Trivalent dihedrants and bi-dihedrants
Mi-Mi Zhang , Jin-Xin Zhou
Department of Mathematics, Beijing Jiaotong University, Beijing, 100044, China
Abstract
A Cayley (resp. bi-Cayley) graph on a dihedral group is called a dihedrant (resp. bi-dihedrant).
In 2000, a classification of trivalent arc-transitive dihedrants was given by Marušič and Pisanski, and several years later, trivalent non-arc-transitive dihedrants of order 4p or 8p (p a prime) were classified by Feng et al. As a generalization of these results, our first result presents a classification of trivalent non-arc-transitive dihedrants. Using this, a complete classification of trivalent vertex-transitive non-Cayley bi-dihedrants is given, thus completing the study of trivalent bi-dihedrants initiated in our previous paper [Discrete Math. 340 (2017) 1757–1772]. As a by-product, we generalize a theorem in [The Electronic Journal of Combinatorics 19 (2012) #P53].
**Key words: Cayley graph, non-Cayley, bi-Cayley, dihedral group, dihedrant, bi-dihedrant
**
1 Introduction
In this paper we describe an investigation of trivalent Cayley graphs on dihedral groups as well as vertex-transitive trivalent bi-Cayley graphs over dihedral groups. To be brief, we shall say that a Cayley (resp. bi-Cayley) graph on a dihedral group a dihedrant (resp. bi-dihedrant).
Cayley graphs are usually defined in the following way. Given a finite group G and an inverse closed subset S⊆G∖{1}, the Cayley graph Cay(G,S) on G with respect to S is a graph with vertex set G and edge set {{g,sg}∣g∈G,s∈S}. For any g∈G, R(g) is the permutation of G defined by R(g):x↦xg for x∈G. Set R(G):={R(g) ∣ g∈G}. It is well-known that R(G) is a subgroup of Aut(Cay(G,S)). We say that the Cayley graph Cay(G,S) is normal if R(G) is normal in Aut(Cay(G,S)) (see [19]).
In 2000, Marušič and Pisanski [13] initiated the study of automorphisms of dihedrants, and they gave a classification of trivalent arc-transitive dihedrants. Following this work, highly symmetrical dihedrants have been extensively studied, and one of the remarkable achievements is the complete classification of 2-arc-transitive dihedrants (see [6, 12]). In contrast, however, relatively little is known about the automorphisms of non-arc-transitive dihedrants. In [1], the authors proved that every trivalent non-arc-transitive dihedrant is normal. However, this is not true. There exist non-arc-transitive and non-normal dihedrants. Actually, in [22, 25], the automorphism groups of trivalent dihedrants of order 4p and 8p are determined for each prime p, and the result reveals that every non-arc-transitive trivalent dihedrant of order 4p or 8p is either a normal Cayley graph, or isomorphic to the so-called cross ladder graph. For an integer m≥2, the cross ladder graph, denoted by CL4m, is a trivalent graph of order 4m with vertex set V0∪V1∪…V2m−2∪V2m−1, where Vi={xi0,xi1},
and edge set {{x2ir,x2i+1r},{x2i+1r,x2i+2s}∣i∈Zm,r,s∈Z2} (see Fig. 1 for CL4m).
It is worth mentioning that the cross ladder graph plays an important role in the study of automorphisms of trivalent graphs (see, for example, [5, 25, 21]). Motivated by the above mentioned facts, we shall focus on trivalent non-arc-transitive dihedrants. Our first theorem generalizes the results in [22, 25] to all trivalent dihedrants.
Theorem 1.1
Let Σ=Cay(H,S) be a connected trivalent Cayley graph, where H=⟨a,b ∣ an=b2=1,bab=a−1⟩(n≥3). If Σ is non-arc-transitive and non-normal, then n is even and Γ≅CL4⋅2n and Sα={b,ba,ba2n} for some α∈Aut(H).
Recall that for an integer m≥2, the cross ladder graph CL4m has vertex set V0∪V1∪…V2m−2∪V2m−1, where Vi={xi0,xi1}. The multi-cross ladder graph, denoted by MCL4m,2, is the graph obtained from CL4m by blowing up each
vertex xir of CL4m into two vertices xir,0 and xir,1. The edge set is
{{x2ir,s,x2i+1r,t},{x2i+1r,s,x2i+2s,r}∣i∈Zm,r,s,t∈Z2} (see Fig. 2 for MCL20,2).
Note that the multi-cross ladder graph MCL4m,2 is just the graph given in [23, Definition 7]. From [7, Proposition 3.3] we know that every MCL4m,2 is vertex-transitive. However, not all
multi-cross ladder graphs are Cayley graphs. Actually, in [23, Theorem 9], it is proved that MCL4p,2 is a vertex-transitive non-Cayley graph for each prime p>7. Our second theorem generalizes this result to all multi-cross ladder graphs.
Theorem 1.2
The multi-cross ladder graph MCL4m,2 is a Cayley graph if and only if either m is even, or m is odd and 3∣m.
Both of the above two theorems are crucial in attacking the problem of classification of trivalent vertex-transitive non-Cayley bi-dihedrants. Before proceeding, we give some background to this topic, and set some notation.
Let R,L and S be subsets of a group H such that R=R−1, L=L−1 and R∪L does not contain the identity element of H. The bi-Cayley graph BiCay(H,R,L,S) over H relative to R,L,S is a graph having vertex set the union of the right part H0={h0 ∣ h∈H} and the left part H1={h1 ∣ h∈H}, and edge set the union of the right edges {{h0,g0} ∣ gh−1∈R}, the left edges {{h1,g1} ∣ gh−1∈L} and the spokes {{h0,g1} ∣ gh−1∈S}. If ∣R∣=∣L∣=s, then BiCay(H, R, L, S) is said to be an s-type bi-Cayley graph.
In [20] we initiated a program to investigate the automorphism groups of the trivalent vertex-transitive bi-dihedrants. This was partially motivated by the following facts. As one of the most important finite graphs, the Petersen graph is a bi-circulant, but it is not a Cayley graph.
Note that a bi-circulant is a bi-Cayley graph over a cyclic group.
The Petersen graph is the initial member of a family of graphs P(n,t), known now as
the generalized Petersen graphs (see [17]), which can be also constructed as bi-circulants.
Let n≥3, 1≤t<n/2 and set H=⟨a⟩≅Zn.
The generalized Petersen graph P(n,t) is isomorphic to
the bi-circulant BiCay(H, {a,a−1}, {at,a−t}, {1}).
The complete classification of vertex-transitive generalized Petersen graphs has been worked out
in [8, 14]. Latter, this was generalized by Marušič et al. in [13, 15] where all trivalent vertex-transitive
bi-circulants were classified in [13, 15], and more recently, all trivalent vertex-transitive bi-Cayley graphs over abelian groups were classified in [24]. The characterization of trivalent vertex-transitive bi-dihedrants is the next natural step.
Another motivation for us to consider trivalent vertex-transitive bi-dihedrants comes from the work in [16]. By checking the census of trivalent vertex-transitive graphs of order up to 1000 in [16], we find out that there are 981 non-Cayley graphs, and among these graphs, 233 graphs are non-Cayley bi-dihedrants. This may suggest bi-dihedrants form an important class of trivalent vertex-transitive non-Cayley graphs.
In [20], we gave a classification of trivalent arc-transitive bi-dihedrants, and we also proved that every trivalent vertex-transitive [math]- or 1-type bi-dihedrant is a Cayley graph, and gave a classification of trivalent vertex-transitive non-Cayley bi-dihedrants of order 4n with n odd. The goal of this paper is to complete the classification of trivalent vertex-transitive non-Cayley bi-dihedrants.
Before stating the main result, we need the following concepts.
For a bi-Cayley graph Γ=BiCay(H, R, L, S) over a group H, we can assume that the identity 1 of H
is in S (see Proposition 2.3 (2)). The triple (R,L,S) of three subsets R,L,S of a group H is called bi-Cayley triple if R=R−1,L=L−1, and 1∈S. Two bi-Cayley triples (R,L,S) and (R′,L′,S′) of a group H are said to be
equivalent, denoted by (R,L,S)≡(R′,L′,S′), if either (R′,L′,S′)=(R,L,S)α or (R′,L′,S′)=(L,R,S−1)α
for some automorphism α of H. The bi-Cayley graphs corresponding to
two equivalent bi-Cayley triples of the same group are isomorphic (see Proposition 2.3 (3)-(4)).
Theorem 1.3
Let Γ=BiCay(R,L,S) be a trivalent vertex-transitive bi-dihedrant where H=⟨a,b ∣ an=b2=1,bab=a−1⟩ is a dihedral group. Then either Γ is a Cayley graph or one of the following occurs:
- (1)
(R,L,S)≡({b, baℓ+1},{ba, baℓ2+ℓ+1},{1}), where n≥5,
ℓ3+ℓ2+ℓ+1≡0 (mod n), ℓ2≡1 (mod n).
2. (2)
(R,L,S)≡({ba−ℓ, baℓ},{a, a−1},{1}), where n=2m and ℓ2≡−1 (mod m). Furthermore, Γ is also a bi-Cayley graph over an abelian group Zn×Z2.
3. (3)
(R,L,S)≡({b,ba},{b,ba2m},{1}), where n=2(2m+1), m≡1 (mod 3), and the corresponding graph is isomorphic the multi-cross ladder graph MCL4m,2.
