On Abelian Subgroups ofp-Groups

199, 262]280 Ž1998. JA977179

JOURNAL OF ALGEBRA ARTICLE NO.

On Abelian Subgroups of p-Groups Yakov Berkovich* Department of Mathematics and Computer Science, Uni¨ ersity of Haifa, Haifa 31905, Israel Communicated by Walter Feit Received January 27, 1997 IN MEMORY OF MY DEAR FRIEND GRISHA KARPILOVSKY

Ž1940]1997. This is a continuation of our note Ž J. Algebra 186 Ž1996., 120]131.. We generalize and present simplified proofs almost all elementary lemmas from Section 8 of the Odd Order Paper. In many places we use Hall’s enumeration principle and other simple combinatorial arguments. A number of related results are proved as well. Some open questions are posed. Q 1998 Academic Press

Let G be a finite p-group, where p is a prime. Let Q be a group-theoretic property and T the set of all Q-subgroups in G. Suppose that p ¦ < T <. Let G act on T by conjugation. Then one of the G-orbits on T contains one element, say H. Thus H y 1 G. Moreover, if G is a normal subgroup of some larger p-group W, then T admits the action of W by conjugation and, as previously, T contains a one-element W-orbit. We see that, in many cases, counting theorems are more fundamental than the respective theorems on the existence of normal subgroups. In this note we use this approach extensively Žas we know, Sylow was the first who used this argument in analogous situations .. We will show how this argument works in another situation. Suppose that a p-group G contains a subgroup H of maximal class and index p, < G < ) p pq 1. We claim that AutŽ G . is a p-group, where p s p Ž pŽ p 2 y 1.. Žp Ž n. is the set of all prime divisors of n g N.. Note that p ŽAutŽG.. : p for every p-group generated by two elements. Therefore, the result is true if G is of maximal class, so suppose that G is not of maximal class. Assume that a is an automorphism of G of prime order q f p . By w6x and w18x, the number of subgroups of maximal class and order < H < in G is * The author was supported in part by the Ministry of Absorption of Israel. 262 0021-8693r98 $25.00 Copyright Q 1998 by Academic Press All rights of reproduction in any form reserved.

ON ABELIAN SUBGROUPS OF

p-GROUPS

263

divisible by p 2 . Since a p-group of maximal class is generated by two elements, the minimal number of generators of G is 3, and so the number of maximal subgroups in G is 1 q p q p 2 - 2 p 2 . Therefore G contains exactly p 2 subgroups of maximal class and index p. Since q / p, one of these subgroups, say H, is a-invariant. Since AutŽ H . is a p-group, a centralizes H. Then ² a : stabilizes the chain G ) H )  14 Žsince q ¦ p y 1.. By w21x, Lemma 8.1 or Lemma 14, oŽ a . is a power of p}contradiction. Similarly, if < G < is a 2-group of order at least 2 5, H a subgroup of maximal class and index 2 in G, then AutŽ G . is a 2-group Žthe condition < G < G 2 5 is essential .. To prove this, we have to use the fact that a group of automorphisms of a 2-group of maximal class and order at least 2 4 is a 2-group. We retain, as a rule, the notation and definitions from w21x. If A is a maximal normal abelian subgroup of a p-group G, then CG Ž A. s A Ži.e., A is a maximal abelian subgroup of G .. This ‘‘selfcentralizer’’ property allows us to restrict the structure of G if all its maximal abelian normal subgroups are small in some sense. Some more general results in this direction were proved and used in w21x and in the subsequent Thompson’s paper on simple groups all of whose local subgroups are solvable. For example, if p ) 2 and A is a maximal normal elementary abelian subgroup of G, then V 1Ž CG Ž A.. s A Žsee w21x, Lemma 8.3 and w24x, Lemma 10.15.. Alperin w1x generalized that result in the following way: Let A be a maximal normal abelian subgroup of a p-group G such that expŽ A. F p n, where p n ) 2; then V nŽ CG Ž A.. s A. ŽFor another proof of Alperin’s Theorem, due to N. Blackburn, see w28x, Theorem 3.12.1. We will prove the following stronger result Žwhich we consider as the main theorem of this note.: THEOREM 1. Let A - B F G, where A, B are abelian subgroups of a p-group G, expŽ B . F p n and p n ) 2. Let A be the set of all abelian subgroups T of G such that A - T, < T : A < s p and expŽT . F p n. Then < A < ' 1 Žmod p .. Let us show that Theorem 1 implies Alperin’s Theorem. Let A be a maximal normal abelian subgroup of G of exponent at most p n, where p n ) 2. Let x g CG Ž A. y A such that oŽ x . F p n. Then A - B s ² A, x : and B is abelian of exponent at most p n. In the notation of Theorem 1, < A < ' 1Žmod p .. Then there exists T g A such that T y 1 G, contradicting the choice of A. In what follows, we will deduce from Theorem 1 essentially stronger result ŽCorollary 2.. Proof of Theorem 1. We use induction on G. If T g A, then T F CG Ž A.. Therefore, without loss of generality, we may assume that G s CG Ž A., i.e., A F ZŽ G .. If A s  14 , then A consists of all subgroups of G of order p, and so < A < ' 1 Žmod p . by Sylow’s Theorem. Therefore, in what follows, we assume that A )  14 . If < G: A < s

264

YAKOV BERKOVICH

p, then A s  G4 , < A < s 1 and all is done. Therefore, we may assume that < G: A < ) p. Let AX s T g A < T 1 G4 . Since < A < ' < AX < Žmod p ., it suffices to prove that < AX < ' 1 Žmod p .. To prove that AX / B we use induction on < G <. Let A - B F M - G, where M is maximal in G. Then the number of elements of A that contained in M is congruent to 1 Žmod p ., and our claim follows. Set D s ²T N T g AX :. By the definition of AX , DrA is an elementary abelian subgroup of ZŽ GrA.. Therefore, the nilpotency class of D is at most two. We claim that expŽ D . F p n. By construction, D is generated by elements of orders at most p n. Therefore, if x g D, then x s a1 ??? a s , where oŽ a i . F p n for all i. Since D is of class at most 2 Žit is essential that p n ) 2., we have n

n

n

x p s a1p ??? a sp ? c1p ??? c tp ,

Ž 1.

