Pentavalent arc-transitive Cayley graphs on Frobenius groups with soluble vertex stabilizer

Hailin Liu

doi:10.1515/math-2019-0041

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Pentavalent arc-transitive Cayley graphs on Frobenius groups with soluble vertex stabilizer

Hailin Liu

Published/Copyright: May 30, 2019

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$Open Mathematics$

From the journal Open Mathematics Volume 17 Issue 1

Abstract

A Cayley graph Γ is said to be arc-transitive if its full automorphism group AutΓ is transitive on the arc set of Γ. In this paper we give a characterization of pentavalent arc-transitive Cayley graphs on a class of Frobenius groups with soluble vertex stabilizer.

Keywords: Arc-transitive graph; Frobenius group; Cayley graph; Soluble vertex stabilizer

MSC 2010: 20B25; 05C25

1 Introduction

Throughout the paper, graphs considered are simple, connected and undirected. For a graph Γ, we denote the vertex set, edge set, arc set, valency and full automorphism group of Γ by VΓ, EΓ, AΓ, val(Γ) and AutΓ, respectively. Γ is said to be G-vertex-transitive, G-edge-transitive or G-arc-transitive if G ≤ AutΓ is transitive on VΓ, EΓ or AΓ; in particular, if G = AutΓ, then Γ is simply called vertex-transitive, edge-transitive or arc-transitive. An s-arc of Γ is a sequence of vertices (u₀, u₁, …, u_s) such that u_i is adjacent to u_i+1 and u_i–1 ≠ u_i+1 for all possible i. For a subgroup G ⩽ AutΓ, Γ is said to be (G, s)-arc-transitive if G is transitive on the set of s-arcs in Γ. In particular, a 0-arc is called a vertex, and a 1-arc is called an arc for short. As we all know, a graph Γ is G-arc-transitive for some G ≤ AutΓ if and only if G is transitive on VΓ and the vertex stabilizer G_v of v ∈ VΓ in G is transitive on the neighborhood Γ(v) of v.

A graph Γ is called a Cayley graph if there exist a group G and a subset S ⊂ G ∖ {1} with S = S^–1: = {g^–1 | g ∈ S} such that the vertices of Γ may be identified with the elements of G in such a way that x is adjacent to y if and only if yx^–1 ∈ S. The Cayley graph Γ is denoted by Cay(G, S). Throughout this paper, we denote the vertex of Cay(G, S) corresponding to the identity of G by 1.

A graph Γ is a Cayley graph on G if and only if AutΓ contains a subgroup which is regular on vertices and isomorphic to G. It is well-known that a Cayley graph is vertex-transitive. However, a Cayley graph is of course not necessarily arc-transitive. Thus much excellent work has dealt with arc-transitive Cayley graphs. In particular, there are many works about cubic and pentavalent Cayley graphs. For the cubic case, see [1, 2, 3] for cubic symmetric Cayley graphs on finite nonabelian simple groups, which are normal except for A₄₇, see [4] for a characterisation of connected cubic s-transitive Cayley graphs, see [5] for a classification of the connected arc-transitive cubic Cayley graphs on PSL(2, p) where p ⩾ 5 is a prime, and see [6] for a classification of cubic arc-transitive Cayley graphs on a class of Frobenius groups. For the pentavalent case, see [7] for a classification of arc-transitive pentavalent Cayley graphs on finite nonabelian simple groups, see [8] for a construction of 2-arc transitive pentavalent Cayley graph of A₃₉, and see [9] for a characterization of connected core-free pentavalent 1-transitive Cayley graphs.

The objective of this paper is to give a characterization of pentavalent arc-transitive Cayley graphs on a class of primitive Frobenius groups with soluble vertex stabilizer.

A group G is said to be a Frobenius group if G has the form G = W:H such that xy ≠ yx for any x ∈ W ∖ {1} and y ∈ H ∖ {1}. In particular, G is called a primitive Frobenius group if H acts irreducibly on W.

Theorem 1.1

Let G = W:H ≅ Zpd :ℤ_n be a primitive Frobenius group, where p is a prime, and d, n are positive integers. Suppose that Γ is a connected pentavalent arc-transitive Cayley graph on G with soluble stabilizer. Let A = AutΓ. Then one of the following statements holds:

G ≅ D_2p, Γ ≅ G(2p, 5) with p ≡ 1(mod 5) and p > 11, and A = (ℤ_p:ℤ₅):ℤ₂;
G is normal in A, and A₁ ≅ ℤ₅, D₁₀, or F₂₀;
(Γ, G) = (𝓒₁₂, Z22 :ℤ₃), where 𝓒₁₂ is constructed in Example 3.1.

