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
1,2and 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, Mand 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∼=Mif 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 Mor H×M,where H is a Frobenius group with an elementary kernel of orderand a cyclic complement of order.
We need the following lemmas.
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/N)×M/N
CG/N(M/N) ≤
G/N CG/N(M/N)
Aut(M/N).
Let K/N =CG/N(M/N)×M/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 i≤n– 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(l≤m≤ai) 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·,
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,MandM.
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(),AandM.
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(),MorM.
By the same arguments as the proofs of Case ., we can prove thatK/H cannot be isomorphic toAorA.
Assume thatK/H∼=L(). Since|Out(L())|= , we have|G/K| | and|H|= a·,
where a= or . Suppose that|G/K|= . Then|H|= ·. Letχ∈Irr(G) such that
χ() =L(G) = · andθ∈Irr(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
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=
g∈GIG(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(H)×B, 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) = ···. Letχ∈Irr(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, MandM.
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
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) = . Letχ∈Irr(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(),AandM.
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 thatS≤Z(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=
[χH,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|. LetH≤AHsuch that|H/A|= , thenH/A≤Z(G/A). SinceMult(L()) = , we haveG/A∼=H/A×L().
HenceGhas a normal series NMGsuch thatM/N∼=L() and|N|= α·,
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|= ·. Letθ∈Irr(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),q≥7. 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