PRESENTATIONS OF FREE ABELIAN-BY-(NILPOTENT OF CLASS 2) GROUPS

PRESENTATIONS OF FREE ABELIAN-BY-(NILPOTENT OF CLASS 2) GROUPS MARTIN J EVANS 1 Introduction Let F n, M n and U n denote the free group of rank n, the free metabelian group of rank n, and the free abelian-by-(nilpotent of class 2) group of rank n, respectively Thus M n F n F n and U n F n [γ (F n ), γ (F n )], where γ (F n ) [F n, F n ], the third term of the lower central series of F n Consider an arbitrary epimorphism θ: F n F k, where k n A routine argument (see [1, Theorem 33]) using the Nielsen reduction process shows that there exists a free basis y, y,, y n of F n such that ker θ is the normal closure in F n of y, y,, y n k A similar result has been obtained for free metabelian groups by C K Gupta, N D Gupta and G A Noskov [5] Indeed, it follows immediately from [5, Theorem 31] that if k n and θ: M n M k is an epimorphism, then there exists a free basis m, m,, m n of M n such that ker θ is the normal closure in M n of m, m,, m n k In the light of these results it is natural to consider the following question Question Let V n denote the free group of rank n in a variety V, and let θ: V n V k be an arbitrary epimorphism for some k n Does there exist a free basis,,, n of V n such that ker θ is the normal closure of,,, n k in V n? We have seen that this question has an affirmative answer, for all k n,ifvis the variety of all groups or the variety of all metabelian groups The main purpose of this work is to show that the question has a negative answer in general We accomplish this by proving the following theorem THEOREM For each n 2, there exists an epimorphism ψ n : U n+ such that ker ψ n is not the normal closure in U n+ of a single element Let V n again denote the free group of rank n in some variety V We say that x V n is a primitie element of V n if there exist x, x,, x n V n such that x, x,, x n is a basis of V n From Theorem, we shall deduce the following corollary COROLLARY 12 (1) For each n 2, there exists an epimorphism ψ n : U n+ such that ker ψ n does not contain any primitie elements of U n+ (2) For each n 2, there exists an epimorphism β n : F n+ such that ker β n does not contain any primitie elements of F n+ The reader might like to compare the second part of this corollary with the main result of [4], which asserts that there exists an epimorphism γ: F M such that ker γ does not contain any primitive elements of F Received 3 August 1993; revised 2 February 1997 1991 Mathematics Subject Classification 2F5, 2E1 Bull London Math Soc 3 (1998) 136 144

FREE ABELIAN-BY-(NILPOTENT OF CLASS 2) GROUPS 137 In [7], P A Linnell proved the following interesting theorem THEOREM 13 Let G be a finite group which can be generated by k elements, and let n k1 If M and N are normal subgroups of F n such that F n M F n N G, then F n M F n N In Section 6 we shall show that this theorem does not extend to infinite groups G THEOREM 14 For each n 2, there exist normal subgroups M and N of F n+ with F n+ M F n+ N U n and such that F n+ M F n+ N Our notation is standard throughout Given elements x and y in a group G, we have that xy y xy and [x, y] x y xy Operators are usually written on the right In particular, all modules will be right modules However, we shall write module homomorphisms on the left This should cause no confusion We conclude this introduction by listing, for the convenience of the reader, some of the nonstandard notation introduced below n a fixed integer greater than 1 n the free nilpotent group of class 2 and rank n g, g,, g n a free basis of n Λ the integral group ring of n t, t,, t n a basis of the free Λ-module of rank n1 c the commutator [ g, g ] For typographical reasons, it will be convenient on occasion to let a g and b g Thus a, b, g,, g n is a free basis of n, and c [a, b] 2 Free nilpotent groups of class 2 and stably free modules Let n denote the free nilpotent group of class 2 and rank n We begin this section by recording some well-known elementary properties of n The notation introduced here will be used throughout the sequel We fix once and for all an integer n 2 and a free basis g, g,, g n of n Now Z Z( n ), the centre of n, is free abelian and is freely generated by the elements c i,j [g i, g j ], where 1 i j n Moreover, n Z is free abelian of rank n and is freely generated by g Z, g Z,, g n Z Evidently n has a normal series Z G G G n n, where G i Z, g,, g i for i 1,, n Observe that G i n and G Z G i+ G i C for i 1,, n1 Note also that G is free abelian with basis g, c i,j 1 i j n On the other hand, G is not abelian, since [g, g ] c 1 It will simplify our notation if we set c c so that,, gg cg, where c is a non-trivial element of the centre of n A finitely generated module K over a ring R is said to be stably free if K L is a free R-module for some finitely generated free R-module L In[13], J T Stafford shows that if G is a non-abelian poly-(infinite cyclic) group, then the integral group ring G contains a non-cyclic stably free right ideal I such that I G G G Observe that I can be viewed as the kernel of the G-epimorphism η: G G G

