# regular sets in Cayley sum graphs

Fateme sadat seiedali1 ,  Behrooz Khosravi1 and  Zeinab Akhlaghi1,2 1 Faculty
of Mathematics and Computer Science, Amirkabir University of Technology
(Tehran Polytechnic), 15914 Tehran, Iran.  2 School of Mathematics, Institute
for Research in Fundamental Science (IPM) P.O. Box:19395-5746, Tehran, Iran.
(F. Seiedali<EMAIL_ADDRESS>
(Z. Akhlaghi<EMAIL_ADDRESS>
(B. Khosravi<EMAIL_ADDRESS>
###### Abstract.

A subset $C$ of the vertex set of a graph $\Gamma$ is said to be
$(\alpha,\beta)$-regular if $C$ induces an $\alpha$-regular subgraph and every
vertex outside $C$ is adjacent to exactly $\beta$ vertices in $C$. In
particular, if $C$ is an $(\alpha,\beta)$-regular set in some Cayley sum graph
of a finite group $G$ with connection set $S$, then $C$ is called an
$(\alpha,\beta)$-regular set of $G$ and a $(0,1)$-regular set is called a
perfect code of $G$. By Sq$(G)$ and NSq$(G)$ we mean the set of all square
elements and non-square elements of $G$. As one of the main results in this
note, we show that a subgroup $H$ of a finite abelian group $G$ is an
$(\alpha,\beta)$-regular set of $G$, for each $0\leq\alpha\leq|$NSq$(G)\cap
H|$ and $0\leq\beta\leq\mathcal{L}(H)$, where $\mathcal{L}(H)=|H|$, if
Sq$(G)\subseteq H$ and $\mathcal{L}(H)=|$NSq$(G)\cap H|$, otherwise. As a
consequence of our result we give a very brief proof for the main results in
[17, 18]. Also, we consider the dihedral group $G=D_{2n}$ and for each
subgroup $H$ of $G$, by giving an appropriate connection set $S$, we determine
each possibility for $(\alpha,\beta)$, where $H$ is an
$(\alpha,\beta)$-regular set of $G$.

###### Key words and phrases:

Perfect code · Subgroup perfect code · Cayley sum graph · Finite group.
Regular set

###### 2000 Mathematics Subject Classification:

05C25 , 05C69 , 94B25

The third author is supported by a Grant from IPM (no. 1403200013)

##  1\. Introduction

Let $C$ be a subset of ${\mathbb{F}}_{q}^{n}$, where ${\mathbb{F}}_{q}$ is the
finite field of order $q$. Then we call $C$ a code of lenght $n$ over
${\mathbb{F}}_{q}$ and the elements of $C$ are called codwords. For two
vectors $x,y\in{\mathbb{F}}_{q}^{n}$, their Hamming distance $d(x,y)$ is the
number of coordinates in which they differ and for $x\in{\mathbb{F}}_{q}^{n}$,
we define $d(x,C)=\min\\{d(x,y)\ :\ y\in C\\}$. A code $C$ over
${\mathbb{F}}_{q}$ is called $q$-ary $\rho$-covering code if for every vector
$y$ there is a codeword $x\in C$ such that the Hamming distance
$d(x,y)\leq\rho$. The covering radius of a code $C$ is the smallest $\rho$
such that $C$ is $\rho$-covering. Let $x\in{\mathbb{F}}_{q}^{n}$. Denote by
$B_{x,i}$, the number of codewords of distance $i$ from $x$. A code $C$ with
covering radius $\rho$ is called $t$-regular ($0\leq t\leq\rho$) if for all
$i=0,\dots,\rho$, $B_{x,i}$ depends only on $i$ and $d(x,C)$, for all $x$ such
that $d(x,C)\leq t$ (see for instance [9, 12]). A code $C$ is completely
regular if it is $\rho$-regular (see [9]). We refer to [5], for a
comprehensive survey on completely regular codes. In $1$-regular code, we face
two important parameters $\alpha=B_{x,1}$, for $x\in C$ and $\beta=B_{x,1}$,
for $x\in{\mathbb{F}}_{q}^{n}\setminus C$ (we have two other trivial parametrs
$B_{x,0}=1$ for $x\in C$ and $B_{x,0}=0$ for $x\not\in C$, which can be
ignored). It means that for $x\in C$, there exist exactly $\alpha$ elements in
$C$ of distance 1 from $x$ and for all $x\in{\mathbb{F}}_{q}^{n}\setminus C$,
there exist exactly $\beta$ elements in $C$ of distance $1$ from $x$. In
particular, a $1$-regular code with $B_{x,1}=0$, for $x\in C$ and $B_{x,1}=0$,
for $x\in{\mathbb{F}}_{q}^{n}\setminus C$, is called a perfect code. The
combinatorial properties of completely regular codes allow to establish
different relations with other combinatorial structures such as distance-
regular graphs, association schemes and designs. To extend the idea of regular
code in other combinatorial structure, many authors pay attention perfect
codes and regular sets in graphs.

In this paper, all groups are finite and all graphs are simple. Let
$\Gamma=({\bf V}(\Gamma),{\bf E}(\Gamma))$ be a simple graph, where ${\bf
V}(\Gamma)$ and ${\bf E}(\Gamma)$ are the sets of its vertices and the set of
its edges, respectively. For non-negative integers $\alpha,\beta$, a subset
$C$ of ${\bf V}(\Gamma)$ is called an $(\alpha,\beta)$-regular set in
$\Gamma$, if every vertex of $C$ is adjacent to exactly $\alpha$ vertices of
$C$ and every vertex of ${\bf V}(\Gamma)\setminus C$ is a neighbor to exactly
$\beta$ vertices of $C$ (see [6]). A $(0,1)$-regular set is called a perfect
code. Clearly, the definition of regular set in a graph arises from the
definition of the $1$-regular code, by replacing the hamming distance of two
vectors by the distance of the vertices of the graph, as the metric.

For a graph $\Gamma=({\bf V}(\Gamma),{\bf E}(\Gamma))$, a partition of ${\bf
V}(\Gamma)$ with cells $\mathcal{V}=\\{V_{1},\dots,V_{k}\\}$ is called an
equitable $k$-partition, when each cell induces a regular subgraph and any
vertex of $V_{i}$ is adjacent to $b_{ij}$ vertices of $V_{j}$, [11, Section
9.3]. The quotient matrix of the partition $\mathcal{V}$ is defined as
$M=(b_{ij})$. We note that, if the row sums of a $k\times k$ matrix $A$ is
equal to a fixed number, say $r$, then $r$ is an eigenvalue of $A$ [11,
Theorem 9.3.3]. Therefore, if $\Gamma$ is a connected $r$-regular graph, then
$r$ is a simple eigenvalue of the quotient matrix $M$ of $\Gamma$. If $\Gamma$
is a $r$-regular graph, $\mathcal{V}$ an equitable partition of $V(\Gamma)$
and $M$ is its quotient matrix such that all eigenvalues of $M$ except $r$ are
equal to $\mu$, then $\mathcal{V}$ is called $\mu$-equitable. We note that a
non-trivial coarsening of a $\mu$-equitable partition is $\mu$-equitable (see
[4, Corollary 2.3]). So, the study of equitable partition with exactly two
parts is important.

Moreover, an $(\alpha,\beta)$-regular set in a $r$-regular graph $\Gamma$ is
exactly a completely regular code $C$ in $\Gamma$ (see, for example, [19])
such that the corresponding distance partition has exactly two parts, namely
$\\{C,{\bf V}(\Gamma)\setminus C\\}$. An equitable $2$-partition is also
called perfect 2-coloring [10]. The notion of perfect coloring is a common
research subject in coding theory [3, 16].

Let $G$ be a group with identity element $e$ and $S$ be an inverse-closed
subset of $G$, i.e. $S^{-1}=\\{s^{-1}:s\in S\\}=S$, where $e\notin S$. The
Cayley graph  Cay$(G,S)$ of $G$ with respect to the connection set $S$ is
defined to be the graph with vertex set $G$ such that two elements $x$, $y$
are adjacent if $yx^{-1}\in S$, see [1, 8].

An element $x$ of $G$ is called a square if $x=y^{2}$, for some element $y\in
G$. A subset of $G$ whose elements are not square, is called square-free. A
subset $S$ of $G$ is called a normal subset if for every $g\in G$,
$g^{-1}Sg=\\{g^{-1}sg:s\in S\\}=S$. Let $S$ be a normal square-free subset of
$G$. The Cayley sum graph CayS$(G,S)$ of $G$ with respect to the connection
set S is a simple graph with vertex set $G$ and two vertices $x$ and $y$ are
adjacent if $xy\in S$. Since $S$ is a normal square-free subset of $G$, $xy\in
S$ if and only if $yx\in S$, hence CayS$(G,S)$ is an undirected graph without
loops. It is easy to see that CayS$(G,S)$ is $|S|$-regular. The Cayley sum
graph is first defined for abelian groups (see [7]) and then it is generalized
to any arbitrary group in [18]. The perfect code of the cayley graphs of
finite groups are widely studied by some authors (see for instance [14, 22]).
We note that in [20], a perfect code of a Cayley graph of $G$, briefly is
called a perfect code of $G$. In [17, 18], a subset $C$ of $G$ is called a
perfect code of G if there exists a Cayley sum graph of $G$ which admits $C$
as a perfect code and in particular, if a subgroup $H$ of $G$ is a perfect
code of $G$, then $H$ is called a subgroup perfect code of $G$. In the sequel
of this paper, by a perfect code of $G$ we mean a perfect code of a Cayley sum
graph of $G$ with respect to a subset $S$.

In [17], Ma et al. provided necessary and sufficient conditions for a non-
trivial subgroup of an abelian group to be a subgroup perfect code. They
determine whether a subgroup of an abelian group is a perfect code according
to its Sylow 2-subgroup. Moreover, they specified all subgroup perfect codes
of a cyclic group, a dihedral group, and a generalized quaternion group. Also
in [18], they gave a shortened proof for classifying all perfect codes of
abelian groups.

In [20], it is proved that a normal subgroup $H$ is a perfect code of a Cayley
graph of $G$ if and only if it is an $(\alpha,\beta)$-regular set in a Cayley
graph of $G$, for each $0\leq\alpha\leq|H|-1$ and $0\leq\beta\leq|H|$, where
gcd$(2,|H|-1)$ divides $\alpha$.

In [23], for a normal subgroup $H$ of $G$, it is proved that for
$0\leq\alpha\leq|H|-1$ and $0\leq\gamma\leq|H|/2$ such that gcd$(2,|H|-1)$
divides $\alpha$, $H$ is an $(\alpha,2\gamma)$-regular set in a Cayley graph
of $G$.

