On quasiprimitive edge-transitive graphs of odd order and twice prime valency

Theorem 1.1.

Let Γ=(V,E) be a connected graph of odd order and valency 2⁢r for some prime r, and let G≤Aut⁡Γ. Assume that Γ is G-edge-transitive but not (G,2)-arc-transitive. If G is quasiprimitive on V, then either G is almost simple, or soc⁡(G)=Zpk for some odd prime p and integer 1≤k≤r.

Theorem 1.2.

Let Γ=(V,E) be a connected graph of odd order and twice prime valency, and let G≤Aut⁡Γ. Assume that Γ is G-edge-transitive but not (G,2)-arc-transitive. If G is primitive on V, then either Γ is a complete graph, or soc⁡(Aut⁡Γ)=soc⁡(G); in particular, Γ is 2-arc-transitive if and only if Γ is a complete graph.

Theorem 1.3.

Let Γ=(V,E) be a connected graph of odd order and twice prime valency, and let G≤Aut⁡Γ. Assume that Γ is G-edge-transitive but not (G,2)-arc-transitive. If G is almost simple and primitive on V, then either Γ is soc⁡(G)-edge-transitive, or Γ is of valency 4 and isomorphic to one of the graphs in Example 5.1.

Lemma 2.1.

Let G be a transitive permutation group on V. Suppose that |V| is odd, and that G has a transitive minimal normal subgroup N. Then N is the unique transitive minimal normal subgroup of G.

Proof.

Assume that M is a minimal normal subgroup of G with M≠N. Then M∩N=1, and hence N≤𝐂G⁢(M). By [5, Theorem 4.2A], M is semiregular on V; in particular, M has odd order as |M| is a divisor of |V|. By the choice of M, we know that M is characteristically simple; see [10, p. 21, Lemma 4.8]. Then M=T1×T2×⋯×Tk for isomorphic simple groups Ti; see [10, p. 51, Theorem 9.12]. It follows that every Ti has odd order, and so Ti is a cyclic group of prime order. Thus M is abelian. If M is transitive on V, then, by [5, Theorem 4.2A], 𝐂G⁢(M)=M, yielding N≤𝐂G⁢(M)=M, a contradiction. Therefore, M is not a transitive subgroup of G, and the lemma follows. ∎

Lemma 2.2.

Let Δ=(V,A) be a connected digraph and α∈V. Let G≤Aut⁡Δ, and let N be a normal subgroup of G. Assume that G is transitive on V. Then π⁢(Nα)=π⁢(NαΔ⁢(α)), and either π⁢(Gα[1])⊆π⁢(Nα), or N is intransitive on V.

Proof.

Let p be a prime divisor of |Nα| which is not a divisor of |NαΔ⁢(α)|. Let P be a Sylow p-subgroup of Nα. Then 1≠P≤Nα[1]. Let γ∈V∖{α}. By [18, Lemma 2], there are vertices α=α0,α1,…,αt=γ such that (αi-1,αi)∈A for 1≤i≤t. Take g∈G with αg=α1. By (2.1), we conclude that NαΔ⁢(α)≅Nα1Δ⁢(α1). This implies that p∉π⁢(Nα1Δ⁢(α1)), and then P≤Nα1[1]. By induction on t, we get P≤Nγ[1]. It follows that P fixes V point-wise, and so P=1, a contradiction. Then we have π⁢(Nα)⊆π⁢(NαΔ⁢(α)), and hence π⁢(Nα)=π⁢(NαΔ⁢(α)).

Suppose that N is transitive on V. Then G=N⁢Gα. Set X=N⁢Gα[1]. Then X is normal in G, Xα=Nα⁢Gα[1] and Xα[1]=Gα[1]. Note that

Then we have

and the lemma follows. ∎

Lemma 2.3.

Let Δ=(V,A) be a connected digraph, G≤Aut⁡Δ and α∈V. Assume that G is transitive on A and N is a normal subgroup of G. If N is intransitive on V, then |Δ⁢(α)|=|ΔN⁢(B)|⁢|C∩Δ⁢(α)| for each given arc (B,C) of ΔN with α∈B; in this case, G induces a group acting transitively on the arc set of ΔN.