X

where c1 , . . . , c t are elements of D Žthe equality Ž1. is known for s s 2; for s ) 2 it is proved by induction on s .. Since expŽ DX . F expŽ DrZŽ D .. s n p Žrecall that A F ZŽ D .., formula Ž1. is reduced to x p s 1. Therefore, X n expŽ D . F p , as claimed. All elements of A are contained in D. If HrA is a subgroup of order p in DrA, then H is abelian, H 1 G and expŽ H . F expŽ D . F p n, i.e., H g AX . Therefore, we may assume, without loss of generality, that G s D. In particular, GrA is elementary abelian. Let < GrA < s p k . By the previous text, < AX < is the number of subgroups of order p in the elementary abelian group GrA of order p k , i.e., < AX < s 1 q p q ??? qp ky 1 ' 1 Žmod p ., proving the theorem. Kulakoff’s Theorem w32x and its analog for p s 2 Žsee w4x, Section 5. show that < A < ' 1 q p Žmod p 2 . unless CG Ž A.rA is cyclic or a 2-group of maximal class. The subgroup A of Theorem 1 is not necessarily of exponent p n. Theorem 1 is not true for p n s 2 Žlet G be a dihedral group of order 8, A s ZŽ G .; then < A < s 2.. COROLLARY 2. Let N be a normal subgroup of a p-group G, let A be a maximal G-in¨ ariant abelian subgroup of N such that expŽ A. F p n, p n ) 2. Then V nŽ CN Ž A.. s A. Proof. Assume that x g CN Ž A. y A is of order at most p n. Set A N s  T F N < A - T , < T : A < s p, T is abelian, exp Ž T . F p n 4 . Since ² A, x : is abelian of exponent at most p n, the set A N is nonempty. By Theorem 1, < A N < ' 1 Žmod p .. Therefore, there exists T g A N such that T 1 y G, contrary to the choice of A. This generalizes w21x, Lemma 8.3 and the main result in w1x. The following three results are known.

ON ABELIAN SUBGROUPS OF

p-GROUPS

265

LEMMA 3. The number of abelian subgroups of index p in a nonabelian p-group G is one of the numbers 0, 1, p q 1. LEMMA 4 w2x. Suppose that a p-group G contains an abelian subgroup A of index p 2 . Then G contains a normal abelian subgroup of index p 2 . Proof. Let A - M - G, where M is a maximal subgroup of G; then < M: A < s p. Since the number of abelian subgroups of index p in M is congruent to 1 Žmod p . ŽLemma 3. and M 1 G, the result follows. Moreover, by w29x, Theorem Žiii., the number in Lemma 4 is 2 or ' 1 Žmod p .. LEMMA 5 w4x, Section 5. Let G be a noncyclic p-group. Suppose that G is not a 2-group of maximal class. If n g N and n ) 1, then the number of cyclic subgroups of order p n is G is di¨ isible by p. For p ) 2 this result is due to G. A. Miller. COROLLARY 5X Žw4x; see also w25x, Chapter 5, Exercise 9 and w24x, Lemma 10.11.. Let N be a noncyclic normal subgroup of a p-group G. If N has no G-in¨ ariant noncyclic subgroups of order p 2 , then N is a 2-group of maximal class. Corollary 5X is a generalization of Roquette’s Lemma Žsee w28x, Satz 3.7.6.. COROLLARY 5Y . Let p ) 2, N a normal subgroup of a p-group G, A a G-in¨ ariant elementary abelian subgroup of order p 2 in N. If N contains an elementary abelian subgroup B of order p 3, then N contains a G-in¨ ariant elementary abelian subgroup B1 such that A - B1. Proof. It follows from p 2 ¦
266

YAKOV BERKOVICH

Remark 2. Let us prove the following result Žsee w21x, Lemma 8.5.: Suppose that P g Syl p Ž G ., where p ) 2 is the smallest prime divisor of < G <; if P has no elementary abelian subgroups of order p 3, then G is p-nilpotent. Suppose that G is a counterexample. Then G contains a non p-nilpotent minimal nonnilpotent group S Žby Frobenius’ Normal p-Complement Theorem.. We may assume that P l S g Syl p Ž S .. If p Ž S . s  p, q4 , then q ) p q 1, and so q ¦ p 2 y 1, contradicting Remark 1. THEOREM 6. Suppose that a p-group G, p ) 2, contains an elementary abelian subgroup of order p 3. Then the number of such subgroups in G is congruent to 1 Žmod p .. Proof. For H F G, let e3 Ž H . denote the number of elementary abelian subgroups of order p 3 contained in H. We have to prove that e3 Ž G . ' 1 Žmod p .. Let M denote the set of all maximal subgroups of G. It is known that < M < ' 1 Žmod p .. By Hall’s enumeration principle w26x, e3 Ž G . '

Ý

e3 Ž H .

Ž mod p . .

Ž 2.

Hg M

Suppose that the theorem has proved for all proper subgroups of G. Take H g M . By the induction hypothesis, e3 Ž H . s 0 or e3 Ž H . ' 1 Žmod p .. If e3 Ž H . ' 1 Žmod p . for all H g M , then by Ž2., e3 Ž G . ' < M < ' 1 Žmod p ., proving the theorem. Therefore, we may assume that some maximal subgroup of G, say H, has no elementary abelian subgroups of order p 3. Suppose that H contains a subgroup L of order p 4 and exponent p. Let A be a maximal normal abelian subgroup of L. Since A - L and CLŽ A. s A, it follows that < A < s p 3 , contrary to what was proved in the previous paragraph. By assumption, G contains an elementary abelian subgroup E0 of order p 3. Let E0 F F g M . By the induction hypothesis, e3 Ž F . ' 1 mod p, and so G contains an elementary abelian subgroup E of order p 3. Let eX3 Ž G . be the number of normal elementary abelian subgroups of order p 3 in G. Since e3 Ž G . ' eX3 Ž G . Žmod p ., it suffices to prove that eX3 Ž G . ' 1 Žmod p .. Therefore, we may assume that G contains a normal elementary abelian subgroup E1 of order p 3, E1 / E. Set D s EE1. By Fitting’s Lemma, the nilpotency class of D is at most two. Therefore Žsee the proof of Theorem 1., expŽ D . s p. Considering D l H and taking into account that H has no subgroups of order p 4 and exponent p, we conclude that < D < s p 4 . By Lemma 3, e3 Ž D . ' 1 Žmod p .. Hence the number of G-invariant elementary abelian subgroups of order p 3 in D is congruent to 1 Žmod p .; therefore, we may assume that G contains a normal elementary abelian subgroup E2 of order p 3 such that E2 g D

