Characterization of Groups L2(q)by NSE Where q∈ {17,27,29}∗

Changguo SHAO Qinhui JIANG

1 Introduction

Throughout this paper,all groups are finite andGis always a group.We denote byπ(G)the set of prime divisors of|G|,and byπe(G)the set of element orders ofG.Ifris a prime divisor of the order ofG,thenPrdenotes a Sylowr-subgroup ofGandnr(G)denotes the number of Sylowr-subgroups ofG.Letnbe an integer.We denote byϕ(n)the Euler function ofn.Gis called a simpleKn-group ifGis simple such that

The prime graph GK(G)of a groupGis de fined as a graph with the vertex setπ(G).Two distinct primesare adjacent ifGcontains an element of orderpq.Moreover,the connected components of GK(G)are denoted bywheret(G)is the number of connected components ofG.In particular,we de fine byπ1the component containing the prime 2 for a group of even order.

The motivation of this article is to investigate Thompson’s Problem as follows(see[1,Problem 12.37]).

WriteandG2are of the same order type if and only if

Thompson’s ProblemSuppose that G1and G2are of the same order type.If G1is solvable,is it true that G2is also necessarily solvable?

Unfortunately,so far,no one could prove it completely,or even give a counterexample.

Letk∈πe(G)andmk(G)be the number of elements of orderkinG.Let nse(G):=,the set of numbers of elements with the same order.If groupsG1andG2are of the same order type,we clearly see thatand nse(G1)=nse(G2).So it is natural to investigate the Thompson’s Problem by|G|and nse(G).Notice that not all groups can be characterized by nse(G)and|G|.For instance,in 1987,Thompson gave an example as follows:Letandbe two maximal subgroups ofM23,whereM23is a Mathieu group of degree 23.Then nse(G1)=nse(G2)andUnfortunately,

The authors of[2]proved that all simpleK4-groupsGcan be uniquely determined by nse(G)andLater,Asboei et al.[3]characterized sporadic simple groups by nse(G)and|G|.The authors of this paper proved(see[4])that linear groupsL2(q)are characterizable by their ordersand the set nse(L2(q)),ifq=2a−1 or 2a+1 is a prime.On the other hand,some groups can be determined uniquely by the set nse.For instance,it is proved(see[5])thatandL2(9)are uniquely determined by nse(G).Khatami,Khosravi and Akhlaghi[6]proved that simple groupsL2(q)are characterizable uniquely by the set nse(L2(q))ifq∈{7,8,11,13}.Moreover,Zhang and Shi[7],Asboei and Amiri[8]proved thatL2(q)can be characterized uniquely by the set nse(L2(q)),whereq∈{16,25}.In this present paper,by introducing the prime graph of a group as a different method,we go on characterizing linear groupsL2(q)whenq∈{17,27,29}.Our result is the following theorem 1.1.

Theorem 1.1Let G be a group and q∈{17,27,29}.Thenif and only ifnse(G)=nse(L2(q)).

We denotenr(G)bynrandmk(G)bymkif there is no confusion.Further unexplained notation is standard,and readers may refer to[9].

2 Preliminaries

In this section,we give some lemmas which will be used in the sequel.

Lemma 2.1Let G be a group.Ifnse(G)and,then the following statements hold:

(1)

(2)m2=n;

(3)for any

ProofLet 1/=t∈πe(G)andkbe the number of cyclic subgroups ofGwith ordert.ThenIft＞2,thenϕ(t)is even,so ismt.Hencemt/=nsincenis odd.As a result,m2=nand 2∈π(G),as required.

Lemma 2.2(see[10])Let G be a group and m be a positive integer dividing|G|.Ifthen

Lemma 2.3(see[4,Lemma 2.3])Let G be a group and P be a cyclic Sylow p-subgroup of G.Assume further thatand r is an integer such that par∈πe(G).Then mpar=mr(CG(P))mpa.In particular,ϕ(r)mpa|mpar.

Lemma 2.4(see[11])Let G be a group and p∈π(G)be odd.Suppose that P is a Sylow p-subgroup of G and n=psm,where(p,m)=1.If P is not cyclic and s＞1,then the number of elements of order n is always a multiple of ps.

