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Normality of 2-Cayley digraphs
Discrete Mathematics 338 (2015) 41–47
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Discrete Mathematics journal homepage: www.elsevier.com/locate/disc
Normality of 2-Cayley digraphs Majid Arezoomand, Bijan Taeri ∗ Department of Mathematical Sciences, Isfahan University of Technology, Isfahan 84156-83111, Iran
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Article history: Received 30 January 2013 Received in revised form 18 October 2014 Accepted 20 October 2014
Keywords: 2-Cayley digraph Normal 2-Cayley digraph Automorphism group of digraph
abstract A digraph Γ is called a 2-Cayley digraph over a group G, if there exists a semiregular subgroup RG of Aut(Γ ) isomorphic to G with two orbits. We say that Γ is normal if RG is a normal subgroup of Aut(Γ ). In this paper, we determine the normalizer of RG in Aut(Γ ). We show that the automorphism group of each normal 2-Cayley digraph over a group with solvable automorphism group, is solvable. We prove that for each finite group G ̸= Q8 × Zr2 , r ≥ 0, where Q8 is the quaternion group of order 8 and Z2 is the cyclic group of order 2, there exists a normal 2-Cayley graph over G and that every finite group has a normal 2-Cayley digraph. © 2014 Elsevier B.V. All rights reserved.
1. Introduction Graphs come in two principle types directed and undirected graphs. We shall refer to directed graphs as digraphs and use the term graph to refer to undirected graphs. A digraph Γ is a pair (V , E ) of vertices V and edges E where E ⊆ V × V ; the digraph Γ is said to be finite if V is finite. A graph is a digraph with no edges of the form (α, α) and with the property that (α, β) ∈ E implies (β, α) ∈ E. The set of all permutations of V which preserve the adjacency structure of Γ is called the automorphism group of Γ ; it is denoted by Aut(Γ ). In this paper all digraphs have no loops. For the group-theoretic and graph-theoretic terminology not defined here we refer the reader to [3,7]. Let G be a group and S a subset of G not containing the identity element 1. The Cayley digraph Γ = Cay(G, S ) of G with respect to S has vertex set G and arc set {(g , sg ) | g ∈ G, s ∈ S }. If S = S −1 , then Cay(G, S ) can be viewed as an undirected graph, identifying an undirected edge with two directed edges (g , h) and (h, g ). This graph is called Cayley graph of G with respect to S. By a theorem of Sabidussi [11], a digraph Γ is a Cayley digraph over a group G if and only if there exists a regular subgroup of Aut(Γ ) isomorphic to G. There is a natural generalization of Sabidussi’s Theorem: a digraph Γ is called an n-Cayley digraph over a group G if there exists an n-orbit semiregular subgroup of Aut(Γ ) isomorphic to G. 2-Cayley graphs are often called semi-Cayley graphs [10,6], or bi-Cayley graphs [8]. Also a special case of 2-Cayley graphs are called bi-Cayley graphs by some authors, see for example [13,9]. Let Γ be a Cayley digraph over a group G. It is well-known that the right regular representation R(G) of G is a regular subgroup of Aut(Γ ). If R(G) is a normal subgroup of Aut(Γ ), then Γ is called a normal Cayley digraph over G, [12]. In the literatures, the study of normality of Cayley digraphs has been becoming a rather active topic in the algebraic graph theory, which also plays an important role in the investigation of various symmetry properties of digraphs. We encourage the reader to consult [5] for a survey up to 2008. In this paper we define the concept of normality for 2-Cayley digraphs and present some equivalent conditions to the normality of 2-Cayley digraphs. We prove that if Γ is a normal 2-Cayley digraph over a group G such that Aut(G) is solvable,
∗
Corresponding author. E-mail addresses: [email protected] (M. Arezoomand), [email protected] (B. Taeri). URL: http://[email protected]/ (B. Taeri).
http://dx.doi.org/10.1016/j.disc.2014.10.019 0012-365X/© 2014 Elsevier B.V. All rights reserved.
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a
b
Fig. 1. The cube is a 2-Cayley graph over (a) Z4 , (b) Z2 × Z2 .
Fig. 2. The Petersen graph is a 2-Cayley graph over Z5 .
then Aut(Γ ) is solvable. In particular, the automorphism group of each normal 2-Cayley graph over a cyclic group is solvable. Also we prove that every finite group G ̸= Q8 × Zr2 , r ≥ 0, where Q8 is the quaternion group of order 8 and Zn is the cyclic group of order n, has at least one normal 2-Cayley graph, and every finite group G has at least one normal 2-Cayley digraph. 2. Normal 2-Cayley digraphs Recall that a digraph Γ is called a 2-Cayley digraph over a group G if its automorphism group has a 2-orbit semiregular subgroup RG isomorphic to G. By [1, Lemma 2], we may identify Γ with a digraph Cay(G; Ti,j | 1 ≤ i, j ≤ 2) with vertex set G × {1, 2} and ((g , i), (h, j)) is an arc if and only if hg −1 ∈ Ti,j , for some suitable subsets T1,1 , T1,2 , T2,1 and T2,2 of G in which 1 does not belong to T1,1 and T2,2 . Also we can identify the 2-orbit semiregular subgroup RG of Aut(Γ ) isomorphic to G with
{ρg : V → V | g ∈ G, (x, j)ρg = (xg , j)}. Furthermore, Γ is undirected if and only if T1−,11 = T1,1 , T2−,21 = T2,2 and T1−,21 = T2,1 (with the convention that ∅−1 = ∅). Definition. A 2-Cayley digraph Γ over a group G is called normal (over G) if RG is a normal subgroup of Aut(Γ ). We denote the symmetric group on a set Ω by Sym(Ω ) and when |Ω | = n is finite, by Sn . In what follows, we give some examples of normal and non-normal 2-Cayley digraphs. First, note that the normality of a 2-Cayley digraph depends on the underlying group G. To see this, let Γ be the cube. Then Aut(Γ ) ∼ = S2 ≀ S3 , see [7, Exercise 14.26]. If G = ⟨a⟩ ∼ = Z4 , T1,1 = T2,2 = {a, a3 } and T1,2 = T2,1 = {1}, then Cay(G; Ti,j | 1 ≤ i, j ≤ 2) ∼ = Γ (see Fig. 1(a)). So a cube is a 2-Cayley graph over Z4 . Also if H = ⟨a, b | a2 = b2 = (ab)2 = 1⟩ ∼ = Z2 × Z2 , S1,1 = S2,2 = {a, b} and S1,2 = S2,1 = {1}, then Cay(H ; Si,j | 1 ≤ i, j ≤ 2) ∼ = Γ (see Fig. 1(b)). Hence a cube is also a 2-Cayley graph over Z2 × Z2 . Since the only normal subgroup of Aut(Γ ) ∼ = S2 ≀ S3 of order 4 is isomorphic to Z2 × Z2 , the graph Γ is a normal 2-Cayley graph over Z2 × Z2 , but not over Z4 . Example 2.1. Let P be the Petersen graph. Then P = Cay(G, Ti,j | 1 ≤ i, j ≤ 2), where G = ⟨a⟩ ∼ = Z5 , T1,1 = {a, a4 }, T2,2 = 2 3 ∼ {a , a } and T1,2 = T2,1 = {1} (see Fig. 2). It is well-known that Aut(P ) = S5 (see for example [3, Exercise 2.3.4]). Since S5 has no normal subgroup of order 5, P is not normal over Z5 . Example 2.2. Let Γ be a 2-Cayley digraph over a group G such that Aut(Γ ) acts primitively on V (Γ ). Then Γ is not normal over G. In fact, if G is a normal subgroup of Aut(Γ ) then [3, Theorem 1.6A] implies that RG is transitive on V (Γ ), which is impossible. In particular, if Γ = K2n or 2nK1 , where n ≥ 3, then Aut(Γ ) ∼ = S2n acts primitively on V (Γ ) (see for example [3, Exercise 1.5.12]). Now for every semiregular subgroup G of S2n of order n with two orbits, Γ is a non-normal 2-Cayley graph over G.
