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p-Groups of Small Breadth
Journal of Algebra 213, 52᎐68 Ž1999. Article ID jabr.1998.7660, available online at http:rrwww.idealibrary.com on
p-Groups of Small Breadth Gemma Parmeggiani* Uni¨ ersita ` degli Studi di Pado¨ a, Italy
and Bernd Stellmacher Christian-Albrechts-Uni¨ ersitat ¨ zu Kiel, 24118 Kiel, Germany Communicated by Gernot Stroth Received February 23, 1998
Let G be a finite p-group. For x g G the breadth bŽ x . of x is defined as p bŽ x . [ < G : CG Ž x . < , and the breadth bŽ G . of G is the maximum of the breadths of the elements of G. More generally, we define for x g G and an abelian normal subgroup A of G p bA Ž x . [ < A : C A Ž x . < , bA Ž G . [ max bA Ž x . N x g G 4 , BA [ x g G N bA Ž x . s bA Ž G . 4 . Clearly, bŽ G . s 0 is equivalent to G⬘ s 1; and H. G. Knoche wKn1x has shown that bŽ G . s 1 is equivalent to < G⬘ < s p Žsee the Proposition in Section 3.. Knoche also obtained some results for bŽ G . s 2 wKn2x; in particular, he has shown that < G⬘ < F p 3. *This author is grateful to the University of Kiel for the kind hospitality during the preparation of this paper, and to C.N.R. ŽItaly. for the financial support. 52 0021-8693r99 $30.00 Copyright 䊚 1999 by Academic Press All rights of reproduction in any form reserved.
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In this paper we prove the following two theorems. THEOREM A. Let G be a finite p-group. Then bŽ G . s 2, if and only if one of the following holds: Ža. Žb.
< G⬘ < s p 2 , or < G : ZŽ G .< s p 3 and < G⬘ < s p 3.
Moreo¨ er, if bŽ G . s 2 and < G⬘ < ) p 2 , then < ArZŽ G .< s p for e¨ ery maximal abelian normal subgroup A of G. THEOREM B. Let G be a finite p-group with bŽ G . s 3, and let A be a maximal abelian normal subgroup of G. Then one of the following holds: Ža. < G⬘ < s p 3. Žb. < G : ZŽ G .< s p 4 and bAŽ G . g 1, 24 . Žc. < G⬘ < s p 4 , D [ Fx g B w A, x x has order p, and A < GrD : Z Ž GrD . < s p 3
and
b Ž GrD . s 2;
moreo¨ er bAŽ G . g 1, 24 and
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1. ELEMENTARY PROPERTIES Let G be a finite p-group and A a normal abelian subgroup of G. We set b [ bŽ G ., bA [ bAŽ G ., and TA [ x g G N bA Ž x . s 1 4 . The proof of the first lemma is elementary and well known: LEMMA 1.
For e¨ ery x g G
w A, x x s w A, ² x : x s w a, x x N a g A4 ; in particular,
w A, u x = w A, ¨ x s w A, u¨ xw A, ¨ x and thus p bAŽ u.qbAŽ ¨ . s w A, u¨ xw A, ¨ x - p bAŽ u ¨ .qbAŽ ¨ . . LEMMA 3. Let x, y g TA , T s ² x, y :, and T s TCG Ž A.rCG Ž A.. Then one of the following holds: Ža. T _ CG Ž A. : TA and w A, x x s w A, T x; moreo¨ er CAŽ x . / CAŽ y . if T is not cyclic. Žb. T _ CG Ž A. : TA and CAŽ x . s CAŽT .; moreo¨ er w A, x x / w A, y x if T is not cyclic. Žc. CAŽT . s CAŽ x . l CAŽ y . and w A, T x s w A, xy x s w A, x x = w A, y x. Proof. If T is cyclic, then CAŽ x . s CAŽ y ., and Žb. holds. Thus, we may assume that T is non-cyclic. Assume first that xy g TA . Then by Lemma 2, w A, x x s w A, y x or CAŽ x . s CAŽ y .. This gives Ža. resp. Žb.. Assume now that xy f TA , i.e., bAŽ xy . ) bAŽ x . s bAŽ y .. Then clearly CAŽ xy . s CAŽ x . l CAŽ y . and w A, xy x s w A, x x = w A, y x. This is Žc.. LEMMA 4. Let x, u, ¨ g G such that u¨ y1 f CG Ž A.. Suppose that bAŽ x . ) 1. Then at least one of the elements u, ¨ , u¨ , ux, ¨ x, u¨ x is not in TA .
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Proof. Assume that ⍀ [ u, ¨ , u¨ , ux, ¨ x, u¨ x 4 : TA . Let T1 s ² u¨ y1 , u:, T2 s ² u¨ y1 , ¨ x :, and T3 s ² u, ¨ x :. Then case Ža. or Žb. of Lemma 3 applies to each of the subgroups T1 , T2 , and T3 . Fix i, j g 1, 2, 34 such that i / j and
Let x g G and a g A. Then the following hold:
Ža. w a, Ma x l M x x F w A, x x. Žb. w A, Dx x F w A, x x s w Dx , x x; in particular, if b s bA then
w a, G x F w xy1 , U xw A, x xw A, U x . Proof. Let a g A and y g Ma x . Then y s cb for some c g CG Ž ax . and b g A. It follows y
a y x y s Ž ax . s Ž ax .
cb
b
s Ž ax . s ax b
and w a, y x s w b, xy1 xw xy1 , y x. In particular w xy1 , y x g w A, G x, and Žc. follows. Assume that y g M x . Then w xy1 , y x g w A, xy1 x s w A, x x and thus w a, y x g w A, x x, and Ža. follows. Now Ža. and w M x , x x s w A, x x yield
w a, Dx x F w a, M x l Ma x x F w A, x x F w Dx , x x F w M x , x x s w A, x x and thus w A, Dx x F w A, x x s w Dx , x x.
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Assume that b s bA and let z g BA . Then G s Ma z for every a g A, i.e., G s Dz , and the previous inequality gives w A, G x s w A, Dz x F w A, z x. Thus w A, G x s w A, z x has order p bAŽ z . s p bA , and the proof of Žb. is complete. For the proof of Žd. note that for all x, y, z g G
w z, xy x s w z, y xw z, x x s w z, y xw z, x xw z, x, y x . y
It follows that
w a, G x s w a, Ma x U x F w A, U x a, CG Ž ax . w A, U x s a, CG Ž ax . w A, U x . Ž ). For c g CG Ž ax . we get w a, c x s w xy1 , c x and thus a, CG Ž ax . F w xy1 , G x s xy1 , M x U F w xy1 , U xw xy1 , A xw A, U x . Together with the above relation Ž). this gives Žd.. LEMMA 6. x g BA .
