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Finite Nonsolvable Groups in Which Only Two Nonlinear Irreducible Characters Have Equal Degrees
JOURNAL OF ALGEBRA ARTICLE NO.
184, 538]560 Ž1996.
0273
Finite Nonsolvable Groups in which Only Two Nonlinear Irreducible Characters Have Equal Degrees* Yakov Berkovich Department of Mathematics and Computer Science, The Uni¨ ersity of Haifa, 31905 Haifa, Israel
and Lev Kazarin Department of Mathematics, Yarosla¨ l State Uni¨ ersity, Yarosla¨ l 150000, Russia Communicated by Walter Feit Received May 23, 1995
DEDICATED TO PROFESSOR AVINOAM MANN ON THE OCCASION OF HIS
60TH BIRTHDAY
Y. Berkovich et al. w Proc. Amer. Math. Soc. 115 Ž1992., 955]959x classified finite groups in which the degrees of the nonlinear irreducible characters are distinct. Theorem 24.7 from wY. Berkovich, J. Algebra 184 Ž1996., 584]603x contains the classification of solvable groups in which only two nonlinear irreducible characters have equal degrees Ž D 1 -groups.. In this paper we obtain the classification of nonsolvable D 1-groups, completing the classification of D 1-groups. Our proof depends on the classification of finite simple groups. The results of the important paper w Illinois J. Math. 33, No. 1 Ž1988., 103]131x on rational simple groups play a key role as well. Q 1996 Academic Press, Inc.
*The first author was supported in part by the Ministry of Absorption of Israel. The second author was supported in part by ISF. 538 0021-8693r96 $18.00 Copyright Q 1996 by Academic Press, Inc. All rights of reproduction in any form reserved.
FINITE NONSOLVABLE GROUPS
539
1 Let IrrŽ G . be the set of all irreducible characters of G, Irr1Ž G . the set of the nonlinear characters in IrrŽ G ., and LinŽ G . the set of the linear characters of G. Let cd 1Ž G . denote the set of degrees of the characters in Irr1Ž G .. DEFINITION. Let k be a nonnegative integer. A group G is said to be a D k-group if
L2 Ž5. and L2 Ž7. are the only nonsol¨ able D 1-groups.
In the following we denote by Ž A, B . a Frobenius group with complement A and kernel B. We are indebted to Avinoam Mann for useful discussions and help. 2 In what follows Žsee the proof of Theorem 2.1. we use the following Remarks. 1. Let Z be a subgroup of a nonabelian p-group G such that < CG Ž Z .< s p 2 . We will prove that G is a p-group of maximal class. We use induction on < G <. Obviously, ZŽ G . - CG Ž Z . and CG Ž CG Ž Z .. s CG Ž Z .. Therefore < NG Ž CG Ž Z ..< s p 3. Then < CG r ZŽG.Ž CG Ž Z .rZŽ G ..< s p 2 , so, by induction, GrZŽ G . is abelian of order p 2 or a p-group of maximal class, and the result follows. 2. Let H be a proper subgroup of a group G such that all elements of G y H are involutions. Suppose that G is not an elementary abelian 2-group. We will prove that < G : H < s 2. Assume that < G : H < ) 2. Without loss of generality we may assume that H s ² G y InvŽ G .:, where InvŽ G . is the set of all involutions in G. Then H eG and GrH is a noncyclic elementary abelian 2-group. Since G is not an elementary abelian 2-group, there exists an element a g H such that oŽ a. ) 2. Let x, y g G y H be such that xH / yH. Since xa, ya are involutions we have xax s ay1 ,
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yay s ay1 . Therefore axy s xya and Ž axy . 2 s a2 / 1, contradicting axy f H Žsince xy f H, xy is an involution.. In this section we will prove the following important component of the Main Theorem: THEOREM 2.1.
If G is a nonsol¨ able D 1-group, then G9 s G.
Proof. Suppose that G9 - G. Then G contains a normal subgroup H of a prime index p. Ži. p s 2. Assume that p ) 2. By wBZ, Proposition 25.9x and the Remark following it, there exists x g Irr1Ž G . such that p ¦ x Ž1.. Then, by Clifford theory, x H g IrrŽ H ., and Ž x H . G s x 1 q ??? qx p , where IrrŽŽ x H . G . s x 1 , . . . , x p 4 , and x 1 , . . . , x p have the same degree x Ž1., a contradiction, since p ) 2. Our claim follows. Žii. GrG9 is a cyclic 2-group. It follows from Ži. that GrG9 is a 2-group. Assume that GrG9 is noncyclic. Then there exists a normal subgroup F of G such that GrF is elementary abelian of type Ž2, 2.. As in Ži., there exists in Irr1Ž G . a character x g Irr1Ž G . of odd degree. By Clifford theory, x F s u g IrrŽ F .. Again, by Clifford theory, u G s eŽ x 1 q ??? qx t ., where e, t are natural numbers, e < x Ž1.. Since e <2, we have e s 1. Since x 1Ž1. s ??? s x t Ž1., we have t F 2 Ž G is a D 1-group.. It follows from 4 x Ž1. s 4u Ž1. s < G : F < u Ž1. s u G Ž1. s t x 1Ž1. s t x Ž1. that t s 4, a contradiction. Thus, GrG9 is a cyclic 2-group. Žiii. < G : G9 < s 2. As above, there exists x g Irr1Ž G . such that x Ž1. is odd. Then u s xG9 g IrrŽ G9.. Since all irreducible constituents of u G have the same degree x Ž1., and, by Clifford theory, ² x , u G : s 1 then
k Ž G9 . y 2 2
q2?2s
k Ž G9 . q 6 2
.
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FINITE NONSOLVABLE GROUPS
Let k G Ž G9. be the number of G-classes containing in G9. Then k G Ž G9 . s
k Ž G9 . y 2 2
q2s
k Ž G9 . q 2 2
.
Thus, k Ž G . y k G Ž G9. s Ž k Ž G9. q 6.r2 y Ž k Ž G9. q 2.r2 s 2. Let G y G9 s M j N, where M, N are G-classes. Put < G < s g, < M < s grm, < N < s grn. Then gr2 s grm q grn, or 1r2 s 1rm q 1rn. Suppose that m F n. Let x g M, y g N. Since < G : G9 < s 2 then oŽ x . and oŽ y . are even so m and n are even, and hence m s n s 4. By the above, < CG Ž x .< s 4. Therefore P g Syl 2 Ž G . is of maximal class ŽRemark 1.. Since P l G9 is noncyclic in view of solvability of groups of odd order, G y G9 contains an element y of order expŽ P ., so y g N. Therefore, < CG Ž y .< s 4. Thus, expŽ P . s 4 and P is dihedral of order 8. In particular, G y G9 contains an involution z. Since < G : CG Ž z .< s 4 then the fraction of involutions in G is greater than 14 . This contradicts Theorem 28.3 from wBZx.
3 Let G be a nonsolvable D 1-group, H a maximal normal subgroup of G. It follows from Theorem 2.1 and wBCHx that GrH is a simple D 1-group. In this section we consider two cases when GrH is isomorphic to L2 Ž5. or L2 Ž7.. LEMMA 3.1. Let 14 - H eG and GrH isomorphic to L2 Ž5. or L2 Ž7.. If the irreducible constituents of lG ha¨ e distinct degrees for each nonprincipal l g IrrŽ H ., then H is sol¨ able. Proof. We may assume that H ) 14 . Take a nonprincipal l g IrrŽ H . and put Irr Ž lG . s x 1 , . . . , x k 4 ,
lG s e1 x 1 q ??? qe k x k ,
e1 ) ??? ) e k ) 0 Žall e i are pairwise distinct since all x i have pairwise distinct degrees.. Let
x Hi s e i Ž l1 q ??? ql t .
Ž l1 s l .
be the Clifford decomposition. Then < G : H < ty1 s IG Ž l . : H s e12 q ??? qe k2 .
Ž ).
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BERKOVICH AND KAZARIN
Here IG Ž l. is the inertia group of l in G, and e1 , . . . , e k are degrees of irreducible projective representations of IG Ž l.rH. Ža. Let GrH ( L2 Ž5.. In this case e i g 1, 2, 3, 4, 5, 64 Žrecall that GrH has SLŽ2, 5. as a representation group, and the degrees of irreducible projective representations of subgroups of SLŽ2, 5. are equal to one of the numbers 1, 2, 3, 4, 5, 6. and cd Ž SL Ž 2, 5 . . s 1, 2, 3, 4, 5, 6 4 ,
t g 1, 5, 6, 10, 12, 15, 20, 30, 60 4
Žnumber t equals an index of a subgroup in GrH .. Ž1a. t s 1. Then, by Ž*., e12 q ??? qe k2 s 60. Since 5 2 q 4 2 q 3 2 q 2 2 q 12 - 60, e1 s 6. Since 6 2 q 5 2 ) 60, e2 - 5. Since 6 2 q 3 2 q 2 2 q 12 - 60, e2 s 4. Since 6 2 q 4 2 q 3 2 ) 60, e3 - 3. Since 6 2 q 4 2 q 2 2 q 12 - 60, we obtain a contradiction. Ž2a. t s 5. Then, by Ž*., e12 q ??? qe1k s 12 and IG Ž l.rH ( A 4 , so e i g 1, 2, 34 . As in Ž1a., we obtain a contradiction. Ž3a. t s 6. Then e12 q ??? qe k2 s 10, IG Ž l.rH ( DŽ10., a dihedral group of order 10, e i g 1, 24 , and we obtain a contradiction in view of Ž*.. Ž4a. t s 10. Then e12 q ??? qe k2 s 6, IG Ž l.rH ( S3 , e i g 1, 24 . Therefore e12 q e22 6, a contradiction. Ž5a. t s 12. Then e12 ??? qe k2 s 5, IG Ž l.rH ( C Ž5., e i s 1, and a contradiction follows. Ž6a. t s 15. Then IG Ž l.rH is elementary abelian of order 4, k s 1, e1 s 2, lG s 2 x 1 , x Ž1. s 30 lŽ1.. Ž7a. t s 20. Then IG Ž l.rH ( C Ž3., k s 1, e1 s 1, and a contradiction follows. Ž8a. t s 30. Then IG Ž l.rH ( C Ž2., e i s 1, k s 1, and a contradiction follows. Ž9a. t s 60. Then IG Ž l. s H, k s 1, e1 s 1, lG s x 1 , x 1Ž1. s 60 lŽ1.. Thus, if x g IrrŽ G . y IrrŽ GrH . and l g IrrŽ x H ., then lG s 2 x and x Ž1. s 30 lŽ1., or lG s x and x Ž1. s 60 lŽ1..
543
FINITE NONSOLVABLE GROUPS
Denote by V 1 , . . . , V n4 the set of the G-orbits on the set IrrŽ H . y 1 H 4 , and suppose Žin view of Ž1a. ] Ž9a.. that < V i < s 15 Ž i s 1, . . . , m4 ,
< V j < s 60 Ž j s m q 1, . . . , n . .
Take l i g V i Ž i s 1, . . . , n.. Then 60 < H < s < G < s < G : H < q
x Ž 1.
Ý
2
x gIrr Ž G .yIrr Ž GrH . m
s 60 q 30 2
is1
< H < s 1 q 15
n
Ý li Ž 1. 2 q 60 2 Ý
m
jsmq1 n
Ý li Ž 1. 2 q 60 Ý 1s1
2
l j Ž 1. , 2
l j Ž 1. .
jsmq1
Hence < H < ' 1 Žmod 15.. By Thompson’s Theorem on minimal simple groups, the order of such a group is divisible by 3 or 5. Hence H is solvable and in this case the lemma follows. Žb. GrH ( L2 Ž7.. We keep the same notation as in Ža.. Then e i g 1, 2, 3, 4, 6, 7, 84 . Now, SLŽ2, 7. is the only representation group of GrH and cd Ž SL Ž 2, 7 . . s 1, 3, 4, 6, 7, 8 4 , t g 1, 7, 8, 14, 21, 24, 28, 42, 56, 84, 168 4 . As in Ža. we may prove that one of the following possibilities holds: Ž1b. Ž2b.
t s 42, k s 1, e s 2. t s 168, k s 1, e s 1.
Let t s 42, k s 1, e s 2. For appropriate l g IrrŽ H ., lG s 2 x , where x g IrrŽ G .. In this case x Ž1. s 84lŽ1., and l belongs to the G-orbit of length 42. Let t s 168, k s 1, e s 1. Then for appropriate l g IrrŽ H ., lG s x g IrrŽ G . and l belongs to the G-orbit of length 168. Denote by V 1 , . . . , V n4 the set of G-orbits on the set IrrŽ H . y 1 H 4 . Suppose that < V i < s 42, i g 1, . . . , m, < V j < s 168, j s m q 1, . . . , n.