138 MARTIN J EVANS given by η(i, α) α for all i I and α G Here, of course, we are making use of the isomorphism I G G G We shall need an explicit description of this epimorphism η in case G is a free nilpotent group of class 2 Such a description is obtained in Lemma 23 below However, we wish to emphasize that this is already implicit in [13] Let R be a ring and let σ be an automorphism of R Recall that the skew Laurent extension S R[x, x ; σ] is the ring that is additively isomorphic to the free R- module with basis xi i and has multiplication defined by rxi xir σi for all r R and i The following lemma is little more than a restatement of [13, Theorem 12] LEMMA 21 Let σ be an automorphism of a Noetherian domain R, and let S R[x, x ; σ] Suppose that there exists a non-unit r R and some s R such that S rs(xs)s and sr σ rr Define an S-homomorphism θ: S S S by θ(s,s )rs (xs)s for all (s, s ) S S Then θ is an epimorphism, and ker θ is a non-cyclic S-module satisfying ker θ S S S Throughout the sequel we shall let a g and b g Thus a, b, g,, g n is a free basis of n, and c [a, b] LEMMA 22 (compare [13, Lemma 28]) Let θ: G G G be the G - homomorphism defined by θ(α, α ) (ab 1)α (bab1)α for all (α, α ) G G Then θ is an epimorphism, and ker θ is a non-cyclic G -module satisfying ker θ G G G Proof Let S G, and note that S R[b, b ; σ] where R G and σ is the automorphism of R defined by r σ b rb for all r R Since G is a free abelian group of finite rank, Hilbert s basis theorem implies that R is Noetherian Furthermore, it is easy to see that R is a domain We aim to apply Lemma 21 with x b, r ab 1 and s ab1 Therefore, to complete the proof, it suffices to show that ab 1 is a non-unit of G, that ab 1 and bab1 together generate G as a right ideal and that (ab1) (ab 1) (ab 1) G Let ε denote the augmentation map ε: G, and note that (ab 1)ε Since uε 1 for all units u G, we deduce that ab 1 is a non-unit of G Now (ab 1) (ab1)b (bab1) (bab 1)b 1, and so (ab 1) G (bab1) G G Recall that G is a free abelian group that has a free basis a, c i,j 1 i j n and that c c [a, b] Clearly,, ca, c i,j 1 i j n is also a free basis of G and so we may define a ring homomorphism ξ: G G by setting caξ 1 and c i,j ξ c i,j for 1 i j n Since ca ab, it is easy to see that ker ξ (ab 1) G Now (ab1) (ab 1) ξ (ca1) (ca1) ξ (c 1) (c1), and it follows that (ab1) (ab 1) (ab 1) G The proof is complete LEMMA 23 (compare [13, Theorem 212]) Let η: n n n be the n -homomorphism defined by η(β, β ) (ab 1) β (bab1) β for all (β, β ) n n Then η is an epimorphism, and ker η is a non-cyclic n -module satisfying ker η n n n