Recall thatGis a 2-Frobenius group ifGhas a normal seriessuch thatG/HandKare Frobenius groups withK/HandHbeing Frobenius kernels,respectively.

Lemma 2.5(see[12,Theorem 2])If G is a 2-Frobenius group of even order,then t(G)=2and G has a normal seriessuch thatand K/H are cyclic.In particular,and G is solvable.

Lemma 2.6(see[13,Theorem A])Let G be a group such that t(G)≥2.Then G has one of the following structures:

(a)G is a Frobenius or 2-Frobenius group.

(b)G has a normal such thatand G1/N is a nonabelian simple group.

Lemma 2.7Let G be a simple group.If{2,3,7,13},thenor L2(27);if

ProofThis follows immediately by[14,Theorem 2],[15,Corollary 1]and[16,Theorem].

Lemma 2.8(see[2,Lemma 2.5])Let G be a group with a normal series:Suppose that whereIf P≤L and |K|,then the following statements hold:

(1)

(2)that is,for some positive integer t.Furthermore,

3 Proof of Theorem 1.1

Proof of Theorem 1.1The necessity is obvious,so we only prove the sufficiency.Letn∈πe(G)andkbe the number of cyclic subgroups of orderninG.Thenmn=k·ϕ(n).In particular,

We divide the proof into three cases.

Case 1nse(G)={1,153,272,306,612,816,288}=nse(L2(17)).

By Lemma 2.1,we see that 2∈π(G)andm2=153.Further,Lemma 2.2 indicates that{2,3,7,13,17,19,43,307,613}.If 13∈π(G),then Lemma 2.2 implies thatm13=272 andyieldingwhich is not an integer,a contradiction.Similarly,we haveSuppose 307∈π(G).Thenm307=306 andby Lemma 2.2.We claim thatP307acts fixed-point-freely on Ω2:={all elements of order 2 inG}.If not,then 307·2∈πe(G).By Lemma 2.3,we havewhich leads tom2·307=306 or 612.However,both cases indicate thata contradiction to Lemma 2.2.Analogously,we obtain that 19,43,613/∈π(G),and thusπ(G)⊆{2,3,17}.Now we prove that the equality holds.

Assume that exp(P2)=2s.Then by(3.1)we haveϕ(2s)|m2s,yieldings≤6.Analogously,if 3∈π(G),then exp(P3)≤33withm3=272,m9=816,m27=612 or 288;if 17∈π(G),thenand exp(P17)=17.Further,if exp(P3)=32or 33,thenP3is cyclic by Lemma 2.4.

Suppose thatNotice thatand exp(P17)=17.We show thatIf not,a contradiction to our assumption.Further,by(3.1).As a result,withi≥3 becauseP17acts fixedpoint-freely on Ω2i:={all elements of order 2iinG},again a contradiction to Lemma 2.2.This shows that exp(P2)≤22.In this case,Lemma 2.2 implies thatleading toHowever,a contradiction.Therefore,π(G)={2,3,17},as required.

We prove that|P17|=17.Assume that this is false.Then.If exp(P3)=33,thenP3is cyclic,implyingHence 17Letbe a group of order 17.Thenby Sylow’s theorem,implyingIt follows by Lemma 2.3 thatNote thatm27=612 or 288,which is a contradiction.Suppose that exp(P3)=3.By Lemma 2.2,|P3|=3.Moreover,By the same argument as above there is a contradiction,leading to exp(P3)=32andMoreover,and thus 17LetThenBy Sylow’s theorem,we obtain thatand thus 3·17∈πe(G).Howevera contradiction to Lemma 2.2.Henceas required.

If there is some primesuch thatthenby Lemma 2.3.Further,r=2 andm34=m17.However,contradicting Lemma 2.2.As a result,t(G)≥2.

