollary 4.4.

Let $n\geqslant 5$ be an integer.

  * $(1)$

If $\varphi(n)$ is not a power of $2$, then for every odd prime divisor of
$\varphi(n)$, there exists an undirected $p$-integral Cayley graph $\Gamma$
over $\mathbb{Z}_{n}$ with valency $\frac{\varphi(n)}{p}$;

  * $(2)$

If $\varphi(n)$ is a power of $2$, then there exists an undirected
$2$-integral Cayley graph $\Gamma$ over $\mathbb{Z}_{n}$ with valency
$\frac{\varphi(n)}{2}$.

In particular, for every prime $p$ and integer $k\geqslant 1$ with
$p^{k}\geqslant 5$, there exists a connected $p$-integral circulant graph of
order $p^{k}$ and valency $p^{k-2}(p-1)$.

###### Proof.

Let $A=\mathrm{Aut}(\mathbb{Z}_{n})$ and $\tau$ be the element of $A$ which
maps each element to its inverse.

Suppose that $\varphi(n)$ is not a power of $2$ and $p$ is an odd prime
divisor of $\varphi(n)$. Since $p\,\,\big{|}\,\,\varphi(n)$ and $A$ is an
abelian group of order $\varphi(n)$, there exists a subgroup $K$ of $A$ such
that $|A:K|=p$. Put $H:=\langle K,\tau\rangle$. If $\langle\tau\rangle\nleq K$
then $K$ is a proper subgroup of $H$ and $|H:K|=2$. Since
$p=|A:K|=|A:H||H:K|=2|A:H|$, we have $p=2$ and $A=H$, a contradiction. Hence
$\langle\tau\rangle\leqslant K$. By Proposition 4.3, there exists an inverse-
closed generating set $S$ (put $k=d=1$ in the proposition) of $G$ such that
$|S|=|K|=\frac{\varphi(n)}{p}$ and $\deg(\Gamma)=p$, where
$\Gamma=\mathrm{Cay}(\mathbb{Z}_{n},S)$. This proves $(1)$.

Suppose $\varphi(n)$ is a power of $2$, that is, $\varphi(n)=2^{k}$ for some
$k\geqslant 3$. Hence $A$ is an abelian $2$-group of order $2^{k}$. We know
that there exists a maximal subgroup $M\neq 1$ of $A$ containing
$\langle\tau\rangle$. Since $A$ is abelian, $|A:M|$ must be a prime, which
means $|A:M|=2$ and the $(2)$ follows from Proposition 4.3 as above.

For every prime $p$ and integer $k\geqslant 1$ with $p^{k}\geqslant 5$, by
$(1)$, $(2)$ and $\varphi(p^{k})=p^{k-1}(p-1)$, there exists a connected
$p$-integral circulant graph of order $p^{k}$ and valency $p^{k-2}(p-1)$. This
complete the proof. ∎

Next we are going to give a way to construct the subgroup $K$, given in
Proposition 4.3, which contains $\langle\tau,T_{d}\rangle$ for given $d\mid
n$. Let $n=p_{1}^{m_{1}}p_{2}^{m_{2}}\ldots p_{k}^{m_{k}}\geqslant 5$, where
$2\leqslant p_{1}<\ldots<p_{k}$ are primes and $m_{i}$’s are positive
integers. Recall that

  * (1)

$\mathbb{Z}_{n}\cong\mathbb{Z}_{p_{1}^{m_{1}}}\times\ldots\times\mathbb{Z}_{p_{k}^{m_{k}}}$;

  * (2)

$\mathrm{Aut}(\mathbb{Z}_{n})\cong\mathbb{Z}_{n}^{*}\cong
Z_{p_{1}^{m_{1}}}^{*}\times\ldots\times\mathbb{Z}_{p_{k}^{m_{k}}}^{*}$, where
if $p_{i}\neq 2$ then $\mathbb{Z}_{p_{i}^{m_{i}}}^{*}$ is a cyclic group of
order $p_{i}^{m_{i}-1}(p_{i}-1)$, and if $p_{i}=2$ then
$\mathbb{Z}_{2^{0}}^{*}=\mathbb{Z}_{2^{1}}^{*}=1$,
$\mathbb{Z}_{2^{2}}\cong\mathbb{Z}_{2}$, and
$\mathbb{Z}_{2^{m_{i}}}^{*}\cong\mathbb{Z}_{2^{m_{i}-2}}\times\mathbb{Z}_{2}$
for $m_{i}\geqslant 3$.

Suppose $G=\langle a\rangle\cong\mathbb{Z}_{n}$, $A=\mathrm{Aut}(G)$ and $K$
be a subgroup of $A$ containing $\langle\tau,T_{d}\rangle$. Let
$G=P_{1}\times\ldots\times P_{k}$, where $P_{i}=\langle
a_{i}\rangle\cong\mathbb{Z}_{p_{i}^{m_{i}}}$ and $a=a_{1}\cdots a_{k}$. Then
we may assume that $A=A_{1}\times\ldots\times A_{k}$, where
$A_{i}=\mathrm{Aut}(P_{i})$. If $\sigma=(\sigma_{1},\ldots,\sigma_{k})\in A$,
then $a^{\sigma}=a_{1}^{\sigma_{1}}\cdots a_{k}^{\sigma_{k}}$. Notice that
$\sigma=(\sigma_{1},\ldots,\sigma_{k})\in T_{d}$ if and only if for each
$1\leqslant i\leqslant k$ and each $1\leqslant l\leqslant n$ with $(l,n)=d$,
we have $(a_{i}^{l})^{\sigma_{i}}=a_{i}^{l}$.

Now we deal with the following cases:

Case I. $p_{1}=2$ and $1\leqslant m_{1}\leqslant 2$, or $p_{1}\neq 2$. In this
case, for each $i$, $A_{i}$ is a cyclic group of order
$p_{i}^{m_{i}-1}(p_{i}-1)$. Let $A_{i}=\langle\pi_{i}\rangle$. Then
$A=\langle\pi_{1},\ldots,\pi_{k}\rangle$ and so
$\tau=\pi_{1}^{l_{1}}\pi_{2}^{l_{2}}\cdots\pi_{k}^{l_{k}}$ for some non-
negative integers $l_{1},\ldots,l_{k}$. This implies that for each $i$,
$l_{i}=0$ or $\frac{p_{i}^{m_{i}-1}(p_{i}-1)}{2}$. Since $a^{\tau}=a^{-1}$,
$l_{i}=\frac{p_{i}^{m_{i}-1}(p_{i}-1)}{2}$ for each $i$.

Case II. $p_{1}=2$ and $m_{1}\geqslant 3$. In this case,
$A_{1}=\langle\sigma\rangle\times\langle\theta\rangle\cong\mathbb{Z}_{2^{m_{1}-2}}\times\mathbb{Z}_{2}$,
where

$\sigma:a_{1}\mapsto a_{1}^{5},~{}~{}~{}\theta:a_{1}\mapsto a_{1}^{-1},$

and for $i\neq 1$, $A_{i}=\langle\pi_{i}\rangle$, where $\pi_{i}$ has order
$p_{i}^{m_{i}-1}(p_{i}-1)$. Since $a^{\tau}=a^{-1}$, we have
$\tau=\sigma^{2^{m_{1}-3}}\theta\pi_{2}^{l_{2}}\ldots\pi_{k}^{l_{k}}$, where
$l_{i}=\frac{p_{i}^{m_{i}-1}(p_{i}-1)}{2}$.

By the above discussions, to construct a subgroup $K$ of $A$ containing
$\langle\tau,T_{d}\rangle$, it is enough to find a generator of any group
$A_{i}$, say $\pi_{i}$, put $\tau$ as one the above cases and use the fact
that $T_{d}\leqslant K\leqslant\langle\pi_{1},\ldots,\pi_{k}\rangle$. We
illustrate this by the following example.

