Classification of Cayley Rose Window Graphs
Angsuman Das
[email protected]
Arnab Mandal
[email protected]
Department of Mathematics,
Presidency University, Kolkata
86/1, College Street, Kolkata - 7000073
West Bengal, India
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.
keywords:
vertex-transitive , regular subgroup , rose window graph.
MSC:
[2010] 05C75, 05E18
††journal:
1 Introduction
Rose window graphs were introduced in [6] in the following way:
Definition 1.1**.**
Given natural numbers n≥3 and 1≤a,r≤n−1, the Rose Window graph Rn(a,r) is defined to be the graph with vertex set V={Ai,Bi:i∈Zn} and four kind of edges: AiAi+1 (rim edges), AiBi (inspoke edges), Ai+aBi (outspoke edges) and BiBi+r (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:
-
Rn(2,1).
2. 2.
R2m(m±2,m±1)**
3. 3.
R12m(±(3m+2),±(3m−1))* and R12m(±(3m−2),±(3m+1)).*
4. 4.
R2m(2b,r), where b2≡±1(mod m), 2≤2b≤m, and r∈{1,m−1} is odd.
A similar characterization for vertex-transitive graphs was proved in [1]:
Theorem 1.2**.**
[1]** A rose window graph Rn(a,r) is vertex-transitive if and only if it belongs to one of the following families:
-
Rn(a,r), where r2≡±1(mod n) and ra≡±a(mod n).
2. 2.
R4m(2m,r), where r is odd and (r2+2m)≡±1(mod 4m).
3. 3.
R2m(m±2,m±1)**
4. 4.
R12m(±(3m+2),±(3m−1))* and R12m(±(3m−2),±(3m+1)).*
5. 5.
R2m(2b,r), where b2≡±1(mod m), 2≤2b≤m, and r∈{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 Rn(a,r) is Cayley if and only if one of the following holds:
-
Rn(a,r), where r2≡±1(mod n) and ra≡±a(mod n).
2. 2.
R4m(2m,r), where r is odd and (r2+2m)≡1(mod 4m).
3. 3.
R2m(m±2,m±1)* where m is a multiple of 2 or 3.*
4. 4.
R12m(±(3m+2),±(3m−1))* and R12m(±(3m−2),±(3m+1)) where m≡0(mod 4).*
5. 5.
R2m(2b,r), where b2≡±1(mod m), 2≤2b≤m, and r∈{1,m−1} is odd.∎
Before stating the proof, we note a few generic automorphisms and other properties of Rn(a,r). Other automorphisms, specific to any particular family of rose window graphs, will be introduced whenever they are needed.
-
Define τ:V→V by τ(Ai)=A−i and τ(Bi)=B−i. Clearly τ is an automorphism with τ2=id and hence Rn(a,r)≅Rn(−a,r).
2. 2.
Rn(a,r)=Rn(a,−r).
3. 3.
Define ρ:V→V by ρ(Ai)=Ai+1 and ρ(Bi)=Bi+1; and μ:V→V by μ(Ai)=A−i and μ(Bi)=B−a−i. Clearly ρ and μ are automorphisms. As ρn=μ2=id and μρμ=ρ−1, we have ⟨ρ,μ⟩≅Dn.
4. 4.
If (n,r)=1, then ζ:V→V given by ζ(Ai)=B−ir−1 and ζ(Bi)=A−ir−1 is an automorphism and hence Rn(a,r)≅Rn(ar−1,r−1).
Remark 1.1**.**
In view of the first two observations, it is enough to study Rn(a,r) for 1≤a,r≤⌈2n⌉.
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 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 [Rn(a,r): r2≡±1(mod n) and ra≡±a(mod n)]
If r2≡±1(mod n) and ra≡±a(mod n), then δ:V→V given by δ(Ai)=Bri and δ(Bi)=Ari is an automorphism. For proof, see Lemma 2 [6] or Lemma 3.7 [1]. If r2≡1(mod n), then δ2=id and if r2≡−1(mod n), then δ2=τ, i.e., δ is of order 4.
Theorem 2.1**.**
If r2≡1(mod n) and ra≡±a(mod n), then Rn(a,r) is a Cayley graph.
**Proof: **Since Rn(a,r)=Rn(a,−r), without loss of generality, we can assume that ra≡−a(mod n). Consider ρ and δ as defined above. We have ρn=δ2=id and δρδ=ρr. Define
[TABLE]
[TABLE]
Clearly, H is a subgroup of Aut(Rn(a,r)). It suffices to show that H acts regularly on Rn(a,r). For that we observe that
ρj(Ai)=Ai+j and ρj(Bi)=Bi+j, and
ρjδ(Ai)=Bri+j and ρjδ(Bi)=Ari+j.
As gcd(r,n)=1, the map i↦ri+j is a bijection on {0,1,…,n−1}. Thus H acts transitively on Rn(a,r). It is also clear from the construction of H, that for any pair of vertices in Rn(a,r), there exists a unique element in H which maps one to the other. Hence, Rn(a,r) is a Cayley graph.∎
Lemma 2.1**.**
If r2≡−1(mod n) and ra≡±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 pi∣n and pi+1∤n. Then r2≡−1(mod pi) and r2≡−1(mod p). Again, pi∣a(r±1), i.e., p∣a(r±1). If p∣(r±1), then r2≡1(mod p), a contradiction, as −1≡1(mod p). Thus for all odd prime factors p of n, we have pi∣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∤n. Because if 4∣n, then r2≡−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∤a and hence a is odd and n=2a.∎
Theorem 2.2**.**
If r2≡−1(mod n) and ra≡±a(mod n), then Rn(a,r) is a Cayley graph.
