Non-Isomorphic Cayley Graphs of Metacyclic Groups of Order 8p with the Same Spectrum

Abstract

The spectrum of a graph $\mathrm{\Gamma }$, denoted by $Spec\left(\mathrm{\Gamma }\right)$, is the multiset of eigenvalues of its adjacency matrix. A Cayley graph $Cay(G,S)$ of a finite group G is called Cay-DS (Cayley graph determined by its spectrum) if, for any other Cayley graph $Cay(G,T)$, $Spec\left(Cay\right(G,S\left)\right)=Spec\left(Cay\right(G,T\left)\right)$ implies $Cay(G,S)\cong Cay(G,T)$. A group G is said to be Cay-DS if all Cayley graphs of G are Cay-DS. An interesting open problem in the area of algebraic graph theory involves characterizing finite Cay-DS groups or constructing non-isomorphic Cayley graphs of a non-Cay-DS group that share the same spectrum. The present paper contributes to parts of this problem of metacyclic groups $M_{8p}$ of order $8p$ (with center of order 4), where p is an odd prime, in terms of irreducible characters, which are first presented. Then some new families of pairwise non-isomorphic Cayley graph pairs of $M_{8p}$ ($p\ge 5$) with the same spectrum are found. As a conclusion, this paper concludes that $M_{8p}$ is Cay-DS if and only if $p=3$.

1. Introduction

Let $\mathrm{\Gamma }$ be a connected graph. The (adjacency) spectrum of $\mathrm{\Gamma }$ is the multiset of eigenvalues of the adjacency matrix of $\mathrm{\Gamma }$, denoted by $Spec\left(\mathrm{\Gamma }\right)$. Two graphs are called isospectral if they share the same adjacency spectrum. It is obvious that two graphs are isospectral whenever they are isomorphic. Naturally, the problem of whether the converse setting holds comes to mind. A graph $\mathrm{\Gamma }$ is said to be determined by its spectrum (or DS for short) if, for any graph ${\mathrm{\Gamma }}^{\prime }$, $\mathrm{\Gamma }\cong {\mathrm{\Gamma }}^{\prime }$ whenever $Spec\left(\mathrm{\Gamma }\right)=Spec\left({\mathrm{\Gamma }}^{\prime }\right)$. In 1956, the following problem originated from chemistry was proposed:

Problem 1

([1]). Which graphs are DS?

Initially, it was widely accepted that every graph is DS until 1957, when Collatz et al. [2] constructed a pair of isospectral yet non-isomorphic trees. Schwenk [3] later established a stronger result: almost all trees are isospectral and, consequently, fail to be DS. Brooks et al. [4] generalized the Seidel switching, which is an operation on graphs preserving spectral properties under specific regularity conditions, to produce a family of isospectral 6-regular graphs. Meanwhile, loops and multiple edges are allowed in such graphs, which inherently prevents the definition of their complement. A nice advantage of Seidel switching lies in its application to simple graphs, which maintain isospectral complements of the original and of the resulting graph. Seress [5] demonstrated this property by constructing a family of isospectral 8-regular simple graphs on n vertices with isospectral complements. In subsequent years, the DS problem has attracted significant attention; one may refer to [6,7,8,9,10,11,12,13,14] for more details.

A particularly interesting class of graphs is known as Cayley graphs, a distinguished class of highly symmetric graphs. In algebraic graph theory, Cayley graphs over finite groups represent one of the most prosperous topics. Thus, they attract sustained and widespread interest of scholars. Let G be a finite group with identity element e. For an inverse-closed subset $S\subseteq G\setminus \left\{e\right\}$, the Cayley graph of G with respect to S, denoted by $Cay(G,S)$, is defined as the graph with the vertex set G, such that x, y are adjacent if and only if $x{y}^{-1}\in S$. It is well known that the Cayley graph $Cay(G,S)$ is connected if and only if S generates G. Throughout this paper, we consider only connected Cayley graphs. This paper contributes towards the DS problem for Cayley graphs. For a certain group G, the Cayley graph $Cay(G,S)$ is called Cay-DS if, for any other Cayley graph $Cay(G,T)$, $Spec\left(Cay\right(G,S\left)\right)=Spec\left(Cay\right(G,T\left)\right)$ implies $Cay(G,S)\cong Cay(G,T)$. The group G is Cay-DS if all Cayley graphs of G are Cay-DS. A DS Cayley graph is always Cay-DS, but the converse fails in general as suggested in [6]. Naturally, the following problems are considered.

Problem 2

(Cay-DS problem [6]). $\left(1\right)$ Which Cayley graphs are Cay-DS?

$\left(2\right)$ Which finite groups are Cay-DS?

Elspas and Turner [15] pioneered and constructed several pairs of isospectral but non-isomorphic circulant graphs. Babai [7] later generalized this by proving that, for any integer $k\ge 2$ and prime $p\ge 64k$, the dihedral group $D_{2p}$ admits k pairwise isospectral but non-isomorphic Cayley graphs. Abdollahi et al. [6,16] made much significant contribution to this problem, including providing infinite families of non-Cay-DS solvable groups and exhibiting $\left({\displaystyle \genfrac{}{}{0pt}{}{\frac{p-1}{2}}{\lfloor \frac{d}{2}\rfloor -3}}\right)$ ($p\ge 23$ is prime) pairs of isospectral but non-isomorphic Cayley graphs of $D_{2p}$. Huang et al. [17] demonstrated that a circulant graph whose order is either a prime power or the product of two distinct primes is Cay-DS if its generating set satisfies certain conditions. Lubotzky et al. [18] employed Bruhat–Tits building theory to present some isospectral but non-isomorphic Cayley graphs over $PSL(d,q)$. The Cay-DS problems of dicyclic groups $T_{4p}$ and dihedral groups $D_{6p}$ were investigated by Tang et al. [19] and Yang et al. [20], respectively. Despite these advances, a complete resolution to the Cay-DS problem remains open.

Let $M_{2ns}=\langle a,b:a^n=b^{2s}=e,b^{-1}ab=a^{-1}\rangle$ be a metacyclic group of order $2ns$$(n\ge 3,s\ge 1)$, which is non-abelian. In particular, $M_{2ns}=D_{2n}$ is a dihedral group if $s=1$; $M_{2ns}=T_{4n}$ is a dicyclic group if $s=2$. As we mentioned above, isospectral but non-isomorphic Cayley graphs of a dihedral group and a dicyclic group were investigated. Moreover, various spectra of a commuting graph, non-commuting graph, and commuting conjugacy class graph of the metacyclic groups $M_{2ns}$ were already obtained [21,22,23]. Our purpose here is for the case $s=4$ and the adjacency spectra of the Cayley graphs of metacyclic groups $M_{8n}$. The aim of the present paper is to contribute to parts of the Cay-DS problem of metacyclic groups $M_{8p}$ of order $8p$, where p is an odd prime, in terms of irreducible characters. After presenting some preliminary results, it first derives the irreducible characters of $M_{8n}$. Then as an application, it establishes a system of equations concerning the spectrum of Cayley graphs of $M_{8n}$. Next, building upon the experience of [6,19], it constructs an extensive family of pairwise non-isomorphic yet isospectral Cayley graphs pairs of $M_{8p}$ with $p\ge 5$. As a conclusion, this result shows that $M_{8p}$ is Cay-DS if and only if $p=3$, which offers new insights into the Cay-DS problem.

2. Preliminary Results

In this section, notations and technical results that will be used later are presented. For a detailed description of group representation theory, we refer to [24,25].

Lemma 1

([24]). The number of distinct irreducible characters of a finite group G coincides with the number of conjugacy classes of G.

Definition 1.

Let G be a finite group and let $\mathbb{C}$ be the field of complex numbers. Let ψ, φ be two functions from G to $\mathbb{C}$. Then the inner product of ψ, φ is defined by $\langle \psi ,\phi \rangle ={\displaystyle \frac{1}{\left|G\right|}}{\sum }_{g\in G}\psi \left(g\right)\overline{\phi \left(g\right)}.$

Lemma 2

([24]). Let G be a finite group and χ be a character of G. Then χ is irreducible if and only if $\langle \chi ,\chi \rangle =1$.

Definition 2.

Let G be a group. For arbitrary two elements $a,b\in G$, the commutator of a and b is defined by $[a,b]=a^{-1}{b}^{-1}ab$. The derived subgroup (or commutator subgroup) of G, denoted by $G^{\prime }$, is the subgroup generated by all commutators, i.e., $G^{\prime }=\langle [a,b]:a,b\in G\rangle$.

Lemma 3

([24]). The linear characters of a finite group G are exactly the lifts to G of the irreducible characters of the group $G/G^{\prime }$.

Lemma 4

([24]). The irreducible characters of the cyclic group $C_n=\langle a:a^n=1\rangle$ are ${\chi }_h$ with ${\chi }_h\left(a^k\right)={\epsilon }^{hk}$, where $h,k\in \{0,1,\dots ,n-1\}$, and $\epsilon =e^{{\displaystyle \frac{2\pi i}{n}}}$.

Lemma 5

([24]). Let G be a finite group and let ρ be a representation of G of degree two. Then ρ is irreducible if there exist two elements $g,h\in G$, such that $\rho \left(g\right)\rho \left(h\right)\ne \rho \left(h\right)\rho \left(g\right)$.

