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R E S E A R C H

Open Access

A new characterization of

Mathieu

-groups by

the order and one irreducible character

degree

Haijing Xu

1

, Yanxiong Yan

1,2

and Guiyun Chen

1*

*Correspondence:

gychen@swu.edu.cn

1School of Mathematics and

Statistics, Southwest University, Chongqing, 400715, China Full list of author information is available at the end of the article

Abstract

The main aim of this article is to characterize the finite simple groups by less character quantity. In fact, we show that eachMathieu-groupGcan be determined by their largest and second largest irreducible character degrees.

MSC: 20C15

Keywords: finite group; simple group; character degree

1 Introduction and preliminary results

Classifying finite groups by the properties of their characters is an interesting problem in group theory. In , Huppert conjectured that each finite non-abelian simple group

Gis characterized by the setcd(G) of degrees of its complex irreducible characters. In [–], it was shown that many non-abelian simple groups such asL(q) andSz(q) satisfy

the conjecture. In this paper, we manage to characterize the finite simple groups by less character quantity. LetGbe a finite group;L(G) denotes the largest irreducible character degree ofGandS(G) denotes the second largest irreducible character degree ofG. We characterize the five Mathieu groupsGby the order ofGand its largest and second largest irreducible character degrees. Our main results are the following theorems.

Theorem A Let G be a finite group and let M be one of the following Mathieu groups:M, Mand M.Then G∼=M if and only if the following conditions are fulfilled:

() |G|=|M|; () L(G) =L(M).

Theorem B Let G be a finite group.Then G∼=Mif and only if|G|=|M|and S(G) = S(M).

Theorem C Let G be a finite group.If|G|=|M|and L(G) =L(M), then either G is isomorphic to Mor H×M,where H is a Frobenius group with an elementary kernel of orderand a cyclic complement of order.

We need the following lemmas.

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Lemma  Let G be a non-solvable group.Then G has a normal seriesHK G such that K/H is a direct product of isomorphic non-abelian simple groups and|G/K| |

|Out(K/H)|.

Proof LetGbe a non-solvable group. ThenGhas a chief factorM/Nsuch thatM/Nis a direct product of isomorphic non-abelian simple groups. HenceCG/N(M/N)∩M/N=

Z(M/N) = , and so

M/N∼= CG/N(M/NM/N

CG/N(M/N) ≤

G/N CG/N(M/N)

Aut(M/N).

Let K/N =CG/N(M/NM/N and H/N =CG/N(M/N). Then G/K ≤Out(M/N) and

K/H∼=M/Nis a direct product of isomorphic non-abelian simple groups. Thus H KGis a normal series, as desired.

Lemma  Let G be a finite solvable group of order pa

p

a

 · · ·pann,where p,p, . . . ,pnare

distinct primes.If kpn+ piaifor each in– and k> ,then the Sylow pn-subgroup is

normal in G.

Proof LetNbe a minimal normal subgroup ofG. Then|N|=pmforGis solvable. Ifp=pn,

by induction onG/N, we see that normality of the Sylowpn-subgroup inG. Now suppose

thatp=pifor somei<n. Now considerG/N. By induction, the Sylowpn-subgroupP/Nof

G/Nis normal inG/N. ThusPG. LetQbe a Sylowpn-subgroup ofP. ThenP=NQ. By

Sylow’s theorem,|P:NP(Q)|=pli(lmai) andpn|pli– . But this means thatkpn+ |

pai, and thenk=  by assumption. HenceQPandQG.

2 Proof of theorems

Proof of TheoremA We only need to prove the sufficiency. We divide the proof into three cases.

Case .M=M

In this case, we have|G|= ··· andL(G) = . We first show thatGis

non-solvable. Assume the contrary. By Lemma , we know that the Sylow -subgroup ofGis normal inG. LetNbe the -Sylow subgroup ofG. SinceNis abelian, we haveχ()| |G/N|

for all χ∈Irr(G). ButL(G) =  and |G/N|, a contradiction. Therefore, Gis non-solvable.

SinceGis non-solvable, by Lemma , we get thatGhas a normal series HKG

such thatK/His a direct product of isomorphic non-abelian simple groups and|G/K| |

|Out(K/H)|. As|G|= ···, we haveK/H=A

,A,L() orM.

We first assume thatK/H∼=A. Since|Out(A)|= , we have|G/K| | and|H|= t·

·, wheret=  or . Letχ∈Irr(G) such thatχ() =L(G) =  andθ∈Irr(H) such that [χH,θ]= . Thenθ() =  by the Clifford theorem (see Theorem . in []). On the other

hand, since|H|= t··, we haveHis solvable. LetNbe a Sylow -subgroup ofH. Then

NHby Lemma . Henceθ()| |H/N|= t·, a contradiction.

