Some simple groups which are determined by the set of their character degrees. II

Some Simple Groups which are determined by the Set of their Character Degrees II

BERTRAMHUPPERT(*)

To Guido Zappa on his 90th birthday

In part I of this paper we began to study the following conjecture.

CONJECTURE. LetHbe a simple nonabelian group. If Gis any group such thatGandHhave the samesetof character degrees of irreducible characters overC, thenGHA, whereAis abelian.

In part I we proved this conjecture ifHis a Suzuki group Sz(q) or (using an unpublished resultby F. LuÈbeck) if HˆSL(2;2^f) for somef. In this paper we study the case HˆPSL(2;p^f), where p is odd. As PSL(2;5)PSL(2;4)A₅, we can assume thatp^f >5.

The proof follows the same pattern as in part I. Also references to lemma 1 to lemma 6 refer to part I.

In section 4, we give a proof, which relies on the following lemma.

LEMMA1. Let p be a prime. We say that a group G has property (p), if the following holds:

(1) Ifx2IrrG, thenx…1†is a power of p or prime to p.

(2) There exists somex2IrrG such thatx…1† ˆp^d>1.

If G is a simple group with property (p), then either GPSL(2;p^d), where p^d>3;or GPSL(2;2^f), where pˆ2^f 1on pˆ2^f ‡1;or pˆ3 and2^f ˆ8.

(As a simple group has some even degree, so forpˆ2 condition (2) can be omitted).

(*) Indirizzo dell'A.: Weinbietstrasse 26, 67117 Limburgerhof, Germania.

PROOF. a) PSL(2,q) has the degrees q, q‡1, q 1 and if 26 jq the odd one of q1

2 . The cases, where q‡1 on q 1 is a power of some prime, are the cases stated above. If the odd degree q1

2 is a prime- powerp^a, t henp^a and 2p^aare degrees, so condition (p) is notsatisfied.

b) We have to show that all other simple groups do not have property (p) for any prime p. All quasi-simple groups, which have some prime-power degree, have been determined in Malle-ZaleskiõÆ [10]. We have to go through the list of such groups, following the numbering in [10]. All statements about degrees of sporadic and other small simple groups come from the Atlas.

(I have to thank G. Malle for very detailed information).

(1) LetGbe a simple group of Lie type of characteristicpand letxbe the Steinberg character ofGwithx…1† ˆ jGj_pˆq^d.

The listin Curtis-Iwahori-Kilmoyer [2] on p. 111 gives in mostcases a degree, which is divisible byp, butnota power ofp. The exceptions are the following:

PSL(2,q), as mentioned in a).

B_l…2† ˆSp(2l;2) andpˆ2. Butby [2], p. 114 B_l(2) has the degree 2^{(l 1)2}(2^{l 1}‡1)(2^l‡1)

23 ;

which is even for l3, butnota power of 2. Observe that B₂(2)ˆSp(4;2)S⁰₆.

By [2], p. 114F₄…2†has the degree 23²5²1317.

G2…3†has the degrees 2⁶and 3⁶, butalso 56 and 21.

B¹₂…q† ˆSz(q) has the degree (q 1) 2ⁿ, whereqˆ2^2n‡1(Suzuki [11]).

G¹₂(q) has the degreeq(q² q‡1), whereqˆ3^2n‡1(Ward [12], p. 87).

(Observe that asA8PSL…4;2) the degree 2⁶ofA8has been handled.) (2) The caseGˆPSL(2;q) has been considered in section a).

(3) Let GˆPSL(n;q), where q>2, n is an odd prime such that (n;q 1)ˆ1 and x…1† ˆqⁿ 1

q 1 is a power of a prime (which happens sometimes). By Carter [1], 13:8 for n4 the group PSL…n;q) has the

degree

q²…qⁿ 1†

q 1

…q^{n 3} 1†

q² 1 : This is a multiple of the prime-powerqⁿ 1

q 1, butalso a multiple ofq.

