Arch. Math., Vol. 38, 193--203 ( 1 9 8 2 )

Generalisation

0003-889X/82/3803-0011 $ 01.50-b 0.20/0 9 1982 Birkh~user Verlag, Basel

of some theorems

on one-relator groups

By BEI~JAI~Ir~ BAUI~I[SLAG

1. Introduction. Let I be a set, let H(o (i e I) be the restricted direct product of free groups, and let H ~- * H(0 (i e I) be their free product. Let R be a cyclically reduced element of H of length at least two ,and let G = H / { R ) H. Thus G is obtained from a one-relator group b y adding relations to make certain sets of generators commute with other sets of generators. In [1] an analogue of the Freiheitssatz was established for G. I t is the purpose of this paper to give further examples of how theorems on onerelator groups can be extended to G. Thus we show Theorem 2. Let I consist o/at least three elements. Then the centre o/G is trivial. Theorem 3. G has elements of ]inite order i / a n d only i / R is a proper power. Theorem 4. We suppose that /tee generators /or each o/ the direct components o/ each H(o have been chosen, and these generators appear with positive exponents only in the word R. Then G is residually solvable. These Theorems are generalisations of the corresponding Theorems for one-relator groups which appear respectively in [2], [3] and [4]. The proofs of our Theorems depend strongly on the Freiheitssatz proved in [1]. I thank S. J. Pride for some helpful discussions. 2. Notation. will denote the group with generators X and defining relators Y. I f X is any subset of an arbitrary group G, X ~ will denote the normal closure of X in G, and gp(X) will denote the subgroup generated b y X. I f R, S, T . . . . are elements of G, we shall denote G/ { R, S, T . . . . } ~ by G / { R, S, T, . . . ), and G/ { R } by G/ R. H will denote the free product 9 H(l)(i ~ I), and if J C I, H ( J ) = . H(O(i ~ J). Each H(~) will be the direct product of free groups H(~, J). In the sequel let R be a fixed element of H of length at least two, and G will stand for H / R unless otherwise stated. I f J C I is such that R ~ H (J), we call H (J) a J - M c ~ n ~ subgroup, or more briefly a Magnus subgroup. For convenience we shall call a free product of direct products of tree groups a P D F group, and a group with an additional relator ~ as described in the IntroducArchiv dev Mathematik 38

13

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tion, a one-relator P D F group. If, furthermore, the hypotheses of Theorem 4 hold, we speak of a positive one-relator P D F group. ~ (G) will denote t h e / - t h member of the derived series of an arbitrary group G. Any group G is said to be residually solvable if the intersection of the members of its derived series is 1. I f F is a subgroup of an arbitrary group G, -~ is said to b e solvably separable if, for a n y g e G \ F , there exists an integer 1 such t h a t g q~F ~ ( G ) . More details concerning these concepts appear in [4] and [5], while the paper [1] will be referred to frequentlY. General ideas on free products, ~ groups etc. appear in [7].

3. Centre and torsion. 1. Lemma. Let G be an H N N extension o / S , G ---- (t, S ; t - l L t = U>. Let z be an element o / G which commutes with a non-trivial Tower oI t. T h e n z ---- got r (r e 7/), where g o e L n U. P r o o f . Let z = got ~' g l t ~ ... t~"gn be in H_NN normal form -- see page 181 of [7] - and suppose it commutes with t ~ (1 > 0). I f go ~ L, t-~zt~ when expressed in normal form begins with t-~go and hence is not equal to z. I f go e L but not in U, t~zt -Z is similarly easily seen to differ from z. I f go e L ~ U, suppose 81 ---- 1 and gl 9 1. Let gr ---- trgo t-r, and suppose t h a t gr-1 belongs to U b u t gr does not. Then t l z t -~ ---t~-rg~tr+~gl, is clearly not equal to z. Ifg~ belongs to U, then t~zt -~ -Yl' ~~ + ~ "y l is also clearly not equal to z. A similar argument m a y be used for el ---- - - 1. Similar arguments hold when g l ---~g2 ----- "'" = gi-1 ---- 1 but g~ ~ 1. r

p

9

9

9

=

9

9

9

Corollary. I] L 4= S, then z lies in L n U. I] in addition S has no centre, neither has G. P r o o f . F r o m the Lemma, z -----got r. Let s be an element of S outside L. Then gotrst-rgol --~ 8 i m p l i e s t r s t - r .~- g ~ l sgo, which is impossible if r =~ 0.

