Locally finite groups with finitely many isomorphism classes of derived subgroups

Journal of Algebra 393 (2013) 102–119

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Journal of Algebra www.elsevier.com/locate/jalgebra

Locally finite groups with finitely many isomorphism classes of derived subgroups ✩ Patrizia Longobardi a , Mercede Maj a , Derek J.S. Robinson b,∗ a b

Dipartimento di Matematica, Università di Salerno, 84084 Fisciano (Salerno), Italy Department of Mathematics, University of Illinois at Urbana-Champaign, Urbana, IL 61801, USA

a r t i c l e

i n f o

Article history: Received 10 March 2013 Available online 31 July 2013 Communicated by E.I. Khukhro

a b s t r a c t Groups which have at most n isomorphism classes of derived subgroups (Dn -groups) are studied. A complete classification of locally finite D3 -groups into nine types is obtained. © 2013 Elsevier Inc. All rights reserved.

MSC: 20F14 Keywords: Derived subgroups Isomorphism types

1. Introduction By a derived subgroup in a group G is meant the derived (or commutator) subgroup of a subgroup of G. It is natural to enquire how important the derived subgroups are within the lattice of all subgroups. Recently there has been interest in imposing restrictions on the number of derived subgroups in a group and investigating the resulting effect on the structure of the group – see [2–4]. In this article we are concerned with groups for which the set of isomorphism types of derived subgroups is small. If n is a positive integer, let

Dn ✩ This research had its origins during June–July 2011 while the third author was a visitor at the University of Salerno under the auspices of GNSAGA–INdAM (Italy). He acknowledges splendid hospitality from the Dipartimento di Matematica. Corresponding author. E-mail addresses: [email protected] (P. Longobardi), [email protected] (M. Maj), [email protected] (D.J.S. Robinson).

*

0021-8693/$ – see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.jalgebra.2013.06.036

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denote the class of groups whose derived subgroups fall into at most n isomorphism classes. Clearly D1 is the class of abelian groups. The class D2 was investigated in [6], where a classification of locally finite D2 -groups in terms of finite fields was given. In the present work our principal aim is to give a complete classification of locally finite D3 -groups. Not surprisingly, this is a much more challenging task than for D2 -groups, and it emerges that there are nine different families of groups. The complexity of the classification suggests that it may not be possible to classify locally finite D4 -groups in a useful way, even although all these groups are known to be soluble. 1.1. Results We begin in Section 2 with some general results about the classes Dn , including the important fact that the derived subgroup of a locally finite Dn -group is finite. Also locally finite Dn -groups are shown to be soluble for n  4. Thereafter our objective is to pin down the structure of locally finite D3 -groups and then to demonstrate that all groups possessing this structure are D3 -groups. The analysis falls naturally into two cases, nilpotent groups in Section 3, and non-nilpotent groups in Sections 4–6. The classification of periodic nilpotent D3 -groups appears in Theorem 2, where an essential component is shown to be the so-called generalized Camina p-groups. This is the most technically challenging part of our analysis. In the non-nilpotent case it turns out that the groups are semidirect products of the form X  O p (G ) where X is an abelian p  -group. The analysis proceeds by dealing first with the groups for which O p (G ) is abelian: once again finite fields play an essential part here. The group structure is displayed in Theorem 3. If O p (G ) is non-abelian, this subgroup is an extra-special p-group, with exponent p if p > 2, and the X -module structure of [ P , X ] P  / P  plays a decisive role. The definitive result is Theorem 4. Finally, in Section 7 it is proved that all the groups allowed by Theorems 3 and 4 are actually D3 -groups, which is Theorem 5. Notation. G  , G  , G (i ) : terms of the derived series of a group G. Z (G ): the center of G. Z i (G ): a term of the upper central series of G. γi (G ): a term of the lower central series of G. r (G ): the Prüfer rank of a group. d(G ): the minimum number of generators of a group. G  H : the central product of G and H , with an amalgamated central subgroup A. A

A [m]: the subgroup {a ∈ A | am = 1} where A is an abelian group. π (G ): the set of primes dividing the order of a finite group G. 2. Some general results on the classes Dn We begin with an easy, but useful, observation. Lemma 1. Let G = H Z where H  G and Z  Z (G ). Then G ∈ Dn if and only if H ∈ Dn . This is because a commutator in G is also a commutator in H . The next result provides an important structural restriction on locally finite groups in one of the classes Dn . Proposition 1. If G is a locally finite Dn -group, then G  is finite.

 1: write H 1 = Proof. Assume that G  is infinite. Then there exist a1 , b1 in G such that [a1 , b1 ] = a1 , b1 . Then H 1 is finite, so G   H 1 and there exist a2 , b2 in G such that [a2 , b2 ] ∈ / H 1 . Put H 2 =  H 1 , a2 , b2 , which is finite; then H 1 < H 2 . By repeating this procedure we can construct an infinite

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ascending chain of finite subgroups { H i } such that H 1 < H 2 < · · · < H n < · · · . However, H i and H j cannot be isomorphic if i = j, so we have reached a contradiction. 2 Lemma 2. A soluble Dn -group has derived length at most n. Proof. Let G ∈ Dn have derived length d. The derived subgroups 1 = G (d) , G (d−1) , . . . , G (1) = G  have respective derived lengths 0, 1, . . . , d − 1, so no two of them are isomorphic. Hence d  n. 2 The next result shows that a sufficiently strong restriction on isomorphism types of derived subgroups implies solubility of the group. Theorem 1. A locally finite D4 -group is soluble with derived length at most 4. Proof. By Lemma 2 it is enough to prove the result for a finite group. Let G be a finite insoluble D4 -group of least order. By the Burnside p–q Theorem there exist three distinct primes p , q, r dividing |G |. Let P , Q , R denote Sylow p , q, r-subgroups of G. Suppose that G is p-nilpotent; then G = X  O p  (G ) with X a p-group. Since X = 1, minimality shows that O p  (G ) is soluble and hence G is soluble. Therefore G cannot be p, q or r-nilpotent. Burnside’s Criterion shows that there exist x ∈ N G ( P )\C G ( P ), y ∈ N G ( Q )\C G ( Q ) and z ∈ N G ( R )\C G ( R ). Now x, P  = P  [ P , x] = 1,  y , Q  = Q  [ Q , y ] =  1 and z, R  = R  [ R , z] =  1. These derived subgroups are p , q, r-groups respectively, which means that there are at least five non-isomorphic derived subgroups in G, a contradiction. 2 On the other hand, the alternating group A 5 is a D5 -group since its non-trivial derived subgroups are A 5 , Z5 , Z3 , V 4 . So Theorem 1 is the end of this particular story. 3. Periodic locally nilpotent groups Our aim in this section is to classify periodic locally nilpotent D3 -groups. First notice that if G is a periodic locally nilpotent D3 -group, then G  is finite by Proposition 1, and from this it follows easily that G is actually nilpotent. We begin with some generalities about nilpotent D3 -groups. The following simple fact is used repeatedly. Lemma 3. Let G be a nilpotent D3 -group. Then either G ∈ D2 or G has a subgroup H , with H  not isomorphic to G  , such that H  is cyclic of prime or infinite order, and for any non-abelian subgroup S of G, either S   G  , or S   H  and S   Z ( S ). Proof. Assume G ∈ / D2 . Then G is not abelian, so there exist c ∈ Z 2 (G ) \ Z (G ) and d ∈ G such that 1 = [c , d] ∈ Z (G ). Hence c , d = [c , d]. If [c , d] has infinite order, put H = c , d, so that H  = [c , d]. If [c , d] has finite order m and p is a prime dividing m, write m = pn with n ∈ Z, and set H = cn , d. Thus H  = [c , d]n , which has order p. So in either case H ∈ D2 (see [6, Theorem 1]) and H   G  since G ∈ / D2 . If S is a non-abelian subgroup and S   G  , then S   H   Z or Z p since G ∈ D3 . Consequently S   Z ( S ) since S is nilpotent. 2 Next we obtain some initial structural information about nilpotent D3 -groups. Lemma 4. Let G be a nilpotent D3 -group. Then G has nilpotent class at most 3 and G  is abelian. Proof. Clearly we can assume that G is finitely generated. We argue first that G   Z (G ). Indeed suppose this is false and there exists x ∈ G such that [G  , x] = 1. By Lemma 2 the subgroup G  is abelian and hence x, G   = [G  , x]. Now there are non-trivial derived subgroups [G  , x] < G  < G  , so two of these must be isomorphic. Note that G   G  or [G  , x]: for otherwise G  /G  is finite

