Characteristic Subgroups and Complete

Characteristic Subgroups and Complete Automorphism Groups Joan L. Dyer; Edward Formanek American Journal of Mathematics, Vol. 99, No. 4. (Aug., 1977), pp. 713-753. Stable URL: http://links.jstor.org/sici?sici=0002-9327%28197708%2999%3A4%3C713%3ACSACAG%3E2.0.CO%3B2-H American Journal of Mathematics is currently published by The Johns Hopkins University Press.

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CHARACTERISTIC SUBGROUPS AND COMPLETE

AUTOMORPHISM GROUPS*.

FORMANEK** By JOANL. DYERand EDWARD

Introduction. Let A ( G ) denote the automorphism group of the group G, and let G+A(G) be the homomorphism which assigns to each g E G the inner automorphism x b g x g - l . G is said to be complete if G-+A(G) is an isomorphism. In [6] we proved that A ( F ) is complete if F is a free group of finite rank greater than one. In this paper we attempt to establish the completeness of A ( G ) when G = F / C and C is a characteristic subgroup of F. We assume that the center of G is trivial. This implies that G+A(G) is injective, as is A ( G ) - + A ( A ( G )=A'(G), ) and we will regard G and A ( G ) as subgroups of A 2 ( ~ )The . action of A ( G ) on G is then given by conjugation within A (G). In this situation Burnside [4, p. 951 has shown that A ( G ) is complete if and only if G is a characteristic subgroup of A(G), that is, if and only if G is a normal subgroup of A'(G). Our approach is to characterize G in terms of other normal subgroups of A2(G).We have been able to do this for a fairly large class of groups G F/C, C characteristic in F (Theorem E), but by no means for all such groups. We have had more success showing that G ' is a normal subgroup of A'(G) (Theorem A). The main results are the following, where yiG denotes the j-th term of the lower central series of G, G ' = y,G, and TiG= { g E G l g k E yiG for some O# k E 2 ) . F is always a free group of finite rank n > 2. THEOREM A. Let G = F/C, where C is a characteristic subgroup of F. Suppose that G is residually torsion-free nilpotent and has trivial center. Then (1) K = Ker{A(G)+A(G/G1)) (2) K2 = Ker{A (G)+A (G/t,G)), and (3) GK2 Manuscript received October 19, 1974. *This paper was written while both authors were at the Institute for Advanced Study, Princeton. **Research supported by National Science Foundation grant GP-36418x1. American Journal of Mathematics, Vol. 99, No. 4, pp. 71g753

Copyright O 1977 by Johns Hopkins University Press.

714

JOAN L. DYER AND EDWARD FORMANEK.

are normal subgroups of A'(G). If i n addition K / G is residually nilpotent, then (4) G ' is a normal subgroup of A'(G). In view of Burnside's theorem cited above, we have the following corollary

to Theorem A:

COROLLARY A. Let G,K be as above with K / G residually nilpotent. If the center of K / G ' is G / G', then A ( G ) is complete.

Theorem A guarantees that K and G ' are normal subgroups of A'(G). Therefore the preimage in K of the center of K / G ' is also normal in A 2 ( ~ ) . Under the hypothesis of Corollary A, G is this preimage. Since G is normal in A2(G), by Burnside's theorem A ( G ) is complete. We note that [K, GI < G ' by the definition of K so G / G f is always contained in the center of K/G'. The residual nilpotence of K / G is implied by the statement: If a € A (G), then a E G (that is, a is inner) if and only if the automorphism induced by a on G/yiG is inner for all j (see Corollary 5.7). This follows from the fact that the groups

form a descending central series for K, and the statement above says that n GK, = G. We are therefore led to the investigation of two properties (where a primitive element of G = F/ C is the image of a primitive element of F):

Property I. If a E A ( G ) and for every primitive x e G, a ( x ) and x are conjugate in G, then a E G. Property II. If x is a primitive element of G which is not conjugate to y E G, then the images of x and y are not conjugate in G/TiG for some integer

i. In using Properties I and I1 we are following Grossman [8, cf. Theorem 11. The result we obtain is THEOREM B. Let G = F/ R ', where R is a normal subgroup of F such that F I R is torsion-free. Then (1) G is torsion-free and has trivial center, and (2) G has Property I. If i n addition F I R is residually torsion-free nilpotent, then (3) G is residually torsion-free nilpotent, and (4) G has Property II.

R

< F' and

CHARACTERIST~C SUBGROUPS.

715

Part (1) follows from Auslander-Lyndon [ l , Theorem 21, and part (3) is due to Hartley [ l l , Theorem D2], As a corollary to Theorem B we have THEOREM C. Let G = F I R ' , where R is a normal subgroup of F such that R < F' and F/ R is residually torsion-free nilpotent. Then K / G is residually nilpotent. As remarked earlier, we seek to characterize G in terms of normal subgroups of A'(G). This motivates

R

THEOREM D. Let G = F/ R ', where R is a characteristic subgroup of F, F I R is residually torsion-free nilpotent. Then

< F', and

G = { ~ E K[ a~, K ]

< G').

The next theorem is our main result. THEOREM E. Let G = F / R ', where R is a characteristic subgroup of F, R < F', and F/ R is residually torsion-free nilpotent. Then A ( G ) is complete. Theorem E is a consequence of Theorems B to D and Corollary A, for G is residually torsion-free nilpotent with trivial center by Theorem B and by Theorem C, K / G is residually nilpotent. Finally by Theorem D, G / G ' is the center of K/G'. Thus Corollary A applies, and A ( G ) is complete. The main part of this paper is devoted to establishing Theorems A to D, and is organized as follows. In sections 1 and 2 we prove Theorem A, first by showing that K and K, are normal in A,(G) and then by studying K/K, as a left A(G)/K-module as in 16, section 31. The determination of A ( G L ( ~ , z ) b) y Hua and Reiner [12] plays a decisive role. The remainder of the paper is concerned with groups G of the form F I R ' . In section 3 we describe Magnus' representation of F/ R ' as a certain group of 2 X2 matrices [14] and its connection with Fox's free differential calculus [7]. The group structure of F I R ' is thereby related to that of the integral group ring Z[F/ R]. We prove Theorem B (1) and (3) in a form which will enable us to reduce the remainder of B and Theorem D to the case in which F I R is a torsion-free and nilpotent group. We then cite some facts about the group rings of such groups. In section 4 we prove that G has Property I, and in section 5 that G has Property 11; Theorem C is also proved in section 5. Theorem D is proved in the final section 6. In sections 4 to 6 the fundamental tool is the Magnus-Fox representation of F/ R ', We use a decimal numbering in sections 1 through to 6, so that Lemma 3.4