**Proof: **Let α=ρ2;β=ρδ2;γ=μδ. Clearly, α,β,γ∈Aut(Rn(a,r)). It can be easily checked that βα=α−1β;γα=α−rγ and γ2=α2a−1β. 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 γ=μδ flips Ai’s and Bj’s, and α,β maps Ai’s to Aj’s and Bi’s to Bj’s, k2−k1 must be [math], i.e., k1=k2. Thus, we have
[TABLE]
If j2−j1=1, then αi1−i2=β=ρδ2. But αi1−i2(A0)=A2(i1−i2) (even index) and ρδ2(A0)=A1 (odd index). Hence, j2−j1=0, i.e., j1=j2. This implies αi1−i2=id and as a result i1=i2. Thus the elements of H are distinct and ∣H∣=n/2×2×2=2n.
We claim that H acts transitively on Rn(a,r). It suffices to show that the stabilizer of A0 in H, StabH(A0)={id}.
Let αiβjγk∈StabH(A0), i.e., αiβjγk(A0)=A0. Since, γ flips Ai’s and Bj’s, and α,β do not, we have k=0. Thus, αiβj(A0)=A0. If j=1, then αiβ(A0)=αiρδ2(A0)=ρ1+2iδ2(A0)=A0, i.e., A1+2i=A0, a contradiction, as the parity of indices on both sides does not match. Thus, j=0 and we have αi(A0)=A0. But this implies A2i=A0, i.e., i=0. Hence StabH(A0)={id}.
Finally, in view of Remark 1.2, H acts regularly on Rn(a,r) and hence Rn(a,r) is a Cayley graph.∎
3 Family-2 [R4m(2m,r): r is odd and (r2+2m)≡±1(mod 4m)]
Proposition 3.1**.**
If n is divisible by 4, r is odd, a=n/2 and (r2+n/2)≡±1(mod n), then
gcd(r,n)=1.
If γ:V→V be defined by γ(Ai)=Bri and γ(Bi)=A(r+a)i, then γ∈Aut(Rn(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∈N. Thus n=4ls,a=2ls and r=lt. Now (r2+n/2)≡±1(mod n) implies l2t2+2ls≡±1(mod 4ls), which in turn implies l∣(l2t2+2ls±1), i.e., l∣1, a contradiction. Thus gcd(r,n)=1.
γ, as defined above, has been shown to be in Aut(Rn(a,r)) in Lemma 3.8 [1]. ∎
Proposition 3.2**.**
If n is divisible by 4, r is odd, a=n/2 and (r2+n/2)≡1(mod n), then
r−1=r+a* (mod n)*
ζ∈Aut(Rn(a,r))* (defined before) takes the following form: ζ(Ai)=B−(r+a)i and ζ(Bi)=A−(r+a)i, and ζ4=id.*
Proof: r(r+a)≡r2+ar≡1−a+ar≡1+a(r−1)≡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 ζ 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 (r2+n/2)≡1(mod n), then Rn(a,r) is a Cayley graph.
**Proof: **Let α=ρ2,β=ρμ and σ=γζ2, where γ and ζ are as defined in Propositions 3.1 and 3.2. It can be easily checked that σ(Ai)=B(r+a)i and σ(Bi)=Ari; αn/2=β2=σ2=id; βαβ=α−1,σασ=αr,(βσ)2=α2a−r+1. Define
[TABLE]
[TABLE]
We claim that the elements in H are distinct. If not, suppose
[TABLE]
i.e.,
[TABLE]
Now, as σ flips Ai’s and Bj’s, and α,β maps Ai’s to Aj’s and Bi’s to Bj’s, k1−k2 must be [math], i.e., k1=k2. Thus, we have
[TABLE]
Since, α maintains the parity of indices and β flips the parity of indices of Ai’s and Bi’s, j2−j1 is even, i.e., j1=j2. This implies αi1−i2=id and as a result i1=i2. Thus the elements of H are distinct and ∣H∣=n/2×2×2=2n.
We claim that H acts transitively on Rn(a,r). In order to prove it, we show that the orbit of A0, OA0, under the action of H is the vertex
set of Rn(a,r). By orbit-stabilizer theorem, we get
[TABLE]
As the number of vertices in Rn(a,r) is 2n and ∣H∣=2n, it is enough to
show that StabH(A0)={id}. Let αiβjσk be an arbitrary element of H which stabilizes A0, i.e., αiβjσk(A0)=A0, with 0≤i<n/2,0≤j,k≤1. Now, as σ flips Ai’s and Bj’s, and α,β maps Ai’s to Aj’s and Bi’s to Bj’s, k=0. Thus αiβj(A0)=A0, i.e., α−i(A0)=βj(A0). Since, α maintains the parity of indices and β flips the parity of indices of Ai’s and Bi’s, j=0 and hence i=0. Thus StabH(A0)={id}.
Finally, in view of Remark 1.2, H acts regularly on Rn(a,r) and hence Rn(a,r) is a Cayley graph.∎
In Family 2, if (r2+n/2)≡−1(mod n), we will show that Rn(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 Rn(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 Rn(a,r) has two edge orbits, then one orbit consists of rim and hub edges, and the other consists of spoke edges. If Rn(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 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 Rn(a,r) sending every rim edge to a hub edge and vice-versa if and only if one of the following holds:
-
a=n/2, r2≡1(mod n) and ra≡±a(mod n);
2. 2.
a=n/2, r2≡±1(mod n) and ra≡±a(mod n);
3. 3.
n* is divisible by 4, gcd(n,r)=1, a=n/2 and (r2+n/2)≡±1(mod n).*
Corollary 3.3** (Corollary 3.9,[1]).**
If n is divisible by 4, r is odd, a=n/2 and (r2+n/2)≡±1(mod n), then the automorphism group of Rn(a,r) has two edge orbits and the full automorphism group of Rn(a,r), Aut(Rn(a,r))=⟨ρ,μ,γ⟩, where γ is as defined in Proposition 3.1.
Theorem 3.3**.**
If n is divisible by 4, r is odd, a=n/2 and (r2+n/2)≡−1(mod n), then Rn(a,r) is not a Cayley graph.
**Proof: **As evident from Corollary 3.3, the full automorphism group of Rn(a,r) is given by
[TABLE]
One can easily check the relations between the generators starting from the definition and conclude that ∣Aut(Rn(a,r))∣=n×2×4=8n. If possible, let Rn(a,r) be a Cayley graph with a regular subgroup H of Aut(Rn(a,r)) and ∣H∣=2n.
