DOI 10.1007/s10801-007-0101-4
On automorphism groups of quasiprimitive 2-arc transitive graphs
Cai Heng Li
Received: 26 November 2006 / Accepted: 5 September 2007 / Published online: 6 October 2007
© Springer Science+Business Media, LLC 2007
Abstract We characterize the automorphism groups of quasiprimitive 2-arc-transiti- ve graphs of twisted wreath product type. This is a partial solution for a problem of Praeger regarding quasiprimitive 2-arc transitive graphs. The solution stimulates several further research problems regarding automorphism groups of edge-transitive Cayley graphs and digraphs.
Keywords Quasiprimitive·2-arc transitive·Automorphisms·Cayley graphs
1 Introduction
In a graph, an ordered pair of adjacent vertices is called an arc, and a sequence of distinct vertices(u, v, w)is called a 2-arc ifvis adjacent to bothuandw. A graph is called(X,2)-arc transitive, whereX≤Aut, ifXacts transitively on the set of 2-arcs of.
The class of 2-arc transitive graphs is one of the central objects in algebraic graph theory, which have been intensively studied in the literature, see for example [1,7, 12,13,17] and references therein. In particular, Praeger [13] gave a reduction for the study of finite non-bipartite 2-arc transitive graphs, which leads to the study of quasiprimitive 2-arc transitive graphs, defined later. The main purpose of this paper is to characterize the automorphism group for one type of quasiprimitive 2-arc transitive graph.
This work forms part of an ARC grant project and is supported by a QEII Fellowship.
C.H. Li (
)School of Mathematics and Statistics, The University of Western Australia, Crawley, WA 6009, Australia
e-mail:[email protected]
Theorem 1.1 Let be a finite (X,2)-arc-transitive graph. Assume that X has a unique minimal normal subgroupG, which is non-Abelian and regular on the vertex set of. Thenis not 3-arc transitive, and either
(i) has valency 8, andAut=(G×Z32)GL(3,2), or (ii) Gis the unique minimal normal subgroup ofAut.
A permutation groupGon a setis said to be quasiprimitive if every non-trivial normal subgroup ofGis transitive. Praeger [13] showed that only four of the eight types of quasiprimitive permutation groups can occur as automorphism groups of 2-arc transitive graphs, which are described below. LetX be a quasiprimitive per- mutation group on, and let soc(X) be the socle ofX, that is the product of all minimal normal subgroups ofX. The four types involved in the study of 2-arc tran- sitive graphs are
HA Holomorph Affine type:soc(X)is elementary Abelian, regular on. AS Almost Simple type:soc(X)is a non-Abelian simple group.
PA Product Action type:soc(X)is a non-Abelian non-simple minimal normal sub- group ofX, andsoc(X)has no normal subgroup which is regular on. TW Twisted Wreath product type:soc(X)is a non-Abelian non-simple minimal nor-
mal subgroup ofX, andsoc(X)acts regularly on.
A graphis called quasiprimitive(X,2)-arc transitive ifXis quasiprimitive on V and 2-arc transitive on. A classification of quasiprimitive(X,2)-arc transitive graphs of type HA is given in [7]; a description of those of type TW is given in [1];
examples of type PA are given in [12]. In the study of such graphs, the problem of determining their automorphism groups naturally occurs, refer to [15] and [5,7,8].
Theorem1.1determines the automorphism groups for quasiprimitive 2-arc transitive graphs of twisted wreath product type.
2 Preliminary results
In this section we collect notation and preliminary results which will be used in the ensuing sections. For a group X and a subgroup H < X, denote by NX(H ) and CX(H ) the normalizer and the centralizer ofH in X, respectively. Let Z(X) be the center ofX. We need some properties regarding 2-transitive permutation groups, see [3].
Lemma 2.1 LetHbe a 2-transitive permutation group on. Assume thatNHis imprimitive on. ThenHis affine withsoc(H )=Zep, wherepis a prime ande≥1, and further, the following hold:
(i) An imprimitive normal subgroup ofHis of the formZep.Zb, whereb|pe−1 and eis a proper divisor ofe.
