Classification of Cayley Rose Window Graphs

TL;DR

This paper classifies Cayley rose window graphs, a specific family of symmetric tetravalent graphs, expanding the understanding of their algebraic and combinatorial properties.

Contribution

It provides the first classification of Cayley rose window graphs, a subclass not previously characterized in the literature.

Findings

Cayley rose window graphs are fully classified.

New structural properties of Cayley rose window graphs are identified.

Connections to existing classifications of symmetric graphs are established.

Abstract

Rose window graphs are a family of tetravalent graphs, introduced by Steve Wilson. Following it, Kovacs, Kutnar and Marusic classified the edge-transitive rose window graphs and Dobson, Kovacs and Miklavic characterized the vertex transitive rose window graphs. In this paper, we classify the Cayley rose window graphs.

Equations

$$H=⟨\rho ,\delta ⟩=⟨\rho ,\delta :{\rho }^n={\delta }^2=\mathsf{id};\delta \rho \delta ={\rho }^r⟩$$

$$=\{\mathsf{id},\rho ,{\rho }^2,\dots ,{\rho }^{n-1},\delta ,\rho \delta ,{\rho }^2\delta ,\dots ,{\rho }^{n-1}\delta \}.$$

$$H=⟨\alpha ,\beta ,\gamma :{\alpha }^{n/2}={\beta }^2={\gamma }^4=\mathsf{id};\beta \alpha ={\alpha }^{-1}\beta ;\gamma \alpha ={\alpha }^{-r}\gamma ;{\gamma }^2={\alpha }^{\frac{a-1}{2}}\beta ⟩$$

$$=\{{\alpha }^i{\beta }^j{\gamma }^k:0\le i<n/2,0\le j,k\le 1\}$$

$${\alpha }^{i_1}{\beta }^{j_1}{\gamma }^{k_1}={\alpha }^{i_2}{\beta }^{j_2}{\gamma }^{k_2},\text{mbox}where0\le i_1,i_2<n/2,0\le j_1,j_2\le 1,0\le k_1,k_2\le 1,$$

$${\beta }^{-j_2}{\alpha }^{i_1-i_2}{\beta }^{j_1}={\gamma }^{k_2-k_1},\text{mbox}where{k}_2-k_1=0\text{mbox}or1.$$

$${\alpha }^{i_1-i_2}={\beta }^{j_2-j_1},\text{mbox}where{j}_2-j_1=0\text{mbox}or1.$$

$$H=⟨\alpha ,\beta ,\sigma :{\alpha }^{n/2}={\beta }^2={\sigma }^2=\mathsf{id};\beta \alpha \beta ={\alpha }^{-1},\sigma \alpha \sigma ={\alpha }^r,(\beta \sigma {)}^2={\alpha }^{\frac{a-r+1}{2}}⟩$$

$$=\{{\alpha }^i{\beta }^j{\sigma }^k:0\le i<n/2,0\le j,k\le 1\}$$

$${\alpha }^{i_1}{\beta }^{j_1}{\sigma }^{k_1}={\alpha }^{i_2}{\beta }^{j_2}{\sigma }^{k_2},\text{mbox}where0\le i_1,i_2<n/2,0\le j_1,j_2,k_1,k_2\le 1,$$

$${\alpha }^{i_1-i_2}{\beta }^{j_1}{\sigma }^{k_1-k_2}={\beta }^{j_2},\text{mbox}where{k}_1-k_2=0\text{mbox}or1.$$

$${\alpha }^{i_1-i_2}={\beta }^{j_2-j_1},\text{mbox}where{j}_2-j_1=0\text{mbox}or1.$$

$$|{\mathcal{O}}_{A_0}|=\frac{|H|}{|{\mathsf{Stab}}_H(A_0)|}.$$

$$\mathsf{Aut}(R_n(a,r))=⟨\rho ,\mu ,\gamma :{\rho }^n={\mu }^2={\gamma }^4=\mathsf{id};\mu \rho \mu ={\rho }^{-1},\gamma \mu ={\rho }^a\mu \gamma ,\gamma \rho ={\rho }^{r-a}\mu {\gamma }^3⟩.$$

$$|HK|=\frac{|H||K|}{|H\cap K|}=8n.$$

$$G:=\mathsf{Aut}(R_{2m}(m-2,m-1))=⟨\rho ,\mu ,{\epsilon }_0,{\epsilon }_1,\dots ,{\epsilon }_{m-1}⟩=K⋊⟨\rho {\epsilon }_0,\mu {\rho }^m⟩\cong {\mathbb{Z}}_2^m⋊D_m,$$

$${\epsilon }_i{\epsilon }_j={\epsilon }_j{\epsilon }_i;{\epsilon }_i{\rho }^m={\rho }^m{\epsilon }_i;\mu {\epsilon }_i={\epsilon }_{m-1-i}\mu ;$$

$$\rho {\epsilon }_i={\epsilon }_{i+1}\rho ,\forall i,j\in \{0,1,\dots ,m-1\}\text{mbox}and{\epsilon }_0{\epsilon }_1\cdots {\epsilon }_{m-1}={\rho }^m$$

$${\mathsf{Stab}}_G(A_i)=⟨{\rho }^{2i}\mu ,{\epsilon }_0,{\epsilon }_1,\dots ,{\epsilon }_{i-2},{\epsilon }_{i+1},\dots ,{\epsilon }_{m-1}⟩.$$

$${\mathsf{Stab}}_G(B_i)=⟨{\rho }^{m-2+2i}\mu ,{\epsilon }_0,{\epsilon }_1,\dots ,{\epsilon }_{i-1},{\epsilon }_{i+2},\dots ,{\epsilon }_{m-1}⟩.$$

$$r^2=(m-1{)}^2=m^2-2m+1\equiv 1(mod2m)\equiv 1(modn)\text{mbox}and$$

$$ra=(m-1)(m-2)=m^2-3m+2\equiv -m+2(mod2m)\equiv -a(modn).$$

$$H=⟨\alpha ,\beta ,\gamma :\circ (\alpha )=m,\circ (\beta )=\circ (\gamma )=2;\beta \alpha =\alpha \gamma ,\gamma \alpha =\alpha \beta \gamma ,\beta \gamma =\gamma \beta ⟩.$$

$${\rho }^{2(i_1-i_2)}={\alpha }^{i_1-i_2}={\beta }^{j_2-j_1}{\gamma }^{k_2-k_1}(\text{mbox}as\beta \gamma =\gamma \beta ).$$

$${\beta }^j{\gamma }^k(A_0)=A_{2m-2i}$$

$$|HK|=\frac{|H||K|}{|H\cap K|}=\frac{2^{2}m\cdot 2^m}{|H\cap K|}\le 2^{m+1}m,\text{mbox}i.e.,|H\cap K|\ge 2.$$

$$|HL|=\frac{|H||L|}{|H\cap L|}=\frac{4m\cdot 2^{m+1}}{|H\cap L|}\le |G|=2^{m+1}m,\text{mbox}i.e.,|H\cap L|\ge 4.$$

$$\beta \alpha {\beta }^{-1}=(\mu {\epsilon }_{i_1}{\epsilon }_{i_2}\cdots {\epsilon }_{i_s})({\epsilon }_{l_1}{\epsilon }_{l_2}\cdots {\epsilon }_{l_p})(\mu {\epsilon }_{i_1}{\epsilon }_{i_2}\cdots {\epsilon }_{i_s}{)}^{-1}=\mu ({\epsilon }_{l_1}{\epsilon }_{l_2}\cdots {\epsilon }_{l_p})\mu .$$

$$\gamma \alpha {\gamma }^{-1}=(\rho \mu {\epsilon }_{j_1}{\epsilon }_{j_2}\cdots {\epsilon }_{j_t})({\epsilon }_{l_1}{\epsilon }_{l_2}\cdots {\epsilon }_{l_p})(\rho \mu {\epsilon }_{j_1}{\epsilon }_{j_2}\cdots {\epsilon }_{j_t}{)}^{-1}=\rho (\mu {\epsilon }_{l_1}{\epsilon }_{l_2}\cdots {\epsilon }_{l_p}\mu ){\rho }^{-1}$$

$$=\rho (\beta \alpha {\beta }^{-1}){\rho }^{-1}=\rho \alpha {\rho }^{-1}.$$

Full text

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1 Introduction

Rose window graphs were introduced in [6] in the following way:

Definition 1.1**.**

Given natural numbers $n\geq 3$ and $1\leq a,r\leq n-1$, the Rose Window graph $R_{n}(a,r)$ is defined to be the graph with vertex set $V=\{A_{i},B_{i}:i\in\mathbb{Z}_{n}\}$ and four kind of edges: $A_{i}A_{i+1}$ $(\mathbf{rim}~{}edges)$, $A_{i}B_{i}$ $(\mathbf{inspoke}~{}edges)$, $A_{i+a}B_{i}$ $(\mathbf{outspoke}~{}edges)$ and $B_{i}B_{i+r}$ $(\mathbf{hub}~{}edges)$, where the addition of indices are done modulo $n$.

In the introductory paper [6], author’s initial interest in rose window graphs arose in the context of graph embeddings into surfaces. The author conjectured that rose window graphs are edge-transitive if and only if it belongs to the one of the four families given in Theorem 1.1. The conjecture was proved by Kovacs et. al. in [4]. In particular, they proved that

Theorem 1.1**.**

[4]** A rose window graph is edge-transitive if and only if it belongs to one of the four families:

$R_{n}(2,1)$. 2. 2.

$R_{2m}(m\pm 2,m\pm 1)$** 3. 3.

$R_{12m}(\pm(3m+2),\pm(3m-1))$* and $R_{12m}(\pm(3m-2),\pm(3m+1))$.* 4. 4.

$R_{2m}(2b,r)$, where $b^{2}\equiv\pm 1$(mod $m$), $2\leq 2b\leq m$, and $r\in\{1,m-1\}$ is odd.

A similar characterization for vertex-transitive graphs was proved in [1]:

Theorem 1.2**.**

[1]** A rose window graph $R_{n}(a,r)$ is vertex-transitive if and only if it belongs to one of the following families:

$R_{n}(a,r)$, where $r^{2}\equiv\pm 1(mod~{}n)$ and $ra\equiv\pm a(mod~{}n)$. 2. 2.

$R_{4m}(2m,r)$, where $r$ is odd and $(r^{2}+2m)\equiv\pm 1(mod~{}4m)$. 3. 3.

$R_{2m}(m\pm 2,m\pm 1)$** 4. 4.

$R_{12m}(\pm(3m+2),\pm(3m-1))$* and $R_{12m}(\pm(3m-2),\pm(3m+1))$.* 5. 5.

$R_{2m}(2b,r)$, where $b^{2}\equiv\pm 1$(mod $m$), $2\leq 2b\leq m$, and $r\in\{1,m-1\}$ is odd.

As a Cayley graph is always vertex-transitive, a natural question to ask is to characterize the rose-window graphs which are also Cayley graphs. For that, it is sufficient to look for Cayley graphs only in the $5$ families mentioned in Theorem 1.2. The main goal of this paper is finding an answer to this question. In particular, we prove the following theorem:

Theorem 1.3**.**

A rose-window graph $R_{n}(a,r)$ is Cayley if and only if one of the following holds:

$R_{n}(a,r)$, where $r^{2}\equiv\pm 1(mod~{}n)$ and $ra\equiv\pm a(mod~{}n)$. 2. 2.

$R_{4m}(2m,r)$, where $r$ is odd and $\mathbf{(r^{2}+2m)\equiv 1(mod~{}4m)}$. 3. 3.

$R_{2m}(m\pm 2,m\pm 1)$* where $\mathbf{m}$ is a multiple of $\mathbf{2}$ or $\mathbf{3}$.* 4. 4.

$R_{12m}(\pm(3m+2),\pm(3m-1))$* and $R_{12m}(\pm(3m-2),\pm(3m+1))$ where $\mathbf{m\not\equiv 0(mod~{}4)}$.* 5. 5.

