Automorphism groups of relatively free groups

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Math. Proc. Camb. Phil. Soc. (1999), 127, 411 Printed in the United Kingdom

c 1999 Cambridge Philosophical Society

Automorphism groups of relatively free groups By R. M. BRYANT Department of Mathematics, University of Manchester Institute of Science and Technology, Manchester, M60 1QD and V. A. ROMAN’KOV Department of Mathematics, Omsk State University, 644077 Omsk, Russia (Received 28 October 1998)

1. Introduction For each positive integer n, let Fn be a free group of rank n with basis (in other words, free generating set) {x1 , . . . , xn }. If θ is an automorphism of Fn then {x1 θ, . . . , xn θ} is also a basis of Fn and every basis of Fn has this form. For any variety of groups V, let V(Fn ) denote the verbal subgroup of Fn corresponding to V. (See [10] for information on varieties and related concepts.) Let Gn = Fn /V(Fn ). Then Gn is a relatively free group of rank n in the variety V. By a basis of Gn we mean a subset S such that every map of S into Gn extends, uniquely, to an endomorphism of Gn . Write xi = xi V(Fn ) for i = 1, . . . , n. Then {x1 , . . . , xn } is a basis of Gn . If λ is an automorphism of Gn then {x1 λ, . . . , xn λ} is also a basis of Gn and every basis of Gn has this form. Any automorphism θ of Fn induces an automorphism θ of Gn in which xi θ = (xi θ)V(Fn ) for i = 1, . . . , n. Thus every basis of Fn induces a basis of Gn . The converse however is not always true; in general, there are automorphisms of Gn which are not induced by automorphisms of Fn . The most important positive results are for the variety A of all abelian groups and the variety AA of all metabelian groups (for any varieties U and V, UV denotes the variety of all groups with a normal subgroup in U and factor group in V). It is well known that, for all n, every automorphism of Fn /Fn0 can be lifted to an automorphism of Fn . (This is because the automorphism group of Fn /Fn0 is generated by ‘elementary’ automorphisms and each of these can be lifted.) The much deeper result of [1] and [11] shows that, for n 3, every automorphism of Fn /Fn00 can be lifted to an automorphism of Fn . (As usual, Fn0 denotes the derived group of Fn and Fn00 the second derived group.) However, there are many negative results for other varieties – see the introduction to [5] and the papers cited therein. An automorphism of Gn which is induced by an automorphism of Fn is called tame. If g1 , . . . , gm are distinct elements of Gn such that {g1 , . . . , gm } is contained in a basis of Gn then (g1 , . . . , gm ) is called a primitive system of Gn . If (f1 , . . . , fm ) is a primitive system of Fn then (f1 , . . . , fm ) induces a primitive system (f1 V(Fn ), . . . , fm V(Fn )) of Gn . But, in general, not every primitive system of Gn is induced by a primitive system of Fn .

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R. M. Bryant and V. A. Roman’kov

In this paper we consider two variations on the theme discussed above and we obtain results which apply to a wide class of relatively free groups. Suppose that k > n. Then, in the obvious way, we may regard the free group Fn with basis {x1 , . . . , xn } as a subgroup of the free group Fk with basis {x1 , . . . , xk }. More generally, since V(Fk ) w Fn = V(Fn ), we may regard Gn as a subgroup of Gk . If λ is an automorphism of Gn we can ask whether, for some k > n, there exists an automorphism θ of Fk such that xi θ = xi λ for i = 1, . . . , n. In fact, does there exist k > n such that every basis (more generally, every primitive system) of Gn is induced by a primitive system of Fk ? It can be deduced from the main result of [9] that the answer is positive if V is nilpotent, that is, if V is a subvariety of Nc for some positive integer c, where Nc denotes the variety of all groups which are nilpotent of class at most c. Also, by theorem 1 of [5], the answer is positive if AA ⊆ V ⊆ Nc A for some c. In this paper we generalize these two results by proving the following theorem for all nilpotentby-abelian varieties. Theorem A. Let V be a subvariety of Nc A, where c > 1. Let n be a positive integer and write k = 2n (n + 1) + 2c. Then every primitive system of Fn /V(Fn ) is induced by some primitive system of Fk . We are more interested in the existence of k than its magnitude: thus we have made no attempt to find the least possible k. It would however be interesting if we could take k to have the form k = n + e where e is independent of n. The main step in proving Theorem A is the proof of the same result for subvarieties of AA. There are related questions for a free group F of infinite rank. Although there are examples of varieties V where F/V(F ) has non-tame automorphisms (see [3]), this phenomenon appears not to be as widespread as in the finite rank case. By [6], if V ⊆ Nc for some c then every automorphism of F/V(F ) is tame. By [4] the same fact holds for V = AA and, by [5], it also holds for varieties V satisfying AA ⊆ V ⊆ Nc A for some c. Here we generalize these results as follows. Theorem B. Let V be a subvariety of Nc A, where c > 1. Let F be a free group of infinite rank. Then every automorphism of F/V(F ) is tame. As for Theorem A, the main step is for the subvarieties of AA. Theorem B has some immediate consequences through the results of [2] and [7] (see these papers for the definitions of the terms used in the following result). Corollary. Let Fω be a free group of countably infinite rank and let V be a subvariety of Nc A, where c > 1. Then Fω /V(Fω ) has the basis cofinality property and the small index property. The automorphism group Aut (Fω /V(Fω )) is not the union of a countable chain of proper subgroups. Also, Aut (Fω /V(Fω )) has no proper normal subgroup of index less than 2ℵ0 and it is a perfect group. Proof. By Theorem B, every automorphism of Fω /V(Fω ) is tame. Hence, by lemma 1·7 of [2], Fω /V(Fω ) has the basis cofinality property. (This could also be proved directly using a method similar to the proof of Theorem B.) Therefore, by theorem 1·3 of [2], Fω /V(Fω ) has the small index property and Aut (Fω /V(Fω )) is not the union of a countable chain of proper subgroups. The remaining results follow from the small index property by theorem C of [7].