716


is the fourth result in section 3. To aid the reader we include a small "table of contents". Theorem A (1)= Theorem 1.6

Theorem A (2) = Theorem 2.4

Theorem A (3)= Theorem 2.8

Theorem A (4)= Theorem 2.9

Theorem B (1) = Theorem 3.5

Theorem B (2) = Theorem 4.1

Theorem B (3) = Corollary 3.9

Theorem B (4)= Theorem 5.5

Theorem C = Corollary 5.7

Theorem D = Theorem 6.6

For motivation and background the reader is referred to 161. The treatment there illustrates the approach we have taken here but does not have the complications of the present paper. We thank Gilbert Baumslag for his interest and we are grateful to him for raising the questions about automorphism groups which motivated our work. 1. The Invariance of K . Throughout this paper, F denotes a free g m p of finite rank n 2 2. In sections 1 and 2 C denotes a fixed characteristic subgroup of F , and G = F/C. W e assume that the center of G is trivial and that G is residually torsion-free nilpotent. We write [x, y] = xyx-ly-' and xY= yxy -l, SO that x(yZ)=(xZ)Y.We will view G and A (G) as subgroups of A ~ ( G )For . a € A (G), g E G,

and similarly for the action of A'(G) on its subgroups A(G) and G. Let yjG be the j-th term of the lower central series of G, defined inductively by

and set

Then 5 G is the smallest subgroup of G such that G/yjG is torsion-free and nilpotent of class at most j - 1. Alternatively, yjG/yjG is the torsion subgroup

CHARACTERISTIC SUBGROUPS.

of G / y i G . By hypothesis, the series

is a properly descending fully invariant filtration of G: that is, n YiG= 1 (because G is residually torsion-free nilpotent), YiG#Yi+,G (because G is centerless), [yiG,ykG]< %+&G,and for any endomorphism a : G+G, a(yiG)< v,G. The elements x,, . . . ,xn will denote a fixed free generating set for F, and we will not distinguish between elements of F and their images in G under the natural map F+ F/ C = G. The symmetric group Sn of -permutations of (1,. . .,n ) is a subgroup of A (G), where a Sn is identified with the automorphism a E A ( G ) given by a:xic-,xTi ( i = l , ...,n). For i = 1,. ..,n, the automorphism ui is defined by

and we set 0 = u , ~ ~ ~ u n : x i ~ x i - ' . LEMMA1.1. Let D be a characteristic subgroup of F such that F/D is torsion-free. (1) If F > D > y2F, then D = F. (2) If y2F > D > y3F, then D = y2F. (3) If y3F > D > y,F, then D = y,F. Proof. (1) Since D is characteristic, D = Fmy2F for some integer m#O. Since F / D is torsion-free, m = 1 and so D = F. (2) The cosets containing the commutators {[xi,xi] i < i} are a basis for the free abelian group y2F/y3F. Let

/

be a nontrivial element of D/y3F with the smallest number of nonzero exponents e(i,1). Modulo y3F, conjugation by ui inverts [x,,x,] if i € { r , ~ a}nd fixes [x,, x,] otherwise. If there are two nonzero exponents e ( p, q) and e(r,s) with, say, p @ { r , s ) , then d(updup-') is a nontrivial element of D l y 3 F with fewer nonzero exponents than d, contradicting the choice of d. Hence Exp,xqIe(p'q)E D for some p, q with e ( p, q) # 0 . Since F / D is torsion-free, [xp,xq]E D . Since the images of [xp,x,] under the symmetric group Sn< A ( F )

718


generate y2F modulo y3F and D is characteristic in F, it follows that D = y2F. (3) The cosets containing the commutators { [ x i ,[xi,xk]]1 i > j < k) are a basis for the free abelian group y3F/y4F. Let

be a nontrivial element of D / y 4 F with the smallest number of nonzero exponents. Since conjugation by al inverts [xi, [xi,xk]] modulo y4F if xl occurs exactly once in the sequence xi,xi, xk and fixes [xi, [xi,xk]] modulo y4F otherwise, it follows from our choice of d that dot= d ". Under the action of S, we may therefore assume that either

or that

d=n([xi,

1) and

e(2)fO.

When n = 2, (**) is the only possibility and D = y3F follows as in (2) above. Thus we may further assume that n > 3. In case (**), let a E A ( F ) be defined by

Then d - ld a ( d - 'd ")-"3is a nontrivial element of D which has the form (*), so we may assume (*) occurs. Define ,b E A ( F ) by

Then

and belongs to D / y 4 F , so each commutator with two equal entries belongs to D. Modulo y4F

which therefore also belongs to D. Since

CHARACTERISTIC

SUBGROUPS.

[IS, p. 293 (7)], it follows that

Interchange x, and x, to obtain

This equation combined with (***) shows that [x,, [x,,x,]]~€D. NOWD also contains each commutator with three distinct entries, so D = y3F. COROLLARY 1.2. C < y4F, and so G / y4G

-

F/ y4F.

Proof. Suppose C { y4F and let i E {1,2,3) satisfy

By 1.1, Cyi+,F has finite index yiF and so

This contradicts the standing hypothesis that G = F/ C is centerless and residually torsion-free nilpotent. Incidentally, C is not necessarily contained in y,F, as illustrated by taking C = F". Define, for j > 1,

and set K = K,, K ( F ) = K,(F). Note that yiF= yiF. Since G = F / C , where C is characteristic in F, there is an induced homomorphism A (F)+A ( G ) under which the image of Ki(F) lies in K,. Since C < y,F there is an induced homomorphism A(G)-+A(F/yiF) (i < 4) and the composition

-

is the natural induced homomorphism. Since A (F)+A(F/F') is onto, it follows that A ( F ) / K ( F ) A (F/F1) GL (n, 2 ) [IS, p. 1681. Likewise, A (F)-+A(F/ Y,F) is onto, so A (F)/K,(F) =

720

JOAN L. DYER AND EDWARD

FORMANEK.

