# On integral mixed Cayley graphs over non-abelian finite groups admitting an
abelian subgroup of index 2

Angelot Behajaina<EMAIL_ADDRESS>Université
Paris-Saclay, CNRS, Laboratoire de Mathématiques d’Orsay, 91405 Orsay, France
and  François Legrand<EMAIL_ADDRESS>Normandie Univ., UNICAEN,
CNRS, Laboratoire de Mathématiques Nicolas Oresme, 14000 Caen, France

###### Abstract.

Recently, several works by a number of authors have provided characterizations
of integral undirected Cayley graphs over generalized dihedral groups and
generalized dicyclic groups. We generalize and unify these results in two
different ways. Firstly, we work over arbitrary non-abelian finite groups
admitting an abelian subgroup of index 2. Secondly, our main result actually
characterizes integral mixed Cayley graphs over such finite groups, in the
spirit of a very recent result of Kadyan–Bhattarcharjya in the abelian case.

###### Key words and phrases:

Cayley graphs, integral graphs

###### 2010 Mathematics Subject Classification:

05C25, 05C50

##  1\. Introduction

In this note, we consider mixed graphs with no loop and no multiple edge,
i.e., couples $(V,E)$, where $V$ is a non-empty finite set and $E$ is a subset
of $(V\times V)\setminus\\{(v,v)\,:\,v\in V\\}$ (see §2 for the basic
terminology used in the sequel). A mixed graph is integral if the eigenvalues
of its Hermitian adjacency matrix, a matrix introduced independently by Liu–Li
and Guo–Mohar in [LL15] and [GM17], respectively, are all in $\mathbb{Z}$. The
aim of the present note is to contribute to the study of the integrality of a
special class of mixed graphs, namely mixed Cayley graphs. For a subset $S$ of
a finite group $G$ with $1\not\in S$, by the mixed Cayley graph over $G$ with
respect to $S$, we mean the mixed graph $(G,\\{(g,h)\in G^{2}:g^{-1}h\in
S\\})$, denoted by ${\rm{Cay}}(G,S)$.

Motivated by works of Harary–Schwenk [HS74] and Ahmady–Alon–Blake–Shparlinski
[AABS09], many authors studied the integrality of undirected Cayley graphs,
i.e., of mixed Cayley graphs ${\rm{Cay}}(G,S)$ which assume $S^{-1}=S$ (for
results on the integrality of other kinds of mixed graphs, see, e.g., [BC76,
Wat79, WS79] and, for a survey of some of the known results on eigenvalues of
mixed Cayley graphs, see [LZ22]). For $G$ abelian, several characterizations
of integrality were obtained by Bridges–Mena [BM82], Klotz–Sander [KS10], and
Alperin–Peterson [AP12]. Moreover, Lu–Huang–Huang focused on the case where
$G$ is dihedral in [LHH18], and their results were extended to generalized
dihedral groups by Huang–Li in [HL21]. Furthermore, Cheng–Feng–Huang
considered in [CFH19] the case where $G$ is a dicyclic group, and their
results were extended to generalized dicyclic groups in our previous work
[BL22].

However, the integrality of mixed Cayley graphs which are not necessarily
undirected seems to be a very recent topic, and we are aware of only one
result of Kadyan–Bhattarcharjya [KB21], who characterized integral mixed
Cayley graphs over abelian groups. Let us say that their characterization was
extended by Huang–Lu–Mönius in [HLM22], who determined the splitting field of
the characteristic polynomial of the Hermitian adjacency matrix of any mixed
Cayley graph in the abelian case.

Such characterizations of integrality usually involve the classification of
the (complex) irreducible representations of the underlying finite group $G$
and, via a result of Babai [Bab79], are derived from computing the eigenvalues
of some endomorphisms of some vector spaces whose dimensions are those of the
irreducible representations of $G$. After abelian groups, it is then natural
to study the case where $G$ is non-abelian and where every irreducible
representation of $G$ has dimension at most 2. By a result of Amitsur [Ami61,
Theorem 3], the latter can happen only in the next two situations: (1) $G$ has
an abelian subgroup of index 2, and (2)
$G/Z(G)\cong(\mathbb{Z}/2\mathbb{Z})^{3}$.

Here we work over arbitrary non-abelian finite groups $G$ as in (1), and our
main result (Theorem 3.1) characterizes integral mixed Cayley graphs over such
finite groups $G$, in terms of the irreducible representations of the
underlying finite group. In particular, Theorem 3.1 generalizes and unifies
various results from the works [LHH18, CFH19, HL21, BL22] quoted above in two
different ways. As a first step, we classify all irreducible representations
(up to equivalence) of any of our finite groups $G$ (see §3.2.1).

Theorem 3.1 has several consequences. In the undirected case, our main result
takes a simple form (see Corollary 3.5) and we derive a convenient sufficient
condition, involving the Boolean algebra of the underlying abelian subgroups,
for an undirected Cayley graph over any of our groups $G$ to be integral (see
Corollary 3.6). But Theorem 3.1 also applies to non-necessarily undirected
Cayley graphs. For example, in the case of generalized dihedral/dicyclic
groups, our main result also takes a simple form (see Corollary 3.8) and, in
particular, it yields the first integrality results for non-necessarily
undirected Cayley graphs in the non-abelian case (see also Corollaries 3.9 and
3.11 for results devoted to the directed case). Our results are illustrated by
several explicit examples (see Examples 3.7, 3.10 and 3.12).

##  2\. Preliminaries

We collect the basic material on representations of finite groups and graphs
that will be used in the sequel. For a subset $S$ of a finite group $G$ and a
non-zero integer $n$, we set $S^{n}=\\{s^{n}\,:\,s\in S\\}$.

###  2.1. Representations of finite groups

A (complex) representation of a given finite group $G$ is a group homomorphism
$\rho:G\rightarrow{\rm{GL}}(V)$, where $V$ is a finite dimensional complex
vector space. The dimension ${\rm{dim}}(\rho)$ of $\rho$ is the dimension of
$V$ and the character of $\rho$ is the map
$\chi_{\rho}:G\rightarrow\mathbb{C}$ defined by
$\chi_{\rho}(g)={\rm{Tr}}(\rho(g))$ for $g\in G$.

A subspace $W$ of $V$ is invariant under $\rho$ if $\rho(g)(W)=W$ for $g\in
G$, and $\rho$ is irreducible if $V\not=\\{0\\}$ and if $\\{0\\},V$ are the
only invariant subspaces of $V$. Letting $L^{2}(G)$ be the vector space of all
functions $f:G\rightarrow\mathbb{C}$, together with the Hermitian inner
product $(\cdot|\cdot)$ given by

$(f_{1}|f_{2})=\frac{1}{|G|}\sum_{g\in G}f_{1}(g)\overline{f_{2}(g)}$ (2.1)

for $f_{1},f_{2}\in L^{2}(G)$, the representation $\rho$ is irreducible if and
only if $(\chi_{\rho}|\chi_{\rho})=1$.

Two representations $\rho_{1}:G\rightarrow{\rm{GL}}(V_{1})$ and
$\rho_{2}:G\rightarrow{\rm{GL}}(V_{2})$ of $G$ are equivalent if there is an
isomorphism $T:V_{1}\rightarrow V_{2}$ such that, for every $g\in G$, we have
$T\circ\rho_{1}(g)=\rho_{2}(g)\circ T$. Classically, $\rho_{1}$ and $\rho_{2}$
are equivalent if and only if $\chi_{\rho_{1}}=\chi_{\rho_{2}}$. The group $G$
has only finitely many inequivalent irreducible representations, and their
number equals the number of conjugacy classes of $G$. Moreover, if
$\rho_{1},\dots,\rho_{n}$ are the inequivalent irreducible representations of
$G$, then

$|G|=({\rm{dim}}(\rho_{1}))^{2}+\cdots+({\rm{dim}}(\rho_{n}))^{2}.$ (2.2)

In particular, if $G$ is abelian, then every irreducible representation of $G$
has dimension 1.

For a finite set $S$ and a function $f:S\rightarrow\mathbb{C}$, we set

$f(S)=\sum_{s\in S}f(s).$

A subset $S$ of $G$ is integral if $\chi_{\rho}(S)\in\mathbb{Z}$ for every
irreducible representation $\rho$ of $G$. For $G$ abelian, we shall also need
the characterization of integral subsets of $G$ in terms of the Boolean
algebra of $G$. To that end, we let $\mathcal{F}_{G}$ be the set of all
subgroups of $G$. The Boolean algebra $\mathbb{B}(G)$ is the set whose
elements are obtained by arbitrary finite intersections, unions, and
complements of elements of $\mathcal{F}_{G}$. The next lemma has been
established by Alperin–Peterson in [AP12]:

######  Lemma 2.1.

A subset $S$ of a finite abelian group $G$ is integral if and only if
$S\in\mathbb{B}(G)$.