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Example 2.3. Let Γ = B(H (11)) be the incidence graph of the doubly transitive Hadamard 2-(11, 5, 2) design. Then Aut(Γ ) = PSL(2, 11) o Z2 ∼ = PGL(2, 11) and Γ is a Cayley graph over the dihedral group D22 of order 22, (see [4, Table 1]). Since Z11 is a subgroup of D22 of index 2, Γ is a 2-Cayley graph over Z11 . Since PGL(2, 11) has no normal subgroup isomorphic to Z11 , Γ is not normal over Z11 . Now we want to find the structure of the normalizer of RG in Aut(Γ ). From now on, the vertex set, edge set and automorphism group of Γ are denoted by V , E and A, respectively, and RG is the subgroup defined above. Also if G is a group acting on a set Ω , then for each α ∈ Ω the stabilizer of α in G, and the orbit of α under G are denoted by Gα and α G , respectively. For a group G and its subgroup H, we denote the normalizer and the centralizer of H in G by NG (H ) and CG (H ), respectively and we put CG (G) = Z (G). We first describe the stabilizer of (1, 1) in NSym(V ) (RG ). Lemma 2.4. Let Γ be a 2-Cayley digraph over a group G and M = NSym(V ) (RG ). Then ϕ ∈ M(1,1) if and only if there exist g ∈ G and σ ∈ Aut(G) such that for all x ∈ G, (x, 1)ϕ = (xσ , 1) and (x, 2)ϕ = (gxσ , 2). Proof. Let α ∈ M(1,1) . Then α ∈ Sym(V ), (1, 1)α = (1, 1) and for every x ∈ G there exists a unique x′ ∈ G such that −1 α −1 ρx α = ρx′ . So there exists σ ∈ Sym(G) such that for every x ∈ G, α −1 ρx α = ρxσ . Thus for all x ∈ G, (x, 1)α = (1, 1)α ρx α ρ xσ σ = (1, 1) = (x , 1). Moreover, for g1 , g2 ∈ G,
((g1 g2 )σ , 1) = (g1 g2 , 1)α = (1, 1)α
−1 ρ
g1 ρg2 α
ρg σ ρg σ
= (1, 1)
1
2
= (g1σ g2σ , 1),
which implies that (g1 g2 )σ = g1σ g2σ . This shows that σ ∈ Aut(G). Since for each x ∈ G, (x, 1)α = (xσ , 1), we have (G × {1})α = G × {1} and therefore (1, 2)α = (g , 2), for some g ∈ G. Hence (x, 2)α = (1, 2)ρx α = (1, 2)αρxσ = (gxσ , 2), for all x ∈ G. This completes the proof of one direction. Conversely, suppose that ϕ : V → V is a map in which there exist g ∈ G and σ ∈ Aut(G) such that for all x ∈ G, (x, 1)ϕ = (xσ , 1) and (x, 2)ϕ = (gxσ , 2). Since (1, 1)ϕ = (1, 1), we have to show that ϕ ∈ M. Clearly ϕ is 1–1 and onto. Let −1 y ∈ G and (x, j) ∈ V . Then (x, j)ϕ ρy ϕ = (xyσ , j) = (x, j)ρyσ . Thus for all y ∈ G, ϕ −1 ρy ϕ = ρyσ , which means that ϕ ∈ M. This completes the proof.
Let Γ = Cay(G, S ) be the Cayley digraph over a group G with respect to S, and R(G) be the right regular representation of G. It is well-known that NAut(Γ ) (R(G)) = R(G) o Aut(G, S ), where Aut(G, S ) = {σ ∈ Aut(G) | S σ = S }, [12, Proposition 1.3]. We wish to find a similar result for 2-Cayley digraphs. First, let us define some subsets of the automorphism group of a 2-Cayley digraph Γ = Cay(G; Ti,j | 1 ≤ i, j ≤ 2). Let X be the set of all maps ψ : V → V , where (x, 1)ψ = (xσ , 1) and (x, 2)ψ = (gxσ , 2), for some g ∈ G and σ ∈ Aut(G) such that T1σ,1 = T1,1 , T2σ,2 = g −1 T2,2 g, T1σ,2 = g −1 T1,2 , and T2σ,1 = T2,1 g.
Also, let Y be the set of all maps ϕ : V → V , where (x, 1)ϕ = (xθ , 2) and (x, 2)ϕ = (hxθ , 1), for some h ∈ G and θ ∈ Aut(G) such that T1θ,1 = T2,2 , T2θ,2 = h−1 T1,1 h, T1θ,2 = h−1 T2,1 and T2θ,1 = T1,2 h, with the convention that if one of the pair sets T1,1 , T2,2 or T1,2 , T2,1 is empty and the other is non-empty, we put Y = ∅. Also if in the above equalities, one of the subsets Ti,j is empty, then we omit the equality including Ti,j . Note that X and Y are disjoint and X = M(1,1) ∩ A, where M = NSym(V ) (RG ), is a subgroup of A(1,1) . Now we are ready to describe the normalizer of RG in A.
Theorem 1. Let Γ = Cay(G; Ti,j | 1 ≤ i, j ≤ 2) be a 2-Cayley digraph over a group G, and X , Y be the sets defined above. Then NA (RG ) = LRG , where L = X ∪ Y . Furthermore, RG ∩ L = {1}. Proof. Let M = NSym(V ) (RG ). Since RG ∼ = G is a semiregular subgroup of Sym(V ), by [3, Exercise 4.2.7], CSym(V ) (RG ) ≤ M is transitive. Thus M is also transitive. For a permutation τ ∈ S2 , we define τ : V → V by (x, j)τ = (x, jτ ). Clearly τ ∈ M. ρ
ρ
( i j)
Setting H = {τ | τ ∈ S2 } we have H ≤ CSym(V ) (RG ). For x, y ∈ G, we have (x, i) x−1 y = (y, i) and (x, i) x−1 y = (y, j), i ̸= j. Thus RG H = HRG is transitive on V . Now [3, Exercise 1.4.1] implies that M = M(1,1) HRG . Therefore, by Dedekind’s Lemma, NA (RG ) = NSym(V ) (RG ) ∩ A = M ∩ A = M(1,1) HRG ∩ A = (M(1,1) H ∩ A)RG . To complete the proof, we must show that L = M(1,1) H ∩ A. Recall from Lemma 2.4 that a map ϕ : V → V is an element of M(1,1) if and only if for all x ∈ G, (x, 1)ϕ = (xσ , 1) and (x, 2)ϕ = (gxσ , 2), for some g ∈ G and σ ∈ Aut(G). Let ψ ∈ M(1,1) H ∩ A. Then ψ ∈ A and ψ = ϕτ , for some ϕ ∈ M(1,1) and τ ∈ H. So there exist g ∈ G and σ ∈ Aut(G) such that for all x ∈ G, (x, 1)ψ = (xσ , 1τ ) and (x, 2)ψ = (gxσ , 2τ ). We consider two cases: Case I. τ = idS2 . In this case (x, 1)ψ = (xσ , 1) and (x, 2)ψ = (gxσ , 2), for all x ∈ G. Since ψ ∈ A, it is easily seen that T1σ,1 = T1,1 , T2σ,2 = g −1 T2,2 g, T1σ,2 = g −1 T1,2 and T2σ,1 = T2,1 g, for instance t ∈ T1,2 ⇔ ((1, 1), (t , 2)) ∈ E
⇔ ((1, 1)ψ , (t , 2)ψ ) = ((1, 1), (gt σ , 2)) ∈ E ⇔ gt σ ∈ T1,2 . This shows that ψ ∈ X .
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Case II. τ = (12). In this case (x, 1)ψ = (xσ , 2) and (x, 2)ψ = (gxσ , 1), for all x ∈ G. Since ψ ∈ A, it is easy to see that T1σ,1 = T2,2 , T2σ,2 = g −1 T1,1 g, T1σ,2 = g −1 T2,1 and T2σ,1 = T1,2 g, for instance t ∈ T1,2 ⇔ ((1, 1), (t , 2)) ∈ E
⇔ ((1, 1)ψ , (t , 2)ψ ) = ((1, 2), (gt σ , 1)) ∈ E ⇔ gt σ ∈ T2,1 . Hence ψ ∈ Y . Thus we have proved that M(1,1) H ∩ A ⊆ X ∪ Y . Now we prove the reverse inclusion. If ψ ∈ X , then there exist g ∈ G and σ ∈ Aut(G) such that T1σ,1 = T1,1 , T2σ,2 = g −1 T2,2 g, T1σ,2 = g −1 T1,2 , and T2σ,1 = T2,1 g; and for all x ∈ G, (x, 1)ψ = (xσ , 1)
and (x, 2)ψ = (gxσ , 2). Hence ψ ∈ M(1,1) and also ψ preserves the adjacency, which means that ψ ∈ L. Similarly if ϕ ∈ Y , we can see that ϕ ∈ L. Hence the equality M(1,1) H ∩ A = X ∪ Y holds. Finally, we show that RG ∩ L is trivial. Let ϕ ∈ RG ∩ L. Then there exist g ∈ G and ψ ∈ L such that ϕ = ρg = ψ . So (g , 1) = (1, 1)ρg = (1, 1)ψ = (1, 1τ ). Therefore g = 1 and so ϕ = 1. This completes the proof. From now on, we assume that L is the set defined in Theorem 1. It is easy to see that if α = ψρx , for some ρx ∈ RG and ψ ∈ L, then ρx and ψ are uniquely determined by α . By the definition of X and Y , each element ϕ of L depends on an element g ∈ G and σ ∈ Aut(G). In the following lemma, we denote the corresponding element g ∈ G and σ ∈ Aut(G) of ϕ by gϕ and σϕ , respectively. Note that in general L is not a group, see Example 2.6. In the following lemma we study the problem that when L is a group? Lemma 2.5. L is a group if and only if Y = ∅ or for each ϕ ∈ L, gϕ = 1. Proof. Recall that X = M(1,1) ∩ A is a group, where M = NSym(V ) (RG ). First let L be a group and ϕ ∈ L = X ∪ Y . Suppose that ϕ ∈ Y . Then (1, 2)ϕ = (gϕ 1σϕ , 1) = (gϕ , 1) and hence −1 −1 (gϕ , 1)ϕ = (1, 2). Since ϕ −1 ∈ L and X and Y are disjoint, we conclude that ϕ −1 ∈ Y . Moreover, (gϕ , 1)ϕ = (gϕθ , 2),
where θ = σϕ −1 and so gϕθ = 1, which implies that gϕ = 1. Now let ϕ ∈ X . Then for all x ∈ G, (x, 1)ϕ = (xσϕ , 1) and (x, 2)ϕ =
(gϕ xθϕ , 2). Choose an element ψ ∈ Y . Then (x, 1)ϕψ = (xσϕ σψ , 2) and (x, 2)ϕψ = ((gϕ )σψ xσϕ σψ , 1). Since L is a group, ϕψ ∈ Y and by the first part, (gϕ )σψ = 1, which implies that gϕ = 1. Conversely, if Y = ∅ then L = X is a group. So suppose that Y ̸= ∅ and for each ϕ ∈ L, gϕ = 1. It is easy to see that for all ψ1 , ψ2 ∈ X and ϕ1 , ϕ2 ∈ Y , ψ1 ψ2−1 , ϕ1 ϕ2−1 ∈ X , and ϕ1 ψ1−1 , ψ1 ϕ1−1 ∈ Y . This shows that for each α1 , α2 ∈ L, we have α1 α2−1 ∈ L, and thus L is a group. In the following examples we show that the structure of L depends on the subsets Ti,j and in general L is not a group. Example 2.6. Let G = ⟨a⟩ ∼ = Z3 . Then Aut(G) = {id, π}, where π maps ak to a2k , k = 0, 1, 2. Consider the 2-Cayley graph Cay(G; Ti,j | 1 ≤ i, j ≤ 2), where T1,1 = T2,2 = ∅, T1,2 = {1, a} and T2,1 = {1, a2 }. Let α ∈ X . Then there exist σ ∈ Aut(G) and g ∈ G such that for all x ∈ G, (x, 1)α = (xσ , 1) and (x, 2)σ = (gxσ , 2). Furthermore, T1σ,2 = g −1 T1,2 and T2σ,1 = T2,1 g. From these equations we conclude that α ∈ {α1 , α2 }, where
α1 : (x, 1) → (x, 1), α2 : (x, 1) → (xπ , 1),
(x, 2) → (x, 2), (x, 2) → (axπ , 2).
