The finite edge-primitive pentavalent graphs

Abstract

A graph is edge-primitive if its automorphism group acts primitively on edges. Weiss (in J. Comb. Theory Ser. B 15, 269–288, 1973) determined edge-primitive cubic graphs. In this paper, we classify edge-primitive pentavalent graphs. The same classification of those of valency 4 is also done.

1 Introduction

Let G be a group acting on a set Ω. Denote by G _α the subgroup of G fixing the point α. G is said to be semiregular if G _α =1 for each α∈Ω, and G is said to be regular if G is transitive and semiregular. A non-empty subset Δ of Ω is called a block for G if for each g∈G either Δ ^g=Δ or Δ ^g∩Δ=∅. Clearly, the set Ω and the singletons {α} (α∈Ω) are blocks for G, called the trivial blocks. Any other block is said to be non-trivial. Suppose that Δ is a non-trivial block for G. Then {Δ ^g∣g∈G} is the system of imprimitivity of G containing Δ. A transitive group G is primitive if G has no non-trivial blocks on Ω.

Throughout this paper, we consider undirected finite graphs without loops or multiple edges. As usual, the notation X=(V,E) denotes a graph with vertex set V and edge set E, and Aut(X) denotes its automorphism group. If two vertices u ,v∈V are adjacent, {u,v} denotes the edge between u and v. By X ₁(v), we mean the neighborhood of a vertex v in X, consisting of vertices which are adjacent to v.

Let X=(V,E) be a graph and G≤Aut(X). Then X is said to be G-locally primitive if the vertex stabilizer G _v acts primitively on X ₁(v) for each v∈V. A graph X is said to be G-vertex-transitive or G-edge-transitive if G acts transitively on V or E, respectively. If G is replaced by Aut(X), the graph X is simply said to be vertex-transitive or edge-transitive.

An s-arc in a graph is an ordered (s+1)-tuple (v ₀,v ₁,…,v _s−1,v _s ) of vertices of the graph X such that v _i−1 is adjacent to v _i for 1≤i≤s, and v _i−1≠v _i+1 for 1≤i≤s−1. A 0-arc is a vertex and a 1-arc is also called an arc for short. A graph X is said to be (G,s)-arc-transitive if G≤Aut(X) is transitive on the set of s-arcs in X. A (G,s)-arc-transitive graph is said to be (G,s)-transitive if it is not (G,s+1)-arc-transitive. A graph X is said to be s-arc-transitive or s-transitive if the graph is (Aut(X),s)-arc-transitive or (Aut(X),s)-transitive. A graph X is G-edge-primitive if G≤Aut(X) acts primitively on the set of edges of X, and X is edge-primitive if it is Aut(X)-edge-primitive.

Weiss [9] determined all edge-primitive cubic graphs, which are the complete bipartite graph K _3,3, the Heawood graph of order 14, the Biggs–Smith cubic distance-transitive graph of order 102 and the Tutte–Coxeter graph of order 30 (also known as Tutte’s 8-cage or the Levi graph). Giudici and Li [3] systematically analyzed edge-primitive graphs via the O’Nan–Scott Theorem to determine the possible edge and vertex actions of such graphs, and determined all G-edge-primitive graphs for G an almost simple group with socle PSL(2,q), where q is a prime power and q≠2, 3. Recently, the authors [4] classified edge-primitive tetravalent graphs, which are the complete graph K ₅, the co-Heawood graph of order 14 (the complement graph of the Heawood graph with respect to the complete bipartite graph K _7,7), the complete bipartite graph K _4,4, and three coset graphs defined on the almost simple groups Aut(PSL(3,3)), Aut(M₁₂) and Aut(G ₂(3)), respectively. In [6], edge-primitive 4-arc-transitive graphs are classified. In this paper, we give a classification of edge-primitive graphs of valency 5.

Theorem 1.1

Let X be an edge-primitive pentavalent graph with an edge e={u,v} and let A=Aut(X). Then X is s-transitive with s≥2, and X, s, A, A _v and A _e are listed in Table  1. Furthermore, such a graph X is uniquely determined by its number of vertices.