Assume first thatis a Frobenius group with the kernelKand a complementH.Ast(G)=2,we see thatπ1={2,3}andπ2={17}as there is no element of order 17rfor each primerdistinct from 17.Then either|H|=17 or|K|=17.If the latter holds,then|H||16 and|G||16·17,a contradiction.Hence|H|=17.Moreover,is also a Frobenius group with a kernelK3and a complementH,yielding toHowever,which is a contradiction.LetGbe a 2-Frobenius group.ThenGhas a normal seriessuch that|K/H|=17 and|G/K|||Aut(K/H)|.Henceand thusis also a Frobenius group with a Frobenius kernelH3and a complementC17.By the same reasoning as above,this is also a contradiction.Hence by Lemma 2.6,Ghas a normal seriessuch thatK/His a simpleK3-group sinceBy Lemma 2.7,we getMoreover,Lemma 2.8 implies thatand,we havet=1 and thusSo we obtain thatSincewe getH=1 andNote thatThen we haveG=K·2 orG=K.IfG=K·2,then by[9],we obtain thata contradiction.Hence

Case 2nse(G)={1,351,728,2106,4536}=nse(L2(27)).

By Lemma 2.1,we see that 2∈π(G)andm2=351.Notice that 1+728=36,1+2106=72·43,1+4536=13·349.Thenby Lemma 2.2.Assume that 43∈π(G).Then Lemma 2.2 implies thatm43=2106.Further,exp(P43)=43 andwhich leads to thatP43is a cyclic group of order 43.Hencewhich is not an integer,a contradiction.Similarly,349/∈π(G)and thusπ(G)⊆{2,3,7,13}.We show that the equality holds.

Assume exp(P2)=2s.Then by(3.1)we obtain thatleading tos≤4.If the equality holds,thenwhich is contrary to the fact thatby Lemma 2.2.Thus exp(P2)≤23.Furthermore,On the other hand,ifthenaccording to Lemma 2.2.Further,by a similar argument as above.Assume that 13∈ π(G).Then,yielding thatπ(G)={2,3,7,13},as required.Supposethenwhich also implies thatπ(G)={2,3,7,13}.Hence we may assume that 13/∈π(G)andNote thatGis neither a 2-group nor a{2,7}-group becauseAs a result,3∈π(G).If 7∈π(G),thenGis a{2,3,7}-group with|πe(G)|≤4·5·2=40.Therefore,

indicatingb=3,since,otherwise,3|11·13,a contradiction.Moreover,implyingIt follows that eithera=5,k3=5 ora=6,k3=10 asa≤6.However,in these two cases,no integerk2satis fies(3.4).Hencek1=27.Then(3.3)is equivalent to

Thusimplyingk2=1 ork2=8.If the latter holds,thenk3=0.However,k1=27,k2=8,k3=0 is not a solution of(3.2).Hencek2=1.However,in this case,(3.5)is equivalent to

This shows that,and thusa=2.Moreover,2k3=3b−4−13,contradicting the factb≤6.Therefore,b=1 or 2.The similar argument as above will also deduce a contradiction.

The remaining case isAsGis not a 2-group,we see thatMoreover,

Easily,5≤b≤6 implies thatk1=0 and thus(3.7)is equivalent to

leading to 3|11·13,again a contradiction.As a result,π(G)={2,3,7,13},as required.

Recall that|P13|=13.We claim thatfor eachs∈π(G)distinct from 13.Otherwise,Lemma 2.3 indicates thatButa contradiction to Lemma 2.2.Hencet(G)≥2.Assume first thatGis a Frobenius group.Thent(G)=2 withπ1={2,3,7}andπ2={13}.WriteSuppose first that 13||K|.SinceKis nilpotent,we obtain thatwhereK13is the Sylow 13-subgroup ofK,a contradiction.Hence 13||H|and thus 2,3,7∈π(K).LetK7be a Sylow 7-subgroup ofK.Thenis also a Frobenius group with a kernelK7and a complementH.This implies that 1contrary to the fact thatSuppose further thatGis a 2-Frobenius group.Then by Lemma 2.5 we see thatGhas a normal serieswith|K/H|=13 and|G/K||12,leading to 7||H|.SinceHis nilpotent,we getm7=6 or 48,again a contradiction.