Also for each element β ∈ Y , there exist g ∈ G and σ ∈ Aut(G) such that for all x ∈ G, (x, 1)β = (xσ , 2), (x, 2)β = (gxσ , 1), T1σ,2 = g −1 T2,1 and T2σ,1 = T1,2 g. Hence β ∈ {β1 , β2 }, where
β1 : (x, 1) → (x, 2), β2 : (x, 1) → (xπ , 2),
(x, 2) → (a2 x, 1) (x, 2) → (xπ , 1).
Thus L = {α1 , α2 , β1 , β2 } and by Lemma 2.5, L is not a group. Example 2.7. Let G = ⟨a⟩ ∼ = Z5 , then Aut(G) = {σi | i = 1, . . . , 4}, where σi maps ak to aki for all k = 0, . . . , 4. By Example 2.1, the Petersen graph P is a 2-Cayley graph P = Cay(G; Ti,j | 1 ≤ i, j ≤ 2), where T1,1 = {a, a4 }, T2,2 = {a2 , a3 } and T1,2 = T2,1 = {1}. So we can consider the vertices of P as {(ai , 1), (ai , 2) | i = 0, . . . , 4}. Now, by a similar argument to that of Example 2.6 we can see that, the elements of L are
ψ1 ψ2 ψ3 ψ4
: (x, 1) → (x, 1), : (x, 1) → (xσ4 , 1), : (x, 1) → (xσ2 , 2), : (x, 1) → (xσ3 , 2),
(x, 2) → (x, 2), (x, 2) → (xσ4 , 2), (x, 2) → (xσ2 , 1), (x, 2) → (xσ3 , 1).
By Lemma 2.5, L is a group. Furthermore, ψ42 = ψ2 , ψ43 = ψ3 , and ψ44 = ψ1 , which implies that L ∼ = Z4 . Also by Theorem 1, NAut(P ) (RG ) = RG o L ∼ = Z5 o Z4 .
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Let Γ = Cay(G, S ) be a Cayley digraph of G with respect to S. We know that the following statements are equivalent, [12, Propositions 1.3 and 1.5]: (i) R(G) E Aut(Γ ), (ii) Aut(Γ ) = R(G) o Aut(G, S ), (iii) Aut(Γ )1 = Aut(G, S ). The 2-Cayley version of this result is the following. Proposition 2. Let Γ = Cay(G; Ti,j | 1 ≤ i, j ≤ 2) be a 2-Cayley digraph over G. Then (1) RG E A if and only if A = LRG , (2) if RG E A, then A(1,1) = X , and the converse holds if A is not transitive on V . Proof. By Theorem 1, it is clear that (1) holds. Now we prove (2). Set N = NA (RG ) and M = NSym(V ) (RG ). First, suppose that RG E A. Then A(1,1) = N(1,1) ≤ M(1,1) . Thus X ≤ A(1,1) ≤ M(1,1) ∩ A = X , which implies that A(1,1) = X . Conversely, suppose that X = A(1,1) and A is not transitive on V . So A has at least two distinct orbits on V . Set O1 := (1, 1)A and O2 = (1, 2)A . We claim that O1 and O2 are the only orbits of A, O1 = (1, 1)RG and O2 = (1, 2)RG . If O1 ∩ O2 ̸= ∅, then there exists α ∈ A which maps (1, 1) to (1, 2) and so ⟨RG , α⟩ acts transitively on V , a contradiction. Hence O1 ∩ O2 = ∅. On the other hand V = (1, 1)RG ∪ (1, 2)RG ⊆ O1 ∪ O2 ⊆ V , which implies that O1 ∪ O2 = V . Since O1 ∩ O2 = ∅ and (1, i)RG ⊆ Oi , i = 1, 2, O1 = (1, 1)RG , O2 = (1, 2)RG and O1 and O2 are all orbits of A. Hence A acts on O1 . Since RG and A are transitive on O1 , we have A = RG A(1,1) . Hence A = RG X ⊆ RG L = NA (RG ) ⊆ A implies that A = NA (RG ). Hence RG E A. This completes the proof. Corollary 2.8. Let Γ be a normal 2-Cayley digraph over a finite group G. Then A(1,1) = L if and only if Γ is not vertex-transitive. Proof. Since Γ is normal over G, by Proposition 2, A = RG L, A(1,1) = X ⊆ L and |A| = |G| |L|. If A(1,1) = L then |A : A(1,1) | = |G| ̸= |V |, which means that Γ is not vertex-transitive. Conversely, suppose that Γ is not vertex-transitive. By the proof of Proposition 2, A = RG A(1,1) . Moreover, RG ∩ A(1,1) = {1} which implies that |A| = |G| |A(1,1) |. Since |A| = |G| |L|, |L| = |A(1,1) |. Now since A(1,1) ⊆ L we have A(1,1) = L, as desired. Now we want to find an equivalent condition to the vertex transitivity of normal 2-Cayley digraphs relating to L. First we need the following lemma. Lemma 2.9. Let Γ be a 2-Cayley digraph over group G. If H = H0 RG ≤ A, where 1 ̸= H0 < H, then H is transitive on V if and only if one of the following holds: (1) H0 is transitive on V (2) V = g ∈G Rρg where R ⊂ V is an orbit of H0 . Proof. Suppose that H is transitive on V but H0 is not. So there exists an orbit R := (x, i)H0 ⊂ V , for some (x, i) ∈ V . Let (y, j) ∈ V \ R. SinceH acts transitively on V , there exist ϕ0 ∈ H0 and 1 ̸= ρg ∈ RGsuch that (x, i)ϕ0 ρg = (y, j). So (y, j) ∈ Rρg and thus V \ R ⊆ 1̸=g ∈G Rρg . It follows that V ⊆ R ∪ 1̸=g ∈G Rρg . Hence V ⊆ g ∈G Rρg . The reverse inclusion is obvious. Therefore V = g ∈G Rρg . Conversely, first suppose that H0 acts transitively on V . Then H also acts transitively on V . Now let V = g ∈G Rρg , where R = (x, i)H0 ⊂ V . Let u = (g1 , i1 ) and v = (g2 , i2 ) be two distinct vertices of Γ . Then there exist g , h ∈ G and ϕ, ψ ∈ H0
−1 −1 such that u = (x, i)ϕρg and v = (x, i)ψρh . So uρg ϕ ψρh = v . On the other hand ρg−1 ϕ −1 ψρh ∈ H, which implies that H is transitive on V , as desired.
Theorem 3. Let Γ be a normal 2-Cayley digraph over a group G and suppose that L is a group. Then Γ is vertex-transitive if and only if V = g ∈G Rρg , where R ⊂ V is an orbit of L. Proof. Since Γis normal, by Proposition 2, A = RG o L. Now by Lemma 2.9, if Γ is vertex-transitive then L is transitive ρg on V or V = g ∈G R , where R ⊂ V is an orbit of L. If the first case holds then for each (x, 1) ∈ V , where x ̸= 1, there exists an α ∈ L such that (1, 1)α = (x, 1). So α ∈ X and there exist a ∈ G and σ ∈ Aut(G) such that (g , 1)α = (g σ , 1) and (g , 2)α = (ag σ , 2) for all (g , j) ∈ V . Hence (x, 1) = (1, 1)α = (1σ , 1) = (1, 1), which implies that x = 1, a contradiction. Therefore the second case holds. The converse is clear by Lemma 2.9. 3. 2-Cayley digraphs with solvable automorphism group In this section we see that if Aut(G) is solvable, then NA (RG ) is solvable. The following lemma is a stronger version of this result. We appreciate the referee for pointing out this lemma. Lemma 3.1. Let G ≤ Sym(Ω ) be semiregular with at most 4 orbits and Aut(G) is solvable. Then NSym(Ω ) (G) is solvable. In particular, a permutation group containing a semiregular normal subgroup with at most 4 orbits and with solvable automorphism group, is solvable.