Lemma 2.4.

Let Δ=(V,A) be a connected digraph, and G≤Aut⁡Δ. Assume that G is transitive on V and N is an intransitive normal subgroup of G. If |Δ⁢(α)|=|ΔN⁢(B)| for some N-orbit B and α∈B, then N is semiregular on V.

Proof.

Assume that |Δ⁢(α)|=|ΔN⁢(B)| for some N-orbit B and α∈B. Then Δ⁢(α)∩B=∅, and |Δ⁢(α)∩C|=1 for every C∈ΔN⁢(B). It follows that Nα fixes Δ⁢(α) point-wise, and hence NαΔ⁢(α)=1. Then we get Nα=1 by Lemma 2.2. By (2.1), Nαg=(Nα)g=1 for each g∈G. Since G is transitive on V, we have Nβ=1 for all β∈V, that is, N is semiregular on V. ∎

Lemma 2.5.

Assume that Γ=(V,E) is a connected G-arc-transitive graph. Let {α,β}∈E. Then there is a 2-element x in G{α,β}∖Gα⁢β such that G=〈x,Gα〉.

Lemma 2.6.

Let Cay⁡(R,S) be a Cayley digraph. Then the following statements hold.

1. A digraph Δ is isomorphic to Cay⁡(R,S) if and only if 𝖠𝗎𝗍⁡Δ has a regular subgroup isomorphic to R.

2. Cay⁡(R,S) is connected if and only if 〈S〉=R, and Cay⁡(R,S) is a graph if and only if S=S-1.

3. 𝐍𝖠𝗎𝗍⁡Cay⁡(R,S)⁢(R)=R:𝖠𝗎𝗍⁡(R,S), and if 〈S〉=R, then 𝖠𝗎𝗍⁡(R,S) acts faithfully on S.

Hypothesis 3.1.

Γ=(V,E) is a connected graph of twice prime valency 2⁢r, G≤Aut⁡Γ, and Γ is G-vertex-transitive, G-edge-transitive but not (G,2)-arc-transitive.

Lemma 3.2.

Suppose that Hypothesis 3.1 holds and that Γ is G-arc-transitive. Let N be a transitive normal subgroup of G, and let α∈V. Then one of the following holds.

1. N is regular on V.

2. Nα is a nontrivial 2 -group.

3. r=max⁡π⁢(Nα)>2, and

   1. either N is transitive on E,

   2. or N has two orbits E1 and E2 on E such that (V,Ei) are N-arc-transitive graphs of valency r; in this case, |V| is even.

Proof.

By Lemma 2.2, π⁢(Nα)=π⁢(NαΓ⁢(α)). If NαΓ⁢(α)=1, then Nα=1, and thus part (1) of this lemma follows. In the following, we suppose that NαΓ⁢(α)≠1.

Since NαΓ⁢(α) is normal in GαΓ⁢(α) and GαΓ⁢(α) is transitive on Γ⁢(α), all NαΓ⁢(α)-orbits on Γ⁢(α) have equal length, say l. Then l=2, r or 2⁢r. If l=2, then NαΓ⁢(α) is a 2-group, and so is Nα by Lemma 2.2, thus part (2) of this lemma occurs. Assume that r=2 and l=2⁢r=4. Then |Γ⁢(α)|=4. Since Γ is not (G,2)-arc-transitive, GαΓ⁢(α) is not 2-transitive on Γ⁢(α). Thus we have NαΓ⁢(α)≤GαΓ⁢(α)≤D8. Then Nα is a 2-group by Lemma 2.2, and part (2) of this lemma occurs. Thus we suppose that l>2 and r>2 in the following.

Assume that GαΓ⁢(α) is primitive on Γ⁢(α). Then NαΓ⁢(α) is a transitive normal subgroup of GαΓ⁢(α). It follows that N acts transitively on the arc set of Γ. Since GαΓ⁢(α) is not 2-transitive, by [7, Theorem 1.51], GαΓ⁢(α)≅A5 or S5, and r=5. Then NαΓ⁢(α)≅A5, and r=5=max⁡π⁢(Nα) by Lemma 2.2, and so (i) of part (3) occurs.