Table 1 s-transitive edge-primitive pentavalent graphs

From Theorem 1.1, we have the following corollary.

Corollary 1.2

All finite edge-primitive pentavalent graphs are 2-arc-transitive.

Remark

Let X be an edge-primitive graph with an edge e={u,v} and let A=Aut(X). Weiss classified such graphs of valency 3 in 1973. However, since then there is no much progress in this line for small valencies. In this paper, we first reduce A to an almost simple group when X has valency 5 and Theorem 1.1 follows from the classification of finite primitive groups with solvable stabilizers given in [6]. The method does not work for valency greater than 5 because A _e can be non-solvable.

2 A reduction

Let X=(V,E) be a G-edge-primitive graph of valency 5 with an edge e={u,v}. Then 2|E|=5|V|, and G is a primitive permutation group on E. By [3, Lemmas 3.1 and 3.4], X is connected and G-arc-transitive. Thus, 5 | |G _v |, but $5^{2}\,{\not|}\,\,\,|G_{v}|$. In particular, X is G-locally primitive. Let N=Soc(G)=T ^k, the socle of G. Then T is a simple group, N is transitive on E, and hence N has at most two orbits on V. If N has two orbits on V, denote by V ₁ and V ₂ these orbits. In this case, X is bipartite with V ₁ and V ₂ as its bipartition sets.

Lemma 2.1

The socle N is a minimal normal subgroup of G and is not semiregular on V, and the graph X is N-locally primitive. If X≠K _5,5, T is non-abelian simple and if further k≥2, T is semiregular on V.

Proof

Let $1\not=M\lhd G$. Suppose that M is semiregular on V. Then M _v =1 and M is transitive on E, implying that M has at most two orbits on V. Thus, |V|=|M| or 2|M|. The edge-primitivity of G implies that M is transitive on E. It follows that |E| | |M| and so |E| | |V|, which is impossible because $|E|={5|V|\over 2}$. Thus, M is not semiregular on V. Note that |X ₁(v)|=5. Since M◁G and X is G-arc-transitive, M _v is transitive on X ₁(v), and hence primitive on X ₁(v). Further, X is M-locally primitive. In particular, by taking M=N we see that N is not semiregular on V and X is N-locally primitive. If G has two distinct minimal normal subgroups, say N ₁ and N ₂, then N ₁×N ₂≤G. By taking M=N ₁ or N ₂, X is N ₁- and N ₂-locally primitive. This implies that 5 | |(N ₁) _v | and |(N ₂) _v |, forcing that 5² | |G _v |, a contradiction. Thus, N is a minimal normal subgroup of G.

To prove the second part, let $X\not=K_{5,5}$. Suppose T is abelian. Then N is abelian and hence regular on E. It follows that $|N|=|E|=\frac{5|V|}{2}$. Recall that N is not semiregular. If N has one orbit on V then N is regular on V, a contradiction. It follows that N has two orbits on V, that is, V ₁ and V ₂, and for v∈V ₁, we have $N_{v}\not=1$. Since N is abelian, N _v fixes every vertex in V ₁, forcing X=K _5,5, a contradiction. Thus, N is non-abelian. To finish the proof, we further let k≥2. Suppose that T is not semiregular on V. Write N=T×L, where L=T ^k−1. By the minimality of N in G, L is not semiregular on V.

Assume that N is transitive on V. Since T is not semiregular on V, $T_{w}\not=1$ for every w∈V, and by the minimality of N in G, $L_{w}\not=1$. By the vertex-transitivity and locally primitivity of N, we have 5 | |T _w | and 5 | |L _w |. It follows that 5² | |N _w |, which is impossible.