The next somewhat technical lemma is crucial for our study, which derives an expression for the spectrum of Cayley graphs in terms of irreducible characters of the finite group G.

Lemma 6

([7]). Let G be a finite group of order n with identity element e. Let ${\chi }_1,{\chi }_2,\dots ,{\chi }_h$ be all irreducible characters of G with the degree $d_k={\chi }_k\left(e\right)$$(k=1,\dots ,h)$. Then for any inverse-closed subset S of $G\setminus \left\{e\right\}$, the spectrum of the Cayley graph $Cay(G,S)$ can be arranged as

$$Spec\left(Cay(G,S)\right)=\{{\left[{\lambda }_{11}\right]}^{d_1},\dots ,{\left[{\lambda }_{1d_1}\right]}^{d_1},\dots ,{\left[{\lambda }_{h1}\right]}^{d_h}\dots ,{\left[{\lambda }_{h{d}_h}\right]}^{d_h}\}.$$

Furthermore, for any $k\in \{1,\dots ,h\}$ and any positive integer m,

$$\begin{array}{c}\hfill {\lambda }_{k1}^m+{\lambda }_{k2}^m+\dots +{\lambda }_{k{d}_k}^m=\sum _{s_1,s_2,\dots ,s_m\in S}{\chi }_k\left(\prod _{l=1}^m{s}_l\right).\end{array}$$

(1)

Using Equation (1) for $m=1,2,\dots ,d_k$, one obtains a system of equations, whose roots are ${\lambda }_{k1},{\lambda }_{k2},\dots ,{\lambda }_{k{d}_k}$. Now we apply the results to the metacyclic group $M_{8n}$, and we next always assume that

Hypothesis 1.

$M_{8n}=\langle a,b:a^n=b^8=e,b^{-1}ab=a^{-1}\rangle$ is a metacyclic group of order $8n$$(n\ge 3)$. Moreover, $M_{8n}=\{b^s{a}^t:s=0,1,\dots ,7;t=0,1,\dots ,n-1\}$.

It is straightforward to verify that

Lemma 7.

For all $b^s{a}^t,b^{s_1}{a}^{t_1},b^{s_2}{a}^{t_2}\in M_{8n}$, the following hold:

$\left(1\right)$$b^s{a}^t=a^t{b}^s$ if $s=0,2,4,6$; $b^s{a}^t=a^{-t}{b}^s$ if $s=1,3,5,7$.

$\left(2\right)$$b^{s_1}{a}^{t_1}{b}^{s_2}{a}^{t_2}=b^{s_1+s_2}{a}^{t_2+t_1}$ if $s_2=0,2,4,6$; $b^{s_1}{a}^{t_1}{b}^{s_2}{a}^{t_2}=b^{s_1+s_2}{a}^{t_2-t_1}$ if $s_2=1,3,5,7$.

$\left(3\right)$${\left(b^s{a}^t\right)}^{-1}=b^{-s}{a}^{-t}$ if $s=0,2,4,6$; ${\left(b^s{a}^t\right)}^{-1}=b^{-s}{a}^t$ if $s=1,3,5,7$.

Lemma 8.

Let $\epsilon =e^{2\pi i/n}$, $\omega =e^{\pi i/4}$, where $i^2=-1$.

$\left(1\right)$ If n is odd, then the $2n+6$ conjugacy classes of $M_{8n}$ are listed as follows:

$$\begin{array}{c}\left\{b^s\right\}(s=0,2,4,6),\{b^s{a}^t,b^s{a}^{-t}\}(s=0,2,4,6;t=1,2,\dots ,{\displaystyle \frac{n-1}{2}}),\hfill \\ \{b^s{a}^t:t=0,1,\dots ,n-1\}(s=1,3,5,7).\hfill \end{array}$$

The character table of $M_{8n}$ is shown in

Table 1. Character table of $M_{8n}$ for odd n.

${\mathit{g}}_{\mathit{i}}$	${\mathit{b}}^{\mathit{s}}{\mathit{a}}^{\mathit{t}}$$(\mathit{s}=0,2,4,6)$	${\mathit{b}}^{\mathit{s}}{\mathit{a}}^{\mathit{t}}$$(\mathit{s}=1,3,5,7)$
${\chi }_h$	${\omega }^{hs}$	${\omega }^{hs}$
${\psi }_{jk}$	${\omega }^{ks}({\epsilon }^{jt}+{\epsilon }^{-jt})$	0

, where ${\chi }_h$ and ${\psi }_{jk}$ are irreducible characters with respect to degree one and two, $h=0,1,\dots ,7$; $j=1,2,\dots ,{\displaystyle \frac{n-1}{2}}$; $k=0,1,2,3$.

$\left(2\right)$ If n is even, then the $2n+12$ conjugacy classes of $M_{8n}$ are listed as follows:

$$\begin{array}{c}\left\{b^s\right\}(s=0,2,4,6),\{b^s{a}^{\frac{n}{2}}\}(s=0,2,4,6),\hfill \\ \{b^s{a}^t,b^s{a}^{-t}\}(s=0,2,4,6;t=1,2,\dots ,\frac{n}{2}-1),\hfill \\ \{b^s{a}^t:t=1,3,5,\dots ,n-1\}(s=1,3,5,7),\hfill \\ \{b^s{a}^t:t=0,2,4,\dots ,n-2\}(s=1,3,5,7).\hfill \end{array}$$

The character table of $M_{8n}$ is shown in

Table 2. Character table of $M_{8n}$ for even n

${\mathit{g}}_{\mathit{i}}$	${\mathit{b}}^{\mathit{s}}{\mathit{a}}^{\mathit{t}}$$(\mathit{s}=0,2,4,6)$	${\mathit{b}}^{\mathit{s}}{\mathit{a}}^{\mathit{t}}$$(\mathit{s}=1,3,5,7)$
${\chi }_{hm}$	${\omega }^{hs}{(-1)}^{mt}$	${\omega }^{hs}{(-1)}^{mt}$
${\psi }_{jk}$	${\omega }^{ks}({\epsilon }^{jt}+{\epsilon }^{-jt})$	0

, where ${\chi }_h$ and ${\psi }_{jk}$ are irreducible characters with respect to degree one and two, $h=0,1,\dots ,7$; $m=0,1$; $j=1,2,\dots ,{\displaystyle \frac{n}{2}}-1$; $k=0,1,2,3$.

Proof. 

For arbitrary two elements $x=b^s{a}^t$, $g=b^u{a}^v\in M_{8n}$, it follows that

$$x^g={\left(b^u{a}^v\right)}^{-1}{b}^s{a}^t\left(b^u{a}^v\right)=b^s{a}^{-{(-1)}^{s}v+{(-1)}^{u}t+v}=\left\{\begin{array}{c}{b}^s{a}^{{(-1)}^{u}t}, if s=0,2,4,6;\hfill \\ b^s{a}^{{(-1)}^{u}t+2v}, if s=1,3,5,7.\hfill \end{array}\right.$$

Therefore, if n is odd, there are $2n+6$ conjugacy classes, as follows:

$$\begin{array}{c}\left\{b^s\right\}(s=0,2,4,6),\{b^s{a}^t,b^s{a}^{-t}\}(s=0,2,4,6;t=1,2,\dots ,{\displaystyle \frac{n-1}{2}}),\hfill \\ \{b^s{a}^t:t=0,1,\dots ,n-1\}(s=1,3,5,7).\hfill \end{array}$$

If n is even, there are $2n+12$ conjugacy classes, as follows:

$$\begin{array}{c}\left\{b^s\right\}(s=0,2,4,6),\{b^s{a}^{\frac{n}{2}}\}(s=0,2,4,6),\hfill \\ \{b^s{a}^t,b^s{a}^{-t}\}(s=0,2,4,6;t=1,2,\dots ,{\displaystyle \frac{n}{2}}-1),\hfill \\ \{b^s{a}^t:t=1,3,5,\dots ,n-1\}(s=1,3,5,7),\hfill \\ \{b^s{a}^t:t=0,2,4,\dots ,n-2\}(s=1,3,5,7).\hfill \end{array}$$

Since $[a,b]=a^{-1}{b}^{-1}ab=a^{-2}$, then the derived subgroup $M_{8n}^{\prime }=\langle a^{-2}\rangle =\langle a^2\rangle$. Therefore, if n is odd, then $M_{8n}^{\prime }=\langle a\rangle ,$ and $M_{8n}/M_{8n}^{\prime }=\{b^s\langle a\rangle :s=0,1,\dots 7\}\cong C_8.$ Let $\omega =e^{\pi i/4}$, where $i^2=-1$. Then Lemma 3 yields that the 8 linear characters of $M_{8n}$ are ${\chi }_h\left(b^s{a}^t\right)={\omega }^{hs}$, where $h=0,1,2,\dots ,7$. Since $\langle {\chi }_h,{\chi }_h\rangle =1$, then Lemma 2 shows that ${\chi }_h$ is irreducible.