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Proof. The image of the induced action of NSym(Ω ) (G) on the set of G-orbits on Ω is solvable (being a permutation group on at most 4 points). Now since G is semiregular, the kernel of this action restricted to any G-orbit on Ω is a split extension of G by a subgroup of Aut(G) (see, for example, [3, Exercise 2.5.5]). Since Aut(G) is solvable and G/Z (G) is isomorphic to a subgroup of Aut(G), we conclude that G also is solvable. This completes the proof. . Since G ∼ = RG , the following theorem is a direct consequence of Lemma 3.1. Theorem 4. Let Γ be a 2-Cayley digraph over a group G. If Aut(G) is solvable then NA (RG ) is solvable.
The converse of Theorem 4 is not true. For example, let Γ = Cay(G; Ti,j | 1 ≤ i, j ≤ 2), where G is an elementary abelian 2-group of order 8 with generators a, b, c, T1,1 = {a, b, c }, T2,2 = ∅ and T1,2 = T2,1 = {1}. Let Γ1 = Cay(G, T1,1 ). Then by the proof of Lemma 4.1, Aut(Γ ) ∼ = Aut(Γ1 ). On the other hand, Γ1 is the cube and by [7, Exercise 14.26], Aut(Γ1 ) ∼ = S2 ≀ S3 . Hence A = Aut(Γ ) ∼ S ≀ S is a solvable group of order 48, and so NA (RG ) is solvable. But Aut(G) ∼ = 2 3 = GL(3, 2) is non-solvable. The following corollary can be obtained immediately from Theorem 4. Corollary 3.2. Let Γ be a normal 2-Cayley digraph over a group G such that Aut(G) is solvable. Then A is solvable. In particular the automorphism group of every normal 2-Cayley digraph over a cyclic group is solvable. Let Γ be a normal 2-Cayley digraph over a group G. If N is a characteristic subgroup of G with index 2 such that Aut(N ) is solvable, then A is solvable. To see this, note that RN := {ρg | g ∈ N } is a semiregular subgroup of A isomorphic to N with 4 orbits. Since Aut(N ) is solvable, by Lemma 3.1, NSym(V (Γ )) (RN ) is solvable. Also RG E A and RN is a characteristic subgroup of RG , which implies that RN E A. Hence A = NA (RN ) ≤ NSym(V (Γ )) (N ) and is solvable. q n −1
Example 3.3. Let Γ = B(PG(n − 1, q)) be the point–hyperplane incidence graph of PG(n − 1, q) and p = q−1 be a prime. Then Aut(Γ ) = P Γ L(n, q) o Z2 and Γ is a Cayley graph over the dihedral group D2p of order 2p (see [4, Table 1]). Now Γ is a 2-Cayley graph over Zp . Also Γ is non-normal, since P Γ L(n, q) is not solvable. 4. Finite groups with a normal 2-Cayley (di)graph Let G be a finite group. Recall that we say G has a normal Cayley (di)graph if there exists a subset S of G not containing the identity element such that Cay(G, S ) is normal. It is proved in [2] that every finite group G has a normal Cayley graph unless G∼ = Z4 × Z2 or G ∼ = Q8 × Zr2 , r ≥ 0, and that every finite group has a normal Cayley digraph. The following question arises naturally: Which finite groups have normal 2-Cayley (di)graphs? In what follows, we give a partial answer to the question. In the following lemma we construct normal 2-Cayley digraphs from normal Cayley digraphs. Lemma 4.1. Let Γ1 = Cay(G, S ) be a normal Cayley digraph with S ̸= ∅. Then Γ = Cay(G; Ti,j | 1 ≤ i, j ≤ 2), where T1,1 = S, T2,2 = ∅ and T1,2 = T2,1 = {1G } is a normal 2-Cayley digraph. Proof. First we show that A = Aut(Γ ) ∼ = Aut(Γ1 ). For each σ ∈ Aut(Γ1 ), we define σ ′ : V → V by (g , i)σ = (g σ , i), for all ′ (g , i) ∈ V . Clearly σ is a bijection and furthermore σ ′ ∈ A, since ′
((x, 1), (y, 1)) ∈ E ⇔ (x, y) ∈ E (Γ1 ) ⇔ (xσ , yσ ) ∈ E (Γ1 ) ⇔ ((xσ , 1), (yσ , 1)) ∈ E , and for i ̸= j,
((x, i), (y, j)) ∈ E ⇔ x = y ⇔ xσ = yσ ⇔ ((xσ , i), (yσ , j)) ∈ E . Now we define ϕ : Aut(Γ1 ) → A by σ ϕ = σ ′ . Clearly ϕ is well-defined and 1–1. Since for all σ1 , σ2 ∈ Aut(Γ1 ), (σ1 σ2 )′ = σ1′ σ2′ , ϕ is a group homomorphism. Now we show that ϕ is onto. Let us identify V (Γ1 ) = G with G × {1}. Suppose that ψ ∈ A, and there exists (x0 , 1) ∈ V such that (x0 , 1)ψ ∈ V \ V (Γ1 ). Then there exists y0 ∈ G such that (x0 , 1)ψ = (y0 , 2). So (x0 , 2)ψ = (y0 , 1) and for all s ∈ S, (sx0 , 1)ψ = (y0 , 1). This implies that (sx0 , 1) = (x0 , 2) for all s ∈ S, a contradiction. Hence the restriction of ψ to V (Γ1 ) is an automorphism of Γ1 . So we may assume that there exists σ ∈ Aut(Γ1 ) such that for all x ∈ G, (x, 1)ψ = (xσ , 1). On the other hand for all x ∈ G, (x, 2)ψ ∈ V \ V (Γ1 ) and ((x, 2), (x, 1)) ∈ E, which show that ((x, 2)ψ , (xσ , 1)) ∈ E. Therefore (x, 2)ψ = (xσ , 2). So σ ϕ = ψ , and thus ϕ is onto. Hence A ∼ = Aut(Γ1 ). Now R(G)ϕ = RG is a normal subgroup of A. Now we are ready to show that every finite group has a normal 2-Cayley digraph, and every finite group G ̸= Q8 × Zr2 , r ≥ 0, has a normal 2-Cayley graph. Theorem 5. Every finite group G ̸= Q8 × Zr2 , r ≥ 0, has at least one normal 2-Cayley graph, and every finite group has at least one normal 2-Cayley digraph. Proof. By [2, Theorem 1], every finite group G ̸= Z4 × Z2 , Q8 × Zr2 , r ≥ 0 has a normal Cayley graph, and every finite group has at least one normal Cayley digraph. By Lemma 4.1, every finite group G ̸= Z4 × Z2 , Q8 × Zr2 , r ≥ 0 has a normal 2-Cayley graph, and every finite group has at least one normal 2-Cayley digraph. Consider the 2-Cayley graph Γ = Cay(Z4 × Z2 , T1,1 , T1,2 , T2,1 , T2,2 ), where Z4 = ⟨a⟩, Z2 = ⟨b⟩, T1,1 = {b, a2 b}, T2,2 = {b} and T1,2 = T2,1 = {1}. Then Γ is a disconnected
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graph with two components isomorphic to the ladder graph L4 with 8 vertices. The automorphism group of L4 is isomorphic to S2 × S2 . Hence Theorem 14.6 of [7] implies that A ∼ = (S2 × S2 ) ≀ S2 . One can check that A has a normal subgroup of A isomorphic to Z4 × Z2 . So Γ is normal over Z4 × Z2 . This completes the proof. We have the following result about normality of 2-Cayley graphs over groups G := Q8 × Zr2 , r ≥ 0: Let Γ = Cay(G, T1,1 , T1,2 , T2,1 , T2,2 ). If T1,2 = T2,1 , then Γ is non-normal. In fact, since Γ is a graph and T1,2 = T2,1 , Ti,j ’s, 1 ≤ i, j ≤ 2 are inverse closed and so are union of some conjugacy classes of G. Now the map ϕ : V → V , where (x, i)ϕ = (x−1 , i) is an automorphism of Γ which fixes (1, 1). If Γ is normal, by Proposition 2, ϕ ∈ X and so there exists σ ∈ Aut(G) such that for all x ∈ G, (x−1 , 1) = (x, 1)ϕ = (xσ , 1). So x−1 = xσ for all x ∈ G, which implies that G is abelian, a contradiction. Now the following question arises naturally. Question. Is there any normal 2-Cayley graph over Q8 × Zr2 , for each r ≥ 0? Acknowledgments The authors gratefully appreciate two anonymous referees for constructive comments and recommendations which definitely helped to improve the readability and quality of the paper. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13]
M. Arezoomand, B. Taeri, On the characteristic polynomial of n-Cayley digraphs, Electron. J. Combin. 20 (3) (2013) #P57, 1–14. W. Changqun, W. Dianjun, X. Mingyao, Normal Cayley graphs of finite groups, Sci. China 4 (3) (1998) 242–251. J.D. Dixon, B. Mortimer, Permutation Groups, Springer-Verlag, New York, 1996. S.F. Du, R.J. Wang, M.Y. Xu, On the normality of Cayley digraphs of groups of order twice a prime, Australas. J. Combin. 1 (1998) 227–234. Y.Q. Feng, Z.P. Lu, M.Y. Xu, Automorphism groups of Cayley digraphs, in: Applications of Group Theory to Combinatorics, CRC Press/Balkema, 2008, pp. 13–25. X. Gao, Y. Lou, The spectrum of semi-Cayley graphs over abelian groups, Linear Algebra Appl. 432 (2010) 2974–2983. F. Harary, Graph Theory, Addison-Welsey Publishing Company, 1969. I. Kovács, A. Malnič, D. Marušič, Š. Miklavič, One-matching bi-Cayley graphs over abelian groups, European J. Combin. 30 (2009) 602–616. Z.P. Lu, C.Q. Wang, M.Y. Xu, Semisymmetric cubic graphs constructed from bi-Cayley graphs of An, Ars Combin. 80 (2006) 177–187. M.J.D. Resmini, D. Jungnickel, Strongly regular semi-Cayley graphs, J. Algebraic Combin. 1 (1992) 171–195. G. Sabidussi, Vertex-transitive graphs, Monatsh. Math. 68 (1964) 426–438. M.Y. Xu, Automorphism groups and isomorphisms of Cayley digraphs, Discrete Math. 182 (1998) 309–319. X. Zou, J.X. Meng, Some algebraic properties of bi-Cayley graphs, Acta Math. Sinica (Chin. Ser.) 50 (5) (2007) 1075–1080.