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since G is finitely generated and nilpotent, which implies that G  is finite and G  = G  – clearly impossible. Therefore G   [G  , x], so that G   G  /C G  (x) and C G  (x) must be trivial. This yields the contradiction G  = 1. Assume that G has class greater than 3. For any x ∈ G we have x, G   = G  [G  , x]  G  . If x, G    G  , then G  /G  [G  , x] is finite, which, since G is nilpotent, implies that G  /G  is finite and G  is finite. Therefore G  = G  [G  , x], which implies that G  = [G  , x] and hence G  = 1, a contradiction. Consequently for any x ∈ G we must have x, G    G  , i.e., G   G  [G  , x]. Therefore G  [G  , x]/G  is finite and G  [G  , x] is abelian. Since G   Z (G ), it follows that [G  , x, y ] is finite for all x, y ∈ G. Therefore [G  , G , G ] is periodic and hence finite. Thus G  is finite and hence G  [G  , x] is finite for all x ∈ G. We therefore have G  = G  [G  , x], i.e., [G  , x]  G  for all x ∈ G. Hence [G  , G ]  G   Z (G ) and γ4 (G ) = 1. This implies that G  is abelian. 2 We look for further restrictions on the structure of a nilpotent D3 -group. Lemma 5. Let G be a nilpotent group in the class D3 . Then either G  is locally cyclic or it can be generated by at most 3 elements. Proof. By Lemma 4 the subgroup G  is abelian and G has class at most 3. If G ∈ D2 , then G  is cyclic and we have the result, so assume that G ∈ / D2 . First suppose there exist a, x ∈ G such that [a, x, x] = 1 and consider the subgroup S = a, x. Then S  = [a, x], [a, x, x], [a, x, a] since G has class at most 3. Now S   Z ( S ), so it follows via Lemma 3 that S   G  and G  is a 3-generator group, as required. We may therefore assume that [a, x, x] = 1 for all a, x ∈ G, i.e., G is a 2-Engel group. Suppose G has class 3, so there exist a, x, y ∈ G such that [a, x, y ] = 1; then [a, x, y ] = [a, y , x]−1 has order 3 by the structure of 2-Engel groups [9, 7.13, 7.14]. Hence [a, x], y  = [a, x, y ] has order 3 and, if H is the subgroup of Lemma 3, then H   [a, x], y  has order 3. If G   (a, y ) = [a, y ], then G  is cyclic, as required. Otherwise [a, y ]  H  has order 3 and a, [a, x], y  = [a, y ], [a, x, y ] is finite and non-cyclic: for [a, y ] ∈ / [a, x, y ] since [a, y ] ∈ / Z (G ). Thus G  = [a, y ], [a, x, y ] is a 2-generator group. Hence we can assume G is nilpotent of class 2. Let a, x, y ∈ G. If a, x, y  is not cyclic, then





G   a, x, y  = [a, x], [a, y ], [x, y ]

is 3-generated. Thus we may assume that a, x, y  is cyclic for all a, x, y ∈ G. Let a, b, x, y ∈ G and set H = a, b, x, y . Then





H  = a, b, x , b, x, y  , [a, y ] , and so H  is a 3-generator group. We may suppose that H  is cyclic, since otherwise H   G  . It follows that [a, b], [x, y ] is cyclic and consequently G  is locally cyclic. 2 For periodic nilpotent D3 -groups much more can be said. Lemma 6. Let G be a periodic nilpotent group which belongs to D3 but not to D2 . Then G  is a finite p-group for some prime p and one of the following holds: (a) |G  | = p 2 ; (b) G  is elementary abelian of order p 3 and G p  Z (G ): moreover, if G has nilpotent class 3, then p is odd. Proof. In the first place G  is finite by Proposition 1. Since G ∈ D3 and it is non-abelian, exactly one of its Sylow subgroups is non-abelian, let us say the Sylow p-subgroup, so G  is a finite p-group. The p  -elements belong to the center, so we may assume G is a p-group. Also G has nilpotent class at most 3 and G  is a 3-generator abelian group by Lemmas 4 and 5. Let us assume that |G  | > p 2 .

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By Lemma 3 there is a subgroup H of G such that H  is cyclic of order p and for any non-abelian subgroup S either S  = G  or S   H  . For any s, v , w ∈ G we have [s, v ], w  = [s, v , w ] = G  since G  = [G  , G ]. Hence either [s, v , w ] = 1 or [s, v ], w   H  . Thus in any event [s, v , w ] p = 1. Next let S = s p , [s, v ], v ; then









S  = s p , v , [s, v , v ]  s, v 

p 

 s, v  , s, v  ,

and (s, v  ) p is contained in the Frattini subgroup of s, v  . Hence, if S  = s, v  , then s, v  = [s, v  , s, v ] and thus s, v  = 1. If, on the other hand, S  = s, v  , then | S  | divides p. Therefore in any case [s p , v ] p = 1. Next [s p , v ] = [s, v ] p if p > 2, since [s, v , s] p = 1 and G has class at most 3. If p = 2, then [s2 , v ] = [s, v ]2 [s, v , s] and thus [s, v ]4 = 1. So in any case we have 2

[s, v ] p = 1 for all s, v , ∈ G. Since G  is abelian, (G  ) p = 1 and, as |G  | > p 2 , it follows that G  cannot be cyclic.  1: thus G is not a 2-Engel group and its Suppose that there exist a, t ∈ G such that [a, t , t ] = nilpotent class is 3. Then a, t   [a, t ], [a, t , t ], a subgroup which does not have order p since [a, t ] ∈ / [a, t , t ]. Thus 2





G  = a, t  = [a, t ], [a, t , t ], [a, t , a] . Suppose that [a, t ] p = 1. If p > 2, then [a p , t ] = [a, t ] p and a p , [a, t ], t  = [a, t ] p , [a, t , t ] = G  , so this derived subgroup has order p. Hence [a, t , t ] ∈ [a, t ] p  and similarly [a, t , a] ∈ [a, t ] p . Thus G  = [a, t ], a contradiction. We conclude that [a, t ] p = 1 and G  is elementary abelian of order p 3 . In addition G p  Z (G ) since [a p , t ] = [a, t ] p = 1 for p > 2. It remains to exclude the possibility that [a, t , t ] =  1 and p = 2, which requires a separate treatment. Consider the following derived subgroups:

















D 1 = a2 , [a, t ], t = [a, t ]2 [a, t , a], [a, t , t ] , D 2 = t 2 , [a, t ], a = [a, t ]2 [a, t , t ], [a, t , a] ,









D 3 = t 2 , [a, t ], at = [a, t ]2 [a, t , t ], [a, t , t ][a, t , a] . None of these can equal G  since [a, t ] ∈ / [a, t , a], [a, t , t ]. Therefore all the D i have order 1 or 2. We show that there are four possible situations:

(i)

[a, t ]2 = [a, t , a] = 1;

(ii)

[a, t ]2 = [a, t , t ] = [a, t , a];

(iii)

[a, t ]2 = [a, t , t ], [a, t , a] = 1;

(iv)

[a, t ]2 = 1, [a, t , t ] = [a, t , a].

Since | D 1 | = 1 or 2 and [a, t , t ] = 1, we deduce that either [a, t ]2 = [a, t , t ][a, t , a] or else [a, t ]2 = [a, t , a]. In the first case D 3 = [a, t , a], [a, t , t ]. Hence either [a, t , a] = 1 and [a, t ]2 = [a, t , t ], which is situation (iii), or else [a, t , a] = [a, t , t ] and [a, t ]2 = 1, which is situation (iv). Otherwise [a, t ]2 = [a, t , a] and D 2 = [a, t , t ], [a, t , a] = [a, t , t ]. Therefore either [a, t , a] = 1 and [a, t ]2 = 1, which is (i), or else [a, t , a] = [a, t , t ] = [a, t ]2 , which is (ii). Thus the claim is established. In situation (i) [a, t ]2 = 1 and G  = [a, t ], [a, t , t ] has order 4. In situations (ii) and (iii) G  = [a, t ] is cyclic. In situation (iv) G  = [a, t ], [a, t , t ] has order 4. Thus a contradiction arises in every case.