Suppose that b F bA q 1. Then w A, x x is normal in G for e¨ ery
Proof. Let x g BA . Note that Ma x normalizes w A, ax x s w A, x x for every a g A. Hence we can assume Gx / G. This gives b s bA q 1 and M x s Dx . Now Lemma 5Žb. shows that w A, x x s w A, M x x. From < G : M x < s p it follows that M x is normal in G, so also w A, M x x is normal in G. LEMMA 7. satisfying
Let bA ) 1. Suppose that w G, z x F w A, G x for all z g G bA Ž z . ) 1
and
2 bA Ž z . G bA .
Ž ).
Then G⬘ s w CG Ž A., G x. In particular, if in addition A s CG Ž A. and b s bA then < G⬘ < s p bA . Proof. Let x g BA and y g G. Then
w A, x x F w A, y xw A, yx x , and we are in one of the following two cases: ŽI. ŽII.
y or yx satisfy Ž). for all y g G, or y, yx g TA for some y g G, and bA s 2.
In the first case w G, y x F w A, G x or w G, yx x F w A, G x. Since also w G, x x F w A, G x we conclude that w G, y x F w A, G x and thus G⬘ s w A, G x s w CG Ž A., G x. Assume now case ŽII.. Then w G, z x F w A, G x for every z g G _ ŽTA j CG Ž A.. and thus w G, z x F w CG Ž A., G x for every z g G _ TA . Suppose
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that there exist u, ¨ g G such that w u, ¨ x f w CG Ž A., G x; i.e., w G, u x g w CG Ž A., G x and w G, ¨ x g w CG Ž A., G x. Then also
w u¨ , ¨ x , w ux, ¨ x , w ¨ x, u x , w u¨ x, ¨ x f CG Ž A . , G ; y1
moreover, w u¨ y1 , ¨ x s w u, ¨ x¨ f w CG Ž A., G x implies u¨ y1 f CG Ž A.. As we have seen above, this implies u, ¨ , u¨ , ux, ¨ x, u¨ x g TA , which contradicts Lemma 4. Lemma 5Žb. gives the additional statement for b s bA and A s CG Ž A.. 2. THE CASE bA F 2 In this section we use the same notation as in Section 1. In particular, A is an abelian normal subgroup of G. LEMMA 8.
Suppose that bA s 1. Then either
Ži. < A : A l ZŽ G .< s p and CAŽ x . s CAŽ y . for e¨ ery x, y g G _ Ž CG A., or Žii.
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Assume Žii. holds. Then D s w A, G x is a normal subgroup of order p such that bŽ GrD . - bŽ G .. LEMMA 9. ² BA : s G.
Suppose that bA s 2. Then either
Proof. Note that G s BA j TA j CG Ž A . . Assume that G / ² BA :. Let x g BA . Then t, tx, xy1 t g TA for every t g G _ ² BA :. If p / 2, then t 2 g TA and
w A, t 2 x s w A, txxy1 t x s p. Hence Lemma 3Ža. or Žb. applies to ² tx, xy1 t : s ² t, x : and x g TA , a contradiction. Thus p s 2. Moreover again by Lemma 3
w A, x x s w A, t x = w A, tx x . Hence w A, ² G _ ² BA ::x F w A, x x. On the other hand ² G _ ² BA :: s G and thus
Suppose that bA s 2. Then one of the following holds:
Ža. < A : A l ZŽ G .< s p 2 , and there exist x, y g BA such that w A, x x l w A, y x s 1. Žb.
w A, x x l w A, y x s 1. Then Lemma 2 gives C [ CA Ž x . s CA Ž y . . We want to show that C s A l ZŽ G .. Thus, we may assume that there exists w g G such that C g CAŽ w .. If a g C _ CAŽ w . then
w a, xw x s w a, w x s w a, yw x / 1. Hence w A, xw x l w A, yw x / 1 and H [ w A, xw xw A, yw x has order at most p 3. Thus
w A, y x F H w A, x x F H w A, y x
Ž ).
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gives that w A, x x g H and w A, y x g H. Since C g CAŽ xw . and C g CAŽ yw ., Lemma 2 yields that
w A, c x l w A, d x / 1
for every c g x, y 4 , d g xw, yw 4 .
In particular from Ž). it follows that w A, xw x / w A, yw x. Then
w A, x x l H s w A, x x l w A, xw x s w A, x x l w A, yw x s w A, xw x l w A, yw x and similarly w A, y x l H s w A, xw x l w A, yw x. This contradicts w A, y x l w A, x x s 1. We assume now
w A, u x l w A, ¨ x / 1
for all u, ¨ g BA .
Ž )).
We choose u, ¨ g BA such that W [ w A, u xw A, ¨ x is maximal. Suppose first that < W < s p 2 . Then the maximality of W gives
w A, u x s w A, ¨ x
for all u, ¨ g BA .
Let t g TA . If tu g TA , then w A, u x s w A, t x = w A, tu x, and if tu g BA , then w A, u x s w A, tu x. Thus in both cases w A, t x F w A, u x. We conclude that W s w A, G x, and Žc. holds. Suppose now that < W < s p 3. Then R [ w A, u x l w A, ¨ x has order p. Assume first that R s Fx g B A w A, x x. Then R is a normal subgroup of order p in G, and Lemma 8 applies to ArR and GrR. This gives Žc. since
Suppose that b s 1. Then < G⬘ < s p.
Proof. For every x g G _ A G s CG Ž x . A s Dx s Gx .
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Hence Lemma 5Žb. gives
w G, A x s ² w G, x x N x g G _ A: s G⬘. Proof of Theorem A. Suppose that b s 2. If bA s 2 for some maximal abelian normal subgroup A of G, then Lemma 7 and Lemmas 5Žb. yield Ža.. Thus, we may assume that bA s 1 for every maximal abelian normal subgroup A of G. We discuss the two cases of Lemma 8. Assume case Ži.. For z g A _ ZŽ G . we get CG Ž a. s A and thus < G : ZŽ G .< F p 3. Now b s 2 gives < G : ZŽ G .< s p 3 and
Suppose that bA s b s 3. Then < G⬘ < s p 3.
Proof. Let x g BA . We have G s ACG Ž x . s M x s Dx . By Lemma 5Žb., w A, G x s w A, x x. Let y g CG Ž x . such that bAŽ y . s 2. Then either G s Gy or Dy s M y . In the first case w G, y x F w A, G x by Lemma 5Žc.. In the second case Lemma 5Žb. gives
w A, x x F A, M y s w A, y x , which contradicts
Let bA s b y 1 s 2. Suppose that there exist x, y g BA such
w A, x x l w A, y x s 1.
Then either < G : ZŽ G .< s p , or p s 2 and < G⬘ < s 2 4 . 4
Proof. Let G be a minimal counterexample. From Lemma 9 and Lemma 10 we get: Ž1. G s ² BA : and < A : ZŽ G .< s p 2 ; in particular, CAŽ z . s ZŽ G . for every z g BA . Since < G : ZŽ G .< ) p 4 Ž1. gives < G : A < G p 3 and thus Ž2. Mu l Mw / A for u, w g BA . By Lemma 6, w A, z x is normal in G for every z g BA . Thus
w A, x, y x s 1 s w A, y, x x . It follows that w A, x xw A, y x F ZŽ G ..
