JOURNAL OF ALGEBRA 199, 262᎐280Ž. 1998 ARTICLE NO. JA977179
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
View metadata, citation and similar papers Receivedat core.ac.uk January 27, 1997 brought to you by CORE provided by Elsevier - Publisher Connector IN MEMORY OF MY DEAR FRIEND GRISHA KARPILOVSKY Ž.1940᎐1997
This is a continuation of our note ŽŽJ. Algebra 186 1996.. , 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. ᮊ 1998 Academic Press
Let G be a finite p-group, where p is a prime. Let ⌰ be a group-theo- retic property and ᑮ the set of all ⌰-subgroups in G. Suppose that p¦<<ᑮ. Let G act on ᑮ by conjugation. Then one of the G-orbits on ᑮ contains one element, say H. Thus H G. Moreover, if G is a normal y1 subgroup of some larger p-group W, then ᑮ admits the action of W by conjugation and, as previously, ᑮ 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 extensivelyŽ as 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, p 1 2 < * The author was supported in part by the Ministry of Absorption of Israel. 262 0021-8693r98 $25.00 Copyright ᮊ 1998 by Academic Press All rights of reproduction in any form reserved. ON ABELIAN SUBGROUPS OF p-GROUPS 263 divisible by p2. 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 2 2 of maximal subgroups in G is 1 q p q p - 2 p . Therefore G contains exactly p2 subgroups of maximal class and index p. Since q / p, one of these subgroups, say H,is ␣-invariant. Since AutŽ.H is a -group, ␣ centralizes H. Then ²:␣ stabilizes the chain G ) H ) Ä41Ž since q ¦ p y 1. . Bywx 21 , Lemma 8.1 or Lemma 14, oŽ.␣ is a power of pᎏcontradiction. Similarly, if < CAGŽ.sA Ži.e., A is a maximal abelian subgroup of G .. This ‘‘self- centralizer’’ 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 inwx 21 and in the subsequent Thompson’s paper on simple groups all of whose local sub- groups are solvable. For example, if p ) 2 and A is a maximal normal elementary abelian subgroup of G, then ⍀1ŽCAG Ž ..sA Žseewx 21 , Lemma 8.3 andwx 24 , Lemma 10.15. . Alperinwx 1 generalized that result in the following way: Let A be a maximal normal abelian subgroup of a p-group nn G such that expŽ.A F p , where p ) 2; then ⍀ nGŽCA Ž ..sA. Ž For another proof of Alperin’s Theorem, due to N. Blackburn, seewx 28 , Theorem 3.12.1.Ž We will prove the following stronger result which we consider as the main theorem of this note. : THEOREM 1. Let A - B F G, where A, B are abelian subgroups of a nn p-group G, expŽ.B F p and p ) 2. Let ᑛ be the set of all abelian n subgroups T of G such that A - T, < If T g ᑛ, then T F CAGŽ.. Therefore, without loss of generality, we may assume that G s CAGŽ., i.e., A F ZG Ž..If AsÄ41 , then ᑛ consists of all subgroups of G of order p, and so <<ᑛ ' 1Ž. mod p by Sylow’s Theorem. Therefore, in what follows, we assume that A ) Ä41.If < n Since ²:A, x is abelian of exponent at most p , the set ᑛ N is nonempty. By Theorem 1, <<ᑛ NN' 1Ž. mod p . Therefore, there exists T g ᑛ such that T G, contrary to the choice of A. 1y This generalizeswx 21 , Lemma 8.3 and the main result in wx 1 . 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 42.wx Suppose that a p-group G contains an abelian subgroup A of index p2. Then G contains a normal abelian subgroup of index p2. Proof. Let A - M - G, where M is a maximal subgroup of G; then < LEMMA 5wx 4 , Section 5. Let G be a noncyclic p-group. Suppose that G is not a 2-group of maximal class. If n g ގ and n ) 1, then the number of n cyclic subgroups of order p is G is di¨isible by p. For p ) 2 this result is due to G. A. Miller. X COROLLARY 5Žwx 4 ; see also wx 25 , Chapter 5, Exercise 9 andwx 24 , Lemma 10.11. . Let N be a noncyclic normal subgroup of a p-group G. If N has no 2 G-in¨ariant noncyclic subgroups of order p , then N is a 2-group of maximal class. X Corollary 5 is a generalization of Roquette’s LemmaŽ seewx 28 , Satz 3.7.6. . Y COROLLARY 5 . Let p ) 2, N a normal subgroup of a p-group G, Aa 2 G-in¨ariant elementary abelian subgroup of order p in N. If N contains an 3 elementary abelian subgroup B of order p , then N contains a G-in¨ariant elementary abelian subgroup B11 such that A - B . 