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On the Nilpotent Length of Polycyclic Groups JOURNAL OF ALGEBRA 203, 125]133Ž. 1998 ARTICLE NO. JA977321 On the Nilpotent Length of Polycyclic Groups Gerard Endimioni* C.M.I., Uni¨ersite de Pro¨ence, UMR-CNRS 6632, 39, rue F. Joliot-Curie, 13453 Marseille Cedex 13, France Communicated by J. Tits Received October 30, 1996 Let G be a polycyclic group. We prove that if the nilpotent length of each finite quotient of G is bounded by a fixed integer n, then the nilpotent length of G is at most n. The case n s 1 is a well-known result of Hirsch. As a consequence, we obtain that if the nilpotent length of each 2-generator subgroup is at most n, then the nilpotent length of G is at most n. A more precise result in the case n s 2 permits us to prove that if each 3-generator subgroup is abelian-by-nilpotent, then G is abelian-by-nilpotent. Furthermore, we show that the nilpotent length of G equals the nilpotent length of the quotient of G by its Frattini subgroup. Q 1998 Academic Press 1. INTRODUCTION AND RESULTS We say that a group G is n-step nilpotent if it has a subnormal series Ä41sG01eGe??? eGns G of length n with each quotient nilpotent. In fact, substituting the core of Ž. Giiin G for G , we can assume that Giis0,1,...,n is a normal series. The least integer n such that G is n-step nilpotent is the nilpotent length of G. First, we shall prove in this paper the following THEOREM 1. Let G be a polycyclic group. If there exists a positi¨e integer n such that each finite quotient of G is n-step nilpotent, then G is n-step nilpotent. For n s 1, this result is due to K. A. Hirschwx 3, 5.4.18 . Later it was improved in the following stronger form: if each finite quotient of a finitely generated soluble group G is nilpotent, then G is nilpotentwx 3, 15.5.3 . On * E-mail: [email protected]. 125 0021-8693r98 $25.00 Copyright Q 1998 by Academic Press All rights of reproduction in any form reserved. 126 GERARD ENDIMIONI the other hand, thanks to an example due to the referee, we shall see that Theorem 1 above is false for n s 2 with ``polycyclic'' replaced by ``finitely generated soluble.'' Theorem 1 permits us to transfer properties of finite groups to polycyclic groups. For instance, it follows fromwx 2, 4 that a finite group is n-step nilpotent if all its 2-generator subgroups are n-step nilpotent. Therefore, as a consequence of the theorem, we obtain: COROLLARY. If there exists a positi¨e integer n such that e¨ery 2-generator subgroup of a polycyclic group G is n-step nilpotent, then G is n-step nilpotent. Let FŽ.G denote the Frattini subgroup of a group G. It is well known that if G is polycyclic and if GrFŽ.G is abelian Ž or even nilpotent. , then Gis nilpotentŽŽ see Lemma 4 i.. below . We extend this result in this way: THEOREM 2. The nilpotent length of a polycyclic group G equals the nilpotent length of GrFŽ.G . For each integer c g N, denote by Nc the variety of nilpotent groups of nilpotency class at most c and put N` s D cG 0 Nc.So N` is the class of nilpotent groups. Let c ,...,c be elements of N Ä4` . As usual N ??? N 1 nccj 1n is the class of groups G with a subnormal series Ä41sG01eGe??? eGns G such that G G N for k 1,...,n. kkr y1g ck s Now consider a polycyclic group G whose each finite quotient belongs to the class N ??? N . Does G belong to N ??? N ? By Theorem 1 above, cc1 n cc1 n the answer to this question is positive if c1 s ??? s cn s `. Since poly- cyclic groups are residually finite, the answer also is positive if c1,...,cn g N, because in this case, N ??? N is a variety. Probably the answer cc1 n remains positive in the general case, but we are unable to prove it. On the other hand, we can conclude for n s 2: THEOREM 3. Let c12, c be elements of N j Ä4` . If each finite quotient of a polycyclic group G belongs to N N , then G belongs to N N . cc12 cc12 Inwx 1 , it has been proved that if each 3-generator