For PSL…3;q) we have the degreesq²‡q‡1 andq(q²‡q‡1).

(4) LetGˆPSU(n;q), wherenis an odd prime such that (n;q‡1)ˆ1 andx…1† ˆqⁿ‡1

q‡1 is the power of a prime. By the Lusztig- Srinivasan cor- respondence between degrees of PSL(n;q) and PSU(n;q), replacingqby

q, asnis odd, we obtain from (3) forn4 the degree q²…qⁿ‡1†

q‡1

…q^{n 3} 1†

q² 1 for PSU(n;q).

PSU(3;q) has the degreesq² q‡1 andq(q² q‡1) (Klemm [9]).

(5) LetGˆSp…2n;q),n>1,qˆr^kwith an odd primer,kn a power of 2 andx…1† ˆqⁿ‡1

2 a power of a prime. By [1], p. 111 Sp(2n,q) has the degree mˆq(q^{n 1}‡1)

q‡1

…qⁿ‡1†

2 :

Asnis a power of 2, so n 1 is odd and hence q^{n 1}‡1

q‡1 ˆ1 q‡q² . . .‡q^{n 2}

is integral. Hencemis divisible byqandqⁿ‡1 2 . (6) LetGˆSp(2n;3),n>1 a prime,x…1† ˆ3ⁿ 1

2 a power of a prime.

By [2], p. 111 Sp(2n;3) has the degree 3(3ⁿ 1)

2

(3^{n 1}‡1)

2 :

(7) LetGˆA_n, wherenˆp^d‡110 andx(1)ˆp^d. ThenA_nhas also the degree.

(n 1)n 2

2 ˆp^d(p^d 1)

2 :

Observe thatA₈PSL(4;2) has the degrees 7 and 2⁶, butalso 14. The cases

A5PSL(2;5) and A6PSL(2;9) have been considered in a).

(11) Let GˆSp(6;2) and x(1)ˆ7. Then G has the degree 21. (The degrees 2⁶ and 3³ of Sp(6;2) are treated in (1) and (17).)

(12)A₆PSL(2;9) has already been mentioned.

(14)M₁₁ andM₁₂ have the degree 11, but also 44 resp. 66.

(15)M11 andM12 have the degree 2⁴, butalso 44 resp. 66.

PSL(3, 3) has the degree 2⁴, butalso 12.

(16)M24, Co2and Co3have the degree 23, but also 1123 resp. 9923 resp. 801923.

(17)A9, Sp(6;2) and²F4(2)⁰have the degree 3³, butalso 237 resp. 21 resp. 2313.

(18) PSU(3, 3) has the degree 2⁵, butalso 28.

(19)G2(3) has the degree 2⁶, butalso 2313.

Hence all cases have been considered.

4. The group PSL(2,p^f), wherep6ˆ2.

Forp^f >5 we have

cd PSL(2;p^f)ˆ f1;p^f 1;p^f;p^f ‡1;rg;

where

rˆp^f 1

2 if p^f 1 (mod 4) rˆp^f‡1

2 if p^f 1 (mod 4):

Observe thatris odd.

Ifp^f 6ˆ9, then SL(2;p^f) is the Schur covering group of PSL(2;p^f). The degrees of the properly projective irreducible representations of PSL(2;p^f) arep^f 1,p^f ‡1 ands, where

sˆp^f ‡1

2 if p^f 1 (mod 4) sˆp^f 1

2 if p^f 1 (mod 4):

(see Dornhoff [3], p. 228). Butobserve thatPSL(2;9)A6 has a Schur multiplier of order 6 and has some more degrees of projective re- presentations (Atlas, p. 5).

THEOREM3. Suppose that p>2, p^f >5and

cdGˆcd PSL(2;p^f)ˆ f1;p^f 1;p^f;p^f ‡1;rg;

where

rˆ

p^f 1

2 if p^f 1 (mod 4) p^f ‡1

2 if p^f 1 (mod 4):

8>

>>

<

>>

>:

Then

GPSL(2;p^f)A;

where A is abelian.