A(F/y,F) [2,Theorem 11. Because of (*), A(G)-+A(F/y,F) is also onto for i = 2,3 and we obtain PROPOSITION 1.3. (1) A ( G ) / K = A ( F ) / K( F ) = G L ( n , Z ) . (2) A ( G ) I K , = A ( F ) / K z ( F ) = A ( F / y , F ) . (3) [2,Lemmas 3 and 51 K / K2 = K ( F ) /K2(F) is a free abelian group with basis represented by the automorphim Kii ( i # j,l< i,j < n ) and Kiil ( i# j < 1 # i, 1 9 i,j, 1 9 n), where

Note that the isomorphism with G L ( n , Z )in (1)is fixed by our choice of generators x,, ...,xo for F and G , and that all the other isomorphisms are the natural ones associated with the projection F+F/ C = G. PROPOSITION 1.4. (1) n Kj= 1. (2) [K,,Ki19Ki+j. (3) [Ki,YjGl < Yi+iG. (4) K / K i is torsion-free. Proof. ( l ) ,(2),and (3)are well-known properties of automorphism groups which stabilize a central series (cf. [lo,p. 14-16] or [15,p. 391-3921),while (4) 17 follows from the fact that G / y j + , Gis torsion-free. PROPOSITION 1.5. [6, Theorem B]. The kernel of any homomorphism A (F)+ GL ( m ,Z) contains K ( F ) , provided m < rank F. We now prove that KQ A ~ ( G )which , is (1)of Theorem A. Proof. It suffices to show that K < K for any 9, E A 2 ( G ) .Since F < K ( F ) , by 1.5 F is contained in the kernel of the homomorphism

Thus


Since

72 1

n Kiq= 1, it follows that there exists an integer r 2 1 such that

Then G' \< [Krq,GI \< G. But [KTq, GI is normal in A ( G ) ( =A ( G ) q ) ,which means that [Krq,GI is a characteristic subgroup of G. Thus there is an integer m > 0 such that [KTv, GI = Gm.G', and so

If m ZO, G < KT+ since K'P/Kr+lVis torsion-free (1.4 (4)). This contradicts (*), so m = 0 and G' = [ Y V , GI. By the definition of K, this implies that Krq < K. Since Krq > yr(Kq)(1.4 (2)), K q K / K is a nilpotent normal subgroup of A ( G ) / K = G L ( n , Z ) . The center { + I ) of G L ( n , Z ) is its maximal nilpotent normal subgroup, and the automorphism 0 :xi tjxi-' of G represents the coset which corresponds to - I under our isomorphism A ( G ) / K= GL( n ,2). Thus KT < K(0). Suppose KV { K. Then there is a k E K such that k0 E Kq. Since G < K q , G' < [G,Kq] and for each x E G,

Therefore G \< [ G ,K T ]< [KTV, K T ]< KT+ This implies that G < KT+ since Kq/Kr+,q is torsion-free, which contradicts (*). COROLLARY 1.7. There is an induced homomorphism r : A2(G)-+ A ( G L( n ,2))given by the composition

The vertical map is the induced map (1.6), the first (horizontal) isomorphism is the natural one, and the second isomorphism is defined in terms of the fixed basis x,, ... ,x, (cf. 1.3 (1)). 2. The Invariance of K2 and Theorem A. In this section we complete the proof of Theorem A, showing first that K2 and GK, are normal in A'(G) and then, with the additional hypothesis that K / G is residually nilpotent, that G ' is normal in A2(G). In case G = F, K2(F)=K (F)' 12, Lemma 51 and so

722


K 2 ( F ) 4 A 2 ( Fbecause ) K ( F ) a A 2 ( F )For . general G we do not know if K,= K', nor do we have any evidence which suggests that they are equal. Suppose g, E A ' ( G ) . By 1.7, g, induces an automorphism ~ ( g , )G: L ( n ,Z)+ G L ( n , Z )via the fixed isomorphism between A ( G ) / K and GL(n,Z) (1.3 ff). Hua and Reiner have determined A ( G L ( n ,2 ))[12, or see 2.1 below], and their determination permits us to assume that ~ ( g , belongs ) to an explicit finite set. We take advantage of this several times. We first show that Gq < K2 for any g, E A 2 ( G ) .Our proof unfortunately is computational, but given that Gq < K, it is relatively easy to show that ~ , d A ~ ( c g We ) . then view K / K , as a module over A ( G ) / K = GL(n,Z). Arguments from [6] are next applied to show that G K , ~ A ~ ( Gand ) , that G ' a A 2 ( G ) when K / G is residually nilpotent. We begin by quoting the Hua-Reiner Theorem. 2.1. (Hua and Reiner [12, Theorem 41). Let A = A ( G L ( n , Z ) ) THEOREM and let J be the subgroup of inner automorphisms. (For X E GL(n,Z),X t is the transpose of X.) (1) If n is even, n 2 4, then the automorphisms R , = identity,

and R, = R3R2 represent the four cosets of J in A. (2) If n is odd, n 2 3, then R , and R, represent the two cosets of J in A. (3) When n=2, the four cosets of J in A are represented by R,, R3, R5, and R,= R3R,, where R, is defined in terms of the generators

The next two lemmas show that if cp E A 2 ( G ) ,then G q # K,. We will use the basis

for the free abelian group K / K 2 (1.3 (3)).Note that


723

Recall that 8 E A (G ) is defined by 8 :xi ts xi - l, and that Sn < A (G ) where m E Sn acts on G by m : xi ts xTi. It follows that

LEMMA2.2. Let (P E A'(G) satisfy there exists k E K such that

T((P) = Ri

for some i = 1,2,. . .,6. Then

(where r is defined in 1.7 and the Ri in 2.1). 8v

Proof. Since 8 represents the only nontrivial central element of A ( G ) / K , 8 mod K. There exist uniquely defined integers e(i,j),f (i,j, I ) such that

(where the products are taken over the basis elements (*) of K modulo K,). For each k E K , 8 ~ k 8 ~ = O k O = k - 1 m ~ d Kso2 ,that

We may therefore replace (P by kp, for some suitable k E K so as to obtain that each of the exponents e(i,j),f (i,j,l) in (1) is either 0 or 1. Note that r ( k ( ~ ) = 7 ((P) = Ri for some i. We claim that all the e(i, j ) are now equal, and all the f (i,j,l) are now equal. To see this, suppose .rr E Sn < A(G). The image of m in GL(n,Z) is a permutation matrix, and such a matrix is mapped either to itself or to its negative by each R,. Thus mq-.ir or 8mmodK. Conjugate (1) by m - q and compute modulo K,K to obtain

Since S, acts transitively on { Kii) and on {K,il" I), all the e (i,j) are equal mod 2 and all the f (i,j, 1) are equal mod 2. Thus

where e,f E {O,l).

We will now show that f = O in (2). Assume, conversely, that f = 1. Then

724


-

n = rankF > 3 since there are no Kip's when n = 2. With an E A (G) defined by un:xn++xn-', xi-xi ( i i n ) , we have anq-unmodK if r(cp)=Rl or R, and unq 80, modK if r(9)= R, or R4. Hence there is a d E K such that dun =

anv (80,)'

ifr(cp)=R1orR, if r (cp)= R3 or R4.