Automorphism groups of relatively free groups

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2. Notation and preliminaries We first fix some notation which will be used throughout the paper. We write Fω for a free group of countably infinite rank with basis {x1 , x2 , . . . } and, for each positive integer n, Fn denotes the subgroup generated by {x1 , . . . , xn }. Thus Fn is a free group with basis {x1 , . . . , xn }. More generally, F denotes a free group with basis {xi : i ∈ I} where I is an arbitrary (not necessarily countable) index set. If Σ is a subset of I we write FΣ for the subgroup of F generated by {xi: i ∈ Σ}. For a given variety V, we write Gω = Fω /V(Fω ), Gn = Fn /V(Fn ), G = F/V(F ) and GΣ = FΣ /V(FΣ ). These are relatively free groups in the variety V. Since V(Fω )wFn = V(Fn ) and V(F ) w FΣ = V(FΣ ), we may regard Gn as a subgroup of Gω and GΣ as a subgroup of G. For each basis element xi of Fω or F we write xi for the corresponding basis element of Gω or G (that is, xi = xi V(Fω ) or xi = xi V(F )). There are two important cases for which we use special notation. For the variety A of all abelian groups the relatively free groups are free abelian groups which we denote by Aω , An , A and AΣ . The basis element corresponding to xi will be denoted by ai . For the variety AA of all metabelian groups we obtain free metabelian groups which we denote by Mω , Mn , M and MΣ . In this case the basis elements of Mω or M corresponding to xi will again be denoted by xi ; no confusion should arise in practice. For any group H an endomorphism θ of H will be called an IA-endomorphism if θ induces the identity map on H/H 0 ; that is (hθ)H 0 = hH 0 for all h ∈ H. An IA-automorphism is an automorphism which is an IA-endomorphism. If λ is an endomorphism of the relatively free group Gn we shall, when convenient, also regard λ as an endomorphism of Gω by taking xi λ = xi for all i > n. Similarly, for any positive integer k with k > n, we may regard λ as an endomorphism of Gk . More generally, if Σ ⊆ Ω ⊆ I, every endomorphism of GΣ may be regarded as an endomorphism of GΩ or of G. These remarks apply, in particular, to F , A and M . Note also that if λ is an automorphism or an IA-endomorphism then so too are the extensions of λ. Since V(F ) is a fully-invariant subgroup of F , every endomorphism θ of F induces an endomorphism θ of G in which xi θ = (xi θ)V(F ) for each i. If θ is an automorphism or an IA-endomorphism then so too is θ. If λ is an endomorphism of G then λ can be lifted to an endomorphism θ of F : we can define θ by taking xi θ, for each i, to be any element such that (xi θ)V(F ) = xi λ. If λ is an IA-endomorphism then it can be lifted to an IA-endomorphism, but if λ is an automorphism it cannot, in general, be lifted to an automorphism. We shall often consider the cases Fn , Fω and FΣ instead of F . If U is a subvariety of V then there is a natural epimorphism F/V(F ) → F/U(F ) and the remarks of the previous paragraph apply just as for the epimorphism F → G. If N is a normal subgroup of a group H and u, v ∈ H we write u ≡ v (mod N ) as an alternative to uN = vN . In the proof of Theorem B we shall need to make use of the ‘finitary lifting property’ as defined in [3] and [6]. For the convenience of the reader we give the relevant definitions here. An automorphism ξ of F is called finitary if there is a finite subset Ω of I such that xi ξ = xi for all i ∈ I\Ω. (This property is dependent on the choice of basis for the

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free group.) Let V be a variety of groups and let U be a subvariety of V. Suppose that I is infinite and that Γ and ∆ are subsets of I such that Γ w ∆ is empty, ∆ is finite and I\Γ is infinite. Suppose that λ is an automorphism of F/V(F ) which fixes xj for all j ∈ Γ and induces the identity automorphism on F/U(F ). We say that the triple (Γ, ∆, λ) can be lifted with respect to U if there exists a finitary automorphism ξ of F such that xj ξ = xj for all j ∈ Γ, (xi ξ)V(F ) = xi λ for all i ∈ ∆ and such that ξ induces the identity automorphism on F/U(F ). We say that V has the finitary lifting property with respect to U if, for every F where I is infinite, every triple (Γ, ∆, λ) of the above form can be lifted with respect to U. (We say that V has the finitary lifting property if it has this property with respect to the variety of all groups of order 1.) Theorem 3 of [3] shows that if V has the finitary lifting property with respect to U and if I is infinite then every automorphism of F/V(F ) which induces the identity automorphism on F/U(F ) is induced by an automorphism of F . 3. An application of Fox derivatives For a given positive integer n we consider the integral group ring ZFn and use the partial derivatives introduced by Fox [8]. In our notation these are defined as follows. For j = 1, . . . , n, the (left) Fox derivative associated with xj is the linear map Dj : ZFn → ZFn satisfying the conditions Dj (xj ) = 1, Dj (xi ) = 0 for i j and Dj (uv) = Dj (u) + uDj (v) for all u, v ∈ Fn . As is well known, Dj induces a linear map dj : ZMn → ZAn . For the sake of completeness, we briefly explain the details. There are natural epimorphisms π: Mn → An , π 0 : Fn → An and π 00 : Fn → Mn which extend to ring epimorphisms π: ZMn → ZAn , π 0 : ZFn → ZAn and π 00 : ZFn → ZMn . The kernels of π 0 and π 00 are the ideals of ZFn generated by the elements u − 1 with u ∈ Fn0 and u ∈ Fn00 , respectively. By an easy calculation, Dj ([u, v]) = u−1 (v −1 − 1)Dj (u) + u−1 v −1 (u − 1)Dj (v), for all u, v ∈ Fn . (Here, [u, v] denotes the commutator u−1 v −1 uv.) It follows that Dj (w) ∈ ker (π 0 ) for all w ∈ Fn00 . Hence Dj (ker (π 00 )) ⊆ ker (π 0 ) and Dj induces a linear map dj : ZMn → ZAn . From the definition we have dj (xj ) = 1, dj (xi ) = 0