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Assume that w A, z x g ZŽ G . for some z g BA . Then w A, z, x x / 1 / w A, z, y x and thus w A, z x s w A, z, x x = w A, z, y x F ZŽ G ., a contradiction. Hence w A, z x F ZŽ G . for all z g BA , and Ž1. gives: Ž3. w A, G x F ZŽ G .. Next we show: Ž4.
G⬘ s w A, G x.
Assume that Ž4. is false. Then by Ž1. there exist u, ¨ g BA such that w u, ¨ x f w A, G x. Thus, Lemma 5Žc. gives ¨ f Gu and u f G¨ . Hence Mu s Du and M¨ s D¨ . Now Lemma 5Žb. yields
w A, Mu x F w A, u x
and
w A, M¨ x F w A, ¨ x .
This implies
w A, x xw A, y x F w A, G x F w A, u xw A, ¨ x and thus w A, u x l w A, ¨ x s 1. But now Mu l M¨ s A, which contradicts Ž2.. Ž5.
p / 2.
Assume that p s 2. Since G is a counterexample we get from Ž4. that w A, x xw A, y x - w A, G x. Set D s M x l M y . We first show: For every d g D there exists a g A _ Z Ž G . such that d g Ma x . Ž ) . If d is a counterexample, then Ma x² d : s Mab x² d : s Mb x² d : s G, where A s ZŽ G .² a, b : and a, b, ab f ZŽ G .. Now d g M x w xy1 , ² d :x F w A, x x. Hence
Ž ax .
y1
, ² d : F w xy1 , ² d : x
ay1
yields
w A, d x F w A, x xw A, d x
and similarly wŽ bx .y1 , ² d :x F w A, x xw A, d x. Thus from Lemma 5Žd. we get
w A, G x s w b, G xw a, G x s
¦ Ž ax .
y1
, ² d : w A, ax xw A, d x , Ž bx .
s w A, x xw A, d x which contradicts
y1
, ² d : w A, bx xw A, d x
;
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Clearly, Ž). also holds with y in place of x. Moreover, from Ž). and Lemma 5Ža. follows w a, d x g w A, x x. Thus Ž1. and w A, x x l w A, y x s 1 imply y f D and similarly x f D; in particular G s D ² x, y :. Since w A, G x g w A, x xw A, y x there exists d g D such that w A, d x g w A, x xw A, y x. If d g BA , then CAŽ d . s ZŽ G . by Ž1.. From Ž). and Lemma 5Ža. we get 1 / w a, d x g w a, M x l Ma x x F w A, x x and thus w A, d x l w A, x x / 1. Similarly, considering y in place of x, w A, d x l w A, y x / 1. Hence w A, x x l w A, y x s 1 yields w A, d x F w A, x xw A, y x, a contradiction. Assume now that bAŽ d . s 1. Then there exist a, b g A such that CAŽ d . s ZŽ G .² a: and A s CAŽ d .² b :. If d g Mb x or d g Mab x , then by Lemma 5Ža., w b, d x s w ab, d x g w A, x x, and
w A, d x s ² w b, d x : F w A, x x , a contradiction. Hence, we have d f Mb x and d f Mab x , and Lemma 5Žd. yields w a, G x F w A, x xw A, d x. Since 1 / w a, y x g w a, G x and
GrA is elementary abelian of order p 3.
By Ž4., GrA is abelian, and by Ž3., w A, G x F ZŽ G .. Thus w a, g p x s w a p , g x for every a g A and g g G. Assume that G p g A. Then also A p g ZŽ G ., and Ž1. shows that ArZŽ G . is cyclic. Let A s ZŽ G .² a:. Then w A, y x s ²w a, y x: is a cyclic group of order p 2 since w a, y x p s w a, y p x, w A, y p x s ²w a, y p x: is a group of order p. Hence y p g TA l D, D as in Ž5.. But now w A, y p x l w A, x x F w A, y x l w A, x x s 1, and Lemma 5Ža. implies y p f Ma x
for all a g A _ CA Ž y p . .
This contradicts < G : Ma x < F p. We have shown that GrA is elementary abelian. Then GrA has order p 3 since p is odd and G is a minimal counterexample. Ž7. There exist a, b g A such that Ma x / G / Mb y . Assume that G s Ma x for all a g A. Then Lemma 5Ža. gives w A, G x s w A, x x which contradicts w A, y x g w A, x x. Hence G / Ma x for some a g A, and the same argument shows the assertion for y. According to Ž7. we may choose our notation such that M x / G / M y . We now derive a final contradiction using Ž5. and Ž6.. In GrA there exist exactly p q 1 subgroups of order p 2 containing ² xA:. On the other hand,
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xA g Ma xrA for every a g A _ ZŽ G . and < Ma xrA < G p 2 . Hence, since < A : ZŽ G .< s p 2 , we may choose our notation additionally such that M x F Ma x
and
M y F Mb y for some a, b g A _ Z Ž G . .
In particular, by Lemma 5Ža.
w a, M x x F w A, x x
and
b, M y F w A, y x .
Assume that w A, M x x g w A, x x. Then w c, M x x g w A, x x for every c g A _ ² a: ZŽ G . and thus M x g Mc x
for every c g A _ ² a: Z Ž G . .
Fix c g A _ ² a: ZŽ G . and set d s ac. Since p G 3 then dc f ² a: ZŽ G .. Assume M x s Ž M x l Mc x . j Ž M x l Md x . j Ž M x l Mdc x . . Then M xrA is a group of order p 2 which is the union of three subgroups of order p. Hence p 2 s < M xrA < s 3 Ž p y 1 . q 1, a contradiction to Ž5.. Thus there exists w g M x _ Ž Mc x j Md x j Mdc x . . It follows that G s Mc x² w : s Mdcy1 c x² w : s Mdc x² w : , and Lemma 5Žd. implies
w A, G x s w ² dcy1 , d : , G x s w A, x xw A, w x since w x, w x g w A, x x. Moreover, w a, w x g w a, M x x F w A, x x yields
Suppose that bA s b y 1 s 2. Then one of the following
Ža. < G : ZŽ G .< s p 4 , and there exist x, y g BA such that w A, x x l w A, y x s 1. Žb.
