Recognizing the Automorphism Groups of Mathieu Groups Through Their Orders and Large Degrees of Their Irreducible Characters∗

Yanxiong YANLiangcai ZHANGHaijing XUGuiyun CHEN

1 Introduction

All groups considered are finite groups and all characters are complex characters.Let G be a group and Irr(G)the set of all irreducible complex characters of G.Also,we denote the set of character degrees of G by cd(G)={χ(1)|χ ∈ Irr(G)}.In this paper,we will refer to character degrees as degrees.We use cd∗(G)to denote the multi-set of degrees of irreducible characters,i.e.,each element of this set cd∗(G)can occur many times upon the number of characters of the same degree.In particular,|cd∗(G)|=|Irr(G)|.H ·M denotes the non-split extension of H by M and H:M the split extension of H by M.For any group G,L1(G)and L2(G)denote the largest and the second-largest irreducible character degrees of G,respectively.All the other notations and terminologies are standard(cf.[1]).

It is a well-known fact that characters of a group can give some important information about the group’s structure.For example,Chen[2]proved that a non-abelian simple group can be uniquely determined by its character table.In[3],Huppert posed the following conjecture.

Huppert’s ConjectureLet H be any non-abelian simple group,and G a group such that cd(G)=cd(H).Then GH×A,where A is an abelian group.

Huppert conjectured that each non-abelian simple group G is characterized by cd(G),the set of degrees of its complex irreducible characters.In[3–5],he con firmed that the conjecture holds for the simple groups,such as L2(q)and Sz(q).Moreover,he also proved that the conjecture follows for 19 out of 26 sporadic simple groups,and a few others(cf.[3–5]).In[6–7],Daneshkhah,et al.showed that the conjecture holds for another three sporadic simple groups Co1,Co2and Co3.Xu,et al.attempted to characterize the finite simple groups by less information of its characters,and for the first time successfully characterized the simple K3-groups and sporadic simple groups by their orders and one or both of its largest and secondlargest irreducible character degrees(cf.[8–10]).For convenience,we summarize some results of these articles which will be used later in the following Proposition 1.1.

Proposition 1.1(cf.[9])LetGbe a finite group andMa Mathieu group.Then the following assertions hold:

(i)IfMis one ofM11,M12andM23,thenGMif and only if|G|=|M|andL1(G)=L1(M).

(ii)IfM=M24,thenGM24if and only if|G|=|M24|andL2(G)=L2(M24).

(iii)IfM=M22,thenGM22orH×M11,whereHis aFrobeniusgroup with an elementary kernel of order8and a cyclic complement of order7,if and only if|G|=|M22|andL1(G)=L1(M22).

In this article,we continue this investigation,and show that the automorphism groups of Mathieu groups can also be characterized by their orders and at most two irreducible character degrees of their character tables.

We obtain the following main Theorem 1.1.

Theorem 1.1LetGbe a finite group and|G|=|Aut(M)|,whereMis a Mathieu group.Then the following assertions hold:

(1)IfM=M11orM23,thenGAut(M)if and only ifL1(G)=L1(Aut(M)).

(2)IfM=M24,thenGAut(M)if and only ifL2(G)=L2(Aut(M)).

(3)IfM=M12,thenGis isomorphic to one of the groupsAut(M12),2·M12and2×M12if and only ifL1(G)=L1(Aut(M))andL2(G)=L2(Aut(M)).

(4)IfM=M22,thenGAut(M)or2 ·Mif and only ifL1(G)=L1(Aut(M))andL2(G)=L2(Aut(M)).

2 Preliminaries

In this section,we consider some results which will be applied for our further investigations.

Lemma 2.1LetGbe a finite solvable group of order,whereq1,q2,···,qsare distinct primes.If(kqs+1)for eachi≤s−1andk>0,then the Sylowqs-subgroupis normal inG.