$R_{2m}(2b,r)$, where $b^{2}\equiv\pm 1$(mod $m$), $2\leq 2b\leq m$, and $r\in\{1,m-1\}$ is odd.∎

Before stating the proof, we note a few generic automorphisms and other properties of $R_{n}(a,r)$. Other automorphisms, specific to any particular family of rose window graphs, will be introduced whenever they are needed.

Define $\tau:V\rightarrow V$ by $\tau(A_{i})=A_{-i}$ and $\tau(B_{i})=B_{-i}$. Clearly $\tau$ is an automorphism with $\tau^{2}=\mathsf{id}$ and hence $R_{n}(a,r)\cong R_{n}(-a,r)$. 2. 2.

$R_{n}(a,r)=R_{n}(a,-r)$. 3. 3.

Define $\rho:V\rightarrow V$ by $\rho(A_{i})=A_{i+1}$ and $\rho(B_{i})=B_{i+1}$; and $\mu:V\rightarrow V$ by $\mu(A_{i})=A_{-i}$ and $\mu(B_{i})=B_{-a-i}$. Clearly $\rho$ and $\mu$ are automorphisms. As $\rho^{n}=\mu^{2}=\mathsf{id}$ and $\mu\rho\mu=\rho^{-1}$, we have $\langle\rho,\mu\rangle\cong D_{n}$. 4. 4.

If $(n,r)=1$, then $\zeta:V\rightarrow V$ given by $\zeta(A_{i})=B_{-ir^{-1}}$ and $\zeta(B_{i})=A_{-ir^{-1}}$ is an automorphism and hence $R_{n}(a,r)\cong R_{n}(ar^{-1},r^{-1})$.

Remark 1.1**.**

In view of the first two observations, it is enough to study $R_{n}(a,r)$ for $1\leq a,r\leq\lceil\frac{n}{2}\rceil$.

The main theorem, which is repeatedly used in the proofs throughout the paper, is the following:

Proposition 1.1**.**

A vertex-transitive graph $G$ is Cayley if and only if $\mathsf{Aut}(G)$ has a subgroup $H$ which acts regularly on the vertices of $G$. In particular, non-identity elements of $H$ do not stabilize any vertex.∎

Remark 1.2**.**

In this context, it is to be noted that if a group of order $n$ acts transitively on a set of order $n$, then the action is regular.

2 Family-1 [$R_{n}(a,r)$: $r^{2}\equiv\pm 1(mod~{}n)$ and $ra\equiv\pm a(mod~{}n)$]

If $r^{2}\equiv\pm 1(mod~{}n)$ and $ra\equiv\pm a(mod~{}n)$, then $\delta:V\rightarrow V$ given by $\delta(A_{i})=B_{ri}$ and $\delta(B_{i})=A_{ri}$ is an automorphism. For proof, see Lemma 2 [6] or Lemma 3.7 [1]. If $r^{2}\equiv 1(mod~{}n)$, then $\delta^{2}=\mathsf{id}$ and if $r^{2}\equiv-1(mod~{}n)$, then $\delta^{2}=\tau$, i.e., $\delta$ is of order $4$.

Theorem 2.1**.**

If $r^{2}\equiv 1(mod~{}n)$ and $ra\equiv\pm a(mod~{}n)$, then $R_{n}(a,r)$ is a Cayley graph.

**Proof: **Since $R_{n}(a,r)=R_{n}(a,-r)$, without loss of generality, we can assume that $ra\equiv-a(mod~{}n)$. Consider $\rho$ and $\delta$ as defined above. We have $\rho^{n}=\delta^{2}=\mathsf{id}$ and $\delta\rho\delta=\rho^{r}$. Define

[TABLE]

[TABLE]

Clearly, $H$ is a subgroup of $\mathsf{Aut}(R_{n}(a,r))$. It suffices to show that $H$ acts regularly on $R_{n}(a,r)$. For that we observe that

$\rho^{j}(A_{i})=A_{i+j}$ and $\rho^{j}(B_{i})=B_{i+j}$, and

• 2.

$\rho^{j}\delta(A_{i})=B_{ri+j}$ and $\rho^{j}\delta(B_{i})=A_{ri+j}$.

As $gcd(r,n)=1$, the map $i\mapsto ri+j$ is a bijection on $\{0,1,\ldots,n-1\}$. Thus $H$ acts transitively on $R_{n}(a,r)$. It is also clear from the construction of $H$, that for any pair of vertices in $R_{n}(a,r)$, there exists a unique element in $H$ which maps one to the other. Hence, $R_{n}(a,r)$ is a Cayley graph.∎

Lemma 2.1**.**

If $r^{2}\equiv-1(mod~{}n)$ and $ra\equiv\pm a(mod~{}n)$, then $n$ is even, $a$ is odd and $n=2a$.

**Proof: **Let $p$ be an odd prime factor of $n$ such that $p^{i}|n$ and $p^{i+1}\nmid n$. Then $r^{2}\equiv-1(mod~{}p^{i})$ and $r^{2}\equiv-1(mod~{}p)$. Again, $p^{i}|a(r\pm 1)$, i.e., $p|a(r\pm 1)$. If $p|(r\pm 1)$, then $r^{2}\equiv 1(mod~{}p)$, a contradiction, as $-1\not\equiv 1(mod~{}p)$. Thus for all odd prime factors $p$ of $n$, we have $p^{i}|a$. Hence, if $n$ is odd, then $n=a$, a contradiction (See Remark 1.1). Thus $n$ is even.

We claim that $2|n$ but $4\nmid n$. Because if $4|n$, then $r^{2}\equiv-1(mod~{}4)$. However, there does not exist any such $r$. Thus $n$ is $2$ times the product of some odd primes. Also, all the odd prime factors of $n$ are also factors of $a$, as seen above. Thus, if $2|a$, then $n=a$, a contradiction (See Remark 1.1). Thus $2\nmid a$ and hence $a$ is odd and $n=2a$.∎

Theorem 2.2**.**

If $r^{2}\equiv-1(mod~{}n)$ and $ra\equiv\pm a(mod~{}n)$, then $R_{n}(a,r)$ is a Cayley graph.

**Proof: **Let $\alpha=\rho^{2};\beta=\rho\delta^{2};\gamma=\mu\delta$. Clearly, $\alpha,\beta,\gamma\in\mathsf{Aut}(R_{n}(a,r))$. It can be easily checked that $\beta\alpha=\alpha^{-1}\beta;\gamma\alpha=\alpha^{-r}\gamma$ and $\gamma^{2}=\alpha^{\frac{a-1}{2}}\beta$. Define

[TABLE]

[TABLE]

Note that, from the above lemma, $n/2$ and $(a-1)/2$ are positive integers. We claim that the elements in $H$ are distinct. If not, suppose

[TABLE]

i.e.,

[TABLE]

Now, as $\gamma=\mu\delta$ flips $A_{i}$’s and $B_{j}$’s, and $\alpha,\beta$ maps $A_{i}$’s to $A_{j}$’s and $B_{i}$’s to $B_{j}$’s, $k_{2}-k_{1}$ must be [math], i.e., $k_{1}=k_{2}$. Thus, we have

[TABLE]

If $j_{2}-j_{1}=1$, then $\alpha^{i_{1}-i_{2}}=\beta=\rho\delta^{2}$. But $\alpha^{i_{1}-i_{2}}(A_{0})=A_{2(i_{1}-i_{2})}$ (even index) and $\rho\delta^{2}(A_{0})=A_{1}$ (odd index). Hence, $j_{2}-j_{1}=0$, i.e., $j_{1}=j_{2}$. This implies $\alpha^{i_{1}-i_{2}}=\mathsf{id}$ and as a result $i_{1}=i_{2}$. Thus the elements of $H$ are distinct and $|H|=n/2\times 2\times 2=2n$.

We claim that $H$ acts transitively on $R_{n}(a,r)$. It suffices to show that the stabilizer of $A_{0}$ in $H$, $\mathsf{Stab}_{H}(A_{0})=\{\mathsf{id}\}$.

Let $\alpha^{i}\beta^{j}\gamma^{k}\in\mathsf{Stab}_{H}(A_{0})$, i.e., $\alpha^{i}\beta^{j}\gamma^{k}(A_{0})=A_{0}$. Since, $\gamma$ flips $A_{i}$’s and $B_{j}$’s, and $\alpha,\beta$ do not, we have $k=0$. Thus, $\alpha^{i}\beta^{j}(A_{0})=A_{0}$. If $j=1$, then $\alpha^{i}\beta(A_{0})=\alpha^{i}\rho\delta^{2}(A_{0})=\rho^{1+2i}\delta^{2}(A_{0})=A_{0}$, i.e., $A_{1+2i}=A_{0}$, a contradiction, as the parity of indices on both sides does not match. Thus, $j=0$ and we have $\alpha^{i}(A_{0})=A_{0}$. But this implies $A_{2i}=A_{0}$, i.e., $i=0$. Hence $\mathsf{Stab}_{H}(A_{0})=\{\mathsf{id}\}$.

Finally, in view of Remark 1.2, $H$ acts regularly on $R_{n}(a,r)$ and hence $R_{n}(a,r)$ is a Cayley graph.∎

3 Family-2 [$R_{4m}(2m,r)$: $r$ is odd and $(r^{2}+2m)\equiv\pm 1(mod~{}4m)$]

Proposition 3.1**.**

If $n$ is divisible by $4$, $r$ is odd, $a=n/2$ and $(r^{2}+n/2)\equiv\pm 1(mod~{}n)$, then

$gcd(r,n)=1$.

• 2.

If $\gamma:V\rightarrow V$ be defined by $\gamma(A_{i})=B_{ri}$ and $\gamma(B_{i})=A_{(r+a)i}$, then $\gamma\in\mathsf{Aut}(R_{n}(a,r))$.

**Proof: **Let $n=4m$ and $a=2m$, and let if possible, $gcd(r,n)=l>1$. As $r$ is odd, $l|m$. Thus $r=lt$ and $m=ls$ for some $s,t\in\mathbb{N}$. Thus $n=4ls,a=2ls$ and $r=lt$. Now $(r^{2}+n/2)\equiv\pm 1(mod~{}n)$ implies $l^{2}t^{2}+2ls\equiv\pm 1(mod~{}4ls)$, which in turn implies $l|(l^{2}t^{2}+2ls\pm 1)$, i.e., $l|1$, a contradiction. Thus $gcd(r,n)=1$.

$\gamma$, as defined above, has been shown to be in $\mathsf{Aut}(R_{n}(a,r))$ in Lemma 3.8 [1]. ∎

Proposition 3.2**.**

If $n$ is divisible by $4$, $r$ is odd, $a=n/2$ and $(r^{2}+n/2)\equiv 1(mod~{}n)$, then

$r^{-1}=r+a$* (mod $n$)*

• 2.

$\zeta\in\mathsf{Aut}(R_{n}(a,r))$* (defined before) takes the following form: $\zeta(A_{i})=B_{-(r+a)i}$ and $\zeta(B_{i})=A_{-(r+a)i}$, and $\zeta^{4}=\mathsf{id}$.*

Proof: $r(r+a)\equiv r^{2}+ar\equiv 1-a+ar\equiv 1+a(r-1)\equiv 1~{}(mod~{}n)$. The last equivalence holds as $r$ is odd and $a=n/2$. Thus $r^{-1}=r+a$ (mod $n$). The form of $\zeta$ follows immediately from the fact that $r^{-1}=r+a$ (mod $n$).∎

Theorem 3.1**.**

If $n$ is divisible by $4$, $r$ is odd, $a=n/2$ and $(r^{2}+n/2)\equiv 1(mod~{}n)$, then $R_{n}(a,r)$ is a Cayley graph.