zsz -1 =

P r o o f o f T h e o r e m 2. I f one of the H(0 does not contribute generators t h a t appear in •, then G is a proper free product and hence has no centre. Thus without loss of generality we m a y assume t h a t each H(O has generators t h a t appear in R. Furthermore, we m a y as well assume b y virtue of the Corollary to L e m m a 1 t h a t each of the free generators of each H( i, J) do appear in the relator R. We suppose R = x l x 2 ... x~ in reduced free product form; then each xt is expressible as a product ylY2 ... y ~ with each Yr in some H(~,~). Let ~(y~) be the length of y~ with respect to the free generators already chosen, and define

(z~) = 0 (yl) + ~ (y2) § ..- § 0 (y~), and (R) -~- ~(Xl) § ~(X2) §

"'" § ~(Xrn)-

We proceed b y induction on 0 (R). I f 6 (R) __<2 - - we remark t h a t for this theorem g need not be restricted to free product length-at least 2 - - then G is a proper free product and the result follows.

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Generallsation of some theorems on one-relator groups

195

I f each H(~) is cyclic, we m a y use (2) since then G is a one-relator group with at least three generators. Otherwise for at least one i, either there is an H(l, f) which is free of r a n k at least two, or there are two cyclic direct factors of H(~). I n either of these cases one of the generators t m a y appear with zero exponent sum in R or else, b y adjoining a root of one of the generators, we can embed G in a one-relator PDI~ group G and using a Tietze transformation produce a generator t t h a t appears with zero exponent sum in the relator R. (For this and the sequel of this proof, [1] provides further elucidation.) Furthermore, a positive power of t will belong to G. Applying the usual Magnus reduction as described in [1], ~ is itself the HIN~ extension (t, G1; t - l L t ~ U~, where G1 is itself a P D F group with one relator P with at least three factors and ~ (P) ~ 6 (R). Suppose now t h a t G has a centre and z 4 : 1 belongs to it. Then some power of t commutes with z and so b y L e m m a 1, z ~ trgo, where go e L n U. Now b y the Freiheitssatz, L is a free product of the groups E, By, Ce, Cq+l . . . . . Ca-i, where Cy denotes H~ and By denotes the free product of the groups H~i) (i e I \ ( 0 , k}). I t is easy to see t h a t r = 0, for ff b =~ 1 is an element of By n G, then (g~lt-~)b(trgo) = b is not possible, since t-rbt r e JBy+r. Thus z ----go. Note t h a t L is a free product with non-trivial elements of G in two distinct free factors. I t is then easy to choose elements which do not commute with go using the free product normal form. Hence the centre of G is 1. I n a similar way, using the Torsion Theorem for H N N extensions [7], we can prove 3. Theorem. A P D F group with one relator has an element of ]inite order i] and only i/the relator is o/the/orm S n where n > 1.

4. The first part of Theorem 4. We restate Theorem 4 in the slightly stronger form needed for our inductive argument:

I / G is a positive one.relator P D F group, then for any Magnus subgroup M of G, A)

For all natural numbers l, M n ~(G) ---- (Sz(M),

B)

M is solvably separable in G. Note that since 1 -~ H (O) is a Magnus subgroup o/G, this implies that G is residually solvable. I n this section we prove A), using

5. L e m m a (F. Levin [6]). Let A be a solvable group o] length l and let w(x, A) be a word in the unknown x and the elements o / A in which x definitely appears and furthermore appears only with positive exponents. Then there is an embedding o / A into a solvable group o/solvability length 1 which contains a solution to the equation w (x, A) -~ 1. Now let M be a Magnus subgroup of G; then G ---- (M * X ) / R , where X is a P D F group which contributes at least one generator to R. Since R is a "positive relator", we m a y m a p each of the generators of X onto the generator x of an infinite cyclic group and when the natural homomorphism is applied, the image of R will be a reduced word in M . gp(x) in which x appears with positive exponent only. Thus if we consider R now as a word in M/(~z (M) and x, b y L e m m a 5, we can embed M/~z (M) 13"

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in a solvable group of length l in which R (x) ~ 1 has a solution. The kernel K of this homomorphism of G into this solvable group contains cSz(G) and since K n M ~- ~z (M),

M n (~l (G) ---- ~ (M).