for i j

and dj (uv) = dj (u) + (uπ)dj (v)

for all u, v ∈ Mn .

Let z1 , . . . , zm be elements of Mn and write z = (z1 , . . . , zm ). The Jacobian matrix J(z) is defined to be the m × n matrix over ZAn whose (i, j) entry is dj (zi ) for i = 1, . . . , m, j = 1, . . . , n. For elements a and b of a group H, ba denotes the conjugate a−1 ba. Since Mn0 is abelian, it may be regarded as a right Z(Mn /Mn0 )-module in the usual way, where


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the module action comes from conjugation in Mn . The epimorphism π: Mn → An induces an isomorphism from Mn /Mn0 to An . So we may regard Mn0 as a right ZAn module. For w ∈ Mn0 and s ∈ ZAn , we write ws to denote the image of w under the action of s (a notation which is consistent with our notation for conjugation). In the next section we shall use similar notation for Mω0 regarded as a right ZAω -module. There is a price to pay for using left Fox derivatives in connection with right modules: we need to use the involutory mapping of ZAn defined as follows. For s ∈ ZAn write s = Σi mi si , where mi ∈ Z and si ∈ An for each i and define ∗ s∗ = Σi mi s−1 i . Thus s 7→ s is an involutory linear mapping from ZAn to ZAn . For w ∈ Mn0 and s ∈ ZAn , it is easily verified that dj (ws ) = s∗ dj (w).

(3·1)

Also note that for w ∈ Mn0 and u ∈ Mn we have dj (wu) = dj (w) + dj (u).

(3·2)

Let u ∈ Mn and write u = u(x1 , . . . , xn ). Let w ∈ Mn0 and j ∈ {1, . . . , n}. Set u0 = u(x1 , . . . , xj−1 , wxj , xj+1 , . . . , xn ) and write ε(2) · · · vk xε(k) u = v1 xε(1) j v2 xj j vk+1 ,

where ε(1), . . . , ε(k) ∈ {1, −1} and where v1 , . . . , vk+1 are products of elements of {x1 , . . . , xj−1 , xj+1 , . . . , xn } and their inverses. Then u0 = v1 (wxj )ε(1) · · · vk (wxj )ε(k) vk+1 s = ws v1 xε(1) · · · vk xε(k) j j vk+1 = w u,

where s ∈ ZAn , namely s=

k X

ε(i)((v1 xε(1) · · · vi−1 xjε(i−1) vi xjδ(i) )−1 π) j

i=1

where δ(i) = 12 (ε(i)−1). It is well known and straightforward to check that s∗ = dj (u), that is, s = (dj (u))∗ . If w1 , . . . , wn ∈ Mn0 then, using the fact that elements of Mn0 commute, we obtain u(w1 x1 , . . . , wn xn ) = w1s1 · · · wnsn u,

(3·3)

where sj = (dj (u))∗ for j = 1, . . . , n. Lemma 3·1 (Roman’kov [12], Timoshenko [13]). Let z1 , . . . , zm be elements of Mn , where 1 6 m 6 n, and write z = (z1 , . . . , zm ). Then z is primitive in Mn if and only if the Jacobian matrix J(z) has a right inverse over ZAn ; that is, if and only if there is an n × m matrix B over ZAn such that J(z)B = Em , where Em is the m × m identity matrix over ZAn . Theorem 3·2. Let z1 , . . . , zm be elements of Mn , where 1 6 m 6 n. Suppose that there is an IA-endomorphism η of Mn such that zi η = xi for i = 1, . . . , m. Then there is an IA-automorphism ζ of Mn such that xi ζ = zi for i = 1, . . . , m.

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Proof. Since η induces the identity automorphism on An we can write xi η = wi xi where wi ∈ Mn0 for i = 1, . . . , n. Also, for i = 1, . . . , m, we can write zi = ui xi where ui ∈ Mn0 . Set ui = ui (x1 , . . . , xn ). Then, for i = 1, . . . , m, ui η = ui (x1 , . . . , xn )η = ui (w1 x1 , . . . , wn xn ). By (3·3), s

ui η = w1 i,1 · · · wnsi,n ui , where si,j = (dj (ui ))∗ for i = 1, . . . , m and j = 1, . . . , n. For i = 1, . . . , m, s

xi = zi η = (ui η)(xi η) = w1 i,1 · · · wnsi,n ui wi xi s

= w1 i,1 · · · wnsi,n wi zi .