4. (4)
(R,L,S)≡({b,ba},{ba24ℓ,ba12ℓ−1},{1}), where n=48ℓ and ℓ≥1.
Moreover, all of the graphs arising from (1)-(4) are vertex-transitive non-Cayley.
2 Preliminaries
All groups considered in this paper are finite, and all graphs are finite, connected, simple and undirected.
For the group-theoretic and graph-theoretic terminology not defined here we refer the
reader to [3, 18].
2.1 Definitions and notations
For a positive integer, let Zn be the cyclic group of order n
and Zn∗ be the multiplicative group of Zn consisting of numbers coprime to n.
For two groups M and N, N⋊M denotes a semidirect product of N by M.
For a subgroup H of a group G, denote CG(H) the centralizer of H in G and by NG(H) the normalizer of H of G.
Let G be a permutation group on a set Ω and α∈Ω.
Denote by Gα the stabilizer of α in G.
We say that G is semiregular on Ω if Gα=1 for every α∈Ω
and regular if G is transitive and semiregular.
For a finite, simple and undirected graph Γ, we use V(Γ), E(Γ), A(Γ), Aut(Γ) to denote its vertex set, edge set, arc set and full automorphism group, respectively. For any subset B of V(Γ), the subgraph of Γ induced by B
will be denoted by Γ[B]. For any v∈V(Γ) and a positive integer i no more than the diameter of Γ, denote by Γi(v) be the set of vertices at distance i from v. Clearly, Γ1(v) is just the neighborhood of v. We shall often abuse the notation by using Γ(v) to replace Γ1(v).
A graph Γ is said to be vertex-transitive, and arc-transitive (or symmetric) if Aut(Γ) acts transitively on V(Γ) and A(Γ), respectively. Let Γ be a connected vertex-transitive
graph, and let G≤Aut(Γ) be vertex-transitive on Γ. For a
G-invariant partition B of V(Γ), the quotient graph
ΓB is defined as the graph with vertex set B such that, for
any two different vertices B,C∈B, B is adjacent to C if and only if
there exist u∈B and v∈C which are adjacent in Γ. Let N
be a normal subgroup of G. Then the set B of orbits of N in
V(Γ) is a G-invariant partition of V(Γ). In this case, the
symbol ΓB will be replaced by ΓN. The original graph Γ is
said to be a N-cover of ΓN if Γ and ΓN have the same valency.
2.2 Cayley graphs
Let Γ=Cay(G,S) be a Cayley graph on G with respect to S. Then Γ is vertex-transitive due to R(G)≤Aut(Γ). In
general, we have the following proposition.
Proposition 2.1
[2, Lemma 16.3]*
A vertex-transitive graph Γ is isomorphic to a Cayley
graph on a group G if and only if its automorphism group has a
subgroup isomorphic to G, acting regularly on the vertex set of
Γ.*
In 1981, Godsil [9] proved that the normalizer of R(G) in Aut(Cay(G,S)) is R(G)⋊Aut(G,S), where Aut(G,S) is the group of automorphisms of G fixing the set S set-wise. This result has been successfully used in characterizing various families of Cayley graphs Cay(G,S) such that R(G)=Aut(Cay(G,S)) (see, for example, [9, 10]).
Recall that a Cayley graph Cay(G,S) is
said to be normal if R(G) is normal in Aut(Cay(G,S)) (see [19]).
Proposition 2.2
[19, Proposition 1.5]*
The Cayley graph Γ=Cay(G,S) is normal if and only if
A1=Aut(G,S), where A1 is the stabilizer of the identity 1
of G in Aut(Γ).*
2.3 Basic properties of bi-Cayley graphs
In this subsection, we let Γ be a connected bi-Cayley graph BiCay(H,R,L,S)
over a group H. It is easy to prove some basic properties of such a Γ, as in [24, Lemma 3.1].
Proposition 2.3
The following hold.
- (1)
H* is generated by R∪L∪S.*
2. (2)
Up to graph isomorphism, S can be chosen to contain the identity of H.
3. (3)
For any automorphism α of H, BiCay(H, R, L, S)≅BiCay(H, Rα, Lα, Sα).
4. (4)
BiCay(H, R, L, S)≅BiCay(H, L, R, S−1).
Next, we collect several results about the automorphisms of bi-Cayley graph Γ=BiCay(H, R, L, S).
For each g∈H, define a permutation as follows:
[TABLE]
Set R(H)={R(g) ∣ g∈H}. Then R(H) is a semiregular subgroup of Aut(Γ) with H0 and H1 as its two orbits.
For an automorphism α of H and x,y,g∈H, define two permutations of V(Γ)=H0∪H1 as follows:
[TABLE]
Set
[TABLE]
Proposition 2.4
[26, Theorem 1.1]*
Let Γ=BiCay(H,R,L,S) be a connected bi-Cayley graph over the group H.
Then NAut(Γ)(R(H))=R(H)⋊F if I=∅ and
NAut(Γ)(R(H))=R(H)⟨F, δα,x,y⟩ if
I=∅ and δα,x,y∈I. Furthermore, for any δα,x,y∈I, we have the following:*
- (1)
⟨R(H), δα,x,y⟩* acts transitively on V(Γ);*
2. (2)
if α has order 2 and x=y=1, then Γ is isomorphic to the Cayley graph Cay(Hˉ, R∪αS), where Hˉ=H⋊⟨α⟩.
3 Cross ladder graphs
The goal of this section is to prove Theorem 1.1.
Proof of Theorem 1.1 Suppose that Σ=Cay(H,S) is a connected trivalent Cayley graph which is neither normal nor arc-transitive, where H=⟨a,b ∣ an=b2=1,bab=a−1⟩(n≥3). Then S is a generating subset of H and ∣S∣=3. So S must contain an involution of H outside ⟨a⟩. As Aut(H) is transitive on the coset b⟨a⟩, we may assume that S={b,x,y} for x,y∈H∖⟨b⟩.
Suppose first that x is not an involution. Then we must have y=x−1. Since S generates H, one has ⟨a⟩=⟨x⟩, and so bxb=x−1. Then there exists an automorphism of H sending b,x to b,a respectively. So we may assume that S={b,a,a−1}.
Now it is easy to check that Σ is isomorphic to the generalized Petersen graph P(n,1). Since Σ is not arc-transitive, by [8, 14], we have ∣Aut(Σ)∣=2∣H∣, and so Σ would be a normal Cayley graph of H, a contradiction.
Therefore, both x and y must be involutions. Suppose that x∈⟨a⟩. Then n is even and x=an/2.
Again since S generates H, one has y=baj, where 1≤j≤n−1 and either (j,n)=1 or (j,n)=2 and 2n is odd.
Note that the subgroup of Aut(H) fixing b is transitive on the set of generators of ⟨a⟩ and that
⟨an/2⟩ is the center of H. There exists α∈Aut(H) such that
[TABLE]
Without loss of generality, we may assume that S={b,ba,a2n} or {b,ba2,a2n}.
If S={b,ba2,a2n}, we shall prove that Σ≅P(n,1). Note that the generalized Petersen graph P(n,1) has vertex set {ui,vi ∣ i∈Zn} and edge set {{ui,ui+1},{vi,vi+1},{ui,vi} ∣ i∈Zn}.
Define a map from V(Σ) to V(P(n,1)) as follows:
[TABLE]
where 0≤i≤2n−1.
It is easy to see that φ is an isomorphism form Σ to P(n,1). Since Σ is not arc-transitive, by [8, 14], we have ∣Aut(Σ)∣=2∣H∣, and so Σ would be a normal Cayley graph of H, a contradiction.
If S={b,ba,a2n}, then Σ has a connected subgraph Σ1=Cay(H,{b,ba}) which is a cycle of length 2n, and Σ is just the graph obtained from Σ1 by adding a 1-factor such that each vertex g of Σ1 is adjacent to its antipodal vertex a2ng. Then R(H)⋊Z2≅Aut(Σ1)≤Aut(Σ), and then since Σ is assumed to be not arc-transitive, Aut(Σ) will fix the 1-factor {{g,a2ng} ∣ g∈H} setwise. This implies that Aut(Σ)≤Aut(Σ1) and so Aut(Σ)=Aut(Σ1). Consequently, we have Σ is a normal Cayley graph of H, a contradiction.
Similarly, we have y∈/⟨a⟩. Then we may assume that x=bai and y=baj for some 1≤i,j≤n−1 and i=j.
Then S={b,bai,baj}⊆b⟨a⟩. This implies that Σ is a bipartite graph with ⟨a⟩ and b⟨a⟩ as its two partition sets. Since Σ is not arc-transitive, Aut(Σ)1 is intransitive on the neighbourhood S of 1, and since Σ is not a normal Cayley graph of H, there exists a unique element, say s∈S, such that Aut(Σ)1=Aut(Σ)s.
Considering the fact that Aut(H) is transitive on b⟨a⟩, without loss of generality, we may assume that Aut(Σ)1=Aut(Σ)b and Aut(Σ)1 swaps bai and baj. Then for any h∈H, we have
[TABLE]
Direct computation shows that
[TABLE]
Let Aut(Σ)1∗ be the kernel of Aut(Σ)1 acting on S. Take an α∈Aut(Σ)1∗. Then α fixes every element in S.