Let us also recall that, for $G$ abelian, the minimal elements of
$\mathbb{B}(G)$ are called atoms. As proved by Alperin–Peterson [AP12], every
element of $\mathbb{B}(G)$ is the union of some atoms of $G$, and every atom
is of the form $\\{x\in G\,:\,\langle x\rangle=\langle g\rangle\\}$ ($g\in
G$). In particular, we have:

######  Lemma 2.2.

Let $G$ be a finite abelian group and $S\in\mathbb{B}(G)$. Then $S=S^{j}$ for
every non-zero integer $j$ which is coprime to every element order in $G$.

###  2.2. Graphs

A mixed graph (with no loop and no multiple edge) is a couple $\Gamma=(V,E)$,
where $V$ and $E$ are a non-empty finite set and a subset of $(V\times
V)\setminus\\{(v,v)\,:\,v\in V\\}$, respectively. If $\Gamma$ fulfills:
$(v_{1},v_{2})\in E$ if and only if $(v_{2},v_{1})\in E$ for $v_{1}\not=v_{2}$
(resp., if $(v_{1},v_{2})\in E$, then $(v_{2},v_{1})\not\in E$ for
$v_{1}\not=v_{2}$), then $\Gamma$ is an undirected graph (resp., a directed
graph). Writing $V=\\{v_{1},\dots,v_{|V|}\\}$, we consider the Hermitian
adjacency matrix $A(\Gamma)$ of $\Gamma$, which is the $|V|\times|V|$-matrix
whose $(k,j)$-entry $a_{kj}$ is defined by

$a_{kj}=\left\\{\begin{array}[]{ll}1&\mbox{if}\,\,\,(v_{k},v_{j})\in
E\,\,\,{\rm{and}}\,\,\,(v_{j},v_{k})\in E,\\\ {\bf
i}&\mbox{if}\,\,\,(v_{k},v_{j})\in E\,\,\,{\rm{and}}\,\,\,(v_{j},v_{k})\not\in
E,\\\ -{\bf i}&\mbox{if}\,\,\,(v_{k},v_{j})\not\in
E\,\,\,{\rm{and}}\,\,\,(v_{j},v_{k})\in E,\\\
0&\mbox{otherwise}\end{array}\right.$

for $1\leq k,j\leq|V|$, where ${\bf i}=\sqrt{-1}$. If $\Gamma$ is undirected,
then $A(\Gamma)$ is the “usual” adjacency matrix of $\Gamma$. The mixed graph
$\Gamma$ is integral if the eigenvalues of $A(\Gamma)$, which are always real
numbers, are in $\mathbb{Z}$.

Given a finite group $G$ and a subset $S$ of $G\setminus\\{1\\}$, the mixed
Cayley graph ${\rm{Cay}}(G,S)$ over $G$ with respect to $S$ is the mixed graph
$(G,\\{(g,h)\in G^{2}:g^{-1}h\in S\\})$. Setting $T=\\{s\in
S\,:\,s^{-1}\not\in S\\},$ the mixed Cayley graph ${\rm{Cay}}(G,S)$ is
undirected (resp., directed) if and only if $T=\emptyset$ (resp., $T=S$), in
which case we say that ${\rm{Cay}}(G,S)$ is an undirected Cayley graph (resp.,
a directed Cayley graph).

In the next section, we shall be interested in the integrality of mixed Cayley
graphs, and hence, shall make use of the following result, which follows from
Babai’s work [Bab79]:

######  Lemma 2.3.

Let $G$ be a finite group, let $S\subseteq G\setminus\\{1\\}$, and let
$T=\\{s\in S\,:\,s^{-1}\not\in S\\}$. Let $\rho_{1},\dots,\rho_{n}$ be the
inequivalent irreducible representations of $G$. Then the eigenvalues of the
Hermitian adjacency matrix of the mixed Cayley graph ${\rm{Cay}}(G,S)$ are
those of

$\sum_{s\in S\setminus T}\rho_{j}(s)+{\bf i}\sum_{s\in T}\rho_{j}(s)-{\bf
i}\sum_{s\in T}\rho_{j}(s^{-1})\,\,\,{\rm{for}}\,\,\,j\in\\{1,\dots,n\\}.$

###### Proof.

Denote the regular representation of $G$ by $\rho^{{\rm reg}}$. By the proof
of [Bab79, Theorem 3.1], letting $\alpha:G\rightarrow\mathbb{C}$ denote the
map defined by

$\alpha(g)=\left\\{\begin{array}[]{ll}1&\mbox{if}\,\,\,g\in S\setminus T,\\\
{\bf i}&\mbox{if}\,\,\,g\in T,\\\ -{\bf i}&\mbox{if}\,\,\,g^{-1}\in T,\\\
0&\mbox{otherwise},\end{array}\right.$

the Hermitian adjacency matrix of ${\rm Cay}(G,S)$ is

$\sum_{g\in G}\alpha(g)\rho^{{\rm reg}}(g)=\sum_{s\in S\setminus T}\rho^{{\rm
reg}}(s)+{\bf i}\sum_{s\in T}\rho^{{\rm reg}}(s)-{\bf i}\sum_{s\in
T}\rho^{{\rm reg}}(s^{-1}).$

Recall that $\rho^{{\rm reg}}$ is equivalent to
$m_{1}\rho_{1}\oplus\dots\oplus m_{n}\rho_{n}$, where $m_{j}$ is the dimension
of $\rho_{j}$ for every $j\in\\{1,\dots,n\\}$. Therefore, we deduce that the
eigenvalues of the Hermitian adjacency matrix of ${\rm Cay}(G,S)$ are those of

$\sum_{s\in S\setminus T}\rho_{j}(s)+{\bf i}\sum_{s\in T}\rho_{j}(s)-{\bf
i}\sum_{s\in T}\rho_{j}(s^{-1})\,\,\,{\rm{for}}\,\,\,j\in\\{1,\dots,n\\},$

as needed for the lemma. ∎

##  3\. Characterization of integrality

###  3.1. Main result

The main aim of this section is Theorem 3.1, which characterizes integral
mixed Cayley graphs over any given non-abelian finite group admitting an
abelian subgroup of index 2. First, we observe that non-abelian finite groups
$G$ admitting an abelian subgroup $A$ of index 2 (and order at least 3 without
loss of generality) are exactly those admitting a presentation

$G=\langle A,x\,|\,x^{2}=y,xax^{-1}=f(a)\,\,(a\in A)\rangle,$ (3.1)

where $y$ is any given element of $A$ and where $f$ is any given automorphism
of $A$ of order 2 and fixing $y$. Therefore, for this section, let $A$ be a
finite abelian group of order at least $3$, let $f\in{\rm{Aut}}(A)$ be of
order 2, and let $y\in A$ be such that $f(y)=y$. We define the finite group
$G$ by the presentation given in (3.1). Note that $G$ is the disjoint union of
$A$ and $xA$. From now on, we consider the subgroup $B=\\{f(a)a^{-1}\,:\,a\in
A\\}$ of $A$.

######  Theorem 3.1.

Let $S\subseteq G\setminus\\{1\\}$, and let $T=\\{s\in S\,:s^{-1}\not\in
S\\}.$ Set $S\setminus T=S_{1}\cup xS_{2}$ and $T=T_{1}\cup xT_{2}$ with
$S_{1},S_{2},T_{1},T_{2}\subseteq A$. Then ${\rm{Cay}}(G,S)$ is integral if
and only if the next two conditions hold:

(1)
$\rho(S_{1})+\rho(x)\rho(S_{2})-2{\rm{Im}}(\rho(T_{1}))-2{\rm{Im}}(\rho(x)\rho(T_{2}))\in\mathbb{Z}$
for every one-dimensional representation $\rho:G\rightarrow\mathbb{C}^{*}$ of
$G$,

(2) for every one-dimensional representation $\pi:A\rightarrow\mathbb{C}^{*}$
of $A$ with $B\not\subseteq{\rm{ker}}(\pi)$, we have
$\delta(\pi)\in\mathbb{Z}$ and there exists $\zeta\in\mathbb{Z}$ such that
$\delta(\pi)^{2}-4\epsilon(\pi)=\zeta^{2}$, where