It is natural to ask about regular sets in other graphs, specially Cayley sum
graphs. In the sequel of this paper, by an $(\alpha,\beta)$-regular set of $G$
we mean an $(\alpha,\beta)$-regular set in a Cayley sum graph of $G$ and we
study the following problem for abelian groups and dihedral groups $G=D_{2n}$,
for $n\geq 3$:

Problem: Describe all subgroups $H$ of $G$ and parameters $\alpha$ and $\beta$
such that $H$ is an $(\alpha,\beta)$-regular set of $G$.

To answer to the above problem, for each subgroup $H$ of $G$ and
$(\alpha,\beta)$ such that $H$ is an $(\alpha,\beta)$-regular set of $G$, we
show that there exists a connection set $S$ of size
$\alpha+\beta((|G|/|H|)-1)$ (Corollary 2.5) such that $H$ is an
$(\alpha,\beta)$-regular set in ${\rm CayS}(G,S)$. We note that this graph has
valency $|S|$ and it is worth mentioning that the connection set $S$ is not
unique necessarily, and sometimes the number of them can be large (for example
in the abelian groups).

Let $G$ be an abelian group and $H$ a subgroup of $G$. In the second section,
we determine all the possibilities for $\alpha$ and $\beta$, such that $H$ is
an $(\alpha,\beta)$-regular set of $G$. As a consequence we give a shorter
proof for the main results in [17, 18].

In the third section, we study this problem for each subgroup $H$ of the
dihedral group $G$. In the proofs of the theorems in this section, for each
subgroup $H$ of $G$ and $(\alpha,\beta)$ such that $H$ is an
$(\alpha,\beta)$-regular set of $G$, we introduce a connection set $S$.

Obviously, $G$ is a perfect code in the empty Cayley sum graph
CayS$(G,\emptyset)$. As a result, any group is a subgroup perfect code of
itself. Similarly, any subgroup $H$ is a $(0,0)$-regular set in ${\rm
CayS}(G,\emptyset)$. As this case is trivial, in the rest of the paper we do
not consider the case when $(\alpha,\beta)=(0,0)$. It is easy to see that
every non-trivial element of $G$ is non-square if and only if $G$ is an
elementary abelian $2$-group. So a Cayley sum graph CayS$(G,S)$ is complete if
and only if G is an elementary abelian $2$-group and $S=G\setminus\\{e\\}$.
This also means that the trivial subgroup $\\{e\\}$ of $G$ is a subgroup
perfect code if and only if $G$ is an elementary abelian $2$-group.

Let $G$ be a finite group. Throughout the paper, the identity element of $G$
is denoted by $e$. The number of elements of $G$ is denoted by $|G|$ and is
called the order of $G$. The order of an element $g\in G$, denoted by $o(g)$,
is the smallest natural number $n$ such that $g^{n}=e$. Let $H$ be a subgroup
of $G$ and $a\in H$, then $Ha=\\{ha\ :\ h\in H\\}$ is called a right coset of
$H$ in $G$. The index of a subgroup $H$ in a group $G$ is the number of
distinct right cosets of $H$ in $G$, which is denoted by $[G:H]$. A right
transversal of $H$ in $G$ is a subset of G which contains exactly one element
from each right coset of H. If the order of $H$ is the largest odd divisor of
the order of $G$, then $H$ is called a Hall $2^{\prime}$-subgroup of $G$. Let
$|G|=p^{n}m$ where $p\nmid m$, then a subgroup of $G$ of order $p^{n}$ is
called a Sylow $p$-subgroup of $G$. An elementary abelian $p$-group is an
abelian group in which every non-trivial element has order $p$, where $p$ is a
prime. The direct product of two groups $G$ and $H$ is $G\times H=\\{(g,h)\ :\
g\in G,\ h\in H\\}$, with the group operation given by
$(g_{1},h_{1}).(g_{2},h_{2})=(g_{1}g_{2},h_{1}h_{2})$, where the coordinate-
wise operation are the operations in $G$ and $H$. Two elements $a$ and $b$ of
$G$ are conjugate if there is an element $g\in G$ such that $b=gag^{-1}$. The
conjugacy class of $a\in G$ is the set of all conjugates of $a$ in $G$, which
is denote by $a^{G}$. Let $X\subseteq G$ be a non-empty subset of $G$, then
the smallest subgroup of $G$, containing $X$ is denoted by $\langle X\rangle$.

By Sq$(G)$ and NSq$(G)$ we mean the set of all square and non-square elements
of $G$. If $G$ is an abelian group, then we denote by ${\bf E}_{2}(G)$, the
greatest direct factor of $G$ isomorphic to an elementary abelian $2$-group.
By $C_{n}$ we mean the cyclic group of order $n$.

##  2\. REGULAR SETS IN ABELIAN GROUPS

######  Lemma 2.1.

[18, Lemma 2.1] Let $S$ and $H$ be a normal square-free subset and a subgroup
of a finite group $G$, respectively. The following are equivalent:

  1. (1)

H is a perfect code of CayS$(G,S)$.

  2. (2)

$S\cup\\{e\\}$ is a right transversal of $H$ in $G$.

  3. (3)

$[G:H]=\left|S\right|+1$ and $\left(S\cup SS^{-1}\right)\cap H=\\{e\\}$.

######  Lemma 2.2.

[18, Theorem 3.1] Let $G$ be an abelian group with non-trivial Sylow
$2$-subgroup $P$, and let $H$ be a non-trivial subgroup of $G$. Then $H$ is a
subgroup perfect code of $G$ if and only if one of the following occurs:

  1. (1)

$P\subseteq H$;

  2. (2)

$[G:H]=|P|$ and $P$ is elementary abelian;

  3. (3)

$P\cap H$ is a non-trivial subgroup perfect code of $P$, and either $[G:H]$ is
a power of $2$ or $P\cap H$ has a non-square element in $G$.

######  Lemma 2.3.

[17, Lemma 3.1] Let $A=C_{2^{m_{1}}}\times C_{2^{m_{2}}}\times\cdots\times
C_{2^{m_{k}}}\times A_{2^{\prime}}$ be an abelian group and $A_{2^{\prime}}$
be the Hall $2^{\prime}$-subgroup of $A$. Suppose $H$ is a subgroup of $A$.
Then $H$ is a subgroup perfect code of $A$ if and only if either $H$ is a
subgroup isomorphic to

$C_{2^{m_{1}-1}}\times C_{2^{m_{2}-1}}\times\cdots\times C_{2^{m_{k}-1}}\times
A_{2^{\prime}}$

or $H$ has a non-square element.

In the following Lemmas we discuss the regular subsets of a Cayley sum graph
of $G$, which is the Cayley sum graph version of [21, Lemma 2.7].

######  Lemma 2.4.

Let $G$ be a group and $H$ be a subgroup of $G$. Then $H$ is a
$(0,\beta)$-regular set of $G$ if and only if $G$ has $\beta$ pairwise
disjoint subsets $T_{i}$, $1\leq i\leq\beta$, such that for each $i$,
$T_{i}\cup\\{e\\}$ is a right transversal of $H$ in $G$ and
$\bigcup\limits_{i=1}^{\beta}T_{i}$ is a normal square-free subset of $G$. In
particular, if $H$ is an $(\alpha,\beta)$-regular set in CayS$(G,S)$, then
$|S\cap Hx|=\beta$, for each $x\in G\setminus H$.

###### Proof.

Suppose that $T_{1},...,T_{\beta}$ are pairwise disjoint subsets of NSq$(G)$,
such that for each $i$, where $1\leq i\leq\beta$, $T_{i}\cup\\{e\\}$ is a
right transversal of $H$ in $G$ and let $S=\bigcup\limits_{i=1}^{\beta}T_{i}$
is a normal subset of $G$. Let $x\in G\setminus H$. Then for each
$i\in\\{1,\dots,\beta\\}$ we have $|Hx\cap T_{i}|$=1, and so there exists
$h_{i}\in H$ such that $h_{i}x\in T_{i}$. Thus, $x$ is adjacent to at least
$\beta$ elements, $h_{1},\dots,h_{\beta}\in H$ in $\rm{CayS}(G,S)$. If there
exists some $h\in H\setminus\\{h_{1},\dots,h_{\beta}\\}$ such that $hx\in S$,
then there exists $i\in\\{1,\dots,\beta\\}$, such that $hx\in T_{i}$. Thus,
$Hx\cap T_{i}$ contains more than one element, which is a contradiction, as
$T_{i}$ is a right transversal of $H$ in $G$. As $S\cap H=\emptyset$, we get
that $H$ is a $(0,\beta)$-regular set in CayS$(G,S)$.

Conversely, assume that $H$ is a $(0,\beta)$-regular set in CayS$(G,S)$, for
some normal square-free subset $S\subseteq G$. It means that for every $x\in
G\setminus H$ there exist exactly $\beta$ distinct elements
$h_{1},h_{2},...,h_{\beta}$ in $H$ such that $h_{i}x\in S$, for
$i=1,\dots,\beta$. Hence, $|S\cap Hx|=\beta$, for all $x\in G\setminus H$,
which means that $S$ is the union of $\beta$ pairwise disjoint subsets
$T_{i}$, where $1\leq i\leq\beta$, such that for each $i$, $T_{i}\cup\\{e\\}$
is a right transversal of $H$ in $G$.

######  Corollary 2.5.

Let $H$ be a subgroup of $G$. If $H$ is an $(\alpha,\beta)$-regular set of
$G$, then there is a normal square-free subset $S$ (the connection set of the
corresponding Cayley sum graph) such that $|S\cap H|=\alpha$ and
$|S\cap(G\setminus H)|=\beta([G:H]-1)$. Obviously, $\beta\leq|H|$ and
$|S|=\alpha+\beta([G:H]-1)$.

Moreover, there exist $\beta$ pairwise disjoint subsets $T_{i}$, $1\leq
i\leq\beta$, such that $S\cap(G\setminus
H)=\bigcup\limits_{i=1}^{\beta}T_{i}$, where for each $i$, $T_{i}\cup\\{e\\}$
is a right transversal of $H$ in $G$.

######  Lemma 2.6.

Let $H$ be a normal subgroup of $G$. Then $H$ is an $(\alpha,\beta)$-regular
set of $G$ if and only if $H$ is an $(\alpha,0)$-regular set of $G$ and a
$(0,\beta)$-regular set of $G$.