Remarks on Theorem 1.1

The Cayley graph Γ in part (ii) is called a normal Cayley graph, introduced in [10].
H acts irreducibly on W if and only if n does not divide p^m – 1 where m < d. We call such n a primitive divisor of p^d – 1, refer to [11, Proposition 2.3].

After this introductory section, some preliminary results are given in Section 2 and a few examples are given in Section 3. Then we complete the proof of Theorem 1.1 in Section 4.

2 Preliminary Results

In this section we give some preliminary results, which will be used in the subsequent sections.

Let Γ be a G-vertex-transitive graph. Then, for α ∈ VΓ, the stabilizer G_α is a core-free subgroup in G, that is, ∩g∈GGαg = 1. Set H = G_α and D = {x | α^x ∈ Γ(α)}. Then D is a union of several double cosets HxH. Moreover, Γ is isomorphic to the coset graph Cos(G, H, D) defined over {Hx | x ∈ G} with edge set {{Hg₁, Hg₂} | g2g1−1 ∈ D}.

The following statements for coset graphs are well known.

Γ is undirected if and only if D = D^–1: = {x^–1 | x ∈ D}.
Γ is connected if and only if 〈H, D〉 = G.
Γ is G-arc-transitive if and only if D = HgH for g ∈ G with g² ∈ H; moreover, g can be chosen as a 2-element such that g ∈ N_G(H ∩ H^g).

For a Cayley graph Γ = Cay(G, S). Let Aut(G, S) = {α ∈ Aut(G) | S^α = S}. Then we have the following basic result.

Lemma 2.1

([12, Lemma 2.1]) Let Γ = Cay(G, S) be a Cayley graph. Then the normalizer N_AutΓ(G) = G:Aut(G, S).

The soluble vertex stabilizer for arc-transitive graphs of valency 5 is known.

Lemma 2.2

[14] Let Γ be a pentavalent (G, s)-transitive graph for some G ≤ AutΓ and s ≥ 1. Let v ∈ VΓ. If G_v is soluble, then |G_v| | 80 and s ≤ 3. Furthermore, one of the following holds:

s = 1, G_v ≅ ℤ₅, D₁₀ or D₂₀;
s = 2, G_v ≅ F₂₀, or F₂₀ × ℤ₂;
s = 3, G_v ≅ F₂₀ × ℤ₄.

We give two basic results for pentavalent graphs.

Lemma 2.3

Let Γ = Cay(G, S) be a connected pentavalent graph with soluble stabilizer. Assume that AutΓ contains a subgroup X such that Γ is X-arc-transitive and G ⊴ X. Then X₁ ≅ ℤ₅, D₁₀, or F₂₀.

Proof

Since Γ is connected, G = 〈S〉, and thus Aut(G, S) acts faithfully on S. So Aut(G, S) ≲ S₅. By Lemma 2.1, X ≤ N_AutΓ(G) = G:Aut(G, S). Thus X₁ ⩽ Aut(G, S) ≲ S₅. Note that X₁ is transitive on S, so X₁ ≅ ℤ₅, D₁₀, or F₂₀.□

We say a vertex-transitive graph Γ is a normal cover of its quotient graph Γ_N if Γ and Γ_N have the same valency, where N ⊲ AutΓ is not transitive on VΓ.

Lemma 2.4

Let Γ be a connected pentavalent X-arc-transitive graph with soluble stabilizer, and let N ⊲ X such that X/N is insoluble, where X ≤ AutΓ. Then Γ is a normal cover of Γ_N.

Proof

Let u ∈ VΓ, and let B = u^N be an orbit of N acting on VΓ. Let K be the kernel of X acting on VΓ_N. Then K_u is soluble as K_u ⊴ X_u. By the Frattini argument, we have that K = NK_u. Note that K/N ≅ NK_u/N ≅ K_u/(N ∩ K_u), so K/N is soluble. Since X/N is insoluble, X/K ≅ (X/N)/(K/N) is insoluble. Thus Γ is a normal cover of Γ_N.□

The next lemma gives a classification of locally primitive Cayley graphs on abelian groups.