$\left\\{\begin{array}[]{lll}\alpha(\pi)=\pi(T_{1})-\pi(T_{1}^{-1}),\\\
\beta(\pi)=\pi(yf(T_{2}))-\pi(T_{2}^{-1}),\\\
\gamma(\pi)=\pi(f(T_{1}))-\pi(f(T_{1}^{-1})),\\\
\delta(\pi)=\pi(f(S_{1}))+\pi(S_{1})+{\bf
i}(\alpha(\pi)+\gamma(\pi)),\end{array}\right.$ (3.2)

and $\epsilon(\pi)$ equals

$\pi(S_{1})\pi(f(S_{1}))-\pi(S_{2})\pi(S_{2}^{-1})+{\bf
i}(\pi(S_{1})\gamma(\pi)+\pi(f(S_{1}))\alpha(\pi)-\beta(\pi)\pi(S_{2})+\overline{\beta(\pi)}\pi(S_{2}^{-1}))\\\
-\alpha(\pi)\gamma(\pi)-\beta(\pi)\overline{\beta(\pi)}.$ (3.3)

###  3.2. Proof of Theorem 3.1

To prove our main result, we shall make use of Lemma 2.3, and so prove that
the eigenvalues of

$\sum_{s\in S\setminus T}\rho(s)+{\bf i}\sum_{s\in T}\rho(s)-{\bf i}\sum_{s\in
T}\rho(s^{-1}),$

for $\rho$ ranging over the irreducible representations of $G$, are in
$\mathbb{Z}$ if and only if (1) and (2) in the statement of the theorem hold.
To that end, we first determine all irreducible representations of $G$, up to
equivalence.

####  3.2.1. Irreducible representations of $G$

The next classification, for which we could not find an explicit reference in
the literature, generalizes and unifies those for generalized dihedral groups
and generalized dicyclic groups from [HL21, BL22].

(1) One-dimensional representations. Let $\rho:G\rightarrow\mathbb{C}^{*}$ be
a one-dimensional representation of $G$. Then
$\rho(a)=\rho(x)\rho(a)\rho(x)^{-1}=\rho(xax^{-1})=\rho(f(a))$ and so
$\rho(f(a)a^{-1})=1$ for $a\in A$, i.e., $\rho$ is trivial on $B$. Hence,
every one-dimensional representation of $G$ gives rise to a one-dimensional
representation of $A/B$.

Conversely, let $\pi:A/B\rightarrow\mathbb{C}^{*}$ be a one-dimensional
representation of $A/B$. Then there are exactly two inequivalent one-
dimensional representations $\rho_{1}$ and $\rho_{2}$ of $G$ giving rise to
$\pi$ and, denoting the reduction modulo $B$ of $a\in A$ by $\overline{a}$ and
fixing a squareroot $\sqrt{\pi(\overline{y})}$ of $\pi(\overline{y})$ in
$\mathbb{C}$, they are actually given by

$\bullet$ $\rho_{1}(x)=\sqrt{\pi(\overline{y})}$ and
$\rho_{1}(a)=\pi(\overline{a})$ for $a\in A$,

$\bullet$ $\rho_{2}(x)=-\sqrt{\pi(\overline{y})}$ and
$\rho_{2}(a)=\pi(\overline{a})$ for $a\in A$.

In particular, $G$ has exactly $2(A:B)$ inequivalent one-dimensional
representations. Note that $(A:B)$ is the index of $B$ in $A$.

(2) Two-dimensional representations. Let $\pi:A\rightarrow\mathbb{C}^{*}$ be a
one-dimensional representation of $A$ with $B\not\subseteq{\rm{ker}}(\pi)$.
Let $R_{\pi}:G\rightarrow{\rm{GL}}_{2}(\mathbb{C})$ be the map defined by

$R_{\pi}(a)=\begin{pmatrix}\pi(a)&0\\\
0&\pi(f(a))\end{pmatrix}\quad{\rm{and}}\,\,\,R_{\pi}(xa)=\begin{pmatrix}0&\pi(yf(a))\\\
\pi(a)&0\end{pmatrix}\,\,\,(a\in A).$ (3.4)

As $f$ has order 2 and fixes $y$, the map $R_{\pi}$ is a representation of
$G$. In fact, $R_{\pi}$ is the representation $\mathrm{Ind}_{A}^{G}(\pi)$ of
$G$ induced by $\pi$.

######  Lemma 3.2.

The representation $R_{\pi}$ is irreducible.

###### Proof.

As recalled in §2.1, it suffices to show

$(\chi_{R_{\pi}}|\chi_{R_{\pi}})=1,$ (3.5)

where $(\cdot|\cdot)$ is defined in (2.1). By (3.4), we have

$(\chi_{R_{\pi}}|\chi_{R_{\pi}})=\frac{1}{2|A|}\sum_{a\in
A}(\pi(a)+\pi(f(a)))(\pi(a^{-1})+\pi(f(a)^{-1}))=1+\frac{1}{|A|}\sum_{a\in
A}\pi(f(a)a^{-1}).$ (3.6)

Now, since $B\not\subseteq{\rm{ker}}(\pi)$, there exists $b\in A$ with
$\pi(f(b)b^{-1})\not=1$. Since

$\sum_{a\in A}\pi(f(a)a^{-1})=\sum_{a\in
A}\pi(f(ab)(ab)^{-1})=\pi(f(b)b^{-1})\sum_{a\in A}\pi(f(a)a^{-1}),$

we get

$\sum_{a\in A}\pi(f(a)a^{-1})=0.$ (3.7)

It then remains to combine (3.6) and (3.7) to get (3.5), as needed for the
lemma. ∎

######  Lemma 3.3.

For $i\in\\{1,2\\}$, let $\pi_{i}:A\rightarrow\mathbb{C}^{*}$ be a one-
dimensional representation of $A$ with $B\not\subseteq{\rm{ker}}(\pi_{i})$.
Then $R_{\pi_{1}}$ and $R_{\pi_{2}}$ are equivalent if and only if
$\pi_{2}=\pi_{1}$ or $\pi_{2}=\pi_{1}\circ f$.

###### Proof.

First, assume $\pi_{2}=\pi_{1}$ or $\pi_{2}=\pi_{1}\circ f$. Then
$\chi_{R_{\pi_{1}}}=\chi_{R_{\pi_{2}}}$, and hence, the representations
$R_{\pi_{1}}$ and $R_{\pi_{2}}$ are equivalent.

Conversely, assume $R_{\pi_{1}}$ and $R_{\pi_{2}}$ are equivalent. Then
$\chi_{R_{\pi_{1}}}=\chi_{R_{\pi_{2}}}$, and so

$\pi_{1}(a)+\pi_{1}(f(a))=\pi_{2}(a)+\pi_{2}(f(a))\,\,\,(a\in A).$ (3.8)

Fix $a\in A$. As $\pi_{1}(a)$ and $\pi_{2}(a)$ are roots of unity, (3.8) gives

$|1+\pi_{1}(f(a)a^{-1})|=|1+\pi_{2}(f(a)a^{-1})|,$

and so

$\pi_{1}(f(a)a^{-1})=\overline{\pi_{2}(f(a)a^{-1})}\,\,\,{\rm{or}}\,\,\,\pi_{1}(f(a)a^{-1})=\pi_{2}(f(a)a^{-1}).$
(3.9)

First, assume $\pi_{1}(f(a)a^{-1})\not=-1$. If
$\pi_{1}(f(a)a^{-1})=\overline{\pi_{2}(f(a)a^{-1})}$, then, using (3.8), we
get

$\pi_{1}(a)+\pi_{1}(f(a))=\pi_{2}(a)+\pi_{2}(f(a))=\pi_{2}(a)+\pi_{2}(a)\overline{\pi_{1}(f(a)a^{-1})}.$

We then have

$\pi_{1}(a)=\pi_{2}(a)\frac{1+\overline{\pi_{1}(f(a)a^{-1})}}{1+\pi_{1}(f(a)a^{-1})}=\pi_{2}(a)\frac{1}{\pi_{1}(f(a)a^{-1})},$

and so $\pi_{2}(a)=\pi_{1}(f(a))$. If
$\pi_{1}(f(a)a^{-1})\not=\overline{\pi_{2}(f(a)a^{-1})}$, then, by (3.9), we
have $\pi_{1}(f(a)a^{-1})=\pi_{2}(f(a)a^{-1})$ and, using (3.8) once more, we
get

$\pi_{1}(a)+\pi_{1}(f(a))=\pi_{2}(a)+\pi_{2}(f(a))=\pi_{2}(a)+\pi_{2}(a)\pi_{1}(f(a)a^{-1}).$

In particular, $\pi_{2}(a)=\pi_{1}(a)$.