ON ABELIAN SUBGROUPS OF

p-GROUPS

267

Žotherwise, all is done.. Suppose that E l E1 - E2 Žsince < D < s p 4 it follows that < E l E1 < s p 2 .. Then E l E1 F ZŽ EE1 E2 . and EE1 E2rE l E1 is elementary abelian of order p 3. In that case, EE1 E2 is of order p 5 and its class is at most two, and so expŽ EE1 E2 . s p Žsee the proof of Theorem 1.. Then H l EE1 E2 is a subgroup of order p 4 and exponent p in H, contrary to what we have previously proved. Therefore, E l E1 g E2 . Since < EE2 < s < E1 E2 < s p 4 Žsee the proof of the equality < D < s p 4 ., it follows that < E l E2 < s p 2 s < E1 l E2 <. Since E l E2 , E1 l E2 are different maximal subgroups of E2 Žsince E l E1 g E2 ., we conclude that E2 s Ž E l E2 .Ž E1 l E2 . - EE1 s D, contrary to the choice of E2 . This completes the proof of the theorem. Theorem 6 was announced in w9x. This theorem is not true for p s 2 Žlet G be the direct product of the dihedral group of order 8 and the cyclic group of order 2.. For another proof of this theorem, see w31x, Theorem 2.2. QUESTION 1. Classify the p-groups G, p ) 2, containing an abelian self-centralizer A of order p 3. Ob¨ iously < NG Ž A.: A < F p 3 if A is noncyclic, and < NG Ž A.: A < F p 2 if A is cyclic. The case < NG Ž A.: A < s p is of some interest. Let E k Ž G . be the set of normal elementary abelian subgroups of order p k , k g N, in G. THEOREM 7. Suppose that a p-group G, p ) 2, contains an elementary abelian subgroup E of order p 4 . Then the number of elementary abelian subgroups of order p 4 in G is congruent to 1 Žmod p .. In particular, < E 4Ž G .< ' 1 Žmod p . Ž ob¨ iously these two assertions are equi¨ alent.. Proof. Suppose that G is a counterexample of minimal order. Then < G < ) p 5 by Lemma 3. By Theorem 6, G contains a normal elementary abelian subgroup A of order p 3. A. Suppose that G has no normal elementary abelian subgroup B of order p 4 such that A - B. It follows from Theorem 1 that Ži. A is a maximal elementary abelian subgroup of G. Therefore, Žii. if K be a subgroup of G of exponent p, then CK Ž A. F A. Note that a Sylow p-subgroup of AutŽ A.Ž( GLŽ3, p .. is nonabelian of order p 3 and exponent p. Therefore, the following two assertions are true: Žiii. If K is an elementary abelian subgroup of G of order p 4 , then < A l K < s p 2 . In particular, G has no elementary abelian subgroups of order p 5.

268

YAKOV BERKOVICH

Živ. G has no subgroup of order p7 and exponent p. If K is a subgroup of G of order p 6 and exponent p then A - K. Let E - M - G, where M is a maximal subgroup of G. By the induction hypothesis, < E 4 Ž M .< ' 1 Žmod p ., so that M contains a G-invariant elementary abelian subgroup of order p 4 . Without loss of generality, we may assume that E 1 G. If E is the only normal elementary abelian subgroup of G of order p 4 , then < E 4 Ž G .< s 1, contrary to the assumption. Therefore, G contains a normal elementary abelian subgroup E1 of order p 4 different of E. Set K s EE1. Since K is of class at most two and p ) 2, it is of exponent p. Suppose that < K < s p 6 . Then A - K by Živ.. Since CK Ž A. s A by Žii., it follows that E l E1 F ZŽ K . - A. Since KrE l E1 is abelian, it follows that KrA is abelian of order p 3. By the remark before Žiii. on GLŽ3, p . it follows that CK Ž A. ) A, contradicting Žii.. Thus, < K < s p 5. Hence, Žv. If B, B1 are two different normal elementary abelian subgroups of G of order p 4 , then < BB1 < s p 5, i.e., < B l B1 < s p 3. Moreover, B l B1 s ZŽ BB1 . since BB1 is nonabelian by the second part of Žiii.. Thus, K s EE1 is nonabelian of order p 5 and exponent p, E l E1 s ZŽ K . is of order p 3. Therefore, by Žii., A / ZŽ K ., and so Žvi.

A g K.

Suppose that K contains all elements of the set E 4Ž G .. By Lemma 3 Žsince < K < s p 5 ., < E 4Ž K .< ' 1 Žmod 1.. Therefore, < E 4Ž G .< ' < E 4 Ž K .<1 Žmod p ., contrary to the assumption. Thus, there exists E2 g E 4Ž G . such that E2 g K. By Žv., < E l E2 < s p 3 s < E1 l E2 < s < E l E1 <. Assume that E2 l E / E2 l E1. Then E2 s Ž E2 l E .Ž E2 l E1 . F EE1 s K, contrary to the choice of E2 . Hence E2 l E s E2 l E1. Then CG Ž E2 l E . G EE1 s K, and so E2 l K s E l E1 s ZŽ K .. Set L s KE2 s EE1 E2 . Then E2 l K F ZŽ L. and < L < s p 6 . Since LrZŽ K . s ErZŽ K . = E1rZŽ K . = E2rZŽ K . is elementary abelian of order p 3 , the nilpotency class of L is two and since L s EE1 E2 is generated by elements of order p, it follows that expŽ L. s p. By Živ., A - L. Then CLŽ A. ) A Žsince < ZŽ L.< s p 3 s < A <., contrary to Žii.. B. Every normal elementary abelian subgroup of G of order p 3 is contained in some elementary abelian subgroup of G of order p 4 . Then every A g E 3 Ž G . is contained in some B g E 4Ž G . by Theorem 1. Since the number of conjugates Žin G . of every nonnormal subgroup is a multiple of p, it suffices to prove that < E 4Ž G . ' 1 Žmod p .. Set E 3 Ž G . s  A1 , . . . , A r 4 , and

E 4 Ž G . s  B1 , . . . , Bs 4 .