Lemma 2.5

[15, Theorem 1.1] Let Γ be a connected locally primitive Cayley graph of an abelian group of valency at least 3. Then one of the following holds:

Γ = K_n, K_n,n, K_n × ⋯ × K_n;
Γ is the standard double cover of Knl ;
Γ is a normal or bi-normal Cayley graph of an elementary abelian 2-group or a meta-abelian 2-group.

3 Examples

In this section we give some examples of pentavalent arc-transitive graphs.

Example 3.1

Let T = PSL(2, 4). Take a subgroup H ≅ ℤ₅ of T. Let g ∈ N_T(L) ∖ L be an involution such that 〈H, g〉 = T, where L = H^g ∩ H. Let G ≅ Z22 :ℤ₃ be a subgroup of T. Then G is transitive on [T:H], and so the coset graph 𝓒₁₂ = Cos(T, H, HgH) is a pentavalent T-arc-transitive Cayely graph on G.

Example 3.2

Let T = PSL(2, 9). Take a subgroup H ≅ D₁₀ of T. Let g ∈ N_T(L) ∖ L be an involution such that 〈H, g〉 = T, where L = H^g ∩ H. Let G ≅ Z32 :ℤ₄ be a subgroup of T. Then G is intransitive on [T:H], and so the coset graph 𝓒₃₆ = Cos(T, H, HgH) is a pentavalent T-arc-transitive graph but not a Cayely graph on G.

4 Proof of Theorem 1.1

In this section we will prove Theorem 1.1 by a series of lemmas.

Let G = W:H ≅ Zpd :ℤ_n be a primitive Frobenius group, where p is a prime, and d, n are integers. Assume that Γ = Cay(G, S) is a connected pentavalent arc-transitive Cayley graph on G with soluble vertex stabilizer. First of all, we study the case where the full automorphism group A := AutΓ is soluble.

Suppose that G ≅ D_2p. Then we have the following lemma, see [16, Proposition 2.7].

Lemma 4.1

Let G be a dihedral group of order 2p, and let Γ be a connected pentavalent arc-transitive Cayley graph on G. Then Γ ≅ G(2p, 5) with p ≡ 1(mod 5) and p > 11, AutΓ ≅ (ℤ_p:ℤ₅):ℤ₂.

Let F be the Fitting subgroup of A, that is, F is the largest nilpotent normal subgroup of A. Then F⧸ = 1, and C_A(F) ⩽ F as A is soluble.

For a group H and a prime p, we denote the Sylow p-subgroup of H by H_p.

Lemma 4.2

If G ≆ D_2p, then W is normal in A.

Proof

We claim that G ∩ F ≠ 1. Suppose that G ∩ F = 1. Since F ≤ A_u, |F| | |A_u|. It follows that |F| | 80 by Lemma 2.2, where u ∈ VΓ.

Assume that F is transitive on VΓ. Then |G| | |F|. Since H acts irreducibly on W, G ≅ ℤ₅:ℤ₄ or Z24 ℤ₅. Since there exists no connected arc-transitive pentavalent graphs of order 4p for each prime p ≥ 5 by [18, Theorem 1.1], the former case does not occur. By [17], there exists no connected arc-transitive pentavalent graphs of order 80, so the latter case is excluded. Similarly, we also exclude the case where Γ_F ≅ K₂.

Thus Γ is a normal cover of Γ_F. Then |F| divides |G|. Since C_G(F) ⩽ F, C_G(F) = 1. Therefore G acts faithfully on F. It follows that G ≲ Aut(F). Thus 2 | |F|.

Suppose that p = 2. By the previous paragraph, we have that n ⩾ 7. Assume that Φ(F) = 1. Since |F| divides both 80 and |G|, we obtain that F₂ ≅ Z2k and F_2′ ≅ ℤ₅ or 1, where k ⩽ min{4, d}. Note that C_A(F) ⩽ F, so G ≲ GL(k, 2). By Atlas [19], there exists no k and G satisfying the above relation. Thus Φ(F) ⧸ = 1. Since F₅ ≲ ℤ₅, we conclude that Φ(F₂) = Φ(F) ⩽ F₂. Let F₂ = F₂/Φ(F) and H = HΦ(F)/Φ(F). Then H ≅ H. If C_H(F₂)⧸ = 1, then C_H(F₂)⧸ = 1 by [20, p.174, Theorem 1.4], a contradiction. Thus C_H(F₂) = 1, and so H ≲ GL(k, 2), where k ≤ 3. By Atlas [19], F₂ ≅ Z23 and H ≅ ℤ₇. It follows that |F₂| = 16 and d = 3, which is a contradiction.