PROOF. Step 1. We haveG⁰ˆG⁰⁰:

Otherwise letG/Nbe a solvable, minimal nonabelian factor group ofG.

Suppose atfirstthatG/N is aq-group for some primeq. Ifqˆpand x2Irr (G) such thatx(1)ˆp^f 1, then x_N2Irr(N) and by lemma 2 of partI thereforext2Irr(G) for allt2Irr(G=N). As nowp^f 2cdG=N, we obtain the forbidden degree

x(1)t(1)ˆ(p^f 1)p^f:

Ifq6ˆp, we apply the same argument tox2Irr (G) withx(1)ˆp^f. Hence by lemma 4 of partI we can assume thatG/N is a Frobenius group with elementary abelian Frobenius kernelF/N. ThenjG=Fj 2cdG andjF=Nj ˆq^afor some primeq, wherejG=Fjdividesq^a 1.

If

jG=Fj 2 fp^f 1;p^f;p^f ‡1g;

then asp^f >3 no proper multiple ofjG=Fjis in cdG. Hence by lemma 4 of partI, ifx2IrrGandq6 jx(1), thenx(1) dividesjG=Fj.

Suppose atfirstjG=Fj ˆp^f, henceq6ˆp. Ifq6 jp^f 1, we obtain the contradiction thatp^f 1 dividesp^f. Ifq6 jp^f ‡1, we get the contradiction

thatp^f‡1 dividesp^f. Henceqdividesp^f 1 andp^f ‡1, henceqˆ2. But as 26 jr, hence lemma 4 of part I provides the contradiction thatrˆp^f 1 dividesp^f. 2

Suppose nextthatjG=Fj ˆp^f 1. Ifq6ˆp, we takex2IrrGsuch that x(1)ˆp^f to obtain the contradiction that p^f dividesjG=Fj. Ifqˆp, we obtain either the contradiction thatp^f ‡1 dividesp^f 1 orp^f 1 divides p^f ‡1. (Observe thatp^f >3).

There remains only the case that

jG=Fj ˆrˆp^f 1 2 : By lemma 4 of partI, ifc2IrrF, theneither

jG=Fjc(1)2cdG;

soc(1)2,or qdividesc(1).

Letx2IrrGsuch that

x(1)ˆ p^f 1 if jG=Fj ˆp^f ‡1 2 p^f ‡1 if jG=Fj ˆp^f 1

2 8>

><

>>

: As 26 j jG=Fj, so

jG=Fj;x(1)

ˆ1:

Therefore x_F2IrrF and x(1)>2. Hence q divides x(1), so q6ˆp. If t2IrrGandt(1)ˆp^f, then alsot_F2IrrF, butq6 jt(1).

This contradiction showsG⁰ˆG⁰⁰. Step 2. IfG⁰=Mis a chief factor ofGthen

G⁰=MPSL(2;p^f):

AsG⁰ˆG⁰⁰, we have

G⁰=MˆS₁. . .S_k;

whereS_iSis simple and nonabelian. The degrees ofSdivide degrees of G, hence are prime topor powers ofp.

Suppose atfirstthatp6 j jSj. Letx2IrrGandx(1)ˆp^f. AsG⁰ˆG⁰⁰, so x_G⁰ cannotsplitin characters ofG⁰of degrees 1. Hence

x_G0ˆX

i

'_i; '_i2IrrG⁰

and 'i(1)ˆp^r>1. As by our assumption 'i(1) is prime to jG⁰=Mj, so ('i)_M2IrrM and 'it2IrrG⁰ for all t2Irr (G⁰=M). As G⁰=M is a non- abelianp⁰-group, there existst2Irr (G⁰=M) such thatt(1)>1 andp6 jt(1).

Butthen'_ithas a ``mixed'' degree, a contradiction.