2 Proof. It follows from p ¦ Remark 2. Let us prove the following resultŽ seewx 21 , Lemma 8.5. : Suppose that P g Syl pŽ.G , where p ) 2 is the smallest prime divisor of < Proof. For H F G, let eH3Ž.denote the number of elementary abelian 3 subgroups of order p contained in H. We have to prove that eG3Ž.'1 Ž.mod p . Let M denote the set of all maximal subgroups of G. It is known that < eG33Ž.'ÝeH Ž . Žmod p ..2Ž. HgM Suppose that the theorem has proved for all proper subgroups of G. Take H g M. By the induction hypothesis, eH33Ž.s0oreH Ž.'1 Ž.Ž.Ž.mod p .If eH33'1 mod p for all H g M, then byŽ.Ž. 2 , eG'< Abe a maximal normal abelian subgroup of L. Since A - L and CALŽ.s 3 A, it follows that < By assumption, G contains an elementary abelian subgroup E0 of order 3 p. Let E03F F g M. By the induction hypothesis, eFŽ.'1 mod p, and so Gcontains an elementary abelian subgroup E of order p3. X Let eG3Ž.be the number of normal elementary abelian subgroups of 3 X order p in G. Since eG33Ž.'eGŽ.Žmod p ., it suffices to prove that X eG3Ž.'1 Ž mod p .. Therefore, we may assume that G contains a normal 3 elementary abelian subgroup E11of order p , E / E. Set D s EE 1.By Fitting’s Lemma, the nilpotency class of D is at most two. ThereforeŽ see the proof of Theorem 1.Ž. , exp D s p. Considering D l H and taking into account that H has no subgroups of order p4 and exponent p,we 4 conclude that < 4 Ž.otherwise, all is done . Suppose that E l E12- E Žsince < QUESTION 1. Classify the p-groups G, p ) 2, containing an abelian 3 3 self-centralizer A of order p . Ob¨iously< NGŽ. A : A Let ᑟ kŽ.G be the set of normal elementary abelian subgroups of order k p , k g ގ,inG. THEOREM 7. Suppose that a p-group G, p ) 2, contains an elementary abelian subgroup E of order p4. Then the number of elementary abelian subgroups of order p4 in G is congruent to 1Ž. mod p . In particular, <ᑟ4Ž.G<'1Ž.Ž mod pob¨iously these two assertions are equi¨alent. . Proof. Suppose that G is a counterexample of minimal order. Then < Ž.ii if K be a subgroup of G of exponent p, then CAKŽ.FA. Note that a Sylow p-subgroup of AutŽ.ŽA ( GL Ž 3, p ..is nonabelian of order p3 and exponent p. Therefore, the following two assertions are true: Ž.iii If K is an elementary abelian subgroup of G of order p4, then 2 < Ž.iv G has no subgroup of order p7 and exponent p.If K is a subgroup of G of order p6 and exponent p then A - K. Let E - M - G, where M is a maximal subgroup of G. By the induc- tion hypothesis, <ᑟ 4Ž.M <' 1Ž. mod p , so that M contains a G-invariant elementary abelian subgroup of order p4. Without loss of generality, we may assume that E 1 G.If Eis the only normal elementary abelian 4 subgroup of G of order p , then <ᑟ 4Ž.G Ž.vIfB,B1 are two different normal elementary abelian subgroups 4 5 3 of G of order p , then < Ž.vi A g K. Suppose that K contains all elements of the set ᑟ 4Ž.G . By Lemma 3 5 Žsince < ᑟ31Ž.GsÄ4A,..., Ar, and ᑟ 41Ž.G s Ä4B ,...,Bs. ON ABELIAN SUBGROUPS OF p-GROUPS 269 We have to prove that s ' 1Ž. mod p . Suppose that Aiiis contained in ␣ elements of the set ᑟ 4Ž.G , and Bjjcontains  elements of the set ᑟ 3Ž.G . Then the number of pairs Ä4Aij, BAŽŽ.