Now assume that GαΓ⁢(α) is imprimitive on Γ⁢(α). Then GαΓ⁢(α) is isomorphic to a subgroup of S2≀Sr or Sr≀S2, which yields that r=max⁡π⁢(Gα). Recalling that l=2⁢r or r, we have r=max⁡π⁢(Nα). If l=2⁢r, then NαΓ⁢(α) is transitive on Γ⁢(α), which yields that Γ is N-arc-transitive, and so (i) of part (3) occurs. Suppose that l=r. Since N is normal in G and transitive on V, we know that N has exactly two orbits on the arc set of Γ, say A1 and A2, and either A2=A1*≠A1, or Ai=Ai* for each i. These two cases yield (i) and (ii) of part (3), respectively. ∎

Lemma 3.3.

Suppose that Hypothesis 3.1 holds and that Γ is not G-arc-transitive. Let N be a transitive normal subgroup of G, and let α∈V. Then

1. either N is regular on V,

2. or r=max⁡π⁢(Nα) and Γ is N-edge-transitive.

Proof.

Let A be a G-orbit on the arc set of Γ. Then the arc set of Γ is equal to A∪A*. Consider the digraph Δ=(V,A). Then |Δ⁢(α)|=r, and so

by Lemma 2.2. Since NαΔ⁢(α) is normal in GαΔ⁢(α) and GαΔ⁢(α) is transitive on Δ⁢(α), all NαΔ⁢(α)-orbits (on Δ⁢(α)) have equal length. Then either NαΔ⁢(α)=1, or NαΔ⁢(α) is transitive on Δ⁢(α). By Lemma 2.2, the former case yields Nα=1, and so N is regular on V. Assume that NαΔ⁢(α) is transitive on Δ⁢(α). Then r=max⁡π⁢(Nα), and N is transitive on A. This implies that N acts transitively on E, and so part (2) of this lemma occurs. ∎

Lemma 3.4.

Suppose that Hypothesis 3.1 holds and r>2. Let α∈V, and let H be a subgroup of Gα with |H|=r. Assume that N is a transitive normal subgroup of G such that Nα is a nontrivial 2-group. Then Gα[1] is a 2-group, and Γ is N⁢H-arc-transitive.

Proof.

By the assumptions in this lemma and Lemma 3.3, we conclude that Γ is G-arc-transitive, and hence Gα acts transitively on Γ⁢(α). Further, by Lemma 2.2, π⁢(Gα[1])⊆π⁢(Nα), and so Gα[1] is a 2-group. Then H≰Gα[1] as H has odd order r.

Since Nα is normal in Gα, all Nα-orbits on Γ⁢(α) have equal length which is a divisor of 2⁢r. Noting that Nα is a nontrivial 2-group, it follows that the set 𝒞 of Nα-orbits on Γ⁢(α) is a Gα-invariant partition of Γ⁢(α) into subsets of size 2. Recalling that H≰Gα[1] since |𝒞|=r=|H|, we conclude that H acts transitively on 𝒞, and thus Nα⁢H acts transitively on Γ⁢(α). Clearly, NH is transitive on V. Noting that (N⁢H)α=N⁢H∩Gα=(N∩Gα)⁢H=Nα⁢H, it follows that Γ is NH-arc-transitive. ∎

Lemma 3.5.

Suppose that Hypothesis 3.1 holds and |V| is odd. Let N be an intransitive normal subgroup of G. Then either N is semiregular on V, or ΓN is a cycle.

Proof.

Since |V| is odd and G is transitive on V, the group N cannot have just 2 orbits on V. Then N has at least 3 orbits on V. Let A be a G-orbit on the arc set of Γ, and let Δ=(V,A). Then the digraph ΔN is connected and vertex-transitive, which has at least three vertices. Let B be an N-orbit on V and α∈B. By Lemmas 2.3 and 2.4, |ΔN⁢(B)| is a divisor of |Δ⁢(α)|, and if |ΔN⁢(B)|=|Δ⁢(α)|, then N is semiregular on V. Thus we assume next that |ΔN⁢(B)|<|Δ⁢(α)|.