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Normality of 2-Cayley digraphs Majid Arezoomand, Bijan Taeri ∗ Department of Mathematical Sciences, Isfahan University of Technology, Isfahan 84156-83111, Iran
article
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Article history: Received 30 January 2013 Received in revised form 18 October 2014 Accepted 20 October 2014
Keywords: 2-Cayley digraph Normal 2-Cayley digraph Automorphism group of digraph
abstract A digraph Γ is called a 2-Cayley digraph over a group G, if there exists a semiregular subgroup RG of Aut(Γ ) isomorphic to G with two orbits. We say that Γ is normal if RG is a normal subgroup of Aut(Γ ). In this paper, we determine the normalizer of RG in Aut(Γ ). We show that the automorphism group of each normal 2-Cayley digraph over a group with solvable automorphism group, is solvable. We prove that for each finite group G ̸= Q8 × Zr2 , r ≥ 0, where Q8 is the quaternion group of order 8 and Z2 is the cyclic group of order 2, there exists a normal 2-Cayley graph over G and that every finite group has a normal 2-Cayley digraph. © 2014 Elsevier B.V. All rights reserved.
1. Introduction Graphs come in two principle types directed and undirected graphs. We shall refer to directed graphs as digraphs and use the term graph to refer to undirected graphs. A digraph Γ is a pair (V , E ) of vertices V and edges E where E ⊆ V × V ; the digraph Γ is said to be finite if V is finite. A graph is a digraph with no edges of the form (α, α) and with the property that (α, β) ∈ E implies (β, α) ∈ E. The set of all permutations of V which preserve the adjacency structure of Γ is called the automorphism group of Γ ; it is denoted by Aut(Γ ). In this paper all digraphs have no loops. For the group-theoretic and graph-theoretic terminology not defined here we refer the reader to [3,7]. Let G be a group and S a subset of G not containing the identity element 1. The Cayley digraph Γ = Cay(G, S ) of G with respect to S has vertex set G and arc set {(g , sg ) | g ∈ G, s ∈ S }. If S = S −1 , then Cay(G, S ) can be viewed as an undirected graph, identifying an undirected edge with two directed edges (g , h) and (h, g ). This graph is called Cayley graph of G with respect to S. By a theorem of Sabidussi [11], a digraph Γ is a Cayley digraph over a group G if and only if there exists a regular subgroup of Aut(Γ ) isomorphic to G. There is a natural generalization of Sabidussi’s Theorem: a digraph Γ is called an n-Cayley digraph over a group G if there exists an n-orbit semiregular subgroup of Aut(Γ ) isomorphic to G. 2-Cayley graphs are often called semi-Cayley graphs [10,6], or bi-Cayley graphs [8]. Also a special case of 2-Cayley graphs are called bi-Cayley graphs by some authors, see for example [13,9]. Let Γ be a Cayley digraph over a group G. It is well-known that the right regular representation R(G) of G is a regular subgroup of Aut(Γ ). If R(G) is a normal subgroup of Aut(Γ ), then Γ is called a normal Cayley digraph over G, [12]. In the literatures, the study of normality of Cayley digraphs has been becoming a rather active topic in the algebraic graph theory, which also plays an important role in the investigation of various symmetry properties of digraphs. We encourage the reader to consult [5] for a survey up to 2008. In this paper we define the concept of normality for 2-Cayley digraphs and present some equivalent conditions to the normality of 2-Cayley digraphs. We prove that if Γ is a normal 2-Cayley digraph over a group G such that Aut(G) is solvable,
∗
Corresponding author. E-mail addresses: [email protected] (M. Arezoomand), [email protected] (B. Taeri). URL: http://[email protected]/ (B. Taeri).
http://dx.doi.org/10.1016/j.disc.2014.10.019 0012-365X/© 2014 Elsevier B.V. All rights reserved.
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a
b
Fig. 1. The cube is a 2-Cayley graph over (a) Z4 , (b) Z2 × Z2 .
Fig. 2. The Petersen graph is a 2-Cayley graph over Z5 .
then Aut(Γ ) is solvable. In particular, the automorphism group of each normal 2-Cayley graph over a cyclic group is solvable. Also we prove that every finite group G ̸= Q8 × Zr2 , r ≥ 0, where Q8 is the quaternion group of order 8 and Zn is the cyclic group of order n, has at least one normal 2-Cayley graph, and every finite group G has at least one normal 2-Cayley digraph. 2. Normal 2-Cayley digraphs Recall that a digraph Γ is called a 2-Cayley digraph over a group G if its automorphism group has a 2-orbit semiregular subgroup RG isomorphic to G. By [1, Lemma 2], we may identify Γ with a digraph Cay(G; Ti,j | 1 ≤ i, j ≤ 2) with vertex set G × {1, 2} and ((g , i), (h, j)) is an arc if and only if hg −1 ∈ Ti,j , for some suitable subsets T1,1 , T1,2 , T2,1 and T2,2 of G in which 1 does not belong to T1,1 and T2,2 . Also we can identify the 2-orbit semiregular subgroup RG of Aut(Γ ) isomorphic to G with
{ρg : V → V | g ∈ G, (x, j)ρg = (xg , j)}. Furthermore, Γ is undirected if and only if T1−,11 = T1,1 , T2−,21 = T2,2 and T1−,21 = T2,1 (with the convention that ∅−1 = ∅). Definition. A 2-Cayley digraph Γ over a group G is called normal (over G) if RG is a normal subgroup of Aut(Γ ). We denote the symmetric group on a set Ω by Sym(Ω ) and when |Ω | = n is finite, by Sn . In what follows, we give some examples of normal and non-normal 2-Cayley digraphs. First, note that the normality of a 2-Cayley digraph depends on the underlying group G. To see this, let Γ be the cube. Then Aut(Γ ) ∼ = S2 ≀ S3 , see [7, Exercise 14.26]. If G = ⟨a⟩ ∼ = Z4 , T1,1 = T2,2 = {a, a3 } and T1,2 = T2,1 = {1}, then Cay(G; Ti,j | 1 ≤ i, j ≤ 2) ∼ = Γ (see Fig. 1(a)). So a cube is a 2-Cayley graph over Z4 . Also if H = ⟨a, b | a2 = b2 = (ab)2 = 1⟩ ∼ = Z2 × Z2 , S1,1 = S2,2 = {a, b} and S1,2 = S2,1 = {1}, then Cay(H ; Si,j | 1 ≤ i, j ≤ 2) ∼ = Γ (see Fig. 1(b)). Hence a cube is also a 2-Cayley graph over Z2 × Z2 . Since the only normal subgroup of Aut(Γ ) ∼ = S2 ≀ S3 of order 4 is isomorphic to Z2 × Z2 , the graph Γ is a normal 2-Cayley graph over Z2 × Z2 , but not over Z4 . Example 2.1. Let P be the Petersen graph. Then P = Cay(G, Ti,j | 1 ≤ i, j ≤ 2), where G = ⟨a⟩ ∼ = Z5 , T1,1 = {a, a4 }, T2,2 = 2 3 ∼ {a , a } and T1,2 = T2,1 = {1} (see Fig. 2). It is well-known that Aut(P ) = S5 (see for example [3, Exercise 2.3.4]). Since S5 has no normal subgroup of order 5, P is not normal over Z5 . Example 2.2. Let Γ be a 2-Cayley digraph over a group G such that Aut(Γ ) acts primitively on V (Γ ). Then Γ is not normal over G. In fact, if G is a normal subgroup of Aut(Γ ) then [3, Theorem 1.6A] implies that RG is transitive on V (Γ ), which is impossible. In particular, if Γ = K2n or 2nK1 , where n ≥ 3, then Aut(Γ ) ∼ = S2n acts primitively on V (Γ ) (see for example [3, Exercise 1.5.12]). Now for every semiregular subgroup G of S2n of order n with two orbits, Γ is a non-normal 2-Cayley graph over G.