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Take l i g V i for all i. Then 168 < H < s < G < s < G : H < q 84 2
m
n
2
Ý li Ž 1. q 168 2 is1
m
Ý
2
l j Ž 1. ,
jsmq1 n
< H < s 1 q 42 Ý l i Ž 1 . 2 q 168
Ý
is1
jsmq1
2
l j Ž 1. .
The last equality shows that < H < is odd. Hence H is solvable. LEMMA 3.2. Let H be a normal subgroup of a D 1-group G. If GrH is isomorphic to L2 Ž5. or L2 Ž7., then H s 14 . Proof. Suppose that H ) 14 . Without loss of generality we may assume that H is a minimal normal subgroup of G. By Lemma 3.1, H is abelian. Put < H < s q s, q is a prime, s ) 0. We retain the notation V i , n, m, e i from the proof of Lemma 3.1. It follows from Lemma 3.1 that < V i < s < V j < implies e i s e j . Therefore m F 1, n y m F 1. Ži. Let G ( L2 Ž5.. Then q s y 1 g 15, 60, 15 q 604 . Thus, q s g 2 4 , 614 . Let q s s 2 4 . Then t s 15, x Ž1. s 30 for a character x g IrrŽ lG ., where l g IrrŽ H . is nonprincipal. In this case k Ž G . s
Let G be a simple D 1-group. Then
Proof. Assume that AutŽ G . ) InnŽ G .. Take a g AutŽ G . y InnŽ G .. Let x 1 , x 2 be two distinct nonlinear irreducible characters of G of the same degree. Let x g IrrŽ G . y x 1 , x 2 4 . Since G is a D 1-group, x a s x . Therefore, by wFS, Theorem Cx and Brauer’s Permutation Lemma, Ž x 1 . a s x 2 . Any b g AutŽ G . y InnŽ G . acts as a . Therefore, if g s aby1 , then x ig s x i for i s 1, 2, and x g s x for all x g IrrŽ G . y x 1 , x 2 4 . Then, by wFSx, g g InnŽ G .. Thus,
FINITE NONSOLVABLE GROUPS
545
Let G be a nonsolvable D 1-group. Then, by Theorem 2.1, G9 s G. Let H be a maximal normal subgroup of G. Then S s GrH is a nonabelian simple group. Since S is not a D 0-group wBCHx it is a D 1-group. Therefore
4 It follows from Lemma 3.2 that for a simple D 1-group G one has
G ( A n is the alternating group of degree n ) 6. G ( L nŽ p ., n ) 1, and Ž n, p y 1. s 1 for n ) 2.
Žiii.
G ( UnŽ p ., n ) 2, Ž n, p q 1. s 1. G ( PSp 2 mŽ q ., m ) 1, and
Živ. Ž1iv. Ž2iv. Ž3iv. Žv. Žvi. Žvii. Žviii. Žix. Žx. Žxi. Žxii. Žxiii. Žxiv. Žxv.
q s p ) 2, or m ) 2, q s 2 e, e - 3, or
m s 2, q s 2. G ( PV 2 mq1Ž p ., m ) 2, p ) 2. Ž . G ( PVq 2 m 2 , m ) 4. Ž . G ( PVy 2 m 2 , m ) 3. Ž . G ( G 2 q , e - 3, p / 3. G ( E6 Ž p ., Ž p y 1, 3. s 1. G ( F4Ž q ., e - 3, and e s 1 for p s 2. G ( E7 Ž q ., where e s 1, or q s 4. G ( E8 Ž q ., e - 3. G (2 E6 Ž p ., Ž p q 1, 3. s 1. G (2 F4Ž2.9 is the Tits simple group. G is isomorphic to one of 26 sporadic simple groups.
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BERKOVICH AND KAZARIN
This follows from wLPS, pp. 18]19x. Note that L2 Ž2., L2 Ž3., U3 Ž2. are solvable. Further on, 2
F4 Ž 2 . : 2 F4 Ž 2 . 9 s 2, L2 Ž 4 . ( L2 Ž 5 . ( A 5 , L2 Ž 9 . ( Sp4 Ž 2 . 9 ( A 6 , L4 Ž 2 . ( A 8 , L2 Ž 7 . ( L3 Ž 2 . , U4 Ž 2 . ( PSp4 Ž 3 . , G 2 Ž 3 . 9 ( U3 Ž 3 . , 2
G 2 Ž 3 . 9 ( L2 Ž 8 . , L2 Ž q . ( PSp 2 Ž q . ( U2 Ž q . ( V 3 Ž q . ,
y 2 PVq 4 Ž q . ( L 2 Ž q . = L 2 Ž q . , PV 4 Ž q . ( L 2 Ž q . , V 5 Ž q . ( PSp4 Ž q . , y e e PVq 6 ( L 4 Ž q . , PV 6 Ž q . ( U4 Ž q . , PV 2 mq1 Ž 2 . ( Sp 2 m Ž 2 . .
5 In this section we will prove some auxiliary results. Let v be a primitive < G
FINITE NONSOLVABLE GROUPS
547
Žmod f < T <., x ' 1 Žmod m. has the unique solution modulo < G < by the Chinese Remainder Theorem. If s is a solution of that system, then c Ž ss . s w k . Thus, im c s H. Put c Ž ss . s ws . Then c Ž ss st . s c Ž sst . s wst s ws w t s c Ž ss . c Ž st .. Let ss g ker c . Then c Ž ss . s w 1 so s ' 1 Žmod < T <.. Conversely, suppose that s ' 1 Žmod < T <.; then ss g ker c . Therefore, ker c s G1. Thus, G1 ( H. Hence, any automorphism of T can be lifted to an automorGrG phism of QŽ v .. Take ss g G0 . Put c Ž ss . s w k . Then there exists an element g g G such that gy1 tg s t k for t g T. Then T g s T and g g NG ŽT .. Therefore w k g H0 s NG ŽT .rCG ŽT . s Aut G ŽT .. In particular, c Ž G0 . : H0 . Since < G : G0 < F 2, we have < H : H0 < F 2. This implies < H < s
548
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case Oq. , SOnyŽ q . Ž n is even, case Oy. , and GŽ V, q . s OnŽ q . Ž n is even., OnqŽ q . Žcase Oq. , OnyŽ q . Žcase Oy. . In case O with n ) 1, the group SŽ V, q . has a subgroup V Ž ¨ , q . of Ž . yŽ . index 2 and in the cases as above we denote it by V nŽ q ., Vq n q , Vn q , respectively. Note that, if n is odd, then p ) 2. All the above groups in Case O are called orthogonal groups. All the groups Žin four cases. defined above are called classical. A cyclic subgroup T s CG Ž x . Ž x g G . of a classical group G is said to be a Singer cycle if T acts irreducibly on V and the bilinear form Ž , . is T-invariant. From the point of view of Chevalley groups a Singer cycle is a maximal torus of the corresponding group. Singer cycles exist for the following groups wSeix: G s GLm Ž q .
Ž of order q m y 1 . ,
Sp 2 m Ž q .
Ž of order q m q 1 . ,
GUm Ž q .
Ž of order q m q 1 for odd m . ,
SOy 2m
Ž of order q m q 1 . .
It is natural that in the corresponding simple groups L m Ž q . , PSp 2 m Ž q . , Um Ž q . , PVy 2mŽ q. there exist cyclic subgroups of respective orders, which are connected with Singer cycles. If G s L mŽ q ., then G s SL mŽ q .rZŽ SL mŽ q ... Therefore G contains a cyclic subgroup T of order Ž q m y 1.rŽ q y 1.Ž q y 1, m.. The group PSp 2 mŽ q . contains a cyclic subgroup T of order Ž q m q 1.rŽ q y 1, 2.. The group UmŽ q . for odd m contains a cyclic subgroup T of order Ž q m q 1.rŽ q Ž . q 1.Ž q q 1, m... Now, the group PVy 2 m q contains a cyclic subgroup T of order Ž q m q 1.rŽ q m q 1, 4.. In many cases we may use the information on Singer cycles to find some maximal cyclic subgroups of Chevalley groups. We note that PV " 2 l Ž q . : PV 2 lq1 Ž q . , PSp 2 l Ž q . : L2 l Ž q . ,
PV 2 ly1 : PV 2 lq1 Ž q . , Un Ž q . : PSp 2 n Ž q . ,
Uny1 Ž q . : Un Ž q . ,
PV 4 ly1 Ž q . : PVq 4l Ž q. .
549
FINITE NONSOLVABLE GROUPS
LEMMA 6.1. Let G s GŽ q . be a classical group from Lemma 4.1. Then there exists some natural number n 0 and cyclic subgroup T of G such that n s < NG ŽT . : CG ŽT .<, n N n 0 and
Ž a . G s L m Ž q . , n ) 2, and q is a prime, < T < s Ž b . G s L2 Ž q . , q is odd, < T < s
qq1 2
Ž c . G s PSp2 l Ž q . , q is odd, < T < s
qm y 1 qy1
, n 0 s n s m.
, n 0 s 2.
ql q 1 2
, n s 2 l, l ) 1.
Ž d. G s PSp2 l Ž 2 n . , l G 2, n F 2, l, n 4 / 2, 2 4 , < T < s 2 n l q 1, n s 2 l. Ž e . G s PV 2 lq1 Ž q . , l ) 1, q is odd, < T < s Ž f . G s U2 lq1 Ž q . , l ) 1, < T < s Ž f . G s U2 lq1 Ž q . , l ) 1, < T < s
q 2 lq1 q 1 qq1 q 2 ly1 q 1 qq1
ql q 1 2
, n s 2 l.
, n s 2 l q 1. , n s 2 l q 1.
Ž g. G s U2 l Ž q . , l ) 1, < T < s q 2 ly1 q 1, n s 2 l y 1. Ž h . G s PVq2 l Ž 2 . , < T < s 2 ly1 q 1, n s 2 Ž l y 1 . . Ž i . G s PVy2 l Ž 2 . , < T < s 2 l q 1, n s 2 l. 7 In this section we prove some number theoretical lemmas and investigate the classical case. In particular, we prove that all classical simple groups G s GŽ q ., apart from L2 Ž5. and L2 Ž7., are not D 1-groups. LEMMAS 7.1. Let n ) 1 be a natural number. If n k 2 Žmod 4., then w Ž n. 2 G n. If n ' 2 Žmod 4., then 2 w Ž n. 2 G n. Proof. We omit the easy proof of this lemma since it follows from Lemma 7.2. LEMMA 7.2.
Let n be a natural number. If w Ž n. 2 F 2 n, then n g 1, 2, 3, 4, 6, 8, 10, 12, 18 4 .
In particular, n - 19.
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BERKOVICH AND KAZARIN
Proof. Let n s mp. Then w Ž n. s Ž p y 1. w Ž m., if p ¦ m, and w Ž n. s pw Ž m., if p < m. Assume that the lemma is proved for all m - n. Let m ) 1. Denote by p the lowest prime divisor of n. Ži. n s p s. Then w Ž n. 2 s Ž p y 1. 2 p 2 sy2 F 2 p s, or Ž p y 1. 2 p s F 2 p 2 . If s ) 1, then p s 2 and s g 2, 34 , so that n g 4, 84 . If s s 1, then Ž p y 1. 2 F 2 p, and p F 3. Thus, in that case n g 1, 2, 3, 4, 84 . Žii. n is not a prime power, n s mp, p ¦ m. Then w Ž n. 2 s Ž p y 1. 2w Ž m. 2 F 2 n s 2 pm. Ž1ii. Let p s 2. Then m is odd and w Ž m. 2 F 4 m. If m s q s, q ) 2 is a prime; then Ž q y 1. 2 q s F 4 q 2 and s - 3. For s s 1 we have Ž q y 1. 2 F 4 q so that q g 3, 54 , and n g 6, 104 . If s s 2, then Ž q y 1. 2 F 4 and q s 3, n s 18. Suppose that m is not a prime power. Let m s q s t, where q is a prime, s ) 0, t ) 1, and q ¦ t. Then Ž q y 1. 2 q sw Ž t . 2 F 4 q 2 t. Assume t ) 3. Then, by induction, 2 t F w Ž t . 2 . Therefore Ž q y 1. 2 q sw Ž t . 2 F 2 q 2w Ž t . 2 , or Ž q y 1. 2 q s F 2 q. Then q s s 3. We have 4w Ž t . 2 s w Ž m. 2 F 4 m s 12 t, or w Ž t . 2 F 3t. Let t s r u be a prime power, r ) 3. Then Ž r y 1. 2 r 2 uy2 s w Ž t . 2 F 3t s 3r u, or Ž r y 1. 2 r u F 3r 2 . Then u s 1, Ž r y 1. 2 F 3r, a contradiction. s If t s Ł is1 ria i is a standard decomposition in prime powers, s ) 1, then, by the above, 2
wŽ t. s
s
Ł w Ž ria . i
is1
2
s
)
Ł 3ria
i
s 3 s t ) 3t ,
is1
a contradiction. Let t s 3. Then m s 3q s, q ) 3. We have 2
2
4 Ž q y 1 . q 2 sy2 s w Ž m . F 4 m s 12 q s , or Ž q y 1. 2 q s F 3q 2 , which is impossible. Ž2ii. p ) 2. Let 2 m ) w Ž m. 2 . Then, by induction, m s 3 Žrecall that m is odd.. This is impossible since p is the lowest prime divisor of n. Thus 2 m F w Ž m. 2 . Therefore Ž p y 1. 2w Ž m. 2 F pw Ž m. 2 , or Ž p y 1. 2 F p, a contradiction. Žiii. n s pm, p < m, m is not a prime power. We have w Ž n. 2 s p 2w Ž m. 2 F 2 pm, or pw Ž m. 2 F 2 m. If p s 2, then w Ž m. 2 F m so that, by induction, m F 2, a contradiction since m is not a prime power. Let p ) 2. Then w Ž m. 2 - m and, by induction, m s 6. Then p s 2, a contradiction.