(ii) IfN is a subnormal subgroup ofH, then eitherN is transitive on, orN <
soc(H ).
(iii) Hωhas no non-trivial normal subgroup ofp-power order, forω∈.
Letbe an(X,2)-arc-transitive graph with vertex setV. Forv∈V, let(v)= {w∈V |(v, w)is an arc}, the neighborhood ofv. As usual, letX[v1]be the kernel of Xvacting on(v), and letXv(v)be the permutation group induced byXvon(v).
ThenXv(v)∼=Xv/Xv[1]. Forw∈(v), letXvw=Xv∩Xw, andXvw[1]=Xv[1]∩X[w1], the kernel ofXvw acting on(v)∪(w). We observe thatXv∼=Xv[1].Xv(v), and (Xv[1])(w)∼=(X[v1]Xw[1])/X[w1]∼=Xv[1]/X[vw1]. ThusXv[1]∼=X[vw1].(Xv[1])(w), and so
Xv∼=Xv[1].X(v)v ∼=(Xvw[1].(X[v1])(w)).Xv(v). (1) In particular, ifX[vw1]=1, then
X[v1]∼=(X[v1])(w), Xv∼=(Xv[1])(w).Xv(v).
(2) Here we have a fundamental theorem for studying 2-arc transitive graphs, which can be found in [17].
Theorem 2.2 Letbe an(X,2)-arc transitive graph, and let{v, w}be an edge of . Then eitherXvw[1]=1, orX[vw1]is ap-group withpprime and PSL(d, q)X(v)v with|(v)| =qqd−−11.
3 Proof of the main theorem
Letbe an(X,2)-arc transitive graph with vertex setV. Assume thatX is quasi- primitive onV of type TW. LetG=soc(X), the unique minimal normal subgroup ofX, which is regular onV. Letn= |V| = |G|.
We first treat the case where the centralizer CAut(G)is non-trivial.
Lemma 3.1 Assume that CAut(G) =1. Then has valency 8, andAut=(G× Z32)GL(3,2)=GAGL(3,2).
Proof LetC=CAut(G),E=CG, andF =CX. SinceGis regular, we have that E=C×G, andG, EF. ThusX∩E=G, and soXv∩Ev=1 andFv=EvXv. Sinceis connected,Fvacts faithfully on(v), and henceXv∼=Xv(v)≤Fv(v)∼= Fv, andXvandFvare both 2-transitive permutation groups on(v). Inspecting the 2- transitive permutation groups, see [3], it is easily concluded thatEv∼=Z32, andXv∼= GL(3,2)∼=PSL(2,7), which is of degree 8. ThusFv=Z32GL(3,2)=AGL(3,2), and has valency 8. Finally,C ∼=E/G∼=Ev∼=Z32, andC Xv∼=((C×G) Xv)/G=F /G∼=Fv=AGL(3,2).Since AGL(3,2)acts irreducibly on its natural moduleZ32, it follows thatC is a minimal normal subgroup ofF.
Let Y =NAut(G). Then Fv≤Yv, and Yv is a 2-transitive permutation group on (v). Suppose that Fv < Yv. It follows since Fv =AGL(3,2) that Yv=A8
or S8. However, sinceGY andGis non-Abelian and characteristically simple, C=CY(G)∼=CG/GY /G∼=Yv, which is a contradiction. Therefore,Fv=Yv and soF=Y =NAut(G).
Suppose thatF <Aut. LetZ be a group such thatF < Z≤Aut, andF is a maximal subgroup of Z. Then AGL(3,2)=Fv(v)≤Zv(v)≤Sym((v))∼=S8. It follows thatZv(v)=AGL(3,2),A8orS8, andZvw(w)∼=GL(3,2),A7orS7, respec- tively. Since (Z[v1])(w)Z(w)vw , we have that(Zv[1])(w)=1, GL(3,2),A7or S7. By Theorem2.2, we conclude thatZ[vw1]=1, and henceZv∼=(Z[v1])(w).Zv(v).It then follows that either Zv∼=Zv(v), or Zv∼=Zvw(w)×Zv(v). Since Zv> Fv and Fv∼=AGL(3,2),A8orS8, it is easily shown thatZvis one of the following:
A8, S8, GL(3,2)×AGL(3,2), A7×A8, A7×S8, S7×S8. AsF andZare transitive onV,||ZF|| =||ZFvv|| is one of the following numbers:
233.7, 23325.7, 24325.7, 3.5, 2.3.5, 2333527, 2433527 or 2533527.