Moreover dun is an element of order 2 which commutes with O q . Modulo K,, conjugation by an inverts the Kin and the KiP with n E {i,j,l), but fixes the remaining elements of the basis (*). Thus [dun,Ov]= 1 together with (2) implies that

Let Y, E A (G), for i = 1,.. . ,n - 1, be defined by

and define Yn E A ( G ) by

Each product is to be taken in some arbitrary but fixed order. We note that Y, E K (i = 1,.. . ,n) and that each Y, can be expressed as a product of those K,, (r # s < t # r) for which i = r and n E { r , s, t ) . These expressions are independent of the order of the products modulo K, = Ker{A ( G ) - + A( G / c G ) ) ; recall that G/t,G= F/y,F (1.2). Define Y E K by

and note that Y ~IIn,(i,i,llKiPmodK,.Set bi= [Y,xi]. Since each Y, fixes xi ( j Z i), it follows that

Moreover, each Yi fixes y,G modulo y,G, so that for i = 2,. . . ,n - 1

and


Since x, = IIiKi,, there exists c E K2 such that

We now work modulo K,K = ~ e r { A (G)-+A (GI y4G ( y , ~ ) ~ ) . Since [K,, K ] < Kg and a, centralizes K2 modulo K,K;, ( d a J 2= 1 implies that 2

1 = ( c ~ , ~ ~ a ~ ) ~ = ( x , ~ Y~ ]a ~~ () Y~ ~= , [) x ~ ,

= b,

-

"Ya, Ya, mod K3K 2.

(3)

Therefore Ya, Ya,, as an automorphism of G/ y4G ( y , ~ ) is~ the , inner automorphism determined by bne.A computation shows that

The retract

x,I-+x,, x2t+x2, x i + l ( i > 3 ) defines an endomorphism of G / y4G ( y , ~=) F~/ y 4 ( ~y , ~ ) Under ~. this endomorphism, since n >/ 3 , b, !+ [x,, x2] and bi I+1 ( i # n ) . Therefore ( 4 ) implies

This equation is false for any e, and gives the desired contradiction. We have therefore shown that f = 0 in ( 2 ) , or

Finally, we claim that e = 0. Assume that e = 1. Since O q is of order 2 and [ K 2 , 0 ]< K,K;, an argument similar to that used to establish ( 3 ) above now shows that

Therefore the inner automorphism of G / ~ , C ( ~ , G )defined ~ by x,. . .x,xl - 1 .. .x,-' is trivial. Under the retract (5) the image of this element is [x,,x,], which does not commute with x,. This contradicts (6);and so e = 0. LEMMA2.3. I f rp E A ~ ( G ) t,hen G q 4 K2.

Proof. Since G q and K2 are normal subgroups of A ( G ) , by (2.1) and (2.2)

726


we may replace rp by ag, for some a E A ( G ) , if necessary, so as to obtain r(g,)= Ri for some i and Bq=BrnodK,. Let U , V E A ( G ) be defined by

Under the composite homomorphism A (G)+A ( G ) / K of U , V are the matrices

= GL(n, Z),the

respectively (unlabelled entries are zero). Hence

VPf

I

VmodK UmodK BVmod K

if~(q)=R,orR, ifr(g,)=R,orR, if r (rp) = R5 or Re.

Direct calculation shows that [ v , e ] = K,,,

[ BV, B] = K,,-', and [U,B] =K,,-l. Furthermore, for all m E S, and k E K

since r ( 9 ) = Ri for some i, and 8 centralizes K modulo K,K Suppose first that r(rp) = R,, R,, R5, or Re. Then

2.

KZlq= [ V, 81Q)~ ~ , m o d ~ ~ K By (l), it follows that Kilq 3 Kil modK2K for all i 2 1, so xlq = (K21.. .Knl )'-Thus xlq fZ K,.

x1 modK2K

images


Finally, suppose r (rp) = R2 or R,. Then

By (I), it follows that Kilq

Kli modK2K for all i 5 1; and so

Again, x,q @ K,. The next theorem is (2) of Theorem A.

Proof. Let cp EA,(G). We claim first that (Gf)q4 K,. Since Gq 4 K, (2.3), it follows from the definition of K, that [Gq, GI 4 y3G = y3G. The center of G / & G is 4 is even, KR3(Mj

) 1.6. Hence we assume that Proof. If n = 2, GK, = K and K a A 2 ( ~by n > 3. Let g, E A2(G). We must show that (GK,),= GK,. The homomorphism r :A2(G)-+A( G L ( ~2)) , of 1.7 maps A ( G ) to the inner automorphism group J < A (GL (n, 2)).Since A ( G ) normalizes GK,, we may replace g, by ag, for some a E A (G), if necessary, so as to obtain ~ ( 9=)Ri for some i = 1, 2, 3, or 4 (where

CHARACTERISTIC

SUBGROUPS.

729

R, is one of the coset representatives of J in A ( G L ( n , Z ) ) specified in the Hua-Reiner Theorem). Let M < G L (-n , Z ) be- the subgroup defined in 2.7. Since g, induces an automorphism of K, and K, = ( 0 , ai E ZK and a,, a,+, # O . Here, a has degree d , leading coefficient a,, and leading exponent r. (2) If U E Z H and ( 1 - x ) a E Z K , then a=O. (3) If ZK has no zero divisors, then Z H has no zero divisors. In this case the degree (leading exponent) of a nonzero product is the sum of the degrees (leading exponents) of its factors. (4) Suppose z E Z K is not zero and not a zero divisor in Z K . If a E Z H is nonzero, then for any O # ~ E Zthe degree of ( x m - z ) a is at least Iml. In the situation of 3.4 we write Z H = ( Z K ) [ x " ] and call ZH the twisted polynomial ring in x over Z K . We now return to the study of G = F / R 1 . The next theorem includes Theorem B ( l ) , and follows from Theorems 1 and 3 in [ I ] . We present a somewhat different proof below. THEOREM 3.5. Suppose 1 # G= F/ R is torsion-free. Then (1) [ I , Theorem 21. The center of G = F/ R ' is trivial. (2)R / R 1 is a characteristic subgroup of G .

a n d g E G - R /RR/fR 1(so , t h a t g Z 1 ) . Let

Proof. (1) Suppose 1 # r € be the derivation associated with the Magnus representation (3.1). A short computation shows that

a : G-T

a [ g,r] = ( g - i)ar. Here ar#O since r # 1 (3.1 ( 2 ) ) ,and g- 1 is a nonzero divisor on T since

g# 1


733

(3.3 (1)).Hence a [ g,r] #O, so [ g, r] # 1 (3.1 (2)). Since [ g, r] # 1 whenever g @ R / R 1 and 1# r E R/ R', it is clear that G is centerless. (2) Since R / R 1 is abelian, it is locally nilpotent. If g @ R / R ' and r E R / R f , then g# 1 and repeated use of (*) shows that

(where 1g,g,r] = 1g, 1g, r]], etc.).Therefore r # 1 implies [ g, ... ,g, r] f; 1. Thus no subgroup of G which properly contains R / R ' is locally nilpotent, and so the maximal locally nilpotent normal subgroup (Hirsch-Plotkin radical) of G is precisely R / R '. THEOREM 3.6. Suppose G= F/R is residually torsion-free nilpotent, and 1# g E G = F / R'. Then there exists an integer j such that g @ N1/R', where N/R=);.(F/R).