FINITE NONSOLVABLE GROUPS
551
Remark ŽG. Freiman.. The equation w Ž n. 2 - cn has a finite number of solutions for any constant c Žin fact, w Ž n. ) nrln n.. LEMMA 7.3. Let q be a prime, and n G 1 an integer. If w ŽŽ q nq 1 y 1.rŽ q y 1.. F 2Ž n q 1., then one of the following assertions holds: Ža. Žb.
q s 2, n - 4. n s 1, q g 3, 5, 7, 11, 134 .
Proof. Put m s Ž q nq 1 y 1.rŽ q y 1.. If m F 18, the result is obtained by easy checking. Let m ) 18. In view of the lemma one has 2 m - w Ž m. 2 F 4Ž n q 1. 2 , or m s 1 q q q ??? qq n - 2Ž n q 1. 2 . If q s 2, then 2 nq1 2Ž n q 1. 2 q 1, and n - 6. In that case n - 4 as the original inequality shows. In the general case, Ž q nq 1 y 1.rŽ q y 1. - 2Ž n q 1. 2 . If q s 3, then 3 nq 1 - 1 q 4Ž n q 1. 2 , and n - 3. The original inequality yields n s 1. Let q ) 3. Then n s 1 and w Ž1 q q . F 4. Using Lemma 7.2, one obtains q g 5, 7, 114 . In what follows G denotes a nonabelian simple group. LEMMA 7.4. or L2 Ž7..
If G s L nŽ q . is a D 1-group, then G is isomorphic to L2 Ž5.
Proof. Let n ) 2. By Lemma 5.2 and Lemma 7.3, w ŽŽ q n y 1.rŽ q y 1.. F 2 n so that q s 2 and n s 3 or 4. Since L4Ž2. is not a D 1-group wAtlasx, n s 3 and G s L3 Ž2. ( L2 Ž7.. Let n s 2. Then G ( L2 Ž4. ( L2 Ž5. ŽLemma 4.1., or q G 7 is a prime. Now, q g 7, 11, 134 ŽLemma 7.3.. Now, L2 Ž11. and L2 Ž13. are not D 1groups wAtlasx. LEMMA 7.5. Let q be a prime, n ) 1 odd, d s Ž q n q 1.rŽ q q 1.. If w Ž d . F 2 n, then one of the following assertions holds: Ža. Žb.
n s 3, q g 2, 34 . n s 5, q s 2.
Proof. Assume that w Ž d . 2 F 2 d. Then, by Lemma 7.2,
ds
qn q 1 qq1
g 1, 2, 3, 4, 6, 8, 10, 12, 18 4 ,
and q s 2, n F 5, or q s 3 s n. Now assume that w Ž d . 2 ) 2 d. Then d s Ž q n q 1.rŽ q q 1. - 2 n2 . If q s 2, then n - 8. In any case q - 4. If q s 3, then n - 6. The result is obtained by direct calculations.
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LEMMA 7.6. Let q be a prime number, n ) 1 odd. If w Ž q n q 1. F 2 n, then n s 3, q s 2. Proof. If q n q 1 - 19, then q s 2, n s 3. Otherwise, w Ž q n q 1. 2 ) 2Ž q n q 1.. Thus, q n q 1 - 2 n2 . So q s 2 and n - 7. The result is obtained by direct calculations. LEMMA 7.7. G s UnŽ q ., n ) 2 is not a D 1-group. Proof. Assume that n is even. Using Lemma 6.1 and 5.2, one obtains w Ž q ny 1 q 1. F 2Ž n y 1.. Then n y 1 s 3, q s 2 ŽLemma 7.6.. But U4Ž2. is not a D 1-group wAtlasx. Let n be odd. Then G g U3 Ž2., U3 Ž3., U5 Ž2.4 by Lemmas 6.1, 5.2, and 7.5. But U3 Ž2. is solvable, and U3 Ž3., U5 Ž2. are not D 1-groups wAtlasx. LEMMA 7.8.
Let q be an odd prime number and n G 2. The group G g PSp 2 n Ž q . , PV 2 nq1 Ž q . 4
is not a D 1-group. Proof. Using Lemmas 6.1 and 5.2, one obtains w ŽŽ q n q 1.r2. F 4 n. If Ž q n q 1.r2 F 18, then n - 3, q F 5, or n - 4, q s 3. If q s 5, n s 2, then the inequality for w does not hold. So let q n q 1 ) 36. Then, by Lemma 7.2, 16n2 G w ŽŽ q n q 1.r2. ) q n q 1. This implies n - 6, q F 7. If q s 3, n s 5, then the inequality for w does not hold. If q G 5, n s 5, then the inequality for w does not hold. So we may assume that n F 4. Further on, w ŽŽ3 4 q 1.r2. s w Ž41. ) 4 ? 4, and w ŽŽ5 4 q 1.r2. ) 4 ? 4. Therefore n / 4. Let n s 3. Then q s 3 or 5. If q s 5, then w ŽŽ5 3 q . 1 r2. s 36 ) 4 ? 3. Therefore, if n s 3, then q s 3. Let n s 2. An easy check shows that in this case q s 3 as well. Taking into account the list of isomorphisms after Lemma 4.1, we see that G g PSp4Ž3., PSp6 Ž3., PV 7 Ž3.4 . But these groups are not D 1-groups wAtlasx. LEMMA 7.9.
If w Ž2 m q 1. F 4 m, then m F 5.
Proof. Assume that 2 m q 1 ) 18 Žotherwise the lemma holds.. Then w Ž2 m q 1. 2 G 2Ž2 m q 1., or 2 m q 1 F 8 m2 and m F 9. It is easy to check that the inequality holds only for m - 6. LEMMA 7.10.
If G s PSp 2 nŽ2 n . is a D 1-group, then n s 5, n s 1.
Proof. By Lemma 4.1, n F 2. Assume that n s 2. It follows from Lemmas 5.2 and 6.1 that w Ž2 2 n q 1. F 4 n. Then w Ž2 m q 1. F 2 m, where m s 2 n. Then 2 n s m F 3, since PSp4Ž4. is not a D-group wAtlasx, contradicting n ) 1. Thus, n s 1. Then w Ž2 n q 1. F 4 n so n - 6 ŽLemma 7.9.. By wAtlasx, PSp6 Ž2. and PSp 8 Ž2. are not D 1-groups. The lemma is proved.
FINITE NONSOLVABLE GROUPS
LEMMA 7.11.
553
qŽ . Ž . Let G s PV " 2 n 2 be a D 1 -group. Then G s PV 12 2 .
Proof. Using Lemmas 5.2 and 6.1, we have w Ž2 ny 1 q 1. F 4Ž n y 1. for e s q, and w Ž2 n q 1. F 4 n for e s y. By Lemma 7.9, n - 7 in the first case, and n - 6 in the second one. By Lemma 4.1, in the first case n G 5. "Ž . Ž . w x Since PVy 8 2 and PV 10 2 are not D 1 -groups Atlas , the lemma holds in all cases. LEMMA 7.12. L2 Ž5. or L2 Ž7..
If G is a classical simple group, then G is isomorphic to
Proof. In view of Lemmas 7.4, 7.7, 7.8, 7.10, and 7.11 we have to show Ž . that PSp10 Ž2. and PVq 12 2 are not D 1 -groups. Ž . Denote G s PVq 2 . Note that 12 < G < s 2 30 Ž 2 6 y 1 . Ž 2 2 y 1 . Ž 2 4 y 1 . Ž 2 6 y 1 . Ž 2 8 y 1 . Ž 2 10 y 1 . . Ž . Ž . The group PVq 12 2 is contained in PSp12 2 and is defined as the group of the 12 = 12-matrices Y g GL12 Ž2., satisfying YAY t s A, where Y t is the transpose to Y, and As
ž
I , 0
0 I
/
where I is the 6 = 6 identity matrix. Note that the skew symmetric form f over GF Ž2. is invariant under the group PSp12 Ž2.. Since the ground field is GF Ž2., the form f is symmetric. It is more convenient to take a basis in which the matrix of our form is the following one, Fs
ž
I 0 0
0 I 0
0 0 , L
/
where I is the identity 5 = 5-matrix, and Ls
ž
0 1
1 . 0
/
All elements B g PSp12 Ž2. satisfy the equality BF B t s F. Therefore we can find an element of order 31 which is a generator of a Sylow 31-subgroup of PSp12 Ž2.. Consider the element Ts
X 0 0
0 Xyt 0
0 0 I
0
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of the group GL12 Ž2.; here I is the identity 2 = 2-matrix, and X g GŽ2, 5. is a matrix having the characteristic polynomial l5 q l3 q 1. It is clear that T g Sp12 Ž2. s PSp12 Ž2. and Xyt has the characteristic polynomial l5 q l q 1. It is sufficient to choose 0 0 Xs 0 0 1
1 0 0 0 1
0 1 0 0 0
0 0 1 0 0
0 0 0 . 1 1
0
Suppose that the element R11 R s R 21 R 31
R12 R 22 R 32
R13 R 23 R 33
0
normalizes the cyclic subgroup ²T :; here R11 and R 22 are 5 = 5-matrices, R 33 is a 2 = 2-matrix. Then RT s T k R for some natural number k, coprime with the order of T. It follows from this matrix equality that R13 s X k R13 , R 32 s R 32 Xyt ,
k
R 23 s Ž Xyt . R 23 , R11 X s X k R11 , k
R 21 X s Ž Xyt . R 21 ,
R 31 s R 31 X , R12 Xyt s X k R12 , k
R 22 Xyt s Ž Xyt . R 22 .
Since the matrices X y I, X k y I, Xyt y I, Ž Xyt . k y I are nonsingular, it follows from the first four equalities that R13 s R 23 s R 31 s R 32 s 0. Therefore R11 R Rs 21 0
R12 R 22 0
0 0 . R 33
0
By Schur’s Lemma, it follows from irreducibility of X on the 5-dimensional vector space over GF Ž2. that in fact R11 Rs 0 0
0 R 22 0
0 0 , R 33
0
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555
or 0 R s R 21 0
R12 0 0
0 0 . R 33
RX11 R2 s 0 0
0 RX22 0
0 0 , RX33
0
In particular,
0
where RX11 and RX22 are contained in the normalizer of the cyclic subgroup ² X : Žwhich is a Singer subgroup of GLŽ5, 2... Therefore the order of NG Ž²T :.rCG Ž²T :. divides 10. In fact, matrices of the form
ž
I 0 0
0 I 0
0 0 Y
/
commute with T. Therefore 2 ? 10 F 2 NG Ž ²T : . rCG Ž ²T : . - w Ž 31 . s 30. But this contradicts the fact that G is a D 1-group. The same arguments show that PSp10 Ž2. is not a D 1-group.
8 In this section we consider the exceptional groups of Lie type. By wAtlasx, F4Ž2. and 2 F4 Ž2.9 are not D 1-groups. Therefore we have to consider the following cases: Ži. Žii. Žiii. Živ. Žv. Žvi.
G 2 Ž q ., where q s p e, p / 3 is a prime number, and e F 2. E6 Ž p ., where p is a prime number, with Ž p y 1, 3. s 1. F4Ž q ., where q s p e, e F 2, and p ) 2 is a prime number. E7 Ž q ., where q s p is a prime number, or q s 4. E8 Ž q . where q s p e, e F 2, and p is a prime number. 2 E6 Ž p ., where p is a prime number with Ž p q 1, 3. s 1.