subgroup of a finite group G is abelian-by-nilpotent, then G is abelian-by-nilpotent. This result, together with Theorem 3, implies: COROLLARY. If each 3-generator subgroup of the polycyclic group G is abelian-by-nilpotent, then G is abelian-by-nilpotent. An example given inwx 1 shows that the result above is false if ``3-genera- tor'' is replaced by ``2-generator.'' NILPOTENT LENGTH OF POLYCYCLIC GROUPS 127 2. PROOF OF THEOREM 1 Let FŽ.G denote the Hirsch]Plotkin radical of a group G. Recall that ŽŽ.. the upper locally nilpotent series FkkG G0of G is the increasing se- Ž. Ä4 Ž. Ž. quence of normal subgroups defined by F0 G s 1 and Fkq1 G rFk G s FŽŽ..GrFk Gfor k G 0. If H is a normal subgroup of G, a simple induction on k shows that FkkŽ.H s H l F Ž.G . Clearly, if G is polycyclic, ŽŽ.. the factors of the series FkkG G0are nilpotent; moreover, in this case, it is easy to prove that G is k-step nilpotent if and only if FkŽ.G s G. LEMMA 1. Let H be a normal subgroup of a polycyclic group G. If H is k-step nilpotentŽ. k ) 0,then: Ž.i H:Fk ŽG .; Ž. Ž. Ž. ii HFky1 G rFky1 G is nilpotent. Proof. Ž.i The result follows from equalities H s FkkŽ.H s H l F Ž.G . Ž. Ž. Ž. Ž. ii The groups HFky1 G rFky1 G and HrH l Fky1 G are isomor- phic; but HrH l Fky1Ž.G s Fkk Ž.H rF y1 Ž.H Ž. Ž. and so HFky1 G rFky1 G is nilpotent. In the following, the ith term of the lower central series of a group G is denoted by giŽ.G . LEMMA 2. If G is a polycyclic group, there exist integers c and e such that gŽ.g Ž. iiGrq1G has finite exponent e for each integer i G c. Proof. Let hGŽ.be the Hirsch length of G. It is easy to see that if a g Ž.g Ž.g Ž.g Ž. central factor iiG r q1G is finite of exponent eii, then 9G r i9q1G is finite for each i9 G i and its exponent divides ei. It follows that gŽ.g Ž. Ž. iiGrq1Gis finite for each integer i G hG; since the sequence of exponents decreases, ei is constant for all sufficiently large i, as required. LEMMA 3. Let f be a nonzero polynomial in ZwxT . For each prime p, afŽp.y1 denote by a f Ž.p the greatest integer such that f ŽZ _ pZ .: p Z Žclearly, af Ž.p)0. .Then, if P is a nonempty finite set of primes, there exists an infinite set of primes Q such that, for all p g P, q g Q, we ha¨e a Ž p. fqŽ.k0Ž. mod p f . 128 GERARD ENDIMIONI Proof. Put P s Ä4p1,..., pm . For each i g Ä41,...,m, by definition of afiŽ.p, there exists an element xig Z _ p iZ such that fxŽ.ik0 Ž mod afŽpi. pi .. By the Chinese Remainder Theorem, there exists an integer x a f Ž p i. such that x ' xiiŽmod p . for i s 1,...,m. Choose for Q the set of primes in the sequence x pa f Ž p1. ??? pa f Ž p m .k . Ž.q 1mkG0 a f Ž p1. a f Ž p m . Since x and p1 ??? pm are coprime, a classical theorem of Dirich- let asserts that Q is infinite. Moreover, for all pi g P, q g Q, we have a f Ž p i. fqŽ.'fx Žii .k0Ž. mod p and this completes the proof of the lemma. Proof of Theorem 1. The proof is by induction on the Hirsch length hGŽ.of G.If hG Ž.s0, then G is finite and the result is trivial. Now consider a group G such that hGŽ.)0 and suppose the result is true for each group with a Hirsch length - hGŽ.. We may say that G contains a r non-trivial torsion-free abelian normal subgroup A , Z wx3, 5.4.15 . For each integer k ) 0, define the subgroup FAkkŽ.eFeGby F krA s FknŽ.GrA. Since hG Ž.Ž.rA -hG, we have F s G by the induction hypothesis. Now we define inductively a finite sequence of normal subgroups of G ??? GnndG y11d dG Ž.with Gkk: F for k s 1,2,..., n . Put Gnks G and suppose that G q1is defined for an integer k g Ä4 1, 2, . , n y 1 , with Gkq1 : Fkq1. Since Fkq1rFk is nilpotent, there exists an integer c such that g Ž.G F ; besides, taking c sufficiently kckkq1: kk large, we may assume that each quotient g g ikŽ.G q1r iq1 Ž.ŽGikq1 Gck . has finite fixed exponent ekkby Lemma 2. Let P be the set of primes dividing ekkand let <<P be its cardinality.
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