ProofLet N be a minimal normal subgroup of G.Since G is solvable,then we have|N|=qm.If q=qs,by induction on G/N,it is easy to see the normality of the Sylow qs-subgroup in G.Now,assume that q=qifor some i

Lemma 2.2(cf.[8])LetGbe a non-solvable group.ThenGhas a normal series1HKG,such thatK/His a direct product of isomorphic non-abelian simple groups and|G/K|||Out(K/H)|.

Lemma 2.3LetGbe a non-solvable group.Suppose thatGhas a normal series1HKGsuch thatK/HMis a non-abelian simple group withMult(M)=1andHhas a normal series:H=H1H2···Ht=1such that

(1)HiK;

(2)Hi−1/Hiis abelian andAut(Hi−1/Hi)does not contain any simple section isomorphic toM,wherei=2,3,···,t.

ThenKhas a normal series1H1Ksuch thatH1M.Moreover,if|Out(M)|=1,thenGM×Tfor some subgroupT≤G.

ProofBy hypotheses,we have H1/H2K/H2.Considering the conjugate action of K/H2on H1/H2,we can obtain that K/H2/CK/H2(H1/H2)Aut(H1/H2).Since Hi−1/Hiis abelian,we have H1/H2≤ CK/H2(H1/H2).Hence the factor group K/H2CK/H2(H1/H2)M or 1.By assumption,we have that K/H2=CK/H2(H1/H2),i.e.,H1/H2=Z(K/H2).Since Mult(M)=1,one has that K/H2=M×H1/H2.Therefore,K has a normal series H2K1K such that K1/H2M and K/K1H1/H2.

Repeating the process of the above reasoning for the normal series H3H2K1,we can get that K1has a normal series H3K2K1such that K2/H3M,K1/K2H2/H3and K1/K3=M×H2/H3.Repeating the process of the above argument,we can get that K has a normal series 1=Kt+1Kt···K1K,where Kt=M.

It is easy to see that H is solvable by(2),and therefore,K has M as its unique simple normal factor,which implies that H1is a characteristic subgroup of K,so H1G.Therefore,GH1× T since|Out(H1)|=1,as desired.

Remark 2.1Let S be a Mathieu group,and then S is isomorphic to one of M11,M12,M22,M23and M24.By[1],we can obtain|S|,|Out(S)|,Mult(S)(the Schur multiplier)and|Aut(S)|.Applying the software Magma(V2.11-1)(cf.[11]),it is easy to compute the values of L1(Aut(S))and L2(Aut(S)).For convenience,we have tabulated the results in Table 1.

Table 1

Remark 2.2Let S be a K3-simple group,and then S is one of A5,A6,L2(7),L2(8),L2(17),L3(3),U3(3)and U4(2).By[1],we can get|S|,|Out(S)|,Mult(S)(the Schur multiplier)and|Aut(S)|.Using the software Magma(V2.11-1),we can compute the values of L1(Aut(S))and L2(Aut(S)),and we have tabulated them in Table 2.

Table 2

3 Proof of Main Theorem

Remark 3.1A group G is called an almost-simple group related to S if S≤G≤Aut(S),where S is a non-abelian simple group.

Proof of Theorem 1.1By Table 1,if M is isomorphic to one of M11,M23and M24,then|Out(M)|=1,and hence Aut(M)=M.By Proposition 1.1,we see that conclusions(1)and(2)hold in this case.In the following,we only need to discuss that the remaining conclusions hold while M=M12or M22.Obviously,it is enough to prove the sufficiency.We write the proof by what M is.

Case 1We are to prove that the theorem follows if M=M12.

In this case,one has that|G|=27·33·5·11,L1(G)=24·11 and L2(G)=24·32by hypotheses and Table 1.Let χ,β ∈ Irr(G)such that χ(1)=24·11 and β(1)=24·32.

We first assert that G is non-solvable.If G is solvable,then by Lemma 2.1,G11is normal in G,where G11∈ Syl11(G).Hence,χ(1)||G:G11|=27·33·5,a contradiction.Therefore,G is non-solvable,so the assertion is true.