###### Proof.

If $H$ is a normal subgroup of $G$ and $S$ a normal square-free subset of $G$,
then $S^{\prime}=S\cap H$ and $S^{\prime\prime}=S\cap(G\setminus H)$ are
normal square-free subsets of $G$. If $H$ is an $(\alpha,\beta)$-regular set
in ${\rm CayS}(G,S)$, then $H$ is an $(\alpha,0)$-regular set in
CayS$(G,S^{\prime})$ and a $(0,\beta)$-regular set in
CayS$(G,S^{\prime\prime})$. Conversely, if $H$ is an $(\alpha,0)$-regular set
in CayS$(G,S^{\prime})$ and a $(0,\beta)$-regular set in
CayS$(G,S^{\prime\prime})$, for some normal square-free subsets $S^{\prime}$
and $S^{\prime\prime}$, then $H$ is an $(\alpha,\beta)$-regular set in
CayS$(G,S^{\prime}\cup S^{\prime\prime})$.

######  Notation 2.7.

Let $G$ be a group and $H$ be a subgroup of $G$. Then we set

$\mathcal{L}(H)=\min\\{|{\rm NSq}(G)\cap Hx|:x\in G\setminus H\\}.$

######  Theorem 2.8.

Let $H$ be a subgroup of an abelian group $G$. Then, $\mathcal{L}(H)=|H|$ if
Sq$(G)\subseteq H$, and $\mathcal{L}(H)=|$NSq$(G)\cap H|$, otherwise.

###### Proof.

Remind that $\mathcal{L}(H)=\min\\{|{\rm NSq}(G)\cap Hx|:x\in G\setminus
H\\}.$ First, assume that $x\in G\setminus H$ is a square element. In this
case, for each non-square element $y\in H$, $yx$ is a non-square element in
$Hx$, and for each square element $y\in H$, $yx$ is a square element in $Hx$.
Hence, $|$NSq$(G)\cap Hx|=|$NSq$(G)\cap H|$.

Now, assume that $x\in G\setminus H$ is a non-square element. If $Hx$ has a
square element, say $z$, then $Hx=Hz$ and, by the above discussion, we
conclude that $|$NSq$(G)\cap Hx|=|$NSq$(G)\cap Hz|=|$NSq$(G)\cap H|$. If all
elements of $Hx$ are non-square, then $|$NSq$(G)\cap Hx|=|Hx|=|H|$. It follows
that, if Sq$(G)\subseteq H$, then for every $x\in G\setminus H$, we have
$Hx\subseteq{\rm NSq}(G)$ and so $\mathcal{L}(H)=|H|$. Otherwise, there exists
$x\in G\setminus H$ which is square and so $\mathcal{L}(H)=|$NSq$(G)\cap H|$.

The above theorem is the main result of this section which has wide
application for abelian groups. Using this result for given abelian group $G$
and subgroup $H$ of $G$, we can determine all possibilities for $\beta$ such
that $H$ is a $(0,\beta)$-regular set of $G$. Also it has a significant role
in giving very brief proofs for all the results related to abelian groups in
[17, 18].

######  Theorem 2.9.

Let $G$ be a group and $H$ be a subgroup of $G$.

  1. (1)

If $H$ is an $(\alpha,\beta)$-regular set of $G$, then $0\leq\alpha\leq|{\rm
NSq}(G)\cap H|$ and $0\leq\beta\leq\mathcal{L}(H)$.

  2. (2)

If $G$ is abelian, then $H$ is an $(\alpha,\beta)$-regular set of $G$ if and
only if $0\leq\alpha\leq|{\rm NSq}(G)\cap H|$ and
$0\leq\beta\leq\mathcal{L}(H)$.

###### Proof.

(1) Let $H$ be an $(\alpha,\beta)$-regular set in CayS$(G,S)$ for some subset
$S$ of $G$ and $\mathcal{L}(H)=|$NSq$(G)\cap Hx_{0}|$, where $x_{0}\in
G\setminus H$. Then, $S\cap Hx_{0}\subseteq$ NSq$(G)\cap Hx_{0}$. Thus, by
Lemma 2.4, $0\leq\beta=|S\cap Hx_{0}|\leq\mathcal{L}(H)$. As $\alpha=|S\cap
H|$, we have $\alpha\leq|{\rm NSq(G)}\cap H|$.

(2) Now, assume that $G$ is abelian. By Lemma 2.6 we only need to prove that
$H$ is an $(\alpha,0)$-regular set of $G$ and a $(0,\beta)$-regular set of
$G$. Let $0\leq\alpha\leq|{\rm NSq}(G)\cap H|$. Then by choosing $S^{\prime}$
to be an arbitrary subset of ${\rm NSq}(G)\cap H$ of size $\alpha$ we have
that $H$ is an $(\alpha,0)$-regular set in ${\rm CayS}(G,S^{\prime})$. Let
$0\leq\beta\leq\mathcal{L}(H)$. Remark that, by the assumption each coset of
$H$ in $G$, except for $H$, has at least $\mathcal{L}(H)$ non-square elements.
So by choosing $\mathcal{L}(H)$ non-square elements of each coset of $H$,
besides $H$, we are able to construct $\mathcal{L}(H)$ pairwise disjoint
square-free subsets $T_{i}$, for $1\leq i\leq\mathcal{L}(H)$, such that for
each $i$, $T_{i}\cup\\{e\\}$ is a right transversal of $H$ in $G$. As $G$ is
abelian, $S=\bigcup\limits_{i=1}^{\beta}T_{i}$ is a normal subset of $G$. So
by Lemma 2.4, $H$ is a $(0,\beta)$-regular set of $G$.

######  Remark 2.10.

It is worth mentioning that if $H$ is an $(\alpha,\beta)$-regular set of an
abelian group $G$, then there are
$\binom{\mathcal{L}(H)}{\beta}^{[G:H]}\binom{|{\rm NSq}(G)\cap H|}{\alpha}$
subsets $S$ such that $H$ is an $(\alpha,\beta)$-regular set in ${\rm
CayS}(G,S)$, according to the proof of Theorem 2.9.

As we mentioned before, Theorem 2.9 leads to the following Corollary which
makes us conclude Lemmas 2.2 and 2.3, ([17, Theorem 3.1] and [18, Theorem
3.1]), more simply and concisely, and so all the other results related to the
abelian groups in the same papers can be obtained more easily.

######  Corollary 2.11.

Let $G$ be an abelian group and $H$ be a subgroup of $G$. Then $H$ is a
perfect code of $G$ if and only if Sq$(G)\subseteq H$ or $\rm{NSq}(G)\cap
H\not=\emptyset$.

###### Proof.

By Theorem 2.9, $H$ is a perfect code of $G$ if and only if
$\mathcal{L}(H)\geq 1$. On the other hand, $\mathcal{L}(H)\geq 1$ if and only
if ${\rm Sq}(G)\subseteq H$ or ${\rm NSq(G)}\cap H\not=\emptyset$, by Theorem
2.8.

Observe that if $G$ is an abelian group and $H$ is a $(0,\beta)$-regular set
of $G$, then for all $0\leq b\leq\beta$, $H$ is a $(0,b)$-regular set of $G$.
Now, we prove the following theorems.

######  Theorem 2.12.

Let $G$ be an abelian group and $H$ be a subgroup of $G$. Then $H$ is not a
$(0,2)$-regular set of $G$ if and only if one of the following occurs:

  1. (1)

$H\subsetneq$ Sq$(G)$;

  2. (2)

$H$ is trivial and $G$ is an elementary abelian group;

  3. (3)

$G\cong H\times K$, where $H\cong C_{2}$ and $K$ has at least one non-trivial
square element.

###### Proof.

Suppose that $H$ is not a $(0,2)$-regular set of $G$. Theorem 2.9 implies that
$\mathcal{L}(H)=0$ or $1$. If $\mathcal{L}(H)=0$, then ${\rm
Sq}(G)\not\subseteq H$ and ${\rm NSq}(G)\cap H=\emptyset$, by Theorem 2.8.
Thus, $H\subsetneq$ Sq$(G)$. Now, assume that $\mathcal{L}(H)=1$. By Theorem
2.8, we get the following cases:

Case 1: $\mathcal{L}(H)=|H|=1$.

In this case, Sq$(G)\subseteq H=\\{e\\}$ and so every non-trivial element of
$G$ is non-square, implying that the order of each non-trivial element is $2$.
So $G$ is an elementary abelian $2$-group.

Case 2: $\mathcal{L}(H)=|$NSq$(G)\cap H|=1$.

Let NSq$(G)\cap H=\\{x\\}$. As $x^{3}\in{\rm NSq}(G)\cap H$, we conclude that
$x=x^{3}$ and so the order of $x$ is $2$. If $y\in H\setminus\\{x\\}$, then
$y$ is a square and so $xy\in$ NSq$(G)\cap H$, which implies that $y=e$.
Therefore, $H=\\{e,x\\}\cong C_{2}$. We may assume that $G=G_{0}\times
G_{1}\times\cdots\times G_{k}\times A$, where $k\geq 0$ and $G_{i}\cong
C_{2^{\alpha_{i}}}$, for some integers $\alpha_{0},...,\alpha_{k}$, and $A$ is
the Hall $2^{\prime}$-subgroup of $G$. Let $x=(l_{0},l_{1},\dots,l_{k},a)$,
where $a\in A$ and $l_{i}\in G_{i}$, for $i=0,\dots,k$. As $x^{2}=e$, we
conclude that $a=e$ and $o(l_{j})\leq 2$, for each $j=0,\dots,k$. As $x$ is
non-square, there exists $0\leq i_{0}\leq k$ such that $l_{i_{0}}$ is non-
square, and so $l_{i_{0}}$ is a generator of $G_{i_{0}}$ of order $2$. So
without loss of generality, we may assume that $i_{0}=0$ and
$c=l_{i_{0}}=l_{0}$. Then $G_{0}=\\{e,c\\}$ and
$x=(c,l_{1},l_{2},...,l_{k},e)$. Setting $K=\\{e\\}\times
G_{1}\times\cdots\times G_{k}\times A$, we have $H\cap K$ is trivial and
consequently, $G=H\times K$.

######  Theorem 2.13.

Let $G$ be an abelian group and $H$ be a subgroup of $G$. Then $H$ is not a
$(0,3)$-regular set of $G$ if and only if one of the following occurs:

  1. (1)

$H\subsetneq$ Sq$(G)$;

  2. (2)

$G$ is an elementary abelian $2$-group and $|H|\leq 2$;

  3. (3)

$G\cong H\times K$, where $H\cong C_{2}$ and $K$ has at least one non-trivial
square element;

  4. (4)

$G\cong\mathbb{E_{2}}(G)\times T$, where $T\cong C_{4}$, and $H$ is the
subgroup of order $2$ of $T$;

  5. (5)

$G\cong\mathbb{E_{2}}(G)\times K$ and $H=\langle x,y\rangle$ where
$x=(m,k_{1})$ and $y=(m,k_{2})$, such that $e\not=m\in\mathbb{E_{2}}(G)$ and
$k_{1},k_{2}\in K$ have orders at most $2$.