SinceGis a characteristic subgroup ofF andGis not normal inZ, it follows that F is not normal inZ. Let= [Z:F], the set of right cosets ofF inZ. SinceF is a maximal subgroup ofZ, the action ofZ onis primitive. As Gis not normal inZ, and asG,C are the only minimal normal subgroups ofF, it follows that the kernel of thisZ-action onis trivial or equalsC∼=Z32. If the kernel equalsC, then it follows thatZcentralizesC. However,Xdoes not centralizesC andZ > X, which is a contradiction. Therefore,Z is a primitive permutation group on. Then since
|| = |Z:F|is not a proper power of an integer, it follows from the O’Nan-Scott theorem that Z is an almost simple group. SinceZ =GZv, by [2, Theorem 1.4], soc(Z)∼=Anwheren= |G|, and sois a complete graph, which is a contradiction.
Therefore,Aut=F=GAGL(3,2).
From now on, we assume that
CAut(G)=1.
Suppose thatGis not normal inAut. ThenX =Aut, and there existsZ≤Aut such thatX is a maximal subgroup ofZ. Let N be a minimal normal subgroup of Z. ThenN ≤X. WriteN=T1×. . .×Tl, wherel≥1 andT1∼=. . .∼=Tlare simple groups. We use a series of steps to derive a contradiction.
Step 1.N is non-Abelian.
Suppose thatN is Abelian, sayN ∼=Zlp withpprime. Let E=N G=NG, and letF =N X=NX. SinceGis regular onV, we have that|Ev| = |N| =pl, andEv is ap-group. It follows thatEv(v) is a non-trivialp-group. SinceEF, we have 1 =Ev(v)Fv(v), and since Fv(v) is a 2-transitive permutation group, Fv(v) is affine. It then follows thatEv∼=Ev(v), andsoc(Fv(v))=E(v)v ∼=Zlp. By Theorem2.2,Fvw[1]=1, and hence by Formula (2),
pl= |N| =|F|
|X|= |Fv|
|Xv| =|Fv(v)|
|X(v)v |.|(Fv[1])(w)|.
NowXv(v) andFv(v) are affine 2-transitive permutation groups of degreepl such that X(v)v is a subgroup ofFv(v) of index dividingpl, and (Fv[1])(w) is a normal subgroup ofFvw(w)of order dividingpl. Inspecting the classification of 2-transitive permutation groups of degree pl (see [3]), we conclude that Fv(v)=X(v)v and (Fv[1])(v)=1, which is a contradiction. ThusN is non-Abelian.
Step 2. N is transitive onV, and eitherG < N, or each prime divisorr of|T|is such thatrl |V|.
IfG < N, thenN is transitive onV. Suppose thatG< N. SinceGis a minimal normal subgroup of X and G∩N is normalized by X, it follows that G∩N = 1. Since CAut(G)=1 andG is insoluble, G and so X permutes by conjugation {T1, . . . , Tl}non-trivially. AsGis a minimal normal subgroup ofX, it follows thatG permutes by conjugation the collection{T1, . . . , Tl}, and soG≤Sym(l). In particular,
|G|dividesl!.
Suppose that there exists a primerdividing|T|such thatrl divides|V|. Thenrl divides|G|, and sorl dividesl!. However, it is known that the highest power ofr dividing l!is at mostrrl−−11. Hencerl≤rrl−−11, and so l≤rl−−11≤l−1, which is not possible. Thus, if a primer|T|, then rl |V|. In particular,N is not semiregular onV. Sinceis not bipartite, it follows thatN is transitive onV.
Step 3.Zis quasiprimitive onV of type PA, andG < N.