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LEMMA 8.1. The group G s G 2 Ž q . is not a D 1-group. Proof. We have q s p e, e F 2, and p / 3 is a prime. The structure of Sylow r-subgroups of G for primes r / p is known Žsee wSSx.. If r is coprime with the order of the Weyl subgroup W Žnote that < W < s 12. of G, then for r, dividing Ž q 2 y q q 1.Ž q 2 q q q 1., a Sylow r-subgroup T of G is cyclic, and NG ŽT .rCG ŽT . is cyclic of order 6. If d s Ž q 2 y q q 1, Ž q 2 y 1 .Ž q 3 y 1 . Ž q q 1 . . ) 1, then q ' y1 Žmod 3. and d s 3 since 9 ¦ Ž q 2 y q q 1.. Similarly, if q ' 1 Žmod 3., then 9 ¦ Ž q 2 q q q 1.. Thus, G has a cyclic subgroup C of order q 2 " q q 1, such that < NG Ž C . : C < s 6. It follows from Lemma 5.2 that w Ž q 2 " q q 1. F 12. If q 2 " q q 1 F 18, then, taking into account that q / 3, 9 we obtain q s 2 or 4. In the first case G9 s G 2 Ž2.9 ( A 6 is not a D 1-group. In the second case, by wAtlasx, G 2 Ž4. is not a D 1-group either. Therefore q 2 " q q 1 ) 18. By Lemma 7.2, 2
w Ž q 2 " q q 1. ) 2 Ž q 2 " q q 1. . Making use of Lemma 5.2, we conclude that 2Ž q 2 " q q 1. - 12 2 s 144 and q g 5, 74 . Next, 5 2 y 5 q 1 ' 0 Žmod 3., 5 2 q 5 q 1 k 0 Žmod 3., and w Ž31. s 30 ) 2 ? 6 s 12. By Lemma 5.2, G is not a D 1-group. Now, w Ž7 2 y 7 q 1. s w Ž43. s 42 ) 2 ? 6 s 12, and we conclude, in view of Lemma 5.2, that G is not a D 1-group in this case either. LEMMA 8.2. The group G s F4 Ž q . is not a D 1-group. Proof. By Lemma 4.1, q s p e, where e F 2 and a prime p is odd. It follows from wCarx that G contains a cyclic subgroup T of order q 4 q 1 such that < NG ŽT . : CG ŽT .< s 8. Therefore, by Lemma 5.2, w Ž q 4 q 1. F 16. Since q ) 2, we obtain q 4 q 1 ) 18. By Lemma 7.2, w Ž q 4 q 1. 2 ) 2Ž q 4 q 1.. Thus, 2Ž q 4 q 1. - 16 2 and q F 3. Since w Ž3 4 q 1. s 40 ) 16 we see that G is not a D 1-group. LEMMA 8.3. The groups E6 Ž q . and 2 E6 Ž q . are not D 1-groups. Proof. It follows from wWilx and wKonx that the groups E6 Ž q . and 2 E6 Ž q . have a cyclic subgroup T such that CG Ž x . F T for all x g T a , and T is of order Ž q 6 q q 3 q 1.rŽ3, q y 1., if G ( E6 Ž q ., and of order Ž q 6 y q 3 q 1.rŽ3, q q 1., if G (2 E6 Ž2.. By wCarx, < NG ŽT . : T < s 9 in every case. Since Ž3, q y 1. s 1 for E6 Ž q ., and Ž3, q q 1. s 1 for 2 E6 Ž2. by virtue of Lemma 4.1, G has an abelian self-centralizer T0 of order q 6 " q q 1, respectively, and such that NG ŽT0 . is a Frobenius group of order 9 < T0 <. Obviously, T0 is a TI-subgroup of G. It follows from the properties of abelian TI-subgroups
FINITE NONSOLVABLE GROUPS
557
that G has at least Ž q 6 " q 3 .r9 distinct irreducible characters of the same degree. Since G is a D 1-group then q 6 " q 3 F 18, which is impossible. LEMMA 8.4. The group G s E7 Ž q . is not a D 1-group. Proof. The group G has a self-centralizer T such that < T < s Ž q 8 y 1.rŽ q y 1. and < NG ŽT . : T < s 32 Žsee wCar, Table 6x.. That table shows that T is a unique up to conjugacy maximal torus of G of order divisible by Ž q 8 y 1.rŽ q y 1. s Ž q 4 q 1.Ž q 2 q 1.Ž q q 1.. On the other hand, by wSte2x, the group G contains a subgroup L8 Ž q ., and the last one has a cyclic subgroup of order Ž q 8 y 1.rŽ q y 1.. Therefore the subgroup T is cyclic. It follows from Lemma 5.2 that w ŽŽ q 8 y 1.rŽ q y 1. F 64. Since q ) 1, it follows that Ž q 8 y 1.rŽ q y 1. ) 18. Therefore, by Lemma 7.2,
w
ž
q8 y 1 qy1
2
/
) 2 Ž q 4 q 1 .Ž q 2 q 1 . Ž q q 1 . .
Hence 2Ž q 4 q 1.Ž q 2 q 1.Ž q q 1. - 2 12 , and q - 3. But
w Ž 2 8 y 1 . s w Ž 17 ? 5 ? 3 . s 16 ? 4 ? 2 s 126 ) 64. Therefore G is not a D 1-group. LEMMA 8.5. The group G s E8 Ž q . is not a D 1-group. Proof. It follows from wCar, Table 7x that G contains a maximal cyclic torus T of order Ž q 9 y 1.rŽ q y 1. s Ž q 6 q q 3 q 1.Ž q 2 q q q 1. such that < NG ŽT . : T < s 54. It follows from Lemma 5.2 that w ŽŽ q 6 q q 3 q 1.Ž q 2 q q q 1.. - 108. Obviously, Ž q 6 q q 3 q 1.Ž q 2 q q q 1. ) 18. By Lemma 7.2, 2
w Ž Ž q 6 q q 3 q 1 .Ž q 2 q q q 1 . . ) 2 Ž q 6 q q 3 q 1 .Ž q 2 q q q 1 . . Hence, 2Ž q 6 q q 3 q 1.Ž q 2 q q q 1. - 108 2 so q s 2. But
w Ž 2 9 y 1 . s w Ž 73 ? 7 . s 432 ) 2 ? 54, a contradiction which shows that G is not a D 1-group. It remains to find all alternating groups which are D 1-groups.
9 In this section we will prove that the alternating group A n for n ) 5 is not a D 1-group.
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There is the 1]1-correspondence a between irreducible characters of Sn and partitions w lx: n s n1 q ??? n k with natural numbers n i such that n1 G ??? G n k . We denote the character a Žw lx. by x w l x. To each such partition there corresponds the associated partition w lx*: n s nU1 q ??? qnUs of n such that nU1 s k and nUi s < j g N < n j G i4<. Let w lx be a partition of n: n1 q ??? qn k . Suppose that m1 , . . . , m s are all distinct numbers among n1 , . . . , n k . Put bi s < r g 1, . . . , k 4< n r s m i 4< for every i g 1, . . . , s4 . Then we say that the partition w lx is of type t Žw lx. s Ž m1b 1 , . . . , m bs s .. A partition w lx is said to be self-associated if w lx* s w lx. We use in what follows the following well-known result Žsee wKerx, for example.: LEMMA 9.1. Let w lx be a partition of n, and x s x w l x the corresponding irreducible character of Sn . Then Ža. If w lx / w lx*, then x A g IrrŽ A n .. n Žb. If w lx is self-associated, then x A s m 1 q m 2 , where m 1 , m 2 are n distinct irreducible characters of A n of the same degree. Suppose that w l1 x, w l2 x are distinct self-associated partitions of n, and x 1, x 2 corresponding irreducible characters of Sn . Put x Ai n s m 1i q m i2 , i s 1, 2. Then the characters m ij , i, j s 1, 2, are four pairwise distinct irreducible characters of A n . LEMMA 9.2. Let aŽ n. be the number of distinct self-associated partitions of a natural number n. Then A n is a D k-group, where k G aŽ n., unless n - 4. Proof. It is easy to check that the lemma holds for n - 6. So we may assume that n ) 5. Then, by wIsa, Theorem 14.23x 2 f cdŽ Sn . s x Ž1.< x g IrrŽ S n .4 . Now the result follows from the reasoning before the lemma. So it remains to prove that aŽ n. - 2 implies that n is small, and then use wAtlasx. We prove the following stronger result. LEMMA 9.3. Ža. If n G 8, then aŽ n. ) 1. Žb. If n ) 11, then aŽ n. ) 2. Proof. Ža. Suppose n s 2 x is even, n ) 8. Then x ) 4, and the following distinct partitions of n are self-associated: the partition l1e of type Ž x, 2, 1 xy2 . ; the partition l2e of type Ž x y 1, 3, 2, 1 xy4 . .
FINITE NONSOLVABLE GROUPS
559
Suppose n s 2 x y 1 is odd. Then x G 5, and the following distinct partitions of n are self-associated. the partition l1o of type Ž x, 1 xy1 . ; the partition l2o of type Ž x y 2, 3, 3, 1 xy5 . . Žb. Suppose n s 2 x is even. Then x G 6, and we have the following additional self-associated partition w l3o x of type Ž x y 2, 4, 2, 2, 1 xy6 .. Suppose n s 2 x y 1 is odd. Then x G 7, and we have the following additional self-associated partition w l3o x of type Ž x y 3, 4, 3, 2, 1 xy7 .. Thus, Lemma 9.3 shows that A n is not a D 1 group for n ) 7. Since A 6 and A 7 are not D 1-groups wAtlasx, then A 5 is the only nonsolvable alternating D 1-group. With the result proved above we see that A 5 and L2 Ž7. are the only nonsolvable D 1-groups. This finishes the proof of the Main Theorem. Remark. Lemma 9.3 shows that A n is not a D 2-group unless n F 11. It follows from wAtlasx that A 9 and A10 are the only A n that are D 2-groups. QUESTION 1. Classify all nonsol¨ able D 2-groups. QUESTION 2. Find aŽ n., the number of self-associated partitions of a natural number n.
REFERENCES wAtlasx J. H. Conway, R. T. Curtis, S. P. Norton, R. A. Parker, and R. A. Wilson, ‘‘Atlas of Finite Groups,’’ Clarendon, Oxford, 1985. wBer1x Y. Berkovich, Non-solvable groups with a large fraction of involutions, in ‘‘Structure Theory of Set Addition, 7]11 June, 1993 CIRM, Marseille,’’ pp. 187]196. wBer2x Y. Berkovich, Finite solvable groups in which only two nonlinear irreducible characters have equal degrees, J. Algebra 184 Ž1996., 584]603. wBCHx Y. Berkovich, D. Chillag, and M. Herzog, Finite groups in which distinct nonlinear irreducible characters have distinct degrees, Proc. Amer. Math. Soc. 115 Ž1992., 955]959. wBZx Y. Berkovich and E. Zhmud’, ‘‘Characters of Finite Groups,’’ Vols. 1, 2, Amer. Math. Society, Providence, in press. wCarx R. W. Carter, Conjugacy classes of Weyl groups, in ‘‘Seminar on Algebraic Groups and Related Topics,’’ ŽA. Borel et al., Eds.., Lecture Notes in Mathematics, Vol. 131, Springer-Verlag, Berlin, 1970. wFSx W. Feit and G. Seitz, On finite rational groups and related topics, Illinois J. Math. 33, No. 1 Ž1988., 103]131. wIsax I. M. Isaacs, ‘‘Character Theory of Finite Groups,’’ Academic Press, New York, 1976. wKonx A. S. Kondrat’ev, On components of the graph of primes of finite simple groups, Mat. Sb. 180, No. 6 Ž1989., 787]797. wRussianx
560 wKerx wLPSx wSeix wSSx wSte1x wSte2x wWilx
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A. Kerber, ‘‘Representations of Permutation Groups I,’’ Lecture Notes in Mathematics, Vol. 240, Springer-Verlag, 1971. M. W. Liebeck, C. E. Praeger, and J. Saxl, The maximal factorizations of the finite simple groups and their automorphism groups, Mem. Amer. Math. Soc., 86, 432 Ž1990.. G. M. Seitz, On the subgroup structure of classical groups, Comm. Algebra 10, No. 8 Ž1982., 875]885. T. A. Springer and R. Steinberg, Conjugacy classes, in ‘‘Seminar on Algebraic Groups and Related Topics’’ ŽA. Borel et al., Eds.., Lecture Notes in Mathematics, Vol. 131, 1970. E. Stensholt, An application of Steinberg construction on twisted groups, Pacific J. Math. 55, No. 2 Ž1974., 595]618. E. Stenholt, Certain embeddings among finite groups of Lie type, J. Algebra, 69 Ž1978., 136]187. J. S. Williams, Prime graph components of finite groups, J. Algebra, 69 Ž1981., 497]513.
