JOURNAL OF ALGEBRA 203, 125133 1998 ARTICLE NO. JA977321 On the Nlpotent Length of Polycyclc Groups Gerard Endmon* C.M.I., Unerste de Proence, UMR-CNRS 6632, 39, rue F. Jolot-Cure, 13453 Marselle Cedex 13, France Communcated by J. Tts Receved October 30, 1996 Let G be a polycyclc group. We prove that f the nlpotent length of each fnte uotent of G s bounded by a fxed nteger n, then the nlpotent length of G s at most n. The case n 1 s a well-nown result of Hrsch. As a conseuence, we obtan that f the nlpotent length of each 2-generator subgroup s at most n, then the nlpotent length of G s at most n. A more precse result n the case n 2 permts us to prove that f each 3-generator subgroup s abelan-by-nlpotent, then G s abelan-by-nlpotent. Furthermore, we show that the nlpotent length of G euals the nlpotent length of the uotent of G by ts Frattn subgroup. 1998 Academc Press 1. INTRODUCTION AND RESULTS We say that a group G s n-step nlpotent f t has a subnormal seres 14 G0G1 Gn G of length n wth each uotent nlpotent. In fact, substtutng the core of G n G for G, we can assume that Ž G. 0,1,...,n s a normal seres. The least nteger n such that G s n-step nlpotent s the nlpotent length of G. Frst, we shall prove n ths paper the followng THEOREM 1. Let G be a polycyclc group. If there exsts a poste nteger n such that each fnte uotent of G s n-step nlpotent, then G s n-step nlpotent. For n 1, ths result s due to K. A. Hrsch 3, 5.4.18. Later t was mproved n the followng stronger form: f each fnte uotent of a fntely generated soluble group G s nlpotent, then G s nlpotent 3, 15.5.3. On * E-mal: endmon@gypts.unv-mrs.fr. 125 0021-869398 $25.00 Copyrght 1998 by Academc Press All rghts of reproducton n any form reserved.
126 GERARD ENDIMIONI the other hand, thans to an example due to the referee, we shall see that Theorem 1 above s false for n 2 wth polycyclc replaced by fntely generated soluble. Theorem 1 permts us to transfer propertes of fnte groups to polycyclc groups. For nstance, t follows from 2, 4 that a fnte group s n-step nlpotent f all ts 2-generator subgroups are n-step nlpotent. Therefore, as a conseuence of the theorem, we obtan: COROLLARY. If there exsts a poste nteger n such that eery 2-generator subgroup of a polycyclc group G s n-step nlpotent, then G s n-step nlpotent. Let Ž G. denote the Frattn subgroup of a group G. It s well nown that f G s polycyclc and f GŽ G. s abelan Ž or even nlpotent., then G s nlpotent Žsee Lemma 4Ž. below.. We extend ths result n ths way: THEOREM 2. The nlpotent length of a polycyclc group G euals the nlpotent length of GŽ G.. For each nteger c, denote by Nc the varety of nlpotent groups of nlpotency class at most c and put N c 0 N c.so N s the class of nlpotent groups. Let c,...,c be elements of 4 1 n. As usual Nc N 1 cn s the class of groups G wth a subnormal seres 14 G0G1 Gn G such that G G N for 1,...,n. 1 c Now consder a polycyclc group G whose each fnte uotent belongs to the class N N. Does G belong to N N? By Theorem 1 above, c c c c 1 n 1 n the answer to ths ueston s postve f c c. Snce poly- 1 n cyclc groups are resdually fnte, the answer also s postve f c 1,...,cn, because n ths case, N N s a varety. Probably the answer c1 cn remans postve n the general case, but we are unable to prove t. On the other hand, we can conclude for n 2: THEOREM 3. Let c, c be elements of 4. If each fnte uotent of a polycyclc group G belongs to Nc N c, then G belongs to Nc N c. In 1, t has been proved that f each 3-generator subgroup of a fnte group G s abelan-by-nlpotent, then G s abelan-by-nlpotent. Ths result, together wth Theorem 3, mples: COROLLARY. If each 3-generator subgroup of the polycyclc group G s abelan-by-nlpotent, then G s abelan-by-nlpotent. An example gven n 1 shows that the result above s false f 3-generator s replaced by 2-generator.