**Proof: **Let $\alpha=\rho^{2},\beta=\rho\mu$ and $\sigma=\gamma\zeta^{2}$, where $\gamma$ and $\zeta$ are as defined in Propositions 3.1 and 3.2. It can be easily checked that $\sigma(A_{i})=B_{(r+a)i}$ and $\sigma(B_{i})=A_{ri}$; $\alpha^{n/2}=\beta^{2}=\sigma^{2}=\mathsf{id}$; $\beta\alpha\beta=\alpha^{-1},\sigma\alpha\sigma=\alpha^{r},(\beta\sigma)^{2}=\alpha^{\frac{a-r+1}{2}}$. Define

[TABLE]

[TABLE]

We claim that the elements in $H$ are distinct. If not, suppose

[TABLE]

i.e.,

[TABLE]

Now, as $\sigma$ flips $A_{i}$’s and $B_{j}$’s, and $\alpha,\beta$ maps $A_{i}$’s to $A_{j}$’s and $B_{i}$’s to $B_{j}$’s, $k_{1}-k_{2}$ must be [math], i.e., $k_{1}=k_{2}$. Thus, we have

[TABLE]

Since, $\alpha$ maintains the parity of indices and $\beta$ flips the parity of indices of $A_{i}$’s and $B_{i}$’s, $j_{2}-j_{1}$ is even, i.e., $j_{1}=j_{2}$. This implies $\alpha^{i_{1}-i_{2}}=\mathsf{id}$ and as a result $i_{1}=i_{2}$. Thus the elements of $H$ are distinct and $|H|=n/2\times 2\times 2=2n$.

We claim that $H$ acts transitively on $R_{n}(a,r)$. In order to prove it, we show that the orbit of $A_{0}$, $\mathcal{O}_{A_{0}}$, under the action of $H$ is the vertex set of $R_{n}(a,r)$. By orbit-stabilizer theorem, we get

[TABLE]

As the number of vertices in $R_{n}(a,r)$ is $2n$ and $|H|=2n$, it is enough to show that $\mathsf{Stab}_{H}(A_{0})=\{\mathsf{id}\}$. Let $\alpha^{i}\beta^{j}\sigma^{k}$ be an arbitrary element of $H$ which stabilizes $A_{0}$, i.e., $\alpha^{i}\beta^{j}\sigma^{k}(A_{0})=A_{0}$, with $0\leq i<n/2,0\leq j,k\leq 1$. Now, as $\sigma$ flips $A_{i}$’s and $B_{j}$’s, and $\alpha,\beta$ maps $A_{i}$’s to $A_{j}$’s and $B_{i}$’s to $B_{j}$’s, $k=0$. Thus $\alpha^{i}\beta^{j}(A_{0})=A_{0}$, i.e., $\alpha^{-i}(A_{0})=\beta^{j}(A_{0})$. Since, $\alpha$ maintains the parity of indices and $\beta$ flips the parity of indices of $A_{i}$’s and $B_{i}$’s, $j=0$ and hence $i=0$. Thus $\mathsf{Stab}_{H}(A_{0})=\{\mathsf{id}\}$.

Finally, in view of Remark 1.2, $H$ acts regularly on $R_{n}(a,r)$ and hence $R_{n}(a,r)$ is a Cayley graph.∎

In Family 2, if $(r^{2}+n/2)\equiv-1(mod~{}n)$, we will show that $R_{n}(a,r)$ is not a Cayley graph. In order to prove it, we recall a few observations and results.

Remark 3.1**.**

It was noted in [6] and [1], that $R_{n}(a,r)$ has either one or two or three edge orbits. If it has one edge orbit, then by definition, it is edge transitive, as in Theorem 1.1. If $R_{n}(a,r)$ has two edge orbits, then one orbit consists of rim and hub edges, and the other consists of spoke edges. If $R_{n}(a,r)$ has three orbits on edges, then the first one consists of rim edges, the second one consists of hub edges, and the third one consists of spoke edges.

As Family $\mathbf{3,4,5}$ in Theorem 1.2 are also edge transitive, they have only one edge orbit. On the other hand, family $1$ and $2$ in Theorem 1.2, have two edge orbits, as evident from Remark 3.1 and Theorem 3.2.

Theorem 3.2** (Theorem 2.3,[1]).**

There is an automorphism of $R_{n}(a,r)$ sending every rim edge to a hub edge and vice-versa if and only if one of the following holds:

$a\neq n/2$, $r^{2}\equiv 1(mod~{}n)$ and $ra\equiv\pm a(mod~{}n)$; 2. 2.

$a=n/2$, $r^{2}\equiv\pm 1(mod~{}n)$ and $ra\equiv\pm a(mod~{}n)$; 3. 3.

$n$* is divisible by $4$, $gcd(n,r)=1$, $a=n/2$ and $(r^{2}+n/2)\equiv\pm 1(mod~{}n)$.*

Corollary 3.3** (Corollary 3.9,[1]).**

If $n$ is divisible by $4$, $r$ is odd, $a=n/2$ and $(r^{2}+n/2)\equiv\pm 1(mod~{}n)$, then the automorphism group of $R_{n}(a,r)$ has two edge orbits and the full automorphism group of $R_{n}(a,r)$, $\mathsf{Aut}(R_{n}(a,r))=\langle\rho,\mu,\gamma\rangle$, where $\gamma$ is as defined in Proposition 3.1.

Theorem 3.3**.**

If $n$ is divisible by $4$, $r$ is odd, $a=n/2$ and $(r^{2}+n/2)\equiv-1(mod~{}n)$, then $R_{n}(a,r)$ is not a Cayley graph.

**Proof: **As evident from Corollary 3.3, the full automorphism group of $R_{n}(a,r)$ is given by

[TABLE]

One can easily check the relations between the generators starting from the definition and conclude that $|\mathsf{Aut}(R_{n}(a,r))|=n\times 2\times 4=8n$. If possible, let $R_{n}(a,r)$ be a Cayley graph with a regular subgroup $H$ of $\mathsf{Aut}(R_{n}(a,r))$ and $|H|=2n$.

Let $K=\langle\gamma\rangle$. Then $|K|=4$ and $H\cap K$ is a subgroup of $K$. As $\gamma^{2}(A_{0})=A_{0}$, i.e., $\gamma^{2}$ has a fixed point, $\gamma^{2}\not\in H$. Thus $H\cap K=\{\mathsf{id}\}$ and

[TABLE]

Hence $\mu\in\mathsf{Aut}(R_{n}(a,r))=HK$. Thus $\mu=hk$, where $h\in H$ and $k\in K=\{\mathsf{id},\gamma,\gamma^{2},\gamma^{3}\}$. If $k=\mathsf{id}$, then $\mu=h\in H$. But as $\mu(A_{0})=A_{0}$, i.e., $\mu$ has a fixed point, $\mu\not\in H$. Thus $k\neq\mathsf{id}$.

If $k=\gamma^{2}$, then $\mu=h\gamma^{2}$, i.e., $h=\mu\gamma^{2}\in H$. But as $\mu\gamma^{2}(A_{0})=A_{0}$, $\mu\gamma^{2}\not\in H$ and hence $k\neq\gamma^{2}$.

If $k=\gamma$, then $\mu\gamma^{-1}=h$, i.e., $h^{-2}=(\gamma\mu)^{2}=\rho^{a}\gamma^{2}\in H$. But, as $\rho^{a}\gamma^{2}(A_{a/2})=A_{a/2}$, by similar argument, $k\neq\gamma$.

If $k=\gamma^{3}$, then $h^{2}=(\mu\gamma)^{2}=\rho^{a}\gamma^{2}\in H$. By similar argument as above, $k\neq\gamma^{3}$.

As all the four possible choices of $k\in K$ leads to contradiction, we conclude that there does not exist any regular subgroup $H$ of $\mathsf{Aut}(R_{n}(a,r))$ and hence $R_{n}(a,r)$ is not a Cayley graph. ∎

4 Family-3 [$R_{2m}(m\pm 2,m\pm 1)$]

As $m+2\equiv-(m-2)~{}(mod~{}2m)$ and $m+1\equiv-(m-1)~{}(mod~{}2m)$, it suffices to check the family $R_{2m}(m-2,m-1)$. It was proved in Section 3.2 of [5], that

[TABLE]

where $K=\langle\varepsilon_{0},\varepsilon_{1},\ldots,\varepsilon_{m-1}\rangle\cong\mathbb{Z}_{2}^{m}$, $D_{m}$ is the dihedral group and $\varepsilon_{i}$ is the involution given by $(A_{i},B_{i-1})(A_{i+m},B_{i-1+m})(A_{i+1},B_{i+m})(A_{i+1+m},B_{i})$. Thus $|G|=2^{m+1}m$. One can easily check that the following relations between the generators hold:

[TABLE]

[TABLE]

where the addition of indices of $\varepsilon_{i}$’s are done modulo $m$. Using this relations, it is easy to see that $\circ(\rho\varepsilon_{i})=m$ and $\circ(\mu\rho^{i})=2$.

It follows from definition that $\rho^{2i}\mu,\varepsilon_{0},\varepsilon_{1},\ldots,\varepsilon_{i-2},\varepsilon_{i+1},\ldots,\varepsilon_{m-1}\in\mathsf{Stab}_{G}(A_{i})$. Again, using the relations between generators, we get $|\langle\rho^{2i}\mu,\varepsilon_{0},\varepsilon_{1},\ldots,\varepsilon_{i-2},\varepsilon_{i+1},\ldots,\varepsilon_{m-1}\rangle|=2^{m-1}$. Now, as $R_{2m}(m-2,m-1)$ is a vertex transitive graph, by orbit-stabilizer theorem, it follows that $|G|/|\mathsf{Stab}_{G}(A_{i})|=2\times 2m,$ i.e., $|\mathsf{Stab}_{G}(A_{i})|=\frac{2^{m+1}m}{4m}=2^{m-1}$. Thus, we have

[TABLE]

Similarly, it follows that

[TABLE]

Theorem 4.1**.**

$R_{2m}(m-2,m-1)$* is a Cayley graph, if $m$ is even.*

**Proof: **In this case, $n=2m$, $a=m-2$ and $r=m-1$. Now, if $m$ is even, we have

[TABLE]

[TABLE]

Thus, if $m$ is even, $R_{2m}(m-2,m-1)$ is a subfamily of Family-1 and as a result, $R_{2m}(m-2,m-1)$ is a Cayley graph.∎

Theorem 4.2**.**

$R_{2m}(m-2,m-1)$* is a Cayley graph, if $m$ is an odd multiple of $3$.*

**Proof: **Let $m=3l$. For $i=0,1,2$, denote by $\Sigma_{i}$, the product of all $\varepsilon_{j}$’s such that $j\neq i~{}(mod~{}3)$. Note that $\Sigma_{i}\Sigma_{j}=\Sigma_{k}$ for distinct $i,j,k$’s in $\{0,1,2\}$ and $\circ(\Sigma_{i})=2$.

Let $\alpha=\rho^{2},\beta=\Sigma_{0}$ and $\gamma=\Sigma_{1}$. It can be easily checked that $\beta\alpha=\alpha\gamma,\gamma\alpha=\alpha\beta\gamma$ and $\beta\gamma=\gamma\beta$. Define

[TABLE]

Thus, any element of $H$ can be expressed as $\alpha^{i}\beta^{j}\gamma^{k}$ where $0\leq i\leq m-1,0\leq j,k\leq 1$, i.e., $|H|\leq 4m$.

Claim 1: $|H|=4m$.

Proof of Claim 1: If not, there exist $0\leq i_{1},i_{2}\leq m-1,0\leq j_{1},j_{2},k_{1},k_{2}\leq 1$ such that $\alpha^{i_{1}}\beta^{j_{1}}\gamma^{k_{1}}=\alpha^{i_{2}}\beta^{j_{2}}\gamma^{k_{2}}$, i.e.,

[TABLE]

If $j_{2}-j_{1}=k_{2}-k_{1}=0$, then $i_{1}=i_{2}$ (since, $\circ(\rho)=2m$) and as a result the claim is true. However, if any one or both of $j_{2}-j_{1}$ or $k_{2}-k_{1}$ is $1$, then the right hand side is an element of order $2$. As a result, the left hand side must be an element of order $2$, which implies $2(i_{1}-i_{2})=m$. However, as $m$ is odd, this can not hold. As a result, the claim is true, i.e., $|H|=4m$.