As Aut(Σ)h=Aut(Σ)bh for any h∈H, α will fix b(bai)=ai and b(baj)=aj.
Note that Σ(bai)∖{1,ai}={ai−j} and Σ(baj)∖{1,aj}={aj−i}.
Then α also fixes ai−j and aj−i, and then α also fixes bai−j and baj−i.
If ∣Σ2(1)∣=6, then it is easy to check that a−i is the unique common neighbor of b and baj−i. So α also fixes a−i. Now one can see that α fixes every vertex in Σ2(1). If ∣Σ2(1)∣<6 and either ∣Σ1(b)∩Σ1(bai)∣>1 or ∣Σ1(b)∩Σ1(baj)∣>1, then α also fixes every vertex in Σ2(1).
In the above two cases, by the connectedness and vertex-transitivity of Σ, α would fix all vertices of Σ, implying that α=1. Hence, Aut(Σ)1∗=1 and Aut(Σ)1≅Z2. This forces that Σ is a normal Cayley graph of H, a contradiction.
Thus, we have ∣Σ2(1)∣<6 and ∣Σ1(b)∩Σ1(bai)∣=∣Σ1(b)∩Σ1(baj)∣=1. This implies that Σ1(bai)∩Σ1(baj)={1,ai−j}={1,aj−i}, and so ai−j=aj−i. It follows that ai−j is an involution, and hence n is even and ai−j=an/2. So S={b,bai,bai+n/2}. As S generates H, one has ⟨ai,an/2⟩=⟨a⟩.
So either (i,n)=1 or (i,n)=2 and 2n is odd.
Note that the subgroup of Aut(H) fixing b is transitive on the set of generators of ⟨a⟩ and that
⟨an/2⟩ is the center of H. There exists α∈Aut(H) such that
[TABLE]
Let β be the automorphism of H induced by the map a↦a−1,b↦ba.
Then
[TABLE]
If 2n is odd, then the map η:a↦a2+2n,b↦ba2n induces an automorphism of H, and {b,ba,ba2n}η={b,ba2,ba2n}.
So there always exists γ∈Aut(H) such that Sγ={b,ba,ba2n}, completing the proof of the first part of our theorem.
Finally, we shall prove Σ≅CL4⋅2n. Without loss of generality, assume that S={b,ba,ba2n}.
Recall that V(CL4⋅2n)={xir ∣ i∈Z2n,r∈Z2} and E(CL4⋅2n)={{xir,xi+1r}, {x2ir,x2i+1r+1}, ∣ i∈Z2n,r∈Z2}.
Let ϕ be a map from V(Σ) to V(CL4⋅2n) as following:
[TABLE]
where 0≤i≤2n−1 and 1≤j≤2n.
It is easy to check that ϕ is an isomorphism from Σ and X(CL4⋅2n), as desired.
□
4 Multi-cross ladder graphs
The goal of this section is to prove Theorem 1.2. We first show that each MCL4m,2 is a bi-Cayley graph.
Lemma 4.1
The multi-cross ladder graph MCL4m,2 is isomorphic to the bi-Cayley graph BiCay(H,{c,ca},{ca,ca2b},{1}), where
[TABLE]
Proof For convenience, let Γ be the bi-Cayley graph given in our lemma, and let X=MCL4m,2. Let ϕ be a map from V(X) to V(Γ) defined by the following rule:
[TABLE]
where t∈Zm.
It is easy to see that ϕ is a bijection. Furthermore, for any t∈Zm, we have
[TABLE]
This shows that ϕ preserves the adjacency of the graphs, and so it is an isomorphism from X to Γ.
□
Remark 1 Let m be odd, let e=ab and f=ca. Then the group given in Lemma 4.1 has the following presentation:
[TABLE]
Clearly, in this case, H is a dihedral group.
Furthermore, the corresponding bi-Cayley graph given in Lemma 4.1 will be
[TABLE]
Proof of Theorem 1.2 By Lemma 4.1, we may let Γ=MCL4m,2 be just the bi-Cayley graph
BiCay(H,R,L,S), where
[TABLE]
We first prove the sufficiency. Assume first that m is even. Then the map
[TABLE]
induces an automorphism, say α of H of order 2.
Furthermore, Rα={c,ca}α=caLca, Lα={ca,ca2b}α=caRca and Sα={1}α=ca{1}ca=S−1.
By Proposition 2.4, δα,ca,ca∈Aut(Γ) and R(H)⋊⟨δα,ca,ca⟩ acts regularly on V(Γ).
Consequently, by Proposition 2.1, Γ is a Cayley graph.
Assume now that m is odd and 3∣m. In this case, we shall use the bi-Cayley presentation for Γ as in Remark 5.1, that is,
[TABLE]
where
[TABLE]
Let β be a permutation of V(Γ) defined as following:
[TABLE]
where t∈Z3m and i∈Z2.
It is easy to check that β is an automorphism of Γ of order 2. Furthermore, R(e),R(f) and β
satisfy the following relations:
[TABLE]
Let G=⟨R(e2),R(f),β⟩ and P=⟨R(e2),β⟩.
Then R(f)∈/P and G=P⟨R(f)⟩. Since R(e)6β=βR(e)6, we have R(e6)∈Z(P). Since R(e)2β=βR(e)4βR(e)−2, it follows that
[TABLE]
Let N=⟨R(e6)⟩.
Clearly, N is a normal subgroup of G. Furthermore,
[TABLE]
Therefore, ∣P∣=4m and ∣G∣≤8m.
Let
[TABLE]
Then Δij’s (i,j∈Z2) are four orbits of ⟨R(e2),R(f)⟩.
Moreover,
[TABLE]
This implies that G is transitive on V(Γ).
Hence, ∣G∣=8m and so G is regular on V(Γ), and by Proposition 2.1, Γ is a Cayley graph.
To prove the necessity, it suffices to prove that if m is odd and 3∤m, then Γ is a non-Cayley graph.
In this case, we shall use the original definition of Γ=MCL4m,2. Suppose that m is odd and 3∤m.
We already know from [7, Proposition 3.3] that Γ is vertex-transitive. Let A=Aut(Γ).
For m=5 or 7, using Magma [4], Γ is a non-Cayley graph. In what follows, we assume that m≥11.
For each j∈Zm, Cj0=(x2j0,0,x2j+10,0,x2j0,1,x2j+10,1) and Cj1=(x2j1,1,x2j+11,1,x2j1,0,x2j+11,0) are two 4-cycles.
Set F={Cji ∣ i∈Z2,j∈Zm}. From the construction of Γ=MCL4m,2, it is easy to see that in Γ=MCL4m,2 passing each vertex there is exactly one 4-cycle, which belongs to F. Clearly, any two distinct 4-cycles in F are vertex-disjoint. This implies that Δ={V(Cji) ∣ i∈Z2,j∈Zm} is an A-invariant partition of V(Γ). Consider the quotient graph ΓΔ, and let T be the kernel of A acting on Δ. Then ΓΔ≅Cm[2K1], the lexicographic product of a cycle of length m and an empty graph of order 2. Hence A/T≤Aut(Cm[2K1])≅Z2m⋊D2m. Note that between any two adjacent vertices of ΓΔ there is exactly one edge of Γ=MCL4m,2. Then T fixes each vertex of Γ and hence T=1.
So we may view A as a subgroup of Aut(ΓΔ)≅Aut(Cm[2K1])≅Z2m⋊D2m.
For convenience, we will simply use the Cji’s to represent the vertices of ΓΔ.
Then ΓΔ has vertex set
[TABLE]
and edge set
[TABLE]
Let B={{Cj0,Cj1} ∣ j∈Zm}. Then B is an Aut(ΓΔ)-invariant partition of V(ΓΔ). Let K be the kernel of Aut(ΓΔ) acting on B.
Then K=⟨k0⟩×⟨k2⟩×⋯×⟨km−1⟩, where we use ki to denote the transposition (Cj0 Cj1) for j∈Zm. Clearly, K is the maximal normal 2-subgroup of Aut(ΓΔ).
Suppose to the contrary that Γ=MCL4m,2 is a Cayley graph. By Proposition 2.1, A has a subgroup, say G acting regularly on V(Γ). Then G has order 8m, and
[TABLE]
Since m odd, it follows that ∣G∩K∣=4 or 8, and so
G∩K≅Z22 or Z23.
If G∩K≅Z22, then ∣GK/K∣=2m and GK/K=Aut(ΓΔ)/K≅D2m. So GK=Aut(ΓΔ)≅Z2m⋊D2m.
Let M be a Hall 2′-subgroup of G. Then M≅Zm and M is also a Hall 2′-subgroup of Aut(ΓΔ). Clearly, Aut(ΓΔ) is solvable, so all Hall 2′-subgroups of Aut(ΓΔ) are conjugate. Without loss of generality, we may
let M=⟨α⟩, where α is the following permutation on V(ΓΔ):
[TABLE]
Then K⋊⟨α⟩ acts transitively on V(ΓΔ). Clearly, CK(α) is contained in the center of K⋊⟨α⟩.
So CK(α) is semiregular on V(ΓΔ). This implies that
[TABLE]
On the other hand, let L=(G∩K)M. Clearly, G∩K⊴G, so L is a subgroup of G of order 4m. For any odd prime factor p of m, let P be a Sylow p-subgroup of M. Then P is also a Sylow p-subgroup of L, and since M is cyclic, one has M≤NL(P). By Sylow theorem, we have ∣L:NL(P)∣=kp+1∣4 for some integer k. Since 3∤m, one has L=NL(P). It follows that M⊴L and so L=M×(G∩K). This implies that G∩K≤CK(M)=CK(α)≅Z2, a contradiction.