Now assume that N has two orbits on V. We may let v∈V ₁ and u∈V ₂. Suppose that $5\,{\not|}\,\,\,|T_{v}|$ and $5\,{\not|}\,\,\,|T_{u}|$. Then $5\,{\not|}\,\,\,|T_{w}|$ and $5\,{\not|}\,\,\,|T_{x}|$ for every w∈V ₁ and every x∈V ₂. Since X is N-locally primitive, T _w and T _x fix X ₁(w) and X ₁(x) pointwise, respectively. The connectivity of X implies that T _u and T _v fix every vertex in V. Then T _u =T _v =1 and hence T _w =T _x =1 for every w∈V ₁ and every x∈V ₂, contrary to the assumption that T is not semiregular on V. Thus, we may assume that 5 | |T _v | (note that we cannot deduce 5 | |T _v | by the locally primitivity of N when $T_{v}\not=1$ because N has two orbits). By the minimality of N, 5 | |L _u |.

Consider the orbits of L. Let {B ₁,B ₂,…,B _m } and {C ₁,C ₂,…,C _n } be the sets of orbits of L on V ₁ and V ₂, respectively. We may assume that v∈B ₁ and u∈C ₁. Note that B _i and C _i are blocks of N. Since 5 | |L _u |, L _u is transitive on X ₁(u). Thus, X ₁(u)⊆B ₁, and X ₁(x)⊆B ₁ for every x∈C ₁ because B ₁ and C ₁ are orbits of L. The edge-transitivity of N implies that if there is an edge between C _i and B ₁ then X ₁(x)⊆B ₁ for every x∈C _i . The connectivity of X implies that m=1, that is, L is transitive on V ₁. Thus, T _v =T _w for every vertex w∈V ₁ because L commutes with T, which forces that X=K _5,5, a contradiction. □

Lemma 2.2

Let X≠K _5,5. Then G _v is non-solvable and G is 2-arc-transitive.

Proof

Since G is arc-transitive, let X be (G,s)-arc-transitive for some s≥1. Note that a transitive permutation group of prime degree is either solvable or 2-transitive. To prove the lemma, we only need to show that G _v is non-solvable. Suppose to the contrary that G _v is solvable. Then the primitive permutation group $G_{v}^{X_{1}(v)}$ of degree 5 is isomorphic to ℤ₅, D ₁₀ or F ₂₀, and hence $G_{vu}^{X_{1}(v)}$ is a 2-group (the identity subgroup is also viewed as a 2-group). Let $G_{v}^{[1]}$ be the kernel of G _v acting on X ₁(v). By [6, Theorem 1.3], G _v is non-solvable for s≥4. Thus, s≤3 and by [10, Theorems 4.6–4.7], $G_{uv}^{[1]}=G_{u}^{[1]}\cap G_{v}^{[1]}=1$. Then $G_{u}^{[1]}\times G_{v}^{[1]}= G_{u}^{[1]}G_{v}^{[1]}\lhd G_{vu}$, and $G_{v}^{[1]}\cong G_{u}^{[1]}\cong G_{u}^{[1]}/(G_{u}^{[1]}\cap G_{v}^{[1]})\cong (G_{u}^{[1]})^{X_{1}(v)}\lhd G_{vu}^{X_{1}(v)}$, implying that $G_{v}^{[1]}$ is a 2-group. It follows that G _vu is a 2-group and hence G _e is a 2-group. The maximality of G _e in G implies that G _e is a Sylow 2-subgroup of G. Thus, |E| is odd and so is ${1\over2}|V|$. Moreover, if N=T ^k with k≥2 then by Lemma 2.1, T is semiregular on V, which is impossible. Thus k=1, and G is almost simple. By [11, Theorem], if the stabilizer of an arc-transitive automorphism group of a graph with prime valency p is solvable then its order is a divisor of p(p−1)². Thus, |G _v | | 80, which forces that |G _e | | 32. Since |G| is divisible by 5, by [6, Tables 14–20] we have G=PGL(2,9), M₁₀, or PSL(2,31), which is also impossible by the Atlas [1]. □