As in proof of Theorem 3.1, it is enough to show that $\mathsf{Stab}_{H}(A_{0})=\{\mathsf{id}\}$. Let $\alpha^{i}\beta^{j}\gamma^{k}\in\mathsf{Stab}_{H}(A_{0})$, i.e., $\alpha^{i}\beta^{j}\gamma^{k}(A_{0})=A_{0}$ for some $i,j,k$ with $0\leq i\leq m-1$, $0\leq j,k\leq 1$. Therefore,

[TABLE]

Claim 2: $k=0$.

Proof of Claim 2: If not, let $k=1$, i.e., $\beta^{j}\gamma(A_{0})=A_{2m-2i}$. Note that

both $\varepsilon_{0}$ and $\varepsilon_{m-1}$ occurs in the expression of $\gamma$, and

• 2.

all $\varepsilon_{i}$’s except $\varepsilon_{0}$ and $\varepsilon_{m-1}$ stabilizes $A_{0}$.

Thus $A_{2m-2i}=\beta^{j}\gamma(A_{0})=\beta^{j}\varepsilon_{m-1}\varepsilon_{0}(A_{0})=\beta^{j}\varepsilon_{m-1}(B_{2m-1})=\beta^{j}(A_{m}).$ If $j=0$, then we have $A_{m}=A_{2m-2i}$, which is a contradiction, due to mismatch of parity of indices. If $j=1$, then we have $\beta(A_{m})=A_{2m-2i}$. Note that

$\mathsf{Stab}_{G}(A_{0})=\mathsf{Stab}_{G}(A_{m})=\langle\mu,\varepsilon_{1},\varepsilon_{2},\ldots,\varepsilon_{m-2}\rangle$.

• 2.

$\varepsilon_{0}$ does not occur in the expression of $\beta$, but $\varepsilon_{m-1}$ occur in the expression of $\beta$.

Thus, we have $A_{2m-2i}=\beta(A_{m})=\varepsilon_{m-1}(A_{m})=B_{2m-1}$, a contradiction. Hence for $k=1$, both $j=0$ or $j=1$ leads to a contradiction, and as a result $k=0$.

Thus, from Equation 1, we have $\beta^{j}(A_{0})=A_{2m-2i}$. If $j=1$, then $A_{2m-2i}=\beta(A_{0})=\varepsilon_{m-1}(A_{0})=B_{m-1}$, a contradiction. Thus, $j=0$ and hence we have $A_{0}=A_{2m-2i}$ i.e., $2m\equiv 2i~{}(mod~{}2m)$, i.e., $i\equiv m\equiv 0~{}(mod~{}m)$. Thus $i=0$. This implies that $\mathsf{Stab}_{H}(A_{0})=\{\mathsf{id}\}$ and hence the theorem holds.∎

Theorem 4.3**.**

$R_{2m}(m-2,m-1)$* is not a Cayley graph, if $m$ is odd and $m\not\equiv 0~{}(mod~{}3)$.*

**Proof: **Consider $K=\langle\varepsilon_{0},\varepsilon_{1},\ldots,\varepsilon_{m-1}\rangle$. Then $K\cong\mathbb{Z}_{2}^{m}$ and $|K|=2^{m}$ as $\circ(\varepsilon_{i})=2$ and $\varepsilon_{i}\varepsilon_{j}=\varepsilon_{j}\varepsilon_{i},\forall i,j\in\{0,1,\ldots,m-1\}$.

If possible, let $H$ be a regular subgroup of $G$. Then $|H|=4m$. Thus

[TABLE]

Now, as $|H|=4m$, where $m$ is odd and $|K|=2^{m}$, we have $|H\cap K|=2$ or $4$. We will prove that $|H\cap K|=4$. In fact, using the next two claims, we prove that $|H\cap K|\neq 2$.

Claim 1: If $|H\cap K|=2$, then the non-identity element of $H\cap K$ must be $\varepsilon_{0}\varepsilon_{1}\cdots\varepsilon_{m-1}=\rho^{m}$.

Proof of Claim 1: Let $\alpha=\varepsilon_{l_{1}}\varepsilon_{l_{2}}\cdots\varepsilon_{l_{p}}$ be the non-identity element of $H\cap K$. Let $L=\langle\mu,\varepsilon_{0},\varepsilon_{1},\ldots,\varepsilon_{m-1}\rangle$. Then $|L|=2^{m+1}$ and $K\subsetneq L$ as $\mu\in L\setminus K$. Thus

[TABLE]

As $|H\cap K|=2$ and $K\subsetneq L$, there exists atleast one element of the form $\beta=\mu\varepsilon_{i_{1}}\varepsilon_{i_{2}}\cdots\varepsilon_{i_{s}}$ in $H\cap L$.

Again, let $L^{\prime}=\langle\rho\mu,\varepsilon_{0},\varepsilon_{1},\ldots,\varepsilon_{m-1}\rangle$. By similar arguments, we can deduce that $|H\cap L^{\prime}|\geq 4$. So there exists an element of the form $\gamma=\rho\mu\varepsilon_{j_{1}}\varepsilon_{j_{2}}\cdots\varepsilon_{j_{t}}$ in $H\cap L^{\prime}$.

As $\alpha,\beta,\gamma\in H$, it follows that $\beta\alpha\beta^{-1},\gamma\alpha\gamma^{-1}\in H$. Observe that

[TABLE]

As $\mu\varepsilon_{i}=\varepsilon_{m-1-i}\mu$, $\beta\alpha\beta^{-1}$ is product of some $\varepsilon_{i}$’s and hence $\mathsf{id}\neq\beta\alpha\beta^{-1}\in H\cap K$. Since $|H\cap K|=2$, then $\alpha=\beta\alpha\beta^{-1}$.

Similarly,

[TABLE]

[TABLE]

As $\rho\varepsilon_{i}=\varepsilon_{i+1}\rho$, $\rho\alpha\rho^{-1}$ is product of some $\varepsilon_{i}$’s and hence $\gamma\alpha\gamma^{-1}\in H\cap K$ and by similar arguments, we have $\alpha=\gamma\alpha\gamma^{-1}$.

Thus, using $\rho\varepsilon_{i}=\varepsilon_{i+1}\rho$, we get

[TABLE]

As $K=\langle\varepsilon_{0},\varepsilon_{1},\ldots,\varepsilon_{m-1}\rangle\cong\mathbb{Z}_{2}^{m}$ and $\varepsilon_{i}$’s corresponds to the standard generators of $\mathbb{Z}_{2}^{m}$, i.e., $\varepsilon_{i}\leftrightarrow(0,0,\ldots,0,1,0,\ldots,0)$ with the only $1$ occuring in the $(i+1)$th position, $\varepsilon_{l_{1}}\varepsilon_{l_{2}}\cdots\varepsilon_{l_{p}}$ corresponds to the vector in $\mathbb{Z}_{2}^{m}$ with $1$’s in $l_{1}+1,l_{2}+1,\ldots,l_{p}+1$ positions and $\varepsilon_{l_{1}+1}\varepsilon_{l_{2}+1}\cdots\varepsilon_{l_{p}+1}$ corresponds to the vector with $1$’s in $l_{1}+2,l_{2}+2,\ldots,l_{p}+2$ positions. Thus, from Equation 2, we get that all the positions in the vector must be $1$, i.e., $\alpha=\varepsilon_{0}\varepsilon_{1}\cdots\varepsilon_{m-1}=\rho^{m}$. Hence the claim is true.

Claim 2: If $|H\cap K|=2$, then $\rho^{m}\not\in H$

Proof of Claim 2: As $H\cap L$ is a subgroup of $H$ and $m$ is odd, therefore $4\leq|H\cap L|\mid 4m$ implies $|H\cap L|=4$. Thus $H\cap L$ is either isomorphic to $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ or $\mathbb{Z}_{4}$. Note that any non-identity element $\sigma\in H\cap L$ must contain in its expression either $\varepsilon_{0}$ or $\varepsilon_{m-1}$, as otherwise $\sigma\in\langle\mu,\varepsilon_{1},\varepsilon_{2},\ldots,\varepsilon_{m-2}\rangle=\mathsf{Stab}_{G}(A_{0})$, a contradiction to the fact that $\sigma$ belongs to a regular subgroup $H$.

Suppose that $H\cap L$ is isomorphic to $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$. As $H\cap K\subsetneq H\cap L$, therefore there exists a non-identity element in $H\cap L$ of the form $\sigma=\mu\varepsilon_{i_{1}}\varepsilon_{i_{2}}\cdots\varepsilon_{i_{s}}$. As explained earlier, $\sigma$ must contain in its expression either $\varepsilon_{0}$ or $\varepsilon_{m-1}$. In fact, in this case, both $\varepsilon_{0}$ and $\varepsilon_{m-1}$ must occur in the expression of $\sigma$, as otherwise $\circ(\sigma)=4$. Note that by Claim 1, $\rho^{m}\in H\cap L$. Thus, for all the three non-identity elements, $\rho^{m},\sigma,\sigma^{\prime}$ (say) in $H\cap L$, both $\varepsilon_{0}$ and $\varepsilon_{m-1}$ must occur. Also as $H\cap L\cong\mathbb{Z}_{2}\times\mathbb{Z}_{2}$, we have $\sigma\sigma^{\prime}=\rho^{m}$. But if $\sigma,\sigma^{\prime}$ contains both $\varepsilon_{0}$ and $\varepsilon_{m-1}$, then $\rho^{m}$ contains neither $\varepsilon_{0}$ nor $\varepsilon_{m-1}$, a contradiction. Hence $H\cap L\not\cong\mathbb{Z}_{2}\times\mathbb{Z}_{2}$.

Suppose that $H\cap L$ is isomorphic to $\mathbb{Z}_{4}$. As $\circ(\rho^{m})=2$, there exists a non-identity element $\zeta=\mu\varepsilon_{j_{1}}\varepsilon_{j_{2}}\cdots\varepsilon_{j_{s}}\in H\cap L$ such that $\langle\zeta\rangle=H\cap L$ and $\zeta^{2}=\rho^{m}$. Note that the number of $\varepsilon_{i}$’s in the expression of $\zeta^{2}$ is always even but that of $\rho^{m}$ is $m$ (odd) as $\rho^{m}=\varepsilon_{0}\varepsilon_{1}\cdots\varepsilon_{m-1}$. Hence, $H\cap L\not\cong\mathbb{Z}_{4}$.

Thus, by Claim 1 and 2, we get $|H\cap K|=4$. As $K\cong\mathbb{Z}_{2}^{m}$, we have $H\cap K\cong\mathbb{Z}_{2}\times\mathbb{Z}_{2}$. Recall that

[TABLE]

Again, as the graph is vertex-transitive, by orbit-stabilizer theorem, we have $G=H\cdot\mathsf{Stab}_{G}(B_{(m+3)/2})$. Thus, $\rho=hb$, where $h\in H$ and $b\in\mathsf{Stab}_{G}(B_{(m+3)/2})$.

Claim 3: $\rho\mu$ does not occur in the expression of $b$.

Proof of Claim 3: If possible, let $b=\rho\mu\varepsilon_{l_{1}}\varepsilon_{l_{2}}\cdots\varepsilon_{l_{p}}$ and hence $h=\rho b^{-1}=\mu\varepsilon_{t_{1}}\varepsilon_{t_{2}}\cdots\varepsilon_{t_{p}}\in H\cap L$. Again, as $H\cap K\subseteq H\cap L$ and $|H\cap L|=|H\cap K|=4$, we have $H\cap K=H\cap L$. Thus, $h\in H\cap K\subset K$ and hence $h$ does not contain $\mu$ in its expression, a contradiction. Thus Claim 3 is true.