Now, assume $\pi_{1}(f(a)a^{-1})=-1$. Then $\pi_{1}(f(a^{2})a^{-2})=1$ and,
from the previous paragraph (applied with $a^{2}$ instead of $a$), we get
either $\pi_{2}(a^{2})=\pi_{1}(a^{2})$ or $\pi_{2}(a^{2})=\pi_{1}(f(a^{2}))$.
In particular, at least one of the following four conditions holds:

$\bullet$ $\pi_{2}(a)=\pi_{1}(a)$,

$\bullet$ $\pi_{2}(a)=-\pi_{1}(a)$,

$\bullet$ $\pi_{2}(a)=\pi_{1}(f(a))$,

$\bullet$ $\pi_{2}(a)=-\pi_{1}(f(a))$.

First, assume $\pi_{2}(a)=-\pi_{1}(f(a))$. Then
$\pi_{2}(a)=-\pi_{1}(a)\pi_{1}(f(a)a^{-1})=\pi_{1}(a)$, since we have assumed
$\pi_{1}(f(a)a^{-1})=-1$ at the beginning of the paragraph. Now, assume
$\pi_{2}(a)=-\pi_{1}(a)$. Then
$\pi_{1}(f(a))=\pi_{1}(f(a)a^{-1})\pi_{1}(a)=-\pi_{1}(a)=\pi_{2}(a).$

Hence, from the previous two paragraphs, for $a\in A$, either
$\pi_{2}(a)=\pi_{1}(a)$ or $\pi_{2}(a)=\pi_{1}(f(a))$. We now consider a case
distinction. If $\pi_{2}(a)=\pi_{1}(a)$ for all $a\in A$, then
$\pi_{2}=\pi_{1}$. Else, there is $a\in A$ with $\pi_{1}(a)\neq\pi_{1}(f(a))$
and $\pi_{2}(a)=\pi_{1}(f(a))$. In this case we claim that
$\pi_{2}=\pi_{1}\circ f$. Indeed, assume there is $b\in A$ with
$\pi_{2}(b)\neq\pi_{1}(f(b))$, so $\pi_{2}(b)=\pi_{1}(b)$. If
$\pi_{2}(ab)=\pi_{1}(ab)$, then

$\pi_{1}(a)\pi_{1}(b)=\pi_{1}(ab)=\pi_{2}(ab)=\pi_{2}(a)\pi_{2}(b)=\pi_{1}(f(a))\pi_{1}(b),$

so $\pi_{1}(a)=\pi_{1}(f(a))$, a contradiction. If
$\pi_{2}(ab)=\pi_{1}(f(ab))$, a similar computation leads to
$\pi_{2}(a)\pi_{2}(b)=\pi_{2}(a)\pi_{1}(f(b))$ and so
$\pi_{2}(b)=\pi_{1}(f(b))$, which cannot happen either. ∎

As there are exactly $|A|-(A:B)$ inequivalent one-dimensional representations
$\pi:A\rightarrow\mathbb{C}^{*}$ of $A$ with $B\not\subseteq{\rm{ker}}(\pi)$,
we then deduce from Lemmas 3.2 and 3.3 that $G$ has at least $(|A|-(A:B))/2$
inequivalent two-dimensional irreducible representations.

(3) Conclusion. Since

$1^{2}(2(A:B))+2^{2}\left(\frac{|A|-(A:B)}{2}\right)=2|A|=|G|,$

we get from (2.2) that $G$ has exactly $(|A|-(A:B))/2$ inequivalent two-
dimensional irreducible representations, and that $G$ has no further
irreducible representation, up to equivalence. Therefore, we have proved the
following result:

######  Theorem 3.4.

The group $G$ has exactly $2(A:B)+(|A|-(A:B))/2$ inequivalent irreducible
representations

$\rho_{1},\dots,\rho_{(A:B)},\rho_{(A:B)+1},\dots,\rho_{2(A:B)},\rho_{2(A:B)+1},\dots,\rho_{2(A:B)+(|A|-(A:B))/2},$

which are as follows.

(1) Let $\pi_{1},\dots,\pi_{(A:B)}$ be the inequivalent irreducible
representations of $A/B$. Then, given $j\in\\{1,\dots,(A:B)\\}$, we have

$\bullet$ $\rho_{j}(x)=\sqrt{\pi_{j}(\overline{y})}$ and
$\rho_{j}(a)=\pi_{j}(\overline{a})$ for $a\in A$,

$\bullet$ $\rho_{j+(A:B)}(x)=-\sqrt{\pi_{j}(\overline{y})}$ and
$\rho_{j+(A:B)}(a)=\pi_{j}(\overline{a})$ for $a\in A$,

where $\overline{a}$ and $\sqrt{\pi(\overline{y})}$ are the reductions modulo
$B$ of $a\in A$ and a given squareroot of $\pi(\overline{y})$ in $\mathbb{C}$,
respectively.

(2) Let $\psi_{1},\dots,\psi_{|A|-(A:B)}$ be the inequivalent irreducible
representations of $A$ whose kernels do not contain $B$. Assume
$\psi_{j+(|A|-(A:B))/2}=\psi_{j}\circ f$ for
$j\in\\{1,\dots,(|A|-(A:B))/2\\}$. Then, given
$j\in\\{1,\dots,(|A|-(A:B))/2\\}$, we have

$\rho_{2(A:B)+j}(a)=\begin{pmatrix}\psi_{j}(a)&0\\\
0&\psi_{j}(f(a))\end{pmatrix}\quad{\rm{and}}\,\,\,\rho_{2(A:B)+j}(xa)=\begin{pmatrix}0&\psi_{j}(yf(a))\\\
\psi_{j}(a)&0\end{pmatrix}\,\,\,(a\in A).$

####  3.2.2. Proof of Theorem 3.1

We now proceed to the proof of our main result.

(a) One-dimensional representations. Given a one-dimensional representation
$\rho:G\rightarrow\mathbb{C}^{*}$ of $G$, we have

$\rho(S\setminus T)=\rho(S_{1})+\rho(x)\rho(S_{2}),$ ${\bf i}\rho(T_{1})-{\bf
i}\rho(T_{1}^{-1})={\bf i}\rho(T_{1})+\overline{{\bf
i}\rho(T_{1})}=2{\rm{Re}}({\bf i}\rho(T_{1}))=-2{\rm{Im}}(\rho(T_{1})),$ ${\bf
i}\rho(xT_{2})-{\bf i}\rho((xT_{2})^{-1})={\bf i}\rho(xT_{2})+\overline{{\bf
i}\rho(xT_{2})}=-2{\rm{Im}}(\rho(x)\rho(T_{2})).$

Hence, $\rho(S\setminus T)+{\bf i}\rho(T)-{\bf i}\rho(T^{-1})\in\mathbb{Z}$
for every one-dimensional representation $\rho:G\rightarrow\mathbb{C}^{*}$ of
$G$ if and only if (1) holds.

(b) Two-dimensional representations. Let $\pi:A\rightarrow\mathbb{C}^{*}$ be a
one-dimensional representation of $A$ with $B\not\subseteq{\rm{ker}}(\pi)$,
and let $R_{\pi}:G\rightarrow{\rm{GL}}_{2}(\mathbb{C})$ be as in (3.4). First,
we have

$\begin{array}[]{lllll}\displaystyle{\sum_{a\in S\setminus
T}R_{\pi}(a)}&=&\displaystyle{\sum_{a\in S_{1}}R_{\pi}(a)+\sum_{a\in
S_{2}}R_{\pi}(xa)}&=&\displaystyle{\sum_{a\in S_{1}}\begin{pmatrix}\pi(a)&0\\\
0&\pi(f(a))\end{pmatrix}+\sum_{a\in S_{2}}\begin{pmatrix}0&\pi(yf(a))\\\
\pi(a)&0\end{pmatrix}}\\\ &&&=&\begin{pmatrix}\pi(S_{1})&0\\\
0&\pi(f(S_{1}))\end{pmatrix}+\begin{pmatrix}0&\pi(yf(S_{2}))\\\
\pi(S_{2})&0\end{pmatrix}\\\ &&&=&\begin{pmatrix}\pi(S_{1})&\pi(yf(S_{2}))\\\
\pi(S_{2})&\pi(f(S_{1}))\end{pmatrix}\end{array}$

and, similarly,

$\sum_{a\in T}R_{\pi}(a)=\begin{pmatrix}\pi(T_{1})&\pi(yf(T_{2}))\\\
\pi(T_{2})&\pi(f(T_{1}))\end{pmatrix}.$