  6. (6)

$H$ is generated by a non-square element of order $4$;

###### Proof.

Suppose that $H$ is not a $(0,3)$-regular set of $G$. Theorem 2.9 implies that
$\mathcal{L}(H)\in\\{0,1,2\\}$. If $\mathcal{L}(H)=0$ or $1$, then the result
is the same as the previous theorem and $H$ satisfies one of the Cases
(1)-(3). So suppose that $\mathcal{L}(H)=2$.

First, assume that Sq$(G)\subseteq H$, hence $\mathcal{L}(H)=|H|=2$. In this
case, if $|{\rm Sq}(G)|=1$, then $G\cong C_{2}^{\alpha}$, for some $\alpha$
and $H$ is any subgroup of $G$ of order $2$, which is Case (2). If
$|$Sq$(G)|=2$, then, as all elements in the Hall $2^{\prime}$-subgroup of $G$
are square, we deduce that $G=\mathbb{E_{2}}(G)\times T$, where $T\cong C_{4}$
and $H$ is the unique subgroup of order $2$ of $T$, which is Case (4).

Now, suppose that $\mathcal{L}(H)=|$NSq$(G)\cap H|=2$. Let NSq$(G)\cap
H=\\{x,y\\}$. Then $x^{3}\in H$ is a non-square element. Therefore, either
$x^{3}=x$ or $x^{3}=y$.

$\bullet$ First, assume $x^{3}=x$, and so $o(x)=2$. Let $z\in
H\setminus\\{x,y\\}$. Then, $xz\in{\rm NSq}(G)\cap H$ and hence $xz=x$ or
$xz=y$. Thus, either $z=e$ or $z=x^{-1}y$, which means that $|H|=4$. As
$o(x)=2$ and $x$ is non-square, then $H\cong C_{2}\times C_{2}$. Assume
$G=G_{0}\times G_{1}\times\cdots\times G_{n}\times A$, where $n\geq 0$ and
$G_{i}\cong C_{2^{\alpha_{i}}}$, for some integers
$\alpha_{0},...,\alpha_{n}$, and $A$ is the Hall $2^{\prime}$-subgroup of $G$.
Let $x=(l_{0},l_{1},\dots,l_{n},a)$, where $a\in A$ and $l_{i}\in G_{i}$, for
$i=0,\dots,n$. As $x^{2}=e$, we conclude that $a=e$ and $o(l_{j})\leq 2$, for
each $j$. Since $x$ is non-square, there exists $0\leq i_{0}\leq n$ such that
$l_{i_{0}}$ is a non-square element of $G_{i_{0}}$, and so $l_{i_{0}}$ is the
generator of $G_{i_{0}}\cong C_{2}$. As $G=\mathbb{E_{2}}(G)\times K$, for
some subgroup $K$ of $G$, we conclude that there exists
$0\not=m\in\mathbb{E_{2}}(G)$ and $k_{1}\in K$ such that $x=(m,k_{1})$, where
$o(k_{1})\leq 2$. Now, consider $z=x^{-1}y\in H$. As $z$ is square and it has
order $2$ then $z=(e,k)\in\mathbb{E_{2}}(G)\times K$, for some $k\in K$ of
order $2$. So $y=xz=(m,k_{1}k)$ has the form described in Case (5).

$\bullet$ Now, assume $x^{3}=y$. If $z\in H\setminus\\{e,x,y\\}$, then, as
$xz$ is non-square and $z$ is non-trivial, we deduce that $xz=x^{3}=y$ and so
$z=x^{2}$. Therefore $H$ is the cyclic group of order 4, generated by $x$,
which is Case (6).

######  Remark 2.14.

Note that in Case (6), $H$ is not necessarily a direct factor of $G$. For
instance, $G=C_{2}\times C_{8}$ and $H=\langle(1,2)\rangle$ is satisfying Case
(6), however $H$ is not a direct factor of $G$.

##  3\. dihedral groups

Throughout this section we assume $n\geq 3$, $G=D_{2n}=\left\langle
a,b:a^{n}=b^{2}=e,b^{-1}ab=a^{-1}\right\rangle$, the dihedral group of order
$2n$. We note that $o(a^{i}b)=2$, for each $1\leq i\leq n$ and
$a^{s}ba^{k}b^{l}=a^{s-k}b^{l+1}$, for all integers $s,k$ and $l$. The
subgroups of $G$ (see for example [2]) are as follows:

  * •

the cyclic subgroup $H=\left\langle a^{t}\right\rangle$, where $t$ divides
$n$; and so $[G:H]=2t$;

  * •

the subgroup $H=\left\langle a^{t},a^{s}b\right\rangle$, where $t$ divides $n$
and $0\leq s\leq t-1$; which is isomorphic to $C_{2}$ (when $t=n$),
$C_{2}\times C_{2}$ (when $n$ is even and $t=n/2$) or the dihedral group
$D_{2n/t}$ (when $t$ is a proper divisor of $n$ and $t\not=n/2$). In this case
$[G:H]=t$.

We use the above notations without further references.

######  Lemma 3.1.

[15] Suppose that $n=2m$, for some positive integer $m>3$. Then $G=D_{2n}$ has
$m+3$ conjugacy classes as follows: $\\{e\\}$, $\\{a^{m}\\}$,
$b^{G}=\\{a^{2j}b:0\leq j\leq m-1\\},$ $(ab)^{G}=\\{a^{2j+1}b:0\leq j\leq
m-1\\},$ $\\{a^{i},a^{-i}\\}$, where $1\leq i\leq m-1$.

######  Lemma 3.2.

[15] Suppose that $n$ is an odd integer. Then $G=D_{2n}$ has $(n+3)/2$
conjugacy classes as follows: $\\{e\\},$ $b^{G}=\\{a^{j}b:0\leq j\leq n-1\\},$
$\\{a^{i},a^{-i}\\},$ where $1\leq i\leq(n-1)/2$.

Note that, for each $1\leq i\leq n$, $a^{i}b$ is a non-square element in
$D_{2n}$. If $n$ is odd, then $a^{i}$ is square, for $i=1,\dots,n$. Also, if
$n$ is even, then $a^{2i}$ is square and $a^{2i+1}$ is non-square, for
$i=0,\dots,(n-2)/2$.

######  Example 3.3.

For comprehending the topic, we describe some examples of an
$(\alpha,\beta)$-regular set of dihedral group $G=D_{8}$:

$\bullet$ Let $H=\left\langle a^{2}b\right\rangle$. Then $[G:H]=4$ and the set
of cosets of $H$ in $G$ is $\\{H,Ha,Ha^{2},Ha^{3}\\}$. In this case, ${\rm
NSq}(G)\cap H=\\{a^{2}b\\}$, ${\rm NSq}(G)\cap Ha=Ha=\\{a,ab\\}$, ${\rm
NSq}(G)\cap Ha^{2}=\\{b\\}$ and ${\rm NSq}(G)\cap
Ha^{3}=Ha^{3}=\\{a^{3},a^{3}b\\}$. So, $\mathcal{L}(H)=1$ and by Theorem
2.9$,\alpha,\beta\leq 1$. Then, $S=b^{G}\cup(ab)^{G}$ is a normal square-free
subset of $G$ and $|S\cap Hx|=1$, for all $x\in G$. By Lemma 2.4 and Corollary
2.5, $H$ is a $(1,1)$-regular set in CayS$(G,S)$. For more details see Figure
1(a),

$\bullet$ Let $H=\left\langle a^{2},ab\right\rangle$. Then $[G:H]=2$. Note
that, $(ab)^{G}$ is the only normal square-free subset of $G$ contained in $H$
and the square-free conjugacy classes of $G$ contained in $G\setminus H$ are
$\\{a,a^{3}\\}$ and $b^{G}$. Set $S^{\prime\prime}=\emptyset$, $\\{a,a^{3}\\}$
or $b^{G}$ or $\\{a,a^{3}\\}\cup b^{G}$. Setting $S=(ab)^{G}\cup
S^{\prime\prime}$, we have $H$ is a $(2,|S^{\prime\prime}|)$-regular set in
CayS$(G,S)$. For instance $H$ is a $(2,2)$-regular set in ${\rm
CayS}(G,(ab)^{G}\cup b^{G})$, which can be seen in Figure (b).

In these graphs, for both examples, the vertices in $H$ are blank, the edges
between two vertices in $H$ are dotted line and the edges between a vertex of
$H$ and a vertex of $G\setminus H$ are bold.

$e$$a^{2}b$$a$$a^{2}$$a^{3}$$b$$ab$$a^{3}b$

$e$$a^{2}b$$a$$a^{2}$$a^{3}$$b$$ab$$a^{3}b$

Figure 1. (a) (1,1)-regular set (b) (2,2)-regular set

######  Theorem 3.4.

Let $H=\langle a^{s}b\rangle$, where $0\leq s\leq n-1$, be a subgroup of
$G=D_{2n}$. Then $H$ is an $(\alpha,\beta)$-regular set of $G$, if and only if
$(\alpha,\beta)=(1,1)$.

###### Proof.

Let $H$ be an $(\alpha,\beta)$-regular set in ${\rm CayS}(G,S)$, for some
subset $S$ of $G$. Note that $Ha^{k}=\\{a^{k},a^{s-k}b\\}$, for each integer
$k$. Therefore, ${\rm NSq}(G)\cap Ha^{2i}=\\{a^{s-2i}b\\}$, for each integer
$i$. Also, if $n$ is even, then ${\rm NSq}(G)\cap
Ha^{2i+1}=\\{a^{2i+1},a^{s-2i-1}b\\}$ and if $n$ is odd, then ${\rm
NSq}(G)\cap Ha^{2i+1}=\\{a^{s-(2i+1)}b\\}$, for each integer $i$. Therefore,
by Theorem 2.9, $\alpha\leq 1$ and $\beta\leq\mathcal{L}(H)=1$. Now, we claim
that if $\beta=1$, then $\alpha=1$. Let $\beta=1$. By Lemma 2.4, $a^{s-2i}b\in
S$, for each $i$. Therefore, as $S$ is a normal subset of $G$, by Lemmas 3.1
and 3.2, we get that $a^{s}b\in(a^{s-2i}b)^{G}\subseteq S$ (as the parity of
$s$ and $s-2i$ are the same), and so $\alpha=|S\cap H|=1$.

If $\beta=0$, then, by Lemma 2.4, for each $1\leq i\leq n-1$, we have $|S\cap
Ha^{i}|=0$. So $a^{s-2}b\not\in S$. As $a^{s}b\in(a^{s-2}b)^{G}$ and
$S\cap(a^{s-2}b)^{G}=\emptyset$, it follows that $\alpha=0$ and $S=\emptyset$,
which is excluded. Thus, the only possibility is $(\alpha,\beta)=(1,1)$.