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Consequently, if G is not 2-Engel, then p must be odd. This completes the proof when G is not a 2-Engel group. Now assume that G is a 2-Engel group, so [a, s, s] = 1 for all a, s ∈ G. We show that in this case G has nilpotent class 2: let a, x, y ∈ G and suppose that [a, x, y ] = 1. Note that a, x = [a, x] and a, y  = [a, y ]. Furthermore a, x  Z (a, x) and a, y   Z (a, y ). If G  = a, x , then [a, y ] ∈ Z (a, x) and [a, y , x] = 1: however, this impossible because [a, y , x] = [a, x, y ]−1 since G is a 2-Engel group. Thus a, x = [a, x] has order p. Similarly, a, y  = [a, y ] has order p. Now consider the derived subgroup a, [a, y ], x , which equals [a, x], [a, y , x]. If [a, x], [a, y , x] has order p, then [a, x] ∈ [a, y , x], yielding the contradiction [a, x, y ] = 1. Hence G  = [a, x], [a, y , x]  C G (x), so that [a, y , x] = 1, a contradiction showing that G has class 2, as claimed. Next we argue that G  is elementary. Suppose there is a commutator [b, y ] with order greater than p; then G  = b, y  = [b, y ]. Moreover, b p , y  = [b, y ] p  = G  and thus [b p , y ] = [b, y ] p has order p. But this means that G  has order p 2 , a contradiction. Therefore [b, y ] p = 1 for all b, y ∈ G p and, since G  is abelian, it follows that G  = 1. Therefore [G p , G ] = (G  ) p = 1 and G p  Z (G ), which completes the proof. 2 3.1. Examples We show next that all of the situations envisaged in Lemma 6 can occur. Of course, if G  is a group of order p 2 , it is obvious that G ∈ D3 and it is easy to find such groups. There are also finite p-groups with nilpotent classes 2 and 3 which satisfy condition (b): in fact there are two basic types of example. Type 1. Let S = a, x, y  be a finite p-group of class 2 with S  elementary abelian of order p 3 . Type 2. Let S = a, x be a finite p-group of class 3 with p > 2, S  elementary abelian of order p 3 and S p  Z ( S ). It is important to observe that in both of these types if X is a proper non-abelian subgroup of S, then | X  | = p: this follows from [1, Proposition 137.6]. As a consequence all groups of Type 1 or 2 belong to D3 . It will be shown in the next two results that if G is a periodic nilpotent D3 -group and G  is a p-group of order greater than p 2 , there is a subgroup S of one of the above types that is involved in the group in an essential way, namely as a supplement of the center. Lemma 7. Let G be a p-group of nilpotent class 2, with G  elementary abelian of order p 3 . Assume that G ∈ D3 . Then G = Z (G ) S where S is a p-group of Type 1 above, i.e., S is 3-generator with nilpotent class 2 and S  is elementary abelian of order p 3 . Proof. Note that G p  Z (G ). Furthermore, for any non-abelian subgroup X of G we have X  = G  or | X  | = p by Lemma 3. First we make two easy observations. (i) If a, x, s ∈ G and [a, s] = 1, [a, x] = 1, then [s, x] ∈ [a, x]. For a, s, x = [a, x], [s, x] and G cannot have a derived subgroup order p 2 . Therefore |[a, x], [s, x]| = p and [s, x] ∈ [a, x]. (ii) If a, x, v , w ∈ G and [ v , a] = [ w , a] = [ v , x] = [ w , x] = 1 and [a, x] = 1, then [ v , w ] ∈ [a, x]. For  v , w , a, x = [ v , w ], [a, x] cannot have order p 2 , so its order is p and [ v , w ] ∈ [a, x]. Notice that for any a ∈ G we have |G : C G (a)|  |G  | = p 3 . Let us assume for the moment that |G : C G (a)| = p 3 for all a ∈ G \ Z (G ). It will be shown that |G / Z (G )|  p 4 . To establish this let a ∈ G \ Z (G ). Then G   Z (G )  C G (a) G and thus G /C G (a) is elementary abelian of order p 3 . Write













G /C G (a) = xC G (a) × yC G (a) × zC G (a)

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for some x, y , z ∈ G, so that G = C G (a)x, y , z. Now a, x, y   [a, x], [a, y ] and [a, x] ∩ [a, y ] = 1. Hence













G  = a, x, y  = [a, x] × [a, y ] × [x, y ] , since G  has order p 3 . Let s ∈ C G (a); then [s, x] ∈ [a, x] by (i) above. Hence [s, x] = [ai , x] for some integer i and sa−i ∈ C G (x) ∩ C G (a). It follows that s ∈ (C G (x) ∩ C G (a))a and G = (C G (x) ∩ C G (a))a, x, y , z. For any v ∈ C G (a) ∩ C G (x) we have by (i) above that [ v , y ] ∈ [a, y ] and [ v , y ] ∈ [x, y ] by (i)  1. Therefore [ v , y ] = 1 and similarly [ v , z] = 1. Thus C G (a) ∩ C G (x)  above, since [a, y ], [x, y ] = C G (a, x, y , z). In addition, if v , w ∈ C G (a) ∩ C G (x), then [ v , w ] ∈ [a, x] by (ii) above, and also [ v , w ] ∈ [a, y ]. Hence [ v , w ] = 1 and C G (x) ∩ C G (a) is abelian. It follows that C G (a) ∩ C G (x)  Z (G ) and therefore G = Z (G )a, x, y , z, so that |G / Z (G )|  p 4 , as claimed. As a consequence there is a finite subgroup T such that G = Z (G ) T . Hence G  = T  , Z ( T ) = Z (G ) ∩ T and | T : C T (a)| = |G : C G (a)| = p 3 for any a ∈ T \ Z ( T ). This means that T is a generalized Camina group in the sense of Lewis [5]. By Lemma 2.8 of [5] the subgroup T is isoclinic to a Camina group C of class 2, which implies that T   C  and T / Z ( T )  C / Z (C ). But then d(C  ) = 3 and d(C / Z (C ))  4, which is impossible by a result of I.D. Macdonald [7, Theorem 3.2]. By this contradiction there exists a ∈ G \ Z (G ) such that |G : C G (a)| = p or p 2 .  1. Suppose first that |G : C G (a)| = p. Then G /C G (a) is cyclic and G = C G (a)x where [a, x] = For any s ∈ C G (a) we have [s, x] ∈ [a, x] by (i), so [s, x] = [ai , x] for some integer i and sa−i ∈ C G (x) ∩ C G (a), which shows that G = (C G (x) ∩ C G (a))a, x. Moreover, if v , w ∈ C G (x) ∩ C G (a), then [ v , w ] ∈ [a, x] by (ii). But this implies that G  = ((C G (x) ∩ C G (a))a, x) = [a, x], which has order p, a contradiction. It follows that |G : C G (a)| = p 2 and G /C G (a) = xC G (a) ×  yC G (a). Thus we may write G = C G (a)x, y . Now a, x, y   [a, x] × [a, y ] and thus













G  = a, x, y  = [a, x] × [a, y ] × [x, y ] . Now let s ∈ C G (a). Then [s, x] ∈ [a, x] by (i) and sa−i ∈ C G (x) for some integer i. Hence s ∈ (C G (x) ∩ C G (a))a and G = (C G (x) ∩ C G (a))a, x, y . For any v ∈ C G (x) ∩ C G (a) we have [ v , y ] ∈ [a, y ] and [ v , y ] ∈ [x, y ] by (i), so [ v , y ] = 1 and C G (x) ∩ C G (a) centralizes a, x, y . Finally, if v , w ∈ C G (x) ∩ C G (a), then from (ii) we obtain [ v , w ] ∈ [a, x] ∩ [a, y ] = 1. Therefore C G (x) ∩ C G (a)  Z (G ) and G = Z (G )a, x, y . Finally set S = a, x, y , which has the required structure 2 Lemma 8. Let G be a p-group of nilpotent class 3, with G  elementary abelian of order p 3 and p > 2. Assume that G ∈ D3 . Then G = Z (G ) S where S is a p-group of Type 2 above, i.e., S is 2-generator with nilpotent class 3, S  is elementary abelian of order p 3 and S p  Z ( S ). Proof. First notice that G p  Z (G ) by Lemma 6 since |G  | = p 3 . Furthermore, X  = G  or | X  | = p for any non-abelian subgroup X of G by Lemma 3. Suppose that G is a 2-Engel group; then for any a, x, y ∈ G we have













a, [a, y ], x = [a, x], [a, y , x] = [a, x], [a, x, y ] < G  ,

and a, [a, y ], x has order dividing p. Since G has class 3, there exist a, x, y ∈ G such that [a, x, y ] =  1. But then [a, x] = 1 and hence [a, x, y ] = [a, x], which implies that [a, x] = 1, a contradiction. Consequently G is not 2-Engel and there exist a, x ∈ G such that [a, x, x] = 1. Thus a, x = [a, x], [a, x, x], [a, x, a] cannot have order p and hence a, x = G  . Therefore













G  = [a, x] × [a, x, x] × [a, x, a] since |G  | = p 3 .