If G∩K≅Z23, then ∣GK/K∣=m. Furthermore, GK/K≅Zm and GK/K acts on B regularly.
Since G is transitive on V(Γ), there exists g∈G such that (x01,1)g=x11,1, where x01,1,x11,1∈C01.
As V(ΓΔ)={Cji ∣ i∈Z2,j∈Zm}, g fixes the 4-cycle C01=(x01,1,x11,1,x01,0,x11,0). Since B={{Cj0,Cj1} ∣ j∈Zm} is also A-invariant, g fixes {C00,C01} setwise. Since GK/K acts on B regularly, g fixes {Cj0,Cj1} setwise for every j∈Zm.
Observe that {x01,1,x2m−11,1} and {x11,1,x21,1} are the unique edges of Γ between C01 and Cm−11, C01 and C21, respectively. This implies that g will map Cm−11 to C21, contradicting that g fixes {Cj0,Cj1} setwise for every j∈Zm.
□
5 A family of trivalent VNC bi-dihedrants
The goal of this section is to prove the following lemma which gives a new family of trivalent vertex-transitive non-Cayley bi-dihedrants.
To be brief, a vertex-transitive non-Cayley graph is sometimes simply called a VNC graph.
Lemma 5.1
Let H=⟨a,b ∣ an=b2=1,ab=a−1⟩ be a dihedral group, where n=48ℓ and ℓ≥1. Then Γ=BiCay(H,{b,ba},{ba24ℓ,ba12ℓ−1},{1}) is a VNC dihedrant.
Proof We first define a permutation on V(Γ) as follows:
[TABLE]
where r∈Z16ℓ, i∈Z2.
It is easy to check that g is an involution, and furthermore,
for any t∈Z16ℓ,
we have
[TABLE]
This implies that g is an automorphism of Γ. Observing that g maps 11 to b0, it follows that ⟨R(H),g⟩ is transitive on V(Γ), and so Γ is a vertex-transitive graph.
Below, we shall first prove the following claim.
Claim. Aut(Γ)10=⟨g⟩.
Let A=Aut(Γ). It is easy to see that g fixes 10, and so g∈A10. To prove the Claim, it suffices to prove that ∣A10∣=2.
Note that the neighborhood Γ(10) of 10 in Γ is ={11,b0,(ba)0}. By a direct computation, we find that in Γ there is a unique 8-cycle passing through 10, 11 and b0, that is,
[TABLE]
Furthermore, in Γ there is no 8-cycle passing through 10 and (ba)0. So A10 fixes (ba)0.
If A10 also fixes 11 and b0, then A10 will fix every neighbor of 10, and the connectedness and vertex-transitivity of Γ give that A10=1, a contradiction.
Therefore, A10 swaps 11 and b0, and (ba)0 is the unique neighbor of 10 such that A10=A(ba)0.
It follows that {10,(ba)0} is a block of imprimitivity of A acting on V(Γ).
Since Γ is vertex-transitive, every v∈V(Γ) has a unique neighbor, say u such that Au=Av.
Then the set
[TABLE]
forms an A-invariant partition of V(Γ). Clearly, {10,(ba)0}∈B.
Similarly, since C0 is also the unique 8-cycle of Γ passing through 10, 11 and b0, A11 swaps 10 and b0, and (ba12ℓ−1)1 is the unique neighbor of 11 such that A11=A(ba12ℓ−1)1. So
{11,(ba12ℓ−1)1}∈B. Set
[TABLE]
Clearly, B=B0∪B1.
Now we consider the quotient graph ΓB of Γ relative to B. It is easy to see that ⟨R(a)⟩ acts semiregularly on B with B0 and B1 as its two orbits. So ΓB is isomorphic to a bi-Cayley graph over ⟨a⟩.
Set B0={10,(ba)0} and B1={11,(ba12ℓ−1)1}.
Then one may see that the neighbors of B0 in ΓB are: B0R(a),B0R(a−1),B1,B1R(a−12ℓ+2),
and the neighbors of B1 in ΓB are: B1R(a12ℓ+1),B1R(a−12ℓ−1),B0,B0R(a12ℓ−2).
So
[TABLE]
Observe that there is one and only one edge of Γ between B0 and any one of its neighbors in ΓB.
Clearly, A acts transitively on V(ΓB), so there is one and only one edge of Γ between every two adjacent blocks of B.
It follows that A acts faithfully on V(ΓB), and hence we may view A as a subgroup of Aut(ΓB).
Recall that g∈A10=A(ba)0. Moreover, g swaps the two neighbors 11 and b0 of 10. Clearly, 11∈B1 and b0∈B0R(a−1), so g swaps the two blocks B1 and B0R(a−1). Similarly, g swaps the two neighbors (ba)1 and a0 of (ba)0.
Clearly, (ba)1∈B1R(a−12ℓ+2) and a0∈B0R(a), so
g swaps the two blocks B1R(a−12ℓ+2) and B0R(a).
Note that R(ab) swaps the two vertices in B0. So ⟨g,R(ab)⟩ acts transitively on the neighborhood of B0 in ΓB.
This implies that A acts transitively on the arcs of ΓB, and so Γ′ is a tetravalent arc-transitive bi-circulant.
In [11], a characterization of tetravalent edge-transitive bi-circulants is given. It is easy to see that our graph Γ′ belongs to Class 1(c) of [11, Theorem 1.1]. By checking [11, Theorem 4.1], we see that the stabilizer Aut(Γ′)u of u∈V(Γ′) has order 4. This implies that ∣A∣=4∣V(ΓB)∣=8n. Consequently, ∣A10∣=2 and so our claim holds.
Now we are ready to finish the proof. Suppose to the contrary that Γ is a Cayley graph.
By Proposition 2.1, A contains a subgroup, say J acting regularly on V(Γ).
By Claim, J has index 2 in A, and since g∈A10, one has A=J⋊⟨g⟩.
It is easy to check that R(a),R(b) and g satisfy the following relations:
[TABLE]
Suppose that R(H)≰J. Then A=JR(H). Since ∣J:R(H)∣=2, it follows that ∣R(H):J∩R(H)∣=2. Thus, J∩R(H)=⟨R(a)⟩ or ⟨R(a2),R(b)⟩. If R(H)∩J=⟨R(a)⟩, then we have R(b)∈/J, R(a)∈J, and hence A=J∪JR(b)=J∪Jg, implying that JR(b)=Jg. It follows that gR(b)∈J, and then g=R(a)(gR(b))2R(a12ℓ−1)∈J due to R(a)∈J, a contradiction.
If R(H)∩J=⟨R(a2),R(b)⟩, then R(a)∈/J, and again we have A=J∪JR(a)=J∪Jg, implying that JR(a)=Jg.
So, R(a)g,gR(a−1)∈J. Then
[TABLE]
a contradiction.
Suppose that R(H)≤J. Then ∣J:R(H)∣=2 and R(H)⊴J. Since J is regular on V(Γ), by Proposition 2.4, there exists a δα,x,y∈J such that 10δα,x,y=11, where α∈Aut(H) and x,y∈H. By the definition of δα,x,y, we have 11=10δα,x,y=(x⋅1α)1=x1, implying that x=1. Furthermore, we have the following relations:
[TABLE]
where R={b,ba},L={ba24ℓ,ba12ℓ−1},S={1}.
In particular, the last equality implies that x=y due to S={1}. So we have x=y=1. From the proof of Claim we know that B0={10,(ba)0} and B1={11,(ba12ℓ−1)1} are two blocks of imprimitivity of A acting on V(Γ). So we have ((ba)0)δα,1,1=(ba12ℓ−1)1. It follows that (ba)α=ba12ℓ−1, and then from Rα=L we obtain that bα=ba24ℓ. Consequently, we have aα=a36ℓ−1. One the other hand, we have {b,ba}=R=Lα={b,ba24ℓ+1}. This forces that ba=ba24ℓ+1, which is clearly impossible.
□
6 Two families of trivalent Cayley bi-dihedrants
In this section, we shall prove two lemmas which will be used the proof of Theorem 1.3.
Lemma 6.1
Let H=⟨a,b ∣ a12m=b2=1,ab=a−1⟩ be a dihedral group with m odd. Then for each i∈Z12m, Γ=BiCay(H,{b,bai},{ba6m,ba3m−i},{1}) is a Cayley graph whenever ⟨ai,a3m⟩=⟨a⟩.
Proof Let g be a permutation of V(Γ) defined as follows:
[TABLE]
where r∈Zm, k∈Z4 and j∈Z2.
It is easy to check that for any r∈Zm, k∈Z4 and j∈Z2, we have
[TABLE]
[TABLE]
It follows that g∈Aut(Γ). Furthermore, one may check that g and R(a2) satisfy the following relations:
[TABLE]
By the last equality, we have
[TABLE]
It then follows from the second and third equalities that
[TABLE]
Therefore, (R(a2)g)3=R(a6).
Let G=⟨R(a2),R(b),g⟩ and T=⟨R(a6)⟩. Then T⊴G and
[TABLE]
So ∣G∣=24m.