Suppose that A is not the arc set of Γ. Then we have |Δ⁢(α)|=r, and hence |ΔN⁢(B)|=1. Thus ΔN is a directed cycle with length at least 3. Similarly, letting Δ*=(V,A*), we know that ΔN* is a directed cycle. Then ΓN is a cycle by noting that ΓN=ΔN∪ΔN*.

Suppose that A is the arc set of Γ. Then ΓN=ΔN is a connected vertex-transitive graph. Recalling that ΓN has order at least 3, we know that ΓN is of valency at least 2. Thus ΓN has valency |ΔN⁢(B)|=2 or r. Note that the latter is impossible with r≥3 since then we would have a regular graph of odd order and odd valency. Therefore, ΓN is a connected graph of valency 2. Then the lemma follows. ∎

Theorem 3.6.

Suppose that Hypothesis 3.1 holds and |V| is odd. Let N be a minimal normal subgroup of G, and let α∈V. Assume that N is transitive on V. Then N is the unique transitive minimal normal subgroup of G, and one of the following holds.

1. N≅ℤpk for some odd prime p and positive integer k≤r, and Gα is isomorphic to a subgroup of S2≀Sr; if further Γ is not G-arc-transitive, then k<r, and Gα is isomorphic to a transitive subgroup of the symmetric group Sr of degree r.

2. N is a nonabelian simple group.

Proof.

By Lemma 2.1, N is the unique transitive minimal normal subgroup of G. Write N=T1×⋯×Tk, where Ti are isomorphic simple groups.

Case 1. Assume that Ti≅ℤp for all i, where p is a prime. Then N≅ℤpk. Since N is abelian and transitive on V, we know that N is regular on V, and so p is odd. We may identify Γ with a Cayley graph Cay⁡(N,S). Let α be the vertex of Γ corresponding to the identity of N. Since N is normal in G, we have Gα≤𝖠𝗎𝗍⁡(N,S); see Lemma 2.6. Noting that N=〈S〉 as Γ is connected, Gα acts faithfully on S. Further, {{s,s-1}∣s∈S} is a Gα-invariant partition of S into subsets of size 2. It follows that Gα is isomorphic to a subgroup of the wreath product S2≀Sr. Let O be a set of representatives of the partition of S. Then N=〈O〉, and hence k≤|O|=|S|2=r.

Suppose that Γ is not G-arc-transitive. Then Gα has two orbits on S. Without loss of generality, we may choose the above O being a Gα-orbit on S; see [14, Lemma 3.2]. Then Gα acts faithfully on O, and so Gα is isomorphic to a transitive subgroup of Sr. Consider the element x=∏s∈Os of N. It is easily shown that Gα centralizes x. Then 〈x〉 is a normal subgroup of G. Since x∈N, by the minimum of N, either k=1, or x=1. Clearly, if k=1, then k<r. If x=1, then N=〈O∖{s}〉 for s∈O, and thus k≤|O∖{s}|=r-1<r. Thus part (1) of the theorem follows.

Case 2. Assume that T1,…,Tk are nonabelian simple groups. Then N has even order. Since |V|=|N:Nα| is odd, Nα contains a Sylow 2-subgroup of N, where α∈V. Of course, Nα≠1 as |N| has even order. Since every Ti is normal in N and N is transitive on V, all Ti-orbits on V have equal length, which is a divisor of |V|. Thus (Ti)α contains a Sylow 2-subgroup of Ti. We shall show part (2) of this theorem holds.

Suppose to the contrary that k>1. Since T1 and T2 centralize each other, if T1 is transitive on V, then T2 is semiregular or acts trivially on V, which is impossible. Thus T1 is intransitive on V. Noting that |V| is odd and N is transitive on V, it follows that T1 has at least three orbits on V. Thus ΓT1 is a connected graph of valency at least 2.

