ZHANG Li,WANG Gaixia

(School of Microelectronics and Data Science，Anhui University of Technology,Maanshan Anhui,243002,China)

Abstract: The graph Γ is said to be 2-arc-transitive if it has at least one 2-arc and Aut(Γ)is transitive on both the vertices and 2-arcs of Γ.Let G be an almost simple group with soc(G)=Ac for c ≥5,we use the concept of coset graph to construct the graphs with square-free order and(G,2)-arc-transitive.Then by analyzing the subgroup chain structures between vertex-stabilizer subgroups and their automorphism groups,a classification of the(G,2)-arc transitive graphs of square-free order was given.

Key words: symmetric graph;square-free order;2-arc-transitive graph;automorphism group

0 Introduction

All graphs considered in this paper are assumed to be finite,simple and undirected.

Let Γ be a graph with vertex set VΓ and edge set EΓ.We use Aut(Γ) to denote the automorphism group of Γ.For a positive integer s,an s-arc of Γ is an(s+1)-tuple(v0,v1,···,vs)of vertices such that{vi-1,vi}∈EΓ for 1 ≤i ≤s and vi-1≠vi+1for 1 ≤i ≤s-1.Let G ≤AutΓ.The graph Γ is said to be(G,s)-arc-transitive if it has at least one s-arc and G is transitive on both the vertices and the s-arcs of Γ,and Γ is(G,s)-transitive if it is(G,s)-arc-transitive but not(G,s+1)-arc-transitive.For the case where G=Aut(Γ),a(G,s)-arc-transitive graph or a(G,s)-transitive graph is simply called s-arc-transitive or s-transitive,respectively.

Characterizing or classifying finite 2-arc-transitive graphs have been an active topic in algebraic graph theory,which is highly attractive from the group-theoretic and combinatorial viewpoint and has received considerable attentions.Recently,Li et al.[1]gave the classification of 2-arc-transitive graphs with odd order in 2021.Pan et al.[2]gave 2-arc-transitive Cayley graphs on alternating group in 2022.A characterization of 2-arc-transitive partial cubes was given by Xie et al.[3].In particular,the cases of 2-arc transitive graphs admitting a Suzuki simple group[4]and a Ree simple group[5]are classified,and the case of 2-arc transitive graphs admitting a 2-dimensional projective linear group[6]is studied in 1999.

Another motivation stems from our work about graphs with square-free order.We classified some kinds of graphs with square-free order,such as vertex-transitive cubic graphs[7],vertex-transitive and edge-transitive tetravalent graphs[8],edge-transitive graphs[9],arc-transitive graphs with small valency[10],2-arc-transitive on alternating group[11],2-arc-regular graphs[12].In this paper,we use another method different from [11] to obtain the same result about the classifications of graphs with square-free order with symmetric group.The following are our main results.

Theorem 1Let G be an almost simple group with alternating socle and let Γ be a connected(G,2)-arc-transitive graph of square-free order.Then Γ is isomorphic to one of the following graphs:

(i)The complete graphs Kpwith square-free number p;

(ii)The complete graphs Kp,p-pKpwith odd square-free number p;

(iii)Tutte’s 8 cage;

(iv)The odd graphs of square-free order;

(v)One of the graphs in the Table 1.

Table 1 Coset graphs admitting alternative group with square-free order

1 Preliminaries

Let G be a group.A subgroup H ≤G is said to be core free if ∩g∈GHg=1,that is,no non-trivial normal subgroups of G are contained in H.For a subset S ?G and a core free subgroup H of G,the coset digraph Γ:=Cos(G,H,HS H)is defined as the digraph with vertex set VΓ:=[G:H]={Hx|x ∈G}such that Hx is adjacent to Hy if and only if yx-1∈HS H.For such a coset digraph,G can be viewed as a subgroup of Aut(Γ)by the following way:

g:HxHxg,x,g ∈G.

The following two Lemmas collect several well-known properties about coset graphs.

Lemma 1Let Γ=Cos(G,H,HS H)be a coset digraph.Then

(i)Γ is connected if and only if〈H,S〉=G;

(ii)Γ is undirected if and only if HS H=HS-1H;

(iii)Γ is a G-arc-transitive graph if and only if HS H=HgH for some 2-element g ∈G with g2∈H;

(iv)Cos(G,H,HgH)is a connected(G,2)-arc transitive graph if and only if g ∈NG(H ∩Hg),g2∈H,〈H,g〉=G,and H is 2-transitive on[H:H ∩Hg]by the right multiplication.