5. Free product lemmas. The proof of Theorem 4 involves producing a generator of zero exponent sum in R and then looking at a subgroup which is built out of free products. In order to deal with such free products we need lemmas which are similar to those t h a t appear in [4]. Indeed, our proof is indebted to [4]. 6. Lemma. Let (1)

G=A.B u

and suppose that ]or an integer 1 there exists a normal subgroup K o] G with G/K solvable o] length l, with (2)

A n K = ~l(A),

B n K ~- ~ (B).

Then i / S is a subgroup o] A, S N ~z(G) = S n (~(A), and i / S is a subgroup o/ .B, S n ~z(G) = S n ~ ( B ) .

P r o o f . Since G/K is solvable of length l, K D=c$~(G). Hence

S n 6~(G) C S n K C A n K - ~ 6z(A). Hence

S n ~z (A) __CS n ~z (G) = S n ~z (A) and the result follows 7. Lemma. Suppose G satisfies (1) and (2)/or almost all l, and suppose/urthermore that (3)

U is solvably separable in both A and B

and (4)

U ~ ~ (A) ~- U n ~z (B) /or almost all l.

Then i] A and B are residually solvable, so is G. Furthermore, i/ B1 C B is solvably seTarable in B, then B1 is solvabIy separable in G. P r o o f . We show first that B1 is solvably separable in G -- that G is residually solvable then follows on taking B1 ---- 1. So suppose g ~ B1.

Case (i). g e/~. Then b y hypothesis we can find l such that g ~ ~ (B) B1. Suppose g e (Sz(G)B1. Then g ---- kb, k e ~z(G) and b e B1. Since g e B,/~ e ~z(G) n B ---- ~ ( B ) b y Lemma 6, and so we have the contradiction g e ~ ( B ) B 1 . Thus g ~ 0~(G)/~I, as required.

Vol. 38, 1982

Generali~sation of some theorems on one-relator groups

Case (ii). g e A \ U or g of length come alternately from A \ U and B \ m a y choose 1 so t h a t for i = 1, 2, ..., = B / ~ (B), then in ~ = A 9 B, ~

197

greater t h a n 1, g = gx g2 ... g~ where the g~ U. (In particular, if g e A we let gl = g)- We n, g~ ~ U O~(A), g~ ~ U 6I(B). Let A = A/6~(A), ~ B.

Using condition (2) - - if necessary choosing I larger - - we can find K normal in G such t h a t ~//~ is solvable of length l a n d / ~ c~ ~ ----/~ n B -----i, so t h a t / ~ is free. Now i f ~ ~ ~ B x , the result follows. Otherwise, ~ = $$ (s e K, b e B~). Choose 5' such t h a t $ ~ c~, (g), which is possible since free groups are residually solvable. Suppose ~ e c~z,(g)/~x. Then ~ = ~lb~, S l e ~z'(K), b~ e B1, and ~ = ~$ = szSx. Since K n B = i, $ ---- ~ and $ -----$1. B u t this implies $ e c~,(-~), which is a contradiction. Hence the result follows. 8. Lemma. Let G = A , B, where A and B are residually solvable. Let M1 be solvably separable in A and M2 a/ree/actor o / B . Then M = M1 * M2 is solvably separable in G. P r o o f . Suppose B = ~2 * M2 and let Gx = A 9 M2. By proposition 2 of [4], M is solvably separable in Gx. B y L e m m a 7 above (with U = 1), M is solvably separable in G = Gx * Bz. Hence the result follows. We note now t h a t we can reduce the proof of Theorem 4 to the case where each H(i) contributes at least one generator to the relator R. I f this is not the case, then G = G1 * G2 where Gx is a positive relator P D F group and G2 is a P D F group. I t follows t h a t if Gx and G2 are residually free, then so too is G [5]. Let M be a Magnus subgroup of G. Then M = Mx * M2, where Mx is a Magnus subgroup of Gx and M2 is a free factor of G2. Then b y L e m m a 8, M is solvably separable in G. 6. R of len~h 2. Suppose now t h a t R is of free product length 2. I n this case, in view of the remark at the end of the preceding section, we m a y assume G = (H(1) . H(2})/R. Then G is the generahsed free product of H~ 1) and H( 2} with a cycle amalgamated. We let A ----H~I), B -----Hc~} and R = ab -1, a in A, b in B, so G ---A * B, where ab -1 is a positive relator. We shall apply L e m m a 7 and so we need to a=b