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Next we prove that C G ([a, x]) = C G (G  ) = Z 2 (G ). In the first place it is clear that Z 2 (G )  C G (G  )  C G ([a, x]); moreover C G ([a, x]) = C G (G  ) since [a, x, x] and [a, x, a] belong to Z (G ). Let s ∈ C G (G  ); then for any y ∈ G, we have s, y  = [s, y ], [s, y , y ], which cannot equal G  and so has order 1 or p.  1, it follows that [s, y ] = [s, y , y ] and [s, y , y ] = 1. By this contradiction [s, y , y ] = 1, If [s, y , y ] = i.e., s is a right 2-Engel element of G. It follows from the properties of such elements (see [9, 7.13]) that [s, y , z]2 = [s, [ y , z]] = 1 for any y , z ∈ G. Hence [s, y , z] = 1 since p > 2 and s ∈ Z 2 (G ), as required. Now we prove that Z 2 (G ) is abelian. First we observe that ( Z 2 (G )x) = [a, x, x] and ( Z 2 (G )a) = [a, x, a]. Indeed in the first place it is obvious that [a, x, x] ∈ ( Z 2 (G )x) . Moreover ( Z 2 (G )x)  C G (x) since Z 2 (G )x is nilpotent of class  2. Hence ( Z 2 (G )x)  C G (x) ∩ G  = [a, x, a], [a, x, x].  1, it follows that ( Z 2 (G )x) = Since there are no derived subgroups of order p 2 and [a, x, x] = [a, x, x]. By a similar argument ( Z 2 (G )a) = [a, x, a]. Therefore Z 2 (G )  [a, x, x] ∩ [a, x, a] = 1 and Z 2 (G ) is abelian. The final step in the proof involves showing that G = Z (G )a, x. Let v ∈ G; then [a, x, v ] ∈ G  ∩ Z (G ) = [a, x, x] × [a, x, a], and thus

  [a, x, v ] = [a, x, x]i [a, x, a] j = a, x, xi a j for some integers i , j. Hence v (xi a j )−1 ∈ C G ([a, x]) = Z 2 (G ) and v ∈ Z 2 (G )a, x, showing that G = Z 2 (G )a, x. Now let c ∈ Z 2 (G ); then [c , a] ∈ ( Z 2 (G )a) = [a, x, a] and thus [c , a] = [[a, x]i , a] for some integer i. Therefore c [a, x]−i ∈ C G (a) ∩ Z 2 (G ) and c ∈ (C G (a) ∩ Z 2 (G ))[a, x], which implies that G = (C G (a) ∩ Z 2 (G ))a, x. Now let s ∈ C G (a) ∩ Z 2 (G ); then s, [a, x], ax = [s, x], [a, x, ax], which has order dividing p. Note that [a, x, ax] = [a, x, x][a, x, a] = 1. Hence

        [s, x] ∈ [a, x, ax] ∩ Z 2 (G )x = [a, x, ax] ∩ [a, x, x] = 1 and s ∈ C G (x). Consequently,

C G (a) ∩ Z 2 (G )  C G (x) ∩ C G (a) ∩ Z 2 (G ) = Z (G ) since G = (C G (a) ∩ Z 2 (G ))a, x and Z 2 (G ) is abelian. It follows that G = Z (G )a, x. Finally, set S = a, x and observe that S has the required properties. 2 We are now able to state the definitive result on periodic nilpotent D3 -groups. Theorem 2. Let G be a non-abelian periodic nilpotent group. Then G ∈ D3 if and only if one of the following situations holds for some prime p: (a) |G  | = p or p 2 ; (b) G = Z (G ) S where S is a finite p-group of Type 1, i.e., S is 3-generator with nilpotent class 2 and S  is elementary abelian of order p 3 ; (c) G = Z (G ) S where S is a finite p-group of Type 2, i.e., S is 2-generator with nilpotent class 3, p > 2, S  is elementary abelian of order p 3 and S p  Z ( S ). Proof. Let G be a non-abelian periodic nilpotent group. If G ∈ D2 , then |G  | = p for some prime p by [6, Theorem 1]. Let G ∈ D3 \ D2 ; then G  is a p-group for a prime p and the p  -elements lie in Z (G ). Thus we can assume that G is a p-group: now the validity of one of the conditions follows from Lemmas 6–8. As for the converse, if G satisfies (a), obviously G ∈ D3 . Assume that either (b) or (c) holds. By Lemma 1 it suffices to prove that S ∈ D3 . Let X be a proper, non-abelian subgroup of S. Then X  has order p by [1, Proposition 137.6], which shows that there are three isomorphism types of derived subgroups and S ∈ D3 , as required. 2

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4. Non-nilpotent, locally finite D3 -groups: basic results We continue the classification of locally finite D3 -groups by showing in the remaining sections how to divide the non-nilpotent groups into seven different types, giving an overall total of nine. The following result is fundamental for the entire theory. Proposition 2. If G is a locally finite D3 -group, then G  is a finite p-group of nilpotent class at most 2 for some prime p. Proof. Recall from Proposition 1 that G  is finite. It is clearly sufficient to prove the result when G is finite, so we assume this. (i) The first step in the proof is to show that G  is nilpotent. Assume this is false and note that by Lemma 2 the group G is soluble with derived length 3. Hence there are distinct derived subgroups 1 < G  < G  and G  is abelian. Since [G  , G  ] = 1, there is a prime p such that [ P , G  ] = 1 where P is the p-component of G  : write G  = P × Q with Q a p  -group. There is an x ∈ G  such that [ P , x] = 1 and x, P  = [ P , x] is a p-group. Since [ P , x]  G  and G ∈ D3 , it follows that [ P , x] = G  and Q = 1. Hence G  is a p-group. Let C = C G (G  ). If x is a p-element of G, then x, G   is a p-group and [G  , x] < G  . Consequently [G  , x] = 1 and x ∈ C . Thus C contains all the p-elements of G. Next we prove that the p-elements form a subgroup of C , and here we may assume that C is not abelian. If C   G  , then C  = G  and G   C = C G (G  ), which implies that G  is nilpotent. By this contradiction C  = G  , which is a p-group: this establishes the assertion. It follows that there is a unique Sylow p-subgroup P of G. Let F be a finite subgroup of G. By the Schur–Zassenhaus Theorem F = X  ( F ∩ P ) where X is a p  -group. Since X   G  or G  , it follows that X  = 1 and F   F ∩ P , which is a p-group. Therefore G  is a p-group, so it is nilpotent. (ii) Next it will be proved that G  is a p-group for some prime p. Assume this to be false and consider first the case where G is nilpotent. Let p ∈ π (G  ) and write G = P × Q where P = G p and Q = G p  . Then G  = P  × Q  , so Q  = 1 and G  is a p-group. Now assume that G is non-nilpotent and suppose that G  is not of prime power order. Since G   Z (G ), there is a prime p such that P = (G  ) p  Z (G ) and thus [ P , x] = 1 for some x ∈ G. Now x, P  = [ P , x] P  is non-trivial and it cannot be isomorphic with G  since the latter is not a p-group. There is a prime q = p dividing |G  |: put Q = (G  )q . For any y ∈ G, we have  y , Q  = [ Q , y ] Q  , a q-group, so it cannot be isomorphic with x, P  or G  . Hence  y , Q  = 1 and Q  Z (G ). Consequently G / P is nilpotent and the p-elements in G form a subgroup. Arguing as above, we show that F  is a p-group for any finite subgroup F , and hence G  is a p-group.  1 for some (iii) Finally, we show that G  has class at most 2. Suppose this is false and [G  , x] = x ∈ G  . Then 1 = x, G   = [G  , x] < G  since x, G   is nilpotent. This is the final contradiction. 2 The next result provides critical structural information about locally finite D3 -groups. Proposition 3. Let G be a locally finite D3 -group. Then G = X  P where P = O p (G ) is nilpotent and X is an abelian p  -group for some prime p. Moreover X /C X ( P ) is finite. Proof. Proposition 2 shows that G  is a finite p-group for some prime p. Hence G   P = O p (G ) and G / P is an abelian p  -group: moreover P is nilpotent since P  is finite. We argue by induction on |G  | that G splits over P , and here we can assume that G  = 1. Let M be a minimal normal subgroup of G contained in G  . By induction hypothesis there exists Y  G such that G = Y P and Y ∩ P = M. It is enough to show that Y splits over M. If [ M , Y ] = 1, then Y = M × Y p  . Otherwise [ M , Y ] = M, from which it follows that Y splits over M by a simple argument with centralizers (cf. [8, Theorem 3.3]). Finally, let D = C X (G  ). Then X / D is finite and [ P , D , D ] = 1. Let |G  | = pk . For any d ∈ D we have

1 = [ P , d] p = [ P , d p ]. Since D is a p  -group, it follows that [ P , d] = 1 and D = C X ( P ). Therefore X /C X ( P ) is finite. 2 k

k

The next result is well known and will be an essential tool here.

