Huppert’s conjecture for <span class="mathjax-formula">$Fi_{23}$</span>

Huppert's Conjecture for Fi

S. H. ALAVI (*) - A. DANESHKAH(**) - H. P. TONG-VIET(***) T. P. WAKEFIELD (***)

ABSTRACT- LetGdenote a finite group and cd(G) the set of all irreducible character degrees of G. Bertram Huppert conjectured that if H is a finite nonabelian simple group such that cd(G)ˆcd(H), thenGHA, whereAis an abelian group. Huppert verified the conjecture for many of the sporadic simple groups.

We illustrate the arguments by presenting the verification of Huppert's Con- jecture forFi23.

1. Introduction and Notation

LetGbe a finite group, Irr(G) the set of irreducible characters ofG, and denote the set of character degrees of Gby cd(G)ˆ fx(1):x2Irr(G)g.

When context allows, the set of character degrees will be referred to as the set of degrees. In the late 1990s, Bertram Huppert posed a conjecture which, if true, would sharpen the connection between the character degree set of a nonabelian simple group and the structure of the group. In [4], he conjectured the following.

(*) Indirizzo dell'A.: School of Mathematics and Statistics, The University of Western Australia, 35 Stirling Hw, Crawley WA 6009, Australia.

E-mail: alavi@maths.uwa.edu.au

(**) Indirizzo dell'A.: Department of Mathematics, Faculty of Science, Bu-Ali Sina University, Hamedan, Iran.

E-mail: adanesh@basu.ac.ir

(***) Indirizzo dell'A.: School of Mathematical Sciences, University of KwaZu- lu-Natal, Pietermaritzburg 3209, South Africa.

E-mail: tvphihung@gmail.com

(***) Indirizzo dell'A.: Department of Mathematics and Statistics, Youngstown State University, Youngstown, Ohio, United States 44555.

E-mail: tpwakefield@ysu.edu

HUPPERT'S CONJECTURE.Let G be a finite group and H a finite non- abelian simple group such that the sets of character degrees of G and H are the same. Then GHA, where A is an abelian group.

Huppert verifies the conjecture for the Suzuki groups, the family of simple groups PSL₂(q) forq4, and many of the sporadic simple groups.

Huppert's proofs rely upon the completion of the following five steps.

(1) ShowG⁰ˆG⁰⁰. Hence ifG⁰=Mis a chief factor ofG, thenG⁰=MS^k, whereSis a nonabelian simple group.

(2) IdentifyHas a chief factorG⁰=M ofG.

(3) Show that if u2Irr(M) and u(1)ˆ1, then u is stable under G⁰, which implies [M;G⁰]ˆM⁰.

(4) Show thatMˆ1.

(5) Show that GˆG⁰CG(G⁰). As G=G⁰CG(G⁰) is abelian and G⁰H, Huppert's Conjecture is verified.

In addition to his work verifying his conjecture for many of the simple groups of Lie type, Huppert also verified the conjecture for many of the sporadic simple groups. Indeed, he demonstrated that his conjecture holds for the Mathieu groups, Janko groups, and many other sporadic simple groups in his preprints. The only sporadic simple groups not considered by Huppert are the Conway groups, Fischer groups, Monster, and Baby Monster. In [7], the authors establish Huppert's Conjecture for the Mon- ster and Baby Monster. The authors have examined the remaining Fischer groups and Conway groups and have verified the conjecture for these fa- milies of simple groups using arguments similar to those presented in this paper. Thus, Huppert's Conjecture has been verified for all twenty-six sporadic simple groups.

Ifnis an integer then we denote byp(n) the set of all prime divisors of n: IfGis a group, we will write p(G) instead ofp(jGj) to denote the set of all prime divisors of the order of G: If N/G and u2Irr(N);

then the inertia group of u in G is denoted by I_G(u): The set of all irreducible constituents of u^G is denoted by Irr(Gju): Other notation is standard.

2. Preliminaries

In this section, we present some results that we will need for the proof of the Huppert's Conjecture.