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Proof. Let G be a counterexample. We apply Lemma 10. Then case Žb. and by Lemma 12 also case Ža. of Lemma 10 cannot hold. Thus, we are in case Žc. of Lemma 10. As there, let D [ Fx g B A w A, x x . If < D < s p 2 , then D s w A, G x and we get Žb., a contradiction. Thus, we have < D < s p. Set G [ GrD, and note that bAŽ G . s 1. If < A : A l ZŽ G .< / p, then Lemma 8 yields Žb., again a contradiction. Hence, we have < A : A l Z Ž G . < s p. Let a g A such that a f ZŽ G .. Then CG Ž A. s CG Ž a. and < G : CG Ž a.< F p . In particular
< w A, G x < F p 4 . Assume that M x s Dx for some x g BA . Then by Lemma 5Žb., w A, M x x s w A, x x. Since < G : M x < F p we conclude that
s A, M x
s p.
Since < G : M x < F p, we conclude that
p s 2 and < G⬘ < s 2 4 . < G⬘ < s p 3.
Proof. Let G s Grw A, G x, and let U be a maximal abelian normal subgroup of G. Note that A F ZŽ G . F U. Then < G : CG Ž x .< F p for x g BA , in particular: Ž1.
G s UCG Ž x . or x g U; and w x, G x F wU, G x for every x g BA .
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We first show: Ž2.
G⬘ s wU, G x.
Assume that there are ¨ , w g G such that w ¨ , w x f wU, G x. From Ž1. we get ¨ , w, ¨ w g TA . Let x g BA . Then Ž1. gives w x, G x F wU, G x and w z, G x g wU, G x for every z g ¨ x, wx, ¨ wx 4 . Hence ¨ x, wx, ¨ wx f A and from Ž1. we also have ¨ x, wx, ¨ wx f BA . Thus ¨ x, wx, ¨ wx g TA and Lemma 4 yields a contradiction. If
Hence by Ž2. and the Proposition, bŽ G . s 2. Assume first that bU Ž G . s 1 for every such that U. From Ž3. and Lemma 8 we get
and
U s CG Ž u . for every u g U _ Z Ž G . .
Let u g G _ ZŽ G .. Since < G : ZŽ G .< ) p 2 we get from Ž). p 2 F py1 < G : Z Ž G . < s < G : U < s < G : CG Ž u . < F p 2 , i.e., < G : U < s p 2 . Moreover
w u, G x F U, x . Hence wU, G x s wU, x x and, as x g TU , we get bU Ž G . s 1, a contradiction.
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LEMMA 15. Suppose that bA s b y 1 s 2 and
p s 2 and < G⬘ < s 2 4 .
Proof. We may assume that G⬘ ) w A, G x. Lemma 9 implies G s ² BA :. Hence, there exist x, y g BA such that w x, y x f w A, G x. Thus x f M y and y f M x , and we have G s M x² y :
M x s H² x :
and
for H [ M x l M y . We also have for every ¨ g H
w ¨ x, y x f w A, G x
and
w ¨ y, x x f w A, G x .
Lemma 5Žb. and Žc. give Ž1. M z s Dz and w A, M z x s w A, z x for every z g ¨ x, ¨ y N ¨ g H 4 such that bAŽ z . s 2. As w A, M x x s w A, x x by Ž1. we get w A, H x F w A, x x l w A, y x \ R. Moreover < R < s p, since w A, G x s w A, x xw A, y x has order p 3. Next we show: Ž2.
M zX F w A, z x for z g x, y4 .
We have M zX s H⬘w A, z x. Suppose that there exist u, ¨ g H such that w u,¨ x f w A, z x. Assume first that ¨ z g BA . Then Ž1. gives w A, M¨ z x s w A, ¨ z x. But from w A, ¨ x F w A, H x F R F w A, z x we get w A, ¨ z x s w A, z x. Hence M¨ z s M z since
w ¨ , bt x s w ¨ zzy1 , bt x s w ¨ z, bt x z w zy1 , bt x s w zy1 , bt x s w zy1 , t xw zy1 , b x s w zy1 , b x g w A, z x t
t
and so w ¨ , w x s w ¨ , abt x s w ¨ , bt xw ¨ , ax b t g w A, z x. Thus w ¨ , M z x F w A, z x, which contradicts w u, ¨ x f w A, z x.
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We may assume now that ¨ z, uz, u¨ z g TA . But also u, ¨ , u¨ g TA since w A, H x F R. Hence, Lemma 4 gives z g TA , a contradiction. This gives Ž2.. From Ž2. we get that H⬘ F R. Set G [ GrR. Then H is an abelian normal subgroup of G such that G s H² x, y : and < G : H < s p 2 . Note that bH Ž x . s bH Ž y . s 1. Thus G⬘ s w H, ² x :xw H, ² y :xw²w x, y x: has order p 3 , and < G⬘ < s p 4 . Hence, we may assume that p / 2. Note w A, G x s w A, ² x, y :x s w A, ² xy, yy1 x :x. Since
˜ F w A, xy x l w A, x x . H⬘ ˜ G x s w A, G x and either HH˜ s M x or H s H. ˜ In On the other hand w H, ˜ w x w x w x w the first case M x , G s HH, G s A, G , which contradicts x, y x f ˜ w A, G x. Thus, we have H s H. w x Assume that R F A, xy . Then R s w A, xy x l w A, x x, and bH Ž G . s 1 by Lemma 3. Hence Lemma 8 gives < H : ZŽ G .< s p and < G : ZŽ G .< s p 3. This is Žc.. Assume now that R g w A, xy x. Then bAŽ xy . s 2 implies bAŽ xy. s 2. Since w A, H x s w A, H˜ x F R we get that w A, H x s 1 and thus A s H. Moreover, since w A, ² x, y :x s w A, ² x, yy1 x :x, we can also assume R g A, yy1 x .
Ž ).
Let Y F A such that Y s CA Ž x . l CA Ž y .. Then < A : Y < s p 2 since bA Ž xy. s 2, and w Y, G x F R. In particular w Y, xy x s w Y, yy1 x x s 1 by Ž). and thus Y s CA Ž xy . F CA Ž yy1 x . . Since ² xy, yy1 x : A s ² x, y : A this gives CA Ž x . s CA Ž y . s CA Ž ² x, y : . s Z Ž G . , and Žb. holds. Proof of Theorem B. Let A and G be as in Theorem B. If bA s b s 3, then Lemma 11 gives Ža.. Assume that bA s b y 1 s 2. Then Lemma 13 gives Žd., or A and G satisfy the hypothesis of one of Lemma 12, 14, or 15. Hence Ža., Žb., Žc., or Žd. holds. Assume finally that bA s 1. We apply Lemma 8. Suppose that
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Suppose that
REFERENCES wGMMPSx wKn1x wKn2x
N. Gavioli, A. Mann, V. Monti, A. Previtali, and C. M. Scoppola, Groups of prime order with many conjugacy classes, J. Algebra 202 Ž1998., 129᎐141. ¨ H. G. Knoche, Uber den Frobenius’schen Klassenbegriff in nilpotenten Gruppen, Math. Z. 55 Ž1951., 71᎐83. ¨ H. G. Knoche, Uber den Frobeniusschen Klassenbegriff in nilpotenten Gruppen, II, Math. Z. 59 Ž1953., 8᎐16.