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Example 2.3. Let Γ = B(H (11)) be the incidence graph of the doubly transitive Hadamard 2-(11, 5, 2) design. Then Aut(Γ ) = PSL(2, 11) o Z2 ∼ = PGL(2, 11) and Γ is a Cayley graph over the dihedral group D22 of order 22, (see [4, Table 1]). Since Z11 is a subgroup of D22 of index 2, Γ is a 2-Cayley graph over Z11 . Since PGL(2, 11) has no normal subgroup isomorphic to Z11 , Γ is not normal over Z11 . Now we want to find the structure of the normalizer of RG in Aut(Γ ). From now on, the vertex set, edge set and automorphism group of Γ are denoted by V , E and A, respectively, and RG is the subgroup defined above. Also if G is a group acting on a set Ω , then for each α ∈ Ω the stabilizer of α in G, and the orbit of α under G are denoted by Gα and α G , respectively. For a group G and its subgroup H, we denote the normalizer and the centralizer of H in G by NG (H ) and CG (H ), respectively and we put CG (G) = Z (G). We first describe the stabilizer of (1, 1) in NSym(V ) (RG ). Lemma 2.4. Let Γ be a 2-Cayley digraph over a group G and M = NSym(V ) (RG ). Then ϕ ∈ M(1,1) if and only if there exist g ∈ G and σ ∈ Aut(G) such that for all x ∈ G, (x, 1)ϕ = (xσ , 1) and (x, 2)ϕ = (gxσ , 2). Proof. Let α ∈ M(1,1) . Then α ∈ Sym(V ), (1, 1)α = (1, 1) and for every x ∈ G there exists a unique x′ ∈ G such that −1 α −1 ρx α = ρx′ . So there exists σ ∈ Sym(G) such that for every x ∈ G, α −1 ρx α = ρxσ . Thus for all x ∈ G, (x, 1)α = (1, 1)α ρx α ρ xσ σ = (1, 1) = (x , 1). Moreover, for g1 , g2 ∈ G,
((g1 g2 )σ , 1) = (g1 g2 , 1)α = (1, 1)α
−1 ρ
g1 ρg2 α
ρg σ ρg σ
= (1, 1)
1
2
= (g1σ g2σ , 1),
which implies that (g1 g2 )σ = g1σ g2σ . This shows that σ ∈ Aut(G). Since for each x ∈ G, (x, 1)α = (xσ , 1), we have (G × {1})α = G × {1} and therefore (1, 2)α = (g , 2), for some g ∈ G. Hence (x, 2)α = (1, 2)ρx α = (1, 2)αρxσ = (gxσ , 2), for all x ∈ G. This completes the proof of one direction. Conversely, suppose that ϕ : V → V is a map in which there exist g ∈ G and σ ∈ Aut(G) such that for all x ∈ G, (x, 1)ϕ = (xσ , 1) and (x, 2)ϕ = (gxσ , 2). Since (1, 1)ϕ = (1, 1), we have to show that ϕ ∈ M. Clearly ϕ is 1–1 and onto. Let −1 y ∈ G and (x, j) ∈ V . Then (x, j)ϕ ρy ϕ = (xyσ , j) = (x, j)ρyσ . Thus for all y ∈ G, ϕ −1 ρy ϕ = ρyσ , which means that ϕ ∈ M. This completes the proof.
Let Γ = Cay(G, S ) be the Cayley digraph over a group G with respect to S, and R(G) be the right regular representation of G. It is well-known that NAut(Γ ) (R(G)) = R(G) o Aut(G, S ), where Aut(G, S ) = {σ ∈ Aut(G) | S σ = S }, [12, Proposition 1.3]. We wish to find a similar result for 2-Cayley digraphs. First, let us define some subsets of the automorphism group of a 2-Cayley digraph Γ = Cay(G; Ti,j | 1 ≤ i, j ≤ 2). Let X be the set of all maps ψ : V → V , where (x, 1)ψ = (xσ , 1) and (x, 2)ψ = (gxσ , 2), for some g ∈ G and σ ∈ Aut(G) such that T1σ,1 = T1,1 , T2σ,2 = g −1 T2,2 g, T1σ,2 = g −1 T1,2 , and T2σ,1 = T2,1 g.
Also, let Y be the set of all maps ϕ : V → V , where (x, 1)ϕ = (xθ , 2) and (x, 2)ϕ = (hxθ , 1), for some h ∈ G and θ ∈ Aut(G) such that T1θ,1 = T2,2 , T2θ,2 = h−1 T1,1 h, T1θ,2 = h−1 T2,1 and T2θ,1 = T1,2 h, with the convention that if one of the pair sets T1,1 , T2,2 or T1,2 , T2,1 is empty and the other is non-empty, we put Y = ∅. Also if in the above equalities, one of the subsets Ti,j is empty, then we omit the equality including Ti,j . Note that X and Y are disjoint and X = M(1,1) ∩ A, where M = NSym(V ) (RG ), is a subgroup of A(1,1) . Now we are ready to describe the normalizer of RG in A.
Theorem 1. Let Γ = Cay(G; Ti,j | 1 ≤ i, j ≤ 2) be a 2-Cayley digraph over a group G, and X , Y be the sets defined above. Then NA (RG ) = LRG , where L = X ∪ Y . Furthermore, RG ∩ L = {1}. Proof. Let M = NSym(V ) (RG ). Since RG ∼ = G is a semiregular subgroup of Sym(V ), by [3, Exercise 4.2.7], CSym(V ) (RG ) ≤ M is transitive. Thus M is also transitive. For a permutation τ ∈ S2 , we define τ : V → V by (x, j)τ = (x, jτ ). Clearly τ ∈ M. ρ
ρ
( i j)
Setting H = {τ | τ ∈ S2 } we have H ≤ CSym(V ) (RG ). For x, y ∈ G, we have (x, i) x−1 y = (y, i) and (x, i) x−1 y = (y, j), i ̸= j. Thus RG H = HRG is transitive on V . Now [3, Exercise 1.4.1] implies that M = M(1,1) HRG . Therefore, by Dedekind’s Lemma, NA (RG ) = NSym(V ) (RG ) ∩ A = M ∩ A = M(1,1) HRG ∩ A = (M(1,1) H ∩ A)RG . To complete the proof, we must show that L = M(1,1) H ∩ A. Recall from Lemma 2.4 that a map ϕ : V → V is an element of M(1,1) if and only if for all x ∈ G, (x, 1)ϕ = (xσ , 1) and (x, 2)ϕ = (gxσ , 2), for some g ∈ G and σ ∈ Aut(G). Let ψ ∈ M(1,1) H ∩ A. Then ψ ∈ A and ψ = ϕτ , for some ϕ ∈ M(1,1) and τ ∈ H. So there exist g ∈ G and σ ∈ Aut(G) such that for all x ∈ G, (x, 1)ψ = (xσ , 1τ ) and (x, 2)ψ = (gxσ , 2τ ). We consider two cases: Case I. τ = idS2 . In this case (x, 1)ψ = (xσ , 1) and (x, 2)ψ = (gxσ , 2), for all x ∈ G. Since ψ ∈ A, it is easily seen that T1σ,1 = T1,1 , T2σ,2 = g −1 T2,2 g, T1σ,2 = g −1 T1,2 and T2σ,1 = T2,1 g, for instance t ∈ T1,2 ⇔ ((1, 1), (t , 2)) ∈ E
⇔ ((1, 1)ψ , (t , 2)ψ ) = ((1, 1), (gt σ , 2)) ∈ E ⇔ gt σ ∈ T1,2 . This shows that ψ ∈ X .