LEMMA2.1 ([4, Lemma2]). Suppose N/G andx2Irr(G):

(a)Ifx_Nˆu₁‡u₂‡ ‡u_kwithu_i2Irr(N);then k dividesjG=Nj:In particular, ifx(1)is prime tojG=Njthenx_N2Irr(N):

(b) (Gallagher's Theorem) If x_N2Irr(N); then xc2Irr(G)for every c2Irr(G=N):

LEMMA 2.2 ([4, Lemma 3]). Suppose N/G and u2Irr(N): Let IˆI_G(u):

(a) If u^IˆP^k

iˆ1W_i with W_i2Irr(I); then W^G_i 2Irr(G): In particular, W_i(1)jG:Ij 2cd(G):

(b)Ifuextends toc2Irr(I);then(ct)^G2Irr(G)for allt2Irr(I=N):In particular,u(1)t(1)jG:Ij 2cd(G):

(c)Ifr2Irr(I)such thatr_Nˆeu;thenrˆu₀t₀;whereu₀is a character of an irreducible projective representation of I of degreeu(1)whilet₀is the character of an irreducible projective representation of I=N of degree e:

The following lemma will be used to verify Step 1:This is [7, Lemma 3].

LEMMA 2.3. Let G=N be a solvable factor group of G; minimal with respect to being nonabelian. Then two cases can occur.

(a) G=N is an r-group for some prime r: Hence there exists c2Irr(G=N) such that c(1)ˆr^b>1: If x2Irr(G) and r⁴jx(1); then xt2Irr(G)for allt2Irr(G=N):

(b)G=N is a Frobenius group with an elementary abelian Frobenius kernel F=N:Then f ˆ jG:Fj 2cd(G)andjF=Nj ˆr^a for some prime r;

and F=N is an irreducible module for the cyclic group G=F;hence a is the smallest integer such that r^a1(mod f): If c2Irr(F) then either fc(1)2cd(G)or r^adividesc(1)²:In the latter case, r dividesc(1):

(1) If no proper multiple of f is in cd(G); thenx(1) divides f for all x2Irr(G) such that r⁴jx(1); and if x2Irr(G) such that x(1)j⁴f; then r^ajx(1)²:

(2)Ifx2Irr(G)such that no proper multiple ofx(1)is incd(G);then either f dividesx(1)or r^adividesx(1)²:Moreover ifx(1)is divisible by no nontrivial proper character degree in G;then f ˆx(1)or r^ajx(1)²:

Let x2Irr(G):We say thatxisisolatedinGifx(1) is divisible by no proper nontrivial character degree ofG;and no proper multiple ofx(1) is a character degree ofG:In this situation, we also say thatx(1) is anisolated degreeofG:

The next two lemmas will be used to verify Steps 2 and 4: The first lemma appears in [1, Theorems 2;3;4].

LEMMA 2.4. If S is a nonabelian simple group, then there exists a nontrivial irreducible characteruof S that extends toAut(S):Moreover the following holds:

(i) if S is an alternating group of degree at least 7; then S has two consecutive characters of degrees n(n 3)=2and(n 1)(n 2)=2that both extend toAut(S):

(ii)if S is a sporadic simple group or the Tits group, then S has two nontrivial irreducible characters of coprime degrees which both extend to Aut(S):

(iii)if S is a simple group of Lie type then the Steinberg character St_Sof S of degreejSj_pextends toAut(S):

LEMMA2.5 ([1, Lemma5]). Let N be a minimal normal subgroup of G so that NS^k;where S is a nonabelian simple group. Ifu2Irr(S)extends toAut(S);thenu^k 2Irr(N)extends to G:

The following Lemma will be used to verify Step 4:

LEMMA2.6 ([4, Lemma6]). Suppose that M/G⁰ˆG⁰⁰and that for any l2Irr(M) with l(1)ˆ1; l^gˆl for all g2G⁰: Then M⁰ˆ[M;G⁰] and jM=M⁰jdivides the order of the Schur multiplier of G⁰=M:

3. The sporadic simple groupFi₂₃

LEMMA3.1. Let H be the sporadic simple group Fi₂₃: (i) The following degrees are isolated degrees of H:

2¹⁸5²711 2⁵3¹²17 3¹³1323 2¹⁶51323 3¹²1723 3¹²51117:

(ii) Let 16ˆx(1)2cd(H): Then (111723;x(1))>1 and (x(1);

111317)>1:

(iii) H has no proper power degrees nor consecutive degrees.