FREE ABELIAN-BY-(NILPOTENT OF CLASS 2) GROUPS 139 Proof Makes the obvious changes in the proof of [13, Theorem 212], using Lemma 22 in place of [13, Lemma 28] 3 The main lemmas Throughout the remainder of this note we shall write Λ for the integral group ring of n, and let t, t,, t n be a basis of a free Λ-module of rank n1 LEMMA 31 Let ρ: t Λ t Λ Λ be the Λ-homomorphism defined by ρt a1 and ρt b1 Then there exists a Λ-epimorphism φ: t Λ t Λ t Λ t Λ t Λ such that (i) ker φ is not cyclic, and (ii) ρφt, ρφt a1 and ρφt b1 Proof Let α: t Λ t Λ t Λ be the Λ-homomorphism defined by αt t (ca1) and αt t (bca1) Since ca ab and ca ab, Lemma 23 shows that α is an epimorphism and that ker α is non-cyclic We extend α to an epimorphism β: t Λ t Λ t Λ t Λ t Λ by letting βt t (ca1), βt t (bca1) and βt t Clearly ker β ker α, so that ker β is non-cyclic The remainder of the proof shows how we may modify β to obtain an epimorphism φ with the required properties Let w t t a b (ca1)t ((ca1) a b b (a1)) (ca1) and w t a b t (ca ca1) b Note that w, w, t is a basis of t Λ t Λ t Λ since w, w, t can be obtained from t, t, t by applying elementary transformations We claim that ρβw, ρβw a1 and ρβt b1 It is clear that ρβt b1, and an easy calculation, using the fact that ab ca, shows that ρβw a1 It is, of course, possible to verify that ρβw by direct calculation, but the details are rather unpleasant Instead, we observe that w t (t b (a1)) (ca1), where (t t (ca1)) a b, and invite the reader to verify that ρβ b (a1) With this information at hand, we have ρβw ρβt (ρβρβt b (a1)) (ca1) ρt (ca1)(ρβρβt b (a1)) (ca1) (ρt (ρβρβt b (a1))) (ca1) (a1(b (a1)(b1) b (a1))) (ca1) ((a1)b (a1)(b1) b (a1)) (ca1) (1b (b1) b )(a1) (ca1), as claimed Since w, w, t is a basis of t Λ t Λ t Λ, there exists an automorphism γ of t Λ t Λ t Λ such that γt w, γt w and γt t Let φβγ Clearly φ: t Λ t Λ t Λ t Λ t Λ is an epimorphism Moreover, ker φγ (ker β), so that ker φ ker β It follows that ker φ is non-cyclic Finally, we note that ρφt ρβγt ρβw, ρφt ρβγt ρβw a1 and ρφt ρβγt ρβt b1

14 MARTIN J EVANS The proof is complete LEMMA 32 Let ρ: t Λ t Λ t n Λ Λ be the Λ-homomorphism defined by ρt i g i 1, for i 1,, n Then there exists a Λ-epimorphism φ: t Λ t Λ t n Λ t Λ t Λ t n Λ such that (i) ker φ is not cyclic, and (ii) ρφt and ρφt i g i 1, for i 1,, n Proof If n2, then this is simply a restatement of Lemma 31, so we shall assume that n 3 Let φ and ρ be the maps referred to in Lemma 31, and define φ by φt φt, φt φt, φt φt and φt i t i for i 3, 4,, n Observe that φ is the restriction of φ to t Λ t Λ t Λ It follows that φ is an epimorphism and that ker φ ker φ From Lemma 31(i) we deduce that ker φ is not cyclic Part (ii) follows immediately from Lemma 32(ii) since ρ is the restriction of ρ to t Λ t Λ 4 Magnus embedding Let G be any group and let M be the free G-module with basis e,, e k Then it is easy to see that G M ( g m ) * gg,mm forms a group under formal matrix multiplication There is an extremely useful method, due to W Magnus [9], for obtaining representations of groups of the form F k R in such groups of matrices We next record a version of Magnus result as strengthened by V N Remeslennikov and V G Sokolov, who proved the final statement in this lemma in [12] LEMMA 41 Let x, x,, x k be a basis for F k, and let G be any group Let θ: F k G be an epimorphism with kernel R, and let e, e,, e k be a basis for a free G-module M Then the map x i R x i θ e i Moreoer, an element with m k i= e i r i is in the image of ψ if and only if extends to an injectie homomorphism ψ: F k R G M g m of G M k (x i θ1) r i g1 i= Let θ: F n n be the epimorphism defined by x i θ g i for i 1,, n Then ker θ γ (F n ), and applying Lemma 41 we obtain the following representation of U n F n [γ (F n ), γ (F n )]