184, 538]560 Ž1996.
0273
Finite Nonsolvable Groups in which Only Two Nonlinear Irreducible Characters Have Equal Degrees* Yakov Berkovich Department of Mathematics and Computer Science, The Uni¨ ersity of Haifa, 31905 Haifa, Israel
and Lev Kazarin Department of Mathematics, Yarosla¨ l State Uni¨ ersity, Yarosla¨ l 150000, Russia Communicated by Walter Feit Received May 23, 1995
DEDICATED TO PROFESSOR AVINOAM MANN ON THE OCCASION OF HIS
60TH BIRTHDAY
Y. Berkovich et al. w Proc. Amer. Math. Soc. 115 Ž1992., 955]959x classified finite groups in which the degrees of the nonlinear irreducible characters are distinct. Theorem 24.7 from wY. Berkovich, J. Algebra 184 Ž1996., 584]603x contains the classification of solvable groups in which only two nonlinear irreducible characters have equal degrees Ž D 1 -groups.. In this paper we obtain the classification of nonsolvable D 1-groups, completing the classification of D 1-groups. Our proof depends on the classification of finite simple groups. The results of the important paper w Illinois J. Math. 33, No. 1 Ž1988., 103]131x on rational simple groups play a key role as well. Q 1996 Academic Press, Inc.
*The first author was supported in part by the Ministry of Absorption of Israel. The second author was supported in part by ISF. 538 0021-8693r96 $18.00 Copyright Q 1996 by Academic Press, Inc. All rights of reproduction in any form reserved.
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539
1 Let IrrŽ G . be the set of all irreducible characters of G, Irr1Ž G . the set of the nonlinear characters in IrrŽ G ., and LinŽ G . the set of the linear characters of G. Let cd 1Ž G . denote the set of degrees of the characters in Irr1Ž G .. DEFINITION. Let k be a nonnegative integer. A group G is said to be a D k-group if
L2 Ž5. and L2 Ž7. are the only nonsol¨ able D 1-groups.
In the following we denote by Ž A, B . a Frobenius group with complement A and kernel B. We are indebted to Avinoam Mann for useful discussions and help. 2 In what follows Žsee the proof of Theorem 2.1. we use the following Remarks. 1. Let Z be a subgroup of a nonabelian p-group G such that < CG Ž Z .< s p 2 . We will prove that G is a p-group of maximal class. We use induction on < G <. Obviously, ZŽ G . - CG Ž Z . and CG Ž CG Ž Z .. s CG Ž Z .. Therefore < NG Ž CG Ž Z ..< s p 3. Then < CG r ZŽG.Ž CG Ž Z .rZŽ G ..< s p 2 , so, by induction, GrZŽ G . is abelian of order p 2 or a p-group of maximal class, and the result follows. 2. Let H be a proper subgroup of a group G such that all elements of G y H are involutions. Suppose that G is not an elementary abelian 2-group. We will prove that < G : H < s 2. Assume that < G : H < ) 2. Without loss of generality we may assume that H s ² G y InvŽ G .:, where InvŽ G . is the set of all involutions in G. Then H eG and GrH is a noncyclic elementary abelian 2-group. Since G is not an elementary abelian 2-group, there exists an element a g H such that oŽ a. ) 2. Let x, y g G y H be such that xH / yH. Since xa, ya are involutions we have xax s ay1 ,
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yay s ay1 . Therefore axy s xya and Ž axy . 2 s a2 / 1, contradicting axy f H Žsince xy f H, xy is an involution.. In this section we will prove the following important component of the Main Theorem: THEOREM 2.1.
If G is a nonsol¨ able D 1-group, then G9 s G.
Proof. Suppose that G9 - G. Then G contains a normal subgroup H of a prime index p. Ži. p s 2. Assume that p ) 2. By wBZ, Proposition 25.9x and the Remark following it, there exists x g Irr1Ž G . such that p ¦ x Ž1.. Then, by Clifford theory, x H g IrrŽ H ., and Ž x H . G s x 1 q ??? qx p , where IrrŽŽ x H . G . s x 1 , . . . , x p 4 , and x 1 , . . . , x p have the same degree x Ž1., a contradiction, since p ) 2. Our claim follows. Žii. GrG9 is a cyclic 2-group. It follows from Ži. that GrG9 is a 2-group. Assume that GrG9 is noncyclic. Then there exists a normal subgroup F of G such that GrF is elementary abelian of type Ž2, 2.. As in Ži., there exists in Irr1Ž G . a character x g Irr1Ž G . of odd degree. By Clifford theory, x F s u g IrrŽ F .. Again, by Clifford theory, u G s eŽ x 1 q ??? qx t ., where e, t are natural numbers, e < x Ž1.. Since e <2, we have e s 1. Since x 1Ž1. s ??? s x t Ž1., we have t F 2 Ž G is a D 1-group.. It follows from 4 x Ž1. s 4u Ž1. s < G : F < u Ž1. s u G Ž1. s t x 1Ž1. s t x Ž1. that t s 4, a contradiction. Thus, GrG9 is a cyclic 2-group. Žiii. < G : G9 < s 2. As above, there exists x g Irr1Ž G . such that x Ž1. is odd. Then u s xG9 g IrrŽ G9.. Since all irreducible constituents of u G have the same degree x Ž1., and, by Clifford theory, ² x , u G : s 1 then
k Ž G9 . y 2 2
q2?2s
k Ž G9 . q 6 2
.
541
FINITE NONSOLVABLE GROUPS
Let k G Ž G9. be the number of G-classes containing in G9. Then k G Ž G9 . s
k Ž G9 . y 2 2
q2s
k Ž G9 . q 2 2
.
Thus, k Ž G . y k G Ž G9. s Ž k Ž G9. q 6.r2 y Ž k Ž G9. q 2.r2 s 2. Let G y G9 s M j N, where M, N are G-classes. Put < G < s g, < M < s grm, < N < s grn. Then gr2 s grm q grn, or 1r2 s 1rm q 1rn. Suppose that m F n. Let x g M, y g N. Since < G : G9 < s 2 then oŽ x . and oŽ y . are even so m and n are even, and hence m s n s 4. By the above, < CG Ž x .< s 4. Therefore P g Syl 2 Ž G . is of maximal class ŽRemark 1.. Since P l G9 is noncyclic in view of solvability of groups of odd order, G y G9 contains an element y of order expŽ P ., so y g N. Therefore, < CG Ž y .< s 4. Thus, expŽ P . s 4 and P is dihedral of order 8. In particular, G y G9 contains an involution z. Since < G : CG Ž z .< s 4 then the fraction of involutions in G is greater than 14 . This contradicts Theorem 28.3 from wBZx.
3 Let G be a nonsolvable D 1-group, H a maximal normal subgroup of G. It follows from Theorem 2.1 and wBCHx that GrH is a simple D 1-group. In this section we consider two cases when GrH is isomorphic to L2 Ž5. or L2 Ž7.. LEMMA 3.1. Let 14 - H eG and GrH isomorphic to L2 Ž5. or L2 Ž7.. If the irreducible constituents of lG ha¨ e distinct degrees for each nonprincipal l g IrrŽ H ., then H is sol¨ able. Proof. We may assume that H ) 14 . Take a nonprincipal l g IrrŽ H . and put Irr Ž lG . s x 1 , . . . , x k 4 ,
lG s e1 x 1 q ??? qe k x k ,
e1 ) ??? ) e k ) 0 Žall e i are pairwise distinct since all x i have pairwise distinct degrees.. Let
x Hi s e i Ž l1 q ??? ql t .
Ž l1 s l .
be the Clifford decomposition. Then < G : H < ty1 s IG Ž l . : H s e12 q ??? qe k2 .
Ž ).
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BERKOVICH AND KAZARIN
Here IG Ž l. is the inertia group of l in G, and e1 , . . . , e k are degrees of irreducible projective representations of IG Ž l.rH. Ža. Let GrH ( L2 Ž5.. In this case e i g 1, 2, 3, 4, 5, 64 Žrecall that GrH has SLŽ2, 5. as a representation group, and the degrees of irreducible projective representations of subgroups of SLŽ2, 5. are equal to one of the numbers 1, 2, 3, 4, 5, 6. and cd Ž SL Ž 2, 5 . . s 1, 2, 3, 4, 5, 6 4 ,
t g 1, 5, 6, 10, 12, 15, 20, 30, 60 4
Žnumber t equals an index of a subgroup in GrH .. Ž1a. t s 1. Then, by Ž*., e12 q ??? qe k2 s 60. Since 5 2 q 4 2 q 3 2 q 2 2 q 12 - 60, e1 s 6. Since 6 2 q 5 2 ) 60, e2 - 5. Since 6 2 q 3 2 q 2 2 q 12 - 60, e2 s 4. Since 6 2 q 4 2 q 3 2 ) 60, e3 - 3. Since 6 2 q 4 2 q 2 2 q 12 - 60, we obtain a contradiction. Ž2a. t s 5. Then, by Ž*., e12 q ??? qe1k s 12 and IG Ž l.rH ( A 4 , so e i g 1, 2, 34 . As in Ž1a., we obtain a contradiction. Ž3a. t s 6. Then e12 q ??? qe k2 s 10, IG Ž l.rH ( DŽ10., a dihedral group of order 10, e i g 1, 24 , and we obtain a contradiction in view of Ž*.. Ž4a. t s 10. Then e12 q ??? qe k2 s 6, IG Ž l.rH ( S3 , e i g 1, 24 . Therefore e12 q e22 6, a contradiction. Ž5a. t s 12. Then e12 ??? qe k2 s 5, IG Ž l.rH ( C Ž5., e i s 1, and a contradiction follows. Ž6a. t s 15. Then IG Ž l.rH is elementary abelian of order 4, k s 1, e1 s 2, lG s 2 x 1 , x Ž1. s 30 lŽ1.. Ž7a. t s 20. Then IG Ž l.rH ( C Ž3., k s 1, e1 s 1, and a contradiction follows. Ž8a. t s 30. Then IG Ž l.rH ( C Ž2., e i s 1, k s 1, and a contradiction follows. Ž9a. t s 60. Then IG Ž l. s H, k s 1, e1 s 1, lG s x 1 , x 1Ž1. s 60 lŽ1.. Thus, if x g IrrŽ G . y IrrŽ GrH . and l g IrrŽ x H ., then lG s 2 x and x Ž1. s 30 lŽ1., or lG s x and x Ž1. s 60 lŽ1..
543
FINITE NONSOLVABLE GROUPS
Denote by V 1 , . . . , V n4 the set of the G-orbits on the set IrrŽ H . y 1 H 4 , and suppose Žin view of Ž1a. ] Ž9a.. that < V i < s 15 Ž i s 1, . . . , m4 ,
< V j < s 60 Ž j s m q 1, . . . , n . .
Take l i g V i Ž i s 1, . . . , n.. Then 60 < H < s < G < s < G : H < q
x Ž 1.
Ý
2
x gIrr Ž G .yIrr Ž GrH . m
s 60 q 30 2
is1
< H < s 1 q 15
n
Ý li Ž 1. 2 q 60 2 Ý
m
jsmq1 n
Ý li Ž 1. 2 q 60 Ý 1s1
2
l j Ž 1. , 2
l j Ž 1. .
jsmq1
Hence < H < ' 1 Žmod 15.. By Thompson’s Theorem on minimal simple groups, the order of such a group is divisible by 3 or 5. Hence H is solvable and in this case the lemma follows. Žb. GrH ( L2 Ž7.. We keep the same notation as in Ža.. Then e i g 1, 2, 3, 4, 6, 7, 84 . Now, SLŽ2, 7. is the only representation group of GrH and cd Ž SL Ž 2, 7 . . s 1, 3, 4, 6, 7, 8 4 , t g 1, 7, 8, 14, 21, 24, 28, 42, 56, 84, 168 4 . As in Ža. we may prove that one of the following possibilities holds: Ž1b. Ž2b.
t s 42, k s 1, e s 2. t s 168, k s 1, e s 1.
Let t s 42, k s 1, e s 2. For appropriate l g IrrŽ H ., lG s 2 x , where x g IrrŽ G .. In this case x Ž1. s 84lŽ1., and l belongs to the G-orbit of length 42. Let t s 168, k s 1, e s 1. Then for appropriate l g IrrŽ H ., lG s x g IrrŽ G . and l belongs to the G-orbit of length 168. Denote by V 1 , . . . , V n4 the set of G-orbits on the set IrrŽ H . y 1 H 4 . Suppose that < V i < s 42, i g 1, . . . , m, < V j < s 168, j s m q 1, . . . , n.