If n is even, then $M_{8n}^{\prime }=\langle a^2\rangle$, $M_{8n}/M_{8n}^{\prime }=\{b^s\langle a^2\rangle ,b^{s}a\langle a^2\rangle :s=0,1,\dots 7\}\cong C_8\times C_2.$ Hence, the 16 linear characters of $M_{8n}$ are ${\chi }_{hm}\left(b^s{a}^t\right)={\omega }^{hs}{(-1)}^{mt}$, where $h=0,1,2,\dots ,7$ and $m=0,1$. Since $\langle {\chi }_{hm},{\chi }_{hm}\rangle =1$, then Lemma 2 shows that ${\chi }_{hm}$ is irreducible.

Define a map ${\rho }_{jk}$ ($k=0,1,2,3$, $j=1,2,\dots ,\lfloor {\displaystyle \frac{n+1}{2}}\rfloor -1$) from $M_{8n}$ to $GL(2,\mathbb{C})$ by

$${\rho }_{jk}\left(a\right)=\left(\begin{array}{cc}{\epsilon }^j& 0\\ 0& {\epsilon }^{-j}\end{array}\right),{\rho }_{jk}\left(b\right)=\left(\begin{array}{cc}0& {\omega }^k\\ {\omega }^k& 0\end{array}\right),{\rho }_{jk}\left(b^s{a}^t\right)={\rho }_{jk}{\left(b\right)}^s{\rho }_{jk}{\left(a\right)}^t,$$

where $\epsilon =e^{2\pi i/n}$. Then ${\rho }_{jk}$ is a representation of $M_{8n}$ since the following relations hold:

$$\begin{gathered} {\rho }_{jk}{\left(a\right)}^n={\left(\begin{array}{cc}{\epsilon }^j& 0\\ 0& {\epsilon }^{-j}\end{array}\right)}^n=\left(\begin{array}{cc}{\epsilon }^{nj}& 0\\ 0& {\epsilon }^{-nj}\end{array}\right)=I_2, \\ {\rho }_{jk}{\left(b\right)}^8={\left(\begin{array}{cc}0& {\omega }^k\\ {\omega }^k& 0\end{array}\right)}^8=\left(\begin{array}{cc}{\omega }^{8k}& 0\\ 0& {\omega }^{8k}\end{array}\right)=I_2, \\ {\rho }_{jk}\left(a\right){\rho }_{jk}\left(b\right)=\left(\begin{array}{cc}0& {\epsilon }^j{\omega }^k\\ {\epsilon }^{-j}{\omega }^k& 0\end{array}\right)={\rho }_{jk}\left(b\right){\rho }_{jk}\left(a^{-1}\right), \end{gathered}$$

where $I_2$ is the identity matrix of order two. Moreover,

$$\begin{array}{c}\hfill {\rho }_{jk}\left(b\right){\rho }_{jk}\left(a\right)=\left(\begin{array}{cc}0& {\epsilon }^{-j}{\omega }^k\\ {\epsilon }^j{\omega }^k& 0\end{array}\right)\ne {\rho }_{jk}\left(a\right){\rho }_{jk}\left(b\right)\end{array}$$

since $j\ne 0,{\displaystyle \frac{n}{2}}$. Hence, Lemma 5 yields that ${\rho }_{jk}$ is irreducible.

Let ${\psi }_{jk}$ be the character of ${\rho }_{jk}$. Then

$${\psi }_{jk}\left(b^s{a}^t\right)=Tr\left({\rho }_{jk}\left(b^s{a}^t\right)\right)=\left\{\begin{array}{c}{\omega }^{ks}({\epsilon }^{jt}+{\epsilon }^{-jt}),ifs=0,2,4,6,\hfill \\ 0,ifs=1,3,5,7,\hfill \end{array}\right.$$

which means that there are $2n-2$ irreducible characters with degree 2 if n is odd, and $2n-4$ irreducible characters with degree 2 if n is even.

Since $M_{8n}$ has $2n+6$ conjugacy classes if n is odd, $2n+12$ conjugacy classes if n is even, and we have produced $2n+6$ irreducible characters if n is odd, $2n+12$ irreducible characters if n is even, then Lemma 1 yields that we have now found out the complete character table of $M_{8n}$, as shown in Table 1 if n is odd, and shown in Table 2 if n is even. □

Next, we always assume that

Hypothesis 2.

$\left(1\right)$ The set of all spectrum of $Cay(M_{8n},S)$ is denoted by

$$\begin{array}{cc}& Spec\left(Cay(M_{8n},S)\right)\hfill \\ =& \{{[{\lambda }_h]}^1,{[{\mu }_{jk1}]}^2,{[{\mu }_{jk2}]}^2:h=0,1,\dots ,7;j=1,2,\dots ,{\displaystyle \frac{n-1}{2}};k=0,1,2,3\}\hfill \end{array}$$

if n is odd,

$$\begin{array}{cc}& Spec\left(Cay(M_{8n},S)\right)\hfill \\ =& \{{[{\lambda }_{hm}]}^1,{[{\mu }_{jk1}]}^2,{[{\mu }_{jk2}]}^2:h=0,1,\dots ,7;m=0,1;j=1,2,\dots ,{\displaystyle \frac{n}{2}}-1;k=0,1,2,3\}\hfill \end{array}$$

if n is even.

$\left(2\right)$ Let $M_1=\{b^s{a}^t\in M_{8p}:s=0,2,4,6\}$, $M_2=\{b^s{a}^t\in M_{8p}:s=1,3,5,7\}$.

It is straightforward to verify that

Lemma 9.

For $\omega =e^{\pi i/4}$, one obtains that

$${\omega }^0=1,{\omega }^1+{\omega }^7=\sqrt{2},{\omega }^2+{\omega }^6=0,{\omega }^3+{\omega }^5=-\sqrt{2},{\omega }^4=-1.$$

(2)

Let G be a group and let $\chi$ be a character of G. For two subsets $A,B$ of G, denote $\chi \left(A\right)={\sum }_{a\in A}\chi \left(a\right)$, $\chi \left(AB\right)={\sum }_{a\in A,b\in B}\chi \left(ab\right)$, $\chi \left(A^2\right)={\sum }_{a_1,a_2\in A}\chi \left(a_1{a}_2\right)$.

For an inverse-closed subset S of $M_{8n}\setminus \left\{e\right\}$, let

$$S_1=S\cap M_1=\{b^{s_j}{a}^{t_j}:j=1,2,\dots ,q\};S_2=S\cap M_2=\{b^{s_j^{\prime }}{a}^{t_j^{\prime }}:j=1,2,\dots ,r\}.$$

(3)

Then Lemma 7 yields that $S_1^{-1}=S_1$, $S_2^{-1}=S_2$.

In view of Lemmas 6–8, the spectrum of the Cayley graph $Cay(M_{8n},S)$ can be given by

$${\lambda }_h={\chi }_h\left(S_1\right)+{\chi }_h\left(S_2\right)=\sum _{u=1}^q{\omega }^{h{s}_u}+\sum _{u=1}^r{\omega }^{h{s}_u^{\prime }},$$

(4)

$${\lambda }_{hm}={\chi }_{hm}\left(S_1\right)+{\chi }_{hm}\left(S_2\right)=\sum _{u=1}^q{\omega }^{h{s}_u}{(-1)}^{m{t}_u}+\sum _{u=1}^r{\omega }^{h{s}_u^{\prime }}{(-1)}^{m{t}_u^{\prime }},$$

(5)

which means that

$${\lambda }_0=q+r,{\lambda }_1={\lambda }_7,{\lambda }_2={\lambda }_6,{\lambda }_3={\lambda }_5,{\lambda }_4=q-r;$$

(6)

$${\lambda }_{00}=q+r,{\lambda }_{1m}={\lambda }_{7m},{\lambda }_{2m}={\lambda }_{6m},{\lambda }_{3m}={\lambda }_{5m},{\lambda }_{40}=q-r;$$

(7)

and

$${\mu }_{j1}+{\mu }_{j2}=\psi \left(S_1\right)+\psi \left(S_2\right)=\psi \left(S_1\right)=\sum _{u=1}^q{\omega }^{k{s}_u}({\epsilon }^{j{t}_u}+{\epsilon }^{-j{t}_u});$$

(8)

$$\begin{array}{cc}& {\mu }_{j1}^2+{\mu }_{j2}^2=\psi \left(S_1^2\right)+\psi \left(S_2^2\right)+\psi \left(S_1{S}_2\right)+\psi \left(S_2{S}_1\right)=\psi \left(S_1^2\right)+\psi \left(S_2^2\right)\hfill \\ =& \sum _{u_1,u_2=1}^q{\omega }^{k(s_{u_1}+s_{u_2})}({\epsilon }^{j(t_{u_1}+t_{u_2})}+{\epsilon }^{-j(t_{u_1}+t_{u_2})})+\hfill \\ & \sum _{u_1,u_2=1}^r{\omega }^{k(s_{u_1}^{\prime }+s_{u_2}^{\prime })}({\epsilon }^{j(t_{u_1}^{\prime }-t_{u_2}^{\prime })}+{\epsilon }^{j(t_{u_2}^{\prime }-t_{u_1}^{\prime })}).\hfill \end{array}$$

(9)

Hypothesis 3.