By Lemma 2.2,G has a normal series 1HKG such that K/H is a direct product of nonabelian simple groups which are pairwise isomorphic to each other and|G/K|||Out(K/H)|.Since|G|=27·33·5·11,we deduce that K/H can only be isomorphic to one of A5,A6,L2(11),M11and M12.

Subcase 1.1K/HA5.

Otherwise,by Table 2,|G:K|=1 or 2.In this case,we have|H|=2u·32·11,where 4≤u≤5.Since H is solvable,by Lemma 2.1,we have that H11is normal in H.Hence,H11Char H.Since HG,one has that H11G.Therefore,χ(1)||G:H11|=27·33·5,a contradiction.

By the similar arguments as before,we can prove that K/HA6.

Subcase 1.2K/HL2(11).

If K/HL2(11),then we have|G:K|=1 or 2.

If|G:K|=1,then|H|=25·32.Let ϕ ∈ Irr(H)such that[βH,ϕ]0.Then β(1)/ϕ(1)||G:H|=22·3 ·5 ·11,and thus 12| ϕ(1).If ϕ(1)>12,then we have ϕ(1)2>|H|,a contradiction.Hence ϕ(1)=12.Using the software Magma(V2.11-1),it is easy to check that there are only 1045 such groups of order 25·32in small groups(25·32)up to isomorphism(cf.[11]).Moreover,we also know that the set of all irreducible character degrees of H,i.e.,cd(H)can only be equal to one of the following sets:

Hence,12cd(H),a contradiction.

If|G:K|=2,then|K|=26·33·5·11 and|H|=24·32.Let Λ ∈ Irr(H)such that[βH,Λ]0.Then β(1)/Λ(1)||G:H|=23·3·5·11.Thus,we have 6|Λ(1).If Λ(1)>12,then Λ(1)2>|H|,a contradiction.If Λ(1)=12,then Λ(1)2>|H|,a contradiction,too.Therefore,Λ(1)=6.Let λ ∈ Irr(H)such that[χH,λ]0,and one has that χ(1)/λ(1)||G:H|=23·3·5·11,so 2|λ(1).Set e=[χH,λ],t=|G:IG(λ)|.Then we have etλ(1)=24·11.Again,using the Magma,there are only 197 such groups of order 24·32in small groups(24·32)up to isomorphism.Moreover,if λ ∈ cd∗(H),then cd∗(H)can only be equal to one of the following sets:

By the structure of cd∗(H),one has that λ(1)=1 or 2 or 4.Moreover,the following conclusions hold:

(i)If λ(1)=1,then t ≤ 11 by the above cd∗(H).Thus e ≥ 24.But[χH,χH]=e2t ≥28·11>|G:H|=23·3·5·11,a contradiction to[12,Lemma 2.29].

(ii)If λ(1)=2 or 4,then t ≤ 9 by the above cd∗(H).But t=|G:IG(λ)|,and we have t=1 as any maximal subgroup has an index in G ≥ 11(cf.[1]).Hence e=23·11.But[χH,χH]=e2t=26·112>|G:H|=23·3·5·11,which by[12,Lemma 2.29],is a contradiction.

Subcase 1.3K/HM11.

Assume that K/HM11.Since|Out(M11)|=1,then|H|=23·3 ·7.If H is nonsolvable,then HL2(7).In this case,we have GM11× L2(7).On the other hand,since L1(M11)=55 and L1(L2(7))=8 by Table 2,then by the structure of G,we can get that the largest irreducible degree L1(G)=23·5 ·11,a contradiction to χ(1)=24·11.

If H is solvable,since|Out(M11)|=Mult(M11)=1,then by Lemma 2.3 we obtain that GM11× H,where|H|=23·3 ·7.By Table 1,we see that the largest irreducible character degree of M11is 55,and 22cd(M11)by[1].According to the structure of GM11× H,one has that there exists no irreducible character in G of degree 24·11,a contradiction.