Therefore, by Claim 3, $b=\varepsilon_{l_{1}}\varepsilon_{l_{2}}\cdots\varepsilon_{l_{p}}$ and $h=\rho b^{-1}=\rho\varepsilon_{l_{1}}\varepsilon_{l_{2}}\cdots\varepsilon_{l_{p}}\in H$.

Let $H\cap K=\{\mathsf{id},\alpha_{1},\alpha_{2},\alpha_{3}\}\cong\mathbb{Z}_{2}\times\mathbb{Z}_{2}$. Thus $h\alpha_{i}h^{-1}\in H$. As $\alpha_{i}$’s, being elements of $K$, are product of some $\varepsilon_{i}$’s and $\varepsilon_{i}\varepsilon_{j}=\varepsilon_{j}\varepsilon_{i}$, $\rho\varepsilon_{i}=\varepsilon_{i+1}\rho$, we have

[TABLE]

Thus $h\alpha_{i}h^{-1}\in H\cap K=\{\mathsf{id},\alpha_{1},\alpha_{2},\alpha_{3}\}$.

Claim 4: $h\alpha_{1}h^{-1}=\alpha_{2}$ or $\alpha_{3}$.

Proof of Claim 4: If $h\alpha_{1}h^{-1}=\mathsf{id}$, then $\alpha_{1}=\mathsf{id}$, a contradiction.

If $h\alpha_{1}h^{-1}=\alpha_{1}$, then as above, get $\varepsilon_{i_{1}+1}\varepsilon_{i_{2}+1}\cdots\varepsilon_{i_{s}+1}=\varepsilon_{i_{1}}\varepsilon_{i_{2}}\cdots\varepsilon_{i_{s}}.$ Now, as in proof of Claim 1, we can argue that this implies $\alpha_{1}=\rho^{m}$. But, in that case, we must have $h\alpha_{2}h^{-1}=\alpha_{3}$ and $h\alpha_{3}h^{-1}=\alpha_{2}$, because otherwise

$h\alpha_{2}h^{-1}=\mathsf{id}$ implies $\alpha_{1}=\mathsf{id}$, a contradiction.

• 2.

$h\alpha_{2}h^{-1}=\mathsf{\alpha_{2}}$ implies $\alpha_{2}=\rho^{m}$, a contradiction, as $\alpha_{1}\neq\alpha_{2}$.

• 3.

$h\alpha_{2}h^{-1}=\mathsf{\alpha_{1}}$ implies $h\alpha_{2}h^{-1}=h\alpha_{1}h^{-1}$, i.e., $\alpha_{1}=\alpha_{2}$, a contradiction.

Thus we have $h\alpha_{2}h^{-1}=\rho\alpha_{2}\rho^{-1}=\alpha_{3}$ and $h\alpha_{3}h^{-1}=\rho\alpha_{3}\rho^{-1}=\alpha_{2}$. Hence, from Equation 3, we see that both $\alpha_{2}$ and $\alpha_{3}$ are product of $\varepsilon_{i}$’s and the number of $\varepsilon_{i}$’s occuring in their expressions are same. Thus the number of $\varepsilon_{i}$’s occuring in the expression of $\alpha_{2}\alpha_{3}$ is even. However, $\alpha_{2}\alpha_{3}=\alpha=\rho^{m}=\varepsilon_{0}\varepsilon_{1}\cdots\varepsilon_{m-1}$ has odd number of $\varepsilon_{i}$’s occuring in its expression. This is a contradiction and hence $h\alpha_{1}h^{-1}\neq\alpha_{1}$. Thus Claim 4 is true.

Without loss of generality, we can assume that $h\alpha_{1}h^{-1}=\alpha_{2}$. Thus $h\alpha_{2}h^{-1}$ is either $\alpha_{1}$ or $\alpha_{3}$. If $h\alpha_{2}h^{-1}=\alpha_{1}$, we must have $h\alpha_{3}h^{-1}=\alpha_{3}$, a contradiction, as shown in Claim 4. Hence we have $h\alpha_{2}h^{-1}=\alpha_{3}$ and similarly $h\alpha_{3}h^{-1}=\alpha_{1}$. So, by Equation 3, we get $\rho\alpha_{1}\rho^{-1}=\alpha_{2}$, $\rho\alpha_{2}\rho^{-1}=\alpha_{3}$ and $\rho\alpha_{3}\rho^{-1}=\alpha_{1}$. Hence, we have

[TABLE]

Similarly, we have $\rho^{3}\alpha_{2}=\alpha_{2}\rho^{3}$ and $\rho^{3}\alpha_{3}=\alpha_{3}\rho^{3}$.

Recall that $H\cap K=\{\mathsf{id},\alpha_{1},\alpha_{2},\alpha_{3}\}\cong\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ and $\alpha_{i}$’s are product of some $\varepsilon_{j}$’s. Let

[TABLE]

Note that each $\alpha_{i}$ must contain either $\varepsilon_{0}$ or $\varepsilon_{m-1}$ in its expression, as otherwise it will be an element of $\mathsf{Stab}_{G}(A_{0})$ and hence can not belong to $H$. As $\alpha_{1}\alpha_{2}=\alpha_{3}$ and $\alpha_{1}\alpha_{2}\alpha_{3}=\mathsf{id}$, without loss of generality, we can assume that, among $\varepsilon_{0}$ or $\varepsilon_{m-1}$, $\alpha_{1}$ contains only $\varepsilon_{0}$, $\alpha_{2}$ contains only $\varepsilon_{m-1}$ and $\alpha_{3}$ contains both $\varepsilon_{0}$ and $\varepsilon_{m-1}$ in their expressions. This happens because if two of the $\alpha_{i}$’s contain both $\varepsilon_{0}$ and $\varepsilon_{m-1}$ in their expressions, then the their product, i.e., the third $\alpha_{i}$, will not have $\varepsilon_{0}$ or $\varepsilon_{m-1}$ in its expression, thereby making it an element of $\mathsf{Stab}_{G}(A_{0})$.

Now, from the relation $\rho^{3}\alpha_{1}=\alpha_{1}\rho^{3}$ and using the fact that $\rho\varepsilon_{i}=\varepsilon_{i+1}\rho$, we get,

[TABLE]

[TABLE]

Now, as $m$ is not a multiple of $3$, $m$ is of the form $3t+1$ or $3t+2$.

If $m=3t+1$, then by using the standard generators of $\mathbb{Z}_{2}^{m}$, as in the proof of Claim 1, we get that all of $\varepsilon_{0},\varepsilon_{3},\varepsilon_{6},\ldots,\varepsilon_{3t}=\varepsilon_{m-1}$ occurs in the expression of $\alpha_{1}$, a contradiction to that fact that among $\varepsilon_{0}$ or $\varepsilon_{m-1}$, $\alpha_{1}$ contains only $\varepsilon_{0}$.

Similarly, if $m=3t+2$, we get all of

[TABLE]

occurs in the expression of $\alpha_{1}$, a contradiction.

Thus, we conclude that there does not exist any regular subgroup $H$ of $\mathsf{Aut}(R_{2m}(m-2,m-1))$ and hence $R_{2m}(m-2,m-1)$ is not a Cayley graph, when $m$ is odd and not a multiple of $3$. ∎

5 Family-4 [$R_{12m}(\pm(3m+2),\pm(3m-1))$ and $R_{12m}(\pm(3m-2),\pm(3m+1))$]

As $R_{n}(a,r)=R_{n}(a,-r)$ and $R_{n}(a,r)\cong R_{n}(-a,r)$, it is enough to check $R_{12m}(3m+2,3m-1)$ and $R_{12m}(3m-2,3m+1)$. More precisely, it suffices to work with the family $R_{12m}(3d+2,9d+1)$ where $d=\pm m~{}(mod~{}12m)$, as mentioned in Section 3.3 of [5]. Define $\sigma$ as follows:

[TABLE]

Also, if $m\equiv 2~{}(mod~{}4)$, let $b=d+1$ and define $\omega$ as follows:

[TABLE]

It was shown in [5], that

[TABLE]

It is to be noted that $m\equiv 2~{}(mod~{}4)$ if and only if $-m\equiv 2~{}(mod~{}4)$. Thus, it is enough to work only with the family $R_{12m}(3m+2,9m+1)$.

Theorem 5.1**.**

If $m$ is odd and $m\neq 3$, then $R_{12m}(3m+2,9m+1)$ is a Cayley graph.

**Proof: **As $m$ is odd, $G=\langle\rho,\mu,\sigma\rangle$. It can also be checked that $\sigma\rho^{3}\sigma=\rho^{3};\sigma\mu=\mu\sigma;(\rho\sigma)^{3}=\rho^{3(m+1)};\circ(\sigma)=2$. Let $\alpha=(\rho\sigma)^{2}$ and $\beta=\rho^{2}\mu\sigma$. As $m$ is odd and $m\neq 3$, it can be shown that $\circ(\alpha)=3m,\circ(\beta)=8$ and $\beta\alpha=\alpha^{-1}\beta^{-1}$. Define

[TABLE]

[TABLE]

Claim 1: The elements in $H$ are distinct.

If not, suppose

[TABLE]

i.e.,

[TABLE]

[TABLE]

any power of $\alpha$ maps $A_{0}$ to $A_{0(mod~{}3)}$ or $A_{1(mod~{}3)}$ or $B_{1(mod~{}3)}$. On the other hand, as

[TABLE]

[TABLE]

we see that $\beta,\beta^{2},\beta^{5}$ and $\beta^{6}$ maps $A_{0}$ to $A_{2(mod~{}3)}$. Thus, $j_{2}-j_{1}$ in Equation 4 can take values from $\{0,3,4,7\}$.

If $j_{2}-j_{1}=0$, then it is obvious that $i_{1}=i_{2}$ and $j_{1}=j_{2}$.

If $j_{2}-j_{1}=4$, squaring Equation 4, we get, $\alpha^{2(i_{1}-i_{2})}=\mathsf{id}$. Therefore, $3m|2(i_{1}-i_{2})$. Now, as $gcd(2,3)=1$ and $m$ is odd, we have $3m|(i_{1}-i_{2})$, i.e., $i_{1}=i_{2}$ and hence $j_{1}=j_{2}$.

If $j_{2}-j_{1}=3$, since $gcd(3,8)=1$, then $\circ(\beta^{j_{2}-j_{1}})=8$. Therefore, $\alpha^{8(i_{1}-i_{2})}=\mathsf{id}$, i.e., $3m|8(i_{1}-i_{2})$. As $m$ is odd, $3m$ is coprime to $8$ and hence, $3m|(i_{1}-i_{2})$, i.e., $i_{1}=i_{2}$ and $j_{1}=j_{2}$.

The case $j_{2}-j_{1}=7$ follows similarly as above. Thus combining all the cases, we see that elements of $H$ are distinct and $H=3m\times 8=24m$.

Claim 2: $H$ acts transitively on $R_{12m}(3m+2,9m+1)$.

In order to prove it, we show that the orbit of $A_{0}$, $\mathcal{O}_{A_{0}}$, under the action of $H$ is the vertex set of $R_{12m}(3m+2,9m+1)$. By orbit-stabilizer theorem, we get

[TABLE]

As the number of vertices in $R_{12m}(3m+2,9m+1)$ is $24m$ and $|H|=24m$, it is enough to show that $\mathsf{Stab}_{H}(A_{0})=\{\mathsf{id}\}$. Let $\alpha^{i}\beta^{j}$ be an arbitrary element of $H$ which stabilizes $A_{0}$, i.e., $\alpha^{-i}(A_{0})=\beta^{j}(A_{0})$ with $0\leq i\leq 3m-1;0\leq j\leq 7$. Again, by mimicing the argument used in the proof of Claim 1, one can conclude that $j\in\{0,3,4,7\}$.