In addition, by taking $S=(ab)^{G}\cup b^{G}$, we get that $H$ is a
$(1,1)$-regular set of $G$.

######  Theorem 3.5.

Let $n$ be an odd integer, $t>1$ a proper divisor of $n$, and $0\leq s\leq
t-1$. Let $H=\left\langle a^{t},a^{s}b\right\rangle$ be a subgroup of
$G=D_{2n}$. Then $H$ is an $(\alpha,\beta)$-regular set of $G$ if and only if
$(\alpha,\beta)=(|H|/2,|H|/2)$.

###### Proof.

By the assumption, $H$ is isomorphic to $D_{2n/t}$. Let $H$ be an
$(\alpha,\beta)$-regular set in CayS$(G,S)$, for some subset
$\emptyset\not=S\subseteq G$. Since $S\subseteq{\rm NSq}(G)=b^{G}$ and $S$ is
normal, we get that $S=b^{G}$. Note that $H=\\{a^{rt},a^{rt+s}b\ |\
\text{where $1\leq r\leq n/t$}\\}$ and so for every integer
$k\in\\{0,\dots,n-1\\}$ and $l\in\\{0,1\\}$,
$Ha^{k}b^{l}=\\{a^{rt+k}b^{l},a^{s+rt-k}b^{l+1}|\ \text{where $1\leq r\leq
n/t$}\\}.$ We can see that $|b^{G}\cap Ha^{k}b^{l}|=|H|/2$. Therefore,
$\alpha=|S\cap H|=|b^{G}\cap H|=|H|/2$, and $\beta=|S\cap Hx|=|b^{G}\cap
Hx|=|H|/2$, for each $x\in G\setminus H$.

######  Theorem 3.6.

Let $n=2m\geq 4$, for some integer $m$, $t>1$ is a proper divisor of $n$ and
$0\leq s\leq t-1$. Suppose $H=\left\langle a^{t},a^{s}b\right\rangle$ is a
subgroup of $G=D_{2n}$. Then $H$ is an $(\alpha,\beta)$-regular set of $G$,
$(\alpha,\beta)\not=(0,0)$, if and only if one of the following occurs:

1) $\alpha\in\\{0,|H|/2\\}$ and $0\leq\beta\leq|H|$, when $t=2$ and $m$ is
odd;

2) $\alpha\in\\{0,|H|/2\\}$ and $0\leq\beta=2\gamma\leq|H|$, for some integer
$\gamma$, when $t=2$ and $m$ is even;

3) $(\alpha,\beta)=(|H|/2,|H|/2)$, when $t>2$ is even;

4)
$(\alpha,\beta)\in\\{(\eta,\zeta),(\eta+m/t,\zeta+m/t),(\eta+2m/t,\zeta+2m/t)\
:\ 0\leq\eta,\zeta\leq m/t\\}$, when $m$ is odd and $t$ is a divisor of $m$;

5)
$(\alpha,\beta)\in\\{(2\eta,\zeta),(2\eta+m/t,\zeta+m/t),(2\eta+2m/t,\zeta+2m/t)\
:\ 0\leq\eta\leq m/2t,\ 0\leq\zeta\leq m/t\\}$, when $m$ is even and $t$ is
odd.

###### Proof.

For getting the result, first we assume that $H$ is an
$(\alpha,\beta)$-regular set in CayS$(G,S)$, for some subset
$\emptyset\not=S\subseteq G$ and we find some restrictions on
$(\alpha,\beta)$. Then for each $(\alpha,\beta)$ satisfying those restrictions
we give a normal square-free subset $S$ such that $H$ is an
$(\alpha,\beta)$-regular set in CayS$(G,S)$.

By the assumption, $H=\\{a^{rt},a^{rt+s}b\ |\ \text{where $1\leq r\leq
n/t$}\\}$ is isomorphic to $D_{2n/t}$ or $C_{2}\times C_{2}$. We consider the
following cases:

Case 1. Let $t$ be even.

Subcase 1.1. Let $t=2$. Then $[G:H]=2$. Thus, $H$ is normal in $G$ and so by
the proof of Lemma 2.6, we get that $H$ is an $(\alpha,0)$-regular set in
${\rm CayS}(G,S\cap H)$ and $H$ is a $(0,\beta)$-regular set in ${\rm
CayS}(G,S\cap(G\setminus H))$.

First, we find the possibilities for $\alpha$ and $\beta$. Note that,
$H=\left\langle a^{t},a^{s}b\right\rangle$ contains $(a^{s}b)^{G}$, which is
the only normal square-free subset of $G$ contained in $H$, and
$|(a^{s}b)^{G}|=|H|/2$. Then $S^{\prime}=S\cap H$ either is the empty-set or
is equal to $(a^{s}b)^{G}$ and so $\alpha=|S\cap H|\in\\{0,|H|/2\\}$. Also, as
$|G:H|=2$, we have $0\leq\beta=|S\cap Hx|=|S\cap(G\setminus H)|\leq|G\setminus
H|=|H|$, for every $x\in G\setminus H$.

In the sequel, we show that if $m$ is odd, then for each
$\beta\in\\{0,\dots,|H|\\}$, there exists a normal square-free subset of size
$\beta$ in $G\setminus H$, and if $m$ is even, then for each even integer
$\beta\in\\{0,\dots,|H|\\}$ there exists a normal square-free subset of size
$\beta$ in $G\setminus H$. In both cases, we denote them by $S_{\beta}$.

$\bullet$ Assume that $m$ is odd. Then the square-free conjugacy classes of
$G$ contained in $G\setminus H$ are as follows: $\\{a^{m}\\}$,
$(a^{s+1}b)^{G}$ (of size $|H|/2$) and
$\mathcal{S}_{i}=\\{a^{2i+1},a^{-(2i+1)}\\}$, where $0\leq i<(|H|/2-1)/2$. If
$0\leq\beta\leq|H|/2$ is odd, then let
$S_{\beta}=\\{a^{m}\\}\cup\Big{(}\bigcup\limits_{i=1}^{(\beta-1)/2}\mathcal{S}_{i}\Big{)}$.
If $0\leq\beta<|H|/2$ is even, then let
$S_{\beta}=\bigcup\limits_{i=1}^{\beta/2}\mathcal{S}_{i}$. Now, let
$\beta>|H|/2$ and $\gamma=\beta-|H|/2$. If $\gamma$ is even, then let
$S_{\beta}=(a^{s+1}b)^{G}\cup\Big{(}\bigcup\limits_{i=1}^{\gamma/2}\mathcal{S}_{i}\Big{)}$.
If $\gamma$ is odd, then let
$S_{\beta}=\\{a^{m}\\}\cup(a^{s+1}b)^{G}\cup\Big{(}\bigcup\limits_{i=1}^{(\gamma-1)/2}\mathcal{S}_{i}\Big{)}$.

Let $0\leq\beta\leq|H|$. So, in this case $H$ is a $(0,\beta)$-regular set in
${\rm CayS}(G,S_{\beta})$ and $H$ is a $(|H|/2,\beta)$-regular set in ${\rm
CayS}(G,(ab)^{G}\cup S_{\beta})$, (except for $(\alpha,\beta)=(0,0)$), which
is Part (1) of the statement of the theorem.

$\bullet$ Suppose that $m$ is even. Then $a^{m}\in H$ and the square-free
conjugacy classes of $G$ contained in $G\setminus H$ are as follows:
$(a^{s+1}b)^{G}$ (of size $|H|/2$) and
$\mathcal{S}_{i}=\\{a^{2i+1},a^{-(2i+1)}\\}$, where $0\leq i<(|H|/2-1)/2$. As
$S$ is a union of some square-free conjugacy classes of $G$ and all conjugacy
classes of $G\setminus H$ have even size, $\beta=|S\cap(G\setminus H)|$ is
even. Hence, by similar discussion as above for every even number
$0\leq\beta\leq|H|$ we can construct a normal square-free subset $S_{\beta}$
of size $\beta$.

Let $0\leq\beta\leq|H|$ be even. So, in this case $H$ is a $(0,\beta)$-regular
set in ${\rm CayS}(G,S_{\beta})$ and $H$ is a $(|H|/2,\beta)$-regular set in
${\rm CayS}(G,(ab)^{G}\cup S_{\beta})$, (except for $(\alpha,\beta)=(0,0)$),
which is Part (2) of the statement of the theorem.

Subcase 1.2 Let $t\geq 4$. We note that $Ha^{k}=\\{a^{rt+k},a^{s+rt-k}b\ :\
\text{where $1\leq r\leq n/t$}\\}$. In this case, $|$NSq$(G)\cap
Ha^{2l}|=|H|/2$, $|$NSq$(G)\cap Ha^{2l+1}|=|H|$, where $0\leq l\leq m-1$.
Since $Ha^{r}b=Ha^{s-r}$, for each $0\leq r\leq n-1$, we get that
$\mathcal{L}(H)=|H|/2$. Thus, according to Theorem 2.9, $0\leq\alpha\leq|H|/2$
and $0\leq\beta\leq|H|/2$. In the sequel, we prove that the only possibility
is $\alpha=\beta=|H|/2$.

Note that ${\rm NSq}(G)\cap Ha^{2l}\subseteq(a^{s}b)^{G}$ and if $\beta=0$,
then $S\cap Ha^{2}=\emptyset$. By the normality of $S$, we have
$S\cap(a^{s}b)^{G}=\emptyset$. So $S\cap H=\emptyset$, implies that
$\alpha=0$, which is excluded, as $S\not=\emptyset$ .

Let $\beta=|S\cap Ha^{2}|\geq 1$. As NSq$(G)\cap Ha^{2}\subseteq(a^{s}b)^{G}$,
by the normality of $S$, we conclude that $(a^{s}b)^{G}\subseteq S$. On the
other hand, ${\rm NSq(G)}\cap H\subseteq(a^{s}b)^{G}\subseteq S$. Thus,
$\alpha=|S\cap H|=|{\rm NSq}(G)\cap H|=|H|/2$ and $\beta=|{\rm NSq}(G)\cap
Ha^{2}|=|H|/2$. So the only possibility is $\alpha=\beta=|H|/2$. On the other
hand, taking $S=(a^{s}b)^{G}\cup(a^{s+1}b)^{G}=b^{G}\cup(ab)^{G}$, leads to
the conclusion that $|S\cap Ha^{k}|=|H|/2$, for each $0\leq k\leq n-1$ and so
$H$ is a $(|H|/2,|H|/2)$-regular set in ${\rm CayS}(G,S)$, which is Part (3)
of the statement of the theorem.