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Lemma 9. Let A be an abelian group and Γ a group of automorphisms of A with finite order m. If A is uniquely divisible by m, then

A = [ A , Γ ] × C A (Γ ).



Proof. For convenience we include a short proof. Define θ : A → C A (Γ ) by aθ = γ ∈Γ aγ ; then θ is a homomorphism. Note that θ|C A (Γ ) is an isomorphism since it maps a to am , while A, and hence C A (Γ ), is uniquely divisible by m. Thus Im(θ) = C A (Γ ) and Im(θ) = Im(θ 2 ). It follows that A = Im(θ)Ker(θ) = C A (Γ ) × Ker(θ). Next for any a ∈ A we have aθ ≡ am (mod [ A , Γ ]). Thus aθ = 1 if and only if am ∈ [ A , Γ ], which by unique divisibility implies that a ∈ [ A , Γ ]. Therefore Ker(θ) = [ A , Γ ]. 2 This lemma will be applied in the context of a locally finite D3 -group G = X  P where P = O p (G ), with A = P / P  and Γ = X /C X ( P ): note that Γ is a finite abelian p  -group by Proposition 3. Thus we obtain

P / P  = P 1 P / P  × C / P , where

P1 = [P , X]

and

C / P  = C P / P  ( X ).

This notation will be maintained throughout the remainder of the article. Notice that P 1 G and C G. Assume now that G is not nilpotent, so that P 1 = 1. Since G  = P 1 P  , we have P   G  : for otherwise P  = G  and [ P , X ]  P  , which implies that [ P , X ] = 1 since P is nilpotent and X is a finite p  -group. If P is non-abelian, there are three non-isomorphic derived subgroups

1 < P  < G  = P 1 P , and thus any derived subgroup of G must be isomorphic with one of them. At this point it is necessary to split up the discussion according to the degree of commutativity present in P . 5. The case: P = O p ( G ) abelian Continuing the notation established in the preceding section, we consider what happens when P = O p (G ) is abelian. Then by Lemma 9 we have P = P 1 × C where P 1 = [ P , X ] and C = C P ( X ). Since G = X  P , we also have C  Z (G ) and therefore

G = ( X P 1) × C . In the light of Lemma 1, we see that it suffices to describe the structure of the D3 -group X P 1 , the only difference from G being the abelian direct factor C , which is plainly arbitrary. Thus there is no real loss in assuming that C = 1 and P = P 1 = [ P , X ] = G  . The first step in the classification is to describe the structure of P . Lemma 10. If P = [ P , X ] is abelian, then it is finite and either elementary or homocyclic of exponent p 2 . Proof. Assume that P is not elementary, so that P > P p > 1. Now  X , P  = [ P , X ] = P and

 X , P p  = [ P p , X ] = P p . A count of derived subgroups reveals that P p = 1 and there are nonisomorphic derived subgroups 1 < P p < P . From Lemma 9 we obtain P [ p ] = [ P [ p ], X ] × C P [ p ] ( X ) = [ P [ p ], X ], since C P ( X ) = 1, and thus ( X P [ p ]) = P [ p ]. Since P p  P [ p ], it follows that P p = P [ p ] and P is homocyclic. 2 i

i

i

2

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5.1. The sub-case: P elementary abelian We consider next the case where P = P 1 is an elementary abelian group of order pn . Here the X -module structure of P is critical. Suppose first that P is a simple X -module. Then X¯ = X /C X ( P ) is cyclic, say of order m with X¯ = ¯x. Note that r ( P ) = n = | p |m , the order of p modulo m. Should it be the case that P is a simple Y -module for every non-trivial subgroup Y of X , then P will be called strongly X -simple. In the opposite direction, it follows from results in [6] (see Theorem 2) that if P is strongly simple, then G ∈ D2 . We must therefore investigate the situation when P is not strongly X -simple. First we assume that P is X -simple. Let 1 = y¯ ∈ X¯ with y¯ = yC X ( P ), and let S be a simple  y¯ -submodule of P . Then  y , S  = [ S , y ] = S since C P ( y ) = 1 by X -simplicity of P . If P is not a simple  y -module, we have P = S 1 × S 2 × · · · × S k where the S i are  y¯ -simple submodules and k  2. However, if k > 2, there will be at least three non-trivial, non-isomorphic derived subgroups S 1 , S 1 × S 2 , P . Therefore k = 2 and also S 1  S 2 . Hence n = r ( P ) = 2r ( S i ) and r ( S i ) = | p || y¯ | = 1 n = 12 | p |m . Consequently, if d > 1 is a divisor of m, then | p |d = | p |m = n if P is xm/d -simple and 2

| p |d = 12 | p |m = 12 n if P is not xm/d -simple. Since P is a faithful simple X¯ -module, we can identify P with the additive group F + of a field F of order pn and X¯ with a subgroup of the multiplicative group F ∗ with order m. The advantage of this identification is that it furnishes us with a method for constructing D3 -groups of this type. 5.1.1. Construction Let p be a prime and m a positive integer not divisible by p. Write n = | p |m and let F be a field of order pn . Then F ∗ has a (cyclic) subgroup X of order m. Regard A = F + as an X -module via the field multiplication and form the semidirect product

G ( p , m) = X  A . We will say that the pair ( p , m) is 2-allowable if, for any d > 1 dividing m, it is true that

| p |d = pm or

1 2

| p |m .

This is a generalization of the “allowable pairs” introduced in [6] (Section 3). The connection with the property D3 is shown in: Proposition 4. The group G ( p , m) is a D3 -group if and only if ( p , m) is a 2-allowable pair. Proof. Let G = G ( p , m). Note that G  = A = O p (G ) is a simple X -module. If G ∈ D3 , the preceding discussion shows that ( p , m) is a 2-allowable pair. To establish the converse assume that ( p , m) is 2-allowable and let H  G be non-abelian. Thus H = xa, H ∩ A  where 1 = x ∈ X , a ∈ A and H ∩ A = 1. If A  H , then H = x, A  and H  = [ A , x] = A  Z pn , so we assume that H ∩ A < A. Note that r ( A ) = n = | p |m . If | p ||x| = n, then A is a faithful simple x-module and, since H ∩ A is an x-submodule, it coincides with A. Since this is incorrect, | p ||x| must equal 12 n and A = A 1 × A 2 where A i is a 1

n

x

simple x-module and r ( A i ) = | p ||x| = 12 n; thus A i  Z p2 . Hence H ∩ A  A i , where i = 1 or 2, and 1 2n

H ∩ A = [ H ∩ A , x] = [ H ∩ A , xa]  H  . Thus H  = H ∩ A  A i and H   Z p , so that G ∈ D3 .

2

We turn next to the case where P is elementary abelian, but it is not X -simple. If S is a simple X -submodule of P , we have  X , S  = [ S , X ] = S: a count of derived subgroups reveals that P = S 1 × S 2 where S i is a simple X -submodule and S 1  S 2 . Thus r ( P ) = n = 2r ( S i ) and n is even. We claim that S i is even strongly X -simple. To see this let x ∈ X \C X ( S i ) and suppose that T is a proper non-trivial x-submodule of S i . Then 1 < x, T  = [ T , x] = T < S i < G  since C S i (x) = 1. This is impossible.