Let X≠K _5,5 and s≤3. Note that $G_{v}^{[1]}$ is a {2,3}-group and hence solvable. By Lemma 2.2, $G_{v}^{X_{1}(v)}$ is non-solvable and so $G_{v}^{X_{1}(v)}=\mathrm{A}_{5}$ or S₅, which implies that $G_{vu}^{X_{1}(v)}=\mathrm{A}_{4}$ or S₄, respectively. If $G_{v}^{[1]}=1$ then G _v =A₅ or S₅. Now assume $G_{v}^{[1]}\not=1$. Since $G_{v}^{[1]}\lhd G_{vu}$ and $G_{v}^{[1]}\cap G_{u}^{[1]}=1$, $G_{v}^{[1]}$ is transitive on X ₁(u)∖{v}.

We claim that G _v has a normal subgroup A₅ or S₅ and $G_{v}^{[1]}=\mathrm{A}_{4}$ or S₄. To prove it, let $H=\langle G_{z}^{[1]}\mid z\in X_{1}(v)\rangle$. Then H◁G _v . Since $G_{v}^{[1]}\cap G_{u}^{[1]}=1$, the action of $G_{v}^{[1]}$ on X ₁(u)∖{v} is non-trivial and so H has a non-trivial action on X ₁(v). Since $H^{X_{1}(v)}\unlhd G_{v}^{X_{1}(v)}=\mathrm{A}_{5}$ or S₅, we have $H^{X_{1}(v)}=\mathrm{A}_{5}$ or S₅. Then $H_{vu}^{X_{1}(v)}=\mathrm{A}_{4}$ or S₄, and so H contains a non-identity element h of order 3-power such that h∈H _vuw for some w∈X ₁(u) with $w\not=v$. On the other hand, it is easy to show that $[H,G_{v}^{[1]}]\leq G_{v}^{[1]}\cap G_{u}^{[1]}=1$, which implies that $H\cap G_{v}^{[1]}\leq Z(G_{v}^{[1]})$, the center of $G_{v}^{[1]}$. It follows that H commutes with $G_{v}^{[1]}$ and hence h fixes X ₁(u) pointwise because $G_{v}^{[1]}$ is transitive on X ₁(u)∖{v}. Thus, $3\,|\,|G_{u}^{[1]}|$ and $3\,|\,|G_{v}^{[1]}|$. Since $G_{v}^{[1]}\cong G_{v}^{[1]}/G_{uv}^{[1]}\unlhd G_{vu}^{X_{1}(v)}=\mathrm{A}_{4}$ or S₄, we have $G_{v}^{[1]}=\mathrm{A}_{4}$ or S₄. Furthermore, $H\cap G_{v}^{[1]}\leq Z(G_{v}^{[1]})=1$, implying that $G_{v}^{[1]}H=G_{v}^{[1]}\times H$ and H is faithful on X ₁(v). Thus, H=A₅ or S₅, as claimed.

Now it is easy to see that $|G_{v}:G_{v}^{[1]}\times H|=1$ or 2 and we have the following lemma.

Lemma 2.3

Suppose that X≠K _5,5 and X is connected (G,s)-transitive with s≤3. Then either

• (1) s=2, and G _v =A₅ or S₅, or

• (2) s=3, and G _v =A₄×A₅, S₄×S₅, or (A₄×A₅).ℤ₂.

In particular, this lemma tells us that G _v does not have a subnormal subgroup ℤ₅.

Lemma 2.4

Suppose that X≠K _5,5 and X is connected (G,s)-transitive with s≤3. Then G is almost simple.

Proof

Suppose that $1\not=M\lhd N$ is regular on E. Then X is M-edge-transitive, and hence M has at most two orbits on V. Thus, $|M|=|E|={5|V|\over2}$ is divisible by |V| or ${1\over2}|V|$, forcing that M has exactly two orbits on V, X is bipartite, and |M _v |=5. It follows that ℤ₅≅M _v ◁N _v ◁G _v , which is impossible by Lemma 2.3. Thus, N does not have a normal subgroup which is regular on E, and by O’Nan–Scott’s theorem [2, Theorem 4.1A], G is almost simple, or of product action on E.