For i = 1, . . . , m and j = 1, . . . , n, let ti,j = (dj (zi ))∗ . Using the Kronecker δij notation, we have, by (3·2), dj (zi ) = dj (ui xi ) = dj (ui ) + δij . t

t

Therefore ti,j = si,j + δij and so xi = w1i,1 · · · wni,n zi . Using (3·1) and (3·2), we obtain δij = dj (xi ) = d1 (zi )dj (w1 ) + · · · + dn (zi )dj (wn ) + dj (zi ), for i = 1, . . . , m and j = 1, . . . , n. Hence J(z)B = Em , where z = (z1 , . . . , zm ) and B is the n × m matrix with (i, j) entry equal to dj (wi ) + δij . It follows, by Lemma 3·1, that z is primitive in Mn . Therefore there is an automorphism χ of Mn such that xi χ = zi for i = 1, . . . , m. It remains to replace χ by an IA-automorphism. Let χ be the automorphism of An induced by χ. Then χ fixes a1 , . . . , am . Using additive notation for An , we can represent χ by a matrix over the integers of the form Em 0 , C= X1 X2 where Em is the m × m identity matrix. By elementary linear algebra, C may be reduced to the matrix Em 0 0 X2 by adding suitable multiples of the first m rows to the last n−m rows. Then it may be reduced further to the n×n identity matrix by means of elementary row and column operations involving only the last n − m rows and columns. Thus C = C1 C2 · · · Ck , where each Ci is an elementary matrix whose first m rows are the same as those of the identity matrix. It follows that χ = θ1 θ2 · · · θk , where θ1 , . . . , θk are ‘elementary’ automorphisms of An which fix a1 , . . . , am . It is easy to verify that for j = 1, . . . , k there is an automorphism χj of Mn which induces θj on An and fixes x1 , . . . , xm (this holds even at the level of Fn ). Let −1 ζ = χ−1 k · · · χ1 χ. Then xi ζ = xi χ = zi for i = 1, . . . , m. Also, ζ induces the identity automorphism on An , so ζ is an IA-automorphism.


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4. Laws in metabelian varieties Mω0

as a right ZAω -module and Mn0 as a right ZAn -module, as in Section We regard 3, and, for w ∈ Mω0 and s ∈ ZAω , we use the notation ws as in Section 3. Since Mω is metabelian it satisfies the ‘Jacobi identity’ [u, v, w][v, w, u][w, u, v] = 1 for all u, v, w ∈ Mω . (The commutators here are taken as left-normed.) In the case where u = xi , v = xj and w = xk we may write this as [xi , xj ](ak −1) [xj , xk ](ai −1) [xk , xi ](aj −1) = 1.

(4·1)

Lemma 4·1. Let u ∈ Mn0 . Then we may write Y [xi , xj ]si,j , u= i,j n>i>j>1

where si,j ∈ ZAi for all i, j. Proof. Let N be the subgroup of Mn generated by all elements of the form [xi , xj ]s where i, j ∈ {1, . . . , n}, i > j and s ∈ ZAi . It suffices to show that N = Mn0 . Clearly N ⊆ Mn0 and [xi , xj ] ∈ N for all i, j. Thus it suffices to show that N is normal in Mn . Suppose that i, j, k ∈ {1, . . . , n} where i > j and let s ∈ ZAi . It suffices to show −1 that [xi , xj ]sak and [xi , xj ]sak belong to N . This is clear if k 6 i. So suppose that k > i. Write [xi , xj ]sak = [xi , xj ](ak −1)s [xi , xj ]s and −1

[xi , xj ]sak = [xi , xj ](ak −1)t [xi , xj ]s , s where t = −sa−1 k . Note that s, t ∈ ZAk . Since [xi , xj ] ∈ N it suffices to show that [xi , xj ](ak −1)q ∈ N for all q ∈ ZAk . By (4·1),

[xi , xj ](ak −1)q = [xk , xj ](ai −1)q [xk , xi ](1−aj )q . The result follows since (ai − 1)q and (1 − aj )q belong to ZAk . Let H be a metabelian group. Then for every choice of g1 , g2 , . . . in H there is a unique homomorphism Mω → H such that xi 7→ gi for each i. If w ∈ Mn and we write w = w(x1 , . . . , xn ) then w(g1 , . . . , gn ) is the image of w under the given homomorphism. An element w of Mω is called a law of H if wξ = 1 for every homomorphism ξ: Mω → H. Thus, for w ∈ Mn , w is a law of H if and only if w(g1 , . . . , gn ) = 1 for all g1 , . . . , gn ∈ H. Lemma 4·2. Let V be a variety of metabelian groups and let Gn = Fn /V(Fn ) with basis {x1 , . . . , xn } corresponding to the basis {x1 , . . . , xn } of Fn . Let w ∈ Mn and write w = w(x1 , . . . , xn ). Then w is a law of every group of V if and only if w(x1 , . . . , xn ) = 1. Proof. This follows easily from 13·25 of [10]. Lemma 4·3. Let u ∈ Mn0 and write u as in the statement of Lemma 4·1. Suppose that

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u is a law of some metabelian group H. Then, for i = 2, . . . , n, [xi , x1 ]si,1 · · · [xi , xi−1 ]si,i−1

(4·2)

is a law of H. Also, v si,j (ai −1) is a law of H for all v ∈ Mω0 and all i, j satisfying n > i > j > 1. Proof. Suppose k ∈ {1, . . . , n}. Let uk be the element of Mk obtained by substituting 1 in place of xk+1 , . . . , xn in u. Thus Y [xi , xj ]si,j . uk = i,j k>i>j>1