Let
[TABLE]
Then Ωij’s (0≤i,j≤1) are orbits of T and V(Γ)=0≤i,j≤1⋃Ωij. Since 10g=(ba6m)1∈Ω01, a0g=(a3m+1)0∈Ω00 and a1g=(ba3m+1)1∈Ω01, it follows that G is transitive, and so regular on V(Γ). By Proposition 2.1, Γ is a Cayley graph on G, as required.
□
Lemma 6.2
Let H=⟨a,b ∣ a12m=b2=1,ab=a−1⟩ be a dihedral group with m even and 4∤m. Then the following two bi-Cayley graphs:
[TABLE]
are both Cayley graphs.
Proof Let V=H0∪H1. Then V(Γ1)=V(Γ2)=V. We first define two permutations on V as follows:
[TABLE]
[TABLE]
where r∈Z3m and i∈Z2.
It is easy to check that gj∈Aut(Γj) for j=1 or 2. Furthermore, R(a2),R(b) and gj (j=1 or 2) satisfy the following relations:
[TABLE]
For j=1 or 2, let Gj=⟨R(a),R(b),gj⟩.
From the above relations it is east to see that
[TABLE]
has order at most 48m. Observe that 10gj=(ba6m)1∈H1 for j=1 or 2. It follows that Gj is transitive on V(Γj), and so Gj acts regularly on V(Γj). By Proposition 2.1, each Γj is a Cayley graph.
□
7 Vertex-transitive trivalent bi-dihedrants
In this section, we shall give a complete classification of trivalent vertex-transitive non-Cayley bi-dihedrants.
For convenience of the statement, throughout this section, we shall make the following assumption.
Assumption I.
H: the dihedral group D2n=⟨a,b ∣ an=b2=1,bab=a−1⟩(n≥3),
Γ=BiCay(H, R, L, {1}): a connected trivalent 2-type vertex-transitive bi-Cayley graph over the group H (in this case, ∣R∣=∣L∣=2),
G: a minimum group of automorphisms of Γ subject to that R(H)≤G and G is transitive on the vertices but intransitive on the arcs of Γ.
The following lemma given in [20] shows that the group G must be solvable.
Lemma 7.1
[20, Lemma 6.2]*
G=R(H)P is solvable, where P is a Sylow 2-subgroup of G.*
7.1 H0 and H1 are blocks of imprimitivity of G
The case where H0 and H1 are blocks of imprimitivity of G has been considered in [20], and the main result is the following proposition.
Proposition 7.2
[20, Theorem 1.3]*
If H0 and H1 are blocks of imprimitivity of G on V(Γ),
then either Γ is Cayley or one of the following occurs:*
- (1)
(R,L,S)≡({b, baℓ+1},{ba, baℓ2+ℓ+1},{1}), where n≥5,
ℓ3+ℓ2+ℓ+1≡0 (mod n), ℓ2≡1 (mod n);
2. (2)
(R,L,S)≡({ba−ℓ, baℓ},{a, a−1},{1}), where n=2k and ℓ2≡−1 (mod k). Furthermore, Γ is also a bi-Cayley graph over an abelian group Zn×Z2.
Furthermore, all of the graphs arising from (1)-(2) are vertex-transitive non-Cayley.
In particular, it is proved in [20] that if n is odd and Γ is not a Cayley graph, then H0 and H1 are blocks of imprimitivity of G on V(Γ). Consequently, we can get a classification of trivalent vertex-transitive non-Cayley bi-Cayley graphs over a dihedral group D2n with n odd.
Proposition 7.3
[20, Proposition 6.4]*
If n is odd, then either Γ is a Cayley graph, or H0 and H1 are blocks of imprimitivity of G on V(Γ).*
7.2 H0 and H1 are not blocks of imprimitivity of G
In this subsection, we shall consider the case where H0 and H1 are not blocks of imprimitivity of G on V(Γ). We begin by citing a lemma from [20].
Lemma 7.4
[20, Lemma 6.3]*
Suppose that H0 and H1 are not blocks of imprimitivity of G on V(Γ). Let N be a normal subgroup of G, and let K be the kernel of G acting on V(ΓN). Let Δ be an orbit of N. If N fixes H0 setwise, then one of the following holds:*
- (1)
Γ[Δ]* has valency 1, ∣V(ΓN)∣≥3 and Γ is a Cayley graph;*
2. (2)
Γ[Δ]* has valency [math], ΓN has valency 3, and K=N is semiregular.*
The following lemma deals with the case where CoreG(R(H))=1, and in this case we shall see that Γ is just the cross ladder graph.
Lemma 7.5
Suppose that H0 and H1 are not blocks of imprimitivity of G on V(Γ). If CoreG(R(H))=⋂g∈GR(H)=1,
then Γ is isomorphic to the cross ladder graph CL4n with n odd, and furthermore, for any minimal normal subgroup N of G, we have the following:
- (1)
N* is a 2-group which is non-regular on V(Γ);*
2. (2)
N* does not fix H0 setwise;*
3. (3)
every orbit of N consists of two non-adjacent vertices.
Proof Let N be a minimal normal subgroup of G. By Lemma 7.1, G is solvable. It follows that N is an elementary abelian r-subgroup for some prime divisor r of ∣G∣. Clearly, N≰R(H) due to CoreG(R(H))=1. Then ∣NR(H)∣/∣R(H)∣∣∣G∣/∣R(H)∣. From Lemma 7.1 it follows that ∣G∣/∣R(H)∣ is a power of 2, and hence N is a 2-group.
Suppose that N is regular on V(Γ). Then NR(H) is transitive on V(Γ) and R(H) is also a 2-group. Therefore, NR(H) is not transitive on the arcs of Γ. The minimality of G gives that G=NR(H).
Since n is even, R(a2n) is in the center of R(H). Set Q=N⟨R(a2n)⟩. Then Q⊴G and then 1=N∩Z(Q)⊴G. Since N is a minimal normal subgroup of G, one has N≤Z(Q), and hence Q is abelian. It follows that ⟨R(a2n)⟩⊴G, contrary to the assumption that CoreG(R(H))=1. Thus, N is not regular on V(Γ).
(1) is proved.
For (2), by way of contradiction, suppose that N fixes H0 setwise. Consider the quotient graph ΓN of Γ relative to N, and
let K be the kernel of G acting on V(ΓN). Take Δ to be an orbit of N on V(Γ).
Then either (1) or (2) of Lemma 7.4 happens.
For the former, Γ[Δ] has valency 1 and ∣V(ΓN)∣≥3. Then ΓN is a cycle. Moreover, any two neighbors of u∈Δ are in different orbits of N. It follows that the stabilizer Nv of v in N fixes every neighbor of u. The connectedness of Γ implies that Nv=1. Thus, K=N is semiregular and ΓN is a cycle of length ℓ=2∣R(H)∣/∣N∣. So G/N≤Aut(ΓN)≅D2ℓ. If G/N<Aut(ΓN), then ∣G:N∣=ℓ and so ∣G∣=2∣R(H)∣. This implies that R(H)⊴G, contrary to the assumption that CoreG(R(H))=1. If G/N=Aut(ΓN), then ∣G:R(H)∣=4. Since N≤R(H) and since N fixes H0 setwise,
one has ∣G:R(H)N∣=2. It follows that R(H)N⊴G. Clearly, H0 and H1 are just two orbits of R(H)N, and they are also two blocks of imprimitivity of G on V(Γ), a contradiction.
For the latter, Γ[Δ] has valency [math], ΓN has valency 3 and N=K is semiregular.
Let Hˉi be the set of orbits of N contained in Hi with i=1,2. Then ΓN[Hˉ0] and ΓN[Hˉ1] are of valency 2 and the edges between Hˉ0 and Hˉ1 form a perfect matching. Without loss of generality, we may assume that 10∈Δ. Since R(H) acts on H0 by right multiplication, we have the subgroup of R(H) fixing Δ setwise is just R(H)Δ={R(h) ∣ h0∈Δ}. If R(H)Δ≤⟨R(a)⟩, then R(H)Δ⊴R(H), and the transitivity of R(H) on H0 implies that R(H)Δ will fix all orbits of N contained in H0. Since the edges between Hˉ0 and Hˉ1 are independent, R(H)Δ fixes all orbits of N. It follows that R(H)Δ≤N, namely, R(H)N/N acts regularly on Hˉ0. Then ∣R(H)/(R(H)∩N)∣=∣R(H)N/N∣=∣H0/N∣, and so ∣N∣=∣R(H)∩N∣, forcing N≤R(H), a contradiction. Thus, R(H)Δ≰⟨R(a)⟩, and so ⟨R(a)⟩R(H)Δ=R(H). This implies that ⟨R(a),N⟩/N is transitive and so regular on Hˉ0. Similarly, ⟨R(a),N⟩/N is also regular on Hˉ1. Thus, ΓN is a trivalent 2-type bi-Cayley graph over ⟨R(a),N⟩/N. By [24, Lemma 5.3], Hˉ0 and Hˉ0 are blocks of imprimitivity of G/N, and so H0 and H1 are blocks of imprimitivity of G, a contradiction.
So far, we have completed the proof of (2). Then N does not fix H0 setwise, and then NR(H) is transitive on V(Γ). The minimality of G gives that G=NR(H). Let P and P1 be Sylow 2-subgroups of G and R(H), respectively, such that P1≤P. Then N≤P and P=NP1.