(iv) If K is a maximal subgroup of H such thatjH:Kjdivides some character degreex(1)of H then one of the following cases holds.

(a) K2Fi22 andx(1)=jH:Kj is one of the numbers in the set A1; consisting of the following numbers:

2³3³711 2⁴51113 2²3³711 35²713 3⁸ 3²51113

2⁵513 2⁵511 51113 23³1113 2²3⁴13 23³11:

(b)KO^‡₈(3):S3andx(1)=jH:Kjis one of the numbers in the setA2; consisting of the following numbers:

2⁶313 2²3²513 5²713 2²35²7 2⁵513 3²5²7 3⁴5 3³13 3713 2513 3513 2⁵13:

(v)The outer automorphism group and the Schur multiplier of H are trivial.

PROOF. The list of maximal subgroups and their indices of the Fischer groupFi₂₃ is given in Table 1. This table is taken from [6]. The other in-

formation can be found in [3]. p

LEMMA 3.2. Let LˆO^‡₈(3)be the simple orthogonal group. The only nontrivial degree of a proper irreducible projective representation of L which divides one of the numbers inA₂is520:

PROOF. By [3], the Schur multiplier ofLis elementary abelian of order 4:LetL~be the full covering group ofL:ThenZ(L)~ Z²₂:HenceZ(L) is non-~ cyclic and henceL~has no faithful irreducible characters. Thus ifx2Irr(L);~ then x must be an irreducible character of one of the following groups:

O^‡₈(3);2O^‡₈(3);2⁰O^‡₈(3);or 2⁰⁰O^‡₈(3):We observe that ifxis a nontrivial proper projective irreducible character ofL;thenxis an ordinary character of 2O^‡₈(3);2⁰O^‡₈(3);or 2⁰⁰O^‡₈(3):We have thatV^‡₈(3)2O^‡₈(3) and the three double covers of Lare isomorphic via the graph automorphism of order 3 ofL:Hence it suffices to find the faithful irreducible characters of V^‡₈(3) which divide one of the numbers in A₂: We see that the largest number inA2is 2496:Using [2], the faithful irreducible character degrees ofV^‡₈(3) which are less than 2496 are 520;560 and 1456:However only 520 divides one of the numbers inA2:The proof is now complete. p

4. Verifying Huppert's Conjecture forFi23

We assume thatHFi23andGis a group such that cd(G)ˆcd(H):In this section we will show that GHA; where Ais an abelian group, which confirms Huppert's Conjecture for the sporadic simple groupFi₂₃: We will follow five steps of Huppert's method described in the introduc- tion.

4.1 ±Verifying Step 1

By way of contradiction, suppose that G⁰6ˆG⁰⁰: Then there exists a normal subgroupN/Gsuch thatG=Nis solvable and minimal with respect to being non-abelian. By Lemma 2.3,G=Nis anr-group for some primeror G=N is a Frobenius group.

CASE1: G=Nis anr-group. Then there existsc2Irr(G=N) such that c(1)ˆr^b>1: However by Lemma 3.1(iii); G has no nontrivial prime power degrees. Hence this case cannot happen.

CASE 2: G=N is a Frobenius group with Frobenius kernel F=N;

jF=Nj ˆr^a;15f ˆ jG:Fj 2cd(G) andr^a1(mod f):By Lemma 2.3(b);if x2Irr(G) such that x(1) is isolated then either f ˆx(1) or rjx(1): We observe that there is no prime which divides all the isolated degrees listed in Lemma 3.1(i):Thusf must be one of the isolated degrees in Lemma 3.1(i):Hencef is isolated inG:By Lemma 2.3(b) again, ifx2Irr(G) with r⁴jx(1) thenx(1)jf:As f is isolated we deduce that r must divide every nontrivial degreex(1) ofGsuch thatx(1)6ˆf:

Assume first that f ˆ3¹²1723: Then r must divide all of the re- maining isolated degrees in Lemma 3.1(i): However (2¹⁸5²711, 3¹³1323)ˆ1;which is a contradiction. Hence f 6ˆ3¹²1723 so that r2 f3;17;23g:Suppose thatrˆ3:Then 2¹⁸5²711 and 2¹⁶51323 are two isolated degrees of H; which are both coprime to r; so that f must be equal to both of them by Lemma 2.3(b)(2), which is impossible.