######  Example 4.1.

Let $G=\langle a\rangle\cong\mathbb{Z}_{20}$. Then $G=H\times L$, where
$H=\langle a^{5}\rangle\cong\mathbb{Z}_{4}$ and $L=\langle
a^{4}\rangle\cong\mathbb{Z}_{5}$. Let $A=\mathrm{Aut}(G)$,
$A_{1}=\mathrm{Aut}(H)$ and $A_{2}=\mathrm{Aut}(L)$. Then
$A_{1}=\langle\pi_{1}\rangle\cong\mathbb{Z}_{2}$ and
$A_{2}=\langle\pi_{2}\rangle\cong\mathbb{Z}_{4}$, where

$\displaystyle\pi_{1}:$ $\displaystyle~{}H\mapsto H,~{}~{}a^{5}\mapsto
a^{15},$ $\displaystyle\pi_{2}:$ $\displaystyle~{}L\rightarrow
L,~{}~{}~{}a^{4}\mapsto a^{12}.$

Since there exist integers $s_{1},s_{2}$ such that $5s_{1}+4s_{2}=1$,
$a=(a^{5})^{s_{1}}(a^{4})^{s_{2}}$. Take $s_{1}=-3$, $s_{2}=4$ and from the
above discussions, we have $A=A_{1}\times
A_{2}=\langle\pi_{1}\rangle\times\langle\pi_{2}\rangle=\\{\pi_{2}^{j},\pi_{1}\pi_{2}^{j}\mid
0\leqslant j\leqslant 3\\}$, and for each $0\leqslant j\leqslant 3$,

$\displaystyle\pi_{2}^{j}:~{}$ $\displaystyle G\rightarrow G,~{}~{}a\mapsto
a^{-15+16\times 3^{j}},$ $\displaystyle\pi_{1}\pi_{2}^{j}:~{}$ $\displaystyle
G\rightarrow G,~{}~{}a\mapsto a^{-5+16\times 3^{j}}.$

For convenience of later discussion, let $\sigma_{j}=\pi_{2}^{j}$ and
$\theta_{j}=\pi_{1}\pi_{2}^{j}$ for each $0\leqslant j\leqslant 3$. From
above, we have $\tau=\theta_{2}$.

Now we wish to construct an undirected $2$-integral Cayley graph over $G$. To
this end, by Proposition 4.3, it is enough to construct a subgroup $K_{d}$ of
$A$ with index $2$ containing $\langle\tau,T_{d}\rangle$ for some $d|n$.
Clearly, $d\mid n$ if and only if $d=1,2,4,5,10,20$.

First let $d=1$. Then $T_{d}$ is the identity subgroup of $A$. Put
$K_{1}:=\langle\tau,\sigma_{2}\rangle$. Since
$\tau\notin\langle\sigma_{2}\rangle$, we have
$K_{1}=\langle\sigma_{2}\rangle\times\langle\tau\rangle$, and so that
$|K_{1}|=2|\langle\sigma_{2}\rangle|$. On the other hand,
$|\langle\sigma_{2}\rangle|=o(\pi_{2}^{2})=2$. Hence $|K_{1}|=4$ and
$|A:K_{1}|=2$. Now let $1\leqslant k\leqslant 19$ be an integer comprime to
$20$. Then, by the notations of Proposition 4.3,

$\displaystyle S_{k,1}$ $\displaystyle=$
$\displaystyle\\{(a^{k})^{\sigma_{2l^{\prime}}},(a^{-k})^{\sigma_{2l^{\prime}}}\mid
0\leqslant l^{\prime}\leqslant 1\\}$ $\displaystyle=$
$\displaystyle\\{a^{k(-15+3^{2l^{\prime}}16)},a^{-k(-15+3^{2l^{\prime}}16)}\mid
0\leqslant l^{\prime}\leqslant 1\\}.$

Furthermore, $\mathrm{Cay}(G,S_{k,1})$ is a connected $4$-regular $2$-integral
Cayley graph, where
$S_{1,1}=S_{9,1}=S_{11,1}=S_{19,1}=\\{a,a^{19},a^{9},a^{11}\\}$,
$S_{3,1}=S_{7,1}=S_{13,1}=S_{17,1}=\\{a^{3},a^{17},a^{7},a^{13}\\}$.

Now let $d=2$. Then for $1\leqslant l\leqslant 20$, we have $(l,20)=2$ if and
only if $l=2,6,14,18$. Hence, by the above discussion and
$(a^{10})^{\pi_{1}}=a^{10}$, $(a^{8})^{\pi_{2}}=a^{4}$,
$(a^{8})^{\pi_{2}^{2}}=a^{12}$, $(a^{8})^{\pi_{2}^{3}}=a^{16}$, we have
$T_{2}=\\{\sigma_{0}=id_{A},\theta_{0}\\}$. Put
$K_{2}:=\langle\tau,\theta_{0}\rangle$. Then $|K_{2}|=4$, $|A:K_{2}|=2$,
$|K_{2}:T_{2}|=2$, and $\mathrm{Cay}(G,S_{k,2})$, where $k=2,6,14,18$, is a
$2$-regular and $2$-integral Cayley graph, where
$S_{2,2}=S_{18,2}=\\{a^{2},a^{18}\\}$, $S_{6,2}=S_{14,2}=\\{a^{6},a^{14}\\}$.

Now let $d=4$. Then, since $a^{20}=1$ and no power of $\pi_{2}$ maps $a^{16}$
to itself, similar to the previous paragraph, we see that
$T_{4}=\\{\sigma_{0},\theta_{0}\\}$. Hence, similar to the case $d=2$, we
conclude that $\mathrm{Cay}(G,S_{k,4})$, where $k=4,8,12,16$ is $2$-integral.
More precisely, $S_{4,4}=S_{16,4}=\\{a^{4},a^{16}\\}$ and
$S_{8,4}=S_{12,4}=\\{a^{8},a^{12}\\}$.

Now let $d=5$. Since $a^{25}=a^{5}$ and $a^{20}=1$, similar to the above, we
see that $T_{5}=\\{\sigma_{0},\sigma_{1},\sigma_{2},\sigma_{3}\\}$, which
means that $\langle\tau,T_{5}\rangle=A$ and so, by Proposition 4.3, the
corresponding Cayley graphs are integral. So in this case we can not construct
any $2$-integral Cayley graph over $G$.

In the case, $d=10,20$, clearly $T_{d}=A$, which again we can not construct
any $2$-integral Cayley graph over $G$. Hence we have showed that if $S$ is
one of the following sets, then $\mathrm{Cay}(G,S)$, where $G=\langle
a\rangle\cong\mathbb{Z}_{20}$ is $2$-integral:

$\\{a,a^{19},a^{9},a^{11}\\},~{}\\{a^{3},a^{17},a^{7},a^{13}\\},~{}\\{a^{2},a^{18}\\},~{}\\{a^{6},a^{14}\\},\\{a^{4},a^{16}\\},~{}\\{a^{8},a^{12}\\}.$

In the following lemma, we study the $2$-integrality of
$\Gamma=\mathrm{Cay}(G,S)$, where $\emptyset\neq S\subseteq[g]$ for some $g\in
G$.

######  Lemma 4.5.

Let $G$ be a finite group, $g\in G$ be an element of order $n$, $\emptyset\neq
S\subseteq[g]$ and $\Gamma=\mathrm{Cay}(G,S)$. Then $\Gamma$ is $2$-integral
if and only if there exists a subgroup $H$ of $\mathrm{Aut}(\langle g\rangle)$
such that $|\mathrm{Aut}(\langle g\rangle):H|=2$, $S=s^{H}$ for all $s\in S$
and $|S|=\frac{\varphi(n)}{2}$.

###### Proof.

Suppose $\Gamma$ is $2$-integral. Then $\mathrm{Cay}(\langle g\rangle,S)$ is
$2$-integral by Lemma 4.2 and $S\neq[g]$ by [19, Corollay 7.2]. Furthermore,
by [15, Theorem 1], $|A:H|=2$, where $A=\mathrm{Aut}(\langle
g\rangle)\cong\mathbb{Z}_{n}^{*}$ and $H=\\{\sigma\in A\mid S^{\sigma}=S\\}$.
Since for any $\sigma\in A$ there exists an integer $1\leqslant i\leqslant n$
with $(i,n)=1$ such that $g^{\sigma}=g^{i}$, $A$ acts transitively on $[g]$.
Note that $|A|=|[g]|$, which means that $A$ acts regularly on $[g]$. Hence the
action of $H$ on each orbit of $H$ is regular. Let $s\in S$. Then
$|s^{H}|=|H|=\frac{|A|}{2}=\frac{\varphi(n)}{2}$. Since $H$ acts semiregularly
on $S$, we have $|H|$ divides $|S|$. On the other hand, $s^{H}\subseteq
S\subset[g]$, and so $s^{H}=S$.

Conversely, suppose that $S=s^{H}$ for some $s\in S$, where $H$ is a subgroup
of $\mathrm{Aut}(\langle g\rangle)$ such that $|\mathrm{Aut}(\langle
g\rangle):H|=2$ and $|S|=\frac{\varphi(n)}{2}$. Let
$\Sigma=\mathrm{Cay}(\langle g\rangle,S)$. Then, by Proposition 4.3 (put
$d=1$), $\deg(\Sigma)=2$. Now Lemma 4.2 implies that $\Gamma$ is $2$-integral
as desired. ∎

By Lemma 4.5, we can characterize the generating sets of undirected
$2$-integral normal Cayley graphs.