Moreover, given $a\in A$, we have
$(xa)^{-1}=a^{-1}x^{-1}=x^{-1}f(a^{-1})=xy^{-1}f(a^{-1})\in xA$. Combined with
$(S\setminus T)^{-1}=S\setminus T$, this yields
$xS_{2}=(xS_{2})^{-1}=xy^{-1}f(S_{2}^{-1})$, i.e., $f(S_{2}^{-1})=yS_{2}$.
Since $f$ has order 2 and fixes $y$, we eventually get $S_{2}^{-1}=yf(S_{2})$.
In particular, we have

$\sum_{a\in S\setminus
T}R_{\pi}(a)=\begin{pmatrix}\pi(S_{1})&\pi(S_{2}^{-1})\\\
\pi(S_{2})&\pi(f(S_{1}))\end{pmatrix}.$

Furthermore, using once more $(xa)^{-1}=x^{-1}f(a^{-1})$ for $a\in A$, we have

$\displaystyle\sum_{a\in T}R_{\pi}(a^{-1})$ $\displaystyle=\sum_{a\in
T_{1}}R_{\pi}(a^{-1})+\sum_{a\in T_{2}}R_{\pi}(x^{-1}f(a^{-1}))$
$\displaystyle=\sum_{a\in T_{1}}\begin{pmatrix}\pi(a^{-1})&0\\\
0&\pi(f(a^{-1}))\end{pmatrix}+\sum_{a\in T_{2}}\begin{pmatrix}0&1\\\
\pi(y^{-1})&0\end{pmatrix}\begin{pmatrix}\pi(f(a^{-1}))&0\\\
0&\pi(f(f(a^{-1})))\end{pmatrix}$
$\displaystyle=\begin{pmatrix}\pi(T_{1}^{-1})&0\\\
0&\pi(f(T_{1}^{-1}))\end{pmatrix}+\sum_{a\in T_{2}}\begin{pmatrix}0&1\\\
\pi(y^{-1})&0\end{pmatrix}\begin{pmatrix}\pi(f(a^{-1}))&0\\\
0&\pi(a^{-1})\end{pmatrix}$ $\displaystyle=\begin{pmatrix}\pi(T_{1}^{-1})&0\\\
0&\pi(f(T_{1}^{-1}))\end{pmatrix}+\sum_{a\in
T_{2}}\begin{pmatrix}0&\pi(a^{-1})\\\ \pi(y^{-1}f(a^{-1}))&0\end{pmatrix}$
$\displaystyle=\begin{pmatrix}\pi(T_{1}^{-1})&0\\\
0&\pi(f(T_{1}^{-1}))\end{pmatrix}+\begin{pmatrix}0&\pi(T_{2}^{-1})\\\
\pi(y^{-1}f(T_{2}^{-1}))&0\end{pmatrix}$
$\displaystyle=\begin{pmatrix}\pi(T_{1}^{-1})&\pi(T_{2}^{-1})\\\
\pi(y^{-1}f(T_{2}^{-1}))&\pi(f(T_{1}^{-1}))\end{pmatrix}.$

Hence,

$\sum_{a\in S\setminus T}R_{\pi}(a)+{\bf i}\sum_{a\in T}R_{\pi}(a)-{\bf
i}\sum_{a\in T}R_{\pi}(a^{-1})$ (3.10)

equals

$\begin{pmatrix}\pi(S_{1})+{\bf
i}(\pi(T_{1})-\pi(T_{1}^{-1}))&\pi(S_{2}^{-1})+{\bf
i}(\pi(yf(T_{2}))-\pi(T_{2}^{-1}))\\\ \pi(S_{2})+{\bf
i}(\pi(T_{2})-\pi(y^{-1}f(T_{2}^{-1})))&\pi(f(S_{1}))+{\bf
i}(\pi(f(T_{1}))-\pi(f(T_{1}^{-1})))\end{pmatrix}\\\ .$

Using (3.2) and (3.3), we get that (3.10) equals

$\begin{pmatrix}\pi(S_{1})+{\bf i}\alpha(\pi)&\pi(S_{2}^{-1})+{\bf
i}\beta(\pi)\\\ \pi(S_{2})-{\bf i}\overline{\beta(\pi)}&\pi(f(S_{1}))+{\bf
i}\gamma(\pi)\end{pmatrix}\\\ ,$

whose characteristic polynomial equals $X^{2}-\delta(\pi)X+\epsilon(\pi).$
Hence, the eigenvalues of (3.10) are

$\lambda_{1}=\frac{\delta(\pi)-\sqrt{\delta(\pi)^{2}-4\epsilon(\pi)}}{2}\,\,\,{\rm{and}}\,\,\,\lambda_{2}=\frac{\delta(\pi)+\sqrt{\delta(\pi)^{2}-4\epsilon(\pi)}}{2}.$

If $\lambda_{1},\lambda_{2}$ are in $\mathbb{Z}$, then
$\delta(\pi)=\lambda_{1}+\lambda_{2}$ is in $\mathbb{Z}$ and
$\sqrt{\delta(\pi)^{2}-4\epsilon(\pi)}=\lambda_{2}-\lambda_{1}\in\mathbb{Z}$.
Conversely, assume $\delta(\pi)\in\mathbb{Z}$ and
$\sqrt{\delta(\pi)^{2}-4\epsilon(\pi)}\in\mathbb{Z}$. Then
$4\epsilon(\pi)=\delta(\pi)^{2}-(\sqrt{\delta(\pi)^{2}-4\epsilon(\pi)})^{2}\in\mathbb{Z}$.
In particular, we have $\epsilon(\pi)\in\mathbb{Q}$. Since $\epsilon(\pi)$ is
also integral over $\mathbb{Z}$, we actually have
$\epsilon(\pi)\in\mathbb{Z}$. Hence, $\delta(\pi)$ and
$\sqrt{\delta(\pi)^{2}-4\epsilon(\pi)}$ have the same parity, and we get
$\lambda_{1},\lambda_{2}\in\mathbb{Z}$.

In view of the classification from Theorem 3.4, we then get that the
eigenvalues of

$\sum_{s\in S\setminus T}\rho(s)+{\bf i}\sum_{s\in T}\rho(s)-{\bf i}\sum_{s\in
T}\rho(s^{-1})$

are in $\mathbb{Z}$ for every irreducible two-dimensional representation
$\rho$ of $G$ if and only if (2) holds.

###  3.3. Corollaries

The statement of Theorem 3.1 is technical because we consider mixed Cayley
graphs over arbitrary non-abelian finite groups admitting an abelian subgroup
of index 2. As illustrations of our main result, we now consider two special
cases of interest.

We first restrict to the undirected situation, in which case Theorem 3.1 takes
the next simpler form:

######  Corollary 3.5.

Let $S$ be a subset of $G$ with $1\not\in S$ and $S^{-1}=S$. Set $S=S_{1}\cup
xS_{2}$ with $S_{1},S_{2}\in A$. Then the undirected Cayley graph
${\rm{Cay}}(G,S)$ is integral if and only if the next two conditions hold:

(1) $\rho(S_{1})+\rho(x)\rho(S_{2})\in\mathbb{Z}$ for every one-dimensional
representation $\rho:G\rightarrow\mathbb{C}^{*}$ of $G$,

(2) for every one-dimensional representation $\pi:A\rightarrow\mathbb{C}^{*}$
with $B\not\subseteq{\rm{ker}}(\pi)$, we have
$\pi(f(S_{1}))+\pi(S_{1})\in\mathbb{Z}$ and there exists $\zeta\in\mathbb{Z}$
such that

$(\pi(f(S_{1}))-\pi(S_{1}))^{2}+4\pi(S_{2})\pi(S_{2}^{-1})=\zeta^{2}.$

The following corollary allows to generate integral undirected Cayley graphs
over $G$:

######  Corollary 3.6.

Let $S_{1},S_{2}\in\mathbb{B}(A)$ be such that $1\not\in S_{1}$,
$f(S_{1})=S_{1}$ and $f(S_{2})=y^{-1}S_{2}$. Then the undirected Cayley graph
${\rm{Cay}}(G,S_{1}\cup xS_{2})$ is integral.

###### Proof.

First, the undirected Cayley graph ${\rm{Cay}}(G,S_{1}\cup xS_{2})$ is well-
defined. Indeed, since $S_{1}\in\mathbb{B}(A)$, Lemma 2.2 yields that
$S_{1}^{-1}$ and $S_{1}$ coincide. Similarly, we have $S_{2}^{-1}=S_{2}$ and,
as $f(S_{2})=y^{-1}S_{2}$, we get

$(xS_{2})^{-1}=S_{2}^{-1}x^{-1}=x^{-1}f(S_{2}^{-1})=xy^{-1}f(S_{2}^{-1})=xy^{-1}yS_{2}^{-1}=xS_{2}^{-1}=xS_{2},$
(3.11)

thus yielding $(S_{1}\cup xS_{2})^{-1}=S_{1}\cup xS_{2}$, as needed.

