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. ᮊ 1998 Academic Press

1. INTRODUCTION AND RESULTS

We say that a group G is n-step nilpotent if it has a subnormal series

Ä41sG01eGeиии 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 ᮊ 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 ⌽Ž.G denote the Frattini subgroup of a group G. It is well known that if G is polycyclic and if Gr⌽Ž.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 Gr⌽Ž.G .

For each integer c g ގ, denote by Nc the variety of nilpotent groups of nilpotency class at most c and put Nϱ s D cG 0 Nc.So Nϱ is the class of nilpotent groups. Let c ,...,c be elements of ގ Ä4ϱ . As usual N иии N 1 nccj 1n is the class of groups G with a subnormal series

Ä41sG01eGeиии 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 иии N . Does G belong to N иии N ? By Theorem 1 above, cc1 n cc1 n the answer to this question is positive if c1 s иии s cn s ϱ. Since poly- cyclic groups are residually finite, the answer also is positive if c1,...,cn g ގ, because in this case, N иии 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 ގ j Ä4ϱ . 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 ␥iŽ.G .

LEMMA 2. If G is a polycyclic group, there exist integers c and e such that ␥Ž.␥ 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 ␥ Ž.␥ Ž.␥ Ž.␥ Ž. central factor iiG r q1G is finite of exponent eii, then ЈG r iЈq1G is finite for each iЈ G i and its exponent divides ei. It follows that ␥Ž.␥ 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 ޚwxT . For each prime p, ␣fŽp.y1 denote by ␣ f Ž.p the greatest integer such that f Žޚ _ pޚ .: p ޚ Žclearly, ␣f 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

␣ 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 ␣fiŽ.p, there exists an element xig ޚ _ p iޚ such that fxŽ.ik0 Ž mod ␣fŽpi. pi .. By the Chinese Remainder Theorem, there exists an integer x ␣ 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 p␣ f Ž p1. иии p␣ f Ž p m .k . Ž.q 1mkG0

␣ f Ž p1. ␣ f Ž p m . Since x and p1 иии 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

␣ 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 , ޚ 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

иии 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 ␥ Ž.G F ; besides, taking c sufficiently kckkq1: kk large, we may assume that each quotient

␥ ␥ ikŽ.G q1r iq1 Ž.ŽGikq1 Gck . has finite fixed exponent ekkby Lemma 2. Let P be the set of primes dividing ekkand let <

rŽr1.2rr1 fsTyrŽ.Ž.Ž.Ty1Tyy1иии T y 1

Ž.recall that r is the rank of A .If Pkkis emptyŽ that is, e s 1. , we put ␤kks1. Then G is defined by G ␥G , with d c ␤ 1 <

Now, consider a prime q g Q, where Q is given by Lemma 3, for

rŽr 1. 2 r r 1 f s T y r Ž.Ž.Ž.T y 1 T y y 1 иии T y 1 Ž. and P s D is1,...,ny1 PQi is the set of all primes if P is empty . We qq prove by induction that AGkrA is k-step nilpotent Ž.k s n, n y 1,...,1 . q qq q Since hGŽ rA.Ž.-hG, by the induction hypothesis, GrA s AGnrA is qqŽ. n-step nilpotent. Suppose that AGkq1rA is k q 1 -step nilpotent for an integer k Ž.1 F k F n y 1 . Define the subgroup Fk, q by

q q q Fk,qkrA s F Ž.GrA with A e F k,qeG

Ž. qqq and apply Lemma 1 ii to GrA , with H s AGkq1rA: it follows that FGk, qkq1rFk,qkis nilpotent. First, consider the case e s 1. Then ␥Ž.G ␥ Ž.G and so F contains ␥ Ž.G because ckkkq1scq1kq1 k,qckkq1 FGk, qkq1rFk,qis nilpotent. Since ␥G ␥G G, ckkkŽ.q1sdk Ž.q1sk

q q q the quotient A GkkrA is a subgroup of F ,qrA , and so is k-step nilpo- tent, as required. Now, consider the case ek ) 1. In order to obtain a contradiction, suppose that

␥ ␥ Fk,qiŽ.G kq1 / Fk,qiq1 Ž.Gkq1 for i s ckk, c q 1,...,d k.