Ž. jgᑟ3G,Bjgᑟ4G,is1,...,r, js1,...,s.is ␣1q иии q␣rs 1q иии qs.3Ž. 4 We have j ' Žp y 1.Žr p y 1 .' 1 Ž mod p .for j s 1,...,s. By Theo- rem 1, ␣i ' 1Ž. mod p for i s 1,...,r. Therefore, readingŽ. 3 by modulo p, we obtain r ' s Ž.mod p . Since r ' 1 Ž. mod p by Theorem 6, it follows that s ' 1Ž. mod 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 inwx 5 , Theorem 11. For another proof, see wx31 , Theorem 2.2. In the same paper Konvisser and Jonah proved that if p)2, then <ᑟ 55Ž.G <' 1Ž. mod p unless ᑟ Ž.ŽG s л but there exists G such that ᑟ 6Ž.G s 229.wx. COROLLARY 8. Let N be a normal subgroup of a p-group G, p ) 2. 3 Suppose that N has no G-in¨ariant elementary abelian subgroups of order p . Then N has no elementary abelian subgroups of order p3 Ž and hence, the structure of N is completely determined in wx17. . This follows from Theorem 6. If N s G in Corollary 8, we obtain Hobby’s Theoremwx 27 . See also wx 31 and wx 24 , Proposition 10.17. COROLLARY 9. Let N be a normal subgroup of a p-group G, p ) 2. If N 4 contains an elementary subgroup of order p , then N contains a G-in¨ariant elementary abelian subgroup of order p4. This follows from Theorem 7. If N s G in Corollary 9, we obtain Hobby’s result. See alsowx 31 and wx 24 , Proposition 10.17. LEMMA 10.Ž O. J. Schmidtwx 34 , J. A. Gol’fand wx 23 ; see also wx 28 , Satz 3.5.2. . Let S be a minimal nonnilpotent group. Then S is aÄ4 p, q -group, p, q are different primes. Let P g Syl pqŽ.S , Q g Syl Ž.S . One of Sylow subgroups X of S, say Q, is normal in S. The subgroup Q s S is elementary abelian or special, P is cyclic and< P:P l ZSŽ. Proof. By Corollary 8, Q0 has no elementary abelian subgroups of 3 order q . 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:␣ и G be a natural semidirect product. By assumption, 2 2 p)qq1 so that p ¦ q y 1 and < THEOREM 12Ž compare withwx 4 , Theorem 5.4. . LetGbeap-group of m order p , p ) 2, n g ގ, n G 4. Then the number of metacyclic subgroups of n order p inGisdi¨isible by p, unless G is metacyclic or a 3-group of maximal class. Proof. We use induction on < ᑟ. Bywx 6 , the number of elements of the set ᑟ in any element of M1 is '1Ž. mod p . Therefore, arguing, as in the end of the proof of Theorem 7, we obtain < Note, that Theorem 12 is true for n s 3Ž seewx 6, 9, 15 . It is interesting to study <ᑧŽ.G <Ž.mod 2 for 2-groups G. LEMMA 13. Let G s G01) G ) иии ) Gns Ä41 be a chain of normal subgroups of a group G. Set ␣ ␣ A s Ä4g AutŽ.ŽG ¬ xGii .s xG for all x g G iy1y Gi, i s 1,...,n , Ž.i.e., A stabilizes that chain . Let p be a prime number such that p ¦ < Ž.<<< ZGiy1rGi for i s 2,...,n. Then p ¦ A . Proof. We use induction on n. We may assume that n ) 1Ž otherwise the lemma is true.Ž. . Obviously, A is a subgroup of Aut G . Suppose that A contains an element ␣ of order p. Let W s ²:␣ и G be the natural direct ²:␣ и ␣ product. By the induction hypothesis, Gny1 1 W. By assumption, 272 YAKOV BERKOVICH ²:␣ и ²:␣ ²:␣ centralizes Gny1 so that Gny1 s = Gny1. Assume that is ²:␣ Ž. not characteristic in H s = Gny1. Then there exists g Aut H such that ␣Ž² :.s ² :/ ²␣ :. Obviously, ² :1 H and ²␣,  : is a Ž. 2 ²:␣  subgroup of ZH of order p . Then , l Gny1 is a subgroup of Ž. ²:␣ ZGny1 of order p, contradicting the assumption. Thus, is character- ²:␣ ²:␣ ␣ istic in = Gny1. Then 1 W, so that centralizes G. In that case, ␣ s id, which is a contradiction. LEMMA 14Žwx 21 , Lemma 8.1. . If G is a -group, then the stability group A of a chain G s G01) G ) иии ) Gns Ä41 is a -group. Proof. We use induction on n. Suppose that A contains an element a X of order p for some p g . By the induction hypothesis, a centralizes G1. a a p p Let x g G; then x s xy with y g G1. Then x s x s xy . It follows that y s 1 so that a s 1ᎏcontradiction. LEMMA 15. Let G be a -group and suppose that D is a subgroup generated by all elements of G of orders 4, p for all p g . Let ␣ be a X -automorphism of G. If ␣ centralizes D, then ␣ s id. Proof. Suppose that the lemma is false. Without loss of generality, we X may assume that oŽ.