Let K=⟨γ⟩. Then ∣K∣=4 and H∩K is a subgroup of K. As γ2(A0)=A0, i.e., γ2 has a fixed point, γ2∈H. Thus H∩K={id} and
[TABLE]
Hence μ∈Aut(Rn(a,r))=HK. Thus μ=hk, where h∈H and k∈K={id,γ,γ2,γ3}. If k=id, then μ=h∈H. But as μ(A0)=A0, i.e., μ has a fixed point, μ∈H. Thus k=id.
If k=γ2, then μ=hγ2, i.e., h=μγ2∈H. But as μγ2(A0)=A0, μγ2∈H and hence k=γ2.
If k=γ, then μγ−1=h, i.e., h−2=(γμ)2=ρaγ2∈H. But, as ρaγ2(Aa/2)=Aa/2, by similar argument, k=γ.
If k=γ3, then h2=(μγ)2=ρaγ2∈H. By similar argument as above, k=γ3.
As all the four possible choices of k∈K leads to contradiction, we conclude that there does not exist any regular subgroup H of Aut(Rn(a,r)) and hence Rn(a,r) is not a Cayley graph. ∎
4 Family-3 [R2m(m±2,m±1)]
As m+2≡−(m−2) (mod 2m) and m+1≡−(m−1) (mod 2m), it suffices to check the family R2m(m−2,m−1). It was proved in Section 3.2 of [5], that
[TABLE]
where K=⟨ε0,ε1,…,εm−1⟩≅Z2m, Dm is the dihedral group and εi is the involution given by (Ai,Bi−1)(Ai+m,Bi−1+m)(Ai+1,Bi+m)(Ai+1+m,Bi). Thus ∣G∣=2m+1m. One can easily check that the following relations between the generators hold:
[TABLE]
[TABLE]
where the addition of indices of εi’s are done modulo m. Using this relations, it is easy to see that ∘(ρεi)=m and ∘(μρi)=2.
It follows from definition that ρ2iμ,ε0,ε1,…,εi−2,εi+1,…,εm−1∈StabG(Ai). Again, using the relations between generators, we get ∣⟨ρ2iμ,ε0,ε1,…,εi−2,εi+1,…,εm−1⟩∣=2m−1. Now, as R2m(m−2,m−1) is a vertex transitive graph, by orbit-stabilizer theorem, it follows that ∣G∣/∣StabG(Ai)∣=2×2m, i.e., ∣StabG(Ai)∣=4m2m+1m=2m−1. Thus, we have
[TABLE]
Similarly, it follows that
[TABLE]
Theorem 4.1**.**
R2m(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, R2m(m−2,m−1) is a subfamily of Family-1 and as a result, R2m(m−2,m−1) is a Cayley graph.∎
Theorem 4.2**.**
R2m(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 Σi, the product of all εj’s such that j=i (mod 3). Note that ΣiΣj=Σk for distinct i,j,k’s in {0,1,2} and ∘(Σi)=2.
Let α=ρ2,β=Σ0 and γ=Σ1. It can be easily checked that βα=αγ,γα=αβγ and βγ=γβ. Define
[TABLE]
Thus, any element of H can be expressed as αiβjγk where 0≤i≤m−1,0≤j,k≤1, i.e., ∣H∣≤4m.
Claim 1: ∣H∣=4m.
Proof of Claim 1: If not, there exist 0≤i1,i2≤m−1,0≤j1,j2,k1,k2≤1 such that αi1βj1γk1=αi2βj2γk2, i.e.,
[TABLE]
If j2−j1=k2−k1=0, then i1=i2 (since, ∘(ρ)=2m) and as a result the claim is true. However, if any one or both of j2−j1 or k2−k1 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(i1−i2)=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 StabH(A0)={id}. Let αiβjγk∈StabH(A0), i.e., αiβjγk(A0)=A0 for some i,j,k with 0≤i≤m−1, 0≤j,k≤1. Therefore,
[TABLE]
Claim 2: k=0.
Proof of Claim 2: If not, let k=1, i.e., βjγ(A0)=A2m−2i. Note that
both ε0 and εm−1 occurs in the expression of γ, and
all εi’s except ε0 and εm−1 stabilizes A0.
Thus A2m−2i=βjγ(A0)=βjεm−1ε0(A0)=βjεm−1(B2m−1)=βj(Am).
If j=0, then we have Am=A2m−2i, which is a contradiction, due to mismatch of parity of indices. If j=1, then we have β(Am)=A2m−2i. Note that
StabG(A0)=StabG(Am)=⟨μ,ε1,ε2,…,εm−2⟩.
ε0 does not occur in the expression of β, but εm−1 occur in the expression of β.
Thus, we have A2m−2i=β(Am)=εm−1(Am)=B2m−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 βj(A0)=A2m−2i. If j=1, then A2m−2i=β(A0)=εm−1(A0)=Bm−1, a contradiction. Thus, j=0 and hence we have A0=A2m−2i i.e., 2m≡2i (mod 2m), i.e., i≡m≡0 (mod m). Thus i=0. This implies that StabH(A0)={id} and hence the theorem holds.∎
Theorem 4.3**.**
R2m(m−2,m−1)* is not a Cayley graph, if m is odd and m≡0 (mod 3).*
**Proof: **Consider K=⟨ε0,ε1,…,εm−1⟩. Then K≅Z2m and ∣K∣=2m as ∘(εi)=2 and εiεj=εjεi,∀i,j∈{0,1,…,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∣=2m, we have ∣H∩K∣=2 or 4. We will prove that ∣H∩K∣=4. In fact, using the next two claims, we prove that ∣H∩K∣=2.
Claim 1: If ∣H∩K∣=2, then the non-identity element of H∩K must be ε0ε1⋯εm−1=ρm.