.+zCG . ~i s, re. Proof. Since g#1, o # ~ ~ E T = z ~ ~ ~ + . .Moreover sidually torsion-free nilpotent, so there is a j E Z such that the image of ag in is nonzero, where N/ R = Ti (F/ R ), U is the free left Z[F/ N]-module with basis u,, ...,u,,, and the map T+ U is the obvious one. This gives rise to a commutative diagram of Magnus representations

a

The composition G + T+ U is the derivation associated with pN. Hence pN(g) # 1 since the image of ag in U is nonzero (3.1 (2)). Thus g @N f / R ' since N'/R ' = KerpN (3.1 (1)). is a torsion-free nilpotent We turn now to the special case in which group. In this situation, we will write N in place of R so that

734


There is a normal series

in which each Hi/Hi-, is infinite cyclic. This shows that ZG can be obtained from 2 by repeated application of the twisted polynomial ring construction. Let hi E 6 represent a generator of H,/Hi-,. Then every element of G may be written uniquely as hle('). . , hme(m) with e (i)E 2. If x E G is not a proper power, then H,, . .. ,Hm and h,, . . .,hm can be chosen so that x occurs among the hi. Let I be the augmentation ideal of zG; that is

and define r(i), i = 1,.. .,m, to be the largest integer for which hi - 1 E IT(')). Note that r(i) is finite by part (1) of the next theorem, which summarizes some fundamental results of Jennings.

G be a finitely generated torsion-free 3.7 (Jennings [13]). Let THEOREM nilpotent group, and choose h,, .. .,hm E G as above. (1)[13, Theorem 4.31. n I T = 0. (2)113, p. 176-1771. The set of elements of the form

where 0 < e ( i )E Z , is a 2-basis for only if

Z E Such an element belongs to I'

if and

and these form a basis for IT. The same statements hold if the order in which the factors occur is reversed. (3)[ l o , Theorem 7.11. (1+ I T )n G= (4)(1+ I T )n G= i for all r > m . Remark. Theorem 7.1 of [lo] actually asserts only that (1+ I Q r ) nG= where IQ = Ker{e : Q&Q). However, (2) shows that (IQ)'n zG= IT, so that (3) is valid. We conclude this section by using a modified version of the Magnus representation 3.1 (1) together with 3.6 and 3.7 to obtain a set of torsion-free

yrc

735


nilpotent groups which distinguish elements of F/ R ' , when F/ R is residually torsion-free nilpotent. This is Theorem B (3). We remark that the group 3, (4) becomes

Since L 2 , b - l ~

a,: G-ZG of Theorem 3.1(2), to

G,it is invertible and so is not zero and not a zero divisor in

zG. If aibm# 0 , by 3.4(4) the right hand side of (5) has degree at least m l

when viewed as an element of (ZG)[ lml, or r + d = O and Irl> iml. Since the left hand side of (6) is independent of rn, a,b, = 0 for sufficiently large Iml and (6) becomes

The right hand side of (7) belongs to zG,; and so we may apply 3.4(2) in ( ~ ~ , ) [ 2 , : ,to" ]conclude that a,b = 0 . We have now shown that a,b= a,b= . . = a,b=O. The "fundamental formula" of section 3 becomes

C,,

Since CE 3.4(2)applied in (ZG,)[2,"] yields alb = 0 . Thus all whence b = 1 (3.1(2))and

sib are zero,

- ~required. Since a = xlea, a ( x , ) = ~ , ~ x , x , as If n = 2, a = xle and we are done. For n > 3, we may apply the argument above to ,8, x,, xi and to ,l x,,? xi (in , place of a ,x,, x,), for P satisfies (8) and also maps each primitive of G to a conjugate. Therefore there exist integers f,g such that

Apply

ai : G+ZGto this equation to obtain

which implies f = g = 0 , so P = 1 and a = xle. 5. Separating Conjugates. The main object of this section is to complete the proof of Theorem B by showing that if F I R is residually torsion-free and nilpotent, then G = F / R' has property 11: If x is a primitive element of G , and x is not conjugate to y E G, then the images of x and y are not conjugate in G / T i G for some integer j. We do this by first establishing a criterion for the conjugacy of x and y in terms of the solvability of certain equations in Z [ F / R ] using the Magnus representation (3.1). We next show that the nonsolvability of these equations in Z [ F / R ] implies their nonsolvability in Z I F / N ] / I r ,where N / R = y , ( F / R ) for

c=

CHARrlCTERISTIC

SUBGROUPS.

739

a suitable k and I' is a suitable power of the augmentation ideal I = Ker{& : Z [ F / N ] + Z ) . The homomorphism defined by the composition

then carries x, y to nonconjugate elements in a torsion-free nilpotent group (using 3.6 and 3.8). Finally, we deduce Theorem C as a corollary. Since we have assumed that x is a primitive element of G, we may fiirther assume that x = x,, part of the primitive basis x,, .. . ,x, of G , and that a : G+ T is the derivation associated with this basis (cf. 3.1). PROPOSITION 5.1. Let G = F / R'. If G= F / R is torsion-free, then y E G is conjugate to x, i n G if and only if there exist g E G, t E T such that

Proof. If gx, = yg, then (*) is satisfied by t = ag. Conversely, suppose that (*) has a solution. If y= 1, let g E G be any preimage of g E (?,Then

so gx,= yg by 3.1 and 3.2. We may now assume that ij# 1. Let t = Xiiiti, where fundamental formula and (*),

+

(xi

4E

Z By ~ the

+

so that (1- y )(1 C 4 - 1))= g - g3C1. But 1 Cii, (gi - 1) is nonzero since it has augmentation 1, and z jhas infinite order since G is torsion-free. Furthermore, g - gx1 is a difference of two distinct group elements. We may apply 3.3(2) to conclude that

Since the augmentation of the for some integer m > 0 and some element b of left hand side above is 1, it follows that on the right m=O and the sign is positive. Thus 1 ZZi (< - 1)= b and

+

740

JOAK L. DYER AND EDWARD

FORMANEK.