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Put C i = C X ( S i ) and X¯ i = X /C i . Then X¯ i is cyclic of order mi where | p |mi = r ( S i ) = 12 n. Since S i is strongly X -simple, ( p , mi ) is an allowable pair. Notice that C 1 ∩ C 2 = C X ( P ) = 1, so that X¯ is a subdirect product of X¯ 1 and X¯ 2 . This completes our analysis of the case where P is elementary abelian. 5.2. The sub-case: P not elementary abelian Assume next that P is abelian, but not elementary abelian; then Lemma 10 shows that P is homocyclic of exponent p 2 . (We continue to assume that C P ( X ) = 1). Now  X , P p  = [ P p , X ] = [ P , X ] p = P p < P . A count of derived subgroups shows that P p must be strongly X -simple, and beX

cause P / P p  P p , the same holds for P / P p . Our conclusions about the structure of G when P is abelian are summed up in the following theorem. Theorem 3. Let G be a locally finite, non-nilpotent D3 -group with P = O p (G ) abelian. Then, up to an abelian direct factor, G = X  P with X an abelian p  -group and P = [ P , X ] is finite. In addition one of the following holds. (i) P is elementary abelian and a simple X -module: also G /C X ( P )  G ( p , m) where ( p , m) is 2-allowable. (ii) P is elementary abelian and P = S 1 × S 2 where S i is a strongly simple X -module and S 1  S 2 . X

(iii) P is homocyclic of exponent p 2 and P p  P / P p is strongly X -simple. 6. The case: P = O p ( G ) non-abelian We begin by recording basic information on the structure of the finite subgroup P 1 = [ P , X ]. As before G = X  P , and X is an abelian p  -group. It will be seen that the structure of P 1 is simpler than that of locally finite p-groups in D3 in general. Lemma 11. (i) C X ( P 1 P  / P  ) = C X ( P 1 ) = C X ( P ). (ii) If P is non-abelian and P 1 = 1, then | P  | = p. (iii) If P 1 is non-abelian, P 1 = P   P 1 and G  = P 1 . Proof. Let x ∈ C X ( P 1 P  / P  ); then [ P , x] = [ P 1 C , x]  P  since [C , X ]  P  . Thus x ∈ C X ( P / P  ), which implies that [ P , x] = 1 since xC X ( P 1 ) has p  -order. Hence C X ( P 1 / P  )  C X ( P ), while the inclusions C X ( P )  C X ( P 1 )  C X ( P 1 P  / P  ) are obvious. Now assume that P is non-abelian and P 1 = 1. Then there exist x ∈ Z 2 ( P )\ Z ( P ) such that x p ∈ Z ( P ), and y ∈ P such that [x, y ] =  1. Hence 1 = [x p , y ] = [x, y ] p and x, y  = [x, y ] has order p. If P  = G  , then [ P , X ]  P  and P 1 = [ P , X ] = 1. Hence P   G  . Also we have 1 = x, y   P  < G  , so it follows that P  = [x, y ] and | P  | = p. The remaining statements in (iii) are obvious. 2 Next we address the X -module structure of P 1 / P  . Lemma 12. Assume that P 1 is non-abelian. Then the following are true. (i) If x ∈ X \C X ( P 1 ), then C P 1 / P  (x) = 1. (ii) If | P 1 | > p 3 , then P 1 / P  is a strongly simple X -module. Proof. (i) We have x, P 1  = P 1 [ P 1 , x]  P  . If [ P 1 , x]  P  , then Lemma 11(i) shows that [ P , x] = 1. Hence x, P 1  > P  and x, P 1  = G  = P 1 by Lemma 11 again; thus P 1 = P  [ P 1 , x]. Hence [ P 1 / P  , xP  ] = P 1 / P  , which implies that C P 1 / P  (x) = 1.

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(ii) Now assume that | P 1 | > p 3 . Let us suppose that the assertion in (ii) is false, so there exist A and Y ⊆ X \C X ( P 1 ) such that P  < A < P 1 and A Y = A. By (i) we have 1 = (Y A ) = A  [ A , Y ]  A < P 1 = G  . Hence, if A  = 1, then P  = A   (Y A ) , so that P  = (Y A ) : thus [ A , Y ]  P  , in contradiction to (i). Therefore A  = 1 and (Y A ) = [ A , Y ]  P  and |[ A , Y ]| = p. Since A is abelian, A = [ A , Y ] × C A (Y ) by Lemma 9. Moreover C A (Y ) = 1 since otherwise | A | = p and A = P  . But C A (Y )  P  by (i), so C A (Y ) = P  and | A | = p 2 . By Maschke’s Theorem there is a direct decomposition of P 1 / P  into simple Y -modules. By a count of derived subgroups we see that there are just two such submodules and thus | P 1 | = p 3 , a final contradiction. 2 Even more precise information about the structure of the subgroup P 1 is provided in the next result. Lemma 13. If P 1 is non-abelian, it is an extra-special p-group and if p > 2, the exponent of P 1 is p. Proof. To prove that P 1 is extra-special it is enough by Lemma 11 to show that Z ( P 1 )  P  . Now if | P 1 | = p 3 , this is true since P 1 / Z ( P 1 ) cannot be cyclic. Assume that | P 1 | > p 3 . Then Lemma 12 shows that P 1 / P  is a simple X -module, so Z ( P 1 ) = P  since P 1  Z ( P 1 ) < P 1 . p Next assume that p > 2 and P 1 = 1. First suppose that P 1 / P  is a simple X -module. The elements with order dividing p form an X -invariant subgroup S satisfying P   S < P 1 , so S = P  . But this cannot happen in an extra-special p-group. It follows that P 1 / P  is not a simple X -module and thus | P 1 | = p 3 by Lemma 12(ii). Hence there is an X -invariant decomposition P 1 / P  = a P  ×b P  . Let x ∈ X \C X ( P 1 ); then there are relations ax = ai c, b x = b j d where i , j ≡ 0 (mod p ) and c , d ∈ P  . Suppose that |a| = p 2 . Now (a p )x = (ax ) p and a p ∈ P  ; therefore (a p )i j = (ai c ) p = a pi since [a, b]x = [a, b]i j . It follows that j ≡ 1 (mod p ) and [b, x] = d ∈ P  , in contradiction to Lemma 12(i). Hence |a| = p and in the same way |b| = p. Since p > 2, it follows that P p = 1. 2 The structure of G is affected markedly by whether or not C is commutative (where C / P  = C P 1 / P  ( X )). Lemma 14. Assume that P 1 and C are non-abelian. Then the following hold. (i) [C , X ] = 1 = [C , P 1 ] and P = P 1  C , the central product with P  amalgamated. (ii) P 1 / P  is strongly X -simple.

P

Proof. By hypothesis C  = 1, so C  = P  . Thus [C , X ] = 1 since [C , X ]  P  = C  . Hence [ X , C , P ] = 1 = [C , P , X ], and it follows by the Three Subgroup Lemma that [ P , X , C ] = 1, i.e., [ P 1 , C ] = 1. Therefore P = P1  C. P

Note that P   P 1 by Lemma 11. Suppose that P 1 / P  is not strongly X -simple, so there exist x ∈ X \C X ( P 1 ) and P  < S < P 1 with S = S x . Note that SC is x-invariant and

x, SC  = ( SC ) [ SC , x] = S  C  [ S , C ][ S , x][C , x]. Now C  = P  and [C , x]  P  . Hence x, SC  = P  [ S , x]  S < P 1 = G  . Consequently P  [ S , x] = P  and [ S , x]  P  . By Lemma 12(i) this is impossible. 2 Lemma 15. Assume that P 1 is non-abelian and C is abelian. The one of the following is true. (i) C  Z (G ) and G = ( X P 1 )  C . P

(ii) [ P  , X ] = P  and G = Y  P 1 where Y = X C C ( X ) is abelian.