Proof of Claim 1: Let α=εl1εl2⋯εlp be the non-identity element of H∩K. Let L=⟨μ,ε0,ε1,…,εm−1⟩. Then ∣L∣=2m+1 and K⊊L as μ∈L∖K. Thus
[TABLE]
As ∣H∩K∣=2 and K⊊L, there exists atleast one element of the form β=μεi1εi2⋯εis in H∩L.
Again, let L′=⟨ρμ,ε0,ε1,…,εm−1⟩. By similar arguments, we can deduce that ∣H∩L′∣≥4. So there exists an element of the form γ=ρμεj1εj2⋯εjt in H∩L′.
As α,β,γ∈H, it follows that βαβ−1,γαγ−1∈H. Observe that
[TABLE]
As μεi=εm−1−iμ, βαβ−1 is product of some εi’s and hence id=βαβ−1∈H∩K. Since ∣H∩K∣=2, then α=βαβ−1.
Similarly,
[TABLE]
[TABLE]
As ρεi=εi+1ρ, ραρ−1 is product of some εi’s and hence γαγ−1∈H∩K and by similar arguments, we have α=γαγ−1.
Thus, using ρεi=εi+1ρ, we get
[TABLE]
As K=⟨ε0,ε1,…,εm−1⟩≅Z2m and εi’s corresponds to the standard generators of Z2m, i.e., εi↔(0,0,…,0,1,0,…,0) with the only 1 occuring in the (i+1)th position, εl1εl2⋯εlp corresponds to the vector in Z2m with 1’s in l1+1,l2+1,…,lp+1 positions and εl1+1εl2+1⋯εlp+1 corresponds to the vector with 1’s in l1+2,l2+2,…,lp+2 positions. Thus, from Equation 2, we get that all the positions in the vector must be 1, i.e., α=ε0ε1⋯εm−1=ρm. Hence the claim is true.
Claim 2: If ∣H∩K∣=2, then ρm∈H
Proof of Claim 2: As H∩L is a subgroup of H and m is odd, therefore 4≤∣H∩L∣∣4m implies ∣H∩L∣=4. Thus H∩L is either isomorphic to Z2×Z2 or Z4. Note that any non-identity element σ∈H∩L must contain in its expression either ε0 or εm−1, as otherwise σ∈⟨μ,ε1,ε2,…,εm−2⟩=StabG(A0), a contradiction to the fact that σ belongs to a regular subgroup H.
Suppose that H∩L is isomorphic to Z2×Z2. As H∩K⊊H∩L, therefore there exists a non-identity element in H∩L of the form σ=μεi1εi2⋯εis. As explained earlier, σ must contain in its expression either ε0 or εm−1. In fact, in this case, both ε0 and εm−1 must occur in the expression of σ, as otherwise ∘(σ)=4. Note that by Claim 1, ρm∈H∩L. Thus, for all the three non-identity elements, ρm,σ,σ′ (say) in H∩L, both ε0 and εm−1 must occur. Also as H∩L≅Z2×Z2, we have σσ′=ρm. But if σ,σ′ contains both ε0 and εm−1, then ρm contains neither ε0 nor εm−1, a contradiction. Hence H∩L≅Z2×Z2.
Suppose that H∩L is isomorphic to Z4. As ∘(ρm)=2, there exists a non-identity element ζ=μεj1εj2⋯εjs∈H∩L such that ⟨ζ⟩=H∩L and ζ2=ρm. Note that the number of εi’s in the expression of ζ2 is always even but that of ρm is m (odd) as ρm=ε0ε1⋯εm−1. Hence, H∩L≅Z4.
Thus, by Claim 1 and 2, we get ∣H∩K∣=4. As K≅Z2m, we have H∩K≅Z2×Z2. Recall that
[TABLE]
Again, as the graph is vertex-transitive, by orbit-stabilizer theorem, we have G=H⋅StabG(B(m+3)/2). Thus, ρ=hb, where h∈H and b∈StabG(B(m+3)/2).
Claim 3: ρμ does not occur in the expression of b.
Proof of Claim 3: If possible, let b=ρμεl1εl2⋯εlp and hence h=ρb−1=μεt1εt2⋯εtp∈H∩L. Again, as H∩K⊆H∩L and ∣H∩L∣=∣H∩K∣=4, we have H∩K=H∩L. Thus, h∈H∩K⊂K and hence h does not contain μ in its expression, a contradiction. Thus Claim 3 is true.
Therefore, by Claim 3, b=εl1εl2⋯εlp and h=ρb−1=ρεl1εl2⋯εlp∈H.
Let H∩K={id,α1,α2,α3}≅Z2×Z2. Thus hαih−1∈H. As αi’s, being elements of K, are product of some εi’s and εiεj=εjεi, ρεi=εi+1ρ, we have
[TABLE]
Thus hαih−1∈H∩K={id,α1,α2,α3}.
Claim 4: hα1h−1=α2 or α3.
Proof of Claim 4: If hα1h−1=id, then α1=id, a contradiction.
If hα1h−1=α1, then as above, get εi1+1εi2+1⋯εis+1=εi1εi2⋯εis. Now, as in proof of Claim 1, we can argue that this implies α1=ρm. But, in that case, we must have hα2h−1=α3 and hα3h−1=α2, because otherwise
hα2h−1=id implies α1=id, a contradiction.
hα2h−1=α2 implies α2=ρm, a contradiction, as α1=α2.
hα2h−1=α1 implies hα2h−1=hα1h−1, i.e., α1=α2, a contradiction.
Thus we have hα2h−1=ρα2ρ−1=α3 and hα3h−1=ρα3ρ−1=α2. Hence, from Equation 3, we see that both α2 and α3 are product of εi’s and the number of εi’s occuring in their expressions are same. Thus the number of εi’s occuring in the expression of α2α3 is even. However, α2α3=α=ρm=ε0ε1⋯εm−1 has odd number of εi’s occuring in its expression. This is a contradiction and hence hα1h−1=α1. Thus Claim 4 is true.