Suppose that p is odd. Assume that Φ(F) = 1. Then F₂ ≅ Z2k and 2^k | n, where k ⩽ 4. Note that G ≲ GL(k, 2), it follows that W ≲ GL(k, 2), and ℤ_2^k ≲ GL(k, 2). By Atlas [19], there exists no k and W satisfying the above relation. Thus Φ(F)⧸ = 1. Since F₅ ≲ ℤ₅, we conclude that Φ(F₂) = Φ(F) ⩽ F₂. Let W = WΦ(F₂)/Φ(F₂) and F₂ = F₂/Φ(F₂). If C_W(F₂)⧸ = 1, then C_W(F₂)⧸ = 1, refer to [20, p.174, Theorem 1.4], a contradiction. Thus C_W(F₂) = 1, and so W ≲ GL(k, 2), where k ≤ 3. By Atlas [19], this is impossible.

To sum up, F ∩ G⧸ = 1. Since W is minimal in G, W ⩽ F. Note that G ≆ D_2p, so Γ is a cover of Γ_{F_p}, and thus W = F_p. Therefore, W is normal in A. This completes the proof.□

Lemma 4.3

With the hypothesis of Lemma 4.2, then G is normal in A, and A_u ≅ ℤ₅, D₁₀, or F₂₀, where u ∈ VΓ.

Proof

By Lemma 4.2, W is normal in A. Since G ≆ D_2p, n > 2. Then Γ is a normal cover of Γ_W. By Lemma 2.5, either G/W ⊴ A/W or Γ_W ≅ K₆ or K_5,5. In the former case, G ⊴ A. By Lemma 2.3, A_u ≅ ℤ₅, D₁₀, or F₂₀, where u ∈ VΓ. If Γ_W ≅ K₆, then A/W ≤ AutΓ_W ≅ S₆. Note that 5 ⋅ 6||A/W|, so A_u is insoluble, which is a contradiction. Similarly, we can exclude the case where Γ_W ≅ K_5,5. This completes the proof of Lemma 4.3.□

In the remaining section, we study the case where the full automorphism group A is insoluble. Denote by R the radical of A, that is, R is the largest soluble normal subgroup of A.

Suppose that R = 1. Then we have the following lemma.

Lemma 4.4

If R = 1, then (G, Γ) = ( Z22 :ℤ₃, 𝓒₁₂).

Proof

Let N be a minimal normal subgroup of A. Since R = 1, N = T₁ × ⋯ × T_ℓ ≅ T^ℓ, where T_i ≅ T is non-abelian simple. By [21], T is one of the following:

PSL(4,2), PSU(3,8), M11, PSp(4,3), PSL(3,q)(q<9), PSL(2,q)(q>3).

Let W_i = T_i ∩ W, where 1 ⩽ i ⩽ ℓ, and let L = N ∩ H. Since G is a Frobenius group, L is a diagonal subgroup of N. Write L = 〈h₁h₂ ⋯ h_ℓ〉, where 〈h_i〉 ≅ L. Let H_i = 〈h_i〉 and G_i = W_i:H_i, where 1 ⩽ i ⩽ ℓ. Then G_i is a Frobenious group. Let m=|T1||G1|∏i=2ℓ|Ti||Wi|. Since N ∩ G = W:L, it follows that |N||N∩G|||Au|, and m=|N||N∩G|. Thus m | 80.

Suppose that T = PSL(4, 2). By Atlas [19], G_i ≲ A₇ and A₇ has no subgroup with index 2, 5 or 10. So m does not divide |A_u|, which is a contradiction. Similarly, we can exclude the cases where T = PSU(3, 8), M₁₁, PSp(4, 3) and PSL(3, q)(q < 9).

Suppose that T = PSL(2, q) where q = r^e with r a prime. According to [22, Theorem 6.25], G_i is isomorphic to a subgroup of one of the following groups:

Zre:Z(re−1)d,D2(re±1)d,A5andPGL(2,rf),

where d = (2, r – 1), and f | e. In the following, we process our analysis by several cases.

Gi≲Zre:Z(re−1)d.