Suppose that G is of product action on E. Then by O’Nan–Scott’s theorem, $N_{e}=T_{e}^{k}$, $T_{e}\not=1$, and k≥2. Since X≠K _5,5, by Lemma 2.1, T _v =1. It follows that T _e =ℤ₂ and $N_{e}=T_{e}^{k}=\mathbb{Z}_{2}^{k}$, which is impossible because 3 | |N _e | (Lemma 2.3). Thus, G is an almost simple group. □

From Lemma 2.4 we find that if X≠K _5,5 then the group G is almost simple, and the edge stabilizer G _e is a maximal subgroup, and G _e =A₄.ℤ₂, S₄.ℤ₂, (A₄×A₄).ℤ₂, (S₄×S₄).ℤ₂, or ((A₄×A₄).ℤ₂).ℤ₂.

3 Classification

In this section, we prove Theorem 1.1. First, we introduce the so-called coset graph. Let G be a finite group, H a subgroup of G and D=D ⁻¹ a union of several double-cosets of the form HgH with g∉H. The coset graph X=Cos(G,H,D) of G with respect to H and D is defined to have vertex set V= [G:H], the set of the right cosets of H in G, and edge set E={{Hg,Hdg}∣g∈G,d∈D}. Then X is well defined and has valency |D|/|H|. Furthermore, X is connected if and only if D generates G. Note that G acts on V by right multiplication and so we can view G/H _G as a subgroup of Aut(X), where H _G is the largest normal subgroup of G contained in H. We may show that G acts transitively on the arcs of X if and only if D=HgH for some g∈G∖H (see [7, 8]). The following two examples were described in [3, Sect. 8].

Example 3.1

Let p be a prime and let G=PSL(2,p) with p≡±1, ±9 ( mod 40). Then by [3, Proposition 8.5], G has a subgroup H=A₅ and one conjugacy class of maximal subgroups K=S₄ such that K∩H=A₄. Take an involution g∈K∖H. Define the pentavalent PSL(2,p)-graph as Cos(G,H,HgH). Then the PSL(2,p)-graph is edge-primitive and has automorphism group PSL(2,p). Furthermore, any connected G-edge-primitive pentavalent graph is isomorphic to the PSL(2,p)-graph.

Example 3.2

Let p be a prime and let G=PGL(2,p) with p≡±11, ±19 ( mod 40). Then by [3, Proposition 8.5], G has a subgroup H=A₅ and one conjugacy class of maximal subgroups K=S₄ such that K∩H=A₄. Take an involution g∈K∖H. Define the pentavalent PGL(2,p)-graph as Cos(G,H,HgH). Then the PGL(2,p)-graph is an edge-primitive graph and has automorphism group PGL(2,p). Furthermore, any connected G-edge-primitive pentavalent graph is isomorphic to the PGL(2,p)-graph.

Now we construct an edge-primitive graph, which was given by Weiss [12].

Example 3.3

Let G=Aut(J₃)=J₃.ℤ₂. Then by the Atlas [1], G has maximal subgroups $H=\mathbb{Z}_{2}^{4}\rtimes \mbox{$\varGamma$L}(2,4)$ and $K=(\mathbb{Z}_{2}^{4}\rtimes(\mathrm{A}_{4}\rtimes \mathrm{S}_{3})).\mathbb{Z}_{2}$ such that $K\cap H=\mathbb{Z}_{2}^{4}\rtimes(\mathrm{A}_{4}\rtimes \mathrm{S}_{3})$. Define the pentavalent J₃.ℤ₂-graph as Cos(G,H,HgH), where g∈K∖H. Then this is a 4-transitive edge-primitive graph, and has automorphism group J₃.ℤ₂. Furthermore, any connected G-edge-primitive pentavalent graph is isomorphic to the J₃.ℤ₂-graph.

The following two edge-primitive pentavalent graphs are extracted from [5, Sect. 2].