Let K be the kernel of N acting on the set of T1-orbits on V. Then N/K is isomorphic to a subgroup of 𝖠𝗎𝗍⁡ΓT1. Noting that N≠K, we have Ti≰K for some i, and so Ti∩K=1. Then Ti≅Ti⁢K/K≤N/K, and hence N/K is insoluble. Thus 𝖠𝗎𝗍⁡ΓT1 is insoluble. This implies that ΓT1 is not a cycle. If N acts transitively on E, then, by Lemma 3.5, ΓT1 is a cycle, a contradiction. Thus N is intransitive on E.

Suppose that r=2. Then Γ has valency 4. Recall that ΓT1 is not a cycle. Applying (2.2) to the pair (N,T1), we know that ΓT1 has valency d=3 or 4. If d=4, then, by Lemma 2.4, T1 is semiregular on V, a contradiction. Assume that d=3. Let B be a T1-orbit on V. Then (2.2) implies that |Γ⁢(β)∩B|=1 for each β∈B. Then the subgraph of Γ induced by B is a regular graph of valency 1. It follows that |B| is even, which is impossible as |B| is a divisor of |V|. Therefore, r is odd.

Since both |V| and r are odd, by Lemmas 3.2 and 3.3, we conclude that Γ is G-arc-transitive and Nα is a 2-group, and then Nα is a Sylow 2-subgroup of N. Recall that each (Ti)α contains a Sylow 2-subgroup of Ti. It follows that (Ti)α is a Sylow 2-subgroup of Ti, where 1≤i≤k. Thus we have

Let H≤Gα with |H|=r. Note that Tih is a normal subgroup of N, where 1≤i≤k and h∈H. Then Tih=Tj for some j; see [10, p. 51, Theorem 9.12] for example. Thus H acts on {T1,…,Tk} by conjugation. Set X=N⁢H. By Lemma 3.4, Γ is X-arc-transitive. Further, Xα=Nα⁢H, and Nα is the unique Sylow 2-subgroup of Xα.

Let M=〈T1h∣h∈H〉. Then M is a normal subgroup of X, M≤N and Mα is a nontrivial 2-group. Suppose that M is intransitive on V. Then, by Lemma 3.5, the quotient graph ΓM is a cycle. Then 𝖠𝗎𝗍⁡ΓM is soluble, and so is N/K, where K is the kernel of N acting on the set of M-orbits on V. This implies that N=K, and N fixes every M-orbit set-wise, a contradiction. Therefore, M is transitive on V. Suppose that M≠N. Then Ti≰M for some i. Since Ti is simple, we have Ti∩M=1. Noting that both M and Ti are normal in N, we have Ti≤𝐂N⁢(M). By [5, Theorem 4.2A], Ti is semiregular on V, which is impossible. Thus we get M=N. It follows that H acts transitively on {T1,…,Tk} by conjugation. Since H has prime order, H acts regularly on {T1,…,Tk}; in particular, k=r. For h∈H and 1≤i≤r, setting Tih=Tj for some j, we have

It follows that H acts regularly on {(T1)α,…,(Tr)α} by conjugation.

Recall that Γ is X-arc-transitive and Xα=Nα⁢H. Then

where β∈Γ⁢(α). Thus |Xα⁢β|=|Nα|2; in particular, Xα⁢β is a 2-group. Recalling that Nα is the unique Sylow 2-subgroup of Xα, we have Xα⁢β≤Nα. Then Xα⁢β=Nα⁢β, and |Nα:Nα⁢β|=2. By Lemma 2.5, there is a 2-element

such that

Noting that H has odd order, we have N∩H≤Nα∩H=1. Then

Since N is normal in X, we know that N⁢〈x〉 is a subgroup of X, and hence |N⁢〈x〉| is a divisor of |X|. Noting that |N〈x〉|=|N||〈x〉:(N∩〈x〉)|, it follows that |〈x〉:(N∩〈x〉)| is a divisor of |H|=r. Since 〈x〉 is a 2-group, |〈x〉:(N∩〈x〉)| is a power of 2. It follows that |〈x〉:(N∩〈x〉)|=1, so x∈N=T1×⋯×Tr. Write x=x1⁢x2⁢⋯⁢xr, where xi∈Ti and 1≤i≤r. Then

Consider the projections

Setting Li=ϕi⁢(Nα⁢β) for 1≤i≤r, we have

Since Nα⁢β≤Nα=(T1)α×⋯×(Tr)α, we have Li≤(Ti)α for 1≤i≤r. Then Nα⁢β≤L1×⋯×Lr≤Nα. Recalling that |Nα:Nα⁢β|=2, we have

It follows that Li=(Ti)α for all but at most one of i. Without loss of generality, we may let Li=(Ti)α for i>1, and |(T1)α:L1|≤2.