Subcase 1.4If K/HM12,then(3)follows.

If K/HM12,by Table 1,we have|G:K|=1 or 2.

If|G:K|=1,then|H|=2,so H≤Z(G).In this case,we can get that G/HM12.Therefore G is a central extension of Z2by M12.Since Mult(M12)=2,G is isomorphic to one of 2 ·M12Z2·M12(a non-split extension of Z2by M12)and 2:M12Z2× M12(a split extension of Z2by M12).

By[1],it is easy to check that both 2·M12and 2:M12satisfy the conditions|G|=|Aut(M12)|,L1(G)=L1(Aut(M12))and L2(G)=L2(Aut(M12)).

If|G:K|=2,then KM12.Therefore GM12·2=Aut(M12).Hence,(3)follows.This concludes Case 1.

Case 2We are to prove that the theorem follows if M=M22.

By Table 1,we have|G|=28·32·5·7·11.Let χ,β ∈ Irr(G)such that χ(1)=L1(G)=24·5·7 and β(1)=L2(G)=5 ·7 ·11.

We first prove that G is not a solvable group.Assume the contrary,and by Lemma 2.1,we have that the Sylow 11-subgroup G11of G is normal in G.Thus χ(1)||G:G11|=28·32·5·7,a contradiction.Therefore,G is non-solvable.By Lemma 2.2,G has a normal series 1HKG such that K/H is a direct product of non-abelian simple groups which are pairwise isomorphic to each other and|G/K|||Out(K/H)|.As|G|=28·32·5·7·11.We deduce that K/H can only be isomorphic to one of A5,A6,L2(7),L2(8),A7,L2(11),M11,L3(4),A8and M22.

Subcase 2.1K/HA5.

If K/HA5,by Table 2,we have|G:K|=1 or 2.Thus|H|=2v·3·7·11,where 5 ≤ v ≤ 6.If H is solvable,then by Lemma 2.1,the Sylow-11 subgroup H11of H is normal in H,and hence H11Char H.Since HG,one has that H11G.Therefore,χ(1)||G:H11|=28·32·5·7,a contradiction.

If H is non-solvable,by Lemma 2.2,we can get a normal series of H:1NMH such that M/NL2(7)and|H/M|||Out(A5)|=2.Thus|H:M|=1 or 2.In this case,we have|N|=2m·11,where 1 ≤ m ≤ 3.Let Δ ∈ Irr(N)such that[βN,Δ]0,and then β(1)/Δ(1)||G:N|=28−m·32·5 ·7.Hence Δ(1)=11.But we have that Δ(1)2>|N|,a contradiction.

The above argument is effective for the subcases that 11|G/K|.Hence,we can prove that K/H is not isomorphic to one of A6,L2(7),L2(8),A7,L3(4)and A8.

Subcase 2.2K/HL2(11).

If not,we have|G:K|=1 or 2.In the following,we only discuss the condition of|G:K|=1,and we omit the details for|G:K|=2 because the arguments are similar to those proofs for|G:K|=1.Hence,we may assume that|G:K|=1,so|H|=26·3 ·7 in this case.

If H is solvable,then there exists a subgroup T of H such that|H:T|=3.Considering the permutation representation of H on the right cosets of T with the kernel TH,the core of T in H,we get that H/THS3.Then|TH|=26·7 or 25·7.

If|TH|=26·7,let θ∈ Irr(TH)such that[χTH,θ]0,and then χ(1)/θ(1)||G:TH|=22·32·5 ·11.Hence,22·7|θ(1).But θ(1)2>|TH|,a contradiction.