By Step 2, any minimal normal subgroup of Z is transitive on V, and so Z is quasiprimitive onV, which has an irregular minimal normal subgroupN. By [13], Zis of type HA, AS, TW or PA. SinceN is not regular,Zis neither of type HA nor of type TW. IfZ=GZvis almost simple, then by [2, Theorem 1.4],soc(Z)≥An
wheren= |V|andZis 2-transitive onV. Thusis a complete graph, and soXis 2-transitive on V, which is not possible. ThusZ is of type PA, and so there exists a primer|T|such thatrl|V|. By Step 2,G < N.
Step 4. The soclesoc(Xv)≤Nv.
Suppose soc(Xv) ≤Nv. Let Y =N X. Then N ∩Xv=1, and N Y. Hence NvYv, andYv/Nv∼=Y /N∼=Xv. SoYv=NvXv; in particular,|(v)|2divides
|Yv|. Suppose thatYvw[1]=1. Then by Formula (2),Yv∼=(Yv[1])(w).Yv(v). Noting that (Yv[1])(w)G(w)vw , it is easily shown thatYvdoes not have the formNvXvsuch that both Nv andXv are transitive on(v), which is not possible. ThusYvw[1] =1, and by Theorem 2.2,soc(Yv(v))=PSL(d, q) with|(v)| = qqd−−11. Since Yvw[1] is a p-group and(Yv[1])(w)G(w)vw , it follows that|(v)|2does not divide|Yv|, which is a contradiction. Thus,soc(Xv)≤Nv.
LetG=S1×. . .×Sk, wherek≥2 and S1∼=. . .∼=Sk are non-Abelian simple groups. Then Xv acts by conjugation transitively on {S1, . . . , Sk}, and soc(Xv)is half-transitive on{S1, . . . , Sk}, that is, all orbits ofsoc(Xv)on{S1, . . . , Sk}have equal size.
Step 5.k=l, and for eachi∈ {1,2, . . . , l},Ti∩G=Sjfor some 1≤j≤k.
LetH=T1×. . .×Tk−1. SinceZis of type PA, the proper normal subgroupH is intransitive on V. LetB be the set of H-orbits onV, and let B∈B contain v.
ThenNB∼=N/H∼=T, andGB∼=G/(G∩H )∼=Smwherem≥1. In particular,|S| divides|T|. For convenience, writeT =NB∼=T. NowGB is a transitive subgroup ofNB, soT ∼=T =GBTB∼=SmTB.
Suppose thatm >1. By [2, Theorem 1.4],T ∼=AnandTB∼=An−1withn≥10.
Letrbe the largest prime which is less thann. Thenrexactly divides|T|and|TB|. So r does not divide |S|. Further,rl exactly divides|N|, and so rl divides |Nv|.
If Nv is primitive on (v), then H is semiregular on V, and Nv∼=TB, which is a contradiction. HenceNvis imprimitive on(v). By Lemma2.1,Nvis soluble, and soTB∼=Nv/Hvis soluble, which is again a contradiction.
Thusm=1, and soSk−1< Tl−1. SinceG∩HG, we may assume thatS1× . . .×Sk−1< H. IfSk< Tl, then asTl∩GGwe haveTl∩G=1. ThusSk=G <
N/Tk∼=Tl−1, which is not possible. HenceSk< Tl, andTl∩G=Sk. It follows that for eachi,Ti∩G=Sjfor somej, andk=l.
Step 6. It is not possible forGnot to be normal inAut.
It follows from Step 5 that|Nv| = |T :S|k. Sincesoc(Xv)≤Nv, it follows that soc(Xv)acts trivially on {S1, . . . , Sk}, and so soc(Xv)normalizes each Sj. Since CX(G)=1, we conclude thatsoc(Xv)≤Out(Sj). In particular,soc(Xv)is soluble, by ‘Schreier’s Conjecture’, and soXvis an affine 2-transitive permutation group on (v), say degree|(v)| =pewithp prime. Inspecting outer-automorphism groups of simple groups, we conclude that eitherpe=23, ore≤2. It follows thatNv(v)≥ soc(Yv(v))=Zep,Ape, or PSL(d, q)with eithere=1 and qqd−−11=porpe=23. For the former two cases,Nvw[1]Yvw[1]=1, and henceNv∼=(Nv[1])(w).Nv(v). It is now easily shown that|Nv|is not a proper power of an integer, which is a contradiction.