Since u is a law of H, so is uk . For i ∈ {2, . . . , n}, let wi = ui u−1 i−1 ∈ Mi . Then wi is a law of H and wi is equal to (4·2). Suppose i and j satisfy n > i > j > 1 and let v ∈ Mω0 . Write wi = wi (x1 , . . . , xi ). Then wi (x1 , . . . , xj−1 , xj v, xj+1 , . . . , xi )wi (x1 , . . . , xi )−1 = [xi , v]si,j = v si,j (1−ai ) . Hence v si,j (ai −1) is a law of H. 5. Endomorphisms of metabelian groups Let V be a variety of metabelian groups and, for any positive integer n, let Gn = Fn /V(Fn ) with basis {x1 , . . . , xn }. As before, we have natural epimorphisms π: Mn → An and π : ZMn → ZAn in which xi π = ai for i = 1, . . . , n. Similarly, there is an epimorphism Mn → Gn such that xi 7→ xi for i = 1, . . . , n. For any element u of Mn we write u for the image of u in Gn . Recall from Section 2 that, for k > n, every endomorphism of Mn may be regarded as an endomorphism of Mk which fixes xn+1 , . . . , xk . The purpose of this section is to prove the following result. Although it may seem rather technical it is the key to the proof of the main theorems of the paper. Theorem 5·1. Let n and m be positive integers where 1 6 m 6 n and write k = 2m n. Let Gn = Fn /V(Fn ), where V is a variety of metabelian groups. Let y1 , . . . , ym be elements of Mn such that yi Mn0 = xi Mn0 for i = 1, . . . , m. Suppose that φ is an IAendomorphism of Mn such that the induced endomorphism φ of Gn satisfies y i φ = xi for i = 1, . . . , m. Then there exist elements z1 , . . . , zm of Mk and an IA-endomorphism ψ of Mk such that z i = y i for i = 1, . . . , m, zi φψ = xi for i = 1, . . . , m and ψ fixes xm+1 , . . . , xn . Proof. The theorem is trivial if n = 1, so we assume that n > 2. Extend the statement of the theorem, in the obvious way, to apply to m = 0. For m = 0 we have k = n and the theorem is true in this case because we can take ψ to be the identity automorphism of Mn . With n fixed, we argue by induction on m, assuming that m > 1 and that the result is true for m − 1. Let e = 2m−1 n. Then, by the inductive hypothesis, there exist z1 , . . . , zm−1 ∈ Me and an IA-endomorphism η of Me such that z i = y i for i = 1, . . . , m − 1, zi φη = xi for i = 1, . . . , m − 1 and η fixes xm , . . . , xn . In order to simplify the notation for the remainder of the proof, it is convenient


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to renumber the subscripts by interchanging 1 and m. After this change we have z2 , . . . , zm ∈ Me and an IA-endomorphism η of Me such that z i = y i for i = 2, . . . , m, zi φη = xi for i = 2, . . . , m and η fixes x1 , xm+1 , . . . , xn . Since y1 Mn0 = x1 Mn0 and φ and η are IA-endomorphisms we can write y1 φη = ux1 where u ∈ Me0 . But, since y 1 φ = x1 and η fixes x1 , we have y 1 φη = x1 . Therefore u = 1 and so, by Lemma 4·2, u is a law of every group of V. In particular, u is a law of Gk , where k is as in the statement of the theorem. Note that k = 2e. By Lemma 4·1 we may write Y [xi , xj ]si,j , u= i,j e>i>j>1

where si,j ∈ ZAi for all i, j. Let W be the ZAk -submodule of Mk0 generated by all [xi , xj ] with i, j ∈ {2, . . . , e}. For i = 2, . . . , e, write si instead of si,1 and note that for all i, j satisfying e > i > j > 1 we have [xi , xj ]si,j ∈ W . Thus u = [x2 , x1 ]s2 · · · [xe , x1 ]se u0 , where u0 ∈ W . By Lemma 4·3, v si (ai −1) is a law of Gk for all v ∈ Mω0 and i = 2, . . . , e. Define z1 ∈ Mk by z1 = [xe+1 , x2 ]s2 (a2 −1) · · · [x2e−1 , xe ]se (ae −1) y1 .

(5·1)

Hence z 1 = y 1 . It is enough to find an IA-endomorphism ζ of Mk such that z1 φηζ = x1 and ζ fixes x2 , . . . , xn : for then we can take ψ = ηζ. We shall first calculate z1 φη. By (5·1), z1 φη is the product of the elements [xj+e−1 , xj ]sj (aj −1) φη, for j = 2, . . . , e, and the element y1 φη. Let j ∈ {2, . . . , e}. Then, by a routine calculation, [xj+e−1 , xj ]sj (aj −1) φη = [xj+e−1 , xj φη]sj (aj −1) = [xj+e−1 , xj ]sj (aj −1) vj , where vj is a product of elements of the form [xi , xi0 ]sj (aj −1)s with e > i > i0 > 1 and s ∈ ZAk . If i0 1 then [xi , xi0 ]sj (aj −1)s ∈ W. But, for i0 = 1, (4·1) gives [xi , xi0 ]sj (aj −1)s = [xj , x1 ]sj (ai −1)s [xj , xi ]−sj (a1 −1)s , where [xj , xi ]−sj (a1 −1)s ∈ W . Thus vj can be written in the form vj = [xj , x1 ]sj bj vj0 , where bj ∈ ZAk and vj0 ∈ W . Also, as already seen, y1 φη = ux1 = [x2 , x1 ]s2 · · · [xe , x1 ]se u0 x1 . Collecting together similar terms we may write z1 φη = [x2 , x1 ]s2 t2 · · · [xe , x1 ]se te [xe+1 , x2 ]s2 (a2 −1) · · · [x2e−1 , xe ]se (ae −1) vx1 , where t2 , . . . , te ∈ ZAk and v ∈ W .