If n is even, then by a similar argument to the second paragraph, a contradiction occurs. Thus, n is odd.
As H≅D2n, P1≅Z2 and P1 is non-normal in R(H). So N∩R(H)=1. Clearly, ∣V(Γ)∣=4n. If N is semiregular on V(Γ), then N≅Z2 or Z2×Z2, and then ∣G∣=∣R(H)∣∣N∣=2∣R(H)∣ or 4∣R(H)∣. Since CoreG(R(H))=1, we must have
∣G:R(H)∣=4 and G≲Sym(4). Since n is odd, one has n=3 and H≅Sym(3). So G≅Sym(4) and hence G10≅Z2. Then all involutions of G(≅Sym(4)) not contained in N are conjugate. Take 1=g∈G10. Then g is an involution which is not contained in N because N is semiregular on V(Γ). Since R(H)∩N=1, every involution in R(H) would be conjugate to g. This is clearly impossible because R(H) is semiregular on V(Γ). Thus, N is not semiregular on V(Γ). (3) is proved.
Since n is odd, we have ∣V(ΓN)∣>2. Since N is not semiregular on V(Γ), ΓN has valency 2 and Γ[Δ] has valency [math]. This implies that the subgraph induced by any two adjacent two orbits of N is either a union of several cycles or a perfect matching. Thus, ΓN has even order. As Γ has order 4n with n odd, every orbit of N has length 2. It is easy to see that Γ is isomorphic to the cross ladder graph CL4n.
□
The following is the main result of this section.
Theorem 7.6
Suppose that H0 and H1 are not blocks of imprimitivity of G on V(Γ). Then Γ=BiCay(H,R,L,S) is vertex-transitive non-Cayley if and only if one of the followings occurs:
- (1)
(R,L,S)≡({b,ba},{b,ba2m},{1}), where n=2(2m+1), m≡1 (mod 3), and the corresponding graph is isomorphic the multi-cross ladder graph MCL4m,2;
2. (2)
(R,L,S)≡({b,ba},{ba24ℓ,ba12ℓ−1},{1}), where n=48ℓ and ℓ≥1.
Proof The sufficiency can be obtained from Theorem 1.2 and Lemma 5.1. We shall prove the necessity in the following subsection by a series of lemmas. □
7.3 Proof of the necessity of Theorem 7.6
The purpose of this subsection is to prove the necessity of Theorem 7.6. Throughout this subsection, we shall always assume that H0 and H1 are not blocks of imprimitivity of G on V(Γ) and that Γ=BiCay(H,R,L,S) is vertex-transitive non-Cayley. In this subsection, we shall always use the following notation.
Assumption II. Let N=CoreG(R(H)).
Our first lemma gives some properties of the group N.
Lemma 7.7
1<N<⟨R(a)⟩, ∣⟨R(a)⟩:N∣=n/∣N∣ is odd and the quotient graph ΓN of Γ relative to N is isomorphic to the cross ladder graph CL4n/∣N∣.
Proof If N=1, then from Lemma 7.5 it follows that Γ≅CL4n which is a Cayley graph by Theorem 1.1, a contradiction. Thus, N>1.
Since H0 and H1 are not blocks of imprimitivity of G on V(Γ), one has N<R(H).
Consider the quotient graph ΓN. Clearly, N fixes H0 setwise. Recall that H0 and H1 are not blocks of imprimitivity of G on V(Γ) and that Γ is non-Cayley. Applying Lemma 7.4, we see that ΓN is a trivalent 2-type bi-Cayley graph over R(H)/N. This implies that ∣R(H):N∣>2, and since H is a dihedral group, one has N<⟨R(a)⟩.
Again, by Lemma 7.4, R(H)/N acts semiregularly on V(ΓN) with two orbits, Hˉ0 and Hˉ1, where Hˉi is the set of orbits of N contained in Hi with i=1,0. Furthermore, N is just the kernel of G acting on V(ΓN) and N acts semiregularly on V(Γ). Then G/N is also a minimal vertex-transitive automorphism group of ΓN containing R(H)/N. If Hˉ0 and Hˉ1 are blocks of imprimitivity of G/N on V(ΓN), then H0 and H1 will be blocks of imprimitivity of G on V(Γ), which is impossible by our assumption. Thus, Hˉ0 and Hˉ1 are not blocks of imprimitivity of G/N on V(ΓN). Since N=CoreG(R(H)), CoreG/N(R(H)/N) is trivial. Then from Lemma 7.5 it follows that ΓN≅CL∣N∣4n, where ∣N∣n is odd. □
Next, we introduce another notation which will be used in the proof.
Assumption III. Take M/N to be a minimal normal subgroup of G/N.
We shall first consider some basic properties of the quotient graph ΓM of Γ relative to M.
Lemma 7.8
The quotient graph ΓM of Γ relative to M is a cycle of length n/∣N∣.
Furthermore, every orbit of M on V(Γ) is a union of an orbit of N on H0 and an orbit of N on H1, and these two orbits of N are non-adjacent.
Proof Applying Lemma 7.5 to ΓN and G/N, we obtain the following facts:
- (a)
M/N is an elementary abelian 2-group which is not regular on V(ΓN),
2. (b)
M/N does not fix Hˉ0 setwise,
3. (c)
every orbit of M/N on V(ΓN) consists of two non-adjacent vertices of ΓN.
From (b) and (c) it follows that every orbit of M on V(Γ) is just a union of an orbit of N on H0 and an orbit of N on H1, and these two orbits are non-adjacent. Since every orbit of N on V(Γ) is an independent subset of V(Γ), each orbit of M on V(Γ) is also an independent subset.
Recall that ΓN≅CL4m where m=∣N∣n is odd. The quotient graph of ΓN relative to M/N is just a cycle of length m, and so the quotient graph ΓM of Γ relative to M is also a cycle of length m. □
By Lemma 7.8, each orbit of M on V(Γ) is an independent subset. It follows that the subgraph induced by any two adjacent orbits of M is either a perfect matching or a union of several cycles. For convenience of the statement, the following notations will be used in the remainder of the proof:
Assumption IV.
- (1)
Let Δ and Δ′ be two adjacent orbits of M on V(Γ) such that Γ[Δ∪Δ′] is a union of several cycles.
2. (2)
Let Δ=Δ0∪Δ1 and Δ′=Δ0′∪Δ1′, where Δ0,Δ0′⊆H0 and Δ1,Δ1′⊆H1 are four orbits of N on V(Γ).
3. (3)
10∈Δ0.
Since Γ[Δ] and Γ[Δ′] are both null graphs and since Γ[Δ∪Δ′] is a union of several cycles, we have the following easy observation.
Lemma 7.9
Γ[Δi∪Δj′]* is a perfect matching for any 0≤i,j≤1.*
The following lemma tells us the possibility of R (Recall that we assume that Γ=BiCay(H,R,L,{1})).
Lemma 7.10
Up to graph isomorphism, we may assume that R={b,bai} with i∈Zn∖{0} and that b0∈Δ0′. Furthermore, we have
[TABLE]
and 11 is adjacent to (bal)1∈Δ1 for some R(al)∈N.
Proof Recall that N is a proper subgroup of ⟨R(a)⟩ and that n/∣N∣ is odd. Since n is even by Proposition 7.3, it follows that N is of even order, and so the unique involution R(an/2) of ⟨R(a)⟩ is contained in N. As 10∈Δ0 and N≤⟨R(a)⟩ acts on H0 by right multiplication, one has Δ0={h0∣h∈N}. Since Γ[Δ0] is an empty graph, one has an/2∈/R. By Proposition 2.3 (1), we have ⟨R∪L⟩=H, and since R and L are both self-inverse, either R⊆b⟨a⟩ or L⊆b⟨a⟩. By Proposition 2.3 (4), we may assume that R⊆b⟨a⟩.
Recall that Γ[Δi∪Δj′] is a perfect matching for any 0≤i,j≤1. Then 10 is adjacent to r0∈Δ0′ for some r∈R. Since R⊆b⟨a⟩ and Aut(H) is transitive on b⟨a⟩, by Proposition 2.3 (3), we may assume that r=b. So 10 is adjacent to b0∈Δ0′. Since N≤⟨R(a)⟩ acts on Hi with i=0 or 1 by right multiplication, we see that the two orbits Δ0,Δ0′ of N are just the form as given in the lemma. Since S={1}, the edges between H0 and H1 form a perfect matching. This enables us to obtain another two orbits Δ1,Δ1′ of N which have the form as given in the lemma.
By Lemma 7.9, Γ[Δ1∪Δ1′] is a perfect matching. So we may assume that 11 is adjacent to (bal)1∈Δ1 for some R(al)∈N. □
Now we shall introduce some new notations which will be used in the following.
Assumption V.
- (1)
Let T=⟨R(al)⟩ be of order t, where al is given in the above lemma.
2. (2)
Let
[TABLE]
3. (3)
B={BR(h) ∣ h∈H}, where B=Ω0∪Ω1.
4. (4)
Let B′=Ω0′∪Ω1′. Then B′=BR(b).
Lemma 7.11
The followings hold.
- (1)
T≤N.