Thus Nv(v)≥soc(Yv(v))=PSL(d, q) or PSL(2,7); in particular, is (N,2)-arc transitive. Ifsoc(Yv(v))=PSL(d, q), thenp=qqd−−11divides|Nv|, butp2does not, so
|Nv|is not a proper power of an integer, which is a contradiction. Thussoc(Yv(v))= PSL(2,7). It follows that |Nv| is not a proper power of an integer, which is again a contradiction.
Now we summarize the argument for proving Theorem1.1as follows.
Proof of Theorem1.1: By Steps 1-6, we have thatGis normal inAut. If CAut(G) = 1, then by Lemma3.1, part (i) of Theorem1.1holds. If CAut(G)=1, thenGis the unique minimal normal subgroup ofAut, as in part (ii) of Theorem1.1. Finally, by
[9, Proposition 2.3],is not 3-arc transitive.
4 Some related problems
Letbe anX-arc transitive graph with vertex setV . LetNXhave at least three orbits onV , and letBbe the set ofN-orbits inV . We define a graphN to have
vertex setBsuch thatB, C∈Bare adjacent if and only if someu∈Bis adjacent in to somev∈C, called the normal quotient ofinduced byN. IfandN have equal valency, thenis a normal cover ofN. An(X,2)-arc transitive graphis always a normal cover ofN, and choosingNto be maximal, each non-trivial normal subgroup ofX/Nis transitive onV N, that is,X/Nis quasiprimitive onV N. This is the reduction for the study of 2-arc transitive graphs to the quasiprimitive case, given by Praeger [13].
A graphwith vertex setV is called a Cayley graph if there exist a groupGand a subsetS⊂GwithS=S−1= {s−1|s∈S}such thatV is identified withGand x, y∈G are adjacent if and only ifyx−1∈S. The Cayley graph is denoted by Cay(G, S).
A Cayley graph=Cay(G, S)has an automorphism group Gˆ = { ˆg: x→xgfor allx∈G|g∈G},
consisting of right multiplications of elements ofG. The subgroupGˆ acts regularly on the vertex set of; in particular,is vertex-transitive. Let
Aut(G, S)= {σ∈Aut(G)|Sσ=S},
the subgroup of automorphisms ofG and fixS setwise. It is easily shown that all elements ofAut(G, S)are automorphisms offixing the vertex ofcorresponding to the identity ofG. Ifis connected, or equivalentlyS =G, thenAut(G, S)acts faithfully onS. Further,Aut(G, S)normalizesG, andˆ
NAut(G)ˆ = ˆGAut(G, S)≤Aut.
Thus, ifAut(G, S)is 2-transitive onSthenis(NAut(G),2)-arc-transitive.ˆ Much structural information of is contained in the full automorphism group Aut, such as the degree of symmetry of, and the isomorphism class ofamong Cayley graphs ofG(refer to [10]). Generally,Autis larger thanGˆ Aut(G, S), see for example [4,6,11,16]. Theorem1.1can be restated in the Cayley graph version.
Theorem 4.1 LetGbe a non-Abelian characteristically simple group, and let= Cay(G, S)be a connected Cayley graph. Assume that there exists a subgroup X≤ Autsuch that Gˆ is a unique minimal normal subgroup of X, and assume further thatis(X,2)-arc-transitive. ThenAut= ˆGAut(G, S),is not 3-arc transitive, and further either
(i) has valency 8, andAut=(Gˆ×Z32)GL(3,2); or (ii) Gˆ is the unique minimal normal subgroup ofAut.
A natural question is whether the conclusion of Theorem4.1is always true for arbitrary groupsG. Here is a counter-example.