(5·2)

420


Write q = s2 t2 (a2 − 1)a1 + 1 and b = s22 t32 a21 . Let ξ2 be the IA-endomorphism of Mk defined by x1 ξ2 = [x2 , x1 ]−s2 t2 q x1 , xe+1 ξ2 = xe+1 [x2 , x1 ]−b and xi ξ2 = xi for i 1, e + 1. We next calculate z1 φηξ2 . By a re-arrangement of (5·2), z1 φηξ2 is the product of the images under ξ2 of the elements [x3 , x1 ]s3 t3 · · · [xe , x1 ]se te ,

[xe+2 , x3 ]s3 (a3 −1) · · · [x2e−1 , xe ]se (ae −1) v, [xe+1 , x2 ]s2 (a2 −1)

and

[x2 , x1 ]s2 t2 x1 .

Let j ∈ {3, . . . , e}. Then [xj , x1 ]sj tj ξ2 = [xj , x1 ξ2 ]sj tj = [xj , x1 ]sj tj [x2 , x1 ]sj tj s2 t2 q(aj −1)a1 . By (4·1), we have [x2 , x1 ](aj −1) = [xj , x1 ](a2 −1) [x2 , xj ](a1 −1) . Hence we may write 0 [xj , x1 ]sj tj ξ2 = [xj , x1 ]sj tj uj , where t0j ∈ ZAk and uj ∈ W . Also, [xe+2 , x3 ]s3 (a3 −1) · · · [x2e−1 , xe ]se (ae −1) v is fixed by ξ2 . Furthermore, 2

[xe+1 , x2 ]s2 (a2 −1) ξ2 = [xe+1 ξ2 , x2 ]s2 (a2 −1) = [x2 , x1 ]−s2 (a2 −1) b u00 , where u00 ∈ W . Finally, ([x2 , x1 ]s2 t2 x1 )ξ2 = [x2 , x1 ξ2 ]s2 t2 [x2 , x1 ]−s2 t2 q x1 2 2

= [x2 , x1 ]s2 t2 [x2 , x1 ]s2 t2 q(a2 −1)a1 [x2 , x1 ]−s2 t2 q x1 2

= [x2 , x1 ]s2 t2 +s2 t2 q(q−1)−s2 t2 q x1 = [x2 , x1 ]s2 t2 (q−1) x1 3 3

2 2

2

= [x2 , x1 ]s2 t2 (a2 −1) a1 x1 = [x2 , x1 ]s2 (a2 −1) b x1 . It follows that 0

0

z1 φηξ2 = [x3 , x1 ]s3 t3 · · · [xe , x1 ]se te [xe+2 , x3 ]s3 (a3 −1) · · · [x2e−1 , xe ]se (ae −1) v 0 x1 ,

(5·3)

0

where v ∈ W . Comparing (5·2) and (5·3) we see that the effect of ξ2 has been to eliminate [x2 , x1 ]s2 t2 together with the corresponding term [xe+1 , x2 ]s2 (a2 −1) . Note also that ξ2 fixes xi for all i except i = 1 and i = e + 1. 0 If e > 3 we may now use an IA-endomorphism ξ3 to eliminate [x3 , x1 ]s3 t3 together with the corresponding term [xe+2 , x3 ]s3 (a3 −1) , where ξ3 fixes xi for all i except i = 1 and i = e + 2. Continuing in this way, we obtain IA-endomorphisms ξ2 , . . . , ξe such that z1 φηξ2 · · · ξe = wx1 , where w ∈ W . Furthermore, ξ2 · · · ξe fixes x2 , . . . , xn . Finally, let σ be the IAendomorphism of Mk defined by x1 σ = w−1 x1 and xi σ = xi for i 1. Then z1 φηξ2 · · · ξe σ = x1 and ξ2 · · · ξe σ fixes x2 , . . . , xn . The result follows by setting ζ = ξ2 · · · ξe σ. 6. Automorphisms of relatively free metabelian groups In this section we shall obtain Theorems A and B in the case where the variety V is metabelian. Theorem 6·1. Let n be a positive integer and write k = 2n (n+1). Let Gn = Fn /V(Fn ), where V is a variety of metabelian groups. Then every primitive system of Gn is induced by a primitive system of Fk .


421

Proof. It suffices to show that every basis of Gn is induced by a primitive system of Fk . Let (g1 , . . . , gn ) be a basis of Gn . Since (g1 G0n , . . . , gn G0n ) is a basis of Gn /G0n , it 0 follows from lemma 1 of [6] that there is an automorphism τ of Fn+1 /Fn+1 such that 0 0 0 0 if τ is the automorphism of Gn+1 /Gn+1 induced by τ then (xi Gn+1 )τ = gi G0n+1 for 0 i = 1, . . . , n. But every automorphism of Fn+1 /Fn+1 can be lifted to an automorphism of Fn+1 (see Section 1). Hence there is an automorphism ξ of Fn+1 such that the induced automorphism ξ of Gn+1 satisfies xi ξ ≡ gi (mod G0n+1 ) for i = 1, . . . , n. Since (g1 , . . . , gn ) is a primitive system of Gn+1 there is a basis (h1 , . . . , hn+1 ) of −1 Gn+1 such that hi = gi ξ for i = 1, . . . , n. Then, for i = 1, . . . , n, we have hi ≡ gi ξ

−1

≡ xi (mod G0n+1 ).