Let $I=\{0,1,\dots ,n-1\}$. For arbitrary $m\in I$ and a non-empty integer set A, let ${\beta }_A(n,m)$ denote the number of solutions to the congruence equation

$$x-y\equiv m(modn),x,y\in A.$$

Lemma 10.

Let n be odd. Let S and T be two inverse-closed subsets of $M_{8n}\setminus \left\{e\right\}$, and let $S_{M_1}=\{(s,t):b^s{a}^t\in S\cap M_1\}$, $S_{M_2.1}=\{s:b^s{a}^t\in S\cap M_2\}$, $S_{M_2.2}=\{t:b^s{a}^t\in S\cap M_2\}$, $T_{M_1}=\{(s,t):b^s{a}^t\in T\cap M_1\}$, $T_{M_2.1}=\{s:b^s{a}^t\in T\cap M_2\}$, $T_{M_2.2}=\{t:b^s{a}^t\in T\cap M_2\}$. Then the Cayley graphs $Cay(M_{8n},S)$ and $Cay(M_{8n},T)$ are isospectral if $S_{M_1}=T_{M_1}$, $S_{M_2.1}=T_{M_2.1}=\{1,7\}$ or $\{3,5\}$ and ${\beta }_{S_{M_2.2}}(n,m)={\beta }_{T_{M_2.2}}(n,m)$ holds for each $m\in \{0,1,\dots ,n-1\}$.

Proof. 

Let

$$\begin{array}{cc}& Spec\left(Cay(M_{8n},S)\right)\hfill \\ =& \{{[{\lambda }_h]}^1,{[{\mu }_{jk1}]}^2,{[{\mu }_{jk2}]}^2:h=0,1,\dots ,7;j=1,2,\dots ,{\displaystyle \frac{n-1}{2}};k=0,1,2,3\};\hfill \\ & Spec\left(Cay(M_{8n},T)\right)\hfill \\ =& \{{[{\lambda }_h^{\prime }]}^1,{[{\mu }_{jk1}^{\prime }]}^2,{[{\mu }_{jk2}^{\prime }]}^2:h=0,1,\dots ,7;j=1,2,\dots ,{\displaystyle \frac{n-1}{2}};k=0,1,2,3\}.\hfill \end{array}$$

From Equations (4), (6), and (8), ${\lambda }_h={\lambda }_h^{\prime }$ and ${\mu }_{j1}+{\mu }_{j2}={\mu }_{j1}^{\prime }+{\mu }_{j2}^{\prime }$ if $S_{M_1}=T_{M_1}$, $S_{M_2.1}=T_{M_2.1}=\{1,7\}$ or $\{3,5\}$. Moreover, Equation (9) yields that ${\mu }_{j1}^2+{\mu }_{j2}^2={{\mu }_{j1}^{\prime }}^2+{{\mu }_{j2}^{\prime }}^2$ if $\psi \left(S_2^2\right)=\psi \left(T_2^2\right)$, where $S_2=S\cap M_2$ and $T_2=T\cap M_2$. Hence, ${\mu }_{j1}^2+{\mu }_{j2}^2={{\mu }_{j1}^{\prime }}^2+{{\mu }_{j2}^{\prime }}^2$ if ${\beta }_{S_{M_2.2}}(n,m)={\beta }_{T_{M_2.2}}(n,m)$ holds for each $m\in \{0,1,\dots ,n-1\}$. Therefore, $Spec\left(Cay(M_{8n},S)\right)=Spec\left(Cay(M_{8n},T)\right)$ if $S_{M_1}=T_{M_1}$, $S_{M_2.1}=T_{M_2.1}=\{1,7\}$ or $\{3,5\}$ and ${\beta }_{S_{M_2.2}}(n,m)={\beta }_{T_{M_2.2}}(n,m)$ holds for each $m\in \{0,1,\dots ,n-1\}$. □

Lemma 11

([26]). Let p be an odd prime. Then $Cay(M_{8p},S)\cong Cay(M_{8p},T)$ if and only if there exists an element $\alpha \in Aut\left(M_{8p}\right)$, such that $S^{\alpha }=T$.

Lemma 12.

For an odd prime p, the automorphism group of $M_{8p}=\{b^s{a}^t:s=0,1,\dots ,7; t=0,1,\dots ,p-1\}$ is

$$Aut\left(M_{8p}\right)=\{{\alpha }_{f,g,h}:f=1,3,5,7;g=1,2,\dots ,p-1;h=0,1,\dots ,p-1\},$$

where

$${\alpha }_{f,g,h}\left(b^s{a}^t\right)=\left\{\begin{array}{c}{b}^{sf}{a}^{tg+h},ifs=1,3,5,7;\hfill \\ b^{sf}{a}^{tg},ifs=0,2,4,6.\hfill \end{array}\right.$$

(10)

Proof. 

It is well known that a bijection $\alpha$ from $M_{8p}$ to $M_{8p}$ is a homomorphism if and only if $o\left(\alpha \right(a\left)\right)=o\left(a\right)$, $o\left(\alpha \right(b\left)\right)=o\left(b\right)$, and $\alpha \left(b^{-1}\right)\alpha \left(a\right)\alpha \left(b\right)=\alpha {\left(a\right)}^{-1}$. Note that $o\left(a^t\right)=p$, $o\left(b^s\right)=o\left(b^s{a}^t\right)=8$, $o\left(b^2\right)=o\left(b^6\right)=4$, $o\left(b^2{a}^t\right)=o\left(b^6{a}^t\right)=4p$, $o\left(b^4\right)=2$, $o\left(b^4{a}^t\right)=2p$ for all $s=1,3,5,7$ and $t=1,\dots ,p-1$. Meanwhile, ${\left(b^s{a}^t\right)}^{-1}{a}^r\left(b^s{a}^t\right)=a^{-r}$ for all $s=1,3,5,7$ and $r,t=0,1,\dots ,p-1$. Hence,

$$\begin{array}{c}\alpha \left(a\right)\in \{a^t:t=1,\dots ,p-1\},\hfill \\ \alpha \left(b\right)\in \{b^s{a}^t:s=1,3,5,7andt=0,1,\dots ,p-1\},\hfill \end{array}$$

which means that $|Aut\left(M_{8p}\right)|\le 4p(p-1)$.

Define a map ${\alpha }_{f,g,h}$ from $M_{8p}$ to $M_{8p}$ by Equation (10). It is straightforward to verify that ${\alpha }_{f,g,h}$ is a bijection. Since ${\alpha }_{f,g,h}\left(a\right)=a^{tg}$ and ${\alpha }_{f,g,h}\left(b\right)=b^f{a}^h$, it follows that

$$\begin{array}{cc}& {\alpha }_{f,g,h}{\left(b\right)}^s{\alpha }_{f,g,h}{\left(a\right)}^t\hfill \\ =& {\left(b^f{a}^h\right)}^s{a}^{tg}\hfill \\ =& \left\{\begin{array}{c}{b}^{sf}{a}^{tg+h},ifs=1,3,5,7\hfill \\ b^{sf}{a}^{tg},ifs=0,2,4,6\hfill \end{array}\right.\hfill \\ =& {\alpha }_{f,g,h}\left(b^s{a}^t\right),\hfill \end{array}$$

which means that ${\alpha }_{f,g,h}$ is a homomorphism. Then ${\alpha }_{f,g,h}\in Aut\left(M_{8p}\right)$. Therefore,

$$G=\{{\alpha }_{f,g,h}:f=1,3,5,7;g=1,2,\dots ,p-1;h=0,1,\dots ,p-1\}\le Aut\left(M_{8p}\right),$$

which means that $\left|G\right|=4p(p-1)\le |Aut\left(M_{8p}\right)|\le 4p(p-1)$. Hence, $G=Aut\left(M_{8p}\right)$. □

Fact 1.

Let ${\mathrm{\Gamma }}_1$, ${\mathrm{\Gamma }}_2$ be two simple (without loops and multiple edges) connected graphs, and let $\overline{{\mathrm{\Gamma }}_1}$, $\overline{{\mathrm{\Gamma }}_2}$ be their complements, respectively. Then ${\mathrm{\Gamma }}_1$ and ${\mathrm{\Gamma }}_2$ are isospectral but non-isomorphic if and only if $\overline{{\mathrm{\Gamma }}_1}$ and $\overline{{\mathrm{\Gamma }}_2}$ are isospectral but non-isomorphic.

3. Isospectral Cayley Graphs of ${\mathit{M}}_{\mathbf{8}\mathit{p}}$ with $\mathit{p}\ge \mathbf{13}$

In this section, we will always assume that

Hypothesis 4.