Without loss of generality, we can assume that hα1h−1=α2. Thus hα2h−1 is either α1 or α3. If hα2h−1=α1, we must have hα3h−1=α3, a contradiction, as shown in Claim 4. Hence we have hα2h−1=α3 and similarly hα3h−1=α1. So, by Equation 3, we get ρα1ρ−1=α2, ρα2ρ−1=α3 and ρα3ρ−1=α1. Hence, we have
[TABLE]
Similarly, we have ρ3α2=α2ρ3 and ρ3α3=α3ρ3.
Recall that H∩K={id,α1,α2,α3}≅Z2×Z2 and αi’s are product of some εj’s. Let
[TABLE]
Note that each αi must contain either ε0 or εm−1 in its expression, as otherwise it will be an element of StabG(A0) and hence can not belong to H. As α1α2=α3 and α1α2α3=id, without loss of generality, we can assume that, among ε0 or εm−1, α1 contains only ε0, α2 contains only εm−1 and α3 contains both ε0 and εm−1 in their expressions. This happens because if two of the αi’s contain both ε0 and εm−1 in their expressions, then the their product, i.e., the third αi, will not have ε0 or εm−1 in its expression, thereby making it an element of StabG(A0).
Now, from the relation ρ3α1=α1ρ3 and using the fact that ρεi=εi+1ρ, 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 Z2m, as in the proof of Claim 1, we get that all of ε0,ε3,ε6,…,ε3t=εm−1 occurs in the expression of α1, a contradiction to that fact that among ε0 or εm−1, α1 contains only ε0.
Similarly, if m=3t+2, we get all of
[TABLE]
occurs in the expression of α1, a contradiction.
Thus, we conclude that there does not exist any regular subgroup H of Aut(R2m(m−2,m−1)) and hence R2m(m−2,m−1) is not a Cayley graph, when m is odd and not a multiple of 3.
∎
5 Family-4 [R12m(±(3m+2),±(3m−1)) and R12m(±(3m−2),±(3m+1))]
As Rn(a,r)=Rn(a,−r) and Rn(a,r)≅Rn(−a,r), it is enough to check R12m(3m+2,3m−1) and R12m(3m−2,3m+1). More precisely, it suffices to work with the family R12m(3d+2,9d+1) where d=±m (mod 12m), as mentioned in Section 3.3 of [5].
Define σ as follows:
[TABLE]
Also, if m≡2 (mod 4), let b=d+1 and define ω as follows:
[TABLE]
It was shown in [5], that
[TABLE]
It is to be noted that m≡2 (mod 4) if and only if −m≡2 (mod 4). Thus, it is enough to work only with the family R12m(3m+2,9m+1).
Theorem 5.1**.**
If m is odd and m=3, then R12m(3m+2,9m+1) is a Cayley graph.
**Proof: **As m is odd, G=⟨ρ,μ,σ⟩. It can also be checked that σρ3σ=ρ3;σμ=μσ;(ρσ)3=ρ3(m+1);∘(σ)=2. Let α=(ρσ)2 and β=ρ2μσ. As m is odd and m=3, it can be shown that ∘(α)=3m,∘(β)=8 and βα=α−1β−1. Define
[TABLE]
[TABLE]
Claim 1: The elements in H are distinct.
If not, suppose
[TABLE]
i.e.,
[TABLE]
[TABLE]
any power of α maps A0 to A0(mod 3) or A1(mod 3) or B1(mod 3). On the other hand, as
[TABLE]
[TABLE]
we see that β,β2,β5 and β6 maps A0 to A2(mod 3). Thus, j2−j1 in Equation 4 can take values from {0,3,4,7}.
If j2−j1=0, then it is obvious that i1=i2 and j1=j2.
If j2−j1=4, squaring Equation 4, we get, α2(i1−i2)=id. Therefore, 3m∣2(i1−i2). Now, as gcd(2,3)=1 and m is odd, we have 3m∣(i1−i2), i.e., i1=i2 and hence j1=j2.
If j2−j1=3, since gcd(3,8)=1, then ∘(βj2−j1)=8. Therefore, α8(i1−i2)=id, i.e., 3m∣8(i1−i2). As m is odd, 3m is coprime to 8 and hence, 3m∣(i1−i2), i.e., i1=i2 and j1=j2.
The case j2−j1=7 follows similarly as above. Thus combining all the cases, we see that elements of H are distinct and H=3m×8=24m.
Claim 2: H acts transitively on R12m(3m+2,9m+1).
In order to prove it, we show that the orbit of A0, OA0, under the action of H is the vertex set of R12m(3m+2,9m+1). By orbit-stabilizer theorem, we get
[TABLE]
As the number of vertices in R12m(3m+2,9m+1) is 24m and ∣H∣=24m, it is enough to show that StabH(A0)={id}. Let αiβj be an arbitrary element of H which stabilizes A0, i.e., α−i(A0)=βj(A0) with 0≤i≤3m−1;0≤j≤7. Again, by mimicing the argument used in the proof of Claim 1, one can conclude that j∈{0,3,4,7}.
If j=4, then α−i(A0)=β4(A0)=A6m. Thus, −i and hence i is a multiple of 3. [since, αx sends A0 to A0(mod 3), only if x is a multiple of 3] Let −i=3k and therefore A6m=α3k(A0)=Ak(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=4.
If j=3, then α−i(A0)=β3(A0)=B3m+1. As 3m+1≡1(mod 3), we have −i=3k+1 [since, αx sends A0 to B1(mod 3), only if x≡1(mod 3)] Therefore, β3(A0)=B3m+1=α3k+1(A0)=α3k(B1), i.e., B3m+1=B1+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=3.
Using similar arguments as above, it can be shown that j=7.
Thus, we have j=0 and this, in turn, implies i=0. Hence, StabH(A0)={id}.
Finally, in view of Remark 1.2, H acts regularly on R12m(3m+2,9m+1) and hence R12m(3m+2,9m+1) is a Cayley graph, if m is odd and m=3.∎
Theorem 5.2**.**
If m=3, then R12m(3m+2,9m+1), i.e, R36(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≡0(mod 4), then R12m(3m+2,9m+1) is not a Cayley graph.