p-Groups of Small Breadth Gemma Parmeggiani* Uni¨ ersita ` degli Studi di Pado¨ a, Italy
and Bernd Stellmacher Christian-Albrechts-Uni¨ ersitat ¨ zu Kiel, 24118 Kiel, Germany Communicated by Gernot Stroth Received February 23, 1998
Let G be a finite p-group. For x g G the breadth bŽ x . of x is defined as p bŽ x . [ < G : CG Ž x . < , and the breadth bŽ G . of G is the maximum of the breadths of the elements of G. More generally, we define for x g G and an abelian normal subgroup A of G p bA Ž x . [ < A : C A Ž x . < , bA Ž G . [ max bA Ž x . N x g G 4 , BA [ x g G N bA Ž x . s bA Ž G . 4 . Clearly, bŽ G . s 0 is equivalent to G⬘ s 1; and H. G. Knoche wKn1x has shown that bŽ G . s 1 is equivalent to < G⬘ < s p Žsee the Proposition in Section 3.. Knoche also obtained some results for bŽ G . s 2 wKn2x; in particular, he has shown that < G⬘ < F p 3. *This author is grateful to the University of Kiel for the kind hospitality during the preparation of this paper, and to C.N.R. ŽItaly. for the financial support. 52 0021-8693r99 $30.00 Copyright 䊚 1999 by Academic Press All rights of reproduction in any form reserved.
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In this paper we prove the following two theorems. THEOREM A. Let G be a finite p-group. Then bŽ G . s 2, if and only if one of the following holds: Ža. Žb.
< G⬘ < s p 2 , or < G : ZŽ G .< s p 3 and < G⬘ < s p 3.
Moreo¨ er, if bŽ G . s 2 and < G⬘ < ) p 2 , then < ArZŽ G .< s p for e¨ ery maximal abelian normal subgroup A of G. THEOREM B. Let G be a finite p-group with bŽ G . s 3, and let A be a maximal abelian normal subgroup of G. Then one of the following holds: Ža. < G⬘ < s p 3. Žb. < G : ZŽ G .< s p 4 and bAŽ G . g 1, 24 . Žc. < G⬘ < s p 4 , D [ Fx g B w A, x x has order p, and A < GrD : Z Ž GrD . < s p 3
and
b Ž GrD . s 2;
moreo¨ er bAŽ G . g 1, 24 and
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1. ELEMENTARY PROPERTIES Let G be a finite p-group and A a normal abelian subgroup of G. We set b [ bŽ G ., bA [ bAŽ G ., and TA [ x g G N bA Ž x . s 1 4 . The proof of the first lemma is elementary and well known: LEMMA 1.
For e¨ ery x g G
w A, x x s w A, ² x : x s w a, x x N a g A4 ; in particular,
w A, u x = w A, ¨ x s w A, u¨ xw A, ¨ x and thus p bAŽ u.qbAŽ ¨ . s w A, u¨ xw A, ¨ x - p bAŽ u ¨ .qbAŽ ¨ . . LEMMA 3. Let x, y g TA , T s ² x, y :, and T s TCG Ž A.rCG Ž A.. Then one of the following holds: Ža. T _ CG Ž A. : TA and w A, x x s w A, T x; moreo¨ er CAŽ x . / CAŽ y . if T is not cyclic. Žb. T _ CG Ž A. : TA and CAŽ x . s CAŽT .; moreo¨ er w A, x x / w A, y x if T is not cyclic. Žc. CAŽT . s CAŽ x . l CAŽ y . and w A, T x s w A, xy x s w A, x x = w A, y x. Proof. If T is cyclic, then CAŽ x . s CAŽ y ., and Žb. holds. Thus, we may assume that T is non-cyclic. Assume first that xy g TA . Then by Lemma 2, w A, x x s w A, y x or CAŽ x . s CAŽ y .. This gives Ža. resp. Žb.. Assume now that xy f TA , i.e., bAŽ xy . ) bAŽ x . s bAŽ y .. Then clearly CAŽ xy . s CAŽ x . l CAŽ y . and w A, xy x s w A, x x = w A, y x. This is Žc.. LEMMA 4. Let x, u, ¨ g G such that u¨ y1 f CG Ž A.. Suppose that bAŽ x . ) 1. Then at least one of the elements u, ¨ , u¨ , ux, ¨ x, u¨ x is not in TA .
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Proof. Assume that ⍀ [ u, ¨ , u¨ , ux, ¨ x, u¨ x 4 : TA . Let T1 s ² u¨ y1 , u:, T2 s ² u¨ y1 , ¨ x :, and T3 s ² u, ¨ x :. Then case Ža. or Žb. of Lemma 3 applies to each of the subgroups T1 , T2 , and T3 . Fix i, j g 1, 2, 34 such that i / j and
Let x g G and a g A. Then the following hold:
Ža. w a, Ma x l M x x F w A, x x. Žb. w A, Dx x F w A, x x s w Dx , x x; in particular, if b s bA then
w a, G x F w xy1 , U xw A, x xw A, U x . Proof. Let a g A and y g Ma x . Then y s cb for some c g CG Ž ax . and b g A. It follows y
a y x y s Ž ax . s Ž ax .
cb
b
s Ž ax . s ax b
and w a, y x s w b, xy1 xw xy1 , y x. In particular w xy1 , y x g w A, G x, and Žc. follows. Assume that y g M x . Then w xy1 , y x g w A, xy1 x s w A, x x and thus w a, y x g w A, x x, and Ža. follows. Now Ža. and w M x , x x s w A, x x yield
w a, Dx x F w a, M x l Ma x x F w A, x x F w Dx , x x F w M x , x x s w A, x x and thus w A, Dx x F w A, x x s w Dx , x x.
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Assume that b s bA and let z g BA . Then G s Ma z for every a g A, i.e., G s Dz , and the previous inequality gives w A, G x s w A, Dz x F w A, z x. Thus w A, G x s w A, z x has order p bAŽ z . s p bA , and the proof of Žb. is complete. For the proof of Žd. note that for all x, y, z g G
w z, xy x s w z, y xw z, x x s w z, y xw z, x xw z, x, y x . y
It follows that
w a, G x s w a, Ma x U x F w A, U x a, CG Ž ax . w A, U x s a, CG Ž ax . w A, U x . Ž ). For c g CG Ž ax . we get w a, c x s w xy1 , c x and thus a, CG Ž ax . F w xy1 , G x s xy1 , M x U F w xy1 , U xw xy1 , A xw A, U x . Together with the above relation Ž). this gives Žd.. LEMMA 6. x g BA .
Suppose that b F bA q 1. Then w A, x x is normal in G for e¨ ery
Proof. Let x g BA . Note that Ma x normalizes w A, ax x s w A, x x for every a g A. Hence we can assume Gx / G. This gives b s bA q 1 and M x s Dx . Now Lemma 5Žb. shows that w A, x x s w A, M x x. From < G : M x < s p it follows that M x is normal in G, so also w A, M x x is normal in G. LEMMA 7. satisfying
Let bA ) 1. Suppose that w G, z x F w A, G x for all z g G bA Ž z . ) 1
and
2 bA Ž z . G bA .