Case 2. Let $t$ be odd. Thus,

${\rm NSq}(G)\cap H=\\{a^{kt},a^{-kt}:\text{ where $1\leq k\leq m/t$ is
odd}\\}\cup(b^{G}\cap H)\cup((ab)^{G}\cap H).$

In Subcases 2.1 and 2.2, we will determine $(\alpha,\beta)$, when $m$ is even
and odd, respectively. Before that, we prove the following claims, in order to
make the subcases easier to follow. Let
$A_{i}=\\{a^{it}b,a^{it+1}b,\dots,a^{(i+1)t-1}b\\}$, where $0\leq
i\leq(2m/t)-1$.

Claim 1: $|b^{G}\cap H|=|(ab)^{G}\cap H|=m/t$.

$\bullet$ Let $s$ be odd. Then $rt+s$ is even if and only if $r$ is odd; and
$rt+s$ is odd if and only if $r$ is even. Thus by Lemma 3.1, $b^{G}\cap
H=\\{a^{(2i+1)t+s}b\ :\ 0\leq
i\leq(m/t)-1\\}\subseteq\bigcup\limits_{i=0}^{(m/t)-1}A_{2i+1}$ and
$(ab)^{G}\cap H=\\{a^{2it+s}b\ :\ 0\leq
i\leq(m/t)-1\\}\subseteq\bigcup\limits_{i=0}^{(m/t)-1}A_{2i}$. We note that,
$b^{G}\cap A_{2i+1}\cap H=\\{a^{(2i+1)t+s}b\\}$ and $(ab)^{G}\cap A_{2i}\cap
H=\\{a^{2it+s}b\\}$, for each $i$.

$\bullet$ Now, let $s$ be even. Then, similarly, $b^{G}\cap
H\subseteq\bigcup\limits_{i=0}^{(m/t)-1}A_{2i}$, $(ab)^{G}\cap
H\subseteq\bigcup\limits_{i=0}^{(m/t)-1}A_{2i+1}$, $b^{G}\cap A_{2i}\cap
H=\\{a^{2it+s}b\\}$ and $(ab)^{G}\cap A_{2i+1}\cap H=\\{a^{(2i+1)t+s}b\\}$,
for $0\leq i\leq(m/t)-1$.

Thus, in both cases, $|b^{G}\cap H|=|(ab)^{G}\cap H|=m/t$, as claimed. $\Box$

Claim 2: $|b^{G}\cap Hx|=|(ab)^{G}\cap Hx|=m/t$, for every $x\in G\setminus
H$.

To prove our claim we just need to show that $b^{G}\cap(G\setminus
H)=\bigcup\limits_{j=1}^{m/t}T_{j}$ and $(ab)^{G}\cap(G\setminus
H)=\bigcup\limits_{j=1}^{m/t}{\mathcal{T}}_{j}$, for some square-free pairwise
disjoint subsets $T_{j}$ and $\mathcal{T}_{j}$ such that for each $1\leq j\leq
m/t$, $T_{j}\cup\\{e\\}$ and $\mathcal{T}_{j}\cup\\{e\\}$ are right
transversals of $H$ in $G$. In the sequal of this paragraph, we take $i$ to be
an integer in $\\{0,\dots,(m/t)-1\\}$. Remind that $H=\\{a^{rt},a^{rt+s}b\ |\
\text{where $1\leq r\leq n/t$}\\}$. As $|H\cap A_{i}|=1$, for each $0\leq
i\leq(2m/t)-1$, we have $|(A_{2i}\cup A_{2i+1})\setminus H|=2(t-1)$ and so
$|(A_{2i}\cup A_{2i+1})\cap b^{G}\setminus H|=|(A_{2i}\cup
A_{2i+1})\cap(ab)^{G}\setminus H|=t-1$. Set $T_{i+1}=(A_{2i}\cup A_{2i+1})\cap
b^{G}\setminus H$ and $\mathcal{T}_{i+1}=(A_{2i}\cup
A_{2i+1})\cap(ab)^{G}\setminus H$. Then it is enough to prove that if
$x_{1},x_{2}\in T_{i+1}$ or $x_{1},x_{2}\in\mathcal{T}_{i+1}$ are distinct
elements, then $Hx_{1}\not=Hx_{2}$. On the contrary, assume that
$Hx_{1}=Hx_{2}$. If $x_{1},x_{2}\in A_{k}$, where $k\in\\{2i,2i+1\\}$, then
$x_{1}=a^{kt+r_{1}}b$ and $x_{2}=a^{kt+r_{2}}b$, for some $0\leq
r_{1}<r_{2}\leq t-1$. Thus $a^{r_{2}-r_{1}}\in H$, which is not possible. If
$x_{1}\in A_{2i}$ and $x_{2}\in A_{2i+1}$, then $x_{1}=a^{2it+r_{1}}b$ and
$x_{2}=a^{(2i+1)t+r_{2}}b$, for some $0\leq r_{1},r_{2}\leq t-1$. Hence,
$a^{t+r_{2}-r_{1}}\in H$, which means that $a^{r_{2}-r_{1}}\in H$. It follows
that $r_{1}=r_{2}$. On the other hand, the parities of $2it+r_{1}$ and
$(2i+1)t+r_{1}$ are the same, as $a^{2it+r_{1}}b$ and $a^{(2i+1)t+r_{1}}b$
both are in the same conjugacy class $b^{G}$ or $(ab)^{G}$. This implies that
the parities of $r_{1}$ and $t+r_{1}$ are the same. Thus, $t$ is even, a
contradiction. So the proof is complete. $\Box$

Claim 3: For each $1\leq\beta\leq m/t$, let
$\Omega_{\beta}=\bigcup\limits_{i=0}^{\beta-1}\mathfrak{S}_{i}$, where
$\mathfrak{S}_{i}=\\{a^{j},a^{-j}:it+1\leq j\leq(i+1)t-1,\text{ where}\ j$ is
odd$\\}$. Then, $H$ is a $(0,\beta)$-regular set in CayS$(G,\Omega_{\beta})$.

First, we prove that $\mathfrak{S}_{i}\cup\\{e\\}$ is a right transversal of
$H$ in $G$, for $0\leq i\leq m/t-1$. We show that for any pair $it+1\leq
j_{1}\leq j_{2}\leq(i+1)t-1$, we have $Ha^{j_{1}}\not=Ha^{-j_{2}}$, and if
$Ha^{j_{1}}=Ha^{j_{2}}$, then $j_{1}=j_{2}$. If $Ha^{j_{1}}=Ha^{-j_{2}}$, then
$a^{j_{1}+j_{2}}\in H$ and so ${j_{1}+j_{2}}=kt$, for some integer $k$. On the
other hand, $j_{1}+j_{2}$ is even and $t$ is odd, implying that $t$ divides
$(j_{1}+j_{2})/2$, a contradiction, as
$j_{1}/2,j_{2}/2\in[(it+1)/2,((i+1)t-1)/2]$. If $Ha^{j_{1}}=Ha^{j_{2}}$, then
$t$ divides $j_{1}-j_{2}$, implying that $j_{1}=j_{2}$. Also $Hx\not=H$, for
$x\in\mathfrak{S}_{i}$. Therefore, as $|\mathfrak{S}_{i}|=t-1$ we conclude
that $\mathfrak{S}_{i}\cup\\{e\\}$ is a right transversal of $H$ in $G$. Thus,
as $\mathfrak{S}_{i}$ is a normal square-free subset of $G$, by Lemma 2.4, $H$
is a $(0,\beta)$-regular set in CayS$(G,\Omega_{\beta})$, for all
$0\leq\beta\leq m/t$, as wanted. $\Box$

In the following subcases, we use the notation $\Omega_{\beta}$ which is
defined in Claim 3, without further references. Also, by $\Omega_{0}$ we mean
the empty set.

Subcase 2.1. Let $m$ be odd. Now we consider the following cases for $S$.

$\bullet$ Suppose that $S\cap(b^{G}\cup(ab)^{G})=\emptyset$. Therefore,
$S\subseteq
T=\\{a^{m}\\}\cup\left(\bigcup\limits_{i=0}^{(m-3)/2}\\{a^{(2i+1)},a^{-(2i+1)}\\}\right)=\\{a^{m}\\}\cup\Omega_{m/t}\cup\left(\bigcup\limits_{j=0}^{((m/t)-3)/2}\\{a^{t(2j+1)},a^{-t(2j+1)}\\}\right)$.
Remind that $\\{a^{m}\\}$ and $\\{a^{tk},a^{-tk}\\}$, for every odd integer
$k<m/t$, are square-free conjugacy classes of $G$ contained in $H$. Then
$0\leq\alpha=|S\cap H|\leq|T\cap
H|=\left|\\{a^{m}\\}\cup\left(\bigcup\limits_{j=0}^{((m/t)-3)/2}\\{a^{t(2j+1)},a^{-t(2j+1)}\\}\right)\right|=m/t$
and $0\leq\beta=|S\cap Hx|\leq|\Omega_{m/t}\cap Hx|=m/t$, for each $x\in
G\setminus H$, by Claim 3.

Now, we show that, in this case, for each $\alpha,\beta\in\\{0,\dots,m/t\\}$,
there exists a normal square-free subset $S$ such that $H$ is an
$(\alpha,\beta)$-regular set in ${\rm CayS}(G,S)$. If $\alpha$ is an odd
number less than or equal to $m/t$, then by setting
$S_{\alpha}=\\{a^{m}\\}\cup\left(\bigcup\limits_{j=0}^{(\alpha-3)/2}\\{a^{t(2j+1)},a^{-t(2j+1)}\\}\right)$,
we get that $H$ is an $(\alpha,0)$-regular set in CayS$(G,S_{\alpha})$.
Similarly, if $\alpha$ is an even number less than $m/t$, then set
$S_{\alpha}=\bigcup\limits_{j=0}^{(\alpha-2)/2}\\{a^{t(2j+1)},a^{-t(2j+1)}\\}$.
Now, let $S=S_{\alpha}\cup\Omega_{\beta}$. If $S\not=\emptyset$, then $H$ is
an $(\alpha,\beta)$-regular set in ${\rm CayS}(G,S)$.

$\bullet$ Let $b^{G}\subseteq S$ and $S\cap(ab)^{G}=\emptyset$. Therefore,
$b^{G}\subseteq S\subseteq
T=b^{G}\cup\\{a^{m}\\}\cup\left(\bigcup\limits_{i=0}^{(m-3)/2}\\{a^{(2i+1)},a^{-(2i+1)}\\}\right)=b^{G}\cup\\{a^{m}\\}\cup\Omega_{m/t}\cup\left(\bigcup\limits_{j=0}^{((m/t)-3)/2}\\{a^{t(2j+1)},a^{-t(2j+1)}\\}\right)$.
By Claim 1, $m/t=|b^{G}\cap H|\leq|S\cap
H|=\alpha\leq\left|\left(b^{G}\cup\\{a^{m}\\}\cup\left(\bigcup\limits_{j=0}^{((m/t)-3)/2}\\{a^{t(2j+1)},a^{-t(2j+1)}\\}\right)\right)\cap
H\right|=2m/t$ and by Claims 2 and 3, $m/t=|b^{G}\cap Hx|\leq|S\cap
Hx|=\beta\leq|(\Omega_{m/t}\cup b^{G})\cap Hx|=2m/t$, for $x\in G\setminus H$.