LEMMA 42 FREE ABELIAN-BY-(NILPOTENT OF CLASS 2) GROUPS 141 U n is isomorphic to the group of all matrices of the form g n i= t i r i with g n, r, r,, r n Λ and n i= (g i 1)r i g1 Equialently, U n is isomorphic to the group of all matrices of the form g with g G, t Λ t n Λ and ρ g1, where ρ is the Λ-homomorphism defined in Lemma 32 It follows immediately from Lemma 41 that the group of matrices referred to in for i 1,, n t i We shall require one more application of Lemma 41 this lemma is generated by u,, u n where u i g i LEMMA 43 Let F n+ be the free group with basis x, x,, x n, and define an epimorphism π: F n+ n by x π 1 and x i π g i for i 1,, n Let S denote the kernel of π Then F n+ S is isomorphic to the group H of all matrices of the form g n i= t i r i with g n, r, r,, r n Λ and n i= (g i 1)r i Equialently, F n+ S is isomorphic to the group H of all matrices of the form g w with g n, w t Λ t Λ t n Λ and ρφw g1, where ρ and φ are the Λ-homomorphisms referred to in Lemma 32 5 The proofs of Theorem and Corollary 12 Throughout this section we shall identify U n with its matrix representation given in Lemma 42 Thus U n u,, u n Furthermore, H is the group of matrices defined in Lemma 43 LEMMA 51 Let φ: t Λ t Λ t n Λ t Λ t n Λ be the epimorphism obtained in Lemma 32 Then φ induces an epimorphism φ*: HU n defined by g w φ* g for all φw g H w Proof We begin by observing that the image of the function φ* lies in U n For let A g be an arbitrary element of H, so that ρφw g1 Then Lemma 42 w shows that Aφ* since ρ(φw) g1 We omit the routine argument which shows that φ* is a homomorphism, and so it remains to show only that φ* is surjective Now φ is surjective so, for each i 1,, n, there exists w i t Λ t Λ t n Λ such that φw i t i For each i 1,, n, we have that ρφw i ρt i g i 1, and therefore h i g i is an element of H Now w i h i φ* u i for all i 1,, n, and as U n is generated by u,, u n, we deduce that φ* is surjective

142 MARTIN J EVANS Our next lemma is well-known LEMMA 52 Let K be any integral group ring, and let (K)m denote the free Kmodule of rank m If µ: (K)m(K)mis a K-epimorphism, then µ is an isomorphism Proof Kaplansky has shown (see []) that if A and B are mm matrices with elements in the complex group ring G, then AB I m implies that BA I m Since (K)m is a projective K-module, we may deduce that (K)m (K)m ker µ The result now follows from [3, Proposition 21] LEMMA 53 Let φ*: HU n be the epimorphism defined in Lemma 51 Then ker φ* is not the normal closure in H of a single element Proof Let us write Λn+ for t Λ t Λ t n Λ and Λn for t Λ t n Λ Observe that if h ker φ*, then h 1 w for some w Λ n+ Suppose, for a contradiction, that ker φ* is the normal closure of h 1 in H Now Hker φ* is w isomorphic to the n-generator group U n, and consequently H ker φ*, h, h,, h n for some h, h,, h n H Since ker φ* is an abelian group that contains h, we deduce that ker φ* hh h h,h,,h n, and it follows that H h, h, h,, h n Let h i l i w i for i 1,, n, so that H - 1 w, l w,, l n w n It is easy to see that w, w,, w n is a basis of Λn+ Since h ker φ*, we have that φw, and therefore ker φ contains a basis element of Λn+, namely w Clearly Λn+wΛ Λn Moreover, φ: Λn+ Λn is an epimorphism, so we also have that Λn+ker φ Λn Now w ker φ, and so there exists a natural homomorphism µ: Λn+wΛ Λn+ker φ This is a Λ-epimorphism from a copy of Λn to another copy of Λn, and so Lemma 52 shows that µ is an isomorphism In particular, µ is monic and it follows that wλ ker φ However, Lemma 32 asserts that ker φ is not cyclic, and this contradiction completes the proof Proof of Theorem Recall from Lemma 43 that H is isomorphic to F n+ S, where F n+ S n Consequently, there exists an epimorphism δ: U n+ H We define the required epimorphism ψ n : U n+ by ψ n δφ* We must show that ker ψ n is not the normal closure in U n+ of a single element Suppose, for a contradiction, that ker ψ n is the normal closure in U n+ of an element u Then clearly ker φ* is the normal closure in H of uδ This contradicts Lemma 53, and completes the proof Proof of Corollary 12 (i) Since U n is finitely generated and abelian-by-nilpotent, it is residually finite [6, Theorem 1] Moreover, finitely generated residually finite groups are Hopfian [8], so it follows that U n is Hopfian Suppose, for a contradiction, that u ker ψ n is a primitive element of U n+ Let N denote the normal closure of u in U n+, and notice that U n+ N Note also