␣ s q g . Let W s ²:␣ и G. For every p g ␣ there exists P g Syl pŽ.G such that P s P. Since ␣ / id, we may assume that ␣ induces a nonidentity automorphism of P. Let H be a minimal nonnilpotent subgroup in G1 s ²:␣ и P ŽF W .. Without loss of generality, we may assume that ␣ g H. Obviously, H l P g Syl pŽ.H . By Lemma 10, H l P F D. This means that ␣ centralizes H l P so H is nilpotentᎏ contradiction. LEMMA 16. Let U be an abelian subgroup of a p-group G, let ␣ be a X p-automorphism of G. If ␣ centralizes CGŽ. U , then ␣ s id. This is exactly what was proved inwx 21 , Lemma 8.12. COROLLARY 17Ž compare withwx 21 , Lemma 8.12. . Let U be an abelian subgroup of a p-group G. Let ⑀ s 1 if p ) 2 and ⑀ s 2 if p s 2. Let ␣ be a X p-automorphism of G that centralizes ⍀⑀ŽŽ..CUG .Then ␣ s id. Proof. Obviously CUGGŽ.is ␣-invariant. ␣ centralizes CUŽ.by Lemma 15. Now the result follows from Lemma 16. COROLLARY 18Žwx 19 ; see also wx 22 , Lemma 2.3. . Let p, G, and ⑀ be such as in Corollary 17. Let U be a maximal abelian subgroup of G of n X exponent p , where n G ⑀. If a p -automorphism ␣ of G centralizes U, then ␣ s id. ON ABELIAN SUBGROUPS OF p-GROUPS 273 n Proof. Let x be an element of CUGŽ.of order at most p . Since ²:U, x n is abelian of exponent at most p , it follows that x g U by the maximal choice of U. Thus, ⍀⑀ŽŽ..CUG sU, and the result follows from Corollary 17. If in Corollary 18, U is ␣-invariant and ␣ centralizes ⍀⑀Ž.U , then ␣sidŽ. by Lemma 15 and Corollary 18 . PROPOSITION 19. Let B be a subgroup of a p-group G such that CGŽ. B F B. Ž.a If B is nonabelian of order p3, then G is of maximal class. 2 Ž.b ŽM.Suzuki . If<< B s p , then G is of maximal class. 4 Proof. We may assume that < Remark 3. We claim that a p-group G is of maximal class if it contains a subgroup H such that N s NHGŽ.is of maximal class. We may assume that N - G and < QUESTION 2. Study the p-groups G containing a subgroup M of maximal 3 class such that CGŽ. M - M. ŽIf<< M ) p , then G is not necessarily of maximal class.. The case p s 2 is of special interest. PROPOSITION 20Ž compare withwx 24 , Lemma 10.27. . Let A be a sub- 2 group of a p-group G such that CGŽ. A is metacyclic. If<< A F p , then G has no normal subgroups of order p pq1 and exponent p. Proof. We may assume that A g ZGŽ.. Suppose that D is a normal subgroup of G of exponent p. We may assume that < 2 metacyclic, and so it is of order p ŽŽ..since exp H s p . We have CHAD Ž. sH. Therefore, AD is of maximal class by Proposition 19Ž. b . Bywx 20 , AD has no subgroups of order p pq1 and exponent p, so that < QUESTION 3. Study the groups G of Proposition 20 in the case when < Ž.a If p s 2, then G is of maximal class. 