ON ABELIAN SUBGROUPS OF

p-GROUPS

269

We have to prove that s ' 1 Žmod p .. Suppose that A i is contained in a i elements of the set E 4 Ž G ., and Bj contains b j elements of the set E 3 Ž G .. Then the number of pairs  A i , Bj 4 Ž A j g E 3 Ž G ., Bj g E 4 Ž G ., i s 1, . . . , r, j s 1, . . . , s . is

a 1 q ??? qa r s b 1 q ??? qb s .

Ž 3.

We have b j ' Ž p 4 y 1.rŽ p y 1. ' 1 Žmod p . for j s 1, . . . , s. By Theorem 1, a i ' 1Žmod p . for i s 1, . . . , r. Therefore, reading Ž3. by modulo p, we obtain r ' s Žmod p .. Since r ' 1 Žmod p . by Theorem 6, it follows that s ' 1 Žmod p ., completing the proof. Using the method of the proof of Theorem 7, we can give another, shorter, proof of Theorem 6. Theorem 7 was announced in w5x, Theorem 11. For another proof, see w31x, Theorem 2.2. In the same paper Konvisser and Jonah proved that if p ) 2, then < E 5 Ž G .< ' 1 Žmod p . unless E 5 Ž G . s B Žbut there exists G such that E 6 Ž G . s 2 w29x.. COROLLARY 8. Let N be a normal subgroup of a p-group G, p ) 2. Suppose that N has no G-in¨ ariant elementary abelian subgroups of order p 3. Then N has no elementary abelian subgroups of order p 3 Ž and hence, the structure of N is completely determined in w17x.. This follows from Theorem 6. If N s G in Corollary 8, we obtain Hobby’s Theorem w27x. See also w31x and w24x, Proposition 10.17. COROLLARY 9. Let N be a normal subgroup of a p-group G, p ) 2. If N contains an elementary subgroup of order p 4 , then N contains a G-in¨ ariant elementary abelian subgroup of order p 4 . This follows from Theorem 7. If N s G in Corollary 9, we obtain Hobby’s result. See also w31x and w24x, Proposition 10.17. LEMMA 10. ŽO. J. Schmidt w34x, J. A. Gol’fand w23x; see also w28x, Satz 3.5.2.. Let S be a minimal nonnilpotent group. Then S is a  p, q4 -group, p, q are different primes. Let P g Syl p Ž S ., Q g Syl q Ž S .. One of Sylow subgroups of S, say Q, is normal in S. The subgroup Q s SX is elementary abelian or special, P is cyclic and < P: P l ZŽ S .< s p. If q ) 2, then expŽ Q . s q. p is a Zsigmondy prime for the pair ² q, log q < QrZŽ G . l Q <:. If Q is nonabelian, then log q < QrZŽ Q .< is e¨ en. THEOREM 11 Žw21x, Lemma 8.8.. Suppose that Q is a q-group, q ) 2, a an automorphism of Q of prime order p. If p ' 1 Žmod q ., and Q contains a maximal subgroup Q0 such that Q0 has no normal elementary abelian subgroups of order q 3, then p s 1 q q q q 2 and Q is elementary abelian of order q 3.

270

YAKOV BERKOVICH

Proof. By Corollary 8, Q0 has no elementary abelian subgroups of order q 3. Therefore, every subgroup of exponent q in Q0 is generated by two elements and its order is at most q 3. In particular, Q has no elementary abelian subgroups of order q 4 . Let W s ² a : ? G be a natural semidirect product. By assumption, p ) q q 1 so that p ¦ q 2 y 1 and < Q < ) q 2 . Let S be a minimal nonnilpotent subgroup of W. Without loss of generality, we may assume that ² a : s P g Syl p Ž S .. Suppose that < Q < s q 3. Since p ¦ q 2 y 1, it follows that Q is elementary abelian. Since Ž p, q 2 y 1. s 1, it follows that p < q 3 y 1 Žin fact, p ¬
ON ABELIAN SUBGROUPS OF

p-GROUPS

271

n s m y 1. Let n - m y 1. Let M denote the set of all maximal subgroups of G Žobviously, < M < s 4.. Let H g M . If H is of maximal class, then by the induction hypothesis, < M Ž H .< ' 1 Žmod 3.. If H is metacyclic, then < M Ž H .< ' 1 Žmod 3. by Sylow’s Theorem. Any subgroup of index 3 in G is metacyclic or of maximal class Žsince m G 5.. Hence, by Hall’s enumeration principle w26x, < M Ž G .< ' Ý H g M < M Ž H .< ' < M < s 4 ' 1 Žmod 3.. In what follows, G is not a 3-group of maximal class. Žii. Let n s m y 1. Without loss of generality, we may assume that H g M is metacyclic Žotherwise, M Ž G . s 0.. By what we have proved, Ži. and w17x, G contains a normal subgroup R of order p 3 and exponent p. In view of the existence of H, G has no subgroups of order p 4 and exponent p and G is generated by three elements Žsince H is generated by two elements.. If R - F g M , then F is not a 3-group of maximal class Žby w20x, a p-group of maximal class and order at least p pq 2 has no normal subgroups of order p p and exponent p; note that n G 3 q 2. and is not metacyclic. Let E be the set of all normal subgroups of G of order p 3 and exponent p. By w6x or w17x and Ži., < E < ' 1 Žmod p .. Let M1 be the set of all maximal subgroups L of G such that L contains an element of the set E. By w6x, the number of elements of the set E in any element of M1 is ' 1 Žmod p .. Therefore, arguing, as in the end of the proof of Theorem 7, we obtain < M1 < ' 0 Žmod p .. Let T g M y M1. Then by w6x, T is metacyclic or a 3-group of maximal class. The number of subgroups of maximal class and index p in G is divisible by p 2 by w6x or w17x. Since < M1 < ' 1 Žmod p ., it follows that < M y M1 < ' 1 Žmod p .. The number of subgroups of maximal class and index p in G is divisible by p 2 by w6x or w17x. Therefore, < M Ž G .< ' 0 Žmod p .. Žiii. n - m y 1. Applying Hall’s enumeration principle, Ži. and Žii., we complete the proof. Note, that Theorem 12 is true for n s 3 Žsee w6, 9, 15x. It is interesting to study < M Ž G .<Žmod 2. for 2-groups G. LEMMA 13. Let G s G 0 ) G1 ) ??? ) Gn s  14 be a chain of normal subgroups of a group G. Set a