Therefore b= 2, and g- 1 = 2iii(xi- 1). By 3.2, there exists a preimage g E G of gE F s u c h that ag= X i t i = t . Equation (*) now implies that gx, = yg. LEMMA 5.2. Suppose H is residually torsion-free nilpotent. Then for each x, y E H either x E ( y ) or x @ ( y)-ikH for some k .

Proof. Suppose x E n ( y ) % H . We must show that x E ( y). Since n TiH we may assume that y # 1 . Then there exists an integer m such that y @ K H . Since H / TmH is torsion-free, ( y ) n C H = 1. For each i, there exists e ( i ) E Z such that = 1,

If k > O then Ti+,H< TiH, so

Therefore e ( j ) = e , independent of

i > m. Now y "x E n TiH = 1, so x E ( y).

Before proving the next result we make a remark about group rings (cf. 3.3). Suppose y E H , where y has infinite order, and S = { s , } is a set of right coset representatives of ( y ) in H. Each $0 element of Z H has a unique expression as a finite sum

where O f (ii E Z ( y ) and the si are distinct elements of S. If augmentation,

z ~ ( y - 1 ) Z H ifandonlyif

E:

ZH+Z

~ ( ( l ~ ) = O ( i,..., = l k);

is the

(**I

and

z€H+(y-1)ZH

ifandonlyif

E ( N ~ ) = ~

(***)

for one j E {I,.. . ,k} and & ( u i= ) 0 for the remaining i # j . PROPOSITION 5.3. Suppose H is residually torsion-free nilpotent, and let YEH,ZEZH. ( 1 ) If z @ ( 1- y ) Z H , then i@ ( 1- Ij)ZIH/TiH] for some integer i (where : H+ H / yi H is the canonical m a p ) . H / T i H + ( 1- $ ) Z [ H / T i H ]for some j. (2) If z @ H ( 1 - y ) Z H , t h e n i@

+


74 1

Proof. If y = 1 or if z = 0 , ( 1 ) and (2)follow immediately from the fact that H is residually torsion-free nilpotent. We therefore assume that y# 1 and z#O. Let S = { s i ) be a set of right coset representatives of ( y ) in H , and let

where ai E Z ( y ) and si E S, be a representation for z of type (*).

( 1 ) Suppose z @ (y - 1 ) Z H . Then e(ai)#O for some i , by (**) above. Choose j such that y @ yiH and smsl-' @ ( y)TiH for all nz # 1 in ( 1 , . . . ,k ) (5.2). Then y has infinite order and i , , . . . ,ikrepresent distinct right cosets of ( y) in H/.I,.H. Therefore the representation

is of the type (*). Moreover the diagram

) e ( 4 )# O . This implies that i @ ( t j - l ) Z [ H / X , H ] b, y is commutative so & ( a i= (**). ( 2 ) The proof parallels that of ( I ) , using (***) instead of (**). We next prove a consequence of Jennings' Theorem (3.7). THEOREM 5.4. Suppose H is a finitely generated torsion-free nilpotent group. L e t I b e the augmentation ideal of Z H , and suppose y E H is not a proper power. Then (1) n { ( y - l ) Z H + I T ) = ( y - 1 ) Z H . (2) n{H+(y-l)ZH+IT)=H+(y-1)ZH.

Proof. Since y is not a proper power, there exist h,, . ..,itm E H corresponding to a normal series

such that y = hk for some k E ( 1 , . . . ,m ) . Since the subgroup Hk- = ( h , , . . . ,h k -

is normal in H , the set

is a complete set of right coset representatives for ( y ) in H .

742


If we apply 3.7 (2) first to H with the hi taken in the order h,, . ..,h,; then Hk with its generators taken in the order hk= y, hk- ,,... ,hl; and finally to :- 1 on h,, . . . ,hk- in this order, we can conclude that the set of elements

,

where e(k)=O, e>O, e ( j ) > O ( j = l , ...,m; j f k ) is a Z-basis for Z H with the property that (*) represents an element of I' if and only if

(recall that r( j) is the largest integer for which hi - 1E Ir(i)).Moreover, since y - 1 is not a zero divisor it is not difficult to verify that Z H is a direct sum of Z-modules

where the set of all elements (*) with e > 1 forms a basis for ( y - 1)ZH and the set of all elements (*) with e=O forms a basis for V. Since a Z-linear combination of elements (*) belongs to I' if and only if each term in the sum belongs to IT,it follows that

Consider any s - 1, s E S. Using the identity

and induction, it follows that s - 1E V. Thus s E V since 1E V. (1) Suppose z E n {( y - 1)ZH Ir}. For each r, we can write z = ( y - l ) u ( r )+ v(r), where v(r) E V n IT.I t follows that u(r) is independent of r, v(r) is independent of r, and therefore that v = v(r) E n I r = O (3.7(1)). Hence Z E ( y-I)ZH.

+

(2) Suppose z E n { H + ( y - l ) Z H + I T } . For each r, choose h ( r ) ~ H , u (r)E ZH and v(r) E I' such that

Now h(r)= y"r)s for some integer i(r) and some s E S. Since

and S c V n H we may assume h(r) E V. Again by (***), we may also assume

CHARACTERISTIC

SUBGROUPS.

that .v (r)E V n I '. Therefore, for all r,

x=( y - l ) u ( r ) + ( h ( r ) + v ( r ) ) is the expression for x with respect to the direct sum decomposition (**), and u ( r ) is independent of r. Now for any k > 0,

+

+

In particular, h ( m ) - ' h ( m k) E 1 I". Since H n (1+ I m ) = 1 (3.7 (4)),h(r) is independent of r > m . Therefore c ( r ) EI' is independent of r > m , and so z E H ( y - 1)ZH follows as in (1). •

+

The next theorem is Theorem B (4). THEOREM 5.5. Let G = F I R ' , where R < F' and G= F/ R is residually torsion-free nilpotent. Suppose x is a primitive element of G and y E G. If x and y are not conjugate in G , then their images are not conjugate in G / y i G for some j.