3 Ž.b If p ) 2, then<< G s p . Proof. Let the proposition has proved for all groups of order - < Y PROPOSITION 20 . Let B be a nonabelian subgroup of order p3 in a p-group G, p ) 2. Suppose that G has no element x such that²: B, x s B = ²:x.Then: Ž.a CG Ž B . is cyclic. p 1 Ž.b G has no B-in¨ariant subgroups of order pq and exponent p. Proof. Ž.a Obvious. Ž.b Assume that H is a B-invariant subgroup of order p pq1 and exponent p in G. Without loss of generality, we may assume that G s BH. Bywx 20 , G is not of maximal class. Therefore, CBGŽ.gBby Proposition 19Ž. a . In particular, It is essential, in the proof ofŽ. b , that ⍀1 ŽBC .s B. Remark 4. Let H be a normal subgroup of G and H F ⌽Ž.G . Suppose X that, whenever N 1 H and H 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 X induction hypothesis, S s Ä41 . In particular 1 ␣ 1 1 1 so that Ž gy . s wgwy y sygy . Hence ␣ leaves every element of X GrGfixed. Therefore, by Proposition 21, ␣ g InnŽ.G . This means that 1 1 for each w g W, there exists u g G such that wy gw s uy gu for all y1 ggG. Thus, wu g CGWWŽ., and so w g CGGŽ. for each w g W. This means that W s CGGWŽ.. p p Ap-group G is said to be generalized regular if, whenever x s y 1p Ž.Žx,ygG, then xyy.sÄ41.IfGis a generalized regular p-group, then expŽ⍀1 ŽG .. s p. Regular p-groups are generalized regularŽ seewx 28 , Kapi- tel 3. . X PROPOSITION 23Ž compare withwx 36 , p. 73, Lemma. . Let a p -group Q act on a generalized regular p-group G. If Q acts tri¨ially on ⌽Ž.G , then wxG,QF⍀1Ž.G. p Proof. If x g G, then x g ⌽Ž.G . Hence, for any y g Q, we have p p y y p p 1yp x sŽ x. sŽ x. . Therefore, wxx, y s Žxxy .sÄ41Ž since G is gen- eralized regular. , so that wxG, Q F ⍀1Ž.G . LEMMA 24. Let G be a sol¨able group of odd order, p g Ž.G , OGpX Ž. 3 sÄ41. If OpŽ. G has no elementary abelian subgroups of order p , then a Sylow p-subgroup is normal in G. Proof. Let B be a critical Thompson’s subgroup of OGpŽ.Žfor its construction and properties, seewx 3 , Corollary 3.Ž. and set T s ⍀1 B . Since Bis of class at most two and p ) 2, expŽ.T p. Obviously, B G. Since s y1 COGGpŽŽ..FOG p Ž., it follows fromwx 3 , Corollary 3Ž. e that CBG Ž .FB. Therefore, by Lemma 15, CTGŽ.is a normal p-subgroup of G. Since T has 3 3 no elementary abelian subgroups of order p , it follows that < Proof. Without loss of generality we may assume that OGpXŽ.sÄ41. Then P G, where p Syl Ž.ŽG by Lemma 24 . . If A B is a chief factor y1 g p r Ž. of G of order p, then GrCAGrB rBis cyclic of order dividing p y 1, and ON ABELIAN SUBGROUPS OF p-GROUPS 277 X Ž. 2 so GBrBFCAG r B rB. The chief p-factors of G are of orders p or p 2 X Žseewx 17. . Let KrL be a chief factor of G of order p .If KgG,it X follows that G centralizes KrL. Hence we have to consider the case when X KFG. Without loss of generality we may assume that L s Ä41 . Then X KFPFFGŽ., and so P centralizes K. Next, GrCKG Ž.as a p -subgroup of odd order in GLŽ. 2, p , is abelian. Therefore, GЈ centralizes K. PROPOSITION 26. Suppose that Ä41 F A - B F G, where G is a p-group, k B elementary abelian of order p ,2FkF4. Then the number of elementary abelian subgroups of G of order pk that contain A is congruent to 1Ž. mod p , unless p s 2 s k, G is maximal class. Proof. Suppose that the proposition has proved for all p-groups of order - < CAGŽ.is a 2-group of maximal class then G is Ž since, in that case GsCAGŽ... Therefore, without loss of generality, we may assume that AFZGŽ.. Let MA be the set of all maximal subgroups of G containing A. By Sylow’s Theorem, < sGkkŽ.'ÝsH Ž . Žmod p ..4Ž. HgMA If sHkAŽ.Ž.'1 mod p for all H g M , then sGkŽ.Ž.Ž.'1 mod p by 4 . 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 AgT. ThenŽŽ.. the elementary abelian subgroup since A F ZG AT k 1 contains an elementary abelian subgroup F such that A - F, < 2.If.Ž.His dihedral or semidihedral, then sH2 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 fromŽ. 4 by induction Žsince, in that case, the number of elements of maximal class in MA is divisible by 4; seewx 4 Section 5. . Bywx 31 , Proposition 26 is true for k s 5 as well. 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