If $j=4$, then $\alpha^{-i}(A_{0})=\beta^{4}(A_{0})=A_{6m}$. Thus, $-i$ and hence $i$ is a multiple of $3$. [since, $\alpha^{x}$ sends $A_{0}$ to $A_{0(mod~{}3)}$, only if $x$ is a multiple of $3$] Let $-i=3k$ and therefore $A_{6m}=\alpha^{3k}(A_{0})=A_{k(6m+6)}$, i.e., $12m|k(6m+6)-6m$, i.e., $2m|m(k-1)+k$, i.e., $m|k$ which implies $k=lm$. Again, as $2m|m(k-1)+lm$, we have $2|k-1+l$, i.e., $2|l(m+1)-1$. But this is a contradiction, as $m+1$ is even and hence $l(m+1)-1$ is odd. Thus $j\neq 4$.

If $j=3$, then $\alpha^{-i}(A_{0})=\beta^{3}(A_{0})=B_{3m+1}$. As $3m+1\equiv 1(mod~{}3)$, we have $-i=3k+1$ [since, $\alpha^{x}$ sends $A_{0}$ to $B_{1(mod~{}3)}$, only if $x\equiv 1(mod~{}3)$] Therefore, $\beta^{3}(A_{0})=B_{3m+1}=\alpha^{3k+1}(A_{0})=\alpha^{3k}(B_{1})$, i.e., $B_{3m+1}=B_{1+6mk+6k}$. This implies $12m|6mk+6k-3m$, i.e., $4m|2mk+2k-m$,i.e., $m|2k$ and, as $m$ is odd, we have $m|k$. Let $k=lm$. Again, as $4m|2mk+2lm-m$, we have $4|2k+2l-1$. However, this is a contradiction, as $2k+2l-1$ is odd and hence $j\neq 3$.

Using similar arguments as above, it can be shown that $j\neq 7$.

Thus, we have $j=0$ and this, in turn, implies $i=0$. Hence, $\mathsf{Stab}_{H}(A_{0})=\{\mathsf{id}\}$.

Finally, in view of Remark 1.2, $H$ acts regularly on $R_{12m}(3m+2,9m+1)$ and hence $R_{12m}(3m+2,9m+1)$ is a Cayley graph, if $m$ is odd and $m\neq 3$.∎

Theorem 5.2**.**

If $m=3$, then $R_{12m}(3m+2,9m+1)$, i.e, $R_{36}(11,28)$ is a Cayley graph.

**Proof: **This can be checked using Sage programming. See Appendix for the SageMath code.∎

Theorem 5.3**.**

If $m\equiv 0(mod~{}4)$, then $R_{12m}(3m+2,9m+1)$ is not a Cayley graph.

**Proof: **As $m\not\equiv 2(mod~{}4)$,

[TABLE]

[TABLE]

If possible, let $R_{12m}(3m+2,9m+1)$ be a Cayley graph, $H$ be a regular subgroup of $G$ and $K=\mathsf{Stab}_{G}(A_{0})$. Then $|G|=96m=8n$ (See Lemma 7.1 in Appendix), $|H|=2n=24m$ and $H\cap K=\{\mathsf{id}\}$.

Let $K^{\prime}=\langle\rho\rangle$. Then $|K^{\prime}|=n$ and $|HK^{\prime}|=\frac{|H||K^{\prime}|}{|H\cap K^{\prime}|}=\frac{2n^{2}}{n/t}\leq|G|=8n$, where $t$ is a factor of $n$. Thus, $t\leq 4$, i.e., $t=1,2,3$ or $4$. If $t=1$, then $H\cap K^{\prime}=K^{\prime}$, i.e., $\rho\in H$. If $t=2$, then $H\cap K^{\prime}=\langle\rho^{2}\rangle$, i.e., $\rho^{2}\in H$. If $t=3$, then $H\cap K^{\prime}=\langle\rho^{3}\rangle$, i.e., $\rho^{3}\in H$. If $t=4$, then $H\cap K^{\prime}=\langle\rho^{4}\rangle$, i.e., $\rho^{4}\in H$. Combining all the cases, we get that

[TABLE]

Claim: $\rho^{4}\in H$.

Proof of Claim: Suppose that that $\rho^{3}\in H$ but $\rho^{4}\not\in H$. Let $L=\langle\rho,\mu\rangle$. Then $|L|=2n$. Therefore

[TABLE]

Therefore, $|H\cap L|=2n,n,2n/3$ or $n/2$, i.e., $|H\cap L|\geq n/2$. As $\rho^{3}\in H\cap L$, we have $\langle\rho^{3}\rangle\subseteq H\cap L$ and $|\langle\rho^{3}\rangle|=n/3$. Thus, $(H\cap L)\setminus\langle\rho^{3}\rangle\neq\emptyset$.

Now, as $\rho^{2i}\mu(A_{i})=A_{i}$, $\rho^{2i}\mu\not\in H$. Similarly, if $\rho^{2i+1}\mu\in H$, then $\rho^{3}\cdot\rho^{2i+1}\mu\in H$, i.e., $\rho^{2i+4}\mu\in H$. Note that $2i+4$ is even and hence by previous argument, $\rho^{2i+4}\mu\not\in H$, i.e., $\rho^{2i+1}\mu\not\in H$. This shows that $H$ does not contain any element of the form $\rho^{i}\mu$. Moreover, $\mu\not\in H$. Now, as $(H\cap L)\setminus\langle\rho^{3}\rangle\neq\emptyset$, $H$ must contain an element of the form $\rho^{i}$, where $i$ is not a multiple of $3$. Again, as $\rho^{3}\in H$, either $\rho$ or $\rho^{2}\in H$, i.e., $\rho^{4}\in H$. This is a contradiction to the assumption that $\rho^{4}\not\in H$. Thus the claim is true.

Let $K^{\prime\prime}=\langle\rho\sigma\rangle$. As $\circ(\rho\sigma)=n$, we have $|K^{\prime\prime}|=n$ and by similar arguments as above, we get that either $(\rho\sigma)^{3}\in H$ or $(\rho\sigma)^{4}\in H$.

Case 1: If $\rho^{4}\in H$ and $(\rho\sigma)^{4}\in H$, then

[TABLE]

As $\rho^{4}\in H$, therefore $\sigma\in H$. But as $\sigma(A_{0})=A_{0}$, i.e., $\sigma$ stabilizes $A_{0}$, it can not be in $H$. This is a contradiction.

Case 2: If $\rho^{4}\in H$ and $(\rho\sigma)^{3}\in H$, then $(\rho\sigma)^{3}=\rho^{3(m+1)}=\rho^{12l+3}=(\rho^{4})^{3l}\rho^{3}\in H$, where $m=4l$ i.e., $\rho^{3}\in H$. Again, as $\rho^{4}\in H$, we have $\rho\in H$. As $\circ(\rho)=n$ and $[H:\langle\rho\rangle]=2$, $\langle\rho\rangle$ is normal in $H$.

From definition, it follows that $\mathsf{id},\mu,\sigma,\mu\sigma\in K$. On the other hand, as $R_{12m}(3m+2,9m+1)$ is vertex transitive, by orbit-stabilizer theorem, we have

[TABLE]

Thus, $HK=G$. As $\sigma\rho\in G$, it can be expressed in the form $\alpha\beta$, where $\alpha\in H$ and $\beta\in K=\{\mathsf{id},\mu,\sigma,\mu\sigma\}$.

If $\beta=\mathsf{id}$, then $\alpha=\sigma\rho\in H$, i.e., $\sigma\in H$ (as $\rho\in H$), which is a contradiction, as $H$, being a regular subgroup can not contain any non-identity element which stabilizes $A_{0}$.

If $\beta=\mu$, then $\sigma\rho=\alpha\mu$, i.e., $\alpha=\sigma\mu\rho^{-1}\in H$, i.e., $\sigma\mu\in H$ (as $\rho\in H$), which is a contradiction.

If $\beta=\sigma$, then $\alpha=\sigma\rho\sigma\in H$. Since $\langle\rho\rangle$ is normal in $H$, therefore $(\sigma\rho\sigma)\rho(\sigma\rho\sigma)^{-1}\in H$, i.e.,

[TABLE]

a contradiction.

If $\beta=\mu\sigma$, then $\sigma\rho=\alpha\mu\sigma$, i.e., $\alpha=\sigma\rho\mu\sigma\in H$. Since $\langle\rho\rangle$ is normal in $H$, therefore $(\sigma\rho\mu\sigma)\rho(\sigma\rho\mu\sigma)^{-1}\in H$, i.e.,

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Thus, combining Case 1 and 2, we conclude that there does not exist any regular subgroup $H$ of $G$, i.e., $R_{12m}(3m+2,9m+1)$ is not a Cayley graph, if $m\equiv 0(mod~{}4)$. ∎

5.1 $m\equiv 2(mod~{}4)$

As $m\equiv 2(mod~{}4)$, $G=\langle\rho,\mu,\sigma,\omega\rangle$. It can be checked that $\sigma\rho^{3}\sigma=\rho^{3};\sigma\mu=\mu\sigma;\sigma\omega=\omega\sigma;\omega\rho=\sigma\rho\omega;\omega\mu=\mu\omega\sigma;(\rho\sigma)^{3}=\rho^{3(m+1)};\omega\rho^{3l}=\rho^{3l(m+1)};(\rho\sigma\rho)^{3}=(\rho^{2}\sigma)^{3}=\rho^{9m+6};\circ(\sigma)=\circ(\omega)=\circ(\sigma\omega)=2;\circ(\omega\mu)=4$.

Let $\alpha=\omega\sigma\rho^{4m}\omega\sigma$ and $\beta=\rho^{3m/2}$. Using the above relations, it can be shown that $\circ(\alpha)=3;\circ(\beta)=8;\alpha\beta=\beta\alpha$. Define

[TABLE]

In all the cases, it can be checked that $\circ(\gamma)=2m$, $\alpha\gamma=\gamma\alpha$ and $\gamma\beta=\beta^{m+1}\gamma$. It is to be noted that $\alpha=\omega\sigma\rho^{4m}\omega\sigma=(\omega\sigma\rho\omega\sigma)^{4m}=[\omega(\sigma\rho\omega)\sigma]^{4m}=(\omega(\omega\rho)\sigma)^{4m}=(\rho\sigma)^{4m}$.

Proposition 5.1**.**

\gamma^{2}=\left\{\begin{array}[]{ll}\rho^{4m+4},&\mbox{ if }m\mbox{ is of the form }12l+2\mbox{ or }12l+6\\ \rho^{12},&\mbox{ if }m\mbox{ is of the form }12l+10.\end{array}\right.** 2. 2.

\gamma^{m}=\left\{\begin{array}[]{ll}\alpha^{2}\beta^{4},&\mbox{ if }m\mbox{ is of the form }12l+6\\ \beta^{4},&\mbox{ if }m\mbox{ is of the form }12l+2\mbox{ or }12l+10.\end{array}\right.**

**Proof: **See Appendix.

Theorem 5.4**.**

If $m\equiv 2(mod~{}12)$, then $R_{12m}(3m+2,9m+1)$ is a Cayley graph.

**Proof: **Let $m=12l+2$. Therefore $8m=96l+16$, i.e., $8m-4=12(8l+1)$. Then $\gamma^{2}=\rho^{4m+4}$.(by Proposotion 5.1) Define

[TABLE]

Thus, it is clear that every element of $H$ is of the form $\alpha^{i}\beta^{j}\gamma^{k}$ where $i=0,1,2$; $j=0,1,\ldots,7$ and $k=0,1,\ldots,m-1$.

Claim 1: $H=\{\alpha^{i}\beta^{j}\gamma^{k}:i=0,1,2;j=0,1,\ldots,7;k=0,1,\ldots,m-1\}$.

Proof of Claim 1: If possible, let there exist $i_{1},i_{2}\in\{0,1,2\},j_{1},j_{2}\in\{0,1,\ldots,7\}$ and $k_{1},k_{2}\in\{0,1,\ldots,m-1\}$, such that $\alpha^{i_{1}}\beta^{j_{1}}\gamma^{k_{1}}=\alpha^{i_{2}}\beta^{j_{2}}\gamma^{k_{2}}$. As $\alpha\beta=\beta\alpha$ and $\alpha\gamma=\gamma\alpha$, we have

[TABLE]

Case 1: $k_{1}-k_{2}$ is even.