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BERKOVICH AND KAZARIN
Take l i g V i for all i. Then 168 < H < s < G < s < G : H < q 84 2
m
n
2
Ý li Ž 1. q 168 2 is1
m
Ý
2
l j Ž 1. ,
jsmq1 n
< H < s 1 q 42 Ý l i Ž 1 . 2 q 168
Ý
is1
jsmq1
2
l j Ž 1. .
The last equality shows that < H < is odd. Hence H is solvable. LEMMA 3.2. Let H be a normal subgroup of a D 1-group G. If GrH is isomorphic to L2 Ž5. or L2 Ž7., then H s 14 . Proof. Suppose that H ) 14 . Without loss of generality we may assume that H is a minimal normal subgroup of G. By Lemma 3.1, H is abelian. Put < H < s q s, q is a prime, s ) 0. We retain the notation V i , n, m, e i from the proof of Lemma 3.1. It follows from Lemma 3.1 that < V i < s < V j < implies e i s e j . Therefore m F 1, n y m F 1. Ži. Let G ( L2 Ž5.. Then q s y 1 g 15, 60, 15 q 604 . Thus, q s g 2 4 , 614 . Let q s s 2 4 . Then t s 15, x Ž1. s 30 for a character x g IrrŽ lG ., where l g IrrŽ H . is nonprincipal. In this case k Ž G . s
Let G be a simple D 1-group. Then
Proof. Assume that AutŽ G . ) InnŽ G .. Take a g AutŽ G . y InnŽ G .. Let x 1 , x 2 be two distinct nonlinear irreducible characters of G of the same degree. Let x g IrrŽ G . y x 1 , x 2 4 . Since G is a D 1-group, x a s x . Therefore, by wFS, Theorem Cx and Brauer’s Permutation Lemma, Ž x 1 . a s x 2 . Any b g AutŽ G . y InnŽ G . acts as a . Therefore, if g s aby1 , then x ig s x i for i s 1, 2, and x g s x for all x g IrrŽ G . y x 1 , x 2 4 . Then, by wFSx, g g InnŽ G .. Thus,
FINITE NONSOLVABLE GROUPS
545
Let G be a nonsolvable D 1-group. Then, by Theorem 2.1, G9 s G. Let H be a maximal normal subgroup of G. Then S s GrH is a nonabelian simple group. Since S is not a D 0-group wBCHx it is a D 1-group. Therefore
4 It follows from Lemma 3.2 that for a simple D 1-group G one has
G ( A n is the alternating group of degree n ) 6. G ( L nŽ p ., n ) 1, and Ž n, p y 1. s 1 for n ) 2.
Žiii.
G ( UnŽ p ., n ) 2, Ž n, p q 1. s 1. G ( PSp 2 mŽ q ., m ) 1, and
Živ. Ž1iv. Ž2iv. Ž3iv. Žv. Žvi. Žvii. Žviii. Žix. Žx. Žxi. Žxii. Žxiii. Žxiv. Žxv.
q s p ) 2, or m ) 2, q s 2 e, e - 3, or
m s 2, q s 2. G ( PV 2 mq1Ž p ., m ) 2, p ) 2. Ž . G ( PVq 2 m 2 , m ) 4. Ž . G ( PVy 2 m 2 , m ) 3. Ž . G ( G 2 q , e - 3, p / 3. G ( E6 Ž p ., Ž p y 1, 3. s 1. G ( F4Ž q ., e - 3, and e s 1 for p s 2. G ( E7 Ž q ., where e s 1, or q s 4. G ( E8 Ž q ., e - 3. G (2 E6 Ž p ., Ž p q 1, 3. s 1. G (2 F4Ž2.9 is the Tits simple group. G is isomorphic to one of 26 sporadic simple groups.
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BERKOVICH AND KAZARIN
This follows from wLPS, pp. 18]19x. Note that L2 Ž2., L2 Ž3., U3 Ž2. are solvable. Further on, 2
F4 Ž 2 . : 2 F4 Ž 2 . 9 s 2, L2 Ž 4 . ( L2 Ž 5 . ( A 5 , L2 Ž 9 . ( Sp4 Ž 2 . 9 ( A 6 , L4 Ž 2 . ( A 8 , L2 Ž 7 . ( L3 Ž 2 . , U4 Ž 2 . ( PSp4 Ž 3 . , G 2 Ž 3 . 9 ( U3 Ž 3 . , 2
G 2 Ž 3 . 9 ( L2 Ž 8 . , L2 Ž q . ( PSp 2 Ž q . ( U2 Ž q . ( V 3 Ž q . ,
y 2 PVq 4 Ž q . ( L 2 Ž q . = L 2 Ž q . , PV 4 Ž q . ( L 2 Ž q . , V 5 Ž q . ( PSp4 Ž q . , y e e PVq 6 ( L 4 Ž q . , PV 6 Ž q . ( U4 Ž q . , PV 2 mq1 Ž 2 . ( Sp 2 m Ž 2 . .
5 In this section we will prove some auxiliary results. Let v be a primitive < G
FINITE NONSOLVABLE GROUPS
547
Žmod f < T <., x ' 1 Žmod m. has the unique solution modulo < G < by the Chinese Remainder Theorem. If s is a solution of that system, then c Ž ss . s w k . Thus, im c s H. Put c Ž ss . s ws . Then c Ž ss st . s c Ž sst . s wst s ws w t s c Ž ss . c Ž st .. Let ss g ker c . Then c Ž ss . s w 1 so s ' 1 Žmod < T <.. Conversely, suppose that s ' 1 Žmod < T <.; then ss g ker c . Therefore, ker c s G1. Thus, G1 ( H. Hence, any automorphism of T can be lifted to an automorGrG phism of QŽ v .. Take ss g G0 . Put c Ž ss . s w k . Then there exists an element g g G such that gy1 tg s t k for t g T. Then T g s T and g g NG ŽT .. Therefore w k g H0 s NG ŽT .rCG ŽT . s Aut G ŽT .. In particular, c Ž G0 . : H0 . Since < G : G0 < F 2, we have < H : H0 < F 2. This implies < H < s
548
BERKOVICH AND KAZARIN
case Oq. , SOnyŽ q . Ž n is even, case Oy. , and GŽ V, q . s OnŽ q . Ž n is even., OnqŽ q . Žcase Oq. , OnyŽ q . Žcase Oy. . In case O with n ) 1, the group SŽ V, q . has a subgroup V Ž ¨ , q . of Ž . yŽ . index 2 and in the cases as above we denote it by V nŽ q ., Vq n q , Vn q , respectively. Note that, if n is odd, then p ) 2. All the above groups in Case O are called orthogonal groups. All the groups Žin four cases. defined above are called classical. A cyclic subgroup T s CG Ž x . Ž x g G . of a classical group G is said to be a Singer cycle if T acts irreducibly on V and the bilinear form Ž , . is T-invariant. From the point of view of Chevalley groups a Singer cycle is a maximal torus of the corresponding group. Singer cycles exist for the following groups wSeix: G s GLm Ž q .
Ž of order q m y 1 . ,
Sp 2 m Ž q .
Ž of order q m q 1 . ,
GUm Ž q .
Ž of order q m q 1 for odd m . ,
SOy 2m
Ž of order q m q 1 . .
It is natural that in the corresponding simple groups L m Ž q . , PSp 2 m Ž q . , Um Ž q . , PVy 2mŽ q. there exist cyclic subgroups of respective orders, which are connected with Singer cycles. If G s L mŽ q ., then G s SL mŽ q .rZŽ SL mŽ q ... Therefore G contains a cyclic subgroup T of order Ž q m y 1.rŽ q y 1.Ž q y 1, m.. The group PSp 2 mŽ q . contains a cyclic subgroup T of order Ž q m q 1.rŽ q y 1, 2.. The group UmŽ q . for odd m contains a cyclic subgroup T of order Ž q m q 1.rŽ q Ž . q 1.Ž q q 1, m... Now, the group PVy 2 m q contains a cyclic subgroup T of order Ž q m q 1.rŽ q m q 1, 4.. In many cases we may use the information on Singer cycles to find some maximal cyclic subgroups of Chevalley groups. We note that PV " 2 l Ž q . : PV 2 lq1 Ž q . , PSp 2 l Ž q . : L2 l Ž q . ,
PV 2 ly1 : PV 2 lq1 Ž q . , Un Ž q . : PSp 2 n Ž q . ,
Uny1 Ž q . : Un Ž q . ,
PV 4 ly1 Ž q . : PVq 4l Ž q. .
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FINITE NONSOLVABLE GROUPS
LEMMA 6.1. Let G s GŽ q . be a classical group from Lemma 4.1. Then there exists some natural number n 0 and cyclic subgroup T of G such that n s < NG ŽT . : CG ŽT .<, n N n 0 and
Ž a . G s L m Ž q . , n ) 2, and q is a prime, < T < s Ž b . G s L2 Ž q . , q is odd, < T < s
qq1 2
Ž c . G s PSp2 l Ž q . , q is odd, < T < s
qm y 1 qy1
, n 0 s n s m.
, n 0 s 2.
ql q 1 2
, n s 2 l, l ) 1.
Ž d. G s PSp2 l Ž 2 n . , l G 2, n F 2, l, n 4 / 2, 2 4 , < T < s 2 n l q 1, n s 2 l. Ž e . G s PV 2 lq1 Ž q . , l ) 1, q is odd, < T < s Ž f . G s U2 lq1 Ž q . , l ) 1, < T < s Ž f . G s U2 lq1 Ž q . , l ) 1, < T < s
q 2 lq1 q 1 qq1 q 2 ly1 q 1 qq1
ql q 1 2
, n s 2 l.
, n s 2 l q 1. , n s 2 l q 1.
Ž g. G s U2 l Ž q . , l ) 1, < T < s q 2 ly1 q 1, n s 2 l y 1. Ž h . G s PVq2 l Ž 2 . , < T < s 2 ly1 q 1, n s 2 Ž l y 1 . . Ž i . G s PVy2 l Ž 2 . , < T < s 2 l q 1, n s 2 l. 7 In this section we prove some number theoretical lemmas and investigate the classical case. In particular, we prove that all classical simple groups G s GŽ q ., apart from L2 Ž5. and L2 Ž7., are not D 1-groups. LEMMAS 7.1. Let n ) 1 be a natural number. If n k 2 Žmod 4., then w Ž n. 2 G n. If n ' 2 Žmod 4., then 2 w Ž n. 2 G n. Proof. We omit the easy proof of this lemma since it follows from Lemma 7.2. LEMMA 7.2.
Let n be a natural number. If w Ž n. 2 F 2 n, then n g 1, 2, 3, 4, 6, 8, 10, 12, 18 4 .
In particular, n - 19.
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BERKOVICH AND KAZARIN
Proof. Let n s mp. Then w Ž n. s Ž p y 1. w Ž m., if p ¦ m, and w Ž n. s pw Ž m., if p < m. Assume that the lemma is proved for all m - n. Let m ) 1. Denote by p the lowest prime divisor of n. Ži. n s p s. Then w Ž n. 2 s Ž p y 1. 2 p 2 sy2 F 2 p s, or Ž p y 1. 2 p s F 2 p 2 . If s ) 1, then p s 2 and s g 2, 34 , so that n g 4, 84 . If s s 1, then Ž p y 1. 2 F 2 p, and p F 3. Thus, in that case n g 1, 2, 3, 4, 84 . Žii. n is not a prime power, n s mp, p ¦ m. Then w Ž n. 2 s Ž p y 1. 2w Ž m. 2 F 2 n s 2 pm. Ž1ii. Let p s 2. Then m is odd and w Ž m. 2 F 4 m. If m s q s, q ) 2 is a prime; then Ž q y 1. 2 q s F 4 q 2 and s - 3. For s s 1 we have Ž q y 1. 2 F 4 q so that q g 3, 54 , and n g 6, 104 . If s s 2, then Ž q y 1. 2 F 4 and q s 3, n s 18. Suppose that m is not a prime power. Let m s q s t, where q is a prime, s ) 0, t ) 1, and q ¦ t. Then Ž q y 1. 2 q sw Ž t . 2 F 4 q 2 t. Assume t ) 3. Then, by induction, 2 t F w Ž t . 2 . Therefore Ž q y 1. 2 q sw Ž t . 2 F 2 q 2w Ž t . 2 , or Ž q y 1. 2 q s F 2 q. Then q s s 3. We have 4w Ž t . 2 s w Ž m. 2 F 4 m s 12 t, or w Ž t . 2 F 3t. Let t s r u be a prime power, r ) 3. Then Ž r y 1. 2 r 2 uy2 s w Ž t . 2 F 3t s 3r u, or Ž r y 1. 2 r u F 3r 2 . Then u s 1, Ž r y 1. 2 F 3r, a contradiction. s If t s Ł is1 ria i is a standard decomposition in prime powers, s ) 1, then, by the above, 2
wŽ t. s
s
Ł w Ž ria . i
is1
2
s
)
Ł 3ria
i
s 3 s t ) 3t ,
is1
a contradiction. Let t s 3. Then m s 3q s, q ) 3. We have 2
2
4 Ž q y 1 . q 2 sy2 s w Ž m . F 4 m s 12 q s , or Ž q y 1. 2 q s F 3q 2 , which is impossible. Ž2ii. p ) 2. Let 2 m ) w Ž m. 2 . Then, by induction, m s 3 Žrecall that m is odd.. This is impossible since p is the lowest prime divisor of n. Thus 2 m F w Ž m. 2 . Therefore Ž p y 1. 2w Ž m. 2 F pw Ž m. 2 , or Ž p y 1. 2 F p, a contradiction. Žiii. n s pm, p < m, m is not a prime power. We have w Ž n. 2 s p 2w Ž m. 2 F 2 pm, or pw Ž m. 2 F 2 m. If p s 2, then w Ž m. 2 F m so that, by induction, m F 2, a contradiction since m is not a prime power. Let p ) 2. Then w Ž m. 2 - m and, by induction, m s 6. Then p s 2, a contradiction.