A s  a g Aut Ž G . ¬ Ž xGi . s xGi for all x g Giy1 y Gi , i s 1, . . . , n4 , Ž i.e., A stabilizes that chain.. Let p be a prime number such that p ¦ < ZŽ Giy1rGi .< for i s 2, . . . , n. Then p ¦ < A <. Proof. We use induction on n. We may assume that n ) 1 Žotherwise the lemma is true.. Obviously, A is a subgroup of AutŽ G .. Suppose that A contains an element a of order p. Let W s ² a : ? G be the natural direct product. By the induction hypothesis, ² a : ? Gny 1 1 W. By assumption, a

272

YAKOV BERKOVICH

centralizes Gny 1 so that ² a : ? Gny1 s ² a : = Gny1. Assume that ² a : is not characteristic in H s ² a : = Gny 1. Then there exists f g AutŽ H . such that f Ž² a :. s ² b : / ² a :. Obviously, ² b : 1 H and ² a , b : is a subgroup of ZŽ H . of order p 2 . Then ² a , b : l Gny 1 is a subgroup of ZŽ Gny 1 . of order p, contradicting the assumption. Thus, ² a : is characteristic in ² a : = Gny 1. Then ² a : 1 W, so that a centralizes G. In that case, a s id, which is a contradiction. LEMMA 14 Žw21x, Lemma 8.1.. If G is a p-group, then the stability group A of a chain G s G 0 ) G1 ) ??? ) Gn s  14 is a p-group. Proof. We use induction on n. Suppose that A contains an element a of order p for some p g p X . By the induction hypothesis, a centralizes G1. p Let x g G; then x a s xy with y g G1. Then x s x a s xy p. It follows that y s 1 so that a s 1}contradiction. LEMMA 15. Let G be a p-group and suppose that D is a subgroup generated by all elements of G of orders 4, p for all p g p . Let a be a p X-automorphism of G. If a centralizes D, then a s id. Proof. Suppose that the lemma is false. Without loss of generality, we may assume that oŽ a . s q g p X . Let W s ² a : ? G. For every p g p there exists P g Syl p Ž G . such that P a s P. Since a / id, we may assume that a induces a nonidentity automorphism of P. Let H be a minimal nonnilpotent subgroup in G1 s ² a : ? P ŽF W .. Without loss of generality, we may assume that a g H. Obviously, H l P g Syl p Ž H .. By Lemma 10, H l P F D. This means that a centralizes H l P so H is nilpotent} contradiction. LEMMA 16. Let U be an abelian subgroup of a p-group G, let a be a pX-automorphism of G. If a centralizes CG ŽU ., then a s id. This is exactly what was proved in w21x, Lemma 8.12. COROLLARY 17 Žcompare with w21x, Lemma 8.12.. Let U be an abelian subgroup of a p-group G. Let e s 1 if p ) 2 and e s 2 if p s 2. Let a be a pX-automorphism of G that centralizes V e Ž CG ŽU ... Then a s id. Proof. Obviously CG ŽU . is a-invariant. a centralizes CG ŽU . by Lemma 15. Now the result follows from Lemma 16. COROLLARY 18 Žw19x; see also w22x, Lemma 2.3.. Let p, G, and e be such as in Corollary 17. Let U be a maximal abelian subgroup of G of exponent p n, where n G e . If a pX-automorphism a of G centralizes U, then a s id.

ON ABELIAN SUBGROUPS OF

p-GROUPS

273

Proof. Let x be an element of CG ŽU . of order at most p n. Since ²U, x : is abelian of exponent at most p n, it follows that x g U by the maximal choice of U. Thus, V e Ž CG ŽU .. s U, and the result follows from Corollary 17. If in Corollary 18, U is a-invariant and a centralizes V e ŽU ., then a s id Žby Lemma 15 and Corollary 18.. PROPOSITION 19.

Let B be a subgroup of a p-group G such that CG Ž B . F

B. Ža. If B is nonabelian of order p 3, then G is of maximal class. Žb. Ž M. Suzuki . If < B < s p 2 , then G is of maximal class. Proof. We may assume that < G < G p 4 and the proposition has proved for groups of order - < G <. Ža. It is known that a Sylow p-subgroup of AutŽ B . is nonabelian of order p 3. Now, CG Ž B . s ZŽ B . s ZŽ G .. Therefore, NG Ž B .rZŽ G . is nonabelian of order p 3 Žin particular, < NG Ž B .< s p 4 .. Then CG r ZŽG.Ž NG Ž B .r ZŽ G .. - NG Ž B .rZŽ G . Žotherwise < NG Ž B .< ) p 4 .. By the induction hypothesis, we see that GrZŽ G . is of maximal class. Since < ZŽ G .< s p, Ža. is proved. Žb. Since p 2 ¦
274

YAKOV BERKOVICH

metacyclic, and so it is of order p 2 Žsince expŽ H . s p .. We have CA D Ž H . s H. Therefore, AD is of maximal class by Proposition 19Žb.. By w20x, AD has no subgroups of order p pq 1 and exponent p, so that < D < - p pq 1. QUESTION 3. Study the groups G of Proposition 20 in the case when < A < s p, in detail. PROPOSITION 20X . Let G be a metacyclic p-group, containing a nonabelian subgroup B of order p 3. Then Ža. Žb.

If p s 2, then G is of maximal class. If p ) 2, then < G < s p 3.

Proof. Let the proposition has proved for all groups of order - < G <. We may assume that < G < ) p 3. Let B F M - G, where M is maximal in G. Ža. By the induction hypothesis, M is of maximal class. Since G has exactly three maximal subgroups, the number of subgroups of maximal class and index 2 in G is not divisible by 4. Therefore G is of maximal class by w4x, Section 5. Žb. By the induction hypothesis, < M < s p 3, and so < G < s p 4 . Obviously, GrGX is abelian of type Ž p 2 , p .. Let C1rGX , . . . , C prGX be all cyclic subgroups in GrGX of order p 2 . Then C1 , . . . , C p are distinct abelian subgroups of G of index p. By Lemma 3, all maximal subgroups of G are abelian, contradicting the existence of B. PROPOSITION 20Y . Let B be a nonabelian subgroup of order p 3 in a p-group G, p ) 2. Suppose that G has no element x such that ² B, x : s B = ² x :. Then: Ža.