NILPOTENT LENGTH OF POLYCYCLIC GROUPS 127 2. PROOF OF THEOREM 1 Let FŽ G. denote the HrschPlotn radcal of a group G. Recall that the upper locally nlpotent seres ŽF Ž G.. 0 of G s the ncreasng se- uence of normal subgroups defned by F Ž G. 14 and F Ž G. F Ž G. 0 1 FŽGF Ž G.. for 0. If H s a normal subgroup of G, a smple nducton on shows that F Ž H. H F Ž G.. Clearly, f G s polycyclc, the factors of the seres ŽF Ž G.. 0 are nlpotent; moreover, n ths case, t s easy to prove that G s -step nlpotent f and only f F Ž G. G. LEMMA 1. Let H be a normal subgroup of a polycyclc group G. If H s -step nlpotent Ž 0,then:. Ž. HFŽ G;. Ž. HF Ž G. F Ž G. s nlpotent. 1 1 Ž. Proof. The result follows from eualtes H F Ž H. H F Ž G.. Ž. The groups HF Ž G. F Ž G. and HH F Ž G. 1 1 1 are somor- phc; but HH F1Ž G. FŽ H. F1Ž H. and so HF Ž G. F Ž G. s nlpotent. 1 1 In the followng, the th term of the lower central seres of a group G s denoted by Ž G.. LEMMA 2. If G s a polycyclc group, there exst ntegers c and e such that Ž G. Ž G. has fnte exponent e for each nteger c. 1 Proof. Let hg be the Hrsch length of G. It s easy to see that f a central factor Ž G. Ž G. s fnte of exponent e, then Ž G. Ž G. 1 1 s fnte for each and ts exponent dvdes e. It follows that Ž G. Ž G. s fnte for each nteger hg; 1 snce the seuence of exponents decreases, e s constant for all suffcently large, as reured. LEMMA 3. Let f be a nonzero polynomal n T. For each prme p, f Ž p.1 denote by p the greatest nteger such that f p p Ž f clearly, Ž p. 0.. f Then, f P s a nonempty fnte set of prmes, there exsts an nfnte set of prmes Q such that, for all p P, Q, we hae f 0 mod p f Ž p..
128 GERARD ENDIMIONI Proof. Put P p,..., p 4. For each 1,...,m 4 1 m, by defnton of Ž p., there exsts an element x p such that fž x. 0 Ž f mod f Ž p p... By the Chnese Remander Theorem, there exsts an nteger x Ž f Ž p such that x x mod p.. for 1,...,m. Choose for Q the set of prmes n the seuence x p f Ž p. 1 p f Ž p m 1 m. 0. Snce x and p f Ž p. 1 p f Ž p m. 1 m are coprme, a classcal theorem of Drchlet asserts that Q s nfnte. Moreover, for all p P, Q, we have Ž. Ž. f f x 0 mod p f Ž p. and ths completes the proof of the lemma. Proof of Theorem 1. The proof s by nducton on the Hrsch length hg of G. If hg 0, then G s fnte and the result s trval. Now consder a group G such that hg 0 and suppose the result s true for each group wth a Hrsch length hg. We may say that G contans a r non-trval torson-free abelan normal subgroup A 3, 5.4.15. For each nteger 0, defne the subgroup F Ž AF G. by FA F Ž GA.. Snce hga hg, we have Fn G by the nducton hypothess. Now we defne nductvely a fnte seuence of normal subgroups of G GnGn1 G1 Ž wth G F for 1,2,..., n.. Put Gn G and suppose that G1 s defned for an nteger 1, 2,..., n 14, wth G1 F 1. Snce F1F s nlpotent, there exsts an nteger c such that Ž G. c 1 F ; besdes, tang c suffcently large, we may assume that each uotent Ž G1. 1Ž G1. Ž c. has fnte fxed exponent e by Lemma 2. Let P be the set of prmes dvdng e and let P be ts cardnalty. If P s nonempty, put max Ž p.; p P 4, where Ž p. s defned n Lemma 3, wth f f ft ržr1.2 Ž T r 1.Ž T r1 1. Ž T 1. Ž recall that r s the ran of A..If P s empty Ž that s, e 1., we put 1. Then G s defned by G Ž G., wth d c Ž 1. P. d 1 Note that G F, because Ž G. F. c 1