FINITE NONSOLVABLE GROUPS
551
Remark ŽG. Freiman.. The equation w Ž n. 2 - cn has a finite number of solutions for any constant c Žin fact, w Ž n. ) nrln n.. LEMMA 7.3. Let q be a prime, and n G 1 an integer. If w ŽŽ q nq 1 y 1.rŽ q y 1.. F 2Ž n q 1., then one of the following assertions holds: Ža. Žb.
q s 2, n - 4. n s 1, q g 3, 5, 7, 11, 134 .
Proof. Put m s Ž q nq 1 y 1.rŽ q y 1.. If m F 18, the result is obtained by easy checking. Let m ) 18. In view of the lemma one has 2 m - w Ž m. 2 F 4Ž n q 1. 2 , or m s 1 q q q ??? qq n - 2Ž n q 1. 2 . If q s 2, then 2 nq1 2Ž n q 1. 2 q 1, and n - 6. In that case n - 4 as the original inequality shows. In the general case, Ž q nq 1 y 1.rŽ q y 1. - 2Ž n q 1. 2 . If q s 3, then 3 nq 1 - 1 q 4Ž n q 1. 2 , and n - 3. The original inequality yields n s 1. Let q ) 3. Then n s 1 and w Ž1 q q . F 4. Using Lemma 7.2, one obtains q g 5, 7, 114 . In what follows G denotes a nonabelian simple group. LEMMA 7.4. or L2 Ž7..
If G s L nŽ q . is a D 1-group, then G is isomorphic to L2 Ž5.
Proof. Let n ) 2. By Lemma 5.2 and Lemma 7.3, w ŽŽ q n y 1.rŽ q y 1.. F 2 n so that q s 2 and n s 3 or 4. Since L4Ž2. is not a D 1-group wAtlasx, n s 3 and G s L3 Ž2. ( L2 Ž7.. Let n s 2. Then G ( L2 Ž4. ( L2 Ž5. ŽLemma 4.1., or q G 7 is a prime. Now, q g 7, 11, 134 ŽLemma 7.3.. Now, L2 Ž11. and L2 Ž13. are not D 1groups wAtlasx. LEMMA 7.5. Let q be a prime, n ) 1 odd, d s Ž q n q 1.rŽ q q 1.. If w Ž d . F 2 n, then one of the following assertions holds: Ža. Žb.
n s 3, q g 2, 34 . n s 5, q s 2.
Proof. Assume that w Ž d . 2 F 2 d. Then, by Lemma 7.2,
ds
qn q 1 qq1
g 1, 2, 3, 4, 6, 8, 10, 12, 18 4 ,
and q s 2, n F 5, or q s 3 s n. Now assume that w Ž d . 2 ) 2 d. Then d s Ž q n q 1.rŽ q q 1. - 2 n2 . If q s 2, then n - 8. In any case q - 4. If q s 3, then n - 6. The result is obtained by direct calculations.
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LEMMA 7.6. Let q be a prime number, n ) 1 odd. If w Ž q n q 1. F 2 n, then n s 3, q s 2. Proof. If q n q 1 - 19, then q s 2, n s 3. Otherwise, w Ž q n q 1. 2 ) 2Ž q n q 1.. Thus, q n q 1 - 2 n2 . So q s 2 and n - 7. The result is obtained by direct calculations. LEMMA 7.7. G s UnŽ q ., n ) 2 is not a D 1-group. Proof. Assume that n is even. Using Lemma 6.1 and 5.2, one obtains w Ž q ny 1 q 1. F 2Ž n y 1.. Then n y 1 s 3, q s 2 ŽLemma 7.6.. But U4Ž2. is not a D 1-group wAtlasx. Let n be odd. Then G g U3 Ž2., U3 Ž3., U5 Ž2.4 by Lemmas 6.1, 5.2, and 7.5. But U3 Ž2. is solvable, and U3 Ž3., U5 Ž2. are not D 1-groups wAtlasx. LEMMA 7.8.
Let q be an odd prime number and n G 2. The group G g PSp 2 n Ž q . , PV 2 nq1 Ž q . 4
is not a D 1-group. Proof. Using Lemmas 6.1 and 5.2, one obtains w ŽŽ q n q 1.r2. F 4 n. If Ž q n q 1.r2 F 18, then n - 3, q F 5, or n - 4, q s 3. If q s 5, n s 2, then the inequality for w does not hold. So let q n q 1 ) 36. Then, by Lemma 7.2, 16n2 G w ŽŽ q n q 1.r2. ) q n q 1. This implies n - 6, q F 7. If q s 3, n s 5, then the inequality for w does not hold. If q G 5, n s 5, then the inequality for w does not hold. So we may assume that n F 4. Further on, w ŽŽ3 4 q 1.r2. s w Ž41. ) 4 ? 4, and w ŽŽ5 4 q 1.r2. ) 4 ? 4. Therefore n / 4. Let n s 3. Then q s 3 or 5. If q s 5, then w ŽŽ5 3 q . 1 r2. s 36 ) 4 ? 3. Therefore, if n s 3, then q s 3. Let n s 2. An easy check shows that in this case q s 3 as well. Taking into account the list of isomorphisms after Lemma 4.1, we see that G g PSp4Ž3., PSp6 Ž3., PV 7 Ž3.4 . But these groups are not D 1-groups wAtlasx. LEMMA 7.9.
If w Ž2 m q 1. F 4 m, then m F 5.
Proof. Assume that 2 m q 1 ) 18 Žotherwise the lemma holds.. Then w Ž2 m q 1. 2 G 2Ž2 m q 1., or 2 m q 1 F 8 m2 and m F 9. It is easy to check that the inequality holds only for m - 6. LEMMA 7.10.
If G s PSp 2 nŽ2 n . is a D 1-group, then n s 5, n s 1.
Proof. By Lemma 4.1, n F 2. Assume that n s 2. It follows from Lemmas 5.2 and 6.1 that w Ž2 2 n q 1. F 4 n. Then w Ž2 m q 1. F 2 m, where m s 2 n. Then 2 n s m F 3, since PSp4Ž4. is not a D-group wAtlasx, contradicting n ) 1. Thus, n s 1. Then w Ž2 n q 1. F 4 n so n - 6 ŽLemma 7.9.. By wAtlasx, PSp6 Ž2. and PSp 8 Ž2. are not D 1-groups. The lemma is proved.
FINITE NONSOLVABLE GROUPS
LEMMA 7.11.
553
qŽ . Ž . Let G s PV " 2 n 2 be a D 1 -group. Then G s PV 12 2 .
Proof. Using Lemmas 5.2 and 6.1, we have w Ž2 ny 1 q 1. F 4Ž n y 1. for e s q, and w Ž2 n q 1. F 4 n for e s y. By Lemma 7.9, n - 7 in the first case, and n - 6 in the second one. By Lemma 4.1, in the first case n G 5. "Ž . Ž . w x Since PVy 8 2 and PV 10 2 are not D 1 -groups Atlas , the lemma holds in all cases. LEMMA 7.12. L2 Ž5. or L2 Ž7..
If G is a classical simple group, then G is isomorphic to
Proof. In view of Lemmas 7.4, 7.7, 7.8, 7.10, and 7.11 we have to show Ž . that PSp10 Ž2. and PVq 12 2 are not D 1 -groups. Ž . Denote G s PVq 2 . Note that 12 < G < s 2 30 Ž 2 6 y 1 . Ž 2 2 y 1 . Ž 2 4 y 1 . Ž 2 6 y 1 . Ž 2 8 y 1 . Ž 2 10 y 1 . . Ž . Ž . The group PVq 12 2 is contained in PSp12 2 and is defined as the group of the 12 = 12-matrices Y g GL12 Ž2., satisfying YAY t s A, where Y t is the transpose to Y, and As
ž
I , 0
0 I
/
where I is the 6 = 6 identity matrix. Note that the skew symmetric form f over GF Ž2. is invariant under the group PSp12 Ž2.. Since the ground field is GF Ž2., the form f is symmetric. It is more convenient to take a basis in which the matrix of our form is the following one, Fs
ž
I 0 0
0 I 0
0 0 , L
/
where I is the identity 5 = 5-matrix, and Ls
ž
0 1
1 . 0
/
All elements B g PSp12 Ž2. satisfy the equality BF B t s F. Therefore we can find an element of order 31 which is a generator of a Sylow 31-subgroup of PSp12 Ž2.. Consider the element Ts
X 0 0
0 Xyt 0
0 0 I
0
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of the group GL12 Ž2.; here I is the identity 2 = 2-matrix, and X g GŽ2, 5. is a matrix having the characteristic polynomial l5 q l3 q 1. It is clear that T g Sp12 Ž2. s PSp12 Ž2. and Xyt has the characteristic polynomial l5 q l q 1. It is sufficient to choose 0 0 Xs 0 0 1
1 0 0 0 1
0 1 0 0 0
0 0 1 0 0
0 0 0 . 1 1
0
Suppose that the element R11 R s R 21 R 31
R12 R 22 R 32
R13 R 23 R 33
0
normalizes the cyclic subgroup ²T :; here R11 and R 22 are 5 = 5-matrices, R 33 is a 2 = 2-matrix. Then RT s T k R for some natural number k, coprime with the order of T. It follows from this matrix equality that R13 s X k R13 , R 32 s R 32 Xyt ,
k
R 23 s Ž Xyt . R 23 , R11 X s X k R11 , k
R 21 X s Ž Xyt . R 21 ,
R 31 s R 31 X , R12 Xyt s X k R12 , k
R 22 Xyt s Ž Xyt . R 22 .
Since the matrices X y I, X k y I, Xyt y I, Ž Xyt . k y I are nonsingular, it follows from the first four equalities that R13 s R 23 s R 31 s R 32 s 0. Therefore R11 R Rs 21 0
R12 R 22 0
0 0 . R 33
0
By Schur’s Lemma, it follows from irreducibility of X on the 5-dimensional vector space over GF Ž2. that in fact R11 Rs 0 0
0 R 22 0
0 0 , R 33
0
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555
or 0 R s R 21 0
R12 0 0
0 0 . R 33
RX11 R2 s 0 0
0 RX22 0
0 0 , RX33
0
In particular,
0
where RX11 and RX22 are contained in the normalizer of the cyclic subgroup ² X : Žwhich is a Singer subgroup of GLŽ5, 2... Therefore the order of NG Ž²T :.rCG Ž²T :. divides 10. In fact, matrices of the form
ž
I 0 0
0 I 0
0 0 Y
/
commute with T. Therefore 2 ? 10 F 2 NG Ž ²T : . rCG Ž ²T : . - w Ž 31 . s 30. But this contradicts the fact that G is a D 1-group. The same arguments show that PSp10 Ž2. is not a D 1-group.
8 In this section we consider the exceptional groups of Lie type. By wAtlasx, F4Ž2. and 2 F4 Ž2.9 are not D 1-groups. Therefore we have to consider the following cases: Ži. Žii. Žiii. Živ. Žv. Žvi.
G 2 Ž q ., where q s p e, p / 3 is a prime number, and e F 2. E6 Ž p ., where p is a prime number, with Ž p y 1, 3. s 1. F4Ž q ., where q s p e, e F 2, and p ) 2 is a prime number. E7 Ž q ., where q s p is a prime number, or q s 4. E8 Ž q . where q s p e, e F 2, and p is a prime number. 2 E6 Ž p ., where p is a prime number with Ž p q 1, 3. s 1.