######  Corollary 4.6.

Let $G$ be a finite group, $\Omega(G)=\\{[g_{1}],\ldots,[g_{n}]\\}$, $S$ is a
normal subset of $G$ and $\Gamma=\mathrm{Cay}(G,S)$. Then
$S=\bigcup_{i=1}^{n}S_{i}$, where $S_{i}=[g_{i}]\cap S$ for each $1\leqslant
i\leqslant n$. If $\Gamma$ is $2$-integral, then for $S_{i}\neq\emptyset$,

  * $(1)$

there exists $s_{i}\in S$ such that $S_{i}=[s_{i}]$ or
$S_{i}=s_{i}^{H}\subset[s_{i}]$ for some subgroup $H$ of $\mathrm{Aut}(\langle
s_{i}\rangle)$ of index $2$, and $|S_{i}|=\frac{\varphi(n_{i})}{2}$, where
$n_{i}=o(s_{i})$;

  * $(2)$

there exists $1\leqslant i_{0}\leqslant n$ such that $S_{i_{0}}$ is of the
latter form.

###### Proof.

Suppose $\Gamma$ is $2$-integral. By Lemma 4.1, $S=\bigcup_{i=1}^{n}S_{i}$ is
a disjoint union of sets $S_{1},\cdots,S_{n}$, where $S_{i}=[g_{i}]\cap S$ for
each $1\leqslant i\leqslant n$, and if $S_{i}\neq\emptyset$,
$\deg(\Gamma_{i})=1$ or $2$, where $\Gamma_{i}=\mathrm{Cay}(G,S_{i})$. For
$S_{i}\neq\emptyset$, $\deg(\Gamma_{i})=1$ if and only if $S_{i}=[g_{i}]$ by
[5, Corollary 7.2], and $\deg(\Gamma_{i})=2$ if and only if there exists a
subgroup $H$ of $\mathrm{Aut}(\langle g_{i}\rangle)$ such that
$|\mathrm{Aut}(\langle g_{i}\rangle):H|=2$, $S_{i}=s^{H}$ for all $s\in S_{i}$
and $|S_{i}|=\frac{\varphi(o(g_{i}))}{2}$ by Lemma 4.5. If for each $i$ such
that $S_{i}\neq\emptyset$, we have $\deg(\Gamma_{i})=1$, then $S$ is a union
of $[g_{i}]$s and so $\deg(\Gamma)=1$, by [5, Proposition 4.1], which
contradicts the $2$-integrality of $\Gamma$. Hence there exists $i_{0}$ such
that $\deg(\Gamma_{i_{0}})=2$. ∎

##  5 Finite abelian groups admitting a connected $2$-integral Cayley graph
with small valency

Recall that a cyclic group $G=\langle a\rangle$ of order $n$ admits a
connected integral Cayley graph of valency $2$ if and only if $n=3,4,6$ [2,
Lemma 2.7]. Also recall that a Cayley graph $\mathrm{Cay}(G,S)$ over an
abelian group $G$ is integral if and only if $S$ is a union of some $[g]$s,
where $g\in S$ [4, 5].

Let $\mathcal{G}_{k}$ be the set of all finite groups admitting a connected
$2$-integral Cayley graph with valency $k$. In this section, we completely
characterize all finite abelian groups belong to $\mathcal{G}_{k}$ for
$2\leqslant k\leqslant 5$.

###  5.1 $\mathcal{G}_{2}$

######  Theorem 5.1.

Let $G$ be a finite abelian group. Then $G\in\mathcal{G}_{2}$ if and only if
$G\cong\mathbb{Z}_{n}$, where $n=5,8,10,12$. Furthermore, a cycle with order
$n$ is $2$-integral if and only if $n=5,8,10,12$.

###### Proof.

Let $S=\\{x,y\\}$ be an inverse-closed generating set of $G$ and
$\Gamma=\mathrm{Cay}(G,S)$ is $2$-integral. Then either $x^{2}=y^{2}=1$ or
$y=x^{-1}$. In the first case, $S=S_{1}\cup S_{2}$, where $S_{1}=\\{x\\}=[x]$,
$S_{2}=\\{y\\}=[y]$, which means $\mathrm{Cay}(G,S_{i})$, $i=1,2$ are both
integral which contradicts Corollary 4.6. In the later case, we have
$S\subset[x]$ and $G=\langle x\rangle$. Let $n=o(x)$. Then, Lemma 4.5 implies
that $\varphi(n)=4$ and so $n=5,8,10,12$.

Conversely, suppose that $G=\langle x\rangle\cong\mathbb{Z}_{n}$, where
$n=5,8,10,12$. Let $A=\mathrm{Aut}(G)$ and put $H=\langle\tau\rangle$. Then in
each case $|A:H|=2$ and $|S|=\frac{\varphi(n)}{2}$, where $S=x^{H}$. Hence
$\mathrm{Cay}(G,S)$ is $2$-integral, by Lemma 4.5. ∎

###  5.2 $\mathcal{G}_{3}$

To classify all finite abelian groups in $\mathcal{G}_{3}$, we need the
following result.

######  Lemma 5.2.

Let $G$ be a finite group and $S$ be a normal subset of $G$. Then for any
$1\neq x\in Z(G)$ that $S\cap[x]=\emptyset$, $\mathrm{Cay}(G,S)$ and
$\mathrm{Cay}(G,S\cup[x])$ have the same splitting field. In particular, if
$x\in Z(G)\setminus S$ is an involution then $\mathrm{Cay}(G,S)$ and
$\mathrm{Cay}(G,S\cup\\{x\\})$ have the same splitting field.

###### Proof.

Let $\chi\in\mathrm{Irr}(G)$ and $1\neq x\in Z(G)$ such that
$S\cap[x]=\emptyset$. Since $x\in Z(G)$, [12, Exercise 5 of Charter 13]
implies that for each $i$ we have $\chi(x^{i})=\lambda^{i}\chi(1)$, where
$\lambda$ is an $n$th root of unity and $n=o(x)$. Hence
$\frac{\chi([x])}{\chi(1)}=\sum_{1\leqslant i\leqslant n,\\\
(i,n)=1}\lambda^{i}$ is an integer, say $t$, by [12, Lemma 22.15]. Since
$\frac{\chi(S\cup[x])}{\chi(1)}=\frac{\chi(S)+\chi([x])}{\chi(1)}=\frac{\chi(S)}{\chi(1)}+t$,
clearly the splitting field of $\mathrm{Cay}(G,S)$ and
$\mathrm{Cay}(G,S\cup[x])$ are the same, as desired. ∎

######  Theorem 5.3.

Let $G$ be a finite abelian group. Then $G\in\mathcal{G}_{3}$ if and only if
$G$ is isomorphic to $\mathbb{Z}_{n}$ or $\mathbb{Z}_{n}\times\mathbb{Z}_{2}$,
where $n=8,10,12$. Furthermore, a connected cubic abelian Cayley graph
$\Gamma$ is $2$-integral if and only if it is isomorphic to one of the $6$
following Cayley graphs $\mathrm{Cay}(G,S)$, where:

  * $i$.

$G=\langle x\rangle\cong\mathbb{Z}_{n}$ where $n=8,10,12$ and
$S=\\{x,x^{-1},x^{\frac{n}{2}}\\}$;

  * $ii$.

$G=\langle x\rangle\times\langle
y\rangle\cong\mathbb{Z}_{n}\times\mathbb{Z}_{2}$ where $n=8,10,12$ and
$S=\\{x,x^{-1},y\\}$.

###### Proof.

Let $S=\\{x,y,z\\}$ be an inverse-closed generating set of $G$ and
$\Gamma=\mathrm{Cay}(G,S)$ is $2$-integral. We deal with the following cases:

Case I. $x^{2}=y^{2}=z^{2}$. In this case, $S=S_{1}\cup S_{2}\cup S_{3}$,
where $S_{1}=[x]$, $S_{2}=[y]$ and $S_{3}=[y]$, which contradicts Corollary
4.6.

Case II. $y=x^{-1}$ and $z=x^{\frac{n}{2}}$, where $o(x)=n$ is even. Then
$G=\langle x\rangle$ and $S=S_{1}\cup S_{2}$, where
$S_{1}=\\{x,x^{-1}\\}\subseteq[x]$ and $S_{2}=[x^{\frac{n}{2}}]$. By Corollary
4.6, $4=\varphi(n)$ which means $n=8,10,12$. Hence in this case $G=\langle
x\rangle\cong\mathbb{Z}_{n}$, where $n=8,10,12$.