Ž ).
Then G⬘ s w CG Ž A., G x. In particular, if in addition A s CG Ž A. and b s bA then < G⬘ < s p bA . Proof. Let x g BA and y g G. Then
w A, x x F w A, y xw A, yx x , and we are in one of the following two cases: ŽI. ŽII.
y or yx satisfy Ž). for all y g G, or y, yx g TA for some y g G, and bA s 2.
In the first case w G, y x F w A, G x or w G, yx x F w A, G x. Since also w G, x x F w A, G x we conclude that w G, y x F w A, G x and thus G⬘ s w A, G x s w CG Ž A., G x. Assume now case ŽII.. Then w G, z x F w A, G x for every z g G _ ŽTA j CG Ž A.. and thus w G, z x F w CG Ž A., G x for every z g G _ TA . Suppose
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57
that there exist u, ¨ g G such that w u, ¨ x f w CG Ž A., G x; i.e., w G, u x g w CG Ž A., G x and w G, ¨ x g w CG Ž A., G x. Then also
w u¨ , ¨ x , w ux, ¨ x , w ¨ x, u x , w u¨ x, ¨ x f CG Ž A . , G ; y1
moreover, w u¨ y1 , ¨ x s w u, ¨ x¨ f w CG Ž A., G x implies u¨ y1 f CG Ž A.. As we have seen above, this implies u, ¨ , u¨ , ux, ¨ x, u¨ x g TA , which contradicts Lemma 4. Lemma 5Žb. gives the additional statement for b s bA and A s CG Ž A.. 2. THE CASE bA F 2 In this section we use the same notation as in Section 1. In particular, A is an abelian normal subgroup of G. LEMMA 8.
Suppose that bA s 1. Then either
Ži. < A : A l ZŽ G .< s p and CAŽ x . s CAŽ y . for e¨ ery x, y g G _ Ž CG A., or Žii.
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PARMEGGIANI AND STELLMACHER
Assume Žii. holds. Then D s w A, G x is a normal subgroup of order p such that bŽ GrD . - bŽ G .. LEMMA 9. ² BA : s G.
Suppose that bA s 2. Then either
Proof. Note that G s BA j TA j CG Ž A . . Assume that G / ² BA :. Let x g BA . Then t, tx, xy1 t g TA for every t g G _ ² BA :. If p / 2, then t 2 g TA and
w A, t 2 x s w A, txxy1 t x s p. Hence Lemma 3Ža. or Žb. applies to ² tx, xy1 t : s ² t, x : and x g TA , a contradiction. Thus p s 2. Moreover again by Lemma 3
w A, x x s w A, t x = w A, tx x . Hence w A, ² G _ ² BA ::x F w A, x x. On the other hand ² G _ ² BA :: s G and thus
Suppose that bA s 2. Then one of the following holds:
Ža. < A : A l ZŽ G .< s p 2 , and there exist x, y g BA such that w A, x x l w A, y x s 1. Žb.
w A, x x l w A, y x s 1. Then Lemma 2 gives C [ CA Ž x . s CA Ž y . . We want to show that C s A l ZŽ G .. Thus, we may assume that there exists w g G such that C g CAŽ w .. If a g C _ CAŽ w . then
w a, xw x s w a, w x s w a, yw x / 1. Hence w A, xw x l w A, yw x / 1 and H [ w A, xw xw A, yw x has order at most p 3. Thus
w A, y x F H w A, x x F H w A, y x
Ž ).
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gives that w A, x x g H and w A, y x g H. Since C g CAŽ xw . and C g CAŽ yw ., Lemma 2 yields that
w A, c x l w A, d x / 1
for every c g x, y 4 , d g xw, yw 4 .
In particular from Ž). it follows that w A, xw x / w A, yw x. Then
w A, x x l H s w A, x x l w A, xw x s w A, x x l w A, yw x s w A, xw x l w A, yw x and similarly w A, y x l H s w A, xw x l w A, yw x. This contradicts w A, y x l w A, x x s 1. We assume now
w A, u x l w A, ¨ x / 1
for all u, ¨ g BA .
Ž )).
We choose u, ¨ g BA such that W [ w A, u xw A, ¨ x is maximal. Suppose first that < W < s p 2 . Then the maximality of W gives
w A, u x s w A, ¨ x
for all u, ¨ g BA .
Let t g TA . If tu g TA , then w A, u x s w A, t x = w A, tu x, and if tu g BA , then w A, u x s w A, tu x. Thus in both cases w A, t x F w A, u x. We conclude that W s w A, G x, and Žc. holds. Suppose now that < W < s p 3. Then R [ w A, u x l w A, ¨ x has order p. Assume first that R s Fx g B A w A, x x. Then R is a normal subgroup of order p in G, and Lemma 8 applies to ArR and GrR. This gives Žc. since
Suppose that b s 1. Then < G⬘ < s p.
Proof. For every x g G _ A G s CG Ž x . A s Dx s Gx .
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Hence Lemma 5Žb. gives
w G, A x s ² w G, x x N x g G _ A: s G⬘. Proof of Theorem A. Suppose that b s 2. If bA s 2 for some maximal abelian normal subgroup A of G, then Lemma 7 and Lemmas 5Žb. yield Ža.. Thus, we may assume that bA s 1 for every maximal abelian normal subgroup A of G. We discuss the two cases of Lemma 8. Assume case Ži.. For z g A _ ZŽ G . we get CG Ž a. s A and thus < G : ZŽ G .< F p 3. Now b s 2 gives < G : ZŽ G .< s p 3 and
Suppose that bA s b s 3. Then < G⬘ < s p 3.
Proof. Let x g BA . We have G s ACG Ž x . s M x s Dx . By Lemma 5Žb., w A, G x s w A, x x. Let y g CG Ž x . such that bAŽ y . s 2. Then either G s Gy or Dy s M y . In the first case w G, y x F w A, G x by Lemma 5Žc.. In the second case Lemma 5Žb. gives
w A, x x F A, M y s w A, y x , which contradicts
Let bA s b y 1 s 2. Suppose that there exist x, y g BA such
w A, x x l w A, y x s 1.
Then either < G : ZŽ G .< s p , or p s 2 and < G⬘ < s 2 4 . 4
Proof. Let G be a minimal counterexample. From Lemma 9 and Lemma 10 we get: Ž1. G s ² BA : and < A : ZŽ G .< s p 2 ; in particular, CAŽ z . s ZŽ G . for every z g BA . Since < G : ZŽ G .< ) p 4 Ž1. gives < G : A < G p 3 and thus Ž2. Mu l Mw / A for u, w g BA . By Lemma 6, w A, z x is normal in G for every z g BA . Thus
w A, x, y x s 1 s w A, y, x x . It follows that w A, x xw A, y x F ZŽ G ..