Proof. By definition, a primitive element of G = F/ R' is the image of a primitive element of F, so that we may assume that x = x,, where x,, . .. ,xn determines the Magnus representation 3.1. By 5.1, the conjugacy of x, and y is equivalent to the existence of gE F / R and t E T = Z [ F / R ] t , . . Z [ F / R ] t nsuch that

+

+

This equation has a solution if and only if the n inclusions

are valid. Since x, and y are not conjugate in G, at least one of the inclusions in (*) is false. Hence, by 5.3, there is an integer k > 2 such that one of the inclusions

G ; j e ~ / N + ( l ~- ~ ) z [ F / N ] @ € ( I - Q ) Z [ F / N ] (i+l) is false, where N / R = % ( F I R )and : R / R - + F / N is the canonical map. If z jis in F/W for i, is not a proper a proper power, then z j is not conjugate to i, power and we are done. When t j is not a proper power 5.4 applies, so there is

744


an integer r such that at least one of the inclusions

is false (where I = Ker{& : Z[F/N]+Z) is the augmentation ideal of Z[F/N]). Now consider the commutative diagram of Magnus representations

+

where U = Z[F/N]u, - . . +Z[F/N]u, and all the maps are the obvious ones (see 3.1, 3.6, 3.8). Direct computation shows that the images of x, and y in 97,are conjugate in 9,if and only if the inclusions (**) are all valid. Thus x = x, and y are mapped to non-conjugate elements in the torsion-free nilpotent group % (3.8).

COROLLARY 5.6. Let G = F I R ', where G=F I R is residually torsion-free nilpotent. Suppose x is a primitive element of G and y E G. If x and y are not conjugate i n G, then for some integer k their images are not conjugate i n F I N ' , where N/R = C ( F / R ) . COROLLARY 5.7. Suppose G = F I R ', where R < F ' and F / R is residually torsion-free nilpotent. Tnen K/ G is residually nilpotent.

Proof. It suffices to show that n GKi = G, where

since y,K < Ki. Suppose a E n GK, and x is a primitive element of G. Since a E GK,, there exist g, E G, /3, E Ki such that a = gi/3,. Then

Since this is true for all j, a(x) and x are conjugate in G, by 5.5. Therefore a E G (that is, a is inner), by 4.1.


745

6. Characterizing G = F/ R ' as a Subgroup of A (G). In this final section we assume that G = F/ R ', where R is a characteristic subgroup of F, R < F', and G= F I R is residually torsion-free nilpotent. We retain the notation developed above; in particular, x,, ... ,x,, will be a primitive basis of G, and we use N in place of R whenever it is also assumed that G=F I N is nilpotent. Furthermore, any homomorphism of the form A(G)+A(H) will always be that induced by a natural projection G-+G/ D = H, where D is a characteristic subgroup of G. In this situation, A (F)+A ( H ) is defined and the diagram

is commutative. When G = F I R ' , R / R f is characteristic in G (3.5(2)), so that restriction to R / R ' induces the homomorphism A (F/ R ')+A (R/ R '). The goal of this section is the proof of Theorem D, which asserts (under the hypothesis above) that the group

is precisely G. In fact, we will establish somewhat more. Recall that K (F)= Ker{A (F)+A ( F I F ' ) ) , and define

and

Since KF < K, it follows that BF(G)> 8 (G). We shall prove that 9,(G) = G. It is clear that G < 8 (G), so Theorem D follows. Although Q,(G) need not even be a normal subgroup of A(G), under any induced homomorphism A(G)+ A ( H ) the image of 8,(G) is contained in GF(H). We will use this fact to reduce to the case in which G is nilpotent, and then again when we perform an induction over the nilpotence class of We continue to exploit the Magnus representation, and results from sections 4 and 5. The main step in the argument is a determination of those ZF-automorphisms of N/Nf (with the induced module structure) which centralize Im{K (F)+A(N/N1)). When rank F = 2, K ( F )= F (cf. 115, Corollary N4, p. 1691) and so K, = G. Therefore GF ( G ) = G is an immediate consequence of

746

JOAN L. DYER AND EDWARD

FORMANEK.

6.1 [3, Theorem 11. Let F be free of rank 2, and let R be a THEOREM normal subgroup of F such that R < F' and the integral group ring Z [ F / R ] is a domain. Then the kernel of A ( F / R ')+A ( F / F 1 )consists of inner automorphisms. We note that, in 6.1 above, it is not required that R be characteristic in F. In our case, G=F / R is residually torsion-free nilpotent, so 3.4(3) implies that ZGis a domain. We will assume henceforth that rankF= n 2 3; this hypothesis is needed in the proof of 6.6. We next collect the information about Z H , for a finitely generated torsion-free nilpotent group H, which will be required below. THEOREM 6.2. Let H be a finitely generated torsion-free nilpotent group. (1) (Jennings [13, Theorem 3.21). ZH has no zero divisors, and the only units of Z H are + h, h E H. (2) ( P . Hall [9, Theorem 11). ZH is left Noetherian. (3) (Goldie [5, Theorem 4.21). Z H satisfies the left Ore condition: If a,b E Z H , b#O, then there exist p,q E Z H , pZO, such that pa= qb. ( 4 ) (Ore [5, Corollary 2, p. 6041). ZH has a left quotient ring Q which is characterized up to isomorphism by the properties: Z H is a subring of Q, and for every q E Q there exists OZa E Z H such that aq = b E Z H (and we write q = a-lb). (5)( M . Smith [18, Theorems 7.2 and 7.41, Passman [16]). The center of Q is the quotient field of the center of Z H . Statements (2) through (4)remain valid when "left" is replaced by "right". Moreover, the left and right quotient rings in ( 4 ) coincide. A complete discussion of the construction of Q may be found in [5, p. 603-6061, We will require the following consequences of 6.2: LEMMA6.3. Let H be a finitely generated torsion-free nilpotent group, %H its center, and Q the left quotient ring of ZH. (1) The center of Q is the quotient field of Z [ % H ] . (2) Let 0 c lie in the center of Q, and 0 # a E Z H. If c "a E ZH for all ~ E Z then , c= ? z for some Z E % H .

+

Proof. (1) By 6.2(5),it suffices to show that the center of ZH is Z [ % H ] . If

lies in the center of Z H , where OZn, E Z and hi E H, then each hi has only finitely many conjugates in H. Since H is a torsion-free finitely generated nilpotent group, it follows that each hi E % H .

CHARACTERISTIC

SUBGROUPS.