**Proof: **As m≡2(mod 4),
[TABLE]
[TABLE]
If possible, let R12m(3m+2,9m+1) be a Cayley graph, H be a regular subgroup of G and K=StabG(A0). Then ∣G∣=96m=8n (See Lemma 7.1 in Appendix), ∣H∣=2n=24m and H∩K={id}.
Let K′=⟨ρ⟩. Then ∣K′∣=n and ∣HK′∣=∣H∩K′∣∣H∣∣K′∣=n/t2n2≤∣G∣=8n, where t is a factor of n. Thus, t≤4, i.e., t=1,2,3 or 4. If t=1, then H∩K′=K′, i.e., ρ∈H. If t=2, then H∩K′=⟨ρ2⟩, i.e., ρ2∈H. If t=3, then H∩K′=⟨ρ3⟩, i.e., ρ3∈H. If t=4, then H∩K′=⟨ρ4⟩, i.e., ρ4∈H. Combining all the cases, we get that
[TABLE]
Claim: ρ4∈H.
Proof of Claim: Suppose that that ρ3∈H but ρ4∈H. Let L=⟨ρ,μ⟩. Then ∣L∣=2n. Therefore
[TABLE]
Therefore, ∣H∩L∣=2n,n,2n/3 or n/2, i.e., ∣H∩L∣≥n/2. As ρ3∈H∩L, we have ⟨ρ3⟩⊆H∩L and ∣⟨ρ3⟩∣=n/3. Thus, (H∩L)∖⟨ρ3⟩=∅.
Now, as ρ2iμ(Ai)=Ai, ρ2iμ∈H. Similarly, if ρ2i+1μ∈H, then ρ3⋅ρ2i+1μ∈H, i.e., ρ2i+4μ∈H. Note that 2i+4 is even and hence by previous argument, ρ2i+4μ∈H, i.e., ρ2i+1μ∈H. This shows that H does not contain any element of the form ρiμ. Moreover, μ∈H. Now, as (H∩L)∖⟨ρ3⟩=∅, H must contain an element of the form ρi, where i is not a multiple of 3. Again, as ρ3∈H, either ρ or ρ2∈H, i.e., ρ4∈H. This is a contradiction to the assumption that ρ4∈H. Thus the claim is true.
Let K′′=⟨ρσ⟩. As ∘(ρσ)=n, we have ∣K′′∣=n and by similar arguments as above, we get that either (ρσ)3∈H or (ρσ)4∈H.
Case 1: If ρ4∈H and (ρσ)4∈H, then
[TABLE]
As ρ4∈H, therefore σ∈H. But as σ(A0)=A0, i.e., σ stabilizes A0, it can not be in H. This is a contradiction.
Case 2: If ρ4∈H and (ρσ)3∈H, then (ρσ)3=ρ3(m+1)=ρ12l+3=(ρ4)3lρ3∈H, where m=4l i.e., ρ3∈H. Again, as ρ4∈H, we have ρ∈H. As ∘(ρ)=n and [H:⟨ρ⟩]=2, ⟨ρ⟩ is normal in H.
From definition, it follows that id,μ,σ,μσ∈K. On the other hand, as R12m(3m+2,9m+1) is vertex transitive, by orbit-stabilizer theorem, we have
[TABLE]
Thus, HK=G. As σρ∈G, it can be expressed in the form αβ, where α∈H and β∈K={id,μ,σ,μσ}.
If β=id, then α=σρ∈H, i.e., σ∈H (as ρ∈H), which is a contradiction, as H, being a regular subgroup can not contain any non-identity element which stabilizes A0.
If β=μ, then σρ=αμ, i.e., α=σμρ−1∈H, i.e., σμ∈H (as ρ∈H), which is a contradiction.
If β=σ, then α=σρσ∈H. Since ⟨ρ⟩ is normal in H, therefore (σρσ)ρ(σρσ)−1∈H, i.e.,
[TABLE]
a contradiction.
If β=μσ, then σρ=αμσ, i.e., α=σρμσ∈H. Since ⟨ρ⟩ is normal in H, therefore (σρμσ)ρ(σρμσ)−1∈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., R12m(3m+2,9m+1) is not a Cayley graph, if m≡0(mod 4). ∎
5.1 m≡2(mod 4)
As m≡2(mod 4), G=⟨ρ,μ,σ,ω⟩. It can be checked that σρ3σ=ρ3;σμ=μσ;σω=ωσ;ωρ=σρω;ωμ=μωσ;(ρσ)3=ρ3(m+1);ωρ3l=ρ3l(m+1);(ρσρ)3=(ρ2σ)3=ρ9m+6;∘(σ)=∘(ω)=∘(σω)=2;∘(ωμ)=4.
Let α=ωσρ4mωσ and β=ρ3m/2. Using the above relations, it can be shown that ∘(α)=3;∘(β)=8;αβ=βα. Define
[TABLE]
In all the cases, it can be checked that ∘(γ)=2m, αγ=γα and γβ=βm+1γ. It is to be noted that α=ωσρ4mωσ=(ωσρωσ)4m=[ω(σρω)σ]4m=(ω(ωρ)σ)4m=(ρσ)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≡2(mod 12), then R12m(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 γ2=ρ4m+4.(by Proposotion 5.1) Define
[TABLE]
Thus, it is clear that every element of H is of the form αiβjγk where i=0,1,2; j=0,1,…,7 and k=0,1,…,m−1.
Claim 1: H={αiβjγk:i=0,1,2;j=0,1,…,7;k=0,1,…,m−1}.
Proof of Claim 1: If possible, let there exist i1,i2∈{0,1,2},j1,j2∈{0,1,…,7} and k1,k2∈{0,1,…,m−1}, such that αi1βj1γk1=αi2βj2γk2. As αβ=βα and αγ=γα, we have
[TABLE]
Case 1: k1−k2 is even.