Hence r6ˆ3: Assume next that rˆ17: Then 2¹⁸5²711 and 3¹³1323 are isolated degrees ofH;which are both relatively prime to rˆ17;hence we obtain a contradiction as in the previous case. Finally, assumerˆ23:Then 2¹⁸5²711 and 2⁵3¹²17 are isolated degrees of H and both are coprime to rˆ23; which leads to a contradiction as before. ThusG⁰ˆG⁰⁰:

4.2 ±Verifying Step 2

LetMG⁰be a normal subgroup ofGsuch thatG⁰=Mis a chief factor ofG:AsG⁰is perfect,G⁰=Mis non-abelian so thatG⁰=MS^kfor some non- abelian simple groupSand some integerk1:

CLAIM 1: kˆ1: By way of contradiction, assume that k2: By Lemma 2.4, S possesses a nontrivial irreducible character u; which is extendible to Aut(S) and so by Lemma 2.5, u^k 2Irr(G⁰=M) extends to G=M; henceu(1)^k2cd(G);which contradicts Lemma 3.1(iii):This shows thatkˆ1:HenceG⁰=MS:

CLAIM2: Sis not an alternating group of degree at least 7:B y way of contradiction, assume thatSˆAn;n7:B y Lemma 2.4, Shas two non- trivial irreducible characters u1;u2 with u1(1)ˆn(n 3)=2; u2(1)ˆ u₁(1)‡1ˆ(n 1)(n 2)=2 and bothu_iextend to Aut(S);so thatGpossesses two consecutive nontrivial character degrees contradicting Lemma 3.1(iii):

CLAIM3: Sis not a simple group of Lie type. IfSis a simple group of Lie type in characteristicp;andS6ˆ²F4(2)⁰;then the Steinberg character ofSof degreejSj_pextends to Aut(S) so thatGpossesses a nontrivial prime power degree, which contradicts Lemma 3.1(iii):

CLAIM4: SFi₂₃:By Claims 1;2;and 3;Sis a sporadic simple group or the Tits group. We will eliminate all other possibilities forSand hence the claim will follow. By Ito-Michler Theorem (see [4, Lemma 1]) we deduce that every prime divisor ofSmust divide some character degree ofS;and since every character degree ofSG⁰=Mdivides some character degree ofH;we obtain that every prime divisor of S is also a prime divisor ofH; so that p(S)p(H)ˆ f2;3;5;7;11;13;17;23g:Hence we only need to consider the simple groups in Table 2. For each sporadic simple group or the Tits groupS in Table 2, we exhibit a nontrivial irreducible characteruofSsuch thatu extends to Aut(S) and either (111317;u(1))ˆ1 or (112317;u(1))ˆ1;

which contradicts Lemma 3.1(ii):This finishes the proof of Step 2:

4.3 ±Verifying Step 3

Ifu2Irr(M);u(1)ˆ1;thenI_G⁰(u)ˆG⁰:Letu2Irr(M) andIˆI_G⁰(u):

Assume that I5G⁰ andu^IˆP^s

iˆ1e_if_i; wheref_i2Irr(I);iˆ1;2;. . .;s:Let

U=M be a maximal subgroup ofG⁰=McontainingI=Mand lettˆ jU:Ij:

Thenf_i(1)jG⁰:Ijis a character degree ofG⁰by Lemma 2.2(a);so it divides some character degree of G: Thus tf_i(1)jG⁰:Uj divides some character degree ofGand so the indexjG⁰:Ujmust divide some character degree of H:By Lemma 3.1(iv);one of the following cases holds.