Case III. $y=x^{-1}$, $z^{2}=1$ and $z\notin\langle x\rangle$. In this case,
$G=\langle x\rangle\times\langle
z\rangle\cong\mathbb{Z}_{n}\times\mathbb{Z}_{2}$, where $n=o(x)$. Furthermore,
$S=S_{1}\cup S_{2}$, where $S_{1}=\\{x,x^{-1}\\}\subset[x]$ and $S_{2}=[z]$.
Similar to the Case II, we get $\varphi(n)=4$ and so $n=5,8,10,12$.

Hence we have showed that if $\Gamma=\mathrm{Cay}(G,S)$ is $2$-integral, then
$G=\langle x\rangle\cong\mathbb{Z}_{n}$ and $S=\\{x,x^{-1},x^{\frac{n}{2}}\\}$
or $G=\langle x\rangle\times\langle
z\rangle\cong\mathbb{Z}_{n}\times\mathbb{Z}_{2}$ and $S=\\{x,x^{-1},z\\}$,
where $n=8,10,12$. This proves one direction.

For the converse direction, first we suppose that $G=\langle
x\rangle\cong\mathbb{Z}_{n}$ and $S=S_{1}\cup S_{2}$, where
$S_{1}=\\{x,x^{-1}\\}\subseteq[x]$,
$S_{2}=\\{x^{\frac{n}{2}}\\}=[x^{\frac{n}{2}}]$ and $n=8,10,12$. Note that
$S_{1}\cap[x^{\frac{n}{2}}]=\emptyset$, then
$\mathbb{SF}(\mathrm{Cay}(G,S))=\mathbb{SF}(\mathrm{Cay}(G,S_{1}))$ by Lemma
5.2. Use the same argument as in the proof of Theorem 5.1, we have
$\mathrm{Cay}(G,S_{1})$ is $2$-integral and so $\mathrm{Cay}(G,S)$ is
$2$-integral. Hence $G\cong\mathbb{Z}_{n}\in\mathcal{G}_{3}$, where
$n=8,10,12$. Similar arguments show the desired results for the case
$G\cong\mathbb{Z}_{n}\times\mathbb{Z}_{2}$, where $n=8,10,12$. ∎

###  5.3 $\mathcal{G}_{4}$

To characterize all finite abelian groups belong to $\mathcal{G}_{4}$, we need
the following result:

######  Lemma 5.4.

Let $G$ be a finite group, $S_{1},\ldots,S_{k}$ be normal subsets of $G$,
$\langle S_{i}\rangle\cap\langle S_{j}\rangle=\\{1\\}$ for all distinct $i,j$,
and $S=S_{1}\cup\cdots\cup S_{k}$. Let $\Gamma=\mathrm{Cay}(G,S)$ and
$\Gamma_{i}=\mathrm{Cay}(G,S_{i})$, $i=1,\ldots,k$. Then

  * $(1)$

$\mathbb{SF}(\Gamma_{i})\subseteq\mathbb{SF}(\Gamma)$ for all $i$,

  * $(2)$

$\deg(\Gamma_{i})$ divides $\deg(\Gamma)$ for all $i$,

  * $(3)$

if $\deg(\Gamma_{i})=\deg(\Gamma)=2$ for some $i$, then
$\mathbb{SF}(\Gamma)=\mathbb{SF}(\Gamma_{i})$,

  * $(4)$

if $\mathbb{SF}(\Gamma_{i})=\mathbb{F}$ for all $i$, then
$\mathbb{SF}(\Gamma)=\mathbb{F}$.

###### Proof.

Put $G_{i}:=\langle S_{i}\rangle$, $i=1,\ldots,k$. Since $S_{i}$ is a normal
subset of $G$, $G_{i}$ is a normal subgroup of $G$. Let $\lambda$ be an
eigenvalue of $\Gamma_{i}$ and
$\mathrm{Irr}(G)=\\{\chi_{1},\cdots,\chi_{k}\\}$. Then
$\lambda=\frac{\psi(S_{i})}{\psi(1)}$ for some $\psi\in\mathrm{Irr}(G_{i})$.
Let $\psi\uparrow G$ be the induced character from $\psi$. Then $\psi\uparrow
G=d_{1}\chi_{1}+\cdots+d_{k}\chi_{k}$ for some integers $d_{1},\ldots,d_{k}$.
Since $S_{i}$’s are normal subsets of $G$ and $\langle S_{l}\rangle\cap\langle
S_{l^{\prime}}\rangle=\\{1\\}$ for distinct $l,l^{\prime}$, we have
$(\psi\uparrow G)(S_{i})=\frac{1}{|G_{i}|}\psi(S_{i})$ and $(\psi\uparrow
G)(S\setminus S_{i})=0$. Hence $\frac{1}{|G_{i}|}\psi(S_{i})=(\psi\uparrow
G)(S)=d_{1}\chi_{1}(S)+\cdots+d_{k}\chi_{k}(S)$. Since the splitting fields
contain the rational field and the characters degrees are positive integers,
we conclude that $\lambda\in\mathbb{SF}(\Gamma)$. Hence
$\mathbb{SF}(\Gamma_{i})\subseteq\mathbb{SF}(\Gamma)$. This proves $(1)$.
Parts $(2)$ and $(3)$ are direct consequences of $(1)$.

Let $\lambda$ be an eigenvalue of $\Gamma$. Since $S_{i}$s are normal subsets
of $G$, $S$ is also a normal subset of $G$. Hence there exists
$\chi\in\mathrm{Irr}(G)$ such that $\lambda=\frac{\chi(S)}{\chi(1)}$. On the
other hand, $S_{i}$’s are pairwise disjoint and so
$\chi(S)=\chi(S_{1})+\cdots+\chi(S_{k})$. Since for each $i$,
$\chi(S_{i})\in\mathbb{F}$, we conclude that $\lambda\in\mathbb{F}$. Thus
$\mathbb{SF}(\Gamma)\subseteq\mathbb{F}$. Now $(1)$ implies that
$\mathbb{SF}(\Gamma)=\mathbb{F}$ as desired. ∎

######  Theorem 5.5.

Let $G$ be a finite abelian group. Then $G\in\mathcal{G}_{4}$ if and only if
$G$ is isomorphic to one the following groups:

  * $(1)$

$\mathbb{Z}_{n}$, where $n=8,10,12,15,16,20,24,30$.

  * $(2)$

$\mathbb{Z}_{n}\times\mathbb{Z}_{2}^{2}$, where $n=5,8,10,12$.

  * $(3)$

$\mathbb{Z}_{n}\times\mathbb{Z}_{2}$, where $n=8,10,12$.

  * $(4)$

$\mathbb{Z}_{n}\times\mathbb{Z}_{m}$, where $(n,m)$ is one of the pairs
$(3,12)$, $(4,8)$, $(4,10)$, $(4,12),(6,8)$, $(6,10)$, $(6,12)$, $(5,5)$,
$(5,10)$, $(8,8)$, $(10,10)$, $(12,12)$.

Furthermore, a connected $4$-regular abelian Cayley graph $\Gamma$ is
$2$-integral if and only if it is isomorphic to one of the $39$ following
Cayley graphs $\mathrm{Cay}(G,S)$, where

  * $i$.

$G=\langle x\rangle\cong\mathbb{Z}_{n}$ where $n=15,16,20,24,30$, and
$S=\\{x,x^{-1},x^{k},x^{-k}\\}$ where $1\leqslant k\leqslant n$ such that
$(k,n)=1$;

  * $ii$.

$G=\langle x\rangle\cong\mathbb{Z}_{8}$ and $S=\\{x,x^{2},x^{6},x^{7}\\}$;

  * $iii$.

$G=\langle x\rangle\cong\mathbb{Z}_{10}$ and $S=\\{x,x^{2},x^{8},x^{9}\\}$;

  * $iv$.

$G=\langle x\rangle\cong\mathbb{Z}_{12}$ and $S=\\{x,x^{2},x^{10},x^{11}\\}$
or $\\{x,x^{3},x^{9},x^{11}\\}$ or $\\{x,x^{4},x^{8},x^{11}\\}$;

  * $v$.

$G=\langle x\rangle\times\langle
w\rangle\cong\mathbb{Z}_{n}\times\mathbb{Z}_{2}$ where $n=8,10,12$ and
$S=\\{x,x^{-1},x^{\frac{n}{2}},w\\}$;

  * $vi$.

$G=\langle x\rangle\times\langle z\rangle\times\langle
w\rangle\cong\mathbb{Z}_{n}\times\mathbb{Z}_{2}^{2}$ where $n=5,8,10,12$ and
$S=\\{x,x^{-1},z,w\\}$;

  * $vii$.