Now, to get Corollary 3.6, it suffices to show that (1) and (2) in Corollary
3.5 hold. First, let $\rho:G\rightarrow\mathbb{C}^{*}$ be a one-dimensional
representation of $G$. To get (1), we have to show
$\rho(S_{1})+\rho(x)\rho(S_{2})\in\mathbb{Z}$. To that end, note that, as
$S_{j}\in\mathbb{B}(A)$ for $j\in\\{1,2\\}$, we have
$\rho(S_{j})\in\mathbb{Z}$ by Lemma 2.1. In particular, if $\rho(S_{2})=0$,
then $\rho(S_{1})+\rho(x)\rho(S_{2})\in\mathbb{Z}$. Therefore, assume
$\rho(S_{2})\not=0$. Then, by (3.11), we have
$\rho(x)\rho(S_{2})=\rho((xS_{2})^{-1})=\overline{\rho(x)\rho(S_{2})}$, i.e.,
$\rho(x)\rho(S_{2})\in\mathbb{R}$. Since
$\rho(S_{2})\in\mathbb{Z}\setminus\\{0\\}$, we get $\rho(x)\in\mathbb{R}$. But
$\rho(x)$ is also a root of unity, and hence, $\rho(x)\in\\{-1,1\\}$, and so
$\rho(x)\rho(S_{2})\in\mathbb{Z}$. In particular, (1) holds.

Now, let $\pi:A\rightarrow\mathbb{C}^{*}$ be a one-dimensional representation
of $A$ with $B\not\subseteq{\rm{ker}}(\pi)$. Since $f(S_{1})=S_{1}$, we have
$\pi(f(S_{1}))+\pi(S_{1})=2\pi(S_{1})$ and, as $S_{1}\in\mathbb{B}(A)$, we
have $\pi(S_{1})\in\mathbb{Z}$ by Lemma 2.1. Therefore,
$\pi(f(S_{1}))+\pi(S_{1})\in\mathbb{Z}$. Moreover, since $S_{2}^{-1}=S_{2}$,
we have

$(\pi(f(S_{1}))-\pi(S_{1}))^{2}+4\pi(S_{2})\pi(S_{2}^{-1})=4\pi(S_{2})^{2}$

and, as $S_{2}\in\mathbb{B}(A)$, we have $\pi(S_{2})\in\mathbb{Z}$ by Lemma
2.1. Hence, (2) holds. ∎

######  Example 3.7.

Assume that $A$ has exponent $e$, where $e\geq 3$, and that there is an
integer $s$ of order 2 modulo $e$ such that, for every $a\in A$, we have
$f(a)=a^{s}$. Then, in the statement of Corollary 3.6, the assumption
“$f(S_{1})=S_{1}$” is redundant and the one “$f(S_{2})=y^{-1}S_{2}$” reduces
to “$yS_{2}=S_{2}$”. Indeed, from our extra assumption on $f$, we have
$f(S_{j})=S_{j}^{s}$ for $j\in\\{1,2\\}$ and, since $s$ is coprime111In fact,
since $s^{2}\equiv 1\pmod{e}$, we have ${\rm{gcd}}(s,e)=1$. to every element
order in $A$, we may then apply Lemma 2.2 to get $f(S_{j})=S_{j}^{s}=S_{j}$
for $j\in\\{1,2\\}$.

In particular, if $A$ has exponent $e$, where $e\geq 3$, if there is an
integer $s$ of order 2 modulo $e$ such that, for every $a\in A$, we have
$f(a)=a^{s}$, and if $y=1$, then every undirected Cayley graph
${\rm{Cay}}(G,S_{1}\cup xS_{2})$, with $1\not\in S_{1}$ and
$S_{1},S_{2}\in\mathbb{B}(A)$, is integral. In addition to generalized
dihedral groups (for which $s=-1$), the latter applies to, e.g., quasi-
dihedral groups, which assume $A=\mathbb{Z}/2^{n-1}\mathbb{Z}$ ($n\geq 4$) and
$s=2^{n-2}-1$, and modular groups, which assume
$A=\mathbb{Z}/2^{n-1}\mathbb{Z}$ ($n\geq 4$) and $s=2^{n-2}+1$.

We now consider the case where $f(a)=a^{-1}$ for $a\in A$, in which case
$B=A^{2}$ and $y^{2}=1$ (and $A$ has exponent at least 3). If $y=1$, then $G$
is the generalized dihedral group ${\rm{Dih}}(A)$, which is given by the
following presentation:

${\rm{Dih}}(A)=\langle A,x\,|\,x^{2}=1,xax^{-1}=a^{-1}\,(a\in A)\rangle.$

If $y\not=1$ (and so $|A|$ is even), then $G$ is the generalized dicyclic
group ${\rm{Dic}}(A,y)$, which is given by the following presentation:

${\rm{Dic}}(A,y)=\langle A,x\,|\,x^{2}=y,xax^{-1}=a^{-1}\,(a\in A)\rangle.$

The next corollary applies to both ${\rm{Dih}}(A)$ and ${\rm{Dic}}(A,y)$:

######  Corollary 3.8.

Assume $f(a)=a^{-1}$ for every $a\in A$. Let $S$ be a subset of $G$ with
$1\not\in S$, and let $T=\\{s\in S\,:s^{-1}\not\in S\\}.$ Set $S\setminus
T=S_{1}\cup xS_{2}$ and $T=T_{1}\cup xT_{2}$ with
$S_{1},S_{2},T_{1},T_{2}\subseteq A$. Then the mixed Cayley graph
${\rm{Cay}}(G,S)$ is integral if and only if the next three conditions hold:

(1) $S_{1}\in\mathbb{B}(A)$,

(2) for every one-dimensional representation $\pi:A\rightarrow\mathbb{C}^{*}$
of $A$ with $A^{2}\not\subseteq{\rm{ker}}(\pi)$ and $y\in{\rm{ker}}(\pi)$,
there exists $\alpha\in\mathbb{Z}$ such that
$\pi(S_{2})\pi(S_{2}^{-1})-(\pi(T_{1})-\pi(T_{1}^{-1}))^{2}=\alpha^{2},$

(3) for every one-dimensional representation $\pi:A\rightarrow\mathbb{C}^{*}$
of $A$ with $A^{2}\not\subseteq{\rm{ker}}(\pi)$ and $y\not\in{\rm{ker}}(\pi)$,
there exists $\alpha\in\mathbb{Z}$ such that
$4\pi(T_{2})\pi(T_{2}^{-1})-(\pi(T_{1})-\pi(T_{1}^{-1}))^{2}=\alpha^{2}.$

###### Proof.

By Theorem 3.1, it suffices to prove the following four statements, in which
we use the notation from our main result:

(A) We have
$\rho(S_{1})+\rho(x)\rho(S_{2})-2{\rm{Im}}(\rho(T_{1}))-2{\rm{Im}}(\rho(x)\rho(T_{2}))\in\mathbb{Z}$
for every one-dimensional representation $\rho:G\rightarrow\mathbb{C}^{*}$ of
$G$.

(B) The next two conditions are equivalent:

(i) $S_{1}\in{\mathbb{B}}(A)$,

(ii) $\delta(\pi)\in\mathbb{Z}$ for every one-dimensional representation
$\pi:A\rightarrow\mathbb{C}^{*}$ of $A$ with
$A^{2}\not\subseteq{\rm{ker}}(\pi)$.