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Proof. Assume first that [C , X ] = 1. Then [C , X , P ] = 1 = [ P , C , X ], so [ P , X , C ] = 1, i.e., [ P 1 , C ] = 1. Hence [ P , C ] = [ P 1 C , C ] = 1 and C  Z (G ). Note that G = ( X P 1 )C and ( X P 1 ) ∩ C = P  , so we are in situation (i). Now suppose that [C , X ] = 1. Since [C , X ]  P  , it follows that [C , X ] = P  and by Lemma 9 we have C = P  × K where K = C C ( X ). Hence P  = [C , X ] = [ P  , X ]. Also G = X P = X P 1 P  K = ( X K ) P 1 . Notice that Y = X K is abelian and G = Y  P 1 since Y ∩ P 1 = K ∩ P 1 = K ∩ P 1 ∩ P  = 1. Thus (i) holds. 2 The essential group structure when P 1 has order exceeding p 3 has now been uncovered. However, the case where | P 1 | = p 3 is more delicate and requires special arguments. Indeed, if P 1 is a nonabelian group of order p 3 , the X -module P¯ 1 = P 1 / P  need not be X -simple and further analysis is called for. Let us write X¯ for X /C X ( P 1 ). First of all consider the situation when | P 1 | = p 3 and P¯ 1 is X -simple, but not strongly X -simple. Since P¯ 1 is a faithful simple X¯ -module, it may be identified with a field F = GF ( p 2 ) and X¯ with a subgroup of F ∗ . The X -module structure of P¯ 1 is then realized by the field multiplication. Lemma 16. Assume that P 1 is non-abelian of order p 3 and P¯ 1 = P 1 / P  is X -simple, but not strongly X -simple. Then p > 2 and X¯ = X /C X ( P 1 ) is cyclic and acts on P¯ 1 = GF ( p 2 ) via the field multiplication as a subgroup of GF ( p 2 ) of order 2m such that X¯ ∩ GF ( p ) = −1 and 1 < m | p + 1. Proof. Note that X¯ is cyclic since P¯ 1 is X -simple, say X¯ = xC X ( P 1 ). By hypothesis X¯ has a nontrivial subgroup Y¯ =  yC X ( P 1 ) of maximum order such that P¯ 1 is not Y¯ -simple. Then there exists a¯ = a P  ∈ P¯ 1 of order p such that ¯a is Y -invariant; thus a¯ y = a¯ i say. Now b¯ = a¯ x ∈ / ¯a and hence P¯ 1 = ¯a × b¯  since | P¯ 1 | = p 2 . Also b¯ y = a¯ xy = (¯ai )x = b¯ i . Hence y operates on P¯ 1 by raising each element to the ith power. Notice that p > 2 and i ≡ 0 or 1 (mod p ). Writing P  = c  where c = [a, b], 2

we have c x = c i . We show next that i ≡ −1 (mod p ): suppose this is false. Let d ∈ P 1 \ P  ; then d y = di c r where r ∈ Z. Observe that  y , d = [d, y ] y   c , d < P 1 = G  since P¯ 1 is not cyclic. It follows that  y , d  P   Z p and hence that [d, y ] is  y -invariant. Now

y  2 [d, y ] y = di −1 cr = di (i −1) cr (i +i −1) , which must equal [d, y ]i = di (i −1) c ri since d ∈ / P  and d p = 1 by Lemma 13. If r ≡ 0 (mod p ), it follows that i 2 + i − 1 ≡ i (mod p ) and hence i ≡ −1 (mod p ). By this contradiction r ≡ 0 and d y = di for i 

all d ∈ P 1 \ P  . But (ab) y = a y b y yields (ab)i = ai b i and [b, a] 2 = 1, a contradiction which shows that i ≡ −1 (mod p ). Returning to the identification of P¯ 1 with GF ( p 2 ), we conclude that Y¯ = X¯ ∩ GF ( p ) = −1 and | X¯ | is even, say | X¯ | = 2m. Then 2m | p 2 − 1 and m | p + 1. 2 Next we address the situation when P¯ 1 / P  is not X -simple. Lemma 17. Assume that P 1 is non-abelian of order p 3 and P¯ 1 is not a simple X -module. Then X¯ = X /C X ( P 1 ) is cyclic, say X¯ = xC X ( P 1 ). Moreover p > 2 and there exist a, b such that P 1 = a, b, P   and ax = ai , b x = b j where i j ≡ 1 (mod p ). Proof. By hypothesis there is a decomposition P¯ 1 = a P   × b P   into X -invariant factors. Let P  = c  where c = [a, b]. If x ∈ X \C X ( P 1 ), then ax = ai c r and b x = b j c s , where i , j ≡ 0 or 1 (mod p ): thus c x = c i j . Clearly p > 2. Next we show that i j ≡ 1 (mod p ). Notice that 1 = x, a, c  = [a, c , x]x  a, c  < P 1 . Hence x, a, c   P   Z p , so [a, c , x] is cyclic. Now [a, c , x]x = ai −1 cr , c i j −1  and |ai −1 cr | = p. Consequently c i j −1 ∈ ai −1 cr  and c i j −1 = (ai −1 cr )k for some k. Hence a(i −1)k ∈ P  and k ≡ 0 (mod p ). It follows that c i j −1 = 1 and i j ≡ 1 (mod p ). The automorphism of P¯ 1 induced by x is therefore determined by i (mod p ) ∈ Z p . Since this is valid for each x ∈ X \C X ( P 1 ), we deduce that X¯ is cyclic, equal to xC X ( P 1 ) say.

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As before write ax = ai c r and b x = b j c s where i j ≡ 1 (mod p ). Define t ≡ (i − 1)−1 (mod p ) and let

a = ac rt ,

b = bc st .

Then (a )x = ai c r +rt since c x = c. Now

(r + rt )(i − 1) ≡ r (i − 1) + r ≡ ri ≡ rit (i − 1) (mod p ), which shows that r + rt ≡ rit (mod p ) and (a )x = ai c rit = (ac rt )i = (a )i . By a similar argument (b )x = (b ) j . Thus by passing to new generators a , b , we may assume that ax = ai , b x = b j . 2 The final case to be analyzed, when P 1 but not P is abelian, can only occur when G  has small order. Lemma 18. Let G be a locally finite, non-nilpotent D3 -group with G = X  P , P = O p (G ) and P 1 = [ P , X ]. If P 1 is abelian but P is non-abelian, then |G  | has order dividing p 2 . Proof. In the first place the derived subgroups P  , ( X P 1 ) = [ P 1 , X ], G  = P 1 P  are all non-trivial, so two of them must be isomorphic. Clearly P   G  . Suppose that ( X P 1 )  P  ; then [ P 1 , X ]  Z p . Since P 1 is abelian, we can write P 1 = [ P 1 , X ] × C P 1 ( X ) by Lemma 9, and C P 1 ( X )  P 1 ∩ C  P  . Hence P 1  [ P 1 , X ] P  and | P 1 | divides p 2 . Now if | P 1 | = p, then |G  | = | P 1 P  | divides p 2 . If, on the other hand, | P 1 | = p 2 , it follows that P 1 = [ P 1 , X ] P  and P   P 1 . Hence |G  | = | P 1 | = p 2 . Next assume that ( X P 1 )  G  , i.e., [ P 1 , X ] = P 1 P  and thus P   P 1 . Hence P 1 = [ P 1 , X ] and G  = P 1 , so it is enough to prove that | P 1 | divides p 2 . Assume that P 1 is elementary abelian. By Maschke’s Theorem P 1 = P  × A 1 × · · · × A k where A i is a simple X -module and k  1. Since P 1 = [ P 1 , X ], we have P  = [ P  , X ] and A i = [ A i , X ]. Thus ( X P  A 1 ) = P  A 1  P  , which shows that P  A 1 = G  = P 1 , i.e., k = 1. Also ( X A 1 ) = A 1 < P 1 , so A 1  P   Z p . Therefore | P 1 | = p 2 . Finally, consider what happens when P 1 is not elementary and is therefore homocyclic of exponent p p 2 by Lemma 10, when applied to X P 1 . Now [ P 1 , X ] = 1 since P 1 = [ P 1 , X ], and



p 

X P1

p Hence P 1  P  and | P 1 | = p 2 .

 p  p = P 1 , X = [ P 1, X ]p = P 1 < P 1 = G .

2

We are now in a position to state the theorem which describes the group structure when P is a non-abelian group. Theorem 4. Let G be a locally finite, non-nilpotent D3 -group with P = O p (G ) non-abelian. Then G = X  P where X is an abelian p  -group and | P  | = p. In addition one of the following holds where P 1 = [ P , X ] and C / P  = C P / P  ( X ). (i) P 1 is an extra-special group, of exponent p if p > 2, and P 1 / P  is a strongly simple X -module. (ii) p > 2, P 1 is non-abelian with order p 3 and exponent p, C is abelian, and X¯ = X /C X ( P 1 )  GF ( p 2 )∗ acts on P 1 / P  = GF ( p 2 ) via the field multiplication as a subgroup of order 2m such that X¯ ∩ GF ( p ) = −1 and 1 < m | p + 1. (iii) p > 2, P 1 is non-abelian with order p 3 and exponent p, C is abelian, and X¯ = X /C X ( P 1 ) = xC X ( P 1 ), P 1 = a, b,  where ax = ai , b x = b j , i j ≡ 1 (mod p ) and i , j ≡ 0, 1 (mod p ). (iv) |G  | divides p 2 .