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Assume that w A, z x g ZŽ G . for some z g BA . Then w A, z, x x / 1 / w A, z, y x and thus w A, z x s w A, z, x x = w A, z, y x F ZŽ G ., a contradiction. Hence w A, z x F ZŽ G . for all z g BA , and Ž1. gives: Ž3. w A, G x F ZŽ G .. Next we show: Ž4.
G⬘ s w A, G x.
Assume that Ž4. is false. Then by Ž1. there exist u, ¨ g BA such that w u, ¨ x f w A, G x. Thus, Lemma 5Žc. gives ¨ f Gu and u f G¨ . Hence Mu s Du and M¨ s D¨ . Now Lemma 5Žb. yields
w A, Mu x F w A, u x
and
w A, M¨ x F w A, ¨ x .
This implies
w A, x xw A, y x F w A, G x F w A, u xw A, ¨ x and thus w A, u x l w A, ¨ x s 1. But now Mu l M¨ s A, which contradicts Ž2.. Ž5.
p / 2.
Assume that p s 2. Since G is a counterexample we get from Ž4. that w A, x xw A, y x - w A, G x. Set D s M x l M y . We first show: For every d g D there exists a g A _ Z Ž G . such that d g Ma x . Ž ) . If d is a counterexample, then Ma x² d : s Mab x² d : s Mb x² d : s G, where A s ZŽ G .² a, b : and a, b, ab f ZŽ G .. Now d g M x w xy1 , ² d :x F w A, x x. Hence
Ž ax .
y1
, ² d : F w xy1 , ² d : x
ay1
yields
w A, d x F w A, x xw A, d x
and similarly wŽ bx .y1 , ² d :x F w A, x xw A, d x. Thus from Lemma 5Žd. we get
w A, G x s w b, G xw a, G x s
¦ Ž ax .
y1
, ² d : w A, ax xw A, d x , Ž bx .
s w A, x xw A, d x which contradicts
y1
, ² d : w A, bx xw A, d x
;
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Clearly, Ž). also holds with y in place of x. Moreover, from Ž). and Lemma 5Ža. follows w a, d x g w A, x x. Thus Ž1. and w A, x x l w A, y x s 1 imply y f D and similarly x f D; in particular G s D ² x, y :. Since w A, G x g w A, x xw A, y x there exists d g D such that w A, d x g w A, x xw A, y x. If d g BA , then CAŽ d . s ZŽ G . by Ž1.. From Ž). and Lemma 5Ža. we get 1 / w a, d x g w a, M x l Ma x x F w A, x x and thus w A, d x l w A, x x / 1. Similarly, considering y in place of x, w A, d x l w A, y x / 1. Hence w A, x x l w A, y x s 1 yields w A, d x F w A, x xw A, y x, a contradiction. Assume now that bAŽ d . s 1. Then there exist a, b g A such that CAŽ d . s ZŽ G .² a: and A s CAŽ d .² b :. If d g Mb x or d g Mab x , then by Lemma 5Ža., w b, d x s w ab, d x g w A, x x, and
w A, d x s ² w b, d x : F w A, x x , a contradiction. Hence, we have d f Mb x and d f Mab x , and Lemma 5Žd. yields w a, G x F w A, x xw A, d x. Since 1 / w a, y x g w a, G x and
GrA is elementary abelian of order p 3.
By Ž4., GrA is abelian, and by Ž3., w A, G x F ZŽ G .. Thus w a, g p x s w a p , g x for every a g A and g g G. Assume that G p g A. Then also A p g ZŽ G ., and Ž1. shows that ArZŽ G . is cyclic. Let A s ZŽ G .² a:. Then w A, y x s ²w a, y x: is a cyclic group of order p 2 since w a, y x p s w a, y p x, w A, y p x s ²w a, y p x: is a group of order p. Hence y p g TA l D, D as in Ž5.. But now w A, y p x l w A, x x F w A, y x l w A, x x s 1, and Lemma 5Ža. implies y p f Ma x
for all a g A _ CA Ž y p . .
This contradicts < G : Ma x < F p. We have shown that GrA is elementary abelian. Then GrA has order p 3 since p is odd and G is a minimal counterexample. Ž7. There exist a, b g A such that Ma x / G / Mb y . Assume that G s Ma x for all a g A. Then Lemma 5Ža. gives w A, G x s w A, x x which contradicts w A, y x g w A, x x. Hence G / Ma x for some a g A, and the same argument shows the assertion for y. According to Ž7. we may choose our notation such that M x / G / M y . We now derive a final contradiction using Ž5. and Ž6.. In GrA there exist exactly p q 1 subgroups of order p 2 containing ² xA:. On the other hand,
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63
xA g Ma xrA for every a g A _ ZŽ G . and < Ma xrA < G p 2 . Hence, since < A : ZŽ G .< s p 2 , we may choose our notation additionally such that M x F Ma x
and
M y F Mb y for some a, b g A _ Z Ž G . .
In particular, by Lemma 5Ža.
w a, M x x F w A, x x
and
b, M y F w A, y x .
Assume that w A, M x x g w A, x x. Then w c, M x x g w A, x x for every c g A _ ² a: ZŽ G . and thus M x g Mc x
for every c g A _ ² a: Z Ž G . .
Fix c g A _ ² a: ZŽ G . and set d s ac. Since p G 3 then dc f ² a: ZŽ G .. Assume M x s Ž M x l Mc x . j Ž M x l Md x . j Ž M x l Mdc x . . Then M xrA is a group of order p 2 which is the union of three subgroups of order p. Hence p 2 s < M xrA < s 3 Ž p y 1 . q 1, a contradiction to Ž5.. Thus there exists w g M x _ Ž Mc x j Md x j Mdc x . . It follows that G s Mc x² w : s Mdcy1 c x² w : s Mdc x² w : , and Lemma 5Žd. implies
w A, G x s w ² dcy1 , d : , G x s w A, x xw A, w x since w x, w x g w A, x x. Moreover, w a, w x g w a, M x x F w A, x x yields
Suppose that bA s b y 1 s 2. Then one of the following
Ža. < G : ZŽ G .< s p 4 , and there exist x, y g BA such that w A, x x l w A, y x s 1. Žb.
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Proof. Let G be a counterexample. We apply Lemma 10. Then case Žb. and by Lemma 12 also case Ža. of Lemma 10 cannot hold. Thus, we are in case Žc. of Lemma 10. As there, let D [ Fx g B A w A, x x . If < D < s p 2 , then D s w A, G x and we get Žb., a contradiction. Thus, we have < D < s p. Set G [ GrD, and note that bAŽ G . s 1. If < A : A l ZŽ G .< / p, then Lemma 8 yields Žb., again a contradiction. Hence, we have < A : A l Z Ž G . < s p. Let a g A such that a f ZŽ G .. Then CG Ž A. s CG Ž a. and < G : CG Ž a.< F p . In particular
< w A, G x < F p 4 . Assume that M x s Dx for some x g BA . Then by Lemma 5Žb., w A, M x x s w A, x x. Since < G : M x < F p we conclude that
s A, M x
s p.