verify the hypotheses. To establish (2), we m a y m a p B homomorphieally onto an infinite cyclic group in such a way t h a t b is not m a p p e d to 1. For each 1 we m a y embed A/~z(A) into a solvable group of length 1 in such a way t h a t a = b has a solution, b y virtue of L e m m a 5. Thus we can find Kx normal in G such t h a t G/Kx is solvable of length 1 and K1 n A = ~z (A). Similarly we can find K2 such t h a t K~ n B = ~ (B) and G/K2 is solvable of length 1. We see t h a t K -----Kx c~ K2 satisfies condition (2). To check t h a t (3) holds we need to show t h a t a cyclic subgroup of a direct product A of free groups Ai is solvably separable. Let c e A \ {1) and suppose g e A and g ~ gp (c). I f x = [g, c/=~ 1, then since A is residually solvable, we can find a solvable homomorphic image of A in which x is not m a p p e d to 1, and hence g ~ gp (c) ~ (A). I f x = 1, let g = gl ... gn, c = cx ... cn with g~ and c~ elements of A~; then [g~, c~] ---- 1. Since A~ is free, there exists x~ such t h a t g~ x~r', ci x i8'. We m a p each A~ onto a solvable =

=

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ARCH. MATH.

group such that each x~ is mapped to an element of infinite order, and hence map A onto the direct product of solvable groups. It is clear that in the image, g ~ gp (c). Finally, condition (4) holds, since a direct product of free groups is residually torsion free solvable. By Lemma 7, G is residually free. Let M be a Magnus subgroup of G. Then M is 1, A or B. The result follows from Lemma 7. 7. Reduction to each H (i,D contributing to R. We will reduce the proof of Theorem 4 to the case where for every pair i, ] all the free generators of H(~,t) appear in the relator R. We have already shown that we may assume there is at least one generator from each H(0 that appears in R. So consider the subgroup S generated by the free generators of H(~,t) that do appear in the relator R. Then G is obtained from S by repeatedly adjoining elements that commute with certain subgroups. The proof that B) follows for G, assuming A) and B) hold for S, follows from 9. Lemma. Let S be residually solvable and suppose L is a 8olvably sel)arable subgroup o / S and (h (S) n L ---- (~l(L) /or almost all 1. Then if G is the ~ extension

(t, s; It, L] = 1~, (i) L f~ ~ (G) -~ 8z (L) /or almost all l. (fi) I f H is 8olvably separable in S, then H is solvably separable in G. (In particular, G is residually solvable and L is solvably separable in G.) (iii) I f H is solvably separable in S, then gP (t, H) = H* is solvably separable in G. P r o o f . (i) follows almost immediately since the kernel K of the homomorphism obtained on mapping t to 1 and S to S/Oz (S) meets Z in ~ (L) for almost all 1. For (ii) and (iii) we proceed as follows. We define N to be the normal closure in G of S, denote t-~St~ by S~ and put

~ , j = g p ( S , ..., Sj)

(i < i).

We have that

(5)

N~,j+I = N~,t* St+l 55

and (6)

N = w U-~,I.

Let (i'), (ii') be the statements obtained from (i) and (ii) by replacing "G" by "Ni,l, i <-- 0 <_ ]." Clearly (i') follows immediately from (i), which we have already proved. In addition we prove (iv') For almost l, (~(Nr, s) = Nr, 8 f~ (~l(N~,~) for all r, s such that i --< r ~ s ~ ]. We prove (ii') and (iv') by induction on :r = Ill + y'. For :r = 0, (ii') follows by hypothesis while (iv') is immediate. Now let :r = n and suppose (ii') and (iv') hold for :r < n and let 0 < ]. Then Nl,t = N~d-1 * St. Consider the homomorphism inJ5

duced by mapping N~,j-1 to Nt, j-1/Oz(N~,j-1) and S t to SI_I Oz(N~,j-1)/Ot(N~,j-1 ) by conjugation by t-l, and let K be its kernel. Then K n N~,I-1 = 0t(N~,t-1 ) and