␥ Ž.␥ Ž. Ž. Clearly, Fk, qiG kq1rFk,qiq1Gkq1 has finite exponent / 1 dividing ekk. Hence for each i g Ä4c , ckq 1,...,dk, there exists a prime p ig Pk <␥ Ž.␥ Ž.< which divides the index Fk, qiG kq1: Fk,qiq1Gkq1. Since dkks c q ␤k Ž.␤kky1<

C F ␥ G F ␥ G F . e k,qdkkq1Ž.kq1e k,qc Ž. kq1e k

We can consider FkrC as a subgroup of the group of automorphisms of q ArA , of order

rŽr 1.2 r r 1 fqŽ.sq y r Žqy1 .Ž.qy y1иии Ž.q y 1. 130 GERARD´ ENDIMIONI

Since p ␤k divides < F ␥ Ž.G : F ␥ Ž.G <, p ␤kdivides all the k, qckk kq1 k,qdq1 kq1 ␣Žp. more fqŽ.. But q g Q and so, fqŽ.k0 Ž mod p f ., a contradiction since ␤␣Ž.␥ Ž.␥ Ž. kfGp. It follows that Fk,qiG kq1s Fk,qiq1Gkq1 for an integer igÄ4ckk,cq1,...,dkand hence F␥G F␥ G . k,qdkkŽ. kq1 sk,qdq1 Ž.kq1 Since FG F is nilpotent, F contains ␥ Ž.G G . Hence k, qkq1rk,qk,qdkkq1s k qq q AGkkrAis a subgroup of F ,qrA , and so is k-step nilpotent, as required. q q In particular, A G11rA is nilpotent. Thus AG rA also is nilpotent. Denote by d an integer such that A contains ␥dŽ.G1 . Clearly, the nilpotent q q class of A G1rA is bounded by d q r y 1. In other words,

␥drŽ.G1␥␥␥drddиии ␥ dŽ.G иии qs qž/12ž/ž/Ž.ny1

q is included in A for each q g Q. Finally, since Q is infinite,

␥␥␥drddиии ␥ dŽ.G иии Ä41 qž/12ž/ž/Ž.ny1 s and so G is n-step nilpotent.

3. PROOF OF THEOREM 2

For the proof of Theorem 2, we need a further lemma:

LEMMA 4. Let G be a finite group.

Ž.i If H is a normal subgroup of G containing ⌽ Ž.G and if Hr⌽ Ž.Gis nilpotent, then H is nilpotent; Ž.ii If Gr⌽ ŽGisn .-step nilpotent Ž n G 0, .then G is n-step nilpotent. Proof. Ž.i This first part is well knownwx 3, 5.2.15 . Ž.ii Consider a sequence of normal subgroups in G

⌽Ž.G s H01e H e иии e Hns G with each quotient nilpotent. Since H1r⌽Ž.G is nilpotent, it follows from Ž.i that H1 is nilpotent and so G is n-step nilpotent. Proof of Theorem 2. Clearly, it suffices to prove that if G is a polycyclic group such that Gr⌽Ž.G is n-step nilpotent, then G is n-step nilpotent. Let G be such a group and let GrH be a finite quotient Ž.H eG .We denote by F the subgroup of G such that

FrH s ⌽Ž.GrH , with H e F eG. NILPOTENT LENGTH OF POLYCYCLIC GROUPS 131

Obviously, F contains ⌽Ž.G ; thus GrF is n-step nilpotent. But we have GrF , Ž.Ž.Ž.Ž.GrH r FrH s GrH r⌽ GrH and so GrH is n-step nilpotent by Lemma 4Ž. ii . Therefore, each finite quotient of G is n-step nilpotent. It follows from Theorem 1 that G is n-step nilpotent, as required.