$G=\langle x\rangle\times\langle
w\rangle\cong\mathbb{Z}_{n}\times\mathbb{Z}_{m}$ where $(n,m)$ is one of the
pairs $(3,12)$, $(4,8)$, $(4,10)$, $(4,12),(6,8)$, $(6,10)$, $(6,12)$,
$(5,5)$, $(5,10)$, $(8,8)$, $(10,10)$, $(12,12)$, and
$S=\\{x,x^{-1},y,y^{-1}\\}$.

###### Proof.

Let $S=\\{x,y,z,w\\}$ be an inverse closed generating set of $G$ and
$\mathrm{Cay}(G,S)$ is $2$-integral. Then we deal with the following cases:

Case I. $x^{2}=y^{2}=z^{2}=w^{2}=1$. In this case, $G$ is an elementary
abelian $2$-group isomorphic to $\mathbb{Z}_{2}^{n}$, where $n=3$ or $4$.
Furthermore, $S=S_{1}\cup S_{2}\cup S_{3}\cup S_{4}$, where
$S_{1}=\\{x\\}=[x]$, $S_{2}=\\{y\\}=[y]$, $S_{3}=\\{z\\}=[z]$ and
$S_{4}=\\{w\\}=[w]$, which contradicts Corollary 4.6.

Case II. $z^{2}=w^{2}=1$, $z,w\notin\langle x\rangle$ and and $y=x^{-1}$. In
this case, $G=\langle x\rangle\times\langle z\rangle\times\langle
w\rangle\cong\mathbb{Z}_{n}\times\mathbb{Z}_{2}^{2}$, where $n=o(x)$.
Furthermore, $S=S_{1}\cup S_{2}\cup S_{3}$, where
$S_{1}=\\{x,x^{-1}\\}\subseteq[x]$, $S_{2}=\\{z\\}=[z]$ and
$S_{3}=\\{w\\}=[w]$. Now Corollary 4.6 implies that $S_{1}\neq[x]$ and
$\varphi(n)=4$ which means $n=5,8,10,12$.

Case III. $y=x^{-1}$, $z=x^{\frac{n}{2}}$, where $n=o(x)$ is even and
$w^{2}=1$. In this case, $G\cong\mathbb{Z}_{n}\times\mathbb{Z}_{2}$. Also
$S=S_{1}\cup S_{2}\cup S_{3}$, where $S_{1}=\\{x,x^{-1}\\}\subseteq[x]$,
$S_{2}=\\{x^{\frac{n}{2}}\\}=[x^{\frac{n}{2}}]$ and $S_{3}=\\{w\\}=[w]$. Then
by Corollary 4.6, $S_{1}\neq[x]$ and $\varphi(n)=4$, which means $n=8,10,12$.

Case IV. $y=x^{k}$ for some $k\neq 1,-1$, $z=x^{-1}$ and $w=y^{-1}$. In this
case, $y^{2}\neq 1$ and $G=\langle x\rangle\cong\mathbb{Z}_{n}$, where
$n=o(x)$.

First assume that $(k,n)=1$. Then $S\subset[x]$ and Lemma 4.5 implies that
$\varphi(n)=8$, which means $n=15,16,20,24,30$.

Now let $(k,n)=d\neq 1$. Then $S=S_{1}\cup S_{2}$, where
$S_{1}=\\{x,x^{-1}\\}\subseteq[x]$ and
$S_{2}=\\{x^{k},x^{-k}\\}\subseteq[x^{k}]$. If $S_{1}=[x]$ and
$S_{2}\neq[x^{k}]$ then $n=3,4,6$ and $\frac{n}{d}=5,8,10,12$, respectively,
which is impossible. If $S_{1}\neq[x]$ and $S_{2}=[x^{k}]$, then $n=5,8,10,12$
and $\frac{n}{d}=3,4,6$, respectively, which implies $(n,k)=(8,2)$, $(12,2)$,
$(12,3)$ or $(12,4)$. If $S_{1}\neq[x]$ and $S_{2}\neq[x^{k}]$ then
$n,\frac{n}{d}=5,8,10,12$, which implies $(n,k)=(10,2)$.

Case V. $\langle x\rangle\cap\langle y\rangle=1$. Then $G=\langle
x\rangle\times\langle y\rangle\cong\mathbb{Z}_{n}\times\mathbb{Z}_{m}$, where
$n=o(x)$ and $m=o(y)$. We may assume that $n\leqslant m$. In this case,
$S=S_{1}\cup S_{2}$, where $S_{1}=\\{x,x^{-1}\\}\subseteq[x]$ and
$S_{2}=\\{y,y^{-1}\\}\subseteq[y]$. By corollary 4.6, $S_{1}=[x],S_{2}\neq[y]$
or $S_{1}\neq[x],S_{2}=[y]$ or $S_{1}\neq[x],S_{2}\neq[y]$. By a similar
discussion to the above cases, in the first case $n=3,4,6,~{}m=5,8,10,12$, in
the second case $n=5,8,10,12,~{}m=3,4,6$ and in the later case
$n,m=5,8,10,12$. The second case is impossible because $n\leqslant m$. Next we
will show that in the last case $(n,m)$ must be $(5,5)$, $(5,10)$, $(8,8)$,
$(10,10)$ or $(12,12)$. Let $g\in G$ and $o(g)=k\geqslant 2$ and
$C_{k}=\mathrm{Cay}(\langle g\rangle,\\{g,g^{-1}\\})$. By a computation, we
have $\mathbb{SF}(C_{5})=\mathbb{SF}(C_{10})=\mathbb{Q}[\sqrt{5}]$,
$\mathbb{SF}(C_{8})=\mathbb{Q}[\sqrt{2}]$ and
$\mathbb{SF}(C_{12})=\mathbb{Q}(\sqrt{3})$. Hence, by Corollary 4.6 and Lemma
5.4, we have the result as desired. Moreover, Cases $\mathrm{I}$-$\mathrm{V}$
prove one direction.

The proof of converse direction of
$G\cong\mathbb{Z}_{n}\times\mathbb{Z}_{2}^{2}$ where $n=5,8,10,12$, and
$G\cong\mathbb{Z}_{n}\times\mathbb{Z}_{2}$ where $n=8,10,12$ are similar to
the proof of converse direction of Theorem 5.3.

Consider the converse direction of case $G=\langle
x\rangle\cong\mathbb{Z}_{n}$, where $o(x)=n$ is shown in $(1)$. Let
$\Sigma_{k}=\mathrm{Cay}(\langle x\rangle,\\{x,x^{k},x^{-1},x^{-k}\\})$. First
suppose that $n=15,16,20,24,30$ and $(n,k)=1$. Let $\sigma_{k}:x\mapsto x^{k}$
and $\tau:x\mapsto x^{-1}$. Then $H=\langle\tau,\sigma_{k}\rangle$ is a
subgroup of index $2$ in $\mathrm{Aut}(G)$ and
$x^{H}=\\{x,x^{k},x^{-1},x^{-k}\\}$. By Lemmas 4.5, we have
$\mathrm{SF}(\Sigma_{k})=2$. Now suppose that $n=8,10,12$. By a tedious
computation, one can see that $\mathbb{SF}(\Sigma_{k})=\mathbb{Q}(\sqrt{2})$
if $(n,k)=(8,2)$, $(12,2)$ or $(12,4)$,
$\mathbb{SF}(\Sigma_{k})=\mathbb{Q}(\sqrt{3})$ or $\mathbb{Q}(\sqrt{5})$ if
$(n,k)=(12,3)$ or $(10,2)$ respectively.

Finally, consider the converse direction of case $G=\langle
x\rangle\times\langle y\rangle\cong\mathbb{Z}_{n}\times\mathbb{Z}_{m}$, where
$(n,m)$ shown in $(3)$. Let $S=S_{1}\cup S_{2}$ and
$S_{1}=\\{x,x^{-1}\\}\subseteq[x]$ and $S_{2}=\\{y,y^{-1}\\}\subseteq[y]$.
Note that $C_{3},C_{4},C_{6}$ are integral with valency $2$ and
$\mathbb{Z}_{8},\mathbb{Z}_{10},\mathbb{Z}_{12}$ are all in $\mathcal{G}_{2}$
by Theorem 5.1. Hence by Lemma 5.2,
$\mathbb{Z}_{n}\times\mathbb{Z}_{m}\in\mathcal{G}_{4}$, where $(n,m)=(3,12)$,
$(4,8)$, $(4,10)$, $(4,12)$, $(6,8)$, $(6,10)$ and $(6,12)$. On the other
hand, since $\mathbb{SF}(C_{5})=\mathbb{SF}(C_{10})=\mathbb{Q}[\sqrt{5}]$,
$\mathbb{SF}(C_{8})=\mathbb{Q}[\sqrt{2}]$ and
$\mathbb{SF}(C_{12})=\mathbb{Q}(\sqrt{3})$, by Lemma 5.4,
$\mathbb{Z}_{n}\times\mathbb{Z}_{m}\in\mathcal{G}_{4}$, where $(n,m)=(5,5)$,
$(5,10)$, $(8,8)$. This completes the proof. ∎

###  5.4 $\mathcal{G}_{5}$

To characterize all finite abelian groups in $\mathcal{G}_{5}$, we need the
following results.