HencepjjSj. By the theorem of Ito and Michler there existss2IrrS such thatpjs(1). Hences(1) is a power ofplarger than 1. Therefore we can apply lemma 1 of this paper.

Case 1. SupposeSPSL(2;p^m) for somem. Obviouslykˆ1, hence G⁰=MPSL(2;p^m):

Ifc2Irr (G⁰=M) andc(1)ˆp^m, thenc(1) dividesx(1) for somex2IrrG.

Hencex(1)ˆp^f, somf.

Supposem<f. We considerGˆG=C_G(G⁰=M). ThenG⁰PSL(2;p^m) andjG:G⁰jdividesjOutPSL(2;p^m)j ˆ2m.

Take atfirstc₁2IrrG⁰ such thatc₁(1)ˆp^m. If we choosex₁2IrrG such that c₁;(x₁)_G⁰

G⁰>0, thenx₁(1)ˆp^f and (x₁)_G⁰ ˆX^k

iˆ1

c^g₁ⁱ;

wherekdividesjG=G⁰j. Hencex(1)ˆp^f ˆkp^mdivides 2mp^m. Aspis odd, sop^f dividesmp^m.

Take c₂2IrrG⁰ such that c₂(1)ˆp^m 1. As c₂(1) divides some de- gree ofG, sop^m 1 dividesp^2f 1. Thereforemdivides 2f. Asm<f, so m2

3 f. Hence

p^f mp^m2 3fp^2f=3: This produces the contradiction

3^f=3p^f=32 3f: Hencemˆf andG⁰=MPSL(2;p^f).

Case 2. Now we have to exclude the possibility that SPSL(2;2^s), where pˆ2^s 1 or pˆ2^s‡1 or pˆ3 and SPSL(2;8). Obviously kˆ1, hence

G⁰=MPSL…2;2^s):

We putGˆG=CG(G⁰=M). ThenjG:G⁰j dividesjOutPSL(2;2^s)j ˆs. Let c2IrrG⁰ and c(1)ˆpor c(1)ˆ3². Ifx2IrrGandx is above c, then x(1)ˆec(1), whereedividess. Aspdividesx(1), sox(1)ˆp^f dividessc(1).

Ifc(1)ˆp, we obtainsp^{f 1}and then pˆ2^s12^{pf 1}1;

which impliesf ˆ1.

Suppose atfirstthatpˆ2^s‡1. Ift2IrrG⁰andt(1)ˆ2^s 1, thent(1) dividesp 1ˆ2^s orp‡1ˆ2^s‡2. This implies

2^s‡2ˆ2(2^s 1);

a contradiction asp^f ˆpˆ2^s‡1>5.

Nextsuppose thatpˆ2^s 1. Ift2IrrG⁰andt(1)ˆ2^s‡1, we again obtain that 2^s‡1 divides p‡1ˆ2^s or p 1ˆ2^s 2, in both cases a contradiction. Hence there remains only the possibility that pˆ3 and SPSL(2;8). We takec2IrrG⁰such thatc(1)ˆ3² andx2IrrGsuch that x is above c. Then c(1)ˆ3² divides x(1)ˆ3^f and x(1) divides 3c(1)ˆ3³. As 72cd PSL(2;8), so 7 divides some degree ofG. As

cdGcdGˆ f1;3^f 1;3^f ‡1;rg;

whererˆ1

2(3^f 1), sof ˆ3 and

cdGˆ f1;13;26;27;28g:

But there exists2IrrG⁰ such that(1)ˆ8, and 8 does notdivide any degree ofG.

Hence this case is impossible.