If ℓ > 1, then (q + 1)² divides 80 since m divides |A_u|. It follows that (q + 1)² divides 16, and thus q = 3, which is a contradiction. Thus ℓ = 1, namely, N ≅ T. For this case, q + 1 divides 80. It implies that q = 4, 7, 9, 19 or 79. Since W ≅ Zre1 with e₁ ≤ e, we conclude that W≅Z22,Z7,Z32,Z19 or Z79.

Assume that W ≅ Z22 . Then G ≅ Z22 :ℤ₃. By Example 3.1, Γ ≅ 𝓒₁₂. Assume that W ≅ Z32 . Then G ≅ Z32 :ℤ₈ or Z32 :ℤ₄. Since PSL(2, 9) has no subgroup with order 72, we can exclude the former case. For the latter case, by Example 3.2, we can also exclude this case.

Assume that W ≅ ℤ₇, ℤ₁₉ or ℤ₇₉. Then G is isomorphic to one of the following

D14, Z7:Z6, D38, Z19:Z6, Z19:Z18, D158, Z79:Z6, Z79:Z26, Z79:Z78.

Since PSL(2, 19) has no subgroup with order 114 and 342 by MAGMA [23], we can exclude the cases where G ≅ ℤ₁₉:ℤ₆, and ℤ₁₉:ℤ₁₈. Besides, we can also exclude the cases where G ≅ D₁₄, D₃₈ and D₁₅₈ by [16, Proposition 2.7], and where G ≅ ℤ₇:ℤ₆, ℤ₇₉:ℤ₆, ℤ₇₉:ℤ₂₆ and ℤ₇₉:ℤ₇₈ by [16, Theprem 4.2].
Gi≲D2(re±1)d.

By the discussion as above, N ≅ T. If G_i ≲ D2(re−1)d, then q(q + 1) divides 80. It follows that q = 4. Thus G ≅ D₆. By [16, Proposition 2.7], Γ ≅ K₆, AutΓ ≅ S₆, and so |A_u| ∤ 80, which is a contradiction. Thus Gi≲D2(re+1)d. For this case, q(q – 1) divides 80. It follows that q = 5, and then G ≅ D₆. By above discussion, this case is also excluded.
G_i ≲ A₅.

Then G_i ≅ D₆ or D₁₀, and N ≅ T. Since |A5||D10|∤80, we can exclude the case where G ≅ D₁₀. Arguing as Case 2, we can exclude the case where G ≅ D₆.
G_i ≲ PGL(2, r^f).

Let r^=re−f(r2e−1)r2f−1. Then r̂ divides 80. If e – f = 1, then e = 2, f = 1 and r(r² + 1) | 80. It follows that r = 2 and N ≅ T. Hence G ≅ D₆ or D₁₀. However, the case does not occur by Case 3. Thus e > f + 1, and so r = 2. Then r2e−1r2f−1=5, which is impossible. This completes the proof of Lemma 4.4.□

In what follows, we will prove that R = 1.

Lemma 4.5

The radical R = 1.

Proof

Suppose that R⧸ = 1. Set L = R ∩ G. Let L ≠ 1. Since W is minimal and normal in G, W ≤ L. Note that A/R is insoluble, so Γ is a normal cover of Γ_R by Lemma 2.4.

Let G = GR/R. Then G ≅ G/(G ∩ R) is cyclic. Since Γ is a Cayley graph of G, Γ_R is a Cayley graph of G. By Lemma 2.5, we obtain that Γ_R ≅ K₆, or K_5,5. Thus AutΓ_R ≅ S₆, or S₅ ≀ ℤ₂. It follows that |A_u| ∤ 80, a contradiction.

Suppose that L = 1. Let G = GR/R. Then G ≅ G. Let A = A/R. Then A = G A_u, where u ∈ VΓ_R. Arguing as the proof of Lemma 4.4 with A = G A_u in place of A = GA_u, A is almost simple, and G ≅ Z22 :ℤ₃. Therefore, A is almost simple, which is a contradiction.□

The assertion of Theorem 1.1 follows from Lemmas 4.1 and 4.3-4.5.

Acknowledgement

This work was partially supported by the NNSF of China (11861076), the Science and Technology Research Project of Jiangxi Education Department (GJJ180488), and the Doctoral Fund Project of Jiangxi University of Science and Technology (jxxjbs18035).

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Received: 2018-09-20

Accepted: 2019-03-09

Published Online: 2019-05-30

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https://doi.org/10.1515/math-2019-0041

Keywords for this article

Arc-transitive graph; Frobenius group; Cayley graph; Soluble vertex stabilizer

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