2. (2)
Ω0,Ω1,Ω0′,Ω1′* are four orbits of T.*
3. (3)
Γ[Ω0∪Ω1∪Ω0′∪Ω1′]* is a cycle of length 4t.*
4. (4)
B* is a G-invariant partition of V(Γ).*
Proof By Lemma 7.10, we see that R(al)∈N, and so T≤N. (1) holds. Since T=⟨R(al)⟩ is assumed to be of order t, one has T=⟨R(an/t)⟩, and then one can obtain (2). By the adjacency rule of bi-Cayley graph, we can obtain (3).
Set Ω=Ω0∪Ω1∪Ω0′∪Ω1′ and B=Ω0∪Ω1. By Lemma 7.8, Γ[Δ] is a null graph, and so B=Δ∩Ω. Since Γ has valency 3, it follows that Δ∪Δ′ is a block of imprimitivity of G on V(Γ), and hence Ω is also a block of imprimitivity of G on V(Γ) since Γ[Ω] is a component of Γ[Δ∪Δ′]. Since Δ is also a block of imprimitivity of G on V(Γ), B(=Δ∩Ω) is a block of imprimitivity of G on V(Γ).
Then B={BR(h) ∣ h∈H} is a G-invariant partition of V(Γ). □
Lemma 7.12
T<N* and the quotient graph ΓB of Γ relative to B is isomorphic to the cross ladder graph CL2t4n. Moreover, T is the kernel of G acting on B.*
Proof Let KB be the kernel of G acting on B. Clearly, T≤KB. Let B′=Ω0′∪Ω1′. Then B′=BR(b)∈B. Let BR(h)∈B be adjacent to B and BR(h)=B′.
Suppose that Γ[B∪BR(h)] is a perfect matching. Since G is transitive on B, ΓB is a cycle of length t2n. Clearly, G/KB is vertex-transitive but not edge-transitive on ΓB, so G/KB≅D2n/t.
If t=1, then it is easy to see that Γ≅CL4n which is a Cayley graph by Theorem 1.1, a contradiction. If t>1, then since Γ[Ω]=Γ[B∪B′] is a cycle of length 4t, KB acts faithfully on B, and so KB≤Aut(Γ[B∪B′])≅D8t. Since KB fixes B, one has ∣KB∣∣4t, implying that ∣G∣=∣KB∣⋅t2n∣8n. As ∣R(H)∣=2n and R(H) is non-normal in G, one has ∣KB∣=4t due to T≤KB. In view of the fact that KB≤D8t, KB has a characteristic cyclic subgroup, say J, of order 2t. Then we have J⊴G because KB⊴G. Clearly, J is regular on B and J∩N=T, so JR(H) is regular on V(Γ). It follows from Proposition 2.1 that Γ is a Cayley graph, a contradiction.
Therefore, Γ[B∪BR(h)] is not a perfect matching. If N=T, then B=Δ and B′=Δ′ are orbits of M, and then Γ[B∪BR(h)] will be a perfect matching, a contradiction. Thus, N>T.
Now we are going to prove that ΓB≅CL2tn. Since B is adjacent to BR(h), Ωi is adjacent to ΩjR(h) for some i,j∈{0,1}. Then because Ωi and ΩjR(h) are orbits of T, Γ[Ωi∪ΩjR(h)] is a perfect matching. This implies that ΓB is of valency 3, and so KB is intransitive on B. As every Bh∈B is a union of two orbits of T on V(Γ), KB fixes every orbit of T.
Since N is cyclic, the normality of N in G implies that T⊴G. Clearly, Ω0 is adjacent to three pair-wise different orbits of T, so the quotient graph ΓT of Γ relative to T is of valency 3.
Consequently, the kernel of G acting on V(ΓT) is T.
Then KB=T.
Now R(H)/T≅D2n/t is regular on B, and so ΓB is a Cayley graph over R(H)/T. Furthermore, G/T is not arc-transitive on ΓB. Since R(H)/T is non-normal in G/T, ΓB is a non-normal Cayley graph over R(H)/T. If ΓB is arc-transitive, then by [13, Theorem 1], either ∣Aut(ΓB)∣=3k∣R(H)/T∣ with k≤2, or ΓB has order 2⋅p with p=3 or 7. For the former, since G/T is not arc-transitive on ΓB, one has ∣G/T:R(H)/T∣≤2, implying R(H)⊴G, a contradiction. For the latter, we have t2n=6 or 14, implying tn=3 or 7. It follows that T is a maximal subgroup of ⟨R(a)⟩, and so T=N, a contradiction. Therefore, ΓB is not arc-transitive. Since R(H)/T is non-normal in G/T, by Theorem 1.1, one has ΓB≅CL2t4n, as required. □
Proof of Theorem 7.6 By Lemma 7.12, we have ΓB≅CL2t4n. By the definition of CL2t4n, we may partition the vertex set of ΓB in the following way:
[TABLE]
and
[TABLE]
Assume that B00=B and B10=B′. Recall that B=Ω0∪Ω1 and B′=Ω0′∪Ω1′=BR(b).
Moreover, Ω0,Ω1,Ω0′ and Ω1′ are four orbits of T.
Then every Bij∈B is just a union of two orbits of T. For convenience, we may let
[TABLE]
where Ωi0j,Ωi1j are two orbits of T.
For B=B00, we let Ω0=Ω000 and Ω1=Ω010, and for B′=B10, we let
Ω0′=Ω100 and Ω1′=Ω110.
For convenience, in the remainder of the proof, we shall use C4t to denote a cycle of length 4t, and we also call C4t a 4t-cycle.
Recall that Γ[B∪B′]=Γ[B00∪B10]≅C4t, and that the edges between Ω0i0(=Ωi) and Ω1j0(=Ωj′) form a perfect matching for all i,j∈Z2.
Since T⊴G, the quotient graph ΓT of Γ relative to T has valency 3.
So the edges between any two adjacent orbits of T form a perfect matching.
From the construction of ΓB, one may see that there exists g∈G such that
{V0,V1}g={V2i,V2i+1} for each i∈Z2tn.
So for each i∈Z2tn,r∈Z2, we may assume that
Γ[B2ir∪B2i+1r]≅C4t, and Ω(2i)sr∼Ω(2i+1)tr for all s,t∈Z2. (Here Ω(2i)sr∼Ω(2i+1)tr means that Ω(2i)sr and Ω(2i+1)tr are adjacent in ΓB.)
Again, from the construction of ΓB, we may assume that
[TABLE]
for each i∈Z2tn. We draw a local subgraph of ΓB in Figure 3.
Observing that every Vi={Bi0,Bi1} with i∈Z2t2n is a block of imprimitivity of G/KB acting on V(ΓB).
So every Bi0∪Bi1 with i∈Z2t2n is a block of imprimitivity of G acting on V(Γ).
Let E be the kernel of G acting on the block system Λ={Bi0∪Bi1∣i∈Z2t2n}.
Then G/E≅Dtn acts regularly on Λ. Clearly, R(H) is also transitive on Ω, so G/E=R(H)E/E.
By Lemma 7.12, T is a the kernel of G acting on B. So E/T is an elementary 2-group.
From R(H)/(R(H)∩E)≅Dtn it follows that R(H)∩E=⟨R(a2tn)⟩≅Z2t, and so (R(H)∩E)/T is a normal subgroup of G/T of order 2. This implies that Bi1=(Bi0)R(a2tn) for i∈Z2t2n. We may further assume that Ω011=(Ω000)R(a2tn)⊆B01.
So Ω000∪Ω011 is just the orbit of ⟨R(a2tn)⟩ containing 10.
Observing that Ω100∼Ω200 and the edges between them are of the form {g0,(baig)0} with g0∈Ω100, one has Ω200=baiΩ100=bai(Ω000)R(b)=(Ω000)R(a−i).
So Ω201⊆(B01)R(a−i).
Since B10=B′=BR(b)=(B00)R(b), one has B11=(B01)R(b). Recall that 11∈Ω110=Ω1′ and 11 is adjacent to 10∈Ω000=Ω0 and (bal)1∈Ω010=Ω1. As we assume that Ω110∼Ω201, 11 is adjacent to some vertex in Ω201. So Ω201⊆H1 and hence
[TABLE]
So we have the following claim.
Claim 1 L={bal,baktn+2tn−i} and R={bai,b}, where ∣R(al)∣=t, i∈Zn and 0≤k≤t−1.
Let G10∗ be the kernel of G10 acting on the neighborhood of 10 in Γ.
Then G10∗≤E10. Recall that for each i∈Z2tn,r∈Z2,
Γ[B2ir∪B2i+1r]≅C4t and the edges between B2i+10∪B2i+11 and B2i+20∪B2i+21 form a perfect matching.
It follows that E acts faithfully on each Bi0∪Bi1. Clearly, G10∗≤E10, so G10∗ acts faithfully on each Bi0∪Bi1.
Claim 2 If t>2 then G10∗=1, and if t=2 then G10∗≤Z2 and 3∣n.
Assume that t≥2. Since Γ[B00∪B10]≅C4t, G10∗ fixes every vertex in B00, and so fixes every vertex in Ω(−1)00 since Ω(−1)00∼Ω000 (see Figure 3). This implies that G10∗ fixes Ω(−1)10 setwise, and so fixes Ω001 setwise since Ω(−1)10∼Ω001. Consequently, G10∗ also fixes Ω011 setwise. Similarly, by considering the edges between B10∪B11 and B20∪B21, we see that G10∗ fixes both Ω101 and Ω111 setwise. Recall that the edges between Ω0i1 and Ω1j1 form a perfect matching for i,j∈Z2. As Γ[B10∪B11]≅C4t, G10∗ acts faithfully on Ω001 (or Ω011), and so G10∗≤Z2.