(C) The next two conditions are equivalent:

(i) for every one-dimensional representation $\pi:A\rightarrow\mathbb{C}^{*}$
of $A$ with $A^{2}\not\subseteq{\rm{ker}}(\pi)$ and $y\in{\rm{ker}}(\pi)$,
there

exists $\zeta\in\mathbb{Z}$ such that
$\delta(\pi)^{2}-4\epsilon(\pi)=\zeta^{2}$,

(ii) for every one-dimensional representation $\pi:A\rightarrow\mathbb{C}^{*}$
of $A$ such that $A^{2}\not\subseteq{\rm{ker}}(\pi)$ and
$y\in{\rm{ker}}(\pi)$,

there exists $\alpha\in\mathbb{Z}$ satisfying
$\pi(S_{2})\pi(S_{2}^{-1})-(\pi(T_{1})-\pi(T_{1}^{-1}))^{2}=\alpha^{2}.$

(D) The next two conditions are equivalent:

(i) for every one-dimensional representation $\pi:A\rightarrow\mathbb{C}^{*}$
of $A$ with $A^{2}\not\subseteq{\rm{ker}}(\pi)$ and $y\not\in{\rm{ker}}(\pi)$,
there

exists $\zeta\in\mathbb{Z}$ such that
$\delta(\pi)^{2}-4\epsilon(\pi)=\zeta^{2}$,

(ii) for every one-dimensional representation $\pi:A\rightarrow\mathbb{C}^{*}$
of $A$ such that $A^{2}\not\subseteq{\rm{ker}}(\pi)$ and
$y\not\in{\rm{ker}}(\pi)$,

there exists $\alpha\in\mathbb{Z}$ satisfying
$4\pi(T_{2})\pi(T_{2}^{-1})-(\pi(T_{1})-\pi(T_{1}^{-1}))^{2}=\alpha^{2}.$

For later use, we observe that, since $(S\setminus T)^{-1}=S\setminus T$, we
must have

$S_{1}^{-1}=S_{1}\,\,\,{\rm{and}}\,\,\,yS_{2}^{j}=S_{2}^{j}\,\,\,(j\in\\{-1,1\\}).$
(3.12)

In particular, for every one-dimensional representation
$\pi:A\rightarrow\mathbb{C}^{*}$ of $A$ with $y\not\in{\rm{ker}}(\pi)$ (and so
$\pi(y)=-1$), and for $j\in\\{-1,1\\}$, we must have
$\pi(S_{2}^{j})=\pi(yS_{2}^{j})=\pi(y)\pi(S_{2}^{j})=-\pi(S_{2}^{j})$, i.e.,

$\pi(S_{2}^{j})=0.$ (3.13)

Proof of (A): Let $\rho:G\rightarrow\mathbb{C}^{*}$ be a one-dimensional
representation of $G$. Since $f(a)=a^{-1}$ for $a\in A$, we have
$A^{2}\subseteq{\rm{ker}}(\rho)$, and hence, $\rho(a)\in\\{-1,1\\}$ for $a\in
A$. Moreover, since $x^{2}=y$ and $y^{2}=1$, we have $\rho(x)\in\\{-1,1,-{\bf
i},{\bf i}\\}$. If $\rho(x)\in\\{-1,1\\}$, then

$\rho(S_{1})+\rho(x)\rho(S_{2})-2{\rm{Im}}(\rho(T_{1}))-2{\rm{Im}}(\rho(x)\rho(T_{2}))\in\mathbb{Z}.$

Therefore, assume $\rho(x)\in\\{-{\bf i},{\bf i}\\}$. Then $\rho(S_{2})=0$ by
(3.13). Since we also have $\rho(x)\rho(T_{2})\in\mathbb{Z}[{\bf i}]$, we get

$\rho(S_{1})+\rho(x)\rho(S_{2})-2{\rm{Im}}(\rho(T_{1}))-2{\rm{Im}}(\rho(x)\rho(T_{2}))\in\mathbb{Z},$

thus ending the proof of (A).

Proof of (B): Let $\pi:A\rightarrow\mathbb{C}^{*}$ be a one-dimensional
representation of $A$ with $A^{2}\not\subseteq{\rm{ker}}(\pi)$. Using (3.12)
and our assumption on $f$, we can rewrite (3.2) and (3.3) as follows:

$\left\\{\begin{array}[]{lll}\alpha(\pi)=\pi(T_{1})-\pi(T_{1}^{-1}),\\\
\beta(\pi)=\pi(T_{2}^{-1})(\pi(y)-1),\\\
\gamma(\pi)=\pi(T_{1}^{-1})-\pi(T_{1})=-\alpha(\pi),\\\
\delta(\pi)=2\pi(S_{1}),\\\
\epsilon(\pi)=\pi(S_{1})^{2}-\pi(S_{2})\pi(S_{2}^{-1})+{\bf
i}(\overline{\beta(\pi)}\pi(S_{2}^{-1})-\beta(\pi)\pi(S_{2}))+\alpha(\pi)^{2}-\beta(\pi)\overline{\beta(\pi)}.\\\
\end{array}\right.$ (3.14)

If $\delta(\pi)\in\mathbb{Z}$, then, by (3.14), we have
$\pi(S_{1})\in\mathbb{Q}$. Since $\pi(S_{1})$ is integral over $\mathbb{Z}$,
we get that $\pi(S_{1})$ is actually in $\mathbb{Z}$. On the other hand, we
have $\pi(S_{1})\in\mathbb{Z}$ for every one-dimensional representation
$\pi:A\rightarrow\mathbb{C}^{*}$ of $A$ with $A^{2}\subseteq{\rm{ker}}(\pi)$.
Hence, if (ii) in (B) holds, then $\pi(S_{1})\in\mathbb{Z}$ for every one-
dimensional representation $\pi:A\rightarrow\mathbb{C}^{*}$ of $A$, that is,
$S_{1}$ is integral. By Lemma 2.1, we conclude that $S_{1}\in\mathbb{B}(A)$.
Conversely, if $S_{1}\in\mathbb{B}(A)$, then $\pi(S_{1})\in\mathbb{Z}$ for
every one-dimensional representation $\pi:A\rightarrow\mathbb{C}^{*}$ of $A$,
by Lemma 2.1. In particular, for such $\pi$ also fulfilling
$A^{2}\not\subseteq{\rm{ker}}(\pi)$, we get that $\delta(\pi)$, which equals
$2\pi(S_{1})$ by (3.14), is in $\mathbb{Z}$, thus ending the proof of (B).

Proof of (C): Let $\pi:A\rightarrow\mathbb{C}^{*}$ be a one-dimensional
representation of $A$ with $A^{2}\not\subseteq{\rm{ker}}(\pi)$ and
$y\in{\rm{ker}}(\pi)$. Then from (3.14), we have

$\delta(\pi)^{2}-4\epsilon(\pi)=4\pi(S_{2})\pi(S_{2}^{-1})-4(\pi(T_{1})-\pi(T_{1}^{-1}))^{2}.$

If there is $\alpha\in\mathbb{Z}$ with
$\pi(S_{2})\pi(S_{2}^{-1})-(\pi(T_{1})-\pi(T_{1}^{-1}))^{2}=\alpha^{2}$, then
it is clear that there is $\zeta\in\mathbb{Z}$ with
$\delta(\pi)^{2}-4\epsilon(\pi)=\zeta^{2}$. Conversely, assume there is
$\zeta\in\mathbb{Z}$ with $\delta(\pi)^{2}-4\epsilon(\pi)=\zeta^{2}$. Then

$\sqrt{\pi(S_{2})\pi(S_{2}^{-1})-(\pi(T_{1})-\pi(T_{1}^{-1}))^{2}}\in\mathbb{Q}.$

But this number is also integral over $\mathbb{Z}$, and hence, it is in
$\mathbb{Z}$. This completes the proof of (C).

Proof of (D): Let $\pi:A\rightarrow\mathbb{C}^{*}$ be a one-dimensional
representation of $A$ with $A^{2}\not\subseteq{\rm{ker}}(\pi)$ and
$y\not\in{\rm{ker}}(\pi)$. Then, by (3.13) and (3.14), we have

$\delta(\pi)^{2}-4\epsilon(\pi)=-4(\pi(T_{1})-\pi(T_{1}^{-1}))^{2}+16\pi(T_{2})\pi(T_{2}^{-1}),$

and it then remains to proceed as in the proof of (C) to get (D). ∎

If $G={\rm{Dih}}(A)$, then (3) is vacuous and the condition
“$y\in{\rm{ker}}(\pi)$” in (2) is automatic. Moreover, if $T=\emptyset$, one
retrieves the characterization of integral undirected Cayley graphs over
${\rm{Dih}}(A)$ obtained in [HL21, Theorem 3.1]222Of course, via the
statements of the proof of Corollary 3.8, we may also derive [HL21, Theorem
3.1] from Corollary 3.5, and the same applies to the case of generalized
dicyclic groups.. On the other hand, the directed case is covered by the next
statement:

######  Corollary 3.9.

Let $A$ be a finite abelian group of exponent at least $3$ and let $S\subseteq
A\setminus\\{1\\}$ be such that $s^{-1}\in S$ for no $s\in S$. 333For a subset
$S=T_{1}\cup xT_{2}$ of ${\rm{Dih}}(A)\setminus\\{1\\}$ with
$T_{1},T_{2}\subseteq A$ and $s^{-1}\in S$ for no $s\in S$, we necessarily
have $T_{2}=\emptyset$ (and hence, $S=T_{1}\subseteq A$). Indeed, if there is
$a\in T_{2}$, we have $(xa)^{-1}=a^{-1}x^{-1}=x^{-1}a=xa\in xT_{2}\subseteq
S$, which cannot hold. Then the directed Cayley graph
${\rm{Cay}}({\rm{Dih}}(A),S)$ is integral if and only if
$\pi(S)-\pi(S^{-1})\in{\bf i}\mathbb{Z}$ for every one-dimensional
representation $\pi:A\rightarrow\mathbb{C}^{*}$ of $A$ with
$A^{2}\not\subseteq{\rm{ker}}(\pi)$.