$\left(1\right)$ $p\ge 13$ be a prime number;

$\left(2\right)$$S_a=\{0,1,2,6,8,11\}$, $S=\{b{a}^t,b^7{a}^t:t\in \left\{S_a\right\}\}$;

$\left(3\right)$$T_a=\{0,2,4,5,10,11\}$, $T=\{b{a}^t,b^7{a}^t:t\in \left\{T_a\right\}\}$;

$\left(4\right)$$S_a^{\prime }=\{0,1,5,7,8,10,12\}$, $S^{\prime }=\{b{a}^t,b^7{a}^t:t\in \left\{S_a^{\prime }\right\}\}$;

$\left(5\right)$$T_a^{\prime }=\{0,1,2,5,7,9,12\}$, $T^{\prime }=\{b{a}^t,b^7{a}^t:t\in \left\{T_a^{\prime }\right\}\}$;

$\left(6\right)$$\mathbb{U}$ is a partition of the set $\mathbb{M}=\{b^s{a}^t:s=0,2,4,6;t=0,1,2,\dots ,p-1\}\setminus \{e,b^4\}$ determined by the equivalence relation ∼: for $x,y\in \mathbb{M}$, $x\sim y$ if and only if $xy=1$ or $x=y$, i.e.,

$$\begin{array}{c}\hfill \mathbb{U}=\{\left\{a^{\pm t_1}\right\},\left\{b^4{a}^{\pm t_1}\right\},\{b^2{a}^{t_2},b^6{a}^{-t_2}\}:t_1=1,\dots ,{\displaystyle \frac{p-1}{2}};t_2=0,1,2,\dots ,p-1\}.\end{array}$$

Lemma 13.

For arbitrary $\alpha \in Aut\left(M_{8p}\right)$, $S^{\alpha }\ne T$ and ${S^{\prime }}^{\alpha }\ne T^{\prime }$.

Proof. 

Suppose on the contrary that there exists $\alpha ={\alpha }_{f,g,h}\in Aut\left(M_{8p}\right)$, such that $S^{\alpha }=T$, which means that there exist some suitable integers $f\in \{1,3,5,7\}$, $g\in \{1,2,\dots ,p-1\}$ and $h\in \{0,1,2,\dots ,p-1\}$, such that

$$\begin{array}{cc}\hfill T=S^{\alpha }=& \{b^f{a}^h,b^f{a}^{g+h},b^f{a}^{2g+h},b^f{a}^{6g+h},b^f{a}^{8g+h},b^f{a}^{11g+h},\hfill \\ & b^{7f}{a}^h,b^{7f}{a}^{g+h},b^{7f}{a}^{2g+h},b^{7f}{a}^{6g+h},b^{7f}{a}^{8g+h},b^{7f}{a}^{11g+h}\}.\hfill \end{array}$$

Let $r_1=h,r_2=g+h,r_3=2g+h,r_4=6g+h,r_5=8g+h,r_6=11g+h.$ Let ${\left(S^{\alpha }\right)}_a=\{r_j$$(modp):1\le j\le 6\}$. Then ${\left(S^{\alpha }\right)}_a=T_a$. It is straightforward to verify that the set of all triples $(x,y,z)\in T_a^3$ satisfying $2x=y+z$ is $(2,0,4)$, $(2,4,0)$, $(5,10,0)$, and $(5,0,10)$. Therefore, precisely one of y or z must be 0. Meanwhile, the set of all triples $(x,y,z)\in {{\left(S^{\alpha }\right)}_a}^3$ satisfying $2x=y+z$ is $(r_2,r_1,r_3)$, $(r_2,r_3,r_1)$, $(r_4,r_2,r_6)$, and $(r_4,r_6,r_2)$, which contradicts $T_a={\left(S^{\alpha }\right)}_a$. Therefore, $S^{\alpha }\ne T$ holds for arbitrary $\alpha \in Aut\left(M_{8p}\right)$.

Similarly, if there exists $\alpha ={\alpha }_{f,g,h}\in Aut\left(M_{8p}\right)$, such that $S^{\prime \alpha }=T^{\prime }$, then there exist some suitable integers $f\in \{1,3,5,7\}$, $g\in \{1,2,\dots ,p-1\}$, and $h\in \{0,1,2,\dots ,p-1\}$, such that

$$\begin{array}{cc}\hfill T^{\prime }={S^{\prime }}^{\alpha }=& \{b^f{a}^h,b^f{a}^{g+h},b^f{a}^{5g+h},b^f{a}^{7g+h},b^f{a}^{8g+h},b^f{a}^{10g+h},b^f{a}^{12g+h},\hfill \\ & b^{7f}{a}^h,b^{7f}{a}^{g+h},b^{7f}{a}^{5g+h},b^{7f}{a}^{7g+h},b^{7f}{a}^{8g+h},b^{7f}{a}^{10g+h},b^{7f}{a}^{12g+h}\}.\hfill \end{array}$$

Let $r_1=h,r_2=g+h,r_3=5g+h,r_4=7g+h,r_5=8g+h,r_6=10g+h,r_7=12g+h$. Let ${\left({S^{\prime }}^{\alpha }\right)}_a=\{r_j(modp):1\le j\le 7\}$. Then ${\left({S^{\prime }}^{\alpha }\right)}_a=T_a^{\prime }$. After some elementary calculation, one obtains that all four possible solutions to the equation $2x=y+z$ in ${\left({S^{\prime }}^{\alpha }\right)}_a$ are $(r_6,r_5,r_7)$, $(r_6,r_7,r_5)$, $(r_3,r_6,r_1)$ and $(r_3,r_1,r_6)$, while there are six possible solutions in $T_a^{\prime }$: $(1,0,2)$, $(1,2,0)$, $(5,1,9)$, $(5,9,1)$, $(7,12,2)$, and $(7,2,12)$, which contradicts ${\left({S^{\prime }}^{\alpha }\right)}_a=T_a^{\prime }$. Therefore, ${S^{\prime }}^{\alpha }\ne T^{\prime }$ holds for arbitrary $\alpha \in Aut\left(M_{8p}\right)$. □

Combining this result with Lemma 11, one obtains that

Corollary 1.

$Cay(M_{8p},S)\ncong Cay(M_{8p},T)$, $Cay(M_{8p},S^{\prime })\ncong Cay(M_{8p},T^{\prime })$.

Lemma 14.

For arbitrary $\alpha ={\alpha }_{f,g,h}\in Aut\left(M_{8p}\right)$, $S^{\alpha }=S$ (or $T^{\alpha }=T$, or ${S^{\prime }}^{\alpha }=S^{\prime }$, or ${T^{\prime }}^{\alpha }=T^{\prime }$) if and only if $f=1$ or 7, $g=1$ and $h=0$.

Proof. 

If $f=1$ or 7, $g=1$ and $h=0$, then, obviously, $S^{\alpha }=S$, $T^{\alpha }=T$, ${S^{\prime }}^{\alpha }=S^{\prime }$ and ${T^{\prime }}^{\alpha }=T^{\prime }$.

For arbitrary $\alpha ={\alpha }_{f,g,h}\in Aut\left(M_{8p}\right)$, if

$$\begin{array}{cc}S=S^{\alpha }=\hfill & \{b^f{a}^h,b^f{a}^{g+h},b^f{a}^{2g+h},b^f{a}^{6g+h},b^f{a}^{8g+h},b^f{a}^{11g+h},\hfill \\ & b^{7f}{a}^h,b^{7f}{a}^{g+h},b^{7f}{a}^{2g+h},b^{7f}{a}^{6g+h},b^{7f}{a}^{8g+h},b^{7f}{a}^{11g+h}\},\hfill \end{array}$$

then $f=1$ or 7. Additionally, let $r_1=h,r_2=g+h,r_3=2g+h,r_4=6g+h,r_5=8g+h$, $r_6=11g+h$. Let ${\left(S^{\alpha }\right)}_a=\{r_j(modp):1\le j\le 6\}$. Then ${\left(S^{\alpha }\right)}_a=S_a=\{0,1,2,6,8,11\}$. It follows that all possible solutions $(x,y,z)$ to the equation $2x=y+z$ in ${\left(S^{\alpha }\right)}_a$ are $(r_2,r_1,r_3)$, $(r_2,r_3,r_1)$, $(r_4,r_2,r_6)$, and $(r_4,r_6,r_2)$, which means that exactly one of x, y, z must be $r_2$. Meanwhile, all possible solutions $(x,y,z)$ to the equation $2x=y+z$ in $S_a$ are $(1,0,2)$, $(1,2,0)$, $(6,11,1)$, and $(6,1,11)$, and exactly one of x, y, z must be 1. Since ${\left(S^{\alpha }\right)}_a=S_a$, then $r_2=1$ and $\{r_2,r_4\}=\{1,6\}$, which means that $r_4=6$. Therefore, $g+h=1,6g+h=6$. Hence, $g=1$ and $h=0$.

Similarly, if $\alpha ={\alpha }_{f,g,h}\in Aut\left(M_{8p}\right)$ and $T^{\alpha }=T$, or ${S^{\prime }}^{\alpha }=S^{\prime }$, or ${T^{\prime }}^{\alpha }=T^{\prime }$, then $f=1$ or 7, $g=1$ and $h=0$. □

Lemma 15.

The Cayley graphs $Cay(M_{8p},S)$ and $Cay(M_{8p},T)$ are connected 12-regular, isospectral, but non-isomorphic.

Proof. 