(6·1)

It suffices to find an automorphism η of Fk such that xi η = hi for i = 1, . . . , n; then (g1 , . . . , gn ) is induced by the primitive system (x1 ηξ, . . . , xn ηξ). Let µ be the endomorphism of Gn+1 defined by hi µ = xi for i = 1, . . . , n and hn+1 µ = hn+1 . By (6·1), µ is an IA-endomorphism of Gn+1 . Let φ be any IA-endomorphism of Mn+1 which induces µ on Gn+1 . Thus the induced endomorphism φ satisfies hi φ = xi 0 for i = 1, . . . , n. Let y1 , . . . , yn be elements of Mn+1 such that yi ≡ xi (mod Mn+1 ) and y i = hi for each i. By Theorem 5·1, there exist elements z1 , . . . , zn of Mk and an IA-endomorphism ψ of Mk such that z i = hi for i = 1, . . . , n and zi φψ = xi for i = 1, . . . , n (and ψ fixes xn+1 ). Therefore, by Theorem 3·2, there exists an IA-automorphism ζ of Mk such that xi ζ = zi for i = 1, . . . , n. Since k > 4, the main result of [1] or [11] yields that there is an automorphism η of Fk which induces ζ on Mk . Hence, for i = 1, . . . , n, we have xi η = xi ζ = z i = hi , as required. In the terminology described in Section 2, the next result shows that if V is a variety of metabelian groups then V has the finitary lifting property with respect to V w A. (In fact it says a little more than this because we may have V w A A.) Theorem 6·2. Let F be a free group with basis {xi : i ∈ I} where I is infinite and let G = F/V(F ), where V is a variety of metabelian groups. Suppose that Γ and ∆ are subsets of I such that Γ and ∆ are disjoint, ∆ is finite and I\Γ is infinite. Let λ be an IA-automorphism of G which fixes xj for all j ∈ Γ. Then there is a finitary IA-automorphism ξ of F such that xj ξ = xj for all j ∈ Γ and xi ξ = xi λ for all i ∈ ∆. Proof. Without loss of generality we may assume that ∆ is non-empty. Let Σ be a finite subset of I such that ∆ ⊆ Σ and GΣ contains G∆ λ. Let ρ: G → GΣ be the (retraction) homomorphism defined by xi ρ = xi for i ∈ Σ and xi ρ = 1 for i ∈ I\Σ. Let ι: GΣ → G be the (inclusion) homomorphism defined by xi ι = xi for all i ∈ Σ. Then let µ: GΣ → GΣ be defined by µ = ιλ−1 ρ. For all i ∈ I we may write xi λ−1 = xi ui where ui ∈ G0 . Thus, for all i ∈ Σ, xi µ = (xi λ−1 )ρ = (xi ρ)(ui ρ) = xi (ui ρ), where ui ρ ∈ G0Σ . Thus µ is an IA-endomorphism of GΣ . Also, for j ∈ Σ w Γ, we have xj µ = xj λ−1 ρ = xj ρ = xj . Lift µ to an IA-endomorphism φ of MΣ such that φ fixes xj for j ∈ Σ w Γ. For all i ∈ ∆ write gi = xi λ. Thus gi ∈ GΣ and gi µ = gi λ−1 ρ = xi ρ = xi . Hence the induced endomorphism φ satisfies gi φ = xi for all i ∈ ∆.

422


Since µ is an IA-endomorphism, gi ≡ xi (mod G0Σ ) for all i ∈ ∆. Thus, for each i ∈ ∆, there is an element yi of MΣ such that yi ≡ xi (mod MΣ0 ) and y i = gi . We apply Theorem 5·1 with ∆ taking the place of {1, . . . , m}, Σ taking the place of {1, . . . , n} and a finite subset Ω of I\(Γ x Σ) taking the place of {n + 1, . . . , k}. Hence, there exist elements zi of MΣxΩ , for each i ∈ ∆, and an IA-endomorphism ψ of MΣxΩ such that z i = y i = gi for i ∈ ∆, zi φψ = xi for i ∈ ∆ and ψ fixes xj for j ∈ Σ\∆. We have zi φψ = xi for i ∈ ∆ and xj φψ = xj for j ∈ Σ w Γ. Hence, by Theorem 3·2, there is an IA-automorphism ζ of MΣxΩ such that xi ζ = zi for i ∈ ∆ and xj ζ = xj for j ∈ Σ w Γ. Regard ζ as an automorphism of M (fixing xi for i ∈ I\(Σ x Ω)). Then ζ is an IA-automorphism of M such that xj ζ = xj for all j ∈ Γ. But, by the results of section 2·3 of [4], AA has the finitary lifting property with respect to A. Therefore there exists a finitary IA-automorphism ξ of F such that xj ξ = xj for all j ∈ Γ and such that the automorphism ξˆ induced by ξ on M satisfies xi ξˆ = xi ζ for all i ∈ ∆. Hence, for all i ∈ ∆, we have xi ξˆ = zi , giving xi ξ = z i = gi = xi λ. This completes the proof. Theorem 6·3. Let F be a free group of infinite rank and let V be a variety of metabelian groups. Then every automorphism of F/V(F ) is tame. Proof. Write G = F/V(F ) and let λ be an automorphism of G. Since every automorphism of G/G0 is tame (see [6]), there exists an automorphism χ of F which induces an automorphism χ of G such that λχ−1 is an IA-automorphism. By Theorem 6·2, V has the finitary lifting property with respect to VwA. Hence, by theorem 3 of [3], λχ−1 is induced by an automorphism ξ of F . Therefore λ is induced by ξχ. 7. Relatively free nilpotent-by-abelian groups Let F be a free group with basis {xi: i ∈ I}, where I may be finite or infinite. Let V be a variety satisfying V ⊆ Nc A, where c > 2, and write U = V w Nc−1 A. Write U = U(F ) and V = V(F ). Thus U/V is abelian and γc+1 (F 0 ) ⊆ V ⊆ U = γc (F 0 )V ⊆ γ2 (F 0 )V. Lemma 7·1. With the notation described above, let α and α0 be automorphisms of F which induce the identity automorphism on F/U . For all i ∈ I write xi α ≡ xi ui (mod V ),