By the same reason as above, one has thatK/HA.

Suppose thatK/H∼=L(). Since|Out(L())|= , we have|G/K| | and so|H|= a·,

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andet=χ()/θ() = . Since|H|= t·, wherea=  or , we have that |Aut(H/H)|.

Hencet= ,e= . But ()=et= [χ

H,χH] >|G:H|= b···, where ≤b≤, a

contradiction.

IfK/H∼=M, by comparing the orders ofGandM, we have|H|= . ThereforeG= K∼=M.

Case .M=M

In this case, we have|G|= ····· andL(G) = ··. ThenO

(G) = .

If not, then|O(G)|=  andO(G) is abelian. HenceL(G) = ··| |G/O(G)|, a

contradiction.

IfGis solvable, then the Sylow -subgroup ofGis normal inGby Lemma , which leads to a contradiction as above. ThereforeGis non-solvable.

SinceGis non-solvable, by Lemma , we get thatGhas a normal series HKG

such thatK/His a direct product of isomorphic non-abelian simple groups and|G/K| |

|Out(K/H)|. As|G|= ·····, we have thatK/Hcan be isomorphic to one of the simple groups:A,L(),A,L(),L(),A,M,L(),A,MandM.

We first assume thatK/H∼=A. Since|Out(A)|= , we have|G/K| | and|H|= m·

···, wherem=  or . Suppose thatH is non-solvable. By Lemma ,H has a normal series ABHsuch thatB/Ais a direct product of isomorphic non-abelian simple groups and|H/B| | |Out(B/A)|. Since|H|= m····, we haveB/A=L

() and

|H/B| |. Thus|A|= a··, where a. LetNbe a Sylow -subgroup ofA. Then

NAby Lemma . Hence we get a subnormal series ofG,NcharABHKG, which implies thatNG. ButO(G) = , a contradiction. IfHis solvable, then the Sylow

-subgroup ofHis normal inHby Lemma , which leads to a contradiction as before. By the same arguments as the proofs ofK/H∼=A, we show thatK/H cannot be

iso-morphic to one of the simple groups:A,L(),L(),L(),A,M,L(),AandM.

IfK/H∼=M, since|G|= ·····, we have that|H|=  andG=K∼=M.

Case .M=M

In this case,|G|= ··· andL(G) = ·. Since |L(G), by the same arguments

as the proofs of Case ., we have thatO(G) = .

We will show thatGis non-solvable. IfGis solvable, then the Sylow -subgroup ofG

is normal inGby Lemma , a contradiction. Therefore,Gis non-solvable.

By Lemma , we get thatGhas a normal series HKGsuch thatK/H is a direct product of isomorphic non-abelian simple groups and|G/K| | |Out(K/H)|. As|G|= ···, we haveK/H=A

,A,L(),MorM.

By the same arguments as the proofs of Case ., we can prove thatK/H cannot be isomorphic toAorA.

Assume thatK/H∼=L(). Since|Out(L())|= , we have|G/K| | and|H|= a·,

where a=  or . Suppose that|G/K|= . Then|H|= ·. LetχIrr(G) such that

χ() =L(G) = · andθIrr(H) such thate= [χ

H,θ]= . Thenχ() =etθ() = ,

wheret=|G:IG(θ)|. Sinceχ()/θ()| |G/H|, we have thatθ() =  or . Ifθ() = , then

et= . Since|H|= ·, we have thatH has at most eight irreducible characters of degree . Hencet≤. We assert thatIG(θ) =G. If not, thenIG(θ) <G. LetUcontaining

IG(θ) be a maximal subgroup ofG. Then ≤ |G:U| | |G:IG(θ)|= . By checking the

maximal subgroups ofL() (see ATLAS table in []), it is easy to get a contradiction.

HenceIG(θ) =G, and sot=  ande= . Bute·t= [χH,χH] >|G:H|, a contradiction. If

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Irr(O(H)) such that [θO(H),λ]= . Since θ() = , we have ≤ |H:IH(λ)| ≤. ButIG(θ) =

G, which implies that ≤ |G:IG(λ)|=|H:IH(λ)| ≤. LetS=

gGIG(λ)g. ThenSGand

G/SS. By the Jordan-Hölder theorem,Shas a normal series O(H)CDS

such thatD/C∼=L() and|C/O(H)|= ,  or . Letα∈Irr(S) such that [χS,α]= . Since

χ()/α()| |G/S|, we have that |α(). Sinceλg is invariant inS, for eachg Gand

≤ |G:IG(λ)|, we have that each irreducible character is invariant inSandO(H)≤Z(S).