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LEMMA 8.1. The group G s G 2 Ž q . is not a D 1-group. Proof. We have q s p e, e F 2, and p / 3 is a prime. The structure of Sylow r-subgroups of G for primes r / p is known Žsee wSSx.. If r is coprime with the order of the Weyl subgroup W Žnote that < W < s 12. of G, then for r, dividing Ž q 2 y q q 1.Ž q 2 q q q 1., a Sylow r-subgroup T of G is cyclic, and NG ŽT .rCG ŽT . is cyclic of order 6. If d s Ž q 2 y q q 1, Ž q 2 y 1 .Ž q 3 y 1 . Ž q q 1 . . ) 1, then q ' y1 Žmod 3. and d s 3 since 9 ¦ Ž q 2 y q q 1.. Similarly, if q ' 1 Žmod 3., then 9 ¦ Ž q 2 q q q 1.. Thus, G has a cyclic subgroup C of order q 2 " q q 1, such that < NG Ž C . : C < s 6. It follows from Lemma 5.2 that w Ž q 2 " q q 1. F 12. If q 2 " q q 1 F 18, then, taking into account that q / 3, 9 we obtain q s 2 or 4. In the first case G9 s G 2 Ž2.9 ( A 6 is not a D 1-group. In the second case, by wAtlasx, G 2 Ž4. is not a D 1-group either. Therefore q 2 " q q 1 ) 18. By Lemma 7.2, 2
w Ž q 2 " q q 1. ) 2 Ž q 2 " q q 1. . Making use of Lemma 5.2, we conclude that 2Ž q 2 " q q 1. - 12 2 s 144 and q g 5, 74 . Next, 5 2 y 5 q 1 ' 0 Žmod 3., 5 2 q 5 q 1 k 0 Žmod 3., and w Ž31. s 30 ) 2 ? 6 s 12. By Lemma 5.2, G is not a D 1-group. Now, w Ž7 2 y 7 q 1. s w Ž43. s 42 ) 2 ? 6 s 12, and we conclude, in view of Lemma 5.2, that G is not a D 1-group in this case either. LEMMA 8.2. The group G s F4 Ž q . is not a D 1-group. Proof. By Lemma 4.1, q s p e, where e F 2 and a prime p is odd. It follows from wCarx that G contains a cyclic subgroup T of order q 4 q 1 such that < NG ŽT . : CG ŽT .< s 8. Therefore, by Lemma 5.2, w Ž q 4 q 1. F 16. Since q ) 2, we obtain q 4 q 1 ) 18. By Lemma 7.2, w Ž q 4 q 1. 2 ) 2Ž q 4 q 1.. Thus, 2Ž q 4 q 1. - 16 2 and q F 3. Since w Ž3 4 q 1. s 40 ) 16 we see that G is not a D 1-group. LEMMA 8.3. The groups E6 Ž q . and 2 E6 Ž q . are not D 1-groups. Proof. It follows from wWilx and wKonx that the groups E6 Ž q . and 2 E6 Ž q . have a cyclic subgroup T such that CG Ž x . F T for all x g T a , and T is of order Ž q 6 q q 3 q 1.rŽ3, q y 1., if G ( E6 Ž q ., and of order Ž q 6 y q 3 q 1.rŽ3, q q 1., if G (2 E6 Ž2.. By wCarx, < NG ŽT . : T < s 9 in every case. Since Ž3, q y 1. s 1 for E6 Ž q ., and Ž3, q q 1. s 1 for 2 E6 Ž2. by virtue of Lemma 4.1, G has an abelian self-centralizer T0 of order q 6 " q q 1, respectively, and such that NG ŽT0 . is a Frobenius group of order 9 < T0 <. Obviously, T0 is a TI-subgroup of G. It follows from the properties of abelian TI-subgroups
FINITE NONSOLVABLE GROUPS
557
that G has at least Ž q 6 " q 3 .r9 distinct irreducible characters of the same degree. Since G is a D 1-group then q 6 " q 3 F 18, which is impossible. LEMMA 8.4. The group G s E7 Ž q . is not a D 1-group. Proof. The group G has a self-centralizer T such that < T < s Ž q 8 y 1.rŽ q y 1. and < NG ŽT . : T < s 32 Žsee wCar, Table 6x.. That table shows that T is a unique up to conjugacy maximal torus of G of order divisible by Ž q 8 y 1.rŽ q y 1. s Ž q 4 q 1.Ž q 2 q 1.Ž q q 1.. On the other hand, by wSte2x, the group G contains a subgroup L8 Ž q ., and the last one has a cyclic subgroup of order Ž q 8 y 1.rŽ q y 1.. Therefore the subgroup T is cyclic. It follows from Lemma 5.2 that w ŽŽ q 8 y 1.rŽ q y 1. F 64. Since q ) 1, it follows that Ž q 8 y 1.rŽ q y 1. ) 18. Therefore, by Lemma 7.2,
w
ž
q8 y 1 qy1
2
/
) 2 Ž q 4 q 1 .Ž q 2 q 1 . Ž q q 1 . .
Hence 2Ž q 4 q 1.Ž q 2 q 1.Ž q q 1. - 2 12 , and q - 3. But
w Ž 2 8 y 1 . s w Ž 17 ? 5 ? 3 . s 16 ? 4 ? 2 s 126 ) 64. Therefore G is not a D 1-group. LEMMA 8.5. The group G s E8 Ž q . is not a D 1-group. Proof. It follows from wCar, Table 7x that G contains a maximal cyclic torus T of order Ž q 9 y 1.rŽ q y 1. s Ž q 6 q q 3 q 1.Ž q 2 q q q 1. such that < NG ŽT . : T < s 54. It follows from Lemma 5.2 that w ŽŽ q 6 q q 3 q 1.Ž q 2 q q q 1.. - 108. Obviously, Ž q 6 q q 3 q 1.Ž q 2 q q q 1. ) 18. By Lemma 7.2, 2
w Ž Ž q 6 q q 3 q 1 .Ž q 2 q q q 1 . . ) 2 Ž q 6 q q 3 q 1 .Ž q 2 q q q 1 . . Hence, 2Ž q 6 q q 3 q 1.Ž q 2 q q q 1. - 108 2 so q s 2. But
w Ž 2 9 y 1 . s w Ž 73 ? 7 . s 432 ) 2 ? 54, a contradiction which shows that G is not a D 1-group. It remains to find all alternating groups which are D 1-groups.
9 In this section we will prove that the alternating group A n for n ) 5 is not a D 1-group.
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There is the 1]1-correspondence a between irreducible characters of Sn and partitions w lx: n s n1 q ??? n k with natural numbers n i such that n1 G ??? G n k . We denote the character a Žw lx. by x w l x. To each such partition there corresponds the associated partition w lx*: n s nU1 q ??? qnUs of n such that nU1 s k and nUi s < j g N < n j G i4<. Let w lx be a partition of n: n1 q ??? qn k . Suppose that m1 , . . . , m s are all distinct numbers among n1 , . . . , n k . Put bi s < r g 1, . . . , k 4< n r s m i 4< for every i g 1, . . . , s4 . Then we say that the partition w lx is of type t Žw lx. s Ž m1b 1 , . . . , m bs s .. A partition w lx is said to be self-associated if w lx* s w lx. We use in what follows the following well-known result Žsee wKerx, for example.: LEMMA 9.1. Let w lx be a partition of n, and x s x w l x the corresponding irreducible character of Sn . Then Ža. If w lx / w lx*, then x A g IrrŽ A n .. n Žb. If w lx is self-associated, then x A s m 1 q m 2 , where m 1 , m 2 are n distinct irreducible characters of A n of the same degree. Suppose that w l1 x, w l2 x are distinct self-associated partitions of n, and x 1, x 2 corresponding irreducible characters of Sn . Put x Ai n s m 1i q m i2 , i s 1, 2. Then the characters m ij , i, j s 1, 2, are four pairwise distinct irreducible characters of A n . LEMMA 9.2. Let aŽ n. be the number of distinct self-associated partitions of a natural number n. Then A n is a D k-group, where k G aŽ n., unless n - 4. Proof. It is easy to check that the lemma holds for n - 6. So we may assume that n ) 5. Then, by wIsa, Theorem 14.23x 2 f cdŽ Sn . s x Ž1.< x g IrrŽ S n .4 . Now the result follows from the reasoning before the lemma. So it remains to prove that aŽ n. - 2 implies that n is small, and then use wAtlasx. We prove the following stronger result. LEMMA 9.3. Ža. If n G 8, then aŽ n. ) 1. Žb. If n ) 11, then aŽ n. ) 2. Proof. Ža. Suppose n s 2 x is even, n ) 8. Then x ) 4, and the following distinct partitions of n are self-associated: the partition l1e of type Ž x, 2, 1 xy2 . ; the partition l2e of type Ž x y 1, 3, 2, 1 xy4 . .
FINITE NONSOLVABLE GROUPS
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Suppose n s 2 x y 1 is odd. Then x G 5, and the following distinct partitions of n are self-associated. the partition l1o of type Ž x, 1 xy1 . ; the partition l2o of type Ž x y 2, 3, 3, 1 xy5 . . Žb. Suppose n s 2 x is even. Then x G 6, and we have the following additional self-associated partition w l3o x of type Ž x y 2, 4, 2, 2, 1 xy6 .. Suppose n s 2 x y 1 is odd. Then x G 7, and we have the following additional self-associated partition w l3o x of type Ž x y 3, 4, 3, 2, 1 xy7 .. Thus, Lemma 9.3 shows that A n is not a D 1 group for n ) 7. Since A 6 and A 7 are not D 1-groups wAtlasx, then A 5 is the only nonsolvable alternating D 1-group. With the result proved above we see that A 5 and L2 Ž7. are the only nonsolvable D 1-groups. This finishes the proof of the Main Theorem. Remark. Lemma 9.3 shows that A n is not a D 2-group unless n F 11. It follows from wAtlasx that A 9 and A10 are the only A n that are D 2-groups. QUESTION 1. Classify all nonsol¨ able D 2-groups. QUESTION 2. Find aŽ n., the number of self-associated partitions of a natural number n.
REFERENCES wAtlasx J. H. Conway, R. T. Curtis, S. P. Norton, R. A. Parker, and R. A. Wilson, ‘‘Atlas of Finite Groups,’’ Clarendon, Oxford, 1985. wBer1x Y. Berkovich, Non-solvable groups with a large fraction of involutions, in ‘‘Structure Theory of Set Addition, 7]11 June, 1993 CIRM, Marseille,’’ pp. 187]196. wBer2x Y. Berkovich, Finite solvable groups in which only two nonlinear irreducible characters have equal degrees, J. Algebra 184 Ž1996., 584]603. wBCHx Y. Berkovich, D. Chillag, and M. Herzog, Finite groups in which distinct nonlinear irreducible characters have distinct degrees, Proc. Amer. Math. Soc. 115 Ž1992., 955]959. wBZx Y. Berkovich and E. Zhmud’, ‘‘Characters of Finite Groups,’’ Vols. 1, 2, Amer. Math. Society, Providence, in press. wCarx R. W. Carter, Conjugacy classes of Weyl groups, in ‘‘Seminar on Algebraic Groups and Related Topics,’’ ŽA. Borel et al., Eds.., Lecture Notes in Mathematics, Vol. 131, Springer-Verlag, Berlin, 1970. wFSx W. Feit and G. Seitz, On finite rational groups and related topics, Illinois J. Math. 33, No. 1 Ž1988., 103]131. wIsax I. M. Isaacs, ‘‘Character Theory of Finite Groups,’’ Academic Press, New York, 1976. wKonx A. S. Kondrat’ev, On components of the graph of primes of finite simple groups, Mat. Sb. 180, No. 6 Ž1989., 787]797. wRussianx
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A. Kerber, ‘‘Representations of Permutation Groups I,’’ Lecture Notes in Mathematics, Vol. 240, Springer-Verlag, 1971. M. W. Liebeck, C. E. Praeger, and J. Saxl, The maximal factorizations of the finite simple groups and their automorphism groups, Mem. Amer. Math. Soc., 86, 432 Ž1990.. G. M. Seitz, On the subgroup structure of classical groups, Comm. Algebra 10, No. 8 Ž1982., 875]885. T. A. Springer and R. Steinberg, Conjugacy classes, in ‘‘Seminar on Algebraic Groups and Related Topics’’ ŽA. Borel et al., Eds.., Lecture Notes in Mathematics, Vol. 131, 1970. E. Stensholt, An application of Steinberg construction on twisted groups, Pacific J. Math. 55, No. 2 Ž1974., 595]618. E. Stenholt, Certain embeddings among finite groups of Lie type, J. Algebra, 69 Ž1978., 136]187. J. S. Williams, Prime graph components of finite groups, J. Algebra, 69 Ž1981., 497]513.