As γ2=ρ4m+4 and β=ρ3m/2, we have αi2−i1=ρx, i.e., (ρσ)4m(i2−i1)=ρx. This implies that 3 divides 4m(i2−i1), i.e., 3∣m or 3∣(i2−i1). As 3 does not divide m, we have 3∣(i2−i1), i.e., i1=i2. Thus βj1−j2=γk2−k1=(γ2)(k2−k1)/2, i.e.,
[TABLE]
Therefore, 12m divides 3m(j1−j2)/2−2(m+1)(k2−k1), i.e.,
[TABLE]
Thus, m divides 4(m+1)(k2−k1). As gcd(m,4)=2 and gcd(d,d+1)=1, it follows that m/2 divides k2−k1, i.e., k2−k1=2ms. Since, 0≤k2−k1<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(j1−j2), i.e., 2 divides j1−j2. Let j1−j2=2t. As 0≤j1−j2<8, we have t∈{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 k2−k1=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 j1−j2=k1−k2=0, and as a result i1=i2. Thus Claim 1 is true, if Case 1 holds.
Case 2: k1−k2 is odd.
Let k1−k2=2t−1. Then we have αi2−i1=βj1−j2(γ2)tγ−1. As γ2=ρ4m+4 and β=ρ3m/2, we have γαi2−i1=ρx. Now i2−i1=0,1 or 2. Thus either of γ,αγ,α2γ is ρx. But
[TABLE]
[TABLE]
[TABLE]
As each of γ,αγ,α2γ maps some Ai to some Bj, none of them is equal to ρx and hence a contradiction. So k1−k2 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 StabH(A0)={id}. Let αiβjγk(A0)=A0.
Claim 2: k is even.
Proof of Claim 2: If possible, let k be odd, say k=2t+1. Then, as α commutes with β and γ, we have βjγ2tγαi(A0)=A0, i.e., γαi(A0)=β−j(γ2)−t(A0)=ρx(A0)=Ax, as in Case 2 above. Now, i=0,1 or 2 and as γ(A0)=B5m+1 and α2γ(A0)=B10m+1, we have i=1. This implies αβjγ2t+1(A0)=A0, i.e., βj(γ2)tγ(A0)=α2(A0)=A11m, i.e.,
[TABLE]
Hence the claim is true and let k=2t. Therefore,
[TABLE]
As left side of the above equation is ρx(A0) and α(A0)=B10m−1, we conclude that i=0 or 1. If i=1, then we have αβj(γ2)t(A0)=A0. Again as α commutes with β and γ, we have
[TABLE]
Therefore, i=0 and hence we have βj(γ2)t(A0)=A0, i.e.,
[TABLE]
Thus 12 divides 3j(6l+1), i.e., 4∣j(6l+1). However as 6l+1 is odd and j∈{0,1,…,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≤k≤m−1, we have 0≤t≤2m−1<6l+1. Thus the only possible value of t is [math] and hence k=0. Therefore, we have βj(A0)=A0, i.e., ρ3(6l+1)j(A0)=A0. This implies that 12m=12(12l+2) divides 3(6l+1)j, i.e., 8∣j and hence j=0.
Thus we have StabH(A0)={id} and the theorem holds.∎
Theorem 5.5**.**
If m≡6(mod 12), then R12m(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, α=(ρσ)4m=((ρσ)3)4(4l+2)=ρ12(m+1)(4l+2)=ρ12(4l+2)=ρ4m. Also γ2=ρ4m+4. (by Proposition 5.1)
Define
[TABLE]
Thus, it is clear that every element of H is of the form αiβjγk where i=0,1,2; j=0,1,…,7 and k=0,1,…,m−1.
Claim 1: H={αiβjγk:i=0,1,2;j=0,1,…,7;k=0,1,…,m−1}.
Proof of Claim 1: If possible, let there exist i1,i2∈{0,1,2},j1,j2∈{0,1,…,7} and k1,k2∈{0,1,…,m−1}, such that αi1βj1γk1=αi2βj2γk2. As αβ=βα and αγ=γα, we have
[TABLE]
If k1−k2 is odd, say k1−k2=2t−1, then γ=αi1−i2βj1−j2γ2t. As α=ρ4m, the right hand side is of the form ρx. On the other hand, γ(A0)=B5m+1. Thus γ=αi1−i2βj1−j2γ2t. Hence k1−k2 is even, say 2t. Thus, we have ρ4m(i2−i1)=ρ(4m+4)t+32m(j1−j2),\mboxi.e.,
[TABLE]
This implies that 4 divides 9(2l+1)(j1−j2), i.e., 4∣(j1−j2). Now as 0≤j1−j2≤7, we have j1−j2=0 or 4.
Sub-claim 1a: j1−j2=0.
If possible, let j1−j2=4. Then, from Equation 10, we have 12m\mboxdivides4(m+1)t+6m+4m(i1−i2) and hence m∣4(m+1)t, i.e., m∣4t, as gcd(m.m+1)=1. Now, as 0≤4t=2(k1−k2)≤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(i1−i2), i.e., 6∣4(i1−i2) which implies 3∣(i1−i2) i.e., i1=i2. However, this implies that 12m∣6m, a contradiction. Thus Sub-claim 1a is true and j1=j2. Thus Equation 10 reduces to
[TABLE]
Again since gcd(m,m+1)=1, this implies that m∣t. However, as 0≤t≤2m−1, we have t=0 and hence k1=k2. Thus from Equation 11, we get 3∣(i1−i2), i.e., i1=i2. 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 StabH(A0)={id}. Let αiβjγk(A0)=A0. As α=ρ4m,β=ρ3m/2 and γ2=ρ4m+4 are powers of ρ and γ(A0)=B5m+1, if k is odd, αiβjγk(A0)=Bx for some index x. Thus k is even, say k=2t. Thus, we have αiβjγk(A0)=ρ4mi+8(m+1)t+23mj=A0, i.e., 12m divides 4mi+8(m+1)t+23mj, 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 StabH(A0)={id} and the theorem holds.∎
Theorem 5.6**.**
If m≡10(mod 12), then R12m(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 γ2=ρ12. Define
[TABLE]
Thus, it is clear that every element of H is of the form αiβjγk where i=0,1,2; j=0,1,…,7 and k=0,1,…,m−1.