Step 3. If#2IrrM, thenIG⁰(#)ˆG⁰and thereforeM⁰ˆ ‰M;G⁰Š:

SupposeIˆI_G⁰(#)<G⁰for some#2IrrMand

#^IˆX

i

W_i; W_i2IrrI:

ThenW^G_i⁰2IrrG⁰, hence

jG⁰:Ij '_i(1)2cdG⁰:

The proper subgroups ofG⁰=MPSL(2;p^f) are of indices atleastp^f ‡1, with the exceptions (observep^f >5) that

p^f ˆ jG⁰:Ij ˆ7 and I=MS₄; p^f ˆ jG⁰:Ij ˆ11 and I=MA5; p^f ˆ9; jG⁰:Ij ˆ6 and I=MA₅:

(Huppert[6], p. 214.) The lastpossibility is excluded as 6 does notdivide any degree of PSL(2, 9). So in all cases'i(1)ˆ1, hence'iis an extension of#to I. Therefore

('it)^G0 2IrrG⁰

for allt2IrrI=M. SojG⁰:Ij t(1) divides some degree ofG.

IfjG⁰:Ij ˆp^f ‡1, thenI=M is metabelian of order p^f(p^f 1)=2, so there exists t2IrrI=M such that t(1)ˆ1

2(p^f 1). Butthis provides a contradiction as1

2(p^f ‡1)(p^f 1) does notdivide any degree ofG.

In the remaining exceptional cases we can choose t(1)ˆ3 if p^f ˆ7 and I=MS₄; t(1)ˆ4 if p^f ˆ11 and I=MA5: This produces the forbidden degrees 37 resp. 411 ofG⁰.

HenceI_G⁰(#)ˆG⁰for all#2IrrM. Therefore by lemma 6 of partI we obtainM⁰ˆ ‰M;G⁰Š.

Step 4. We havejM=M⁰j 2:

By lemma 6 of partI, jM=M⁰j is bounded by the order of the Schur multiplier of PSL(2;p^f). ThereforejM=M⁰j 2 ifp^f 6ˆ9 (Huppert[6], p.

646). But the Schur multiplier of PSL(2;9)A₆has order 6.

Suppose p^f ˆ9 andjM=M⁰j>2. From the Atlas, p. 5 we obtain the degrees of the irreducible characters ofG⁰=M, which do nothaveM=M⁰in their kernel, namely

3; 6; 9; 15 if jM=M⁰j ˆ3; 6; 12 if jM=M⁰j ˆ6:

In this case we have

cdGˆcd PSL(2;9)ˆ f1;5;8;9;10g:

As 6 does notdivide any degree ofG, sojM=M⁰j>2 is notpossible in the casep^f ˆ9.

ThereforejM=M⁰j 2 in all cases.

Step 5. We claim that cdM f1;2gand henceM⁰⁰ˆE:

Suppose#2IrrMand#(1)>1. If#allows an extension#₀toG⁰, then

#0t2IrrG⁰for allt2Irr (G⁰=M). Takingt(1)ˆp^f ‡1, we obtain (p^f ‡1)#(1)2cdG⁰:

Butas#(1)>1, so (p^f ‡1)#(1) does notdivide any degree ofG. Hence# does not allow any extension toG⁰.

If'2IrrG⁰and ('_M; #)_M>0, then, asI_G⁰(#)ˆG⁰by lemma 3c) of part I, 'ˆ#₀t₀, where#₀ and t₀ are characters of irreducible projective re- presentations ofG⁰,#₀(1)ˆ#(1) and t₀is the character of an irreducible, projective, non ordinary representation of G⁰=MPSL(2;p^f). The de- grees of these representations arep^f 1,p^f ‡1 ands, where

sˆp^f ‡1

2 if p^f 1 (mod 4);

sˆp^f 1

2 if p^f 1 (mod 4):

(See Dornhoff [3], p. 228.) As#(1)>1, so (p^f 1)#(1)62cdG⁰:

There remains only the possibility that s#(1)2cdG⁰, which implies

#(1)2. Hence cdM f1;2g, and thereforeM⁰⁰ˆE(Isaacs [8], p. 202).