FREE ABELIAN-BY-(NILPOTENT OF CLASS 2) GROUPS 143 that U n+ ker ψ n Now N ker ψ n, so there is a natural map α: U n+ N U n+ ker ψ n Observe that α is an epimorphism from a copy of U n to a copy of U n Since U n is Hopfian, we deduce that α is an isomorphism Hence N ker ψ n, and therefore ker ψ n is the normal closure in U n+ of u This contradicts Theorem and completes the proof of (i) (ii) Let α: F n+ + be any epimorphism, and set β n αψ n Ifxis a primitive element of F n+ such that x ker β n, then clearly xα is a primitive element of U n+ such that xα ker ψ n However, by (i), ker ψ n does not contain any primitive elements of U n+, and we deduce that ker β n does not contain any primitive elements of F n+ 6 The proof of Theorem 14 Let x, x,, x n be a basis of F n+, and write γ for γ (F n+ ) We define an epimorphism π: F n+ n by x π 1 and x i π g i for i 1,, n Let S denote the kernel of π so that F n+ S n, and note that Lemma 43 shows that F n+ S is isomorphic to H via an isomorphism θ, say It follows from Lemma 53 that θφ*: F n+ SU n is an epimorphism such that ker θφ* is not the normal closure in F n+ S of a single element Moreover, since ker φ* is abelian, it follows that ker θφ* is also abelian Let N be the normal subgroup of F n+ such that S N and NSker θφ* Then F n+ N, and NS is an abelian group which is not the normal closure in F n+ S of a single element Since NS is abelian, N S, and it follows (see [2]) that N S Moreover, since γ S, it follows that γ N S Now F n+ N, and therefore F n+ γ N n Hence F n+ γ N F n+ S, and we deduce from the Hopficity of n that γ N S Let M be the normal closure of x, γ in F n+ Then F n+ M and γ M S Note that MS is the normal closure of x S in F n+ S Suppose, for a contradiction, that there exists an isomorphism α: F n+ N F n+ M Then (NN)α MM, since NN and MM are the Hirsch Plotkin radicals of F n+ N and F n+ M, respectively [1, Theorem 2] We claim that (SN)α SM Recall that γ N S γ M, so that SN(γ NN)(NN) and SM (γ MM)(MM) Now γ NNγ (F n+ N) and γ MMγ (F n+ M), so it follows that (γ NN)α γ MM We have seen above that (NN)α MM, and it now follows that (SN)α (γ NN)α(NN)α (γ MM)(MM) SM as claimed Thus (SN)α SM, and we observe that α therefore induces an automorphism α*: F n+ SF n+ S given by ( fs)α* ( fnα)s for all fs F n+ S Moreover, since (NN)α MM, we deduce that (NS)α* MS This gives the desired contradiction, since MS is the normal closure of x S in F n+ S, whereas NS is not the normal closure of a single element References 1 L AUSLANDER and E SCHENKMAN, Free groups, Hirsch Plotkin radicals, and applications to geometry, Proc Amer Math Soc 16 (1965) 784 788 2 M AUSLANDER and R C LYNDON, Commutator subgroups of free groups, Amer J Math 77 (1955) 929 931 3 P M COHN, Some remarks on the invariant basis property, Topology 5 (1966) 215 228 4 M J EVANS, Presentations of the free metabelian group of rank 2, Canad Math Bull 37 (1994) no 4, 468 472

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