Equations (4), (6), (8), and (9) yield that

$$\begin{array}{c}\hfill Spec\left(Cay(M_{8p},S)\right)=\{{[\pm 12]}^1,{[\pm 6\sqrt{2}]}^2,{\left[0\right]}^2,{[\pm f_{jk}\left(S_a\right)]}^2:k=0,1,2,3;j=1,2,\dots ,{\displaystyle \frac{n-1}{2}}\},\\ \hfill Spec\left(Cay(M_{8p},T)\right)=\{{[\pm 12]}^1,{[\pm 6\sqrt{2}]}^2,{\left[0\right]}^2,{[\pm f_{jk}\left(T_a\right)]}^2:k=0,1,2,3;j=1,2,\dots ,{\displaystyle \frac{n-1}{2}}\},\end{array}$$

where $f_{jk}\left(A\right)=\sqrt{-{\displaystyle \frac{1}{2}}(2+{\omega }^{2k}+{\omega }^{-2k}){\sum }_{u_1,u_2\in A}({\epsilon }^{j(u_1-u_2)}+{\epsilon }^{j(u_2-u_1)})}.$ Therefore, the Cayley graphs $Cay(M_{8p},S)$ and $Cay(M_{8p},T)$ are isospectral since ${\beta }_{S_a}(p,m)={\beta }_{T_a}(p,m)$ holds for each $m\in \{0,1,\dots ,p-1\}$. Corollary 1 yields that $Cay(M_{8p},S)\ncong Cay(M_{8p},T)$. Therefore, the results hold. □

Lemma 16.

The Cayley graphs $Cay(M_{8p},S^{\prime })$ and $Cay(M_{8p},T^{\prime })$ are connected 14-regular, isospectral, but non-isomorphic.

Proof. 

Equations (4), (6), (8), (9) yield that

$$\begin{array}{c}\hfill Spec\left(Cay(M_{8p},S)\right)=\{{[\pm 14]}^1,{[\pm 7\sqrt{2}]}^2,{\left[0\right]}^2,{[\pm f_{jk}\left(S_a^{\prime }\right)]}^2:k=0,1,2,3;j=1,2,\dots ,{\displaystyle \frac{n-1}{2}}\},\\ \hfill Spec\left(Cay(M_{8p},T)\right)=\{{[\pm 14]}^1,{[\pm 7\sqrt{2}]}^2,{\left[0\right]}^2,{[\pm f_{jk}\left(T_a^{\prime }\right)]}^2:k=0,1,2,3;j=1,2,\dots ,{\displaystyle \frac{n-1}{2}}\},\end{array}$$

where $f_{jk}\left(A\right)=\sqrt{-{\displaystyle \frac{1}{2}}(2+{\omega }^{2k}+{\omega }^{-2k}){\sum }_{u_1,u_2\in A}({\epsilon }^{j(u_1-u_2)}+{\epsilon }^{j(u_2-u_1)})}.$ Therefore, the Cayley graphs $Cay(M_{8p},S)$ and $Cay(M_{8p},T)$ are isospectral since ${\beta }_{S_a}(p,m)={\beta }_{T_a}(p,m)$ holds for each $m\in \{0,1,\dots ,p-1\}$. Corollary 1 yields that $Cay(M_{8p},S^{\prime })\ncong Cay(M_{8p},T^{\prime })$. Therefore, the results hold. □

Corollary 2.

Let U be a $2k$-subset of $M_{8p}$, which consists of the union of arbitrary k members of $\mathbb{U}$. Then

$\left(1\right)$

$Cay(M_{8p},S\cup U)$ and $Cay(M_{8p},T\cup U)$ are connected $12+2k$-regular, isospectral, but non-isomorphic;

$\left(2\right)$

$Cay(M_{8p},S\cup U\cup \left\{b^4\right\})$ and $Cay(M_{8p},T\cup U\cup \left\{b^4\right\})$ are connected $13+2k$-regular, isospectral, but non-isomorphic;

$\left(3\right)$

$Cay(M_{8p},S^{\prime }\cup U)$ and $Cay(M_{8p},T^{\prime }\cup U)$ are connected $14+2k$-regular, isospectral, but non-isomorphic;

$\left(4\right)$

$Cay(M_{8p},S^{\prime }\cup U\cup \left\{b^4\right\})$ and $Cay(M_{8p},T^{\prime }\cup U\cup \left\{b^4\right\})$ are connected $15+2k$-regular, isospectral, but non-isomorphic.

Proof. 

(1) If $Cay(M_{8p},U\cup S)\cong Cay(M_{8p},U\cup T)$, then there exists $\alpha \in Aut\left(M_{8p}\right)$ such that ${(U\cup S)}^{\alpha }=U\cup T$. Since $\alpha$ preserves the order of elements of $M_{8p}$, then $U^{\alpha }=U$, $S^{\alpha }=T$, which is impossible by Lemma 13. Hence, $Cay(M_{8p},U\cup S)\ncong Cay(M_{8p},U\cup T)$ follows from Lemma 11. Meanwhile, Lemma 10 and the proof of Lemma 15 yield that the Cayley graphs $Cay(M_{8p},U\cup S)$ and $Cay(M_{8p},U\cup T)$ are isospectral.

Proceeding as in above, (2), (3), and (4) hold. □

Theorem 1.

Let d be an integer such that $12\le d\le 2p+13$ and $\overline{d}=8p-d-1$. Then, for each d, there exist $\left({\displaystyle \genfrac{}{}{0pt}{}{2p}{\lfloor \frac{d}{2}\rfloor -6}}\right)$ pairs d-regular ($\overline{d}$-regular) Cayley graphs of $M_{8p}$ that are pairwise isospectral but non-isomorphic.

Proof. 

Case 1: $d\in \{12,13,4p+12,4p+13\}$. Then $\left({\displaystyle \genfrac{}{}{0pt}{}{2p}{\lfloor \frac{d}{2}\rfloor -6}}\right)=1$. Let U be a $(4p-2)$-subset of $M_{8p}$, which consists of the union of all $2p-1$ members of $\mathbb{U}$. Then Lemma 15 and Corollary 2 yield that the Cayley graphs $Cay(M_{8p},S)$ and $Cay(M_{8p},T)$ are connected 12-regular, isospectral, but non-isomorphic; $Cay(M_{8p},S\cup \left\{b^4\right\})$ and $Cay(M_{8p},T\cup \left\{b^4\right\})$ are connected 13-regular, isospectral, but non-isomorphic; $Cay(M_{8p},S^{\prime }\cup U)$ and $Cay(M_{8p},T^{\prime }\cup U)$ are connected $(4p+12)$-regular, isospectral, but non-isomorphic; and $Cay(M_{8p},S^{\prime }\cup U\cup \left\{b^4\right\})$ and $Cay(M_{8p},T^{\prime }\cup U\cup \left\{b^4\right\})$ are connected $(4p+13)$-regular, isospectral, but non-isomorphic.

Case 2: $14\le d\le 4p+11$. If d is even (odd, respectively), then assume that U, V are two distinct $(d-12)$-subsets ($(d-13)$-subsets, respectively) of $M_{8p}$, which consist of the union of $({\displaystyle \frac{d}{2}}-6)$ ($({\displaystyle \frac{d-1}{2}}-6)$, respectively) members of $\mathbb{U}$. Let

$$\begin{array}{c}{\mathrm{\Gamma }}_{S,U}=Cay(M_{8p},S\cup U)({\mathrm{\Gamma }}_{S,U}=Cay(M_{8p},S\cup \left\{b^4\right\}\cup U),\mathrm{respectively});\hfill \\ {\mathrm{\Gamma }}_{T,U}=Cay(M_{8p},T\cup U)({\mathrm{\Gamma }}_{T,U}=Cay(M_{8p},T\cup \left\{b^4\right\}\cup U),\mathrm{respectively});\hfill \\ {\mathrm{\Gamma }}_{S,V}=Cay(M_{8p},S\cup V)({\mathrm{\Gamma }}_{S,V}=Cay(M_{8p},S\cup \left\{b^4\right\}\cup V),\mathrm{respectively});\hfill \\ {\mathrm{\Gamma }}_{T,V}=Cay(M_{8p},T\cup V)({\mathrm{\Gamma }}_{T,V}=Cay(M_{8p},T\cup \left\{b^4\right\}\cup V),\mathrm{respectively}).\hfill \end{array}$$

It is straightforward to verify that the Cayley graphs ${\mathrm{\Gamma }}_{S,U}$, ${\mathrm{\Gamma }}_{T,U}$, ${\mathrm{\Gamma }}_{S,V}$, and ${\mathrm{\Gamma }}_{T,V}$ are connected and d-regular.

Lemma 11 yields that if ${\mathrm{\Gamma }}_{S,U}\cong {\mathrm{\Gamma }}_{T,V}$, then there exists a automorphism $\alpha \in Aut\left(M_{8p}\right)$ such that ${(S\cup U)}^{\alpha }=T\cup V$ (${(S\cup \left\{b^4\right\}\cup U)}^{\alpha }=T\cup \left\{b^4\right\}\cup V$, respectively), and then $S^{\alpha }=T$ and $U^{\alpha }=V$. Hence, Lemma 13 shows that ${\mathrm{\Gamma }}_{S,U}\ncong {\mathrm{\Gamma }}_{T,V}$. Similarly, ${\mathrm{\Gamma }}_{S,V}\ncong {\mathrm{\Gamma }}_{T,U}$.