Now,we describe some graphs in terms of coset graphs.The odd graph Okis the graph of which the vertex set consists of(k-1)-subsets of a set of size 2k-1 such that two vertices are adjacent whenever they disjoint.Let GQ(q)be the generalized quadrangule of order q=2f,which has(q2+1)(q+1)points and lines.

Example 1Let G=S8acts naturally on {1,2,···,8} and H=23: PSL(3,2) is a core-free subgroup of G.By Atlas[13],we have NG(PSL(3,2))=PSL(3,2):2.Choose the involution g ∈PSL(3,2):2PSL(3,2),then we can construct graph Γ1:=Cos(G,H,HgH).We can easily get Γ1is a connected(G,2)-arc transitive 8-valent graph of 30 vertices.

Example 2Let G=S8acts naturally on {1,2,···,8} and H=23: PSL(2,7)?(23: S4)Z7is a core-free subgroup of G.By Atlas[13],we have NG(23,S4)=24: S4.Choose the involution g ∈24: S423: S4,then we can construct graph Γ2:=Cos(G,H,HgH).We can easily get Γ2is a connected(G,2)-arc-transitive 7-valent graph of 30 vertices.

Example 3Let G=S7acts naturally on{1,2,···,7}and H=PSL(3,2)is a core-free subgroup of G.By Atlas[13],we have NG(7:3)=7:6.Choose the involution g ∈7:67:3,then we can get Γ3:=Cos(G,H,HgH)is a connected(G,2)-arc-transitive 8-valent graph of 30 vertices.

Example 4Let G=S7acts naturally on {1,2,···,7} and H=PSL(2,7) is a core-free subgroup of G.By Atlas[13],we have NG(S4)=S4×S3.Choose the involution g ∈S4×S3S4,then we can get Γ4:=Cos(G,H,HgH) is a connected(G,2)-arc-transitive 7-valent graph of 30 vertices.

Let Γ be a graph and G be a subgroup of Aut(Γ).Let α be a vertex of Γ.Then the stabilizer Gαinduces an action on the neighborhood Γ(α)of α in Γ.Letdenote the permutation group on Γ(α)induced by Gα,and letbe the kernel of this action，and setThen we have the following equation.

where X·Y means a group extension of X by Y.The following two results are well-known.

Lemma 2If G is transitive on VΓ,then Γ is(G,2)-arc transitive if and only ifis a 2-transitive permutation group.

Lemma 3If Γ is(G,s)-arc transitive with s ≥2,thenis a p-group for some prime p,where{α,β}∈EΓ.Further if Γ is s-transitive with s ≥4 then Γ is of valency p+1 and|(AutΓ)α|=(p+1)pt-1m for s ≠6 and a divisor m of(p-1)2.

All finite 2-transitive permutation groups are precisely known,see[14]for example.Then,by Lemma 2 and Lemma 3,we have Corollary 1.

Corollary 1If Γ is a (G,2)-arc transitive graph,then the stabilizer Gαhas at most two insoluble composition factors.Further,if there are two insoluble factors,then either they are not isomorphic whenis almost simple or they are isomorphic whenis an affine group.

ProofBy Lemma 3,is a p-group.Then by the structure of Gα,up to isomorphism,all insolvable composition factors are involved inand.Note thatThen the 2-transitive permutation groupand its a stabilizer acting on Γ(α)give all possible insolvable composition factors of Gα.Thus our result follows from checking the 2-transitive permutation groups one by one.

In the following sections,we shall first describe the case when Gαis soluble,then give an analysis about the soluble quotients of Gαwhen Gαis insoluble.At last,we give the proofs of the main Theorems.

2 The Structure of Gα

Lemma 4Let Γ be a connected(G,2)-arc-transitive graph of square-free order where G is almost simple with alternating socle.Assume that Gαis soluble,then Γ is isomorphic to K5,Petersen graph,K5,5-5K2or odd graph of order 35.

ProofAt first,we consider the graphs with small valency|Γ(α)|=3 or 4.