Now, for each $m/t\leq\alpha,\beta\leq 2m/t$, we give a normal square-free
subset $S$ of $G$ (with given condition) such that $H$ is an
$(\alpha,\beta)$-regular set in ${\rm CayS}(G,S)$. If $\alpha-(m/t)$ is odd,
then let

$S=\left(\bigcup\limits_{j=0}^{(\alpha-(m/t)-3)/2}\\{a^{t(2j+1)},a^{-t(2j+1)}\\}\right)\cup\\{a^{m}\\}\cup
b^{G}\cup\Omega_{\beta-(m/t)}.$

If $\alpha-(m/t)$ is even, then let

$S=\left(\bigcup\limits_{j=0}^{(\alpha-(m/t)-2)/2}\\{a^{t(2j+1)},a^{-t(2j+1)}\\}\right)\cup
b^{G}\cup\Omega_{\beta-(m/t)}.$

By Claims 1,2,3, we have $|S\cap H|=\alpha$ and $|S\cap Hx|=\beta$, for each
$x\in G\setminus H$.

$\bullet$ If $(ab)^{G}\subseteq S$ and $S\cap b^{G}=\emptyset$, we have the
same possibilities for $\alpha$ and $\beta$ as the above case.

$\bullet$ Suppose $b^{G}\cup(ab)^{G}\subseteq S$. Thus,

$\displaystyle b^{G}\cup(ab)^{G}\subseteq S\subseteq
T=b^{G}\cup(ab)^{G}\cup\\{a^{m}\\}\cup\left(\bigcup_{i=0}^{(m-3)/2}\\{a^{(2i+1)},a^{-(2i+1)}\\}\right)$
$\displaystyle=b^{G}\cup(ab)^{G}\cup\\{a^{m}\\}\cup\Omega_{m/t}\cup\left(\bigcup_{j=0}^{((m/t)-3)/2}\\{a^{t(2j+1)},a^{-t(2j+1)}\\}\right).$

Similarly to the above, by Claims 1,2,3, we have $2m/t\leq\alpha,\beta\leq
3m/t$. Now, for each $2m/t\leq\alpha,\beta\leq 3m/t$, we give a normal square-
free subset $S$ of $G$ such that $H$ is an $(\alpha,\beta)$-regular set in
${\rm CayS}(G,S)$. If $\alpha-(2m/t)$ is odd, then let $S$ be the union of
$(\alpha-(2m/t)-1)/2$ sets of type $\\{a^{t(2j+1)},a^{-t(2j+1)}\\}$ with
$\Omega_{\beta-(2m/t)}\cup b^{G}\cup(ab)^{G}\cup\\{a^{m}\\}$. If
$\alpha-(2m/t)$ is even, then let $S$ be the union of $(\alpha-(2m/t))/2$ sets
of type $\\{a^{-t(2j+1)},a^{t(2j+1)}\\}$ with $\Omega_{\beta-(2m/t)}\cup
b^{G}\cup(ab)^{G}$, as wanted.

Therefore, we see that in the case that $t$ and $m$ are odd, all the
possibilities for $(\alpha,\beta)$ are as described in Part (4) of the
statement of the theorem.

Subcase 2.2 Let $m$ be even. Similarly to the above we consider the following
cases for $S$.

Assume that $S\cap(b^{G}\cup(ab)^{G})=\emptyset$. Thus,

$S\subseteq\bigcup\limits_{i=0}^{m/2-1}\\{a^{(2i+1)},a^{-(2i+1)}\\}=\Omega_{m/t}\cup\left(\bigcup\limits_{j=0}^{(m/(2t))-1}\\{a^{t(2j+1)},a^{-t(2j+1)}\\}\right),$

As $H\cap\Omega_{m/t}=\emptyset$, we deduce that $0\leq\alpha=|S\cap H|\leq
m/t$ is even and $0\leq\beta=|S\cap Hx|\leq|\Omega_{m/t}\cap Hx|=m/t$, by
Claim 3, for $x\in G\setminus H$. Hence, setting
$S=\Omega_{\beta}\cup\left(\bigcup\limits_{j=0}^{\alpha/2}\\{a^{t(2j+1)},a^{-t(2j+1)}\\}\right)$,
$H$ is an $(\alpha,\beta)$-regular set of $G$, for all $0\leq\beta\leq m/t$
and each even numbers $0\leq\alpha\leq m/t$.

Now, assume that $S\cap(b^{G}\cup(ab)^{G})\neq\emptyset$. First, suppose that
$S$ contains exactly one of them. Without loss of generality, suppose that
$b^{G}\subseteq S$. Therefore, $b^{G}\subseteq S\subseteq
T=b^{G}\cup\left(\bigcup\limits_{i=0}^{(m-2)/2}\\{a^{(2i+1)},a^{-(2i+1)}\\}\right)=b^{G}\cup\Omega_{m/t}\cup\left(\bigcup\limits_{j=0}^{((m/t)-2)/2}\\{a^{t(2j+1)},a^{-t(2j+1)}\\}\right)$.
By Claim 1, $m/t=|b^{G}\cap H|\leq|S\cap
H|=\alpha\leq\left|\left(b^{G}\cup\left(\bigcup\limits_{j=0}^{((m/t)-2)/2}\\{a^{t(2j+1)},a^{-t(2j+1)}\\}\right)\right)\cap
H\right|=2m/t$ and $\alpha$ is even. Also, by Claims 2 and 3, $m/t=|b^{G}\cap
Hx|\leq|S\cap Hx|=\beta\leq|(b^{G}\cup\Omega_{m/t})\cap Hx|=2m/t$, for $x\in
G\setminus H$.

Now, for each $m/t\leq\alpha,\beta\leq 2m/t$, where $\alpha$ is even, we give
a normal square-free subset $S$ of $G$ (with given condition) such that $H$ is
an $(\alpha,\beta)$-regular set in ${\rm CayS}(G,S)$. Note that, as $\alpha$
is even in this case, we have $\alpha-(m/t)$ is even. Let $S$ be a union of
$(\alpha-(m/t))/2$ conjugacy classes of type $\\{a^{t(2j+1)},a^{-t(2j+1)}\\}$
with $b^{G}\cup\Omega_{\beta-(m/t)}$. Hence, $H$ is an
$(\alpha,\beta)$-regular set of $G$. For the case that $(ab)^{G}\subseteq S$
and $b^{G}\cap S=\emptyset$, we have the same possibilities for
$(\alpha,\beta)$.

Let $b^{G}\cup(ab)^{G}\subseteq S$. Thus

$\displaystyle b^{G}\cup(ab)^{G}\subseteq S\subseteq
T=b^{G}\cup(ab)^{G}\cup\\{a^{m}\\}\cup\left(\bigcup_{i=0}^{(m-2)/2}\\{a^{(2i+1)},a^{-(2i+1)}\\}\right)$
$\displaystyle=b^{G}\cup(ab)^{G}\cup\\{a^{m}\\}\cup\Omega_{m/t}\cup\left(\bigcup_{j=0}^{((m/t)-2)/2}\\{a^{t(2j+1)},a^{-t(2j+1)}\\}\right).$

Similarly to the above, by Claims 1,2,3, we have $2m/t\leq\alpha,\beta\leq
3m/t$, where $\alpha$ is even. Now, for each $2m/t\leq\alpha,\beta\leq 3m/t$,
where $\alpha$ is even, we give a normal square-free subset $S$ of $G$ such
that $H$ is an $(\alpha,\beta)$-regular set in ${\rm CayS}(G,S)$. Noting that
$\alpha-(2m/t)$ is even, let $S$ be the union of $(\alpha-(2m/t))/2$ conjugacy
classes of type $\\{a^{-t(2j+1)},a^{t(2j+1)}\\}$ with
$\Omega_{\beta-(2m/t)}\cup b^{G}\cup(ab)^{G}$, as wanted.

Therefore, we see that in the case that $t$ is odd and $m$ is even, all the
possibilities for $(\alpha,\beta)$ are as described in Part (5) of the
statement of the theorem.

######  Theorem 3.7.

Let $n\geq 3$ be odd and $t$ be a divisor of $n$. Let $H=\left\langle
a^{t}\right\rangle$ be a subgroup of $G=D_{2n}$. Then $H$ is an
$(\alpha,\beta)$-regular set of $G$ if and only if $t=1$ and
$(\alpha,\beta)=(0,|H|)$.

###### Proof.

Noting that NSq$(G)=b^{G}$ and, as the connection set $S\not=\emptyset$, we
get that $S=b^{G}$. So $\alpha=|S\cap H|=0$. If $t>1$, then
$\beta\leq\mathcal{L}(H)\leq|$NSq$(G)\cap Ha|=0$, which is excluded. Thus,
$t=1$ and $\beta=|H|/2$, as desired.

######  Theorem 3.8.

Let $n=2m$, for some positive integer $m\geq 2$ and $t$ be a divisor of $n$.
Let $H=\left\langle a^{t}\right\rangle$ be a subgroup of $G=D_{2n}$. Then $H$
is an $(\alpha,\beta)$-regular set of $G$, where $(\alpha,\beta)\not=(0,0)$,
if and only if one of the following occurs:

1) $(\alpha,\beta)=(0,|H|)$, when $t=2$.

2) $0\leq\alpha\leq|H|/2$ and $\beta\in\\{0,|H|/2,|H|\\}$, when $t=1$ and $m$
is odd.

3) $\alpha=2\gamma$, for some $0\leq\gamma\leq|H|/4$, and
$\beta\in\\{0,|H|/2,|H|\\}$, when $t=1$ and $m$ is even.

4) $0\leq\alpha\leq|H|/2$ and $\beta\in\\{0,|H|/2\\}$, when $t>1$ and $m$ are
odd.

5) $\alpha=2\gamma$, for some $0\leq\gamma\leq|H|/4$, and
$\beta\in\\{0,|H|/2\\}$, when $t>1$ is odd and $m$ is even.

###### Proof.

Note that in this case $H$ is a normal subgroup of $G$ and so by the proof of
Lemma 2.6, $H$ is an $(\alpha,\beta)$-regular set in ${\rm CayS}(G,S)$, for
some subset $\emptyset\not=S\subseteq G$ if and only if $H$ is an
$(\alpha,0)$-regular set in ${\rm CayS}(G,S\cap H)$ and $H$ is a
$(0,\beta)$-regular set in ${\rm CayS}(G,S\cap(G\setminus H))$.