######  Lemma 5.6.

Let $G$ be a finite group of even order, $S$ be a normal subset of $G$ and
there exists an involution $x\in Z(G)\cap S$. If $\mathrm{Cay}(G,S)$ is
connected with the splitting field $\mathbb{F}$, then

  * $(1)$

$G=\langle S\setminus\\{x\\}\rangle$ or $G=\langle
S\setminus\\{x\\}\rangle\times\langle x\rangle\cong\langle
S\setminus\\{x\\}\rangle\times\mathbb{Z}_{2}$;

  * $(2)$

$\mathrm{Cay}(\langle S\setminus\\{x\\}\rangle,S\setminus\\{x\\})$ is a
connected graph with the splitting filed $\mathbb{F}$.

###### Proof.

Since $\mathrm{Cay}(G,S)$ is connected, $G=\langle
S\rangle=\langle(S\setminus\\{x\\})\cup\\{x\\}\rangle$. Since $x\in Z(G)$,
$G=\langle S\setminus\\{x\\}\rangle\langle x\rangle$. Moreover,
$S\setminus\\{x\\}$ is also conjugate-closed, and since $x$ has order $2$,
$G=\langle S\setminus\\{x\\}\rangle$ or $G=\langle
S\setminus\\{x\\}\rangle\times\langle x\rangle\cong\langle
S\setminus\\{x\\}\rangle\times\mathbb{Z}_{2}$. This proves $(1)$.

Let $\Gamma_{1}=\mathrm{Cay}(G,S\setminus\\{x\\})$. We claim that the
splitting field of $\Gamma_{1}$ is $\mathbb{F}$. First we suppose that
$G=\langle S\setminus\\{x\\}\rangle$. Then Lemma 5.2 implies that
$\mathrm{Cay}(G,S\setminus\\{x\\})$ and $\mathrm{Cay}(G,S)$ have the same
splitting filed and so the splitting filed of $\Gamma_{1}$ is $\mathbb{F}$.
Next we suppose that $G=\langle S\setminus\\{x\\}\rangle\times\langle
x\rangle$. Let $\lambda$ be an eigenvalue of $\mathrm{Cay}(G,S)$. Then
$\lambda=\frac{\chi(S)}{\chi(1)}$, for some $\chi\in\mathrm{Irr}(G)$. Since
$x\in Z(G)$ we have $\chi(x)=\pm\chi(1)$, see [12, Exrecise 5 of Chapter 13],
which implies $\lambda=\frac{\chi(S\setminus\\{x\\})}{\chi(1)}\pm 1$. Let
$H=\langle S\setminus\\{x\\}\rangle$ and $\chi\mid_{H}$ be the restriction of
$\chi$ to $H$. Since $|G:H|=2$ and $\chi(x)\neq 0$, [12, Proposition 20.5]
implies that $\chi|_{H}$ is an irreducible character of $H$. Furthermore,
$\chi(1)=\chi|_{H}(1)$ and
$\chi|_{H}(S\setminus\\{x\\})=\chi(S\setminus\\{x\\})$ which imply
$\lambda=\theta\pm 1$, for some eigenvalue $\theta$ of $\Gamma_{1}$. Now
assume that $\mu$ be an eigenvalue of $\Gamma_{1}$. Then
$\mu=\frac{\eta(S\setminus\\{x\\})}{\eta(1)}$ for some
$\eta\in\mathrm{Irr}(H)$. Let $\rho_{0}$ be the principal character of
$\langle x\rangle$. Then $\zeta:=\eta\times\rho_{0}$ is an irreducible
character of $G$ and
$\zeta(S)=\eta(S\setminus\\{x\\})+\rho_{0}(x)=\eta(S\setminus\\{x\\})+1$.
Hence $\mu=\xi-\frac{1}{\zeta(1)}$, where $\xi=\frac{\zeta(S)}{\zeta(1)}$ is
an eigenvalue of $\mathrm{Cay}(G,S)$. Since $\frac{1}{\zeta(1)}\in\mathbb{Q}$,
we conclude that the splitting filed of $\mathrm{Cay}(G,S)$ and $\Gamma_{1}$
are the same. This proves $(2)$. ∎

######  Lemma 5.7.

Let $\Gamma_{1}$ be a graph with the splitting field $\mathbb{F}$ and
$\Gamma_{2}$ be an integral graph. Then $\Gamma_{1}\square\Gamma_{2}$,
$\Gamma_{1}\otimes\Gamma_{2}$, and $\Gamma_{1}\boxtimes\Gamma_{2}$ have the
same splitting field $\mathbb{F}$. Furthermore, if $\Gamma_{2}$ is a connected
regular graph then $\Gamma_{1}[\Gamma_{2}]$ has also the same splitting field
$\mathbb{F}$.

###### Proof.

Let the eigenvalues of $\Gamma_{1}$ and $\Gamma_{2}$ be
$\lambda_{1}\geqslant\lambda_{2}\geqslant\cdots\geqslant\lambda_{n_{1}}$ and
$\mu_{1}\geqslant\mu_{2}\geqslant\cdots\geqslant\mu_{n_{2}}$, respectively,
where $n_{i}$ is the number of vertices of $\Gamma_{i}$. By [6, Table 4], the
eigenvalues of $\Gamma_{1}\square\Gamma_{2}$, $\Gamma_{1}\otimes\Gamma_{2}$
and $\Gamma_{1}\boxtimes\Gamma_{2}$ are $\lambda_{i}+\mu_{j}$,
$\lambda_{i}\mu_{j}$, and $\lambda_{i}+\mu_{j}+\lambda_{i}\mu_{j}$,
respectively, where $1\leqslant i\leqslant n_{1}$ and $1\leqslant j\leqslant
n_{2}$.

Let $\Gamma_{2}$ be a connected $k$-regular graph. Then eigenvalues of
$\Gamma_{1}[\Gamma_{2}]$ are $k+n_{2}\lambda_{i}$, $1\leqslant i\leqslant
n_{1}$ and $\mu_{j}$ with multiplicity $n_{1}$, where $2\leqslant j\leqslant
n_{2}$. Since eigenvalues of $\Gamma_{2}$ are integers, the result follows
immediately. ∎

######  Corollary 5.8.

Let $\Gamma_{1}=\mathrm{Cay}(G_{1},S_{1})$ and
$\Gamma_{2}=\mathrm{Cay}(G_{2},S_{2})$ be two connected Cayley graphs such
that $\Gamma_{1}$ is $k$-integral and $\Gamma_{2}$ is integral. Then
$\mathrm{Cay}(G_{1}\times G_{2},S)$ is a connected $k$-integral Cayley graph
over $G_{1}\times G_{2}$, where $S$ is one of the following sets

  * $(1)$

$S=\\{(s_{1},1),(1,s_{2})\mid s_{1}\in S_{1},s_{2}\in S_{2}\\}$,

  * $(2)$

$S=\\{(s_{1},s_{2})\mid s_{1}\in S_{1},s_{2}\in S_{2}\\}$,

  * $(3)$

$S=\\{(s_{1},1),(1,s_{2}),(s_{1},s_{2})\mid s_{1}\in S_{1},s_{2}\in S_{2}\\}$,

  * $(4)$

$S=\\{(s_{1},g_{2}),(1,s_{2})\mid s_{1}\in S_{1},s_{2}\in S_{2},g_{2}\in
G_{2}\\}$, where at most one of the graphs $\Gamma_{1}$ or $\Gamma_{2}$ is
bipartite.

In particular, if $|S_{i}|=d_{i}$, $i=1,2$, then in the above cases, $|S|$ is
$d_{1}+d_{2}$, $d_{1}d_{2}$, $d_{1}+d_{2}+d_{1}d_{2}$ and
$d_{1}|G_{2}|+d_{2}$, respectively.

###### Proof.

It is a direct consequence of Lemma 5.7, [9, Corollaries 5.3, 5.6, 5.10, 5.14]
and [1, Theorem 3.1, Theorem 4.1, Theorem 5.3, Theorem 6.1]. ∎

######  Corollary 5.9.

Let $G$ be a finite abelian group of even order. There exists a connected
$(2d+1)$-regular Cayley graph over $G$ with algebraic degree $k$ if and only
if

  * $(1)$

there exists a connected $2d$-regular $k$-integral Cayley graph over $G$, or

  * $(2)$

there exists a $H\leqslant G$ such that $G=H\times\mathbb{Z}_{2}$ and $H$
admits a connected $2d$-regular $k$-integral Cayley graph.