If t>2, then since Γ[B−20∪B−10]≅C4t, G10∗ will fix every vertex in this cycle, and in particular,
G10∗ will fix every vertex in Ω(−1)10. As Ω(−1)10∼Ω001, G10∗ will fix every vertex in Ω001. Since G10∗ acts faithfully on Ω001, one has G10∗=1.
Let t=2. We shall show that 3∣n. Then T=⟨R(a2n)⟩. Recall that (R(H)∩E)/T is a normal subgroup of G/T of order 2. Let M=R(H)∩E. Then M is a normal subgroup of G of order 4. Since R(H) is dihedral, one has M=⟨R(a4n)⟩.
Let C=CG(M). Then R(a)∈C and R(b)∈/C. It follows that C is a proper subgroup of G. Since G/E acts regularly on Λ, C10 fixes every element in Λ. Since C10 centralizes M, C10 fixes every vertex in the orbit Ω000∪Ω011 of M containing 10. Clearly, C10≤G10, so C10/(C10∩G10∗)≤Z2. As we have shown that G10∗ acts faithfully on Ω011, it follows that C10∩G10∗=1 since C10 fixes Ω011 pointwise, and hence C10≤Z2. On the other hand, as G10∗≤Z2, one has ∣G∣∣4⋅4n=16n.
Since C<G and R(a)∈C, one has ∣C∣=kn with k∣8.
Suppose that 3∤n. For any odd prime divisor p of n, let P be a Sylow p-subgroup of ⟨R(a)⟩. Then P is also a Sylow p-subgroup of C. If P is not normal in C, then by Sylow’s theorem, we have ∣C:NC(P)∣=k′p+1∣8 for some integer k′. Since p=3, one has p=7 and k′=1. This implies that ∣C∣=8∣NC(P)∣, and so ∣C∣=8n due to R(a)∈C and C<G. Since C10≤Z2, one has ∣C:C10∣≥4n, and so C is transitive on V(Γ). Moreover, we have CC(P)=NC(P)=⟨R(a)⟩. By Burnside theorem, C has a normal subgroup M such that C=M⋊P. Then the quotient graph ΓM of Γ relative to M would be a cycle of length ∣P∣, and the subgraph induced by each orbit of M is just a perfect matching. This implies that M is just the kernel of G acting on V(ΓM). Furthermore, C/M is a vertex-transitive subgroup of Aut(ΓM). Since ΓM is a cycle, C/M must contain a subgroup, say B/M acting regularly on V(ΓM). Then B will be regular on V(Γ), and so by Proposition 2.1, Γ is a Cayley graph, a contradiction. Therefore, P⊴C, and since C⊴G, one has P⊴G, implying P≤N. By the arbitrariness of P, n/∣N∣ must be even, contrary to Lemma 7.7.
Thus, 3∣n, as claimed.
The following claim shows that t=1 or 2.
Claim 3 t≤2.
By way of contradiction, suppose that t>2. Let C=CG(T). Then ⟨R(a)⟩≤C and R(H)≰C since ∣T∣=t>2. Clearly, C10≤E10. As C10 centralizes T, C10 will fixes every vertex in Ω000 since Ω000 is an orbit of T containing 10. Since Γ[B00∪B10]≅C4t, C10 fixes every vertex in this 4t-cycle, and so C10≤G10∗=1 (by Claim 2). Thus, C acts semiregularly on V(Γ).
If C=⟨R(a)⟩, then by N/C-theorem, we have G/⟨R(a)⟩=G/C≤Aut(T). Since T≤N≤⟨R(a)⟩ is cyclic, Aut(T) is abelian. It then follows that R(H)/C⊴G/C, and hence R(H)⊴G, a contradiction.
If C>⟨R(a)⟩, then ∣C∣=2n because Γ is non-Cayley. Since H0 and H1 are not blocks of imprimitivity of G on V(Γ), C does not fix H0 setwise, and so R(H)C is transitive on V(Γ). Clearly, R(H)∩C=⟨R(a)⟩, so ∣R(H)C∣=∣R(H)∣∣C∣/∣⟨R(a)⟩∣=4n. It follows that R(H)C is regular on V(Γ), contradicting that Γ is non-Cayley.
By Claim 3, we only need to consider the following two cases:
Case 1 t=1.
In this case, by Claim 1, we have R={b,bai} and L={b,ba2n−i}. For convenience, we let n=2ℓ. Then
R={b,bai} and L={b,baℓ−i}.
By Proposition 2.3 (1), the connectedness of Γ implies that ⟨ai,aℓ⟩=⟨a⟩. Then either (i,2ℓ)=1, or i=2k with (k,2ℓ)=1 and ℓ is odd.
Recall that H=⟨a,b∣a2ℓ=b2=1,bab=a−1⟩. For any λ∈Z2ℓ∗, let αλ be the automorphism of H induced by the map
[TABLE]
So if (i,2ℓ)=1, then we have
[TABLE]
and if i=2k with (k,2ℓ)=2 and ℓ is odd,
then we have
[TABLE]
So by Proposition 2.3 (3), we have
[TABLE]
Suppose that ℓ is even. Then (R,L,S)≡({b,ba},{b,baℓ−1},{1}).
Since ℓ is even, one has (2ℓ,ℓ+1)=1 and (ℓ+1)2≡1 (mod 2ℓ).
Then it is easy to check that αℓ+1 is an automorphism of H of order 2 that swaps {b,ba} and {b,baℓ−1}.
By Proposition 2.4, we have δαℓ+1,1,1∈I, and then Γ≅BiCay(H,{b,ba},{b,baℓ+1},{1}) is a Cayley graph, a contradiction.
Now we assume that n=2ℓ with ℓ=2m+1 for some integer m.
Let
[TABLE]
Direct calculation shows that (n,2m−1)=1, and 2m(2m−1)≡2 (mod n).
Then the automorphism α2m−1:a↦a2m−1, b↦b maps
the pair of two subsets ({b,ba},{b,ba2m}) to ({b,ba2m−1},{b,ba2}).
So, we have (R,L,S)≡({b,ba},{b,ba2m},{1}).
By Lemma 4.1 and Theorem 1.2, Γ≅MCL(4m,2) and Γ is non-Cayley if and only if 3∤(2m+1).
Note that 3∤(2m+1) is equivalent to m≡1 (mod 3). So we obtain the first family of graphs in Theorem 7.6.
Case 2 t=2.
In this case, by Claim 1, we have R={b,bai} and L={ba2n,ba43n−i} or {ba2n,ba4n−i}.
We still use the following notation: For any λ∈Z2ℓ∗, let αλ be the automorphism of H induced by the map
[TABLE]
Note that
[TABLE]
By replacing −i by i, we may always assume that
[TABLE]
By Claim 2, we have 3∣n. So we may assume that n=12m for some integer m. Then we have
[TABLE]
Since Γ is connected, by Proposition 2.3, we have ⟨ai,a3m⟩=⟨a⟩.
If m is odd, by Lemma 6.1, Γ will be a Cayley graph which is impossible. Thus, m is even.
It then follows that ⟨ai⟩∩⟨a3m⟩>1 since ⟨ai,a3m⟩=⟨ai⟩⟨a3m⟩=⟨a⟩.
Since ⟨a3m⟩≅Z4, one has ∣⟨ai⟩∩⟨a3m⟩∣=2 or 4. For the former, we would have
∣⟨ai⟩∣=6m, and since m is even, one has 4∣∣⟨ai⟩∣, and hence a3m∈⟨ai⟩, a contradiction.
Thus, we have ∣⟨ai⟩∩⟨a3m⟩∣=4, that is, ⟨ai⟩=⟨a⟩.
So (i,12m)=1, and then αi∈Aut(H) which maps ({b,bai},{ba6m,ba3m−i}) to
({b,ba},{ba6m,ba3m−1}) or ({b,ba},{ba6m,ba−3m−1}).
Then
[TABLE]
If m≡2 (mod 4), then by Lemma 6.2, we see that Γ will be a Cayley graph, a contradiction.
Thus, m≡0 (mod 4). Clearly, (3m−1,12m)=1, and hence the map a↦a3m−1,b↦ba6m induces an automorphism, say β of H. It is easy to check that
[TABLE]
Thus,
[TABLE]
By Proposition 5.1, Γ is a non-Cayley graph. Let m=4ℓ for some integer ℓ. Then n=48ℓ and then we get the second family of graphs in Theorem 7.6. This completes the proof of Theorem 7.6.
□
7.4 Proof of Theorem 1.3
By [20, Theorem 1.2], if Γ is [math]- or 1-type, then Γ is a Cayley graph. Let Γ be of 2-type. Suppose that Γ is a non-Cayley graph. Let G≤Aut(Γ) be minimal subject to that R(H)≤G and G is transitive on V(Γ). If H0 and H1 are blocks of imprimitivity of G on V(Γ), then by Proposition 7.2, we obtain the first two families of graphs of Theorem 1.3. Otherwise, H0 and H1 are not blocks of imprimitivity of G on V(Γ), by Theorem 7.6, we obtain the last two families of graphs of Theorem 1.3. □
Acknowledgements: This work was supported by the National
Natural Science Foundation of China (11671030).