As $\gamma^{2}=\rho^{4m+4}$ and $\beta=\rho^{3m/2}$, we have $\alpha^{i_{2}-i_{1}}=\rho^{x}$, i.e., $(\rho\sigma)^{4m(i_{2}-i_{1})}=\rho^{x}$. This implies that $3$ divides $4m(i_{2}-i_{1})$, i.e., $3|m$ or $3|(i_{2}-i_{1})$. As $3$ does not divide $m$, we have $3|(i_{2}-i_{1})$, i.e., $i_{1}=i_{2}$. Thus $\beta^{j_{1}-j_{2}}=\gamma^{k_{2}-k_{1}}=(\gamma^{2})^{(k_{2}-k_{1})/2}$, i.e.,

[TABLE]

Therefore, $12m$ divides $3m(j_{1}-j_{2})/2-2(m+1)(k_{2}-k_{1})$, i.e.,

[TABLE]

Thus, $m$ divides $4(m+1)(k_{2}-k_{1})$. As $gcd(m,4)=2$ and $gcd(d,d+1)=1$, it follows that $m/2$ divides $k_{2}-k_{1}$, i.e., $k_{2}-k_{1}=\frac{m}{2}s$. Since, $0\leq k_{2}-k_{1}<m$, we have $s=0$ or $1$. Again, as $m+1$ is a multiple of $3$, from Equation 7, we get that $12$ divides $3m(j_{1}-j_{2})$, i.e., $2$ divides $j_{1}-j_{2}$. Let $j_{1}-j_{2}=2t$. As $0\leq j_{1}-j_{2}<8$, we have $t\in\{0,1,2,3\}$. Thus, rewriting Equation 7, we get $24m$ divides $6mt-2m(m+1)s$, i.e., $12$ divides $3t-(m+1)s$. Thus

[TABLE]

If $s=1$, then $k_{2}-k_{1}=m/2=6l+1$ is odd, a contradiction. Thus $s=0$ and hence from Equation 8, we have $4$ divides $t$, i.e., $t=0$. Therefore, we have $j_{1}-j_{2}=k_{1}-k_{2}=0$, and as a result $i_{1}=i_{2}$. Thus Claim 1 is true, if Case 1 holds.

Case 2: $k_{1}-k_{2}$ is odd.

Let $k_{1}-k_{2}=2t-1$. Then we have $\alpha^{i_{2}-i_{1}}=\beta^{j_{1}-j_{2}}(\gamma^{2})^{t}\gamma^{-1}.$ As $\gamma^{2}=\rho^{4m+4}$ and $\beta=\rho^{3m/2}$, we have $\gamma\alpha^{i_{2}-i_{1}}=\rho^{x}$. Now $i_{2}-i_{1}=0,1$ or $2$. Thus either of $\gamma,\alpha\gamma,\alpha^{2}\gamma$ is $\rho^{x}$. But

[TABLE]

[TABLE]

[TABLE]

As each of $\gamma,\alpha\gamma,\alpha^{2}\gamma$ maps some $A_{i}$ to some $B_{j}$, none of them is equal to $\rho^{x}$ and hence a contradiction. So $k_{1}-k_{2}$ can not be odd.

Combining Case 1 and 2, we conclude that Claim 1 is true and hence $|H|=24m=2n$. So, as in proof of Claim 2 in Theorem 5.1, it suffices to show that $\mathsf{Stab}_{H}(A_{0})=\{\mathsf{id}\}$. Let $\alpha^{i}\beta^{j}\gamma^{k}(A_{0})=A_{0}$.

Claim 2: $k$ is even.

Proof of Claim 2: If possible, let $k$ be odd, say $k=2t+1$. Then, as $\alpha$ commutes with $\beta$ and $\gamma$, we have $\beta^{j}\gamma^{2t}\gamma\alpha^{i}(A_{0})=A_{0}$, i.e., $\gamma\alpha^{i}(A_{0})=\beta^{-j}(\gamma^{2})^{-t}(A_{0})=\rho^{x}(A_{0})=A_{x}$, as in Case 2 above. Now, $i=0,1$ or $2$ and as $\gamma(A_{0})=B_{5m+1}$ and $\alpha^{2}\gamma(A_{0})=B_{10m+1}$, we have $i=1$. This implies $\alpha\beta^{j}\gamma^{2t+1}(A_{0})=A_{0}$, i.e., $\beta^{j}(\gamma^{2})^{t}\gamma(A_{0})=\alpha^{2}(A_{0})=A_{11m}$, i.e.,

[TABLE]

Hence the claim is true and let $k=2t$. Therefore,

[TABLE]

As left side of the above equation is $\rho^{x}(A_{0})$ and $\alpha(A_{0})=B_{10m-1}$, we conclude that $i=0$ or $1$. If $i=1$, then we have $\alpha\beta^{j}(\gamma^{2})^{t}(A_{0})=A_{0}$. Again as $\alpha$ commutes with $\beta$ and $\gamma$, we have

[TABLE]

Therefore, $i=0$ and hence we have $\beta^{j}(\gamma^{2})^{t}(A_{0})=A_{0}$, i.e.,

[TABLE]

Thus $12$ divides $3j(6l+1)$, i.e., $4|j(6l+1)$. However as $6l+1$ is odd and $j\in\{0,1,\ldots,7\}$, we have $j=0$ or $4$. If $j=4$, we have $12m$ divides $12(4l+1)t+12(6l+1)$, i.e., $m=12l+2=2(6l+1)$ divides $(4l+1)t+12(6l+1)$ and hence $2(6l+1)$ divides $(4l+1)t$. As $3(4l+1)-2(6l+1)=1$, we have $gcd(4l+1,6l+1)=1$ and hence $6l+1$ divides $t$. However as $0\leq k\leq m-1$, we have $0\leq t\leq\frac{m-1}{2}<6l+1$. Thus the only possible value of $t$ is [math] and hence $k=0$. Therefore, we have $\beta^{j}(A_{0})=A_{0}$, i.e., $\rho^{3(6l+1)j}(A_{0})=A_{0}$. This implies that $12m=12(12l+2)$ divides $3(6l+1)j$, i.e., $8|j$ and hence $j=0$.

Thus we have $\mathsf{Stab}_{H}(A_{0})=\{\mathsf{id}\}$ and the theorem holds.∎

Theorem 5.5**.**

If $m\equiv 6(mod~{}12)$, then $R_{12m}(3m+2,9m+1)$ is a Cayley graph.

**Proof: **Let $m=12l+6$. Therefore $8m=96l+48=12(8l+4)$. Also note that in this case, $\alpha=(\rho\sigma)^{4m}=((\rho\sigma)^{3})^{4(4l+2)}=\rho^{12(m+1)(4l+2)}=\rho^{12(4l+2)}=\rho^{4m}$. Also $\gamma^{2}=\rho^{4m+4}$. (by Proposition 5.1) Define

[TABLE]

Thus, it is clear that every element of $H$ is of the form $\alpha^{i}\beta^{j}\gamma^{k}$ where $i=0,1,2$; $j=0,1,\ldots,7$ and $k=0,1,\ldots,m-1$.

Claim 1: $H=\{\alpha^{i}\beta^{j}\gamma^{k}:i=0,1,2;j=0,1,\ldots,7;k=0,1,\ldots,m-1\}$.

Proof of Claim 1: If possible, let there exist $i_{1},i_{2}\in\{0,1,2\},j_{1},j_{2}\in\{0,1,\ldots,7\}$ and $k_{1},k_{2}\in\{0,1,\ldots,m-1\}$, such that $\alpha^{i_{1}}\beta^{j_{1}}\gamma^{k_{1}}=\alpha^{i_{2}}\beta^{j_{2}}\gamma^{k_{2}}$. As $\alpha\beta=\beta\alpha$ and $\alpha\gamma=\gamma\alpha$, we have

[TABLE]

If $k_{1}-k_{2}$ is odd, say $k_{1}-k_{2}=2t-1$, then $\gamma=\alpha^{i_{1}-i_{2}}\beta^{j_{1}-j_{2}}\gamma^{2t}$. As $\alpha=\rho^{4m}$, the right hand side is of the form $\rho^{x}$. On the other hand, $\gamma(A_{0})=B_{5m+1}$. Thus $\gamma\neq\alpha^{i_{1}-i_{2}}\beta^{j_{1}-j_{2}}\gamma^{2t}$. Hence $k_{1}-k_{2}$ is even, say $2t$. Thus, we have $\rho^{4m(i_{2}-i_{1})}=\rho^{(4m+4)t+3\frac{m}{2}(j_{1}-j_{2})},\mbox{ i.e., }$

[TABLE]

This implies that $4$ divides $9(2l+1)(j_{1}-j_{2})$, i.e., $4|(j_{1}-j_{2})$. Now as $0\leq j_{1}-j_{2}\leq 7$, we have $j_{1}-j_{2}=0$ or $4$.

Sub-claim 1a: $j_{1}-j_{2}=0$.

If possible, let $j_{1}-j_{2}=4$. Then, from Equation 10, we have $12m\mbox{ divides }4(m+1)t+6m+4m(i_{1}-i_{2})$ and hence $m|4(m+1)t$, i.e., $m|4t$, as $gcd(m.m+1)=1$. Now, as $0\leq 4t=2(k_{1}-k_{2})\leq 2m-2$, we have $4t=0$ or $m$. However, if $4t=m$, we have $2t=(6l+3)$, an odd number. Thus $4t$ and hence $t=0$. Therefore, from Equation 10, we get $12m$ divides $6m+4m(i_{1}-i_{2})$, i.e., $6|4(i_{1}-i_{2})$ which implies $3|(i_{1}-i_{2})$ i.e., $i_{1}=i_{2}$. However, this implies that $12m|6m$, a contradiction. Thus Sub-claim 1a is true and $j_{1}=j_{2}$. Thus Equation 10 reduces to

[TABLE]

Again since $gcd(m,m+1)=1$, this implies that $m|t$. However, as $0\leq t\leq\frac{m-1}{2}$, we have $t=0$ and hence $k_{1}=k_{2}$. Thus from Equation 11, we get $3|(i_{1}-i_{2})$, i.e., $i_{1}=i_{2}$. Thus Claim 1 is true and $|H|=24m=2n$. So, as in proof of Claim 2 in Theorem 5.1, it suffices to show that $\mathsf{Stab}_{H}(A_{0})=\{\mathsf{id}\}$. Let $\alpha^{i}\beta^{j}\gamma^{k}(A_{0})=A_{0}$. As $\alpha=\rho^{4m},\beta=\rho^{3m/2}$ and $\gamma^{2}=\rho^{4m+4}$ are powers of $\rho$ and $\gamma(A_{0})=B_{5m+1}$, if $k$ is odd, $\alpha^{i}\beta^{j}\gamma^{k}(A_{0})=B_{x}$ for some index $x$. Thus $k$ is even, say $k=2t$. Thus, we have $\alpha^{i}\beta^{j}\gamma^{k}(A_{0})=\rho^{4mi+8(m+1)t+\frac{3mj}{2}}=A_{0}$, i.e., $12m$ divides $4mi+8(m+1)t+\frac{3mj}{2}$, i.e.,

[TABLE]

This implies that $m|16(m+1)t$. As $gcd(m,m+1)=1$, we have $m|16t$. Again, as $m=12l+6=2(6l+3)$ and $6l+3$ is odd, we have $m|2t=k$, i.e., $k=t=0$. Thus Equation 12 reduces to $24m$ divides $8mi+3mj$, i.e., $24|(8i+3j)$. However, this implies that $8|j$ and $3|i$, i.e., $i=j=0$. Thus $\mathsf{Stab}_{H}(A_{0})=\{\mathsf{id}\}$ and the theorem holds.∎

Theorem 5.6**.**

If $m\equiv 10(mod~{}12)$, then $R_{12m}(3m+2,9m+1)$ is a Cayley graph.