Therefore, the following conclusions hold:

(a) S∼=L()×O(H)if|C/O(H)|= ;

(b) S∼= (·L())×O(H)or(Z×L())×O(H)if|C/O(H)|= .

By checking the character table of ·L() andL(), we see that both conclusions (a)

and (b) are not satisfied with the above conditions. Now, we suppose that|C/O(H)|= .

Then |α(). Since O(H)≤Z(S), one has that C∼=O(HB, whereBis a group

of order . Let β be an irreducible component ofαC andt=|S:IS(β)|. Thenβ() = 

andt|α()/β()|. Since the indexes of the maximal subgroups ofScontainingIS(β)

dividet andt| |Aut(C)|, we have thatt= . Hence [αC,αC] >|S:C|= ···, a

contradiction.

Similarly, we can show that|G/K| = .

Suppose thatK/H∼=M. Since|Out(M)|=Mult(M) = , we haveG∼=H×M, where

|H|= ·. By checking the character table ofM, we see thatGhas no irreducible

char-acter of degreeL(G) = ·, a contradiction.

IfK/H∼=M, since|G|= ···, we conclude that|H|=  andG=K∼=M, which

completes the proof of Theorem A.

Proof of TheoremB We only need to prove the sufficiency.

In this case, we have|G|= ····· andS(G) = ···. LetχIrr(G)

such thatχ() =S(G). IfO(G)= , then|O(G)|= , which implies thatχ()| |G:N|,

a contradiction. HenceO(G) = .

We have to show thatGis non-solvable. Assume the contrary, by Lemma , we have that the Sylow -subgroup is normal inG, a contradiction. Therefore,Gis non-solvable.

SinceGis non-solvable, by Lemma , one has thatGhas a normal series HKG

such thatK/His a direct product of isomorphic non-abelian simple groups and|G/K| |

|Out(K/H)|. As|G|= ·····, thenK/H can be isomorphic to one of the

following simple groups:A,L(),A,L(),L(),A,U(),M,L(),A,M,M, MandM.

We first assume thatK/H∼=A. Since|Out(A)|= , we have|G/K| | and|H|= t··

··, wheret=  or . Letθ ∈Irr(H) such that [χH,θ]= . Sinceχ()/θ()| |G/H|,

it implies that |θ(). IfHis solvable, thenO(H)=  by Lemma , which implies that O(H) =O(G)= , a contradiction. ThusHis non-solvable. Then there exists a normal

series ofH: NMHsuch thatM/Nis a direct product of isomorphic non-abelian simple groups and|H/M| | |Out(M/N)|. As|H|= t····, we haveM/N=L

()

orL(), which implies that | |N|. HenceO(N)=  by Lemma , which implies that O(N) =O(G)= , a contradiction.

By the same arguments as the proof ofK/H∼=A, we show thatK/Hcannot be

isomor-phic to one of the simple groups:L(),A,L(),L(),A,U(),M,L(),A,M

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Suppose thatK/H∼=M. Since|Out(M)|=Mult(M) = , we have thatG∼=H×M,

where|H|= ·. By checking the character table ofM

, it is easy to see that there exists

no irreducible character of degree ··· inG, a contradiction.

IfK/H∼=M, since|G|= ·····, one has that|H|=  andG=K∼=M,

which completes the proof of Theorem B.

Proof of TheoremC We only need to prove the sufficiency.

In this case,|G|= ···· andL(G) = . LetχIrr(G) such thatχ() =L(G) =

··. We assert thatO(G) = . Otherwise, we have that|O(G)|=  andO(G) is

abelian. Henceχ()| |G/O(G)|, a contradiction. Similarly,O(G) =O(G) = .

If G is solvable, then the Sylow -subgroup ofGis normal in Gby Lemma . But

O(G) = , a contradiction. Therefore,Gis non-solvable.

SinceGis non-solvable, by Lemma , we get thatGhas a normal series HKG

such thatK/His a direct product of isomorphic non-abelian simple groups and|G/K| |

|Out(K/H)|. As|G|= ····, we see thatK/His isomorphic to one of the simple

groups:A,L(),A,L(),L(),A,M,L(),AandM.

We first assume thatK/H∼=A. Since|Out(A)|= , we have|G/K| | and|H|= t···

, wheret=  or . IfHis solvable, thenO(H) =O(G)=  by Lemma , a contradiction.

HenceH is non-solvable andHhas a normal series NMH such thatM/N is a direct product of isomorphic non-abelian simple groups and|H/M| | |Out(M/N)|. As

|H|= t···, one has thatM/N∼=L() and|N|= s·, where ≤s≤. LetPbe the

Sylow -subgroup ofN. ThenPis normal inNby Sylow theorem. SincePis also a Sylow -subgroup inGandNis subnormal inG, we havePG, a contradiction.