###### Proof.

Note that the number of involutions of any finite abelian group with even
order is odd. If $\mathrm{Cay}(G,S)$ is connected with algebraic degree $k$
and $|S|=2d$, for some inverse-closed subset $S$ of $G$, then there exists an
involution $x\in G\setminus S$. Hence $\mathrm{Cay}(G,S\cup\\{x\\})$ is
connected $(2d+1)$-regular and with algebraic degree $k$ by Lemma 5.2. If
$G=H\times\mathbb{Z}_{2}$ and $H$ admits a connected $2d$-regular Cayley graph
with algebraic degree $k$, since $K_{2}$ as a Cayley graph over
$\mathbb{Z}_{2}$ is integral, then Corollary 5.8 implies that $G$ admits a
connected $(2d+1)$-regular Cayley graph with algebraic degree $k$.

The converse is clear by Lemma 5.6, because any inverse-closed subset of $G$
with odd number of elements has an involution. ∎

######  Theorem 5.10.

Let $G$ be a finite abelian group. Then $G\in\mathcal{G}_{5}$ if and only if
$G\cong H$ or $H\times\mathbb{Z}_{2}$, where $H$ is one of the groups given in
Theorem 5.5 and in the first case $H\ncong\mathbb{Z}_{15},\mathbb{Z}_{5}^{2}$.
Furthermore, a connected $5$-regular abelian Cayley graph $\Gamma$ is
$2$-integral if and only if it is isomorphic to one of the $108$ following
Cayley graphs $\mathrm{Cay}(G,S)$ where

  * $i$.

$G=\langle x\rangle\cong\mathbb{Z}_{n}$ where $n=16,20,24,30$, and
$S=\\{x,x^{-1},x^{\frac{n}{2}},x^{k},x^{-k}\\}$ where $1\leqslant k\leqslant
n$ such that $(k,n)=1$;

  * $ii$.

$G=\langle x\rangle\cong\mathbb{Z}_{8}$ and
$S=\\{x,x^{2},x^{4},x^{6},x^{7}\\}$;

  * $iii$.

$G=\langle x\rangle\cong\mathbb{Z}_{10}$ and
$S=\\{x,x^{2},x^{5},x^{8},x^{9}\\}$;

  * $iv$.

$G=\langle x\rangle\cong\mathbb{Z}_{12}$ and
$S=\\{x,x^{2},x^{6},x^{10},x^{11}\\}$ or $\\{x,x^{3},x^{6},x^{9},x^{11}\\}$ or
$\\{x,x^{4},x^{6},x^{8},x^{11}\\}$;

  * $v$.

$G=\langle x\rangle\times\langle
w\rangle\cong\mathbb{Z}_{n}\times\mathbb{Z}_{2}$ where $n=8,10,12$ and
$S=\\{x,x^{-1},x^{\frac{n}{2}},w,x^{\frac{n}{2}}w\\}$;

  * $vi$.

$G=\langle x\rangle\times\langle z\rangle\times\langle
w\rangle\cong\mathbb{Z}_{n}\times\mathbb{Z}_{2}^{2}$ where $n=5,8,10,12$ and
$S=\\{x,x^{-1},z,w,s\\}$ where $s$ is any involution of $G$ except for $z$ and
$w$;

  * $vii$.

$G=\langle x\rangle\times\langle
w\rangle\cong\mathbb{Z}_{n}\times\mathbb{Z}_{m}$ where $(n,m)$ is one of the
pairs $(3,12)$, $(4,8)$, $(4,10),(4,12)$, $(6,8)$, $(6,10)$, $(6,12)$,
$(5,10)$, $(8,8)$, $(10,10)$, $(12,12)$, and $S=\\{x,x^{-1},y,y^{-1},s\\}$
where $s$ is any involution of $G$.

  * $viii$.

$G=\langle x\rangle\times\langle
y\rangle\cong\mathbb{Z}_{n}\times\mathbb{Z}_{2}$ where $n=16,20,24,30$, and
$S=\\{x,x^{-1},x^{k},x^{-k},y\\}$ where $1\leqslant k\leqslant n$ such that
$(k,n)=1$;

  * $ix$.

$G=\langle x\rangle\times\langle
y\rangle\cong\mathbb{Z}_{8}\times\mathbb{Z}_{2}$ and
$S=\\{x,x^{2},x^{6},x^{7},y\\}$;

  * $x$.

$G=\langle x\rangle\times\langle
y\rangle\cong\mathbb{Z}_{10}\times\mathbb{Z}_{2}$ and
$S=\\{x,x^{2},x^{8},x^{9},y\\}$;

  * $xi$.

$G=\langle x\rangle\times\langle
y\rangle\cong\mathbb{Z}_{12}\times\mathbb{Z}_{2}$ and
$S=\\{x,x^{2},x^{10},x^{11},y\\}$ or $\\{x,x^{3},x^{9},x^{11},y\\}$ or
$\\{x,x^{4},x^{8},x^{11},y\\}$;

  * $xii$.

$G=\langle x\rangle\times\langle w\rangle\times\langle
y\rangle\cong\mathbb{Z}_{n}\times\mathbb{Z}_{2}^{2}$ where $n=8,10,12$ and
$S=\\{x,x^{-1},x^{\frac{n}{2}},w,y\\}$;

  * $xiii$.

$G=\langle x\rangle\times\langle z\rangle\times\langle w\rangle\times\langle
y\rangle\cong\mathbb{Z}_{n}\times\mathbb{Z}_{2}^{3}$ where $n=5,8,10,12$ and
$S=\\{x,x^{-1},z,w,y\\}$;

  * $xiv$.

$G=\langle x\rangle\times\langle w\rangle\times\langle
y\rangle\cong\mathbb{Z}_{n}\times\mathbb{Z}_{m}\times\mathbb{Z}_{2}$ where
$(n,m)$ is one of the pairs $(3,12)$, $(4,8)$, $(4,10)$, $(4,12)$, $(6,8)$,
$(6,10)$, $(6,12)$, $(5,10)$, $(8,8)$, $(10,10)$, $(12,12)$, and
$S=\\{x,x^{-1},y,y^{-1},y\\}$.

###### Proof.

Note that in any graph, the number of vertices of odd degree is even. Thus $G$
must be a group of even order. And so this theorem is a direct consequence of
Theorem 5.5 and Corollary 5.9. ∎

Similar to Theorem 5.5, one can find all finite abelian groups
$G\in\mathcal{G}_{6}$ and then then it is possible to characterize all finite
abelian groups $G\in\mathcal{G}_{7}$. Recursively, one can find all finite
abelian groups in $\mathcal{G}_{k}$ for any given integer $k$.

Acknowledgments. Research of this paper was carried out while the second
author was visiting Beijing Jiaotong University. He expresses his sincere
thanks to the 111 Project of China (grant number B16002) for financial support
and to the School of Mathematics and Statistics of Beijing Jiaotong University
for their kind hospitality. Also he was partially supported by the Shahid
Rajaee Teacher Training University. Tao Feng is supported by NSFC under Grants
11871095 and 12271023.

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The gravitational potential may be probed through the gravitational redshift
experienced by photons losing energy as they escape deep potential wells. The
shape of the potential is changed as gravity is modified. Stacking together
the spectroscopically measured galaxy velocities in a large sample of optical
clusters, Wojtak et al. (2011) identified a shift of velocity for satellite
galaxies relative to centrals, indicative of the sought-after gravitational
redshift signal. Further analyses incorporating all relevant relativistic
effects and consistent changes in the halo mass function however conclude to
little discriminating power of this technique if only dynamical mass
information for a narrow range of cluster masses is available (e.g. Zhao et
al., 2013). A study using X-ray-based masses obtained for $\sim 2500$ galaxy
clusters detected in the SPIDERS survey also concludes to no preference
between General Relativity and a special case of $f(R)$ gravity (Mpetha et
al., 2021). In all these studies, the amount of the measured velocity shift is
of order $10-20$ km s-1. The future use of high-resolution spectroscopic X-ray
data may probe gravitational redshifts of the gas instead of individual
galaxies (e.g. Broadhurst & Scannapieco, 2000; Sakuma et al., 2017). This
observational technique provides a reduction of the noise inherent to
measurements due to the high number of gas parcels in comparison to galaxies,
at the expense of finely controlling all effects creating $\sim$ few 10 km s-1
line shifts in the spectral emission (such as turbulent and bulk motions).
Only with upcoming instrumentation will this observational technique be
accessible, together with the observational and theoretical improvements in
our understanding of the intra-cluster gas dynamics and chemical enrichment.