Since (Ti)α is a Sylow 2-subgroup of Ti and xi is a 2-element, if i>1, then xi∈Li=(Ti)α as (Ti)α is the unique Sylow 2-subgroup of 𝐍Ti⁢(Li). Recalling that H acts regularly on {(T1)α,…,(Tr)α} by conjugation, by (3.1), we have

Set H={1,h2,…,hr}. Then

This implies that T1=〈x1,(T1)α〉.

Recalling that |(T1)α:L1|≤2 and x1∈𝐍T1⁢(L1), it follows that L1 is normal in 〈x1,(T1)α〉=T1. Since L1 is a 2-group and T1 is nonabelian simple, we have L1=1. Thus |(T1)α|=2. In particular, T1 has a cyclic Sylow 2-subgroup. By [10, p. 420, Theorem 2.8], T1 has a normal subgroup of odd order, a contradiction. This completes the proof. ∎

Corollary 3.7.

Suppose that Hypothesis 3.1 holds and |V|=pk for some odd prime p. If G is quasiprimitive on V, then Γ is arc-transitive, and either

or one of the following holds.

1. p=7, G≅PSL3⁢(2) and Γ is the complete graph 𝖪7.

2. p=11, G≅PSL2⁢(11) and Γ is the complete graph 𝖪11.

3. G≅PSU4⁢(2) or PSU4⁢(2)⁢.2, and Γ has order 27 and valency 10.

Proof.

Assume that G is quasiprimitive on V. Let N=soc⁡(G). By Theorem 3.6, either N is nonabelian simple, or N≅ℤpk with k≤r. Assume that N≅ℤpk. Then we may write Γ=Cay⁡(N,S), and Gα is faithful on S, where α is the vertex corresponding to the identity of N. Note that {x,x-1}h={xh,(xh)-1} for x∈S and h∈Gα. Then Gα induces a transitive action on {{x,x-1}∣x∈S}. Let σ:N→N, y↦y-1. Then σ is an automorphism of Γ which fixes α, and 〈Gα,σ〉 is transitive on S. Thus Γ is 〈G,σ〉-arc-transitive.

Assume that N is nonabelian simple. Note that |N:Nα|=|V|=pk for α∈V. By [8], either N is 2-transitive on V, or N≅PSU4⁢(2) and Nα has exactly three orbits on V with length 1, 10 and 16, respectively. The latter case implies that Γ is a G-arc-transitive graph of order 27 and valency 10, and then part (3) of this lemma follows. Suppose that N is 2-transitive on V. Then Γ=𝖪pk, and Γ is N-arc-transitive. Note that G and hence N is intransitive on the set of 2-arcs of Γ. It follows that N is not 3-transitive on V. Checking the groups listed in [8, Theorem 1] one by one, we conclude that either N≅PSL2⁢(11) and |V|=11, or N≅PSLd⁢(q) and |V|=qd-1q-1, where d is a prime. If N≅PSL2⁢(11) and |V|=11, then part (2) of this lemma holds. Assume that N≅PSLd⁢(q) and pk=|V|=qd-1q-1. Then 2⁢r=qd-1q-1-1=q⁢qd-1-1q-1. Noting that r and d are primes, it follows that (d,q)=(2,4) or (3,2), and r=2 or 3, respectively. For (d,q)=(2,4), we have N≅PSL2⁢(4) and |V|=5; however, as a permutation group of degree 5, the group PSL2⁢(4) is 3-transitive, which is not the case. Thus (d,q)=(3,2), and then part (1) of this lemma follows. ∎