(I)Assume that|Γ(α)|=3.We have Gαis isomorphic to S3,2×S3,S4or 2×S4.Then 32?|Gα|.We can easily get that c ≤8 since|G:Gα|is square-free order with soc(G)=Ac.If c=8,then for any case of Gα,|G:Gα|is not square-free order.If c=7,by easily computation,we can get that Gα?2×S3or S4when G=A7and Gα?S4or 2×S4when G=S7.Then Gα≤M ≤G where M is the maximal subgroup of G.By Atlas[13],we have M=S5or 2×S5,respectively.This exists no element g satisfies Cos(G,Gα,GαgGα)which is connected(G,2)-arc transitive graph of square-free order.If c=6,then by easily computation,we can get that Gα?2×S3or S4when G=A6and Gα?S4or 2×S4when G=S6.By[14],we can easily get that this exists no element g satisfies Cos(G,Gα,GαgGα) which is connected (G,2)-arc transitive graph of square-free order.If c=5,then Gα?S3or 2×S3when G=A5and Gα?2×S3or S4when G=S5.We can easily get that for the latter case when G=S5,Γ is isomorphic to Petersen graph.

(II)Assume that|Γ(α)|=4.We have that Gαis isomorphic to A4,S4,3×A4,(3×A4)·2,S3×S4,32:Q8·S3or[35]·Q8·S3.Then 5 ?|Gα|.We can easily get that 5 ≤c ≤9 since|G:Gα|is square-free order with soc(G)=Ac.Similar to the above case,we can get that Γ is isomorphic to odd graph of order 35,K5,5-5K2or K5.

Then we assume|Γ(α)|≥5.Since Gαis soluble,thenis affine 2-transitive permutation group and=1.It follows that=pe:for prime p.By checking the list of 2-transitive group,we have the following cases.

Case 1Assume≤ΓL(1,pe) and |Γ(α)|=pe,since ΓL(1,pe) ?Zpe-1: Ze,then we have Gα?where.If p=2,then a sylow 3-subgroup of Gαhas rank at most 4.It follows that c ≤17.Then by computation|G:Gα|is not square-free order.Take c=17 for example.It follows formthat e=13,14 or 15.For any case,that is|G:Gα|is not square-free order.If p ≥3,then a sylow 2-subgroup of Gαhas rank at most 4.It follows that c ≤12.Similarly,for this case|G:Gα|is not square-free order.

Case 2Assumeis isomorphic to one of the following groups Q8·Z3,Q8·S3,Q8·S3,Z3×(Q8·2),Z3×(Q8·S3),Z5×(Q8·Z3),Z5×(Q8·S3),Z11×(Q8·S3)or 21+4·L,where L is Z5,D10,(Z5·Z4),A5or S5.We can easily get that|G:Gα|is not square-free order since peis large comparatively.

Lemma 5Let Γ be a connected(G,2)-arc transitive graph of square-free order where G is almost simple with alternating socle.If Gαis unsolvable,then unsolvable quotients of Gαis isomorphic to S3?S2,S4?S2,S3?S3,S4?S3,(S2?S2)?S2,A4or S4.

ProofBy Lemma 3,is a p-group.Thus there are two cases.

Case 1≠1.

(I)Γ is 2 or 3-transitive.Because=PSL(d,q)·O with|Γ(α)|=(qd-1)/(q-1)where q=pffor some prime p,we have

where O ≤Out(PSL(d,q))and

Note that Op(Gα)≤Gαand Gα≤Sc,it follows that when(d,q)=(3,2),(3,2f),(4,2)and(5,2),|G:Gα|is not square-free order.For the other case,we have pd(d-1)f/2It follows that when p ≥11,we have|G:Gα|is not square-free order.When p=2,3,5 or 7,form≤T×O where TO.And for any case,Further,

(II)Suppose Γ is s-transitive for s>3,the structure of Gαcan be explicitly known.Then we can easily get that|G:Gα|is not square-free order.

3 The Proofs of the Main Theorems

In this section,we first prove a Lemma which will be used later.

Lemma 6Let soc(G)=Ac(c ≥5)and G acts naturally on ?={1,2,···,c}.Let H be a subgroup of G with square-free index.If the natural action of H on ?is primitive,then Ac≤H or 5 ≤c ≤8.