Similarly,K/Hcannot be isomorphic to the simple groups:L(),A,L(),L(),A, L() orA.

Assume thatK/H∼=L(). Since|Out(L())|= , we have|G/K| | and|H|= α··,

whereα=  or . Suppose thatH=H. ThenHhas a normal subgroupSsuch thatH/S∼=

L(), where|S|=  or . Obviously, we know thatSZ(H), and thenSG. Letθ∈Irr(S)

such that [χS,θ]= . Thenθ() =  sinceS is abelian. Lete= [χS,θ] and t=|G:IG(θ)|.

Thent=  ande=χ() =  by the Clifford theorem (see Theorem . in []). Bute·t=

H,χH] >|G:H|, a contradiction. HenceH<H. Suppose that|H/H|= . ThenH/H

is central inG/H. Letβ be an irreducible component ofχH, and letθ be an irreducible

component ofβH. Thenθ() =β() =  andθ is extendible toβ. Henceλβis invariant in

Gfor everyλ∈Irr(H/H) ifβis invariant inG. Since|H|= α···, whereα=  or ,

Hhas at most  irreducible characters of degree . Lett=|G:IG(β)|. Thent≤. Since

the index of the maximal subgroup ofUcontainingIG(θ) dividest, we have thatt=  or

 by checking maximal subgroups ofL() (see ATLAS table in []). Ift= , thenH

has exactly  irreducible characters of degree , and one of them, sayδ, is invariant inG. Hence,λδis also invariant inGforλ∈Irr(H/H), which forcest≤, a contradiction. Thereforet=  ande= . But ()= [χ

H,χH] >|G:H|, a contradiction. By the same

reasoning as before, we can prove that|H/H| = m·n, where m and n. If

|H/H|= m·n, wherem=  or , then the Sylow  subgroup ofHis normal inH, and so it is normal inG, a contradiction. Now we assume that | |H/H|. LetHAHsuch that|H/A|= , thenH/AZ(G/A). SinceMult(L()) = , we haveG/A∼=H/A×L().

HenceGhas a normal series NMGsuch thatM/N∼=L() and|N|= α·,

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component ofϕN. Thenϕ() =  andη() =  by the Clifford theorem (see Theorem .

in []). Since|Aut(N)|is not divided by  and , one has thatt=|M:IM(η)|= . Therefore

()= [ϕ

N,ϕN] >|M:N|, a contradiction.

Suppose thatK/H∼=M. Since|Out(M)|=Mult(M) = , we haveG∼=H×M, where

|H|= ·. LetθIrr(H) such that [χ

H,θ]= . Sinceθ()|χ() andθ()| |H|, one has that

θ() = , which implies thatHis a Frobenius group with an elementary kernel of order  and a cyclic complement of order .

Suppose thatK/H∼=M. Since|G|= ····, we have that|H|=  andG=K∼= M, which completes the proof of Theorem C.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

HX carried out the study of the Mathieu groupsM11,M12andM23. YY carried out the study of the Mathieu groupM24. GC carried out the study of the Mathieu groupM22. All authors read and approved the final manuscript.

Author details

1School of Mathematics and Statistics, Southwest University, Chongqing, 400715, China.2Department of Mathematics

and Information Engineering, Chongqing University of Education, Chongqing, 400067, China.

Acknowledgements

Article is supported by the Natural Science Foundation of China (Grant No. 11271301; 11171364), The Fundamental Research Funds for the Central Universities (Grant No. XDJK2009C074), Graduate-Innovation Funds of Science of SWU (Grant No. ky2009013) and Natural Science Foundation Project of CQ CSTC (Grant No. cstc2011jjA00020).

Received: 29 December 2012 Accepted: 15 April 2013 Published: 26 April 2013

References

1. Huppert, B: Some simple groups which are determined by the set of their character degrees, I. Ill. J. Math.44(4), 828-842 (2000)

2. Huppert, B: Some simple groups which are determined by the set of their character degrees, II. Rend. Semin. Mat. Univ. Padova115, 1-13 (2006)

3. Wakefield, TP: Verifying Huppert’s conjecture forPSL3(q) andPSU3(q2). Commun. Algebra37, 2887-2906 (2009)

4. Tong-Viet, HP, Wakefield, TP: On Huppert’s conjecture forG2(q),q7. J. Pure Appl. Algebra216, 2720-2729 (2012)

5. Isaacs, IM: Character Theory of Finite Groups. Academic Press, New York (1976)

6. Conway, JH, Nortion, SP, Parker, RA, Wilson, RA: Atlas of Finite Groups. Oxford University Press, Oxford (1985)

doi:10.1186/1029-242X-2013-209

References

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