CASE. U=M2Fi22:Then for eachi; tf_i(1) divides one of the num- bers inA1:AsU=Mis perfect, the center of U=Mlies in every maximal subgroup ofU=M and so the indices of maximal subgroups ofU=M and those ofFi₂₂are the same. By inspecting the list of maximal subgroups of Fi₂₂in [3], the index of a maximal subgroup ofU=Mdivides no number in A1 so that tˆ1 and hence IˆU: Recall that u^IˆP^s

iˆ1e_if_i; where f_i2Irr(I);iˆ1;2;. . .;s:Assume first thate_j ˆ1 for some j: Thenuex- tends tou_o2Irr(I):Hence by Gallagher's Theorem,tu₀ is an irreducible constituent ofu^Ifor anyt2Irr(I=M) and thent(1)u0(1)ˆt(1) divides one of the numbers inA1:However we can chooset2Irr(I=M)ˆIrr(2^:Fi22) witht(1)ˆ2729376 and obviously this degree divides none of the numbers inA₁:Thereforee_i>1 for alli:We deduce that for eachi;e_iis the degree of a nontrivial proper irreducible projective representation of 2Fi₂₂: Moreover asf_i(1)ˆeiu(1)ˆei;eacheidivides one of the numbers inA1:It follows that for eachi;e_i2³3³711ˆ16632 ande_iis the degree of a nontrivial proper irreducible projective representation of 2Fi₂₂:Using [3], we obtain e_i2 f3³13;23³1113g for all i: As u^IˆP^s

iˆ1e_if_i; and f_i(1)ˆei;we deduce thatjI=Mj ˆP^s

iˆ1e²_i:Letaandbbe the numbers ofe⁰_is which equal 3³13 and 23³1113;respectively. We have

2¹⁸3⁹5²71113ˆ3⁶13²a‡2²3⁶11²13²bˆ13²(3⁶a‡2²3⁶11²b);

which is impossible. Thus this case cannot happen.

CASE. U=MO^‡₈(3):S₃: Then for each i; tf_i(1) divides one of the numbers in A₂: Let M/W/U such that W=MO^‡₈(3): We have that M/I\W/IandM/I\WW:AssumeW6I:ThenI^<₄WIU and tˆ jU:Ij ˆ jU:WIj jWI:Ij:NowjWI:Ij ˆ jW:W\Ijand hencetis divisible by jW:W\Ij and also as W=MO^‡₈(3); t is divisible by the index of some maximal subgroup ofO^‡₈(3):Thus some index of maximal subgroup of O^‡₈(3) divides one of the numbers in A₂: However by in- specting the list of maximal subgroups ofO^‡₈(3) in [3], we see that this is

impossible and thus WIU: Write u^W ˆf1m₁‡f2m₂‡ ‡fsm_s; where m_i2Irr(Wju): As W/I; we obtain that for each i; m_i(1) divides one of the members in A2: If f_j ˆ1 for some j then u extends to u₀2Irr(W). Hence by Gallagher's Theorem, tu₀ is an irreducible con- stituent ofu^W for anyt2Irr(W=M) and thent(1)u₀(1)ˆt(1) divides one of the numbers in A₂: However we can choose t2Irr(W=M) with t(1)ˆ716800 and obviously this degree divides none of the numbers in A2:Thereforef_i>1 for alli:We deduce that for eachi;f_iis the degree of a nontrivial proper irreducible projective representation of O^‡₈(3):

Moreover as m_i(1)ˆf_iu(1)ˆf_i; each f_i divides some of the numbers in A₂;and so by Lemma 3.2, we obtain thatf_iˆ520ˆ2³513 for alli:It follows that

jW=Mj ˆ2¹²3¹²5²713ˆX^s

iˆ1

f_i²ˆs2⁶5²13²; which is impossible. Thus we conclude that u isG⁰-invariant.