Since < G : M x < F p, we conclude that
p s 2 and < G⬘ < s 2 4 . < G⬘ < s p 3.
Proof. Let G s Grw A, G x, and let U be a maximal abelian normal subgroup of G. Note that A F ZŽ G . F U. Then < G : CG Ž x .< F p for x g BA , in particular: Ž1.
G s UCG Ž x . or x g U; and w x, G x F wU, G x for every x g BA .
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We first show: Ž2.
G⬘ s wU, G x.
Assume that there are ¨ , w g G such that w ¨ , w x f wU, G x. From Ž1. we get ¨ , w, ¨ w g TA . Let x g BA . Then Ž1. gives w x, G x F wU, G x and w z, G x g wU, G x for every z g ¨ x, wx, ¨ wx 4 . Hence ¨ x, wx, ¨ wx f A and from Ž1. we also have ¨ x, wx, ¨ wx f BA . Thus ¨ x, wx, ¨ wx g TA and Lemma 4 yields a contradiction. If
Hence by Ž2. and the Proposition, bŽ G . s 2. Assume first that bU Ž G . s 1 for every such that U. From Ž3. and Lemma 8 we get
and
U s CG Ž u . for every u g U _ Z Ž G . .
Let u g G _ ZŽ G .. Since < G : ZŽ G .< ) p 2 we get from Ž). p 2 F py1 < G : Z Ž G . < s < G : U < s < G : CG Ž u . < F p 2 , i.e., < G : U < s p 2 . Moreover
w u, G x F U, x . Hence wU, G x s wU, x x and, as x g TU , we get bU Ž G . s 1, a contradiction.
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LEMMA 15. Suppose that bA s b y 1 s 2 and
p s 2 and < G⬘ < s 2 4 .
Proof. We may assume that G⬘ ) w A, G x. Lemma 9 implies G s ² BA :. Hence, there exist x, y g BA such that w x, y x f w A, G x. Thus x f M y and y f M x , and we have G s M x² y :
M x s H² x :
and
for H [ M x l M y . We also have for every ¨ g H
w ¨ x, y x f w A, G x
and
w ¨ y, x x f w A, G x .
Lemma 5Žb. and Žc. give Ž1. M z s Dz and w A, M z x s w A, z x for every z g ¨ x, ¨ y N ¨ g H 4 such that bAŽ z . s 2. As w A, M x x s w A, x x by Ž1. we get w A, H x F w A, x x l w A, y x \ R. Moreover < R < s p, since w A, G x s w A, x xw A, y x has order p 3. Next we show: Ž2.
M zX F w A, z x for z g x, y4 .
We have M zX s H⬘w A, z x. Suppose that there exist u, ¨ g H such that w u,¨ x f w A, z x. Assume first that ¨ z g BA . Then Ž1. gives w A, M¨ z x s w A, ¨ z x. But from w A, ¨ x F w A, H x F R F w A, z x we get w A, ¨ z x s w A, z x. Hence M¨ z s M z since
w ¨ , bt x s w ¨ zzy1 , bt x s w ¨ z, bt x z w zy1 , bt x s w zy1 , bt x s w zy1 , t xw zy1 , b x s w zy1 , b x g w A, z x t
t
and so w ¨ , w x s w ¨ , abt x s w ¨ , bt xw ¨ , ax b t g w A, z x. Thus w ¨ , M z x F w A, z x, which contradicts w u, ¨ x f w A, z x.
p-GROUPS
OF SMALL BREADTH
67
We may assume now that ¨ z, uz, u¨ z g TA . But also u, ¨ , u¨ g TA since w A, H x F R. Hence, Lemma 4 gives z g TA , a contradiction. This gives Ž2.. From Ž2. we get that H⬘ F R. Set G [ GrR. Then H is an abelian normal subgroup of G such that G s H² x, y : and < G : H < s p 2 . Note that bH Ž x . s bH Ž y . s 1. Thus G⬘ s w H, ² x :xw H, ² y :xw²w x, y x: has order p 3 , and < G⬘ < s p 4 . Hence, we may assume that p / 2. Note w A, G x s w A, ² x, y :x s w A, ² xy, yy1 x :x. Since
˜ F w A, xy x l w A, x x . H⬘ ˜ G x s w A, G x and either HH˜ s M x or H s H. ˜ In On the other hand w H, ˜ w x w x w x w the first case M x , G s HH, G s A, G , which contradicts x, y x f ˜ w A, G x. Thus, we have H s H. w x Assume that R F A, xy . Then R s w A, xy x l w A, x x, and bH Ž G . s 1 by Lemma 3. Hence Lemma 8 gives < H : ZŽ G .< s p and < G : ZŽ G .< s p 3. This is Žc.. Assume now that R g w A, xy x. Then bAŽ xy . s 2 implies bAŽ xy. s 2. Since w A, H x s w A, H˜ x F R we get that w A, H x s 1 and thus A s H. Moreover, since w A, ² x, y :x s w A, ² x, yy1 x :x, we can also assume R g A, yy1 x .
Ž ).
Let Y F A such that Y s CA Ž x . l CA Ž y .. Then < A : Y < s p 2 since bA Ž xy. s 2, and w Y, G x F R. In particular w Y, xy x s w Y, yy1 x x s 1 by Ž). and thus Y s CA Ž xy . F CA Ž yy1 x . . Since ² xy, yy1 x : A s ² x, y : A this gives CA Ž x . s CA Ž y . s CA Ž ² x, y : . s Z Ž G . , and Žb. holds. Proof of Theorem B. Let A and G be as in Theorem B. If bA s b s 3, then Lemma 11 gives Ža.. Assume that bA s b y 1 s 2. Then Lemma 13 gives Žd., or A and G satisfy the hypothesis of one of Lemma 12, 14, or 15. Hence Ža., Žb., Žc., or Žd. holds. Assume finally that bA s 1. We apply Lemma 8. Suppose that
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PARMEGGIANI AND STELLMACHER
Suppose that
REFERENCES wGMMPSx wKn1x wKn2x
N. Gavioli, A. Mann, V. Monti, A. Previtali, and C. M. Scoppola, Groups of prime order with many conjugacy classes, J. Algebra 202 Ž1998., 129᎐141. ¨ H. G. Knoche, Uber den Frobenius’schen Klassenbegriff in nilpotenten Gruppen, Math. Z. 55 Ž1951., 71᎐83. ¨ H. G. Knoche, Uber den Frobeniusschen Klassenbegriff in nilpotenten Gruppen, II, Math. Z. 59 Ž1953., 8᎐16.