If|TH|=25·7,let ϑ ∈ Irr(TH)such that[χTH,ϑ]0,and then χ(1)/ϑ(1)||G:TH|=23·32·5 ·11.Thus 14|ϑ(1).If ϑ(1)>14,then ϑ(1)2>|TH|,a contradiction.Assume that ϑ(1)=14.Now,applying the software Magma(V2.11-1),it is easy to check that there are only 197 such groups of order 25·7 in small groups(25·7)up to isomorphism(cf.[11]).Moreover,we can calculate that the sets of all irreducible character degrees of the groups of order 25·7,i.e.,cd(TH),can only be equal to one of{1,2},{1},{1,2,4},{1,7}and{1,4}.

However,by checking each set of cd(TH)above,we know that there exists no such irreducible character of degree 14 of TH,i.e.,14cd(TH),a contradiction.

Therefore,H is non-solvable.Since|H|=26·3 ·7,by Lemma 2.2,H has a normal series:1NMH such that M/NL2(7).Hence,there exist two composite factors of G,i.e.,M1/N1and M2/N2,respectively,such that M1/N1L2(11)and M2/N2L2(7).Considering the action of G/N on M/N,we have that the factor group G/N/CG/N(M/N)Aut(M/N).Let CG/N(M/N)=W/N.Obviously,W/N has a section isomorphic to the simple group L2(11).Let NTSW such that S/TL2(11).Since W has exactly one simple section,we have S/TG/T,so SG.Moreover,it is easy to see that T=TGG.Hence,G/T has a maximal subgroup S/T ×TM/TL2(11)×L2(7).Consequently,we can obtain the following composite group series of G:1QPG such that G/PL2(7)and P/QL2(11).In this case,|Q|=8.Let η ∈ Irr(P)such that[βP,η]0,and then β(1)/η(1)||G:P|=23·3·7.Thus η(1)=55.Let λ ∈ Irr(Q)such that[ηQ,λ]0.Let e=[ηQ,λ]and t=|P:IP(λ)|,and then we have λ(1)=1 and et=55.By checking the maximal subgroups of L2(11)(cf.[1]),we can get that t=1.Hence,e=55.But[ηQ,ηQ]=e2t=552>|P:Q|=22·3 ·5 ·11,a contradiction to[12,Lemma 2.29].

Subcase 2.3K/HM11.

Since|Out(M11)|=1,one has that|H|=24·7 if K/HM11.Since H is solvable and|Out(M11)|=Mult(M11)=1,then by Lemma 2.3,we have that GM11×H.Again,using the software magma(V2.11-1),it is easy to check that there are only 43 such groups of order 24·7 in small groups(24·7)up to isomorphism(cf.[11]).Moreover,we can get the sets of all irreducible character degrees of the groups of order 24·7.In other words,cd(H)can only be equal to one of{1},{1,2},{1,2,4},{1,7}.By[1],we know that the largest irreducible character degree of M11is 55,i.e.,L1(M11)=55 and 22cd(M11).On the other hand,by the structure of GM11×H and as the largest irreducible character degree in the above sets is 7,we have L1(G)=5·7·11,a contradiction to L1(G)=24·5·7.

Subcase 2.4If K/HM22,then(4)follows.

In this case,we can get that|G:K|=1 or 2.

If|G:K|=1,then|H|=2 and H≤Z(G).In this case,we have G/HM22.Therefore G is a central extension of Z2by M22and G is isomorphic to one of 2 ·M22Z2·M22(a non-split extension of Z2by M22)and 2:M22Z2×M22(a split extension of Z2by M22).

If G2 ·M22,by[1],it is easy to check that|G|=|Aut(M22)|,L1(G)=L1(Aut(M22))and L2(G)=L2(Aut(M22)).

If G2:M22,by[1],we get that L1(G)=L1(2:M22)=5·7·11.But L1(G)=24·5·7=560,a contradiction.

If|G:K|=2,then KM12,and thus GM12·2=Aut(M22).Therefore,(4)follows.

This completes the proof of the main Theorem 1.1.

AcknowledgementThe authors are gratefulto the referees for the carefulreading of the manuscript.Their comments and suggestions are very helpful.

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