CG Ž B . is cyclic.

Žb.

G has no B-in¨ ariant subgroups of order p pq 1 and exponent p.

Proof. Ža. Obvious. Žb. Assume that H is a B-invariant subgroup of order p pq 1 and exponent p in G. Without loss of generality, we may assume that G s BH. By w20x, G is not of maximal class. Therefore, CG Ž B . g B by Proposition 19Ža.. In particular, < CG Ž B .< ) p and B g H, i.e., G / H. By Ža., < CG Ž B . l H < s p. It follows that CG Ž B . l H s B l H F ZŽ H .. Let C be a cyclic subgroup of order p 2 in CG Ž B .. Then K s BC l H is a normal noncyclic subgroup of order p 2 in BC, < BC < s p 4 . Since BCrZŽ B . s BrZŽ B . = KrZŽ B . is abelian and p ) 2, it follows that expŽ BC . s p, which is a contradiction Žsince C is cyclic of order p 2 ..

ON ABELIAN SUBGROUPS OF

p-GROUPS

275

It is essential, in the proof of Žb., that V 1Ž BC . s B. Remark 4. Let H be a normal subgroup of G and H F F Ž G .. Suppose that, whenever N 1 H and H X g N, then N 1 G. We claim that H is abelian. Suppose that G is a counterexample of minimal order. Let S be a normal subgroup of H such that HrS is nonabelian but every proper epimorphic image of HrS is abelian. By assumption, S 1 G. By the induction hypothesis, S s  14 . In particular < H X < s p. Let K be a G-invariant subgroup of order p 2 in H. Then H F F Ž G . - CG Ž K ., since < G:CG Ž K .< F p. In particular, K F ZŽ H ., and so H is not of maximal class. By Corollary 5X we may assume that K is noncyclic. Then ZŽ H . contains a subgroup L of order p such that L / H X . Obviously, L 1 G. Then HrL is nonabelian, contrary to the choice of S. Remark 5. Suppose that a p-group G contains an elementary abelian subgroup of order p 3. Let E denote the subgroup generated by all elementary abelian subgroups of order p 3 in G. Suppose that V 1Ž G . g E, i.e., G contains a subgroup L of order p such that L g E. If D is an L-invariant subgroup of exponent p in E then < D < F p p Žby Proposition 19 and w20x, if < D < ) p p , then < CL D Ž L. l D < G p 2 .. PROPOSITION 21 Žcompare with w36x, p. 69, Lemma.. Let G be a nilpotent group and let GrGX be abelian of rank d. Suppose that < GX < d F < GrZŽ G .<. If an automorphism a of G lea¨ es e¨ ery element of GrGX fixed, then a is an inner automorphism of G. Proof. Let GrGX s ² x 1GX : = ??? = ² x d GX :. Since GX F F Ž G ., it follows that G s ² x 1 , . . . , x d :. Let A be the set of all automorphisms of G which leave every element of GrGX fixed; then InnŽ G . F A F AutŽ G . Žsince every coset of GX is G-invariant.. If a g A, then, for i s 1, . . . , d, we obtain x ia s x i yi where yi g GX . Obviously, a is uniquely determined by elements y 1 , . . . , yd . There are < GX < d distinct d-sequences  y 1 , . . . , yd 4 of elements in GX . Hence we have < A < F < GX < d. On the other hand, by what we have just proved and assumption, < GX < d F < GrZŽ G .< s
COROLLARY 22 Žcompare with w36x, Ž4.17... Let G F W, where W is a group and G is a nilpotent subgroup such that GrGX is abelian of rank d and < GX < d F < GrZŽ G .<. If w G, W x F GX , then W s GCW Ž G .. Proof. By assumption G 1 y W. Let w g W. Conjugation by w induces an automorphism a in G. If g g G, then by assumption, w w, g x s y g GX ,

276

YAKOV BERKOVICH

so that Ž gy1 . a s wy1 gy1 w s ygy1. Hence a leaves every element of GrGX fixed. Therefore, by Proposition 21, a g InnŽ G .. This means that for each w g W, there exists u g G such that wy1 gw s uy1 gu for all g g G. Thus, wuy1 g CW Ž G ., and so w g CW Ž G .G for each w g W. This means that W s CW Ž G .G. A p-group G is said to be generalized regular if, whenever x p s y p Ž x, y g G ., then Ž xy1 y . p s  14 . If G is a generalized regular p-group, then expŽ V 1Ž G .. s p. Regular p-groups are generalized regular Žsee w28x, Kapitel 3.. PROPOSITION 23 Žcompare with w36x, p. 73, Lemma.. Let a pX-group Q act on a generalized regular p-group G. If Q acts tri¨ ially on F Ž G ., then w G, Q x F V 1Ž G .. Proof. If x g G, then x p g F Ž G .. Hence, for any y g Q, we have x s Ž x p . y s Ž x y . p. Therefore, w x, y x p s Ž xy1 x y . p s  14 Žsince G is generalized regular., so that w G, Q x F V 1Ž G .. p