4. PROOF OF THEOREM 3

The following lemma is a weak form of Theorem 3:

LEMMA 5. Let c12, c be elements of ގ j Ä4ϱ and let F be a finite normal subgroup of a polycyclic group G. If GrF and each finite quotient of G belong to N N , then G belongs to N N . cc12 cc12 Proof of Lemma 5. It is well known that polycyclic groups are residually finite; thus G contains a normal subgroup of finite index H such that XX HlFsÄ41 . Since GrH belongs to NccN , there exist integers d12, d such X12 that GrH belongs to NddXXN with d is c iif c ig ގ Ž.i s 1, 2 . In the same 12Y Y way, there exist integers d12, d such that GrF belongs to NddY N Y with Y XY 12 diiiscif c g ގ Ž.i s 1, 2 . Put dis max Žd ii, d .. Clearly, GrH and GrF belong to the variety N N . Thus G is in N N and so in N N . dd12 dd12 c1 c 2

Proof of Theorem 3. As we have seen, the result is true for c12s c s ϱ or c12, c g ގ. So it remains to treat two cases.

Case c12s ϱ, c g ގ. For convenience, put c2s c in this case. We argue by induction on the Hirsch length hGŽ.of G. The result being trivial Ž. Ž.␥ Ž. if hG s0, suppose that hG )0. If cq1 G is finite, the result follows ␥ Ž. ␥ Ž. from Lemma 5, with F s cq1 G . Now suppose that cq1 G is infinite. As in the proof of Theorem 1, we can consider a non-trivial torsion-free ޚ r ␥ Ž. abelian normal subgroup A , of cq1 G . In fact, it is easy to see that ␥ Ž. Amay be chosen characteristic in cq1 G and so normal in G. Let q be a q prime. Notice the inequalities hGŽ.Ž.ŽrA -hG and hGrA.Ž.-hG.By induction, GrA belongs to the class Nϱ Nc; hence there exists an integer d ␥␥ŽŽ.. q such that dcq1G is included in A. In the same way, GrA belongs to ␥␥ŽŽ.. q the class Nϱ Ncd. Clearly, this implies that qrcq1Gis included in A Ž.Ž␥␥Ž..Ä4 for each prime q and so dqrcq1Gs1 , as required.

Case c12g ގ, c s ϱ. In this case, put c1s c. We proceed again by induction on hGŽ.. Suppose hG Ž.)0. By Theorem 1, there exists an integer d such that ␥dŽ.G is nilpotent. Moreover, taking d sufficiently large, we may assume that each quotient ␥ ␥ iiŽ.G r q1 Ž.Gi ŽGd . 132 GERARD´ ENDIMIONI has finite exponent e Ž.Ž.see Lemma 2 and that ␥d G has a minimal nilpotency class. If ␥dŽ.G is finite, Lemma 5 shows that G belongs to Nc Nϱ and the proof is finished. So suppose that ␥dŽ.G is infinite. It is well known that the center of an infinite finitely generated nilpotent group is infinite.

Therefore, if Z denotes the center of ␥dŽ.G , we have hG ŽrZ .-hG Ž.. Thus, by induction, GrZ belongs to Nc Nϱ. In other words, there exists an integer dЈ such that Z␥dЈŽ.G rZ is nilpotent of class at most c; moreover, without loss of generality, we can assume that dЈ G d. Since Z␥dЈŽ.G is a central extension of Z by Z␥dЈŽ.G rZ, this group is nilpotent of class at most c q 1. Thus the nilpotency class of ␥dЈŽ.G also is at most c q 1. But ␥dŽ.Gis chosen with a minimal nilpotency class, so this class is bounded by c q 1. Notice that the torsion subgroup of ␥dŽ.G is finite. Therefore, in order to prove that G belongs to Nc Nϱ, we can assume that ␥dŽ.G is torsion-free thanks to Lemma 5. Consider a prime q which does not divide the ␥ Ž.␥ Ž.Ž . Ž q.Ž. exponent e of iiG r q1GiGd. Since hGrZ -hG, it follows by q induction that GrZ is in Nc Nϱ. So, there exists an integer dЉ such that q q Z␥dЉ Ž.GrZ is nilpotent of class at most c; besides, we can choose dЉGd. The group ␥ddŽ.G r␥ Љ Ž.G is finite and its exponent divides a power q of e. Therefore, q and the exponent eЈ of ␥ddŽ.G rZ ␥ ЉŽ.G are coprime. ␥ Ž. eЈ eЈ Let x1,..., xcq1 be elements of d G . We have x1 ,..., xcq1 in q␥Ž. eЈ eЈ q ZdЉ G; hence the commutator wx1 ,..., xcq1 xlies in Z . Using the fact that ␥dŽ.G is nilpotent of class at most c q 1, it is easy to show the relation