Lemma 11 yields that if ${\mathrm{\Gamma }}_{S,U}\cong {\mathrm{\Gamma }}_{S,V}$, then there exists a automorphism $\alpha \in Aut\left(M_{8p}\right)$ such that ${(S\cup U)}^{\alpha }=S\cup V$ (${(S\cup \left\{b^4\right\}\cup U)}^{\alpha }=S\cup \left\{b^4\right\}\cup V$, respectively), and then $S^{\alpha }=S$ and $U^{\alpha }=V$. By Lemma 14, $S^{\alpha }=S$ if and only if $\alpha ={\alpha }_{7,1,0}$ or ${\alpha }_{1,1,0}$. However, $U^{{\alpha }_{7,1,0}}=U^{{\alpha }_{1,1,0}}=U$ since ${\alpha }_{7,1,0}\left(b^s{a}^t\right)=b^{-s}{a}^t$ for $s=0,2,4,6$, which means that ${\mathrm{\Gamma }}_{S,U}\ncong {\mathrm{\Gamma }}_{S,V}$. Similarly, ${\mathrm{\Gamma }}_{T,U}\ncong {\mathrm{\Gamma }}_{T,V}$.

Moreover, Corollary 2 shows that ${\mathrm{\Gamma }}_{S,U}\ncong {\mathrm{\Gamma }}_{T,U}$, ${\mathrm{\Gamma }}_{S,V}\ncong {\mathrm{\Gamma }}_{T,V}$. Hence, d-regular Cayley graphs in the set $\{{\mathrm{\Gamma }}_{S,U},{\mathrm{\Gamma }}_{T,U},{\mathrm{\Gamma }}_{S,V},{\mathrm{\Gamma }}_{T,V}\}$ are pairwise non-isomorphic.

Assume that $U^{\prime }$, $V^{\prime }$ are two distinct $(d-14)$-subsets ($(d-15)$-subsets, respectively) of $M_{8p}$, which consist of the union of $({\displaystyle \frac{d}{2}}-7)$ ($({\displaystyle \frac{d-1}{2}}-7)$, respectively) members of $\mathbb{U}$. Let

$$\begin{array}{c}{\mathrm{\Gamma }}_{S^{\prime },U^{\prime }}=Cay(M_{8p},S^{\prime }\cup U^{\prime })({\mathrm{\Gamma }}_{S^{\prime },U^{\prime }}=Cay(M_{8p},S^{\prime }\cup \left\{b^4\right\}\cup U^{\prime }),\mathrm{respectively});\hfill \\ {\mathrm{\Gamma }}_{T^{\prime },U^{\prime }}=Cay(M_{8p},T^{\prime }\cup U^{\prime })({\mathrm{\Gamma }}_{T^{\prime },U^{\prime }}=Cay(M_{8p},T^{\prime }\cup \left\{b^4\right\}\cup U^{\prime }),\mathrm{respectively});\hfill \\ {\mathrm{\Gamma }}_{S^{\prime },V^{\prime }}=Cay(M_{8p},S^{\prime }\cup V^{\prime })({\mathrm{\Gamma }}_{S^{\prime },V^{\prime }}=Cay(M_{8p},S^{\prime }\cup \left\{b^4\right\}\cup V^{\prime }),\mathrm{respectively});\hfill \\ {\mathrm{\Gamma }}_{T^{\prime },V^{\prime }}=Cay(M_{8p},T^{\prime }\cup V^{\prime })({\mathrm{\Gamma }}_{T^{\prime },V^{\prime }}=Cay(M_{8p},T^{\prime }\cup \left\{b^4\right\}\cup V^{\prime }),\mathrm{respectively}).\hfill \end{array}$$

Proceeding as in above, d-regular Cayley graphs in the set $\{{\mathrm{\Gamma }}_{S^{\prime },U^{\prime }},{\mathrm{\Gamma }}_{T^{\prime },U^{\prime }},{\mathrm{\Gamma }}_{S^{\prime },V^{\prime }},{\mathrm{\Gamma }}_{T^{\prime },V^{\prime }}\}$ are pairwise non-isomorphic. At last, it is routine to verify that, for arbitrary graphs ${\mathrm{\Gamma }}_1\in \{{\mathrm{\Gamma }}_{S,U},{\mathrm{\Gamma }}_{T,U},{\mathrm{\Gamma }}_{S,V},{\mathrm{\Gamma }}_{T,V}\}$, ${\mathrm{\Gamma }}_2\in \{{\mathrm{\Gamma }}_{S^{\prime },U^{\prime }},{\mathrm{\Gamma }}_{S^{\prime },V^{\prime }},{\mathrm{\Gamma }}_{T^{\prime },U^{\prime }},{\mathrm{\Gamma }}_{T^{\prime },V^{\prime }}\}$, one obtains that ${\mathrm{\Gamma }}_1\ncong {\mathrm{\Gamma }}_2$. Therefore, d-regular Cayley graphs in the set $\{{\mathrm{\Gamma }}_{S,U},{\mathrm{\Gamma }}_{T,U},{\mathrm{\Gamma }}_{S,V},{\mathrm{\Gamma }}_{T,V},{\mathrm{\Gamma }}_{S^{\prime },U^{\prime }},{\mathrm{\Gamma }}_{S^{\prime },V^{\prime }},{\mathrm{\Gamma }}_{T^{\prime },U^{\prime }},{\mathrm{\Gamma }}_{T^{\prime },V^{\prime }}\}$ are pairwise non-isomorphic.

Corollary 2 yields that ${\mathrm{\Gamma }}_{S,U}$ and ${\mathrm{\Gamma }}_{T,U}$ are isospectral; ${\mathrm{\Gamma }}_{S,V}$ and ${\mathrm{\Gamma }}_{T,V}$ are isospectral; ${\mathrm{\Gamma }}_{S^{\prime },U^{\prime }}$ and ${\mathrm{\Gamma }}_{T^{\prime },U^{\prime }}$ are isospectral; and ${\mathrm{\Gamma }}_{S^{\prime },V^{\prime }}$ and ${\mathrm{\Gamma }}_{T^{\prime },V^{\prime }}$ are isospectral.

Since there are $\left({\displaystyle \genfrac{}{}{0pt}{}{2p-1}{\lfloor \frac{d}{2}\rfloor -6}}\right)$ distinct $(\lfloor {\displaystyle \frac{d}{2}}\rfloor -6)$-subset of $\mathbb{U}$, $\left({\displaystyle \genfrac{}{}{0pt}{}{2p-1}{\lfloor \frac{d}{2}\rfloor -7}}\right)$ distinct $(\lfloor {\displaystyle \frac{d}{2}}\rfloor -7)$-subset of $\mathbb{U}$, then one can construct

$$\begin{array}{c}\hfill \left({\displaystyle \genfrac{}{}{0pt}{}{2p-1}{\lfloor \frac{d}{2}\rfloor -6}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{2p-1}{\lfloor \frac{d}{2}\rfloor -7}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{2p}{\lfloor \frac{d}{2}\rfloor -6}}\right)\end{array}$$

pairs d-regular Cayley graphs of $M_{8p}$ that are pairwise isospectral but non-isomorphic.

Finally, one obtains the desired results by Fact 1. □

4. Cay-DS Problem of ${\mathit{M}}_{\mathbf{8}\mathit{p}}$

Now we deal with the Cay-DS problem for $3\le p\le 11$ with the help of Magma V2.12-16 [27].

Proposition 1.

$\left(1\right)$ For $p=3$, the metacyclic group $M_{24}$ is Cay-DS.

$\left(2\right)$ For $p=5$, the metacyclic group $M_{40}$ is not Cay-DS. Moreover, let