ProofSuppose P ∈Syl2(G)such that a=(12)(34)∈P and b=(13)(24)∈P.Then≤P and||=4.Let Q ∈Syl2(H),we can conclude Q∩≠1 since|G:H|is square-free order.Thus there is at least one of these involutions(12)(34),(13)(24) and (14)(23) contained in Q.It follows that the minimal degree of H is at most 4.Then H contains a 2-cycle or 3-cycle or the minimal degree is 4.By[15],we have Ac≤H or H is a proper primitive group of Scwhere 5 ≤c ≤8.

Proposition 1Let G be an almost simple group with alternating socle and let Γ be a connected (G,2)-arc-transitive graph of square-free order.Assume soc(G)=Acand Gα≤G acts primitively on ?:={1,2,···,c}.Then Γ is isomorphic to one of the graphs in Table 1 or K6.

ProofBy Lemma 6,we have Ac≤Gαor Gαis a proper primitive group of Scwhere 5 ≤c ≤8.

Case 1Suppose Ac≤Gα,then|G:Gα|=1,2 or 4,this is trivial.

Case 2Suppose Gαis a proper primitive group of Scwhere c ≤8.

(I)Assume c=8.Inspecting the proper primitive permutation groups of degree 8 such that|G:Gα|is square-free order,we can get that G=A8,Gα=23:L3(2)or G=S8,Gα=23:L3(2).

(a)If G=A8,Gα=23:L3(2),we have G is 2-transitive on VΓ.It follows that Γ ?K15.However,this is not possible since G is not 3-transitive.

(a)If G=S7,Gα=PSL(3,2)and|Γ(α)|=8,then Gαβ=7:3 and NG(Gαβ)=7:6.It follows that there exists an involution g ∈NG(Gαβ)NG(Gαβ)such that Γ:=Cos(G,Gα,GαgGα)is a connected(G,2)-arc transitive graph of order 30 with valency 8.Furthermore Γ ?Γ3in Examples.

(b)If G=S7,Gα=PSL(2,7)and|Γ(α)|=7,then Gαβ=S4and NG(Gαβ)=S4×S3.It follows that there exists an involution g ∈NG(Gαβ)NG(Gαβ)such that Γ:=Cos(G,Gα,GαgGα)is a connected(G,2)-arc transitive graph of order 30 with valency 7.Furthermore,Γ ?Γ4in Examples.

(III)Assume c=6.Inspecting the proper primitive permutation groups of degree 6 such that|G:Gα|is square-free order,we can get that(G,Gα)is one of the following pairs: (PGL(2,9),S4),(M10,S4),(PΓL(2,9),S4×Z2),(A6,A5),(S6,S5),(A6,S4),(S6,S4×2)or(S6,S4).For the first three cases,G has a subgroup of index 2 which contains H,say X=S6for G=PΓL(2,9)and X=A6for the other two cases.Thus Γ is a bipartite graph with two bipartition subsets,say U and V,having size 15 respectively.It is easy to see that X acts primitively on both U and V.In particular,X acts transitively on the edges of Γ.We claim that the actions of X on U and V are not permutation equivalent; otherwise,X will have a primitive permutation representation of degree 15 with a 2-transitive subconstituent,which contradicts.Thus we may assume that U consists of 2-subsets of[6]while V is the set of partitions of[6]into three parts with the same size.Let{α,β}be an edge of Γ with α ∈U and β ∈V.Then two possible cases arise.If α is not a part of β,then it is easily shown that Γ(α)=βH={βh|h ∈H}contains 12 partitions of[6],but H can not 2-transitively on Γ(α),a contradiction.Thus α must be a part of β and,in this case,Γ is isomorphic to Tutte’s 8 cage.For the other cases,Γ ?K6.

(IV) Assume c=5.Inspecting the proper primitive permutation groups of degree 5 such that |G :Gα| is square-free order,we can get that G=A5,Gα=D10or G=S5,Gα=5:4.For the former case,it is impossible.For the latter case,we have Γ ?K6.

Proposition 2Let G be an almost simple group with alternating socle and let Γ be a connected (G,2)-arc-transitive graph of square-free order.Assume soc(G)=Acand Gα≤G acts transitively but not primitively on ?:={1,2,···,c}.Then there are no such graphs.