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Case II. τ = (12). In this case (x, 1)ψ = (xσ , 2) and (x, 2)ψ = (gxσ , 1), for all x ∈ G. Since ψ ∈ A, it is easy to see that T1σ,1 = T2,2 , T2σ,2 = g −1 T1,1 g, T1σ,2 = g −1 T2,1 and T2σ,1 = T1,2 g, for instance t ∈ T1,2 ⇔ ((1, 1), (t , 2)) ∈ E
⇔ ((1, 1)ψ , (t , 2)ψ ) = ((1, 2), (gt σ , 1)) ∈ E ⇔ gt σ ∈ T2,1 . Hence ψ ∈ Y . Thus we have proved that M(1,1) H ∩ A ⊆ X ∪ Y . Now we prove the reverse inclusion. If ψ ∈ X , then there exist g ∈ G and σ ∈ Aut(G) such that T1σ,1 = T1,1 , T2σ,2 = g −1 T2,2 g, T1σ,2 = g −1 T1,2 , and T2σ,1 = T2,1 g; and for all x ∈ G, (x, 1)ψ = (xσ , 1)
and (x, 2)ψ = (gxσ , 2). Hence ψ ∈ M(1,1) and also ψ preserves the adjacency, which means that ψ ∈ L. Similarly if ϕ ∈ Y , we can see that ϕ ∈ L. Hence the equality M(1,1) H ∩ A = X ∪ Y holds. Finally, we show that RG ∩ L is trivial. Let ϕ ∈ RG ∩ L. Then there exist g ∈ G and ψ ∈ L such that ϕ = ρg = ψ . So (g , 1) = (1, 1)ρg = (1, 1)ψ = (1, 1τ ). Therefore g = 1 and so ϕ = 1. This completes the proof. From now on, we assume that L is the set defined in Theorem 1. It is easy to see that if α = ψρx , for some ρx ∈ RG and ψ ∈ L, then ρx and ψ are uniquely determined by α . By the definition of X and Y , each element ϕ of L depends on an element g ∈ G and σ ∈ Aut(G). In the following lemma, we denote the corresponding element g ∈ G and σ ∈ Aut(G) of ϕ by gϕ and σϕ , respectively. Note that in general L is not a group, see Example 2.6. In the following lemma we study the problem that when L is a group? Lemma 2.5. L is a group if and only if Y = ∅ or for each ϕ ∈ L, gϕ = 1. Proof. Recall that X = M(1,1) ∩ A is a group, where M = NSym(V ) (RG ). First let L be a group and ϕ ∈ L = X ∪ Y . Suppose that ϕ ∈ Y . Then (1, 2)ϕ = (gϕ 1σϕ , 1) = (gϕ , 1) and hence −1 −1 (gϕ , 1)ϕ = (1, 2). Since ϕ −1 ∈ L and X and Y are disjoint, we conclude that ϕ −1 ∈ Y . Moreover, (gϕ , 1)ϕ = (gϕθ , 2),
where θ = σϕ −1 and so gϕθ = 1, which implies that gϕ = 1. Now let ϕ ∈ X . Then for all x ∈ G, (x, 1)ϕ = (xσϕ , 1) and (x, 2)ϕ =
(gϕ xθϕ , 2). Choose an element ψ ∈ Y . Then (x, 1)ϕψ = (xσϕ σψ , 2) and (x, 2)ϕψ = ((gϕ )σψ xσϕ σψ , 1). Since L is a group, ϕψ ∈ Y and by the first part, (gϕ )σψ = 1, which implies that gϕ = 1. Conversely, if Y = ∅ then L = X is a group. So suppose that Y ̸= ∅ and for each ϕ ∈ L, gϕ = 1. It is easy to see that for all ψ1 , ψ2 ∈ X and ϕ1 , ϕ2 ∈ Y , ψ1 ψ2−1 , ϕ1 ϕ2−1 ∈ X , and ϕ1 ψ1−1 , ψ1 ϕ1−1 ∈ Y . This shows that for each α1 , α2 ∈ L, we have α1 α2−1 ∈ L, and thus L is a group. In the following examples we show that the structure of L depends on the subsets Ti,j and in general L is not a group. Example 2.6. Let G = ⟨a⟩ ∼ = Z3 . Then Aut(G) = {id, π}, where π maps ak to a2k , k = 0, 1, 2. Consider the 2-Cayley graph Cay(G; Ti,j | 1 ≤ i, j ≤ 2), where T1,1 = T2,2 = ∅, T1,2 = {1, a} and T2,1 = {1, a2 }. Let α ∈ X . Then there exist σ ∈ Aut(G) and g ∈ G such that for all x ∈ G, (x, 1)α = (xσ , 1) and (x, 2)σ = (gxσ , 2). Furthermore, T1σ,2 = g −1 T1,2 and T2σ,1 = T2,1 g. From these equations we conclude that α ∈ {α1 , α2 }, where
α1 : (x, 1) → (x, 1), α2 : (x, 1) → (xπ , 1),
(x, 2) → (x, 2), (x, 2) → (axπ , 2).
Also for each element β ∈ Y , there exist g ∈ G and σ ∈ Aut(G) such that for all x ∈ G, (x, 1)β = (xσ , 2), (x, 2)β = (gxσ , 1), T1σ,2 = g −1 T2,1 and T2σ,1 = T1,2 g. Hence β ∈ {β1 , β2 }, where
β1 : (x, 1) → (x, 2), β2 : (x, 1) → (xπ , 2),
(x, 2) → (a2 x, 1) (x, 2) → (xπ , 1).
Thus L = {α1 , α2 , β1 , β2 } and by Lemma 2.5, L is not a group. Example 2.7. Let G = ⟨a⟩ ∼ = Z5 , then Aut(G) = {σi | i = 1, . . . , 4}, where σi maps ak to aki for all k = 0, . . . , 4. By Example 2.1, the Petersen graph P is a 2-Cayley graph P = Cay(G; Ti,j | 1 ≤ i, j ≤ 2), where T1,1 = {a, a4 }, T2,2 = {a2 , a3 } and T1,2 = T2,1 = {1}. So we can consider the vertices of P as {(ai , 1), (ai , 2) | i = 0, . . . , 4}. Now, by a similar argument to that of Example 2.6 we can see that, the elements of L are
ψ1 ψ2 ψ3 ψ4
: (x, 1) → (x, 1), : (x, 1) → (xσ4 , 1), : (x, 1) → (xσ2 , 2), : (x, 1) → (xσ3 , 2),
(x, 2) → (x, 2), (x, 2) → (xσ4 , 2), (x, 2) → (xσ2 , 1), (x, 2) → (xσ3 , 1).
By Lemma 2.5, L is a group. Furthermore, ψ42 = ψ2 , ψ43 = ψ3 , and ψ44 = ψ1 , which implies that L ∼ = Z4 . Also by Theorem 1, NAut(P ) (RG ) = RG o L ∼ = Z5 o Z4 .
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Let Γ = Cay(G, S ) be a Cayley digraph of G with respect to S. We know that the following statements are equivalent, [12, Propositions 1.3 and 1.5]: (i) R(G) E Aut(Γ ), (ii) Aut(Γ ) = R(G) o Aut(G, S ), (iii) Aut(Γ )1 = Aut(G, S ). The 2-Cayley version of this result is the following. Proposition 2. Let Γ = Cay(G; Ti,j | 1 ≤ i, j ≤ 2) be a 2-Cayley digraph over G. Then (1) RG E A if and only if A = LRG , (2) if RG E A, then A(1,1) = X , and the converse holds if A is not transitive on V . Proof. By Theorem 1, it is clear that (1) holds. Now we prove (2). Set N = NA (RG ) and M = NSym(V ) (RG ). First, suppose that RG E A. Then A(1,1) = N(1,1) ≤ M(1,1) . Thus X ≤ A(1,1) ≤ M(1,1) ∩ A = X , which implies that A(1,1) = X . Conversely, suppose that X = A(1,1) and A is not transitive on V . So A has at least two distinct orbits on V . Set O1 := (1, 1)A and O2 = (1, 2)A . We claim that O1 and O2 are the only orbits of A, O1 = (1, 1)RG and O2 = (1, 2)RG . If O1 ∩ O2 ̸= ∅, then there exists α ∈ A which maps (1, 1) to (1, 2) and so ⟨RG , α⟩ acts transitively on V , a contradiction. Hence O1 ∩ O2 = ∅. On the other hand V = (1, 1)RG ∪ (1, 2)RG ⊆ O1 ∪ O2 ⊆ V , which implies that O1 ∪ O2 = V . Since O1 ∩ O2 = ∅ and (1, i)RG ⊆ Oi , i = 1, 2, O1 = (1, 1)RG , O2 = (1, 2)RG and O1 and O2 are all orbits of A. Hence A acts on O1 . Since RG and A are transitive on O1 , we have A = RG A(1,1) . Hence A = RG X ⊆ RG L = NA (RG ) ⊆ A implies that A = NA (RG ). Hence RG E A. This completes the proof. Corollary 2.8. Let Γ be a normal 2-Cayley digraph over a finite group G. Then A(1,1) = L if and only if Γ is not vertex-transitive. Proof. Since Γ is normal over G, by Proposition 2, A = RG L, A(1,1) = X ⊆ L and |A| = |G| |L|. If A(1,1) = L then |A : A(1,1) | = |G| ̸= |V |, which means that Γ is not vertex-transitive. Conversely, suppose that Γ is not vertex-transitive. By the proof of Proposition 2, A = RG A(1,1) . Moreover, RG ∩ A(1,1) = {1} which implies that |A| = |G| |A(1,1) |. Since |A| = |G| |L|, |L| = |A(1,1) |. Now since A(1,1) ⊆ L we have A(1,1) = L, as desired. Now we want to find an equivalent condition to the vertex transitivity of normal 2-Cayley digraphs relating to L. First we need the following lemma. Lemma 2.9. Let Γ be a 2-Cayley digraph over group G. If H = H0 RG ≤ A, where 1 ̸= H0 < H, then H is transitive on V if and only if one of the following holds: (1) H0 is transitive on V (2) V = g ∈G Rρg where R ⊂ V is an orbit of H0 . Proof. Suppose that H is transitive on V but H0 is not. So there exists an orbit R := (x, i)H0 ⊂ V , for some (x, i) ∈ V . Let (y, j) ∈ V \ R. SinceH acts transitively on V , there exist ϕ0 ∈ H0 and 1 ̸= ρg ∈ RGsuch that (x, i)ϕ0 ρg = (y, j). So (y, j) ∈ Rρg and thus V \ R ⊆ 1̸=g ∈G Rρg . It follows that V ⊆ R ∪ 1̸=g ∈G Rρg . Hence V ⊆ g ∈G Rρg . The reverse inclusion is obvious. Therefore V = g ∈G Rρg . Conversely, first suppose that H0 acts transitively on V . Then H also acts transitively on V . Now let V = g ∈G Rρg , where R = (x, i)H0 ⊂ V . Let u = (g1 , i1 ) and v = (g2 , i2 ) be two distinct vertices of Γ . Then there exist g , h ∈ G and ϕ, ψ ∈ H0
−1 −1 such that u = (x, i)ϕρg and v = (x, i)ψρh . So uρg ϕ ψρh = v . On the other hand ρg−1 ϕ −1 ψρh ∈ H, which implies that H is transitive on V , as desired.