NILPOTENT LENGTH OF POLYCYCLIC GROUPS 129 Now, consder a prme Q, where Q s gven by Lemma 3, for f T rž r1.2 Ž T r 1.Ž T r1 1. Ž T 1. and P P Ž Q s the set of all prmes f P s empty. 1,...,n1. We prove by nducton that AGA s -step nlpotent Ž n, n 1,...,1.. Ž. Snce hga hg, by the nducton hypothess, GA AGA n s n-step nlpotent. Suppose that AG A s Ž 1. 1 -step nlpotent for an nteger Ž 1 n 1.. Defne the subgroup F by, F, A FŽ GA. wth A F, G and apply Lemma 1 to GA, wth H AG1A: t follows that F, G1F, s nlpotent. Frst, consder the case e 1. Then Ž G. Ž G. and so F contans Ž G. c 1 c 1 1, c 1 because F, G1F, s nlpotent. Snce Ž G. Ž G. G, c 1 d 1 the uotent A GA s a subgroup of F, A, and so s -step nlpo- tent, as reured. Now, consder the case e 1. In order to obtan a contradcton, suppose that F, Ž G1. F, 1Ž G1. for c, c 1,...,d. Clearly, F Ž G. F Ž G. has fnte exponent Ž 1., 1, 1 1 dvdng e. Hence for each c, c 1,...,d 4, there exsts a prme p P whch dvdes the ndex F Ž G.: F Ž G., 1, 1 1. Snce d c 1 P, there exsts a prme p P such that p dvdes F Ž G.: F Ž G., c 1, d 1 1. Let C denote the normal subgroup of F contanng A such that CA s the centralzer of AA n FA. Note that C contans A and CA s -step nlpotent; snce CA s a central extenson of AA by CA, t s easy to see that CA also s -step nlpotent. Hence F contans C by Lemma 1Ž.,. Moreover, we have clearly A F, ; snce F, A s -step nlpotent, F contans F,. Thus we have the chan C F Ž G. F Ž G. F., d1 1, c 1 We can consder F C as a subgroup of the group of automorphsms of AA, of order fž. ržr1.2 Ž r 1. Ž r1 1. Ž 1..
130 GERARD ENDIMIONI Snce p dvdes F G : F G, p, c 1, d 1 1 dvdes all the Ž f Ž p. more f. But Q and so, f 0 mod p., a contradcton snce Ž p.. It follows that F Ž G. F Ž G. f, 1, 1 1 for an nteger c,c 1,...,d 4 and hence F Ž G. F Ž G.., d 1, d1 1 Snce F G F s nlpotent, F contans Ž G., 1,, d 1 G. Hence A GA s a subgroup of F, A, and so s -step nlpotent, as reured. In partcular, A G1A s nlpotent. Thus AG1A also s nlpotent. Denote by d an nteger such that A contans Ž G. d 1. Clearly, the nlpotent class of A G1A s bounded by d r 1. In other words, ž ž ž n1 /// Ž G. Ž G. dr 1 dr d d d s ncluded n A for each Q. Fnally, snce Q s nfnte, ž ž ž n1 /// Ž G. 14 dr d d d and so G s n-step nlpotent. 3. PROOF OF THEOREM 2 For the proof of Theorem 2, we need a further lemma: LEMMA 4. Let G be a fnte group. Ž. If H s a normal subgroup of G contanng Ž G. and f HŽ G. s nlpotent, then H s nlpotent; Ž. If GŽ G. s n-step nlpotent Ž n 0,then. G s n-step nlpotent. Proof. Ž. Ths frst part s well nown 3, 5.2.15. Ž. Consder a seuence of normal subgroups n G Ž G. H0 H1 Hn G wth each uotent nlpotent. Snce H Ž G. 1 s nlpotent, t follows from Ž. that H1 s nlpotent and so G s n-step nlpotent. Proof of Theorem 2. Clearly, t suffces to prove that f G s a polycyclc group such that GŽ G. s n-step nlpotent, then G s n-step nlpotent. Let G be such a group and let GH be a fnte uotent Ž H G..We denote by F the subgroup of G such that FH Ž GH., wth H F G.