ProofSince Gαis not primitively on ?,let B:={B1,B2,···,Bb}be a maximal Gα-invariant partition of ?,it follows that GBαis primitive and Gα≤M ≤Scwhere M ?Sym(B1)?Sym(B)is a maximal subgroup of Sc.Let K be the kernel of M acting on B,then|M:GαK|is square-free order.It follows that|Sym(B):GBα|=|M/K:GαK/K|is square-free order.Thus we get GBα≤Sym(B)of square-free index and GBαis primitive.By Lemma 6,we can get Alt(B)≤GBαor 5 ≤|B|=b ≤8.

Case 1Suppose Alt(B)≤GBα.We set b=2 for example.Let B={B1,B2}be the maximal Gα-invariant partition of ?and K be the kernel of Gαon B.It follows that KB1?KB2is transitive since Gαis transitive on ?.

(I)Suppose KB1is primitive,then KB1≤Sc/2with index odd square-free order.Thus we have Ac/2≤KB1or 5 ≤c/2 ≤8 and KB1is a proper subgroup of Sc/2.

(a)Suppose Ac/2≤KB1,then KB1?KB2=Ac/2or Sc/2.Since K(B1)?,it follows that K(B1)=1,Ac/2or Sc/2.Thus K=Ac/2,Sc/2,Ac/2×Ac/2,Ac/2×Sc/2or Sc/2×Sc/2.Therefore Gα=Ac/2·2,Sc/2·2,(Ac/2×Ac/2)·2,(Ac/2×Sc/2)·2 or(Sc/2×Sc/2)·2.For Gα=Ac/2·2,Sc/2·2 or Gα=(Ac/2×Ac/2)·2,it follows that|G:Gα|is not square-free order,since Gα≤M ?Sc/2?2.For(Ac/2×Sc/2)·2,then Gα≤M ?Sc/2?2 ≤Sc.If follows that G acts primitively on [G : M] by right multiplication with odd square-free degree.Then we have c=8,Gα=(A4×S4)·2,G=A8.However this is impossible since A8has no subgroup of index 70.For(Sc/2×Sc/2)·2,then Gα?M ?Sc/2?2 ≤Sc.If follows that G acts primitively on[G:Gα]by right multiplication with square-free degree.There is no corresponding graph.

(b)Suppose 5 ≤c/2 ≤8,then 10 ≤c ≤16.For c/2=5,then KB1is a proper primitive group of degree 5.It follows that KB1?D10or 5:4.Then K=5·D10,(5:2)·D10or(5:4)·D10.We can easily get 32since Gα=K·2.That is,|G:Gα|is not square-free order.For c/2=6,then KB1is a proper primitive group of degree 6.It follows that KB1?PSL(2,5)or PGL(2,5).Then K=PSL(2,5),PSL(2,5)·PSL(2,5)or PGL(2,5)·PSL(2,5).We can easily get 32since Gα=K·2.That is,|G:Gα|is not square-free order.For c/2=7,then KB1is a proper primitive group of degree 7.It follows that KB1?PSL(3,2).Then K=PSL(3,2)or PSL(3,2)·PSL(3,2).We can easily get 32since Gα=K·2.That is,|G:Gα|is not square-free.For c/2=8,then KB1is a proper primitive group of degree 8.It follows that KB1?23:PSL(3,2).Then K=PSL(3,2),(22:PSL(3,2))·PSL(3,2)or(23:PSL(3,2))·PSL(3,2).We can easily get 32since Gα=K·2.That is,|G:Gα|is not square-free order.

(II)Suppose KB1is imprimitive,let D:={D11,···,D1d,D21,···,D2d}be a Gα-invariant partition of ?and Dibe{Di1,···,Did}such that KDiis primitive for i=1 and 2.Then Gα≤(Sd11?Sd)?S2≤Scwhere|D11|=d11and c=2dd11.It follows that GDα≤Sd?S2and KD1≤Sd.Thus KD1=Ad,Sdor KD1is a proper primitive group of degree d where 5 ≤d ≤8.

(a)Suppose KD1=Ador Sd,it follows that KD=Ad·2,Sd·2,(Ad×Ad)·2,(Ad×Sd)·2 or(Sd×Sd)·2 and GDα=KD·2.