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Proof. By Proposition 3 we have G = X  P with X an abelian p  -group and | P  | = p by Lemma 11. If P 1 is abelian, Lemma 18 shows that |G  | divides p 2 and we are in situation (iv), so assume that P 1 is non-abelian; then P 1 = P  by Lemma 11. Lemma 13 shows that P 1 is extra-special, and P p = 1 when p > 2. If P¯ 1 / P  is strongly X -simple, we are in situation (i), so suppose this is not the case. Then by Lemma 12 we have | P 1 | = p 3 , while C is abelian by Lemma 14. Finally, apply Lemmas 16 and 17 and deduce that we are in situation (ii) or (iii) respectively. 2 7. Sufficiency We summarize what has been achieved thus far. The locally finite D3 -groups were characterized in the nilpotent case in Theorem 2, where the sufficiency of the conditions therein was also established. In the non-nilpotent case Theorems 3 and 4 listed the possible group structures. To complete the classification of locally finite D3 -groups it remains to establish the converses of Theorems 3 and 4. This is achieved in the following result. Theorem 5. All the types of group appearing in Theorems 3 and 4 are D3 -groups. The following general remark is helpful in proving that a group G = X  P with X abelian is a D3 -group. Since C X ( P ) G and G  ∩ C X ( P ) = 1, we have G   (G /C X ( P )) , so it suffices to prove that G /C X ( P ) is a D3 -group. For this reason we will assume in the following proof that C X ( P ) = 1. Proof of Theorem 5. (a) We begin by establishing the converse of Theorem 3. Let G = X  P be a group of the type allowed in Theorem 3. The claim is that G ∈ D3 . Let G satisfy condition (i). Since we are assuming that C X ( P ) = 1, we have G  G ( p , m) where ( p , m) is a 2-allowable pair. Thus G ∈ D3 by Proposition 4. Now suppose that G satisfies condition (ii) in Theorem 3 and let H  G be non-abelian. Then Y = X ∩ ( H P ) = 1 and H  is a non-trivial Y -submodule of P . Since P = S 1 × S 2 with S i strongly Y

X -simple, we deduce that H   S 1 or S 2 , or else H  = P . Since S 1  S 2 , it follows that G ∈ D 3 . Finally, assume G satisfies condition (iii): thus P is homocyclic of exponent p 2 . Again let H  G be non-abelian and set Y = X ∩ ( H P ) = 1. Now H  ∩ P p = 1 is a Y -submodule of P p = P [ p ], so P p  H  by strong X -simplicity of P p . Since P / P p is also strongly X -simple, H  = P p or P and G ∈ D3 . (b) Next we prove the converse of Theorem 4. Assume that G has the properties listed in this theorem, so that G = X  P where P = O p (G ), X is an abelian p  -group, P 1 = [ P , X ]. If condition (iv) is satisfied, i.e., |G  | = p 2 , then G has at most three isomorphism types of derived subgroup and G ∈ D3 . Thus it is conditions (i)–(iii) that remain to be dealt with: here P 1 is extra-special and P 1 = P  . Assume that G satisfies (i), so that P 1 / P  is strongly X -simple. We argue that G ∈ D3 . First observe that G / P  ∈ D2 : for G  = P 1 P  = P 1 and (G / P  ) = P 1 / P  . Then it is clear from strong X -simplicity that P 1 / P  is the only non-trivial derived subgroup in G / P  (cf. [6, Lemma 3 and Theorem 2]). If H  G, then ( H P  / P  ) = H  P  / P  = 1 or P 1 / P  , i.e., either H   P  or P 1 = H  P  = H  P 1 and hence H  = P 1 . Hence G ∈ D3 . Next assume condition (ii) of Theorem 4 holds. Thus p > 2, P 1 = [ P , X ] is non-abelian of order p 3 and exponent p: also P 1 = P  and C is abelian, where C / P  = C P / P  ( X ). We can assume that P 1 / P  is not strongly X -simple. Note that GF ( p 2 ) = GF ( p )( X ), so P 1 / P  is X -simple. Hence X = x is cyclic, since we are supposing that C X ( P 1 ) = 1. If H  G, then H = xr a, H ∩ P  where a ∈ P . We can assume that y = xr = 1 since otherwise H  P and H   P   Z p . If H ∩ P  C , then H  xr a, C  and again H   P  , so let H ∩ P  C . Since P = P 1 C , this implies that ( H ∩ P )C ∩ P 1  P  . Suppose first that P 1  ( H ∩ P )C . Now P 1 = [ P 1 , ya] P  since P 1 / P  is X -simple and y = 1. Therefore, as P  = P 1 ,





P 1 = [ P 1 , ya] P   ( H ∩ P )C , ya P   H  P  . Thus P 1 = ( P 1 ∩ H  ) P 1 and hence P 1 = P 1 ∩ H  . From this it follows that H  = P 1 .

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Next assume that ( H ∩ P )C ∩ P 1 < P 1 . Then ( H ∩ P )C ∩ P 1 / P  is a proper non-trivial  y -submodule of P 1 / P  , so the latter is not  y -simple. It follows from the module structure of P 1 / P  that y ∈ GF ( p ) and y must induce inversion in P 1 / P  . Also H  K :=  y , P . The argument of Lemma 17 may now be used to find a basis {a P  , b P  } of P 1 / P  such that a y = a−1 , b y = b−1 . We are therefore in the situation of part (iii) of Theorem 4, with y in place of x and i ≡ j ≡ −1 (mod p ). The sufficiency of condition (ii) will be established once we have dealt with (iii), by showing that either | H  | divides p or H  = P 1 . We are left with one final task, of establishing the sufficiency of the condition (iii) in Theorem 4. Here we have P 1 = a, b, X = x where ax = ai , b x = b j and i j ≡ 1 (mod p ). Thus if c = [a, b], then c x = c. Also P 1 = P  and C is abelian. Since [ P  , X ] = 1 and [C , X ]  P  , we have [C , X , X ] = 1 and hence [C , X ] = 1. By the Three Subgroup Lemma [ P , X , C ] = 1, i.e., [ P 1 , C ] = 1. Since C is abelian, it follows that C  Z (G ). Also G = ( X P 1 )C , so by Lemma 1 it is sufficient to prove that X P 1 ∈ D3 . Henceforth we assume that G = X P 1 , so C = P  and P = P 1 . Let H  G be non-abelian and write H =  yd, H ∩ P  where y = xr = 1 and d ∈ P . Thus H   y , P  r r and a y = ai , b y = b j , i r j r ≡ 1 (mod p ). We may replace G by  y , P  and prove that P  and P are the only non-trivial derived subgroups, which permits us to assume that r = 1 and H = xd, H ∩ P . We can also suppose that H ∩ P  P  since otherwise H is abelian. Thus ( H ∩ P ) P  / P  is a non-trivial X -submodule of P / P  . Consider first the case where i ≡ j (mod p ). Then ( H ∩ P ) P  = a, c , b, c  or P . In the last case P  H and H   [ P , xd] = P : for P = [ P , xd] P  since i , j ≡ 1 (mod p ). This implies that H  = P . Assume next that ( H ∩ P ) P  = a, c , the other case being similar. Then H  xd, a, c  and we may assume equality here, provided we prove that | H  | = p. In the first place [a, xd] = a−1 (ai )d and H  = [a, xd]xd . Writing d = au b v , we obtain [a, xd] = i −1 i v a c , and therefore

[a, xd]xd = ai (i −i ) c i (i −1) v +i v = ai (i −i ) c i

2

v

i  = ai −1 c i v = [a, xd]i .

It follows that H  = [a, xd], which has order p since P p = 1. It remains to deal with the case i ≡ j (mod p ). Since i j ≡ 1 (mod p ), this amounts to i ≡ j ≡ −1 (mod p ). Once again consider a non-abelian subgroup H , which, since x2 = 1, may be assumed to have the form xd, H ∩ P  with d ∈ P . If ( H ∩ P ) P  = P , then H  = P . Thus we may assume that ( H ∩ P ) P  / P  has order p and is generated by an element a bm P  . Hence H  xd, a bm , c  and we may assume equality and prove that | H  | = p. Thus



H  = a bm , xd

xd

   a  bm , c

since ( H ∩ P )xd = H ∩ P . Our aim is to prove that [a bm , xd]xd = [a bm , xd]−1 : for this implies that H  = [a bm , xd], which has order p. Writing d = au b v , we have







xd



 au b v = b−m a− a− b−m   = b−m a− a− b−m cmu − v = a−2 b−2m cmu − v −2m .

a bm , xd = b−m a− a bm

xau bm

Therefore

a bm , xd

au b v  = a2 b2m cmu − v −2m = a2 b2m c  v −mu −2m ,

while



a bm , xd

 −1

= b2m a2 c  v −mu +2m = a2 b2m clv −mu −2m ,

which gives the desired result and completes the proof of Theorem 5.

2

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