Vol. 38, 1982

Generalisation of some theorems on one-relator groups

199

x e K (~ Sj implies txt-1 e K n SI_1 -~ ~(iVt,~-l) n $i-1 -~ ~(S~-1), b y (iv'). Also, by (ii'), L is solvably separable in iVl,i-1 and in Sj. Hence we can apply Lemma 7 to conclude (ii'). !qow ff r, s are given with i _~ r ~ s _~ ], then ff i ---- r and ?" -~ s (iv') follows immediately. If~ ~ s, then Nr, s C N~,t_ 1 and with K as in the preceding paragraph the hypotheses of Lemma 6 hold. We can then conclude

iv~,~ c~ ~(iv~,j) = iv~,~ c~ ~z(ivi,~-~) = ~z(iV~, ~) by the induction hypothesis. A similar argument in the case ~ -~ s and i ~ r with iVt,j ~- St * iVt+l,l concludes the proof of (iv'). 2~

We now turn to the proofs of (ii) and (iii). (ii) I f H is solvably separable in S, it suffices to show that H is solvably separable for elements g in iV, since H C iV. Suppose g e iV-I,i. Since H is solvably separable in iV-~,t, there exists 1 such t h a t g ~ HSz(iv~,t). Suppose g e HSz(iv-i,j) for ] > i. Then g = hS, h e l l , ~ e ~(iV-t,l). But then by virtue of (iv'). This contradicts g ~ HS~(iv-~,i). (iii) Let H be solvably separable in S and let H* -~ gp(H, t). Suppose g ~ H * . Since gt r e H* ff and only ff g e H* and r is any integer, we m a y without loss of generality assume that g e iv. Suppose g e hr_l, ~. By adding the relation t ~+~ ---- 1, we have a homomorphism of G onto a cyclic extension of iV-i,t+~. Clearly g ~ g p ( H - l , ..., Hi+l) -~ H-~,t+~. I t suffices to show that g m a y be excluded from H-t, l+l modulo 8~(iV-l,t+~) for some 1. We shall need 10. Proposition (G. Baumslag, Proposition 3 and Lemma 4.5 [4]). Let G = A * B. I,

Let 9: G - > G * be a homomorphism such that K n A = K n B = 1, where K is the kernel o / 9 . Suppose G* is solvable. Let A1 C A and B1 C B be subgroups with A I 9 = B19. Let iv = g p(A1, B1). Then i / g ~ G \ F, there exists l such that g ~ ~ (G) F. Clearly (iii) will follow from the following:

11. Proposition. Let N1 be a given residually solvable group and let JL, H1 be solvably separable subgroups o / i v 1 . Let Oi : iv1 --> ivi be isomorphisms vzith Oi restricted to L, the identity map (i---- 1, 2 . . . . ,2~). Then i/ iv is the free product o/ the ivt with L amalgamated, gp(H1, ..., H21c) ~ H* is solvably separable in IV. I n particular, with H1 ~- 1, iV is residually solvable. P r o o f . We proceed by induction on k. I f k ~- 0, the result follows by hypothesis. I f true for k, consider k -k 1: Then G ---- A 9 B, where A is the tree product of /V1 . . . . . hr2~ and B is the tree product of iV2~+l . . . . , iV2~+~ with L amalgamated. Clearly A --~--B and by induction, A, B, are residually solvable and ff HA = g p ( H 1 , . . . , H2~),

HB = gp(I-Io.~+l . . . . , H2~+~),

then HA, HB respectively are solvably separable in A, B respectively.

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Suppose g e G\H*. Then expressing g in normal form, in the generalised free product A 9 B, g -- g~ ... gn say. Since A and B are solvably separable in A, B Z

respectively, we can find an integer q such that in the natural homomorphism of G onto A/~q (A) * B/~q (B) with L ~q (A)/~q (A) amalgamated with JL~q (B)/~q (B), none of the g~ are mapped into the amalgamated subgroup and any g~ outside HA, Hs lies outside them in the image. Thus we m a y suppose without loss of generality that A and g are isomorphic solvable groups. So we can find a homomorphism : A * B --> A that acts as the identity on A and maps B isomorphically onto A, J5