xi α0 ≡ xi u0i (mod V ),

where ui , u0i ∈ U . Then, for all i ∈ I, xi αα0 ≡ xi u0i ui ≡ xi ui u0i (mod V ). Proof. Since α0 acts as the identity automorphism on F/U it acts as the identity on F 0 V /γ2 (F 0 )V . Hence α0 acts as the identity on γc (F 0 )V /γc+1 (F 0 )V , which is equal to U/V . The result follows as in the proof of the lemma in [5]. The next three results are proved in exactly the same way as the corresponding results in [5]. Proposition 7·2. With the same notation, suppose that {0, 1, . . . , 2c} ⊆ I and write


423

J = {1, . . . , 2c}. Let r(1), . . . , r(2c) ∈ I and let f1 , . . . , fc ∈ F . Then there is an automorphism α of F such that x0 α ≡ x0 [ [xr(1) , xr(2) ]f1 , . . . , [xr(2c−1) , xr(2c) ]fc ] (mod V ), xj α ≡ xj (mod U ) for j ∈ J and xi α = xi for i ∈ I\(J x {0}). Corollary 7·3. With the same notation, suppose that |I| > 2c + 1, that J is a proper subset of I with |J| > 2c and that r ∈ I\J. Let u ∈ U . Then there exists an automorphism α of F such that xr α ≡ xr u (mod V ), xj α ≡ xj (mod U ) for j ∈ J and xi α = xi for i ∈ I\(J x {r}). Corollary 7·4. With the same notation, suppose that I is infinite and let Γ and ∆ be subsets of I such that Γ and ∆ are disjoint, ∆ is finite and I\Γ is infinite. Let θ be an automorphism of F/V which induces the identity automorphism on F/U and which satisfies (xj V )θ = xj V for all j ∈ Γ. Then there is a finitary automorphism ξ of F such that ξ induces the identity automorphism on F/U , xj ξ = xj for all j ∈ Γ and (xi ξ)V = (xi V )θ for all i ∈ ∆. Corollary 7·4 shows that V has the finitary lifting property with respect to U (see the definition in Section 2). We can now prove Theorems A and B, as stated in Section 1. Proof of Theorem A. If c = 1 then the result follows by Theorem 6·1. Thus we assume that c > 2 and argue by induction on c. We use the notation introduced at the beginning of this section with Fk taking the place of F . It suffices to show that every basis of Fn /V(Fn ) is induced by a primitive system of Fk . Let (g1 , . . . , gn ) be a basis of Fn /V(Fn ). Take elements f1 , . . . , fn of Fn such that fi V(Fn ) = gi for each i. Then (f1 U(Fn ), . . . , fn U(Fn )) is a basis of Fn /U(Fn ), so, by the inductive hypothesis, it may be lifted to a primitive system of Fk . Thus there is an automorphism ξ of Fk such that (xi ξ)U = fi U for i = 1, . . . , n. It suffices to find an automorphism η of Fk such that (xi η)V = (fi ξ −1 )V for i = 1, . . . , n; then (g1 , . . . , gn ) is induced by the primitive system (x1 ηξ, . . . , xn ηξ). Since (xi ξ)U = fi U for i = 1, . . . , n, there exist elements u1 , . . . , un of U such that fi ξ −1 = xi ui for each i. By Corollary 7·3, with J = {n + 1, . . . , k}, there exist automorphisms α1 , . . . , αn of Fk such that, for r = 1, . . . , n, αr satisfies xr αr ≡ xr ur (mod V ), xj αr ≡ xj (mod U ) for j ∈ {n + 1, . . . , k} and xi αr = xi for i ∈ {1, . . . , n}\{r}. Let η = α1 · · · αn . Then, by Lemma 7·1, xi η ≡ xi ui (mod V ) for i = 1, . . . , n. The result follows. Proof of Theorem B. If c = 1 then the result follows by Theorem 6·3. Thus we assume that c > 2 and argue by induction on c. We use the notation introduced at the beginning of this section. Let λ be an automorphism of F/V . Then λ induces an automorphism of F/U which, by the inductive hypothesis, may be lifted to an automorphism χ of F . Let χ be the automorphism of F/V induced by χ. Then λχ−1 induces the identity automorphism on F/U . By Corollary 7·4, V has the finitary lifting property with respect to U. Hence, by theorem 3 of [3], λχ−1 is induced by an automorphism ξ of F . Therefore λ is induced by ξχ.

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