eЈ eЈ eЈcq 1 x1 ,..., xcq11swxx ,..., xcq1,

Xc1 eq q␥ Ž. hence wxx1,..., xcq1 lies in Z . Since d G is torsion-free, so is each upper central factor. It follows that ␥ Ž.G Z is torsion-free; thus Xc1 d r x,..., x eqZ implies x ,..., x Z and so x ,..., x q wx1 cq11g Xc1wxcq11g wxcq1g q eqq Ј Z. But wxx1,..., xcq1 gZ and q and e are coprime, so q wxx1,..., xcq1 gZ . This holds for each prime q which does not divide e, ␥ Ž. hence wxx1,..., xcq1 s1. In other words, d G is nilpotent of class at most c, as required.

5. EXAMPLES

The following examples are due to the referee. Example A proves that a finitely generated soluble group in which all finite quotients are 2-step nilpotent need not be 2-step nilpotent. Example B shows that even in the class of finitely generated 2-step nilpotent groups, a group in which all finite quotients are in N11N is not necessarily in N11N . NILPOTENT LENGTH OF POLYCYCLIC GROUPS 133

EXAMPLE A. Let D be a direct product of countably many groups Qi, indexed by the integers, which are all torsion free divisible locally cyclic, and choose in every factor Qiian element x / 1. Let A be a direct product of countably many infinite cyclic groups Ciis ²:y again indexed by the integers. Further choose a one-to-one mapping ␶ of the integers onto the set ⌸ of all primes. We form the split extension F of D by A y1 ␶Žjyi. with the ruling yijijxysx . Finally, we extend F by an infinite cyclic ²: y1 y1 group z putting zxziisxq1,zyziisyq1; denote by G this group. A straightforward computation shows that G s ²:x11, y , z . Moreover, it is not hard to see that F12Ž.G s D and F Ž.G s F; hence G is of nilpotent length 3. But since subgroups of finite index of G contain D, all finite quotients are of nilpotent length at most 2.

EXAMPLE B. Let ޚwx1r2 be the ring of rationnel numbers with 2-power denominator. Denote by L the set of all upper unitriangular 3 = 3-matrices whose entries are in ޚwx1r2 and consider also the subgroup D generated by the 3 = 3-diagonal matrix d with entries 1, 2, 1. Define Eii,jŽ ,js 1, 2, 3. to be the matrix with 1 in the ith row and jth column and 0 elsewhere. It is easily seen that D normalizes L and that the center

ZLDŽ.of LD consists of all unitriangular matrices of the form 1 q ␣E1, 3 Ž␣gޚwx1r2.Ž.Ž. ; in fact, we have ZLDsZLsLЈ. In particular, the subgroup U generated by the matrix 1 q E1, 3 is a normal subgroup of LD. Since LЈrU is divisible, all subgroups of LDrU of finite index contain LЈrU. Therefore, all finite quotient groups of LDrU belong to N11N .On the other hand, LDrU is not metabelian but belongs to the class N21N . Finally, LDrU is finitely generated, namely by the images of d,1qE1, 2 , and 1 q E2, 3.

REFERENCES

1. R. Brandl, On finite abelian-by-nilpotent groups, J. Algebra 86 Ž.1984 , 439᎐444. 2. R. Carter, B. Fischer, and T. Hawkes, Extreme classes of finite soluble groups, J. Algebra 9 Ž.1968 , 285᎐313. 3. D. J. S. Robinson, ‘‘A Course in the Theory of Groups,’’ Springer-Verlag, New York, 1982. 4. J. G. Thompson, Nonsolvable finite groups all of whose local subgroups are solvable, Bull. Amer. Math. Soc. 74 Ž.1968 , 383᎐437.