Theorem 3. Let Γ be a normal 2-Cayley digraph over a group G and suppose that L is a group. Then Γ is vertex-transitive if and only if V = g ∈G Rρg , where R ⊂ V is an orbit of L. Proof. Since Γis normal, by Proposition 2, A = RG o L. Now by Lemma 2.9, if Γ is vertex-transitive then L is transitive ρg on V or V = g ∈G R , where R ⊂ V is an orbit of L. If the first case holds then for each (x, 1) ∈ V , where x ̸= 1, there exists an α ∈ L such that (1, 1)α = (x, 1). So α ∈ X and there exist a ∈ G and σ ∈ Aut(G) such that (g , 1)α = (g σ , 1) and (g , 2)α = (ag σ , 2) for all (g , j) ∈ V . Hence (x, 1) = (1, 1)α = (1σ , 1) = (1, 1), which implies that x = 1, a contradiction. Therefore the second case holds. The converse is clear by Lemma 2.9. 3. 2-Cayley digraphs with solvable automorphism group In this section we see that if Aut(G) is solvable, then NA (RG ) is solvable. The following lemma is a stronger version of this result. We appreciate the referee for pointing out this lemma. Lemma 3.1. Let G ≤ Sym(Ω ) be semiregular with at most 4 orbits and Aut(G) is solvable. Then NSym(Ω ) (G) is solvable. In particular, a permutation group containing a semiregular normal subgroup with at most 4 orbits and with solvable automorphism group, is solvable.
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Proof. The image of the induced action of NSym(Ω ) (G) on the set of G-orbits on Ω is solvable (being a permutation group on at most 4 points). Now since G is semiregular, the kernel of this action restricted to any G-orbit on Ω is a split extension of G by a subgroup of Aut(G) (see, for example, [3, Exercise 2.5.5]). Since Aut(G) is solvable and G/Z (G) is isomorphic to a subgroup of Aut(G), we conclude that G also is solvable. This completes the proof. . Since G ∼ = RG , the following theorem is a direct consequence of Lemma 3.1. Theorem 4. Let Γ be a 2-Cayley digraph over a group G. If Aut(G) is solvable then NA (RG ) is solvable.
The converse of Theorem 4 is not true. For example, let Γ = Cay(G; Ti,j | 1 ≤ i, j ≤ 2), where G is an elementary abelian 2-group of order 8 with generators a, b, c, T1,1 = {a, b, c }, T2,2 = ∅ and T1,2 = T2,1 = {1}. Let Γ1 = Cay(G, T1,1 ). Then by the proof of Lemma 4.1, Aut(Γ ) ∼ = Aut(Γ1 ). On the other hand, Γ1 is the cube and by [7, Exercise 14.26], Aut(Γ1 ) ∼ = S2 ≀ S3 . Hence A = Aut(Γ ) ∼ S ≀ S is a solvable group of order 48, and so NA (RG ) is solvable. But Aut(G) ∼ = 2 3 = GL(3, 2) is non-solvable. The following corollary can be obtained immediately from Theorem 4. Corollary 3.2. Let Γ be a normal 2-Cayley digraph over a group G such that Aut(G) is solvable. Then A is solvable. In particular the automorphism group of every normal 2-Cayley digraph over a cyclic group is solvable. Let Γ be a normal 2-Cayley digraph over a group G. If N is a characteristic subgroup of G with index 2 such that Aut(N ) is solvable, then A is solvable. To see this, note that RN := {ρg | g ∈ N } is a semiregular subgroup of A isomorphic to N with 4 orbits. Since Aut(N ) is solvable, by Lemma 3.1, NSym(V (Γ )) (RN ) is solvable. Also RG E A and RN is a characteristic subgroup of RG , which implies that RN E A. Hence A = NA (RN ) ≤ NSym(V (Γ )) (N ) and is solvable. q n −1
Example 3.3. Let Γ = B(PG(n − 1, q)) be the point–hyperplane incidence graph of PG(n − 1, q) and p = q−1 be a prime. Then Aut(Γ ) = P Γ L(n, q) o Z2 and Γ is a Cayley graph over the dihedral group D2p of order 2p (see [4, Table 1]). Now Γ is a 2-Cayley graph over Zp . Also Γ is non-normal, since P Γ L(n, q) is not solvable. 4. Finite groups with a normal 2-Cayley (di)graph Let G be a finite group. Recall that we say G has a normal Cayley (di)graph if there exists a subset S of G not containing the identity element such that Cay(G, S ) is normal. It is proved in [2] that every finite group G has a normal Cayley graph unless G∼ = Z4 × Z2 or G ∼ = Q8 × Zr2 , r ≥ 0, and that every finite group has a normal Cayley digraph. The following question arises naturally: Which finite groups have normal 2-Cayley (di)graphs? In what follows, we give a partial answer to the question. In the following lemma we construct normal 2-Cayley digraphs from normal Cayley digraphs. Lemma 4.1. Let Γ1 = Cay(G, S ) be a normal Cayley digraph with S ̸= ∅. Then Γ = Cay(G; Ti,j | 1 ≤ i, j ≤ 2), where T1,1 = S, T2,2 = ∅ and T1,2 = T2,1 = {1G } is a normal 2-Cayley digraph. Proof. First we show that A = Aut(Γ ) ∼ = Aut(Γ1 ). For each σ ∈ Aut(Γ1 ), we define σ ′ : V → V by (g , i)σ = (g σ , i), for all ′ (g , i) ∈ V . Clearly σ is a bijection and furthermore σ ′ ∈ A, since ′
((x, 1), (y, 1)) ∈ E ⇔ (x, y) ∈ E (Γ1 ) ⇔ (xσ , yσ ) ∈ E (Γ1 ) ⇔ ((xσ , 1), (yσ , 1)) ∈ E , and for i ̸= j,
((x, i), (y, j)) ∈ E ⇔ x = y ⇔ xσ = yσ ⇔ ((xσ , i), (yσ , j)) ∈ E . Now we define ϕ : Aut(Γ1 ) → A by σ ϕ = σ ′ . Clearly ϕ is well-defined and 1–1. Since for all σ1 , σ2 ∈ Aut(Γ1 ), (σ1 σ2 )′ = σ1′ σ2′ , ϕ is a group homomorphism. Now we show that ϕ is onto. Let us identify V (Γ1 ) = G with G × {1}. Suppose that ψ ∈ A, and there exists (x0 , 1) ∈ V such that (x0 , 1)ψ ∈ V \ V (Γ1 ). Then there exists y0 ∈ G such that (x0 , 1)ψ = (y0 , 2). So (x0 , 2)ψ = (y0 , 1) and for all s ∈ S, (sx0 , 1)ψ = (y0 , 1). This implies that (sx0 , 1) = (x0 , 2) for all s ∈ S, a contradiction. Hence the restriction of ψ to V (Γ1 ) is an automorphism of Γ1 . So we may assume that there exists σ ∈ Aut(Γ1 ) such that for all x ∈ G, (x, 1)ψ = (xσ , 1). On the other hand for all x ∈ G, (x, 2)ψ ∈ V \ V (Γ1 ) and ((x, 2), (x, 1)) ∈ E, which show that ((x, 2)ψ , (xσ , 1)) ∈ E. Therefore (x, 2)ψ = (xσ , 2). So σ ϕ = ψ , and thus ϕ is onto. Hence A ∼ = Aut(Γ1 ). Now R(G)ϕ = RG is a normal subgroup of A. Now we are ready to show that every finite group has a normal 2-Cayley digraph, and every finite group G ̸= Q8 × Zr2 , r ≥ 0, has a normal 2-Cayley graph. Theorem 5. Every finite group G ̸= Q8 × Zr2 , r ≥ 0, has at least one normal 2-Cayley graph, and every finite group has at least one normal 2-Cayley digraph. Proof. By [2, Theorem 1], every finite group G ̸= Z4 × Z2 , Q8 × Zr2 , r ≥ 0 has a normal Cayley graph, and every finite group has at least one normal Cayley digraph. By Lemma 4.1, every finite group G ̸= Z4 × Z2 , Q8 × Zr2 , r ≥ 0 has a normal 2-Cayley graph, and every finite group has at least one normal 2-Cayley digraph. Consider the 2-Cayley graph Γ = Cay(Z4 × Z2 , T1,1 , T1,2 , T2,1 , T2,2 ), where Z4 = ⟨a⟩, Z2 = ⟨b⟩, T1,1 = {b, a2 b}, T2,2 = {b} and T1,2 = T2,1 = {1}. Then Γ is a disconnected
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graph with two components isomorphic to the ladder graph L4 with 8 vertices. The automorphism group of L4 is isomorphic to S2 × S2 . Hence Theorem 14.6 of [7] implies that A ∼ = (S2 × S2 ) ≀ S2 . One can check that A has a normal subgroup of A isomorphic to Z4 × Z2 . So Γ is normal over Z4 × Z2 . This completes the proof. We have the following result about normality of 2-Cayley graphs over groups G := Q8 × Zr2 , r ≥ 0: Let Γ = Cay(G, T1,1 , T1,2 , T2,1 , T2,2 ). If T1,2 = T2,1 , then Γ is non-normal. In fact, since Γ is a graph and T1,2 = T2,1 , Ti,j ’s, 1 ≤ i, j ≤ 2 are inverse closed and so are union of some conjugacy classes of G. Now the map ϕ : V → V , where (x, i)ϕ = (x−1 , i) is an automorphism of Γ which fixes (1, 1). If Γ is normal, by Proposition 2, ϕ ∈ X and so there exists σ ∈ Aut(G) such that for all x ∈ G, (x−1 , 1) = (x, 1)ϕ = (xσ , 1). So x−1 = xσ for all x ∈ G, which implies that G is abelian, a contradiction. Now the following question arises naturally. Question. Is there any normal 2-Cayley graph over Q8 × Zr2 , for each r ≥ 0? Acknowledgments The authors gratefully appreciate two anonymous referees for constructive comments and recommendations which definitely helped to improve the readability and quality of the paper. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13]
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