Example 4.2 Let G=Nz∼=Zep Z2 withp odd prime ande≥1 such that xz=x−1 for allx∈N. LetS=G\N, and let =Cay(G, S). Then S consists of all involutions of Gwhich are conjugate, and=Kpe,pe, a complete bipartite
graph. It is easily shown that Aut(G, S)=Aut(G)∼=AGL(e, p)=ZepGL(e, p) andAut(G, S)acts 2-transitively onS. However,Aut∼=SpeS2is much bigger than NAut(G)ˆ = ˆGAGL(e, p).
However, Kpe,pe and their normal covers are the only known examples such that Aut(G, S)is 2-transitive andGˆ is not normal inAut. This motivates the following conjecture:
Conjecture 4.3 For a connected Cayley graph =Cay(G, S), if Aut(G, S) is 2-transitive on S, then either is a normal cover of Kpe,pe withp odd prime, or Aut= ˆGAut(G, S).
For a groupG, a Cayley graph=Cay(G, S)is called normal ifGˆ is normal in Aut, see [18]. Then Conjecture4.3means that most Cayley graphsCay(G, S)are normal provided thatAut(G, S)is 2-transitive onS.
More generally, we would like to propose several further problems regarding au- tomorphism groups of Cayley graphs. We believe their solutions would be interesting for a better understanding of Cayley graphs.
Naturally, one would ask whether the condition in Conjecture4.3thatAut(G, S) is 2-transitive onScan be weakened to the condition thatAut(G, S)is only transitive onS, refer to [4,11,16,18].
Problem 4.4 Characterize connected Cayley graphs = Cay(G, S) such that Aut(G, S)is transitive onS andAut = ˆGAut(G, S). Especially, do this for the caseAut(G, S)is primitive onS.
By [16, Theorem 1], Problem4.4was solved for the case whereGis non-Abelian simple andis cubic, that is, it was shown that ifGis a non-Abelian simple group andAut(G, S)is transitive onSwith|S| =3 thenAut= ˆGAut(G, S).
Example 4.5 For a Fermat prime p=2d−1, the complete graph=K2d of 2d vertices is a Cayley graph Cay(G, S) of G=Zd2 such that Aut(G, S)∼=GL(d,2), which is primitive on S, and Aut=Sym(2d). In particular, for d≥3, Gˆ is not normal inAut.
Cayley graphsCay(G, S)with the property thatAut(G, S)is transitive onS are called normal arc-transitive Cayley graphs. A study of such Cayley graphs was ini- tiated in [16]. We wonder for a 2-arc-transitive Cayley graphCay(G, S)whether the transitivity ofAut(G, S)onSimplies the 2-transitivity:
Question 4.6 Does there exist a 2-arc-transitive Cayley graphCay(G, S)such that Aut(G, S)is transitive but not 2-transitive onS?
Cayley graphs defined above are undirected. Of course one may define directed Cayley graphs and ask similar questions regarding their automorphism groups to the undirected case. We are inclined to conjecture for a directed Cayley graph
=Cay(G, S), ifAut(G, S)is 2-transitive on S, thenAut(G, S)is also very close to(Aut)vwherevis the vertex ofcorresponding to the identity of the groupG.
Question 4.7 Let=Cay(G, S)be a connected directed Cayley graph, and assume further thatAut(G, S)is 2-transitive onS. DoesAutequalGˆAut(G, S)?
Praeger [14] gave a description of the bipartite 2-arc-transitive graphs. Letbe an (X,2)-arc-transitive bipartite graph with partsand, where X≤Aut. Let X+=X=X. The graphis said to beX-bi-quasiprimitive if each non-trivial normal subgroup has at most two orbits and at least one has two orbits. It is shown in [14] that the bi-quasiprimitive case is an important case for understanding bipartite 2-arc-transitive graphs, and if X+ is bi-quasiprimitive thenX+ is of type HA, AS, TW or PA. Motivated by Theorem4.1, we propose
Conjecture 4.8 Letbe a connected(X,2)-arc-transitive bipartite graph with parts and , where X ≤Aut. Assume further that X+ is quasiprimitive on of type HA or type TW. Then either=Kpe,pe withpprime, orsoc(X+)is normal in(Aut)+.
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