Claim 1: H={αiβjγk:i=0,1,2;j=0,1,…,7;k=0,1,…,m−1}.
Proof of Claim 1: If possible, let there exist i1,i2∈{0,1,2},j1,j2∈{0,1,…,7} and k1,k2∈{0,1,…,m−1}, such that αi1βj1γk1=αi2βj2γk2. As αβ=βα and αγ=γα, we have
[TABLE]
If k1−k2 is odd, say k1−k2=2t−1, then γαi2−i1=βj1−j2γ2t. As γ2=ρ12, the right hand side is of the form ρx, i.e., γαi2−i1=ρx. Now i2−i1=0,1 or 2. Thus either of γ,αγ,α2γ is ρx. But γ(A0)=B9m+5,αγ(A0)=B10m+5, α2γ(B0)=A5m+7.
As each of γ,αγ,α2γ maps some Ai to some Bj, none of them is equal to ρx and hence a contradiction. So k1−k2 is even, say k1−k2=2t. As γ2=ρ12 and β=ρ3m/2, we have αi2−i1=ρx, i.e., (ρσ)4m(i2−i1)=ρx. This implies that 3 divides 4m(i2−i1), i.e., 3∣m or 3∣(i2−i1). As 3 does not divide m, we have 3∣(i2−i1), i.e., i1=i2. Thus ρ23m(j1−j2)=βj1−j2=γk2−k1=(γ2)t=ρ12t, 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 2m∣t. However, as 0≤k2−k1≤m−1, we have 0≤t≤2m−1. Hence t=0 and k1=k2. Also Equation 14 reduces to 8∣(j1−j2). Thus j1=j2. 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 StabH(A0)={id}. Let αiβjγk(A0)=A0.
Claim 2: k is even.
Proof of Claim 2: If possible, let k be odd, say k=2t+1. Then, as α commutes with β and γ, we have βjγ2tγαi(A0)=A0, i.e., γαi(A0)=β−j(γ2)−t(A0)=ρx(A0)=Ax, as in the proof of Claim 1 of this theorem. Now, i=0,1 or 2 and as γ(A0)=B9m+5 and αγ(A0)=B10m+5, we have i=2. This implies α2βjγ2t+1(A0)=A0, i.e., βj(γ2)tγ(A0)=α(A0)=A7m, i.e.,
[TABLE]
Hence the claim is true and let k=2t. Therefore,
[TABLE]
As left side of the above equation is ρx(A0) and α2(A0)=B2m−1, we conclude that i=0 or 2. If i=2, then we have α2βj(γ2)t(A0)=A0. Again as α commutes with β and γ, we have
[TABLE]
Therefore, i=0 and hence we have βj(γ2)t(A0)=A0, i.e.,
[TABLE]
Thus 12 divides 3j(6l+5), i.e., 4∣j(6l+5). However as 6l+5 is odd and j∈{0,1,…,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≤k≤m−1, we have 0≤t≤2m−1<6l+5. Thus the only possible value of t is [math] and hence k=0. Therefore, we have βj(A0)=A0, i.e., ρ3(6l+5)j(A0)=A0. This implies that 12m=24(6l+5) divides 3(6l+5)j, i.e., 8∣j and hence j=0.
Thus we have StabH(A0)={id} and the theorem holds.∎
6 Family-5 [R2m(2b,r): b2≡±1(mod m) and r∈{1,m−1} is odd]
Theorem 6.1**.**
If b2≡±1(mod m) and r∈{1,m−1} is odd, then R2m(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., r2≡1(mod n) and ra≡a(mod n)) and hence, by Theorem 2.1, R2m(2b,r) is a Cayley graph. So we are left with the case when n=2m, a=2b, b2≡±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≡−b(mod m), i.e., 2br≡−2b(mod 2m), i.e., ra≡−a(mod n). Thus, in this case, r2≡1(mod n) and ra≡−a(mod n) holds. Hence, by Theorem 2.1, R2m(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=Aut(R12m(3m+2,9m+1)) , where m≡0(mod 4). Then ∣G∣=96m.
**Proof: **Since, R12m(3m+2,9m+1) is vertex-transitive and its order is 24m and StabG(A0) contains id,μ,σ,μσ, therefore, by orbit-stabilizer theorem, we have ∣G∣≥4×24m=96m. Thus, it is enough to show that ∣G∣≤96m. We also know that
[TABLE]
[TABLE]
Consider the sets X={ρiσρjμk:i∈{0,1,2…,n−1},j∈{0,1,2},k∈{0,1}} and Y={ρiμk:i∈{0,1,2…,n−1},k∈{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 σ are in Y, due to the relations ρn=μ2=id and μρμ=ρ−1. Again, as σμ=μσ and μρ=ρ−1μ, any element in G can be expressed in the form where μ occurs in the extreme right of the expression. Thus it is enough to show that elements in G which involve only ρ and σ are of the form ρiσρj where i∈{0,1,2…,n−1} and j∈{0,1,2}. Again, as σρ3=ρ3σ, it is clear that the power of ρ lying on the right of σ can be made 0,1 or 2. Finally, we deal with elements σρσ and σρ2σ.
As (ρσρ)3=ρ9m+6, we have σρ2σρ2σ=ρ9m+4, i.e.,
[TABLE]
As (ρσ)3=ρ3(m+1), we have (σρσρσ)=ρ3m+2, i.e.,
[TABLE]
Similarly, any other element of G involving ρ and σ can be expressed in the form of elements in X. Thus G=X∪Y and hence
[TABLE]
∎
**Proof of Proposition 5.1 **:
-
For m=12l+2, we have 8m=96l+16, i.e., 8m−4=12(8l+1).
[TABLE]
[TABLE]
[TABLE]
[TABLE]
[TABLE]
[TABLE]
For m=12l+6, we have 8m=96l+48=12(8l+4).
[TABLE]
[TABLE]
[TABLE]
Similarly, for m=12l+10, it can be proved that γ2=ρ12.
2. 2.
The values of γm can be found by raising γ2 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 R36(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()