Step 6. We have cdG=MˆcdG:

As

cdG⁰=Mˆ f1;p^f 1;p^f;p^f ‡1;rg and cdG=McdG, we see immediately that

p^f 1;p^f;p^f ‡12cdG=M:

Takex2IrrGsuch that

x(1)ˆrˆp^f 1 2 : Asris odd and cdM f1;2gby step 5, so

x_MˆX

i

l_i; l_i(1)ˆ1:

HenceM⁰kerx. IfMˆM⁰, thenx2IrrG=Mand x(1)ˆr2cdG=M:

SupposejM=M⁰j ˆ2. AsG⁰ˆG⁰⁰, so

x_G0ˆX

i

r_j; r_j2IrrG⁰; r_j(1)>1:

Ifr_jis faithful onM=M⁰, thenr_j(1) is one of the valuesp^f 1 ors. Butas r_j(1) dividesx(1)ˆr, this is impossible. HenceMkerr_j, sox2IrrG=M also in this case.

Step 7. We now claimG=MˆG⁰=MC_G=M(G⁰=M):

The characters of G⁰=M are all invariantunder G, for otherwise by fusion of characters of G⁰=M we obtain a forbidden degree which is a proper multiple ofp^f orp^f 1, or by fusion of the two characters of degree rˆ1

2(p^f 1) we lose the degreer, which by step 6 is a degree ofG=M. As the characters ofG⁰=Mseparate the conjugacy classes, soGpreserves the conjugacy classes of G⁰=M. Hence by a theorem of Feit and Seitz [4], G induces only inner automorphisms onG⁰=M. (AsG⁰=MPSL(2;p^f), this can be seen directly.) Therefore

G=MˆG⁰=MC_G=M(G⁰=M):

Step 8. We haveMˆE, and hence

GPSL(2;p^f)A;

whereAobviously is abelian:

By st ep 4 we know jM=M⁰j 2. Suppose jM=M⁰j ˆ2. If p^f 6ˆ9, then G⁰=M⁰ is the uniquely determined Schur covering group of G⁰=MPSL(2;p^f), so

G⁰=M⁰SL(2;p^f):

This also is true forp^f ˆ9 andjM=M⁰j ˆ2.

We put C=MˆC_G=M(G⁰=M). If x2G⁰=M⁰ and ordxˆp, y2C=M⁰, thenx^y2xM=M⁰. As ordx^yˆp>2, this impliesx^yˆx. ThereforeC=M⁰ centralizesG⁰=M⁰. Hence by step 7

G=M⁰ˆG⁰=M⁰C=M⁰

is a central product with amalgamated subgroup M=M⁰ˆ hzM⁰i. Take c2IrrC=M⁰ such that c(z)ˆ c(1). If x2IrrG⁰=M⁰ andx(z)ˆ x(1), then

x c2Irr (G⁰=M⁰C=M⁰)

and

x(z)c(z ¹)ˆx(1)c(1):

Hence (z;z ¹)2kerx c. Thereforexcis a character of

(G⁰(M⁰C=M⁰)=h(zM⁰;z ¹M⁰)i G⁰=M⁰C=M⁰ˆG=M⁰:

Hencex(1)c(1)2cdG. The charactersxofG⁰=M⁰withM=M⁰notin their kernel have the degrees p^f 1, p^f ‡1 and s. Hence (p^f 1)c(1), (p^f ‡1)c(1) and sc(1) are in cdG. This implies c(1)ˆ1 and then the contradiction s2cdG. Hence jM=M⁰j ˆ1, so by step 5 MˆM⁰ˆ

ˆM⁰⁰ˆE. p

Added during proof. Recently I was informed of the following results:

LetGbe nonsolvable.

a) If the degree graphD(G) has 3 components, thenGPSL (2;2^f) A;whereAis abelian.

b) IfD(G) has 2 components, then PSL (2;p^f) is the only nonsolvable composition factor ofG.

(M. L. Lewis, D. L. White, J. Algebra266(2003), 51-76 and283(2005), 80-92.)

AsD(G) for solvableGhas atmost2 components, soa) proves theorem 2 of [7].

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Manoscritto pervenuto in redazione il 15 marzo 2005