Consequently,by Lemma 2.6 we see thatGhas a normal serieswhereK/His either a simpleorK4-group and 13||K/H|.Assume first thatK/His a simpleK3-group.Thenby[14,LemmaNote thatThen we obtain that 7||H|.SinceHis nilpotent,we getm7=6 or 48,a contradiction.HenceK/His a simpleK4-group.By Lemma 2.7,we haveIfthenn7(K/H)t=n7by Lemma 2.8.Sincen7=33·13 andn7(K/H)=2·3·13 according to[9,p.8],this is a contradiction.HenceAgain by applying Lemma 2.8,we obtain thatandHencet=1 sinceMoreover,leading toNote thatThis implies thatH=1 and therefore,and.Assume|G/K|=2,and thenG=L2(27)·2.By[9],m2=729,a contradiction.Similarly,the cases|G/K|=3 and|G/K|=6 also imply a contradiction.This showsas wanted.

Case 3nse(G)={1,435,2610,812,1624,3248,840}=nse(L2(29)).

Similar to the proof of Case 1,we obtain thatandm2=435.Moreover,if 3,5,7,29∈π(G),thenm3∈{812,3248},m5=1624,m7=2610 andm29=840.Suppose that exp(P2)=2a.Sinceϕ(2a)|m2aandm2a∈nse(G),along with Lemma 2.2,we geta≤5.Moreover,By the same reasoning,if 3,5,7,29∈ π(G),we obtain that,exp(P5)=5,and exp(P29)=29.

We prove thatπ(G)={2,3,5,7,29}.Assume first 7∈π(G).Thenwhich implies thatπ(G)={2,3,5,7,29},and we are done.As a result,we may assume that

Suppose thatThenwhich leads to

Thenk1must be odd.Ifk1=1,then

has no solutions.Ifk1=3,then(3.11)becomes

leading toa=2,which also is impossible.Hence,yieldinga contradiction.

Suppose thatπ(G)={2,3,5}.If exp(P3)=32andP3is cyclic,thenthenn5=2·7·29,a contradiction.Assume that|P5|=52.By Lemma 2.3,we haveHenceP5acts fixed-point-freely on Ω9:={all elements of order 9 inG}.So we havewhich is contrary toThe same argument implies thatMoreover,P3is non-cyclic.Note thatm3=3248.Thensince,otherwise,contrary to Lemma 2.2.As a result,P3acts fixed-point-freely on Ω5:={all elements of order 5 inG}.Thuswhich is a contradiction since.Hence exp(P3)=3.IfP3is cyclic,thenbecausem3=812 or 3248,also a contradiction.As a consequence,andm3=3248.If 15∈πe(G),thenm15=1624,3248 or 840.By Lemma 2.2,we see thatwhich is a contradiction.ThenP5acts fixedpoint-freely on Ω3:={all elements of order 3 inG}.So we have thatcontradictingThis indicates thatBy the same reasoning,there is also a contradiction,leading to thatAnalogously,and therefore,as required.

Recall thatThenWe prove thatIf not,implyingLetThenby Sylow’s theorem.Henceand thusBy Lemma 2.3,it follows thatcontradictingWe prove thatfor eachdistinct from 29.Otherwise,by Lemma 2.3.This forcesr=2.However,contrary to Lemma 2.2.Consequently,Assume thatandThenby Sylow’s theorem.So we haveand thus 5·29∈πe(G),also a contradiction.Hence

Therefore,Assume first thatGis a Frobenius group.Thust(G)=2 withandπ2={29}.WriteIf 29thencontrary to our assumption.Thusand 7LetK7be a Sylow 7-subgroup ofK.AsKis nilpotent,we see thatThis contradiction implies thatGis a 2-Frobenius group.Moreover,Lemma 2.5 implies thatGhas a normal seriessuch thatandleading toSinceHis nilpotent andwe obtain thatm5=|H5|=4,which contradicts our assumption.Further,Lemma 2.5 indicates thatGis non-solvable and has a normal seriessuch thatis a simple group,29andBecause there is no simpleK3-group whose order is divisible by 29,we see thatK/His a simpleK4orK5-group.By Lemma 2.7,we see thatOn the other hand,Lemma 2.8 implies thatandThust=1 andyielding toNote that there is no element of order 29rforr∈π(G).ThenH=1 and thusMoreover,G=K·2 orG=K.If the former holds,it follows by[9]thatm2=841,a contradiction.HenceG=K∼=L2(29)and the theorem is established.

AcknowledgementThe authors would like to thank the referees for their valuable suggestions.It should be said that we could not have polished the final version of this paper well without their outstanding efforts.

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