747

(2) Since %H is a finitely generated torsion-free abelian group, Z[%H] is a unique factorization domain. We may therefore write c = r-'s where r,s are relatively prime elements of Z[%H]. Since Z H is a free left Z[%H]-module, and smaE rmZH for all m €2, it follows that there is some O# b €Z[%H] such that smbE rmZ[%H] and rmbE smZ[%H] for all m > 0. Therefore b is divisible by rm and s m for m > 0. Since Z[%H] is a unique factorization domain, r and s are units in Z[%H]. By 6.2(1) then, c=r-'s=kzforsomez~%H. We turn now to N / N ' < F / N f = G, where G= F I N is finitely generated, torsion-free and nilpotent. Let a : G-+ T= ZCt, . . . + 2% be the derivation associated with the primitive basis x,, .. . ,xn of G (3.1(2)).When N / N ' is given the left zC-module structure induced by conjugation within G, the restriction of a to N / N f provides us with a zE-linear monomorphism of N / N 1 into T. By 3.2, We denote the image of N / N ' under a by

+

Since N / N ' is a characteristic subgroup of G = F I N ' (3.5(2)),any a E A ( G ) induces an automorphism Z of C. We will also use Z to denote the linear extension of Z € A (C)to an automorphism of ZZThe restriction of a to N / N ' is an automorphism of N / N ' which we denote again by a. The requirement that the diagram

Note that, for any be commutative defines a :f--+f,

gE G and t E G,

In particular, if Z = 1 then a : f+N is a zG-automorphism of N. Since C= F I N is nilpotent, 6.2(4) implies that the free Z(!?-rnodule T is imbedded in the free Q-module T* = Qt, + . . . + Qtn (where Q is the quotient ring of zC).The submodule T is imbedded in

N
3. This establishes (1);note that also

750 -


We claim next that there exist c,, . . . ,cn E Q such that, for each i =Zait, E

N*.

To see this. write

Z=Zaiti=2izla,(aizi)-1Zi. Then by ( I ) ,

a ( X ) = Zizlai (adz,)- lqi 2, in a(Z) is aici where

is independent of i.In this argument, x, was distinguished. The same argument with, say, x, in place of x, establishes (4) and (5) for all j = 1,. ..,n. We claim now that the c, are equal and belong to the center - of Q. Since 4 = (alzi)tl (aizi)ti and a ( $ )= (alzi)cltl (aizi)citibelong to N*, (**) implies that

+

+

and

From these two equations it follows that, for j

> 2,

Let c denote this common value. To show that c is central, and therefore also equal to c, for all j, we show that ?,c = c?,. The fact that c commutes with the other generators follows similarly. Since KB1 is conjugation by x, on ( x , , ~ , ) , equations (3) and (5), together with 6.4 ( I ) , imply that %,(c,) = ?,c,?,-l. On the other hand, since n > 3 , K,,(Xn) = Therefore

1 and define M < F by M / N = %C.Then M is a characteristic proper subgroup of F, and F / M is a torsion-free nilpotent group of class c - 1. Let P E SF ( F / N'). The inductive hypothesis says that C?, ( F / M ' ) < F/ M'. Since 9, ( F I N ' ) is mapped into 8, ( F / M') < F/ M' under the induced homomorphism A ( F / N')+A ( F / M'), P E C?, ( F / N ' ) induces an inner automorphism of F/ M'. Thus there is a g E F I N ' such that @ E Ker{A ( F / N ' ) + A ( F / M ' ) ) . We claim that [@,KF]< N / N t . Since 8, E S F ( F I N ' ) and F/ N' < C?, ( F I N ' ) , [ g p ,K,] < F t / N ' . But the automorphism induced by @ on F / M ' is trivial, so the image of [ @3, K,] in A ( F / M') lies in the center of F / M ' , which is trivial (3.5 (1)).Thus


753

Now M / N = % ( F I N ) , so M f / N ' < N / N ' . We may therefore apply 6.7 to conclude that @, and therefore also P, is inner. CITYUNIVERSITY O F NEWYORK AND

THEUNIVERSITY OF CHICAGO.

REFERENCES.

[ l ] M. Auslander and R. C. Lyndon, "Commutator subgroups of free groups," A m . J . Math. 77 (1955), pp. 929-931. [2] S. Bachmuth, "Induced automorphisms of free groups and free metabelian groups," Trans. Amer. Math. Soc. 122 (1966), pp. 1-17. , E. Formanek, and H. Y. Mochizuki, "IA-automorphisms of two-generator torsion-free [31 groups," submitted. [4] W. Burnside, T h e q of Groups of Finite 07der, Dover, 1955. [5] P. M. Cohn, "Rings of fractions," Amer. Math. Monthly 78 (1971), pp. 59f5-615. [6] J. L. Dyer and E. Formanek, "The automorphism group of a free group is complete," J. London Math. Soc. (2), 10 (1975), pp. 181-190. [7] R. H. Fox, "Free differential calculus I," Ann. of Math. 37 (1953), pp. 547-560. [8] E. K. Grossman, "On the residual finiteness of certain mapping class groups," J. Ladon Math. k c . , (2) 9 (1974), pp. 160-164. [9] P. Hall, "Finiteness conditions for soluble groups," Proc. London Math. Soc. (3), 4 (1954), pp. 419436. [lo] -, The Edmonton Notes on Nilpotent Groups, Queen Mary College mathematical notes, London, 1969. [ l l ] B. Hartley, "The residual nilpotence of wreath products," Proc. London Math. Soc. (3), 20 (1970), p p 365392. [I21 L. K. Hua and I. Reiner, "Automorphisms of the unimodular group," Trans. A m . Math. Soc. 71 (1951), pp. 331348. [I31 S. A. Jennings, "The group ring of a class of infinite nilpotent groups," Canad. J . Math. 7 (1955), pp. 169-187. [14] W. Magnus, "On a theorem of Marshall Hall," Ann. of Math. 40 (1939), pp. 764-768. 1151 -, A. Karrass, and D. Solitar, Combinatorial Group Theory, Wiley (Interscience), 1966. [16] D. S. Passman, "On the ring of quotients of a group ring," PTOC.A m . Math. Soc. 33 (1972), pp. 221-225. [I71 V. N. Remeslennikov and V. G. Sokolov, "Some properties of a Magnus embedding," Algebra i Logika 9 (1970), pp. 566578 (Russian). Translation: Algebra and Logic 9 (1970), pp. 342349. [I81 M. Smith, "Group algebras," J . Alg. 18 (1971), pp. 477499.

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Induced Automorphisms of Free Groups and Free Metabelian Groups S. Bachmuth Transactions of the American Mathematical Society, Vol. 122, No. 1. (Mar., 1966), pp. 1-17. Stable URL: http://links.jstor.org/sici?sici=0002-9947%28196603%29122%3A1%3C1%3AIAOFGA%3E2.0.CO%3B2-E 12

Automorphisms of the Unimodular Group L. K. Hua; I. Reiner Transactions of the American Mathematical Society, Vol. 71, No. 3. (Nov., 1951), pp. 331-348. Stable URL: http://links.jstor.org/sici?sici=0002-9947%28195111%2971%3A3%3C331%3AAOTUG%3E2.0.CO%3B2-A 16

On the Ring of Quotients of a Group Ring D. S. Passman Proceedings of the American Mathematical Society, Vol. 33, No. 2. (Jun., 1972), pp. 221-225. Stable URL: http://links.jstor.org/sici?sici=0002-9939%28197206%2933%3A2%3C221%3AOTROQO%3E2.0.CO%3B2-N

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