NILPOTENT LENGTH OF POLYCYCLIC GROUPS 131 Obvously, F contans G ; thus GF s n-step nlpotent. But we have GF Ž GH. Ž FH. Ž GH. Ž GH. and so GH s n-step nlpotent by Lemma 4. Therefore, each fnte uotent of G s n-step nlpotent. It follows from Theorem 1 that G s n-step nlpotent, as reured. 4. PROOF OF THEOREM 3 The followng lemma s a wea form of Theorem 3: LEMMA 5. Let c, c be elements of 4 and let F be a fnte normal subgroup of a polycyclc group G. If GF and each fnte uotent of G belong to Nc N c, then G belongs to Nc N c. Proof of Lemma 5. It s well nown that polycyclc groups are resdually fnte; thus G contans a normal subgroup of fnte ndex H such that HF1 4. Snce GH belongs to Nc N c, there exst ntegers d 1, d2 such that GH belongs to N N wth d c f c Ž 1, 2. d d. In the same way, there exst ntegers d, d such that GF belongs to N N d d wth d c f c Ž 1, 2.. Put d maxžd, d.. Clearly, GH and GF belong to the varety Nd N d. Thus G s n Nd Nd and so n Nc N c. Proof of Theorem 3. As we have seen, the result s true for c1 c2 or c 1, c2. So t remans to treat two cases. Case c1, c2. For convenence, put c2 c n ths case. We argue by nducton on the Hrsch length hg of G. The result beng trval f hg 0, suppose that hg 0. If Ž G. c1 s fnte, the result follows from Lemma 5, wth F Ž G.. Now suppose that Ž G. c1 c1 s nfnte. As n the proof of Theorem 1, we can consder a non-trval torson-free r abelan normal subgroup A of Ž G. c1. In fact, t s easy to see that A may be chosen characterstc n Ž G. c1 and so normal n G. Let be a prme. Notce the neualtes hga hg and hga Ž. hg.by nducton, GA belongs to the class N N c; hence there exsts an nteger d Ž. such that d c1 G s ncluded n A. In the same way, GA belongs to Ž. the class N N c. Clearly, ths mples that dr c1 G s ncluded n A Ž for each prme. and so Ž Ž G.. 1 4, as reured. dr c1 Case c1, c2. In ths case, put c1 c. We proceed agan by nducton on hg. Suppose hg 0. By Theorem 1, there exsts an nteger d such that Ž G. d s nlpotent. Moreover, tang d suffcently large, we may assume that each uotent Ž G. Ž G. Ž d. 1
132 GERARD ENDIMIONI has fnte exponent e Ž see Lemma 2. and that Ž G. d has a mnmal nlpotency class. If Ž G. d s fnte, Lemma 5 shows that G belongs to Nc N and the proof s fnshed. So suppose that Ž G. d s nfnte. It s well nown that the center of an nfnte fntely generated nlpotent group s nfnte. Therefore, f Z denotes the center of Ž G., we have hgz hg. d Thus, by nducton, GZ belongs to Nc N. In other words, there exsts an nteger d such that Z Ž G. d Z s nlpotent of class at most c; moreover, wthout loss of generalty, we can assume that d d. Snce Z Ž G. d s a central extenson of Z by Z Ž G. d Z, ths group s nlpotent of class at most c 1. Thus the nlpotency class of Ž G. d also s at most c 1. But Ž G. d s chosen wth a mnmal nlpotency class, so ths class s bounded by c 1. Notce that the torson subgroup of Ž G. d s fnte. Therefore, n order to prove that G belongs to N N, we can assume that Ž G. c d s torson-free thans to Lemma 5. Consder a prme whch does not dvde the exponent e of Ž G. Ž G. Ž d.. Snce hgz Ž. hg, 1 t follows by nducton that GZ s n Nc N. So, there exsts an nteger d such that Z d G Z s nlpotent of class at most c; besdes, we