LEMMA 24. Let G be a sol¨ able group of odd order, p g p Ž G ., OpX Ž G . s  14 . If Op Ž G . has no elementary abelian subgroups of order p 3 , then a Sylow p-subgroup is normal in G. Proof. Let B be a critical Thompson’s subgroup of Op Ž G . Žfor its construction and properties, see w3x, Corollary 3. and set T s V 1Ž B .. Since B is of class at most two and p ) 2, expŽT . s p. Obviously, B 1 y G. Since CG Ž Op Ž G .. F Op Ž G ., it follows from w3x, Corollary 3Že. that CG Ž B . F B. Therefore, by Lemma 15, CG ŽT . is a normal p-subgroup of G. Since T has no elementary abelian subgroups of order p 3 , it follows that < T < F p 3 Žrecall that T is of exponent p .. If < T < s p, then < GrCG ŽT .< ¬ p y 1, and therefore CG ŽT . is a normal Sylow p-subgroup of G. Let < T < s p 2 . Then GrCG ŽT . is isomorphic to a subgroup of odd order of AutŽT . ( GLŽ2, p .. Since in that case a Sylow p-subgroup of GrCG ŽT . is normal Žthis follows from the structure of GLŽ2, p .., all is done. Let now < T < s p 3. By assumption, T is nonabelian of exponent p. By Hall’s Lemma on pX-automorphisms of p-groups, CG r T X ŽTrT X . is a normal p-subgroup of GrT X . Since Ž GrT .rCG r T X ŽTrT X . is isomorphic to a subgroup of odd order in GLŽ2, p ., the result follows as in the case < T < s p 2 . PROPOSITION 25 Žw21x, Lemma 8.13.. Let P be a Sylow p-subgroup of a sol¨ able group G of odd order. Suppose that P has no elementary abelian subgroups of order p 3. Then GX centralizes all chief p-factors of G. Proof. Without loss of generality we may assume that OpX Ž G . s  14 . Then P 1 y G, where p g Syl p Ž G . Žby Lemma 24.. If ArB is a chief factor of G of order p, then GrCGrB Ž ArB . is cyclic of order dividing p y 1, and

ON ABELIAN SUBGROUPS OF

p-GROUPS

277

so GX BrB F CG r B Ž ArB .. The chief p-factors of G are of orders p or p 2 Žsee w17x.. Let KrL be a chief factor of G of order p 2 . If K g GX , it follows that GX centralizes KrL. Hence we have to consider the case when K F GX . Without loss of generality we may assume that L s  14 . Then K F P F F Ž G ., and so P centralizes K. Next, GrCG Ž K . as a pX-subgroup of odd order in GLŽ2, p ., is abelian. Therefore, G9 centralizes K. PROPOSITION 26. Suppose that  14 F A - B F G, where G is a p-group, B elementary abelian of order p k , 2 F k F 4. Then the number of elementary abelian subgroups of G of order p k that contain A is congruent to 1 Žmod p ., unless p s 2 s k, G is maximal class. Proof. Suppose that the proposition has proved for all p-groups of order - < G <. The result is true for A s  14 by Lemma 5, Theorems 6 and 7. Let A )  14 . We may assume that B - G. If < CG Ž A.< s p 2 , then G is of maximal class Žby Proposition 19Žb.., < A < s p, and B is the unique elementary abelian subgroup of order p 2 containing A. In that case, all is done. If CG Ž A. is a 2-group of maximal class then G is Žsince, in that case G s CG Ž A... Therefore, without loss of generality, we may assume that A F ZŽ G .. Let MA be the set of all maximal subgroups of G containing A. By Sylow’s Theorem, < MA < ' 1 Žmod p .. For A F H F G, let sk Ž H . denote the number of elementary abelian subgroups of H of order p k containing A. By Hall’s enumeration principle w26x, sk Ž G . '

Ý

sk Ž H .

Ž mod p . .

Ž 4.

Hg MA

If sk Ž H . ' 1 Žmod p . for all H g MA , then sk Ž G . ' 1 Žmod p . by Ž4.. If every elementary abelian subgroup of G of order p k contains A, the result follows from Lemma 5, Theorems 6 and 7. Therefore, we may assume that there exists in G an elementary abelian subgroup T of order p k such that A g T. Then Žthe elementary abelian subgroup since A F ZŽ G .. AT contains an elementary abelian subgroup F such that A - F, < F < s p kq 1 , and L l F is elementary abelian of order at least p k for every maximal subgroup L of G Žand so sk Ž L. ) 0 since A F ZŽ L... As previously, we may assume that sk Ž H . k 1 Žmod p . for some H g MA . Therefore, by the induction hypothesis, p s 2, < A < s 2, and k s 2, H is cyclic, dihedral, or semidihedral, A F ZŽ H .. Then A - F Ž G .. Suppose that G is not a 2-group of maximal class. If H is cyclic, then sk Ž G . s e2 Ž G . s 1 Žthis follows from the classification of 2-groups with cyclic subgroups of index 2.. If H is dihedral or semidihedral, then s2 Ž H . is even. If F is a maximal subgroup of G that does not contain A, then G s A = F and MA has no

278

YAKOV BERKOVICH

subgroups of maximal class. Now the result follows from Ž4. by induction Žsince, in that case, the number of elements of maximal class in MA is divisible by 4; see w4x Section 5.. By w31x, Proposition 26 is true for k s 5 as well. QUESTION 4. Consider the case p s 2 in all propositions of this note. Conjecture. Let A - B - G, where G is a p-group, p ) 2, < A < s p, B is abelian of type Ž p n, p ., n ) 1. Then the number of subgroups of G, that are isomorphic to B and contain A, is congruent to 0 Žmod p . Žwith a small number of exceptions.. Let D 8 be the dihedral group of order 8, C2 the group of order 2. Remark 6. The following assertion is an analog of Theorem 1 for p s 2: Let A - B F G, where G is a 2-group, expŽ B . s 2, < A < G 4. If G has no subgroups isomorphic to D 8 = C2 , then the number of elementary abelian subgroups L - G such that A - L and < L: A < s 2 is odd. We will prove that the preceding assertion is trivial. Indeed, let E be a maximal elementary abelian subgroup of G, < E < ) 4. Suppose that an involution u normalizes E but u f CG Ž E .. Then E contains an involution ¨ such that u¨ / ¨ u. In that case, D s ² u, ¨ : ( D 8 Žindeed, the dihedral group D is contained in the subgroup ² E, ¨ : of exponent 4.. Set ZŽ D . s ² c :; then D l E s ² c : = ² ¨ :. By Proposition 19Žb., < CE Ž u.< ) 2 Žsince ² E, u: is not of maximal class: it contains an elementary abelian subgroup of order 8.. Since D l E does not centralize D, there exists in CE Ž u. an involution w / c. Obviously, w centralizes D s ² D l E, u:. Therefore, ² D, w : ( D 8 = C2 , contrary to the assumption of the theorem. Thus, every involution that normalizes E, also centralizes E. Hence V 1Ž NG Ž E .. s E, and so E is characteristic in NG Ž E .. In that case, NG Ž E . s G and E s V 1Ž G .. Our assertion follows immediately Žsince we may assume without loss of generality that V 1Ž G . s E .. The foregoing reasoning shows that if G is a 2-group such that V 1Ž G . is nonabelian and G has no subgroups isomorphic to D 8 = C2 , then it has no elementary abelian subgroups of order 8. Numerous counting theorems for p-groups Žwhich are, as we saw previously, assertions on the existence of normal subgroups with given structure. were proved or announced in w3]16x Žsome of foregoing results are taken from these papers, but here we offer new proofs., w29, 31, 33x. In particular, A. Mann proved that the number of nonabelian subgroups of order p 3 in a p-group G of order at least p 5 is divisible by p 2 Žand so the