Example 3.4

Let G=Aut(PSL(3,4))=PSL(3,4).D ₁₂. By the Atlas [1], G has a maximal subgroup $K=(\mathbb{Z}_{2}^{4}\rtimes(\mathrm{A}_{4}\rtimes \mathrm{S}_{3})).\mathbb{Z}_{2}$ and a subgroup $H=\mathbb{Z}_{2}^{4}\rtimes \mbox{$\varGamma$L}(2,4)$ such that $K\cap H=\mathbb{Z}_{2}^{4}\rtimes(\mathrm{A}_{4}\rtimes \mathrm{S}_{3})$. Define the pentavalent PSL(3,4).D ₁₂-graph as Cos(G,H,HgH), where g∈K∖H. Then this is a 4-transitive edge-primitive graph and has automorphism group PSL(3,4).D ₁₂. Furthermore, any connected G-edge-primitive pentavalent graph is isomorphic to the PSL(3,4).D ₁₂-graph.

Example 3.5

Let G=Aut(PSp(4,4))=PSp(4,4).ℤ₄. By the Atlas [1], G has a maximal subgroup $K=(\mathbb{Z}_{2}^{6}\rtimes(\mathrm{A}_{4}\rtimes \mathrm{S}_{3})).\mathbb{Z}_{2}$ and a subgroup $H=\mathbb{Z}_{2}^{6}\rtimes \mbox{$\varGamma$L}(2,4)$ such that $K\cap H=\mathbb{Z}_{2}^{6}\rtimes(\mathrm{A}_{4}\rtimes \mathrm{S}_{3})$. Define the pentavalent PSp(4,4).ℤ₄-graph as Cos(G,H,HgH), where g∈K∖H. Then this is a 5-transitive edge-primitive graph and has automorphism group PSp(4,4).ℤ₄. Furthermore, any connected G-edge-primitive pentavalent graph is isomorphic to the PSp(4,4).ℤ₄-graph.

Proof of Theorem 1.1

The graph X has an edge e={v,u}, and is A-edge-primitive, where A=Aut(X). Clearly, K _5,5 is 3-transitive and edge-primitive. Now assume X≠K _5,5. By [3, Lemmas 3.1 and 3.4], X is a connected (A,s)-transitive graph for s≥1.

Let s≤3. By Lemma 2.4, T=Soc(A) is a non-abelian simple group. Note that A _e is a {2,3}-group and hence solvable. By [6, Theorem 1.1], A has a normal subgroup B which is minimal under the condition that B _e =B∩A _e is maximal in B, and the pairs (B,B _e ) are given in [6, Tables 14–20]. Since $B\unlhd A$, the edge-primitivity of A implies that B is edge-transitive and hence edge-primitive by the maximality of B _e in B. Again by [3, Lemma 3.4], X is B-arc-transitive, and by Lemma 2.2, B is 2- or 3-transitive and B _v is non-solvable. Clearly, Soc(B)=T. By Lemma 2.3, B _e =A₄.ℤ₂, S₄.ℤ₂, (A₄×A₄).ℤ₂, (S₄×S₄).ℤ₂, or ((A₄×A₄).ℤ₂).ℤ₂. Checking the pairs (B,B _e ) listed in [6, Tables 14–20], we have T=A₆=PSL(2,9), PSL(2,p) with p a prime (p≡±1 ( mod 8) or p≡±11, ±19 ( mod 40)), or PSL(3,2). Since $5\,{\not|}\,\,\,|\mbox{PSL}(3,2)|$, we have T=PSL(2,9) or PSL(2,p) (p≡±1 ( mod 8), or p≡±11, ±19 ( mod 40)). Since X has valency 5, by [3, Theorem 1.3], X is isomorphic to K ₆, the PSL(2,p)-graph or the PGL(2,p)-graph.

Let s≥4. By [6, Theorem 1.3], X is isomorphic to the J₃.ℤ₂-graph, the PSL(3,4).D ₁₂-graph or the PSp(4,4).ℤ₄-graph. □