can choose d d. The group Ž G. Ž G. d d s fnte and ts exponent dvdes a power of e. Therefore, and the exponent e of G Z Ž G. d d are coprme. e e Let x 1,..., xc1 be elements of d G. We have x 1,..., xc1 n e e Z d G ; hence the commutator x 1,..., xc1 les n Z. Usng the fact that Ž G. s nlpotent of class at most c 1, t s easy to show the relaton d e e e x,..., x x,..., x c 1, 1 c1 1 c1 ec1 hence x,..., x les n Z. Snce Ž G. 1 c1 d s torson-free, so s each upper central factor. It follows that d Ž G. Z s torson-free; thus x ec1 1,..., xc1 Z mples x 1,..., xc1 Z and so x 1,..., xc1 ec1 Z. But x 1,..., xc1 Z and and e are coprme, so x 1,..., xc1 Z. Ths holds for each prme whch does not dvde e, hence x,..., x 1. In other words, Ž G. 1 c1 d s nlpotent of class at most c, as reured. 5. EXAMPLES The followng examples are due to the referee. Example A proves that a fntely generated soluble group n whch all fnte uotents are 2-step nlpotent need not be 2-step nlpotent. Example B shows that even n the class of fntely generated 2-step nlpotent groups, a group n whch all fnte uotents are n N N s not necessarly n N N. 1 1 1 1
NILPOTENT LENGTH OF POLYCYCLIC GROUPS 133 EXAMPLE A. Let D be a drect product of countably many groups Q, ndexed by the ntegers, whch are all torson free dvsble locally cyclc, and choose n every factor Q an element x 1. Let A be a drect product of countably many nfnte cyclc groups C ² y : agan ndexed by the ntegers. Further choose a one-to-one mappng of the ntegers onto the set of all prmes. We form the splt extenson F of D by A wth the rulng y 1 xyx j Žj. j. Fnally, we extend F by an nfnte cyclc ² : 1 1 group z puttng z xzx 1, z yzy 1 ; denote by G ths group. A straghtforward computaton shows that G ² x 1, y 1, z :. Moreover, t s not hard to see that F Ž G. D and F Ž G. F; hence G s of nlpotent length 3. But snce subgroups of fnte ndex of G contan D, all fnte uotents are of nlpotent length at most 2. EXAMPLE B. Let 12 be the rng of ratonnel numbers wth 2-power denomnator. Denote by L the set of all upper untrangular 3 3-matrces whose entres are n 12 and consder also the subgroup D generated by the 3 3-dagonal matrx d wth entres 1, 2, 1. Defne E Ž, j, j 1, 2, 3. to be the matrx wth 1 n the th row and jth column and 0 elsewhere. It s easly seen that D normalzes L and that the center ZŽ LD. of LD conssts of all untrangular matrces of the form 1 E 1, 3 Ž12. ; n fact, we have ZŽ LD. ZŽ L. L. In partcular, the subgroup U generated by the matrx 1 E1, 3 s a normal subgroup of LD. Snce LU s dvsble, all subgroups of LDU of fnte ndex contan LU. Therefore, all fnte uotent groups of LDU belong to N1 N 1.On the other hand, LDU s not metabelan but belongs to the class N2 N 1. Fnally, LDU s fntely generated, namely by the mages of d, 1E 1, 2, and 1 E 2, 3. REFERENCES 1. R. Brandl, On fnte abelan-by-nlpotent groups, J. Algebra 86 Ž 1984., 439444. 2. R. Carter, B. Fscher, and T. Hawes, Extreme classes of fnte soluble groups, J. Algebra 9 Ž 1968., 285313. 3. D. J. S. Robnson, A Course n the Theory of Groups, Sprnger-Verlag, New Yor, 1982. 4. J. G. Thompson, Nonsolvable fnte groups all of whose local subgroups are solvable, Bull. Amer. Math. Soc. 74 Ž 1968., 383437.