ON ABELIAN SUBGROUPS OF

p-GROUPS

279

number of abelian subgroups of order p 3 in G is ' 1 q p Žmod p . by Kulakoff’s Theorem w32x Žsee also w35x. and w6x, Section 5. Next, Mann proved Žsee w16x, Theorem B. that if noncyclic Sylow p-subgroup S of G is of order p m , S is not a 2-group of maximal class, n g N and p n - < S <, then the number of subgroups of order p n in G is ' 1 q p Žmod p 2 .. Note that about all of the papers w3]16x were inspired by w21x, Section 8 and w17]20x.

REFERENCES 1.. J. L. Alperin, Centralizers of abelian normal subgroups of p-groups, J. Algebra 1 Ž1964., 110]113. 2. J. L. Alperin, Large abelian subgroups of p-groups, Trans. Amer. Math. Soc. 117 Ž1965., 10]20. 3. Y. Berkovich, On p-separable groups, J. Algebra 186 Ž1996., 120]131. 4. Y. Berkovich, On p-groups of finite order, Siberian Math. J. 9 Ž1968., 1284]1306 Žin Russian.. 5. Y. Berkovich, The subgroup and normal structure of a finite p-group, So¨ iet Math. Dokl. 106 Ž1971., 71]75. 6. Y. Berkovich, A generalization of theorems of Ph. Hall and Blackburn and applications to irregular p-groups, Math USSR Iz¨ . 35 Ž1971., 815]844. 7. Y. Berkovich, The structure of a group and the structure of its subgroups, So¨ iet Math. Dokl. 9 Ž1968., 301]304. 8. Y. Berkovich, Normal divisors of a finite group, So¨ iet Math. Dokl. 9 Ž1968., 1117]1121. 9. Y. Berkovich, Subgroups, normal divisors and epimorphic images of a finite p-group, So¨ iet Math. Dokl. 10 Ž1969., 499]501. 10. Y. Berkovich, On the number of subgroups of given order in a finite p-groups of exponent p, Proc. Amer. Math. Soc. 4 Ž1990., 875]879. 11. Y. Berkovich, On the number of elements of given order in a finite p-group, Isr. J. Math. 73 Ž1991., 107]112. 12. Y. Berkovich, On the number of subgroups of a given order and exponent p in a finite irregular p-group, Bull. London Math. Soc. 24 Ž1992., 259]266. 13. Y. Berkovich, Counting theorems for finite groups, Arch. Math. 59 Ž1992., 215]222. k 14. Y. Berkovich, On the number of solutions of equation x p s a in a finite p-group, Proc. Ž . Amer. Math. Soc. 116 1992 , 585]590. 15. Y. Berkovich, On the number of subgroups of a given structure in a finite p-group, Arch. Math. 63 Ž1994., 111]118. k 16. Y. Berkovich, On the number of solutions of x p s 1 in a finite p-group, Rend. Lincei, Mat. Appl. 6 Ž1995., 5]12. 17. N. Blackburn, Generalizations of certain elementary theorems on p-groups, Proc. London Math. Soc. Ž3. 11 Ž1961., 1]22. 18 N. Blackburn, Note on a paper of Berkovich, J. Algebra 24 Ž1973., 320]332. 19. N. Blackburn, Automorphisms of finite p-groups, J. Algebra 3 Ž1966., 28]29. 20. N. Blackburn, On a special class of p-groups, Acta. Math. 100 Ž1958., 45]92. 21. W. Feit and J. Thompson, Solvability of groups of odd order, Pacific J. Math. 13 Ž1963., 775]1029. 22. T. M. Gagen, ‘‘Topics in Finite groups,’’ Cambridge University Press, Cambridge, 1976. 23. A. J. Gol’fand, On groups all of whose subgroups are nilpotent, Dokl. Akad. Nauk SSSR 60 Ž1948., 1313]1315 Žin Russian..

280

YAKOV BERKOVICH

24. D. Gorenstein, R. Lyons, and R. Solomin, ‘‘The Classification of Finite Simple Groups, Number 2, Part 1, Chapter G: General Group Theory,’’ Am. Math. Soc., Providence, RI, 1996. 25. G. Gorenstein, ‘‘Finite groups, Harper and Row,’’ New York, 1968. 26. P. Hall, A contribution to the theory of groups of prime power order, Proc. London Math. Soc. Ž2. 36 Ž1933., 29]95. 27. C. Hobby, Abelian subgroups of p-groups, Pacific J. Math. 12 Ž1962., 1343]1345. 28. B. Huppert, ‘‘Endliche Gruppen,’’ Bd. 1, Springer-Verlag, Berlin, 1967. 29. D. Jonah and M. Konvisser, Abelian subgroups of p-groups, an algebraic approach, J. Algebra 34 Ž1975., 386]402. 30. M. Konvisser, Embedding of abelian subgroups in p-groups, Trans. Amer. Math. Soc. 153 Ž1971., 469]481. 31. M. Konvisser and D. Jonah, Counting abelian subgroups of p-groups. A projective approach, J. Algebra 34 Ž1975., 309]330. ¨ 32. A. Kulakoff, Uber die Anzahl der eigentlichen Untergruppen und der Elemente von gegebener Ordnung in p-Gruppen, Math. Ann. 104 Ž1931., 778]793. 33. A. Mann, Conjugacy classes in finite groups, Israel J. Math 31 Ž1978., 78]84. 34. O. J. Schmidt, On groups all of whose subgroups are nilpotent, Mat. Sb. 31 Ž1924., 366]372 Žin Russian.. 35. O. J. Schmidt, The new proof of group-theoretic Kulakoff’s theorem, Mat. Sb. 39 Ž1932., 66]71 Žin Russian.. 36. M. Suzuki, ‘‘Group Theory II,’’ Springer-Verlag, Berlin, 1982.