##  10 Selection function

When the catalogs of galaxy clusters are derived using X-ray data, X-ray
properties of clusters and methods of cluster detection form the selection
function of the survey. The shape of the X-ray emission of the cluster in
combination with the detection method affects the selection: unresolved X-ray
emission (typically covering $2-6^{\prime}$ radius) formed the basis for the
bulk of cluster catalogs based on _ROSAT_ All-Sky survey data, which for
distant clusters covers most of the cluster, while for the nearby ($z<0.1$)
clusters it covers only the core, where X-ray emission scales poorly with
cluster mass. While accounting for the total flux of the cluster is performed,
such selection leads to unbalanced representation of merging and relaxed
clusters. Another drawback of this approach is that 90% of such sources are
not clusters and one has to clean for contamination in the sample and in
cluster X-ray properties. A huge improvement in the purity of cluster
selection came with using the extent of X-ray emission. The selection function
in those surveys was found to depend on the cluster core radii. In order to
assess the limitations of the complexity of the cluster shapes, cloning of
observed clusters has been used to test the cluster detection algorithms.

Scatter in the cluster shape has been found to correlate with scatter in
cluster X-ray luminosity, which affects cluster selection (Käfer et al.,
2019). The full account for covariant scatter on cluster selection is given in
Finoguenov et al. (2020). In terms of the selection, the best area of the
cluster to use seems to be between 1/4 and 1/2 of $R_{500c}$, and efforts have
been made to create the detection procedures for that (Eckert et al., 2020).
Avoiding cluster cores in the detection has been widely used in deep X-ray
surveys (Finoguenov et al., 2010, 2015). The advantage one uses in those
methods is the large information on clusters shapes available at X-rays.

In the forward modelling, for each prediction of mass function, one predicts
the full set of cluster X-ray mass proxies and mimics the selection process.
Detection of galaxy clusters forms the selection function, while the procedure
of measurement of the cluster properties forms the cluster distribution
function. On the procedure of measurements of cluster properties, there is a
large variety of approaches taken. Even to obtain the cluster $L_{X}$ one has
to determine the cluster extent. However, in most cases only the central part
of the cluster is detected and extrapolation of cluster properties is
required. For that, one either uses a beta profile, which is either measured
(Vikhlinin et al., 1999) or is taken from scaling cluster libraries
(Finoguenov et al., 2020), or sets the definition of cluster $L_{X}$ to be
within the zone of likely detection (e.g. 300 kpc for XXL). REFLEX and NORAS
surveys (Böhringer et al., 2004) do not have a scalable definition of $L_{X}$
and corrections have been computed for those clusters (Mantz et al., 2010;
Piffaretti et al., 2011).

The purpose of deriving a selection function is to emulate the behaviour of a
detection chain that led to the construction of a sample, while varying
assumptions on the physics and cosmological model. Expensive end-to-end
simulations are an option which accounts for many of the details (projection
effects, point-source contamination, etc., see e.g. Comparat et al., 2020),
while scaling models are an alternative that bypasses the expensive
computations, but need to formulate strong assumptions on their behaviour.
CODEX, XXL selection functions are somehow half-way between these two
extremes. In order to make sure that certain cluster shapes are not missed in
the current X-ray libraries, studies of clusters selected using different ways
(e.g. SZ effect, galaxy concentrations or lensing) are used. The most obvious
result of those searches so far is finding that clusters without the cool core
are more abundant, compared to the results of REFLEX/NORAS catalogs. Similar
results are obtained using spatially resolved X-ray searches for clusters.

In most cases, identification of galaxy clusters requires detection of cluster
galaxies. Limitations of these data will affect the selection of clusters and
of importance is the covariance of optical and X-ray properties of the sample.
The current results (Farahi et al., 2019) also suggest the anti-covariance,
which implies that cluster entering the sample by upscatter in X-ray
properties, might leave the sample due to downscatter in optical richness. The
scatter of properties in clusters is linked to the physics of clusters. X-ray
luminous clusters have bright cores and are preferentially relaxed. Merger
clusters tend to have high richness (Mulroy et al., 2019). The study of Klein
et al. (2019) has introduced a concept of using cluster identification to
study the chance association between the X-ray source and the optical
counterpart. Using an overlap with Klein et al. sample, Finoguenov et al.
(2020) concluded that chance identification starts to be important below the
10% completeness boundary of the X-ray survey.

##  11 Conclusions and forward look

X-ray clusters characterize well the massive end of halo mass function, and
there are no limitation imposed on future studies to extend the X-ray cluster
searches to redshifts beyond the current record of 2.5. The future studies
will benefit from improved sensitivity in the soft X-ray band (0.1-2 keV)
(e.g. of _Athena_) and spatial resolution better than a few arcseconds, as
anticipated from the Lynx mission. Large catalogs of clusters delivered by
eROSITA all sky will serve a great input for cosmological studies. The best
behaved part of the X-ray emission of clusters is located between 0.2 and
$R_{500c}$, where X-ray (which is necessarily a surface brightness) selection
have best chances to be complete.

Currently, $z<0.5$ X-ray cluster studies have reached the limit of
systematics. This requires both mass calibration and improved modelling of all
baryonic cluster components. A variety of shapes of X-ray emission of clusters
might be associated with a difference in the cluster baryonic content induced
by AGN feedback. In order to understand that, detailed X-ray studies are
required to recover the diversity of gas mass fraction close to the virial
radius. Historically, these searches have been hampered by problems with
robust definitions of massive halos using optical cluster catalogs and
deficiencies in existing X-ray catalogs at low-z. A number of Large Programs
on _XMM-Newton_ are currently underway to progress on this issue. Deep X-ray
and spectroscopic surveys have also been proven to be the best way to address
this (Finoguenov et al., 2009; Connelly et al., 2012). A special topic is
whether the distribution of X-ray properties of clusters follows the power law
dependence on mass with a log-normal scatter, or has a more complicated
functional form. With increasing size of X-ray cluster samples, and increasing
variety of cluster detection methods, one expects to collect more data on this
subject.

Variety of baryonic properties of halos also affects weak lensing cosmology
and present a new research focus on X-ray groups and clusters. For instance
the high precision of _Euclid_ optical and near-infrared measurements will be
sensitive to detailed processes associated with galaxy formation, including
the growth of black holes and their feedback (co-evolution). Semboloni et al.
(2011) predict the importance of the contribution of galaxy groups to the
shear power spectrum down to $10^{13}M_{\odot}$. Debackere et al. (2020)
confirm this finding through modelling the distribution of baryons in groups
following observations of stellar and hot intragroup medium. Baryonic feedback
processes, either from supernovae (SNe) or active galactic nuclei (AGN)
redistribute the gas, which changes the matter power spectrum, in particular
on scales where the sensitivity of cosmic shear is maximal. Semboloni et al.
(2011) showed that ignoring this effect leads to large biases, whereas
marginalising over current model uncertainties weaken the constraining power
of weak lensing shear experiments. The way forward is to constrain feedback
directly and update its implementation in models (e.g. McCarthy et al., 2017).
Groups are the best targets for resolving this issue because on their scale:
on one hand the dominant feedback mechanism must change from stellar
processes, such as galactic winds to AGN-driven processes and imprint telltale
signatures on the warm gas, as the binding energy and the AGN output are
similar; on the other hand, all the components of galaxy groups can be
robustly determined and groups of galaxies are the lowest mass structures that
impact the shear signal on the scales of interest for _Euclid_ cosmology.

##  12 Resources

Any initial analysis of X-ray observations is performed with a mission-
dependent software package, with a common goal of producing calibrated event
list and generating auxiliary responses. For a collection of the software on
various mission we would like to point the reader to
HEASARC191919https://heasarc.gsfc.nasa.gov/. Most of the data products has
been stored using FITS format with the FTOOLS software, available on the same
site, as well as a package for spectral analysis, XSPEC. Detailed study of
line emission shall benefit the largest database of lines to date, available
through another spectral fitting code,
SPEX202020https://www.sron.nl/astrophysics-spex.

Cosmological codes relevant to galaxy clusters are available through
Astropy212121https://docs.astropy.org/en/stable/cosmology/index.html (Astropy
Collaboration et al., 2013) and
COLOSSUS222222http://www.benediktdiemer.com/code/colossus/ (Diemer, 2018).
Cosmology codes in FORTRAN and other languages are available from E. Komatsu’s
website232323https://wwwmpa.mpa-garching.mpg.de/~komatsu/codes.html. A cosmic
variance python module (Lacasa & Grain, 2019) is available
online242424https://github.com/fabienlacasa/PySSC.

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