4.4 ±Verifying Step 4

Mˆ1:By Step 3 and Lemma 2.5, we haveM⁰ˆ[G⁰;M] and jM:M⁰j divides the order of the Schur multiplier of H; which is trivial so that jM:M⁰j ˆ1:ThusMˆM⁰:IfMis abelian then we are done. Hence we assume thatMˆM⁰is nonabelian. LetNMbe a normal subgroup ofG⁰ such thatM=Nis a chief factor ofG⁰:ThenM=NS^kfor some nonabelian simple group S: By Lemma 2.4, there exists an irreducible character t2Irr(S) such that t(1)>1 andt extends to Aut(S):By Lemma 2.5,t^k extends toc2Irr(G⁰);and so by Gallagher's Theorem,cb2Irr(G⁰) for all b2Irr(G⁰=M):Letx2Irr(G⁰=M)ˆIrr(Fi23) such thatx(1) is the largest character degree of H:Thenc(1)x(1)ˆt^k(1)x(1) divides some character degree ofG;which is impossible. HenceMˆ1:

4.5 ±Verifying Step 5

GˆG⁰C_G(G⁰): It follows from Step 4 that G⁰H is a nonabelian simple group. Let CˆC_G(G⁰): ThenG=Cis an almost simple group with socleG⁰:As Out(H)ˆ1;we deduce thatGˆG⁰C:AsG⁰is simple, we have G⁰\Cˆ1 so thatGˆG⁰CˆG⁰C_G(G⁰):It follows thatC_G(G⁰)G=G⁰ is abelian. The proof is now complete.

TABLE1. - Maximal subgroups of the Fischer groupFi23.

Group Index

2Fi22 3⁴1723

2²U6(2)2 3⁷5131723

(2²2^1‡8)(3U4(2))2 3⁸5711131723 2^6‡8(A7S3) 3¹⁰511131723

2¹¹M23 2⁸3³5111723

O^‡₈(3):S3 2⁵111723

S3V7(3) 2⁸3³5111723 S4Sp6(2) 2⁶3⁸511131723

Sp8(2) 2²3⁸111323

S12 2⁸3⁸131723

S4(4):4 2⁸3¹¹7111323 L2(23) 2¹⁵3¹²5²71317 3^1‡8_‡ :2^1‡6:3^1‡2_‡ :2S4 2⁷5²711131723 3^3‡1‡3‡3(2L3(3)) 2¹³5²7111723

TABLE2. - Character degrees of sporadic simple groups and Tits group.

Group Character Degree Group Character Degree

M11 x₉ 3²5 J2 x₆ 2²3²

M12 x₇ 23³ Co3 x₆ 2⁷7

M22 x₂ 37 Fi22 x₅₇ 3⁹713

M23 x₃ 3²5 HS x₇ 5²7

M24 x₇ 2²3²7 Co1 x₂ 2²323

McL x₁₄ 3⁶7 ²F4(2)⁰ x₂₀ 2⁶3³

He x₁₅ 2⁷7² Suz x₄₃ 2¹⁰3⁵

Co2 x₂₂ 3⁶5³

REFERENCES

[1] M. BIANCHIet al.,Character degree graphs that are complete graphs,Proc.

Amer. Math. Soc.,135, no. 3 (2007), pp. 671-676.

[2] W. BOSMA- J. CANNON- C. PLAYOUST,The Magma algebra system. I. The user language,J. Symbolic Comput.,24(1997), pp. 235-265.

[3] J. H. CONWAY- R. T. CURTIS- S. P. NORTON- R. A. PARKER- R. A. WILSON, Atlas of Finite Groups,Oxford University Press, Eynsham, 1985.

[4] B. HUPPERT, Some simple groups which are determined by the set of their character degrees I, Illinois J. of Math.,44(2000), pp. 828-842.

[5] M. ISAACS, Character theory of finite groups, AMS Chelsea Publishing, Providence, RI, 2006.

[6] P. KLEIDMAN- R. A. PARKER- R. A. WILSON,The maximal subgroups of the Fischer group Fi23;J. London Math. Soc., (2)39, no. 1 (1989), pp. 89-101.

[7] H. P. TONG-VIET- T. P. WAKEFIELD,On Huppert's Conjecture for the Monster and Baby Monster,Monatsh. Math. DOI: 10.1007/s00605-011-0309-1 Manoscritto pervenuto in redazione il 15 gennaio 2011.