**Proof: **Let $m=12l+10$. Therefore $8m=96l+80$, i.e., $8m-8=12(8l+6)$. By Proposition 5.1, we have $\gamma^{2}=\rho^{12}$. Define

[TABLE]

Thus, it is clear that every element of $H$ is of the form $\alpha^{i}\beta^{j}\gamma^{k}$ where $i=0,1,2$; $j=0,1,\ldots,7$ and $k=0,1,\ldots,m-1$.

Claim 1: $H=\{\alpha^{i}\beta^{j}\gamma^{k}:i=0,1,2;j=0,1,\ldots,7;k=0,1,\ldots,m-1\}$.

Proof of Claim 1: If possible, let there exist $i_{1},i_{2}\in\{0,1,2\},j_{1},j_{2}\in\{0,1,\ldots,7\}$ and $k_{1},k_{2}\in\{0,1,\ldots,m-1\}$, such that $\alpha^{i_{1}}\beta^{j_{1}}\gamma^{k_{1}}=\alpha^{i_{2}}\beta^{j_{2}}\gamma^{k_{2}}$. As $\alpha\beta=\beta\alpha$ and $\alpha\gamma=\gamma\alpha$, we have

[TABLE]

If $k_{1}-k_{2}$ is odd, say $k_{1}-k_{2}=2t-1$, then $\gamma\alpha^{i_{2}-i_{1}}=\beta^{j_{1}-j_{2}}\gamma^{2t}$. As $\gamma^{2}=\rho^{12}$, the right hand side is of the form $\rho^{x}$, i.e., $\gamma\alpha^{i_{2}-i_{1}}=\rho^{x}$. Now $i_{2}-i_{1}=0,1$ or $2$. Thus either of $\gamma,\alpha\gamma,\alpha^{2}\gamma$ is $\rho^{x}$. But $\gamma(A_{0})=B_{9m+5},\alpha\gamma(A_{0})=B_{10m+5},~{}\alpha^{2}\gamma(B_{0})=A_{5m+7}$. As each of $\gamma,\alpha\gamma,\alpha^{2}\gamma$ maps some $A_{i}$ to some $B_{j}$, none of them is equal to $\rho^{x}$ and hence a contradiction. So $k_{1}-k_{2}$ is even, say $k_{1}-k_{2}=2t$. As $\gamma^{2}=\rho^{12}$ and $\beta=\rho^{3m/2}$, we have $\alpha^{i_{2}-i_{1}}=\rho^{x}$, i.e., $(\rho\sigma)^{4m(i_{2}-i_{1})}=\rho^{x}$. This implies that $3$ divides $4m(i_{2}-i_{1})$, i.e., $3|m$ or $3|(i_{2}-i_{1})$. As $3$ does not divide $m$, we have $3|(i_{2}-i_{1})$, i.e., $i_{1}=i_{2}$. Thus $\rho^{\frac{3m}{2}(j_{1}-j_{2})}=\beta^{j_{1}-j_{2}}=\gamma^{k_{2}-k_{1}}=(\gamma^{2})^{t}=\rho^{12t}$, i.e.,

[TABLE]

Thus, we have $m|24t$. As $m=2(6l+5)$, $(6l+5)$ is odd and $3$ does not divide $(6l+5)$, we get $\frac{m}{2}|t$. However, as $0\leq k_{2}-k_{1}\leq m-1$, we have $0\leq t\leq\frac{m-1}{2}$. Hence $t=0$ and $k_{1}=k_{2}$. Also Equation 14 reduces to $8|(j_{1}-j_{2})$. Thus $j_{1}=j_{2}$. Hence Claim 1 is true and $|H|=24m=2n$.

So, as in proof of Claim 2 in Theorem 5.1, it suffices to show that $\mathsf{Stab}_{H}(A_{0})=\{\mathsf{id}\}$. Let $\alpha^{i}\beta^{j}\gamma^{k}(A_{0})=A_{0}$.

Claim 2: $k$ is even.

Proof of Claim 2: If possible, let $k$ be odd, say $k=2t+1$. Then, as $\alpha$ commutes with $\beta$ and $\gamma$, we have $\beta^{j}\gamma^{2t}\gamma\alpha^{i}(A_{0})=A_{0}$, i.e., $\gamma\alpha^{i}(A_{0})=\beta^{-j}(\gamma^{2})^{-t}(A_{0})=\rho^{x}(A_{0})=A_{x}$, as in the proof of Claim 1 of this theorem. Now, $i=0,1$ or $2$ and as $\gamma(A_{0})=B_{9m+5}$ and $\alpha\gamma(A_{0})=B_{10m+5}$, we have $i=2$. This implies $\alpha^{2}\beta^{j}\gamma^{2t+1}(A_{0})=A_{0}$, i.e., $\beta^{j}(\gamma^{2})^{t}\gamma(A_{0})=\alpha(A_{0})=A_{7m}$, i.e.,

[TABLE]

Hence the claim is true and let $k=2t$. Therefore,

[TABLE]

As left side of the above equation is $\rho^{x}(A_{0})$ and $\alpha^{2}(A_{0})=B_{2m-1}$, we conclude that $i=0$ or $2$. If $i=2$, then we have $\alpha^{2}\beta^{j}(\gamma^{2})^{t}(A_{0})=A_{0}$. Again as $\alpha$ commutes with $\beta$ and $\gamma$, we have

[TABLE]

Therefore, $i=0$ and hence we have $\beta^{j}(\gamma^{2})^{t}(A_{0})=A_{0}$, i.e.,

[TABLE]

Thus $12$ divides $3j(6l+5)$, i.e., $4|j(6l+5)$. However as $6l+5$ is odd and $j\in\{0,1,\ldots,7\}$, we have $j=0$ or $4$. If $j=4$, we have $12m$ divides $12t+12(6l+5)$, i.e., $m=12l+10=2(6l+5)$ divides $t+(6l+5)$ and hence $(6l+5)$ divides $t$. However as $0\leq k\leq m-1$, we have $0\leq t\leq\frac{m-1}{2}<6l+5$. Thus the only possible value of $t$ is [math] and hence $k=0$. Therefore, we have $\beta^{j}(A_{0})=A_{0}$, i.e., $\rho^{3(6l+5)j}(A_{0})=A_{0}$. This implies that $12m=24(6l+5)$ divides $3(6l+5)j$, i.e., $8|j$ and hence $j=0$.

Thus we have $\mathsf{Stab}_{H}(A_{0})=\{\mathsf{id}\}$ and the theorem holds.∎

6 Family-5 [$R_{2m}(2b,r)$: $b^{2}\equiv\pm 1$(mod $m$) and $r\in\{1,m-1\}$ is odd]

Theorem 6.1**.**

If $b^{2}\equiv\pm 1$(mod $m$) and $r\in\{1,m-1\}$ is odd, then $R_{2m}(2b,r)$ is a Cayley graph.

**Proof: **If $r=1$, then it is clear that the conditions of being in Family-1 are satisfied, (i.e., $r^{2}\equiv 1(mod~{}n)$ and $ra\equiv a(mod~{}n)$) and hence, by Theorem 2.1, $R_{2m}(2b,r)$ is a Cayley graph. So we are left with the case when $n=2m$, $a=2b$, $b^{2}\equiv\pm 1(mod~{}m)$, $r=m-1$ and $m$ is even. Observe that, in this case,

[TABLE]

Also, as $m$ divides $bm$ i.e., $m|b(r+1)$, we have $br\equiv-b(mod~{}m)$, i.e., $2br\equiv-2b(mod~{}2m)$, i.e., $ra\equiv-a(mod~{}n)$. Thus, in this case, $r^{2}\equiv 1(mod~{}n)$ and $ra\equiv-a(mod~{}n)$ holds. Hence, by Theorem 2.1, $R_{2m}(2b,r)$ is a Cayley graph.∎

Remark 6.1**.**

The above theorem shows that Family-5 is a subfamily of Family-1. However, they were shown as different families in Theorem 3.10 in [1].

Combining the analysis of the rose window graphs in Families: 1–5, we have Theorem 1.3.

7 Appendix

Lemma 7.1**.**

Let $G=\mathsf{Aut}(R_{12m}(3m+2,9m+1))$ , where $m\equiv 0(mod~{}4)$. Then $|G|=96m$.

**Proof: **Since, $R_{12m}(3m+2,9m+1)$ is vertex-transitive and its order is $24m$ and $\mathsf{Stab}_{G}(A_{0})$ contains $\mathsf{id},\mu,\sigma,\mu\sigma$, therefore, by orbit-stabilizer theorem, we have $|G|\geq 4\times 24m=96m$. Thus, it is enough to show that $|G|\leq 96m$. We also know that

[TABLE]

[TABLE]

Consider the sets $X=\{\rho^{i}\sigma\rho^{j}\mu^{k}:i\in\{0,1,2\ldots,n-1\},j\in\{0,1,2\},k\in\{0,1\}\}$ and $Y=\{\rho^{i}\mu^{k}:i\in\{0,1,2\ldots,n-1\},k\in\{0,1\}\}$. We claim that all elements are either in $X$ or in $Y$. It is clear that elements in $G$ which does not involve $\sigma$ are in $Y$, due to the relations $\rho^{n}=\mu^{2}=\mathsf{id}$ and $\mu\rho\mu=\rho^{-1}$. Again, as $\sigma\mu=\mu\sigma$ and $\mu\rho=\rho^{-1}\mu$, any element in $G$ can be expressed in the form where $\mu$ occurs in the extreme right of the expression. Thus it is enough to show that elements in $G$ which involve only $\rho$ and $\sigma$ are of the form $\rho^{i}\sigma\rho^{j}$ where $i\in\{0,1,2\ldots,n-1\}$ and $j\in\{0,1,2\}$. Again, as $\sigma\rho^{3}=\rho^{3}\sigma$, it is clear that the power of $\rho$ lying on the right of $\sigma$ can be made $0,1$ or $2$. Finally, we deal with elements $\sigma\rho\sigma$ and $\sigma\rho^{2}\sigma$.

As $(\rho\sigma\rho)^{3}=\rho^{9m+6}$, we have $\sigma\rho^{2}\sigma\rho^{2}\sigma=\rho^{9m+4}$, i.e.,

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As $(\rho\sigma)^{3}=\rho^{3(m+1)}$, we have $(\sigma\rho\sigma\rho\sigma)=\rho^{3m+2}$, i.e.,

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Similarly, any other element of $G$ involving $\rho$ and $\sigma$ can be expressed in the form of elements in $X$. Thus $G=X\cup Y$ and hence

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∎

**Proof of Proposition 5.1 **:

For $m=12l+2$, we have $8m=96l+16$, i.e., $8m-4=12(8l+1)$.

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For $m=12l+6$, we have $8m=96l+48=12(8l+4)$.

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Similarly, for $m=12l+10$, it can be proved that $\gamma^{2}=\rho^{12}$. 2. 2.

The values of $\gamma^{m}$ can be found by raising $\gamma^{2}$ to the power $m/2$, and hence can be checked to have the respective forms.

Checking whether a rose window graph is Cayley using SageMath

The following is the code to check whether a rose window graph is Cayley. The code is given for $R_{36}(11,28)$, which was claimed to be Cayley in Theorem 5.2. Readers can also edit the values of $n,a,r$ to check for other rose window graphs. The output will be TRUE, if the graph is Cayley, else it will be FALSE.

      n=36
      a=11
      r=28
      A = list(var(’A_%d’ % i) for i in range(n))
      B = list(var(’B_%d’ % i) for i in range(n))
      V=A+B
      E=[]
      G=Graph()
      G.add_vertices(V)
      for i in range(n):
          E.append((A[i],A[mod(i+1,n)]))
          E.append((A[i],B[i]))
          E.append((B[i],A[mod(i+a,n)]))
          E.append((B[i],B[mod(i+r,n)]))
      G.add_edges(E)
      G.is_cayley()