######  Example 3.10.

As an illustration of our result, we now determine all integral directed
Cayley graphs over the dihedral group with $2p$ elements, where $p$ is any
given odd prime number. The following may be compared with [LHH18, Theorem
4.2], which classifies all integral undirected Cayley graphs over such finite
groups.

First, assume the exponent $e$ of $A$ is not divisible by 4. Then
${\rm{Cay}}({\rm{Dih}}(A),S)$ is integral if and only if $\pi(S)\in\mathbb{R}$
for every one-dimensional representation $\pi:A\rightarrow\mathbb{C}^{*}$ of
$A$ with $A^{2}\not\subseteq{\rm{ker}}(\pi)$.

Indeed, fix such a representation $\pi$ of $A$. If $\pi(S)\in\mathbb{R}$, then
$\pi(S)-\pi(S^{-1})=0\in{\bf i}\mathbb{Z}$. Conversely, assume
$\pi(S)-\pi(S^{-1})\in{\bf i}\mathbb{Z}$. Then there exists $a\in\mathbb{Z}$
such that ${\bf i}a=\pi(S)-\pi(S^{-1})$. Note that $\pi(S)-\pi(S^{-1})$ is an
element of the cyclotomic field $\mathbb{Q}(\zeta_{e})$. Assume $a\not=0$, in
which case we have ${\bf i}\in\mathbb{Q}(\zeta_{e})$. If $e$ is odd, we get a
contradiction (as $e$ and 4 are coprime) and, if $e$ is divisible by 2 (but
not by 4), we get ${\bf i}\in\mathbb{Q}(\zeta_{e})=\mathbb{Q}(\zeta_{e/2})$,
which cannot hold either. Hence, $a=0$, i.e.,
$\pi(S)=\pi(S^{-1})\in\mathbb{R}$.

Now, assume that $A$ is cyclic of prime order $p$ with $p$ odd. Fix a non-
empty subset $S$ of $A$ with $1\not\in S$ and $s^{-1}\in S$ for no $s\in S$,
and assume ${\rm{Cay}}({\rm{Dih}}(A),S)$ is integral. Choosing a root of unity
$\zeta$ of order $p$ and a generator $a$ of $A$, and considering the one-
dimensional representation $\pi$ of $A$ defined by $\pi(a)=\zeta$, we must
have $\pi(S)\in\mathbb{R}$. Setting $S=\\{a^{i_{1}},\dots,a^{i_{n}}\\}$ with
$n\geq 1$ and $i_{j}\in\\{1,\dots,p-1\\}$ for $j\in\\{1,\dots,n\\}$, we then
have

$\zeta^{i_{1}}+\cdots+\zeta^{i_{n}}=\zeta^{-i_{1}}+\cdots+\zeta^{-i_{n}}=\zeta^{p-i_{1}}+\cdots+\zeta^{p-i_{n}}.$

But $\zeta,\dots,\zeta^{p-1}$ are linearly independent over $\mathbb{Q}$ and,
since $1\leq p-i_{j}\leq p-1$ for $j\in\\{1,\dots,n\\}$, we get
$\\{i_{1},\dots,i_{n}\\}=\\{p-i_{1},\dots,p-i_{n}\\}$, which contradicts the
assumption that $s^{-1}\in S$ for no $s\in S$. Hence, the directed Cayley
graph ${\rm{Cay}}({\rm{Dih}}(A),S)$ is not integral if $S\not=\emptyset$, and
it is easily seen that the converse holds.

On the other hand, if the exponent of $A$ is divisible by 4, Corollary 3.9
yields examples of integral directed Cayley graphs over ${\rm{Dih}}(A)$. For
instance, one easily obtains that every directed Cayley graph over the
dihedral group with 8 elements is integral.

On the other hand, if $G={\rm{Dic}}(A,y)$ and $T=\emptyset$ in Corollary 3.8,
one retrieves the characterization of integral undirected Cayley graphs over
generalized dicyclic groups obtained in [BL22, Theorem 3.1]. The counterpart
for directed Cayley graphs is given by the following statement:

######  Corollary 3.11.

Let $A$ be a finite abelian group with even order and exponent at least $3$,
and let $y\in A$ be of order 2. Let
$S\subseteq{\rm{Dic}}(A,y)\setminus\\{1\\}$ be such that $s^{-1}\in S$ for no
$s\in S$. Set $S=T_{1}\cup xT_{2}$ with $T_{1},T_{2}\subseteq A$. Then the
directed Cayley graph ${\rm{Cay}}({\rm{Dic}}(A,y),S)$ is integral if and only
if the following two conditions hold:

(a) $\pi(T_{1})-\pi(T_{1}^{-1})\in{\bf i}\mathbb{Z}$ for every one-dimensional
representation $\pi:A\rightarrow\mathbb{C}^{*}$ of $A$ with
$A^{2}\not\subseteq{\rm{ker}}(\pi)$ and $y\in{\rm{ker}}(\pi),$

(b) for every one-dimensional representation $\pi:A\rightarrow\mathbb{C}^{*}$
of $A$ with $A^{2}\not\subseteq{\rm{ker}}(\pi)$ and $y\not\in{\rm{ker}}(\pi)$,
there exists $\alpha\in\mathbb{Z}$ such that
$4\pi(T_{2})\pi(T_{2}^{-1})-(\pi(T_{1})-\pi(T_{1}^{-1}))^{2}=\alpha^{2}.$

######  Example 3.12.

As an illustration of our result, we determine all integral directed Cayley
graphs over the dicyclic group with 8 elements.

To that end, set $A=\mathbb{Z}/4\mathbb{Z}$ and let $a$ be a generator of $A$,
in which case we necessarily have $a^{2}=y$. Let
$S\subseteq{\rm{Dic}}(A,y)\setminus\\{1\\}$ be such that $s^{-1}\in S$ for no
$s\in S$. Set $S=T_{1}\cup xT_{2}$ with $T_{1},T_{2}\subseteq A$. Given $b\in
A$, we have $(xb)^{-1}=x^{-1}b=xyb$. Hence, as $s^{-1}\in S$ for no $s\in S$,
we have

$T_{1}\in\\{\emptyset,\\{a\\},\\{a^{3}\\}\\}\quad{\rm{and}}\quad
T_{2}\in\\{\emptyset,\\{1\\},\\{a\\},\\{a^{2}\\},\\{a^{3}\\},\\{1,a\\},\\{1,a^{3}\\},\\{a,a^{2}\\},\\{a^{2},a^{3}\\}\\}.$

Now, let $\pi:A\rightarrow\mathbb{C}^{*}$ be a one-dimensional representation
of $A$ with $A^{2}\not\subseteq{\rm{ker}}(\pi)$. Since $A^{2}=\langle
y\rangle$, we have $y\not\in{\rm{ker}}(\pi)$, and hence, Condition (a) in
Corollary 3.11 is empty. We then have $\pi(y)=-1$, which yields
$\pi(a)=\pm{\bf i}$. In particular, we have

$(\pi(T_{1})-\pi(T_{1}^{-1}))^{2}=\left\\{\begin{array}[]{lll}0&{\rm{if}}&T_{1}=\emptyset\\\
-4&{\rm{if}}&|T_{1}|=1\end{array}\right.\quad{\rm{and}}\quad
4\pi(T_{2})\pi(T_{2}^{-1})=\left\\{\begin{array}[]{lll}0&{\rm{if}}&T_{2}=\emptyset\\\
4&{\rm{if}}&|T_{2}|=1\\\ 8&{\rm{if}}&|T_{2}|=2\\\ \end{array}\right..$

In view of Condition (b) of Corollary 3.11, we get that the directed Cayley
graph ${\rm{Cay}}({\rm{Dic}}(A,y),S)$ is integral if and only if either
$T_{1}\in\\{\\{a\\},\\{a^{3}\\}\\}$ and $T_{2}=\emptyset$, or
$T_{1}=\emptyset$ and
$T_{2}\in\\{\emptyset,\\{1\\},\\{a\\},\\{a^{2}\\},\\{a^{3}\\}\\}$.

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