For getting the result, first we assume that $H$ is an
$(\alpha,\beta)$-regular set in CayS$(G,S)$, for some subset
$\emptyset\not=S\subseteq G$ and we find some restrictions on
$(\alpha,\beta)$. Then for each $(\alpha,\beta)$ satisfying those restrictions
we give a normal square-free subset $S$ such that $H$ is an
$(\alpha,\beta)$-regular set in CayS$(G,S)$. We consider the following two
cases:

Case 1. Assume that $t$ is an even divisor of $n$. Thus, $H\subseteq{\rm
Sq}(G)$ and hence $\alpha=0$. Thus, as $S\not=\emptyset$, we must have
$\beta\geq 1$.

First, assume $t>2$. In this case, $a^{2}\notin H$ and $Ha^{2}\subseteq{\rm
Sq}(G)$. So, by Theorem 2.9, $\beta\leq\mathcal{L}(H)\leq|$NSq$(G)\cap
Ha^{2}|=0$. Therefore, for even integer $t>2$, $H$ is not an
$(\alpha,\beta)$-regular set of $G$.

Now, let $t=2$. Then $H=$ Sq$(G)$ and $[G:H]=4$. Therefore, $G\setminus
H=Ha\cup Hb\cup Hab$, where $Hb=b^{G}$ and $Hab=(ab)^{G}$. If $m$ is even,
then $Ha=\\{a^{2j+1},a^{-(2j+1)}:0\leq j\leq m-1\\}$, and in case $m$ is odd,
$Ha=\\{a^{2j+1},a^{-(2j+1)}:0\leq j\leq m-1\\}\cup\\{a^{m}\\}$. As $\beta\geq
1$, we have $S\cap Hb\not=\emptyset$ and $S\cap Hab\not=\emptyset$. Therefore,
by the normality of $S$ we conclude that $Hab\cup Hb\subseteq S$ and since
$|S\cap Hg|=\beta$, for each $g\in G\setminus H$, we get that $S=G\setminus H$
and so $\beta=|H|$. It follows that the only possibility is $\beta=|H|$, as
described in Part (1) of the theorem.

Case 2. Assume that $t$ is an odd divisor of $n$. In this case, ${\rm
NSq}(G)\cap H=H\setminus\langle a^{2t}\rangle$. Then, $0\leq\alpha=|S\cap
H|\leq|{\rm NSq}(G)\cap H|=|H|/2$.

In the following we discuss the possibilities for $\alpha$ and for each
feasible $\alpha$ we introduce a normal square-free subset of $H$ of size
$\alpha$, say $S_{\alpha}$.

$\bullet$ Let $m$ be odd. If $1\leq\alpha\leq|H|/2$ is odd, then we set
$S_{\alpha}$ to be a union of $\\{a^{m}\\}$ with $(\alpha-1)/2$ sets of type
$\\{a^{kt},a^{-kt}\\}$, where $k<m/t$ is odd. If $0\leq\alpha\leq|H|/2$ is
even, then we set $S_{\alpha}$ to be a union of $\alpha/2$ sets of type
$\\{a^{kt},a^{-kt}\\}$, where $k<m/t$ is odd.

$\bullet$ Let $m$ be even. Then $\alpha$ is even, as $S\cap
H\subseteq\bigcup\limits_{i=0}^{((m/t)-2)/2}\\{a^{t(2i+1)},a^{-t(2i+1)}\\}$, a
union of some square-free conjugacy classes of size $2$. Let $S_{\alpha}$ be a
union of $\alpha/2$ sets of type $\\{a^{kt},a^{-kt}\\}$, where $k<m/t$ is odd.

Therefore $H$ is an $(\alpha,0)$-regular set in ${\rm CayS}(G,S_{\alpha})$.
Now, in the following two cases we discuss possibilities for $\beta$ and in
each case we give normal square-free subsets $S$ such that $H$ is an
$(\alpha,\beta)$-regular set in ${\rm CayS}(G,S)$.

$\blacktriangleright$ Let $t=1$.

Then $[G:H]=2$. Therefore, $G\setminus H=Hb=b^{G}\cup(ab)^{G}$. By the
normality of $S$ we conclude that $S\cap(G\setminus H)$ can be equal to either
$\emptyset$, $b^{G}$, $(ab)^{G}$ or $b^{G}\cup(ab)^{G}$. Hence,
$\beta=|S\cap(G\setminus H)|\in\\{0,|H|/2,|H|\\}$.

Thus, setting $S=S_{\alpha}$ (if $\alpha\geq 1$ and $\beta=0$),
$S=S_{\alpha}\cup b^{G}$ or $S_{\alpha}\cup b^{G}\cup(ab)^{G}$ we have $H$ is
an $(\alpha,\beta)$-regular set in ${\rm CayS}(G,S)$, where
$0\leq\alpha\leq|H|/2$ and $\beta\in\\{0,|H|/2,|H|\\}$. Note that by previous
discussion $\alpha$ must be even, when $m$ is even. Thus, we have Parts (2)
and (3) of the theorem.

$\blacktriangleright$ Let $t>1$.

Then, ${\rm NSq}(G)\cap Ha^{i}b=Ha^{i}b\subseteq b^{G}\cup(ab)^{G}$,
$|(ab)^{G}\cap Ha^{i}b|=|b^{G}\cap Ha^{i}b|=|H|/2$ and $|$NSq$(G)\cap
Ha^{2i}|=|$NSq$(G)\cap Ha^{2i+1}|=|H|/2$, for every integer $i$. Hence, if
$\beta\geq 1$, then $b^{G}\cap Hb\subseteq S$ or $(ab)^{G}\cap Hb\subseteq S$,
implying that $\beta\geq|H|/2$ and as $|$NSq$(G)\cap Ha|=|H|/2$, we get that
$\beta\leq\mathcal{L}(H)=|H|/2$, yields to $\beta=|H|/2$. Thus,
$\beta\in\\{0,|H|/2\\}$.

Let $\Omega=\\{a^{k}\ :\ \text{where $k$ is an odd integer not divided by
$t$}\\}\subseteq{\rm NSq}(G)$. Then, setting $S=S_{\alpha}$ (if $\alpha\geq 1$
and $\beta=0$) we have $H$ is an $(\alpha,0)$-regular set in ${\rm
CayS}(G,S)$. Let $S=b^{G}\cup\Omega\cup S_{\alpha}$. Then, as $|Ha^{i}\cap
S|=|Ha^{i}\cap\Omega|=|H|/2$, for $1\leq i\leq t-1$, and $|Ha^{i}b\cap
S|=|Ha^{i}b\cap b^{G}|=|H|/2$, we have $H$ is an $(\alpha,|H|/2)$-regular set
in ${\rm CayS}(G,S)$, where $0\leq\alpha\leq|H|/2$. Remind that, $\alpha$ is
even if $m$ is even. So, we have Parts (4) and (5) of the theorem.

Acknowledgement. The authors gratefully thank the referees for the
constructive comments and recommendations which definitely help to improve the
quality and readability of the paper.

## References

  * [1] M. Amooshahi, B. Taeri, On Cayley sum graphs of non-abelian groups. Graphs Combin., 32, 17–29 (2016).
  * [2] C. R. Stephan, The subgroups of the dihedral group. Mathematics Magazine, 48, 107-107 (1975).
  * [3] E. A. Bespalov, D. S. Krotov, A. A. Matiushev, K. V. Vorob’ev, Perfect 2-colorings of Hamming graphs, J. Combin. Des., 29, 1-30 (2021).
  * [4] R. A. Bailey, P. J. Cameron, A. L. Gavrilyuk, S. V. Goryainov, Equitable partitions of Latin-square graphs, J. Combin. Des., 27, 142-160 (2019).
  * [5] J. Borges, J. Rifa, V. A. Zinoviev, On completely regular codes. Probl. Inf. Transm., 55, 1–45 (2019).
  * [6] D. M. Cardoso, An overview of $(\kappa,\tau)$-regular sets and their applications, Discrete Appl. Math., 269 2-10 (2019).
  * [7] F. R. K. Chung, Diameters and eigenvalues, J. Amer. Math. Soc., 2, 187-196 (1998).
  * [8] M. DeVos , L. Goddyn, B. Mohar, R. Sámal, Cayley sum graphs and eigenvalues of (3, 6)-fullerenes. J. Combin. Theory Ser. B, 99, 358–369 (2009).
  * [9] P. Delsarte, An algebraic approach to the association schemes of coding theory, PhD thesis. Philips Research Laboratories, 1973.
  * [10] D. G. Fon-Der-Flaas, Perfect 2-colorings of a hypercube, Siberian Math. J., 48, 740-745 (2007).
  * [11] C. Godsil, G. Royle, Algebraic graph theory, Graduate Texts in Mathematics, Springer, New York (2001).
  * [12] J. M. Goethals and H. C. A. Van Tilborg, Uniformly packed codes, Philips Research Reports 30, 9–36 (1973).
  * [13] H. Huang, B. Xia, S. Zhou, Perfect codes in Cayley graphs, SIAM J. Discrete Math., 32, 548-559 (2018).
  * [14] Y. Khaefi, Z. Akhlaghi, B. Khosravi. On the subgroup perfect codes in Cayley graphs, Des. Codes Cryptogr., 91, 1-7 (2022).
  * [15] G. James, M. Liebeck, Representations and Characters of Groups, Cambridge University Press, Cambridge (2001).
  * [16] I. Mogilnykh, A. Valyuzhenich, Equitable 2-partitions of the Hamming graphs with the second eigenvalue, Discrete Math., 343 (2020) https://doi.org/10.1016/j.disc.2020.112039.
  * [17] X. Ma, M. Feng, K. Wang, Subgroup perfect codes in Cayley sum graphs, Des. Codes Cryptogr., 88, 1447–1461 (2020).
  * [18] X. Ma, K. Wang, Y. Yang, Perfect codes in Cayley sum graphs, Electron. J. Combin., 29, 1-12 (2022).
  * [19] A. Neumaier, Completely regular codes, Discrete Math., 106/107, 353-360 (1992).
  * [20] Y. Wang, B. Xia, S. Zhou, Regular sets in Cayley graphs, J. Algebr. Comb., 57, 547–558 (2023).
  * [21] X. Wang, S. Xu, S. Zhou, Regular sets in Cayley graphs, J. Algebr. Comb., 59, 753-759 (2024).
  * [22] J. Zhang, S. Zhou, On subgroup perfect codes in Cayley graphs, European J. Combin., 91, 1-5 (2021).
  * [23] J. Zhang, Y. Zhu, A note on regular sets in Cayley graphs, Bull. Aust. Math. Soc.,109, 1-5 (2024).