$$\begin{array}{c}{K}_1=\{y{x}^2,y^{2}x,y^3{x}^2,y^3{x}^3,y^4{x}^2,y^4{x}^3,y^5{x}^2,y^5{x}^3,y^6{x}^4,y^7{x}^2\};\hfill \\ H_1=\{y{x}^4,y^2{x}^3,y^3{x}^3,y^3{x}^4,y^{4}x,y^4{x}^4,y^5{x}^3,y^5{x}^4,y^6{x}^2,y^7{x}^4\};\hfill \\ K_2=\{y,y{x}^3,y^2{x}^3,y^2{x}^4,y^3{x}^2,y^5{x}^2,y^{6}x,y^6{x}^2,y^7{x}^3,y^7\};\hfill \\ H_2=\{y{x}^2,y{x}^4,y^2{x}^3,y^2{x}^4,y^3{x}^3,y^5{x}^3,y^{6}x,y^6{x}^2,y^7{x}^2,y^7{x}^4\};\hfill \\ K_3=\{x^2,x^3,y,y{x}^3,y^2{x}^4,y^3,y^7{x}^3,y^5,y^{6}x,y^7\};\hfill \\ H_3=\{x,x^4,y{x}^2,y{x}^4,y^2{x}^3,y^3{x}^2,y^5{x}^2,y^6{x}^2,y^7{x}^2,y^7{x}^4\};\hfill \\ K_4=\{y{x}^2,y^{2}x,y^2{x}^2,y^3{x}^2,y^3{x}^4,y^5{x}^2,y^5{x}^4,y^6{x}^3,y^6{x}^4,y^7{x}^2\};\hfill \\ H_4=\{y{x}^3,y^2{x}^2,y^2{x}^4,y^3,y^3{x}^3,y^5,y^5{x}^3,y^{6}x,y^6{x}^3,y^7{x}^3\};\hfill \\ K_5=\{x^2,x^3,y^{2}x,y^2{x}^4,y^{3}x,y^{4}x,y^4{x}^4,y^{5}x,y^6{x}^4,y^{6}x\};\hfill \\ H_5=\{x,x^4,y^{2}x,y^2{x}^4,y^3{x}^2,y^4{x}^2,y^4{x}^3,y^5{x}^2,y^6{x}^4,y^{6}x\};\hfill \\ K_6=\{x,x^4,y,y{x}^4,y^3{x}^4,y^4{x}^2,y^4{x}^3,y^5{x}^4,y^7,y^7{x}^4\};\hfill \\ H_6=\{x^2,x^3,y{x}^3,y^3{x}^2,y^3{x}^3,y^{4}x,y^4{x}^4,y^5{x}^2,y^5{x}^3,y^7{x}^3\};\hfill \\ K_7=\{x,x^4,y,y^{2}x,y^2{x}^3,y^{4}x,y^4{x}^4,y^6{x}^2,y^6{x}^4,y^7\};\hfill \\ H_7=\{x,x^4,yx,y^2{x}^3,y^2{x}^4,y^{4}x,y^4{x}^4,y^{6}x,y^6{x}^2,y^{7}x\};\hfill \\ K_8=\{y,y{x}^2,y{x}^3,y^3{x}^3,y^3{x}^4,y^5{x}^3,y^5{x}^4,y^7,y^7{x}^3,y^7{x}^2\};\hfill \\ H_8=\{yx,y{x}^3,y^3{x}^2,y^3{x}^3,y^3{x}^4,y^5{x}^2,y^5{x}^3,y^5{x}^4,y^{7}x,y^7{x}^3\}.\hfill \end{array}$$

Then the Cayley graphs $Cay(M_{40},K_i)$ and $Cay(M_{40},H_i)$ are connected 10-regular, isospectral, but non-isomorphic, where $i=1,2,3,4,5,6,7,8$.

$\left(3\right)$ For $p=7$, the metacyclic group $M_{56}$ is not Cay-DS. Moreover, let

$$\begin{array}{c}{K}_1=\{x^2,x^5,y^2{x}^3,y^3{x}^3,y^4{x}^2,y^4{x}^3,y^4{x}^4,y^4{x}^5,y^5{x}^3,y^6{x}^4\};\hfill \\ H_1=\{x^3,x^4,y^2{x}^5,y^3{x}^3,y^4{x}^2,y^4{x}^3,y^4{x}^4,y^4{x}^5,y^5{x}^3,y^6{x}^2\};\hfill \\ K_2=\{y^2{x}^2,y^2{x}^3,y^2{x}^5,y^3{x}^3,y^4{x}^3,y^4{x}^4,y^5{x}^3,y^6{x}^2,y^6{x}^4,y^6{x}^5\};\hfill \\ H_2=\{y^{2}x,y^2{x}^4,y^2{x}^6,y^3{x}^3,y^4{x}^3,y^4{x}^4,y^5{x}^3,y^{6}x,y^6{x}^3,y^6{x}^6\};\hfill \\ K_3=\{x,x^2,x^5,x^6,y^2{x}^6,y^3{x}^3,y^4{x}^2,y^4{x}^5,y^5{x}^3,y^{6}x\};\hfill \\ H_3=\{x,x^2,x^5,x^6,y^2{x}^2,y^3{x}^3,y^{4}x,y^4{x}^6,y^5{x}^3,y^6{x}^5\};\hfill \\ K_4=\{x^2,x^5,y^{2}x,y^2{x}^2,y^2{x}^6,y^3{x}^3,y^5{x}^3,y^{6}x,y^6{x}^5,y^6{x}^6\};\hfill \\ H_4=\{x,x^6,y^{2}x,y^2{x}^2,y^2{x}^5,y^3{x}^3,y^5{x}^3,y^6{x}^2,y^6{x}^5,y^6{x}^6\}.\hfill \end{array}$$

Then the Cayley graphs $Cay(M_{40},K_i)$ and $Cay(M_{40},H_i)$ are connected 10-regular, isospectral, but non-isomorphic, where $i=1,2,3,4$.

$\left(4\right)$ For $p=11$, the metacyclic group $M_{88}$ is not Cay-DS. Moreover, let

$$\begin{array}{c}{K}_1=\{x^3,x^8,y^{2}x,y^6{x}^{10},y{x}^9,y^{6}x,y^6{x}^8,y^2{x}^{10},y^7{x}^9,y^2{x}^3\};\hfill \\ H_1=\{x^5,x^6,y{x}^9,y^2{x}^4,y^2{x}^6,y^2{x}^7,y^6{x}^4,y^6{x}^5,y^6{x}^7,y^7{x}^9\};\hfill \\ K_2=\{x^5,x^6,y{x}^9,y^{2}x,y^2{x}^5,y^2{x}^{10},y^{6}x,y^6{x}^6,y^6{x}^{10},y^7{x}^9\};\hfill \\ H_2=\{x,x^{10},y{x}^9,y^2{x}^5,y^2{x}^6,y^2{x}^{10},y^{6}x,y^6{x}^5,y^6{x}^6,y^7{x}^9\};\hfill \\ K_3=\{x^3,x^8,y{x}^9,y^2{x}^5,y^4{x}^3,y^4{x}^5,y^4{x}^6,y^4{x}^8,y^6{x}^6,y^7{x}^9\};\hfill \\ H_3=\{x,x^{10},y{x}^9,y^2{x}^6,y^{4}x,y^4{x}^5,y^4{x}^6,y^4{x}^{10},y^6{x}^5,y^7{x}^9\};\hfill \\ K_4=\{y{x}^9,y^2{x}^2,y^2{x}^7,y^2{x}^9,y^4{x}^4,y^4{x}^7,y^6{x}^2,y^6{x}^4,y^6{x}^9,y^7{x}^9\};\hfill \\ H_4=\{y{x}^9,y^2{x}^4,y^2{x}^7,y^2{x}^9,y^4{x}^2,y^4{x}^9,y^6{x}^2,y^6{x}^4,y^6{x}^7,y^7{x}^9\};\hfill \\ K_5=\{x^2,x^5,x^6,x^9,y{x}^9,y^2{x}^5,y^4{x}^2,y^4{x}^9,y^6{x}^6,y^7{x}^9\};\hfill \\ H_5=\{x^2,x^3,x^8,x^9,y{x}^9,y^2{x}^8,y^4{x}^2,y^4{x}^9,y^6{x}^3,y^7{x}^9\};\hfill \\ K_6=\{x^3,x^4,x^7,x^8,y{x}^9,y^2{x}^8,y^4{x}^4,y^4{x}^7,y^6{x}^3,y^7{x}^9\};\hfill \\ H_6=\{x^2,x^4,x^7,x^9,y{x}^9,y^2{x}^9,y^4{x}^4,y^4{x}^7,y^6{x}^2,y^7{x}^9\};\hfill \\ K_7=\{x^5,x^6,y{x}^9,y^2{x}^4,y^4{x}^4,y^4{x}^5,y^4{x}^6,y^4{x}^7,y^6{x}^7,y^7{x}^9\};\hfill \\ H_7=\{x,x^{10},y{x}^9,y^2{x}^7,y^{4}x,y^4{x}^4,y^4{x}^7,y^4{x}^{10},y^6{x}^4,y^7{x}^9\};\hfill \\ K_8=\{y{x}^9,y^2{x}^2,y^2{x}^5,y^2{x}^6,y^4{x}^2,y^4{x}^9,y^6{x}^5,y^6{x}^6,y^6{x}^9,y^7{x}^9\};\hfill \\ H_8=\{y{x}^9,y^{2}x,y^2{x}^4,y^2{x}^7,y^{4}x,y^4{x}^{10},y^6{x}^4,y^6{x}^7,y^6{x}^{10},y^7{x}^9\}.\hfill \end{array}$$

Then the Cayley graphs $Cay(M_{88},K_i)$ and $Cay(M_{88},H_i)$ are connected 10-regular, isospectral, but non-isomorphic, where $i=1,2,3,\dots ,8$.

Combining Theorem 1 with Proposition 1, one obtains the following interesting result.

Proposition 2.

Let $p\ge 3$ be a prime number and let $M_{8p}$ be the metaclic group of order $8p$. Then $M_{8p}$ is Cay-DS if and only if $p=3$.

5. Discussion

The present paper contributes to parts of the Cay-DS problem of metacyclic groups $M_{8p}$ of order $8p$, where p is an odd prime, in terms of irreducible characters. It constructed an extensive family of pairwise non-isomorphic yet isospectral Cayley graph pairs of $M_{8p}$ with $p\ge 5$. Moreover, this paper shows that $M_{8p}$ is Cay-DS if and only if $p=3$, which offers new insights into the Cay-DS problem. At the end of this paper, we point out that determining the complete set of the isospectral but non-isomorphic Cayley graphs of $M_{8p}$ ($p\ge 3$) is still an open problem.