Assume d ≥5,then Gαhas two isomorphic insoluble composition factors,it is impossible.Assume d=4,since|Sd:GDα|is square-free order,it follows that GDα=(S4×S4)·2=S4?S2.Assume d=3,since|Sd:GDα|is square-free order,it follows that GDα=(S3×S3)·2=S3?S2.Both cases are impossible since Gαhas no quotient group isomorphic to one of these groups by section 3.

Assume d=2,then let D={D11,D12,D21,D22}be the Gα-invariant partition of ?and let Dij:={Dij1,Dij2,···,Dijdi}such that(Gα)DijkDijkis primitive.It follows that GDαis primitive.Since|S4:GDα|is square-free order,then GDα≤(S2?S2)?S2.This is impossible by the structure of the soluble quotient group of Gα.

(b)Suppose that KD1is a proper primitive group of degree d where 5 ≤d ≤8.For d=5,then KD1is a proper primitive group of degree 5.It follows that KD1?D10or 5:4.Then KD=5·D10,(5:2)·D10or(5:4)·D10.This is impossible since|S2d:KD|is not square-free order.For d=6,then KD1is a proper primitive group of degree 6.It follows that KD1?PSL(2,5)or PGL(2,5).Then KD=PSL(2,5),PSL(2,5) · PSL(2,5) or PGL(2,5) · PSL(2,5).This is impossible since |S2d:KD| is not square-free order.For d=7,then KD1is a proper primitive group of degree 7.It follows that KD1?PSL(3,2).Then KD=PSL(3,2)or PSL(3,2)·PSL(3,2).This is impossible since|S2d:KD|is not square-free order.For d=8,then KD1is a proper primitive group of degree 8.It follows that KD1?23: PSL(3,2).Then KD=PSL(3,2),(22: PSL(3,2)) · PSL(3,2) or(23:PSL(3,2))·PSL(3,2).This is impossible since|S2d:KD|is not square-free order.

Case 2Suppose 5 ≤|B|≤8 and GBαis a proper primitive group of degree|B|.

(I)Assume|B|=5.Let B={B1,B2,B3,B4,B5}be the maximal Gα-invariant partition of ?.It follows that Gα≤Sym(B1)?S5where|B1|≥2.Checking the list of the proper primitive group of degree 5 with square-free order,we can get GBα?A4or S4and Sym(B)=A5or S5.By the structure of soluble quotient group of Gα,one has Gα=A4×A5,A4×S5or S4×S5.However,for any case,This is impossible.

(II)Assume|B|=6.Then GBα?A5or S5and Sym(B)=A6or S6.It follows that A5is an insoluble composition factor of Gα.By the structure of Gα,we can easily get thatthen|G:Gα|is not square-free order.

(III)Assume|B|=7.Then GBα?PSL(3,2)and Sym(B)=A7or S7.It follows that PSL(3,2)is an insoluble composition factor of Gα.By the structure of Gα,we can easily get thatthen|G:Gα|is not square-free order.

(IV) Assume |B|=8.Then GBα?23: PSL(3,2) and Sym(B)=A8or S8.It follows that PSL(3,2) is an insoluble composition factor of Gα.By the structure of Gα,we can easily get thatthen|G:Gα|is not square-free order.

Using similar method to the above proposition and carefully computations,we can get the following proposition.

Proposition 3Let G be an almost simple group with alternating socle and let Γ be a connected (G,2)-arc-transitive graph of square-free order.Assume soc(G)=Acand Gα≤G acts intransitively on ?:={1,2,···,c}.Then Γ is isomorphic to one of the odd graphs of square-free order.

Now we are ready to prove the main theorems of the paper.

ProofLet Γ be the(G,2)-arc transitive graphs of square-free order,where G is almost simple with alternating socle.The results of propositions 1 ～3 in this section are sufficient to prove the existence of graphs in Theorem 1 as stated in the introduction.And it is easy to see that they are not isomorphic to each other.At last,we determine the full automorphism group for the graphs occurred in Examples in section 2.The methods are similar,we take Γ1as an example.Let A:=Aut(Γ1).Since Γ1is of valency 8,it follows that ifthen p=2,3,5 or 7.However,we can get p ≠5.In fact,ifthen=A8or S8.Then by checking 2-transitive permutation group of degree 8 such thatwe have=PSL(3,2)=.Therefore,we have A=G=S8.