with HA ~ = H~ q~. Thus Proposition 10 holds and the result follows. 8, At least o n e I-I (i'j) is not cyclic. We suppose R - - - - x l . . . x~ in reduced free product form; then each x~ is expressible as a product Yl ... Ym with each y~ e H(L J). We fix a set of free generators in each H(~,Y) and again let ~(y~) be the length of Yi and define 6(x~) ---- ~(Yl) + " " + (~(Ym) and ~(R) ---- ~(Xl) -~- . . . . ~ ~(Xn). Our proof is b y induction on ~ (R). I f ~ (R) ---- 2, then we have derived the result in 6. So assume the result is true for all relators P with 6(P) < ~(R). Suppose now t h a t some H(~,J) is free of rank at least two, say without loss of generality H(0,1), and suppose x, y are two of its free generators, appearing in R with exponent sum u, fl respectively. We replace H( ~ 1) b y a free group with the same generators except that x is replaced b y ~ and x is identified with ~a. The relator R(x, y . . . . ) becomes R ( ~ , y, ...) and thus we have embedded G in a group ~ (c.f. Lemma 3, [1]). Clearly it suffices to show that G is residually solvable. We now apply a Tietze transformation to the presentation of G by replacing y by y~-~. In the resulting word R (~, y~-~ . . . . ), ~ appears with zero exponent sum. We can thus use the methods of w3 of [1]. Thus we put ~ ---- t, and define groups H~O(i ~= 0) b y H~i) = t-~H(OtJ (where j is an arbitrary integer). We let E be the normal closure in H (~ of all the generators except t. We then express R as a word P in terms of the elements of E and the H~r (i ~ 0). We note that P is a positive relator. We fix an integer k in I \ ( 0 } and let ~, ~ be respectively the maximum and minimum ] such t h a t elements of H~.~) appear. We use the convention that the notation H~0 implies that for 0 ~ i ~= k0 ] is an arbitrary integer, while for i = / c , ] ranges from ~ to ~ unless otherwise stated. We put D ---- B , H~~) and D* -~ D/P. Then

---- (D*, t; t - l L t = U> where L and U are defined as in [1]. ~or the purposes of this paper it is convenient to change notation b y putting H~k) ~- Qy and C ---- E * H~~) (0 =~ i =~/c). Note that t normalises C. Thus

D=C*Q~*Q~+I*'"*Qe, U = C , Q~+~... * Q~.

L--C*Q~*'"*Qe-1,

We note that each of Qa and Qe contributes a generator to P, Let s be the normal closure in G of D*, so that GIN is infinite cyclic with generator t h r. In accordance with the convention, for any subgroup X of h r, we write X~ for t-~Xt~; this is consistent with for instance Qi. Then for i _~ ], we define N~,~- ----

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Generalisation of some theorems on one-relator groups

201

gp{D*, ..., D*}. I t is important to observe that (7)

N,,i+I = (N,,~ 9 Qo+j+I)/Pi+I,

(8)

2V~,j = (Q~+l* Nt+ I,~)/ P~ = (Q~+~* Q~+~+I * " " * Q~+j-1 * Dj)/{P~,

(9)

N,,j+I = Nl, i * D~'+I,

where

Pt+l,..., P~},

Aj+I = C * Q a + I + I * ' " * Q q + j ,

Aj+x

(10)

N = U

-~,~,

i=l

(11)

2V-i,,---~ -hr0,2t

(i = 1,2, ...)

For each integer l, there exists K <~ Nt,~+l such that (12)

KnNi,j=~(N,.I)

and

Kc~D~+I=O~(D~+I),

with

2V~,j+I/K

solvable of length I. P r o o f . (7) to (11) follow without difficulty. To see how (12) follows, consider (7). B y Lemma 5, mapping Q~+j+I onto an infinite cycle and N~,S onto N~,/d~(N~,j), we can find a normal subgroup K1 such t h a t N,j+I/K1 is solvable of length 1 and 2Vi,j n K1 = d~(N,,j). Next consider D]+I/5~(D~+I). Map the generators of Qa+l onto the generator of an infinite cyclic group. Then b y Lemma 5 we can find a solution to the equation Pi = 1 in a solvable group of length 1 in which D~+~/d~(D,+I) is embedded. To the resultant group we adjoin a cyclic image of O,+1-1 to obtain a group satisfying Pj-1 = 1 and in which D~+I/O1(D~+~)is embedded. We continue in this way to obtain a group which is a homomorphic image of N,,j+I with kernel K2 such that Nt,3"+I/K2 is solvable of length 1 and D~+ 1 n K2 = ~(D,+~). We finally put K = K1 n K2 to derive (12). With these facts at our disposal, we are now ready to derive our result. We let M = H (J) be a J-Magnus subgroup. We need to show that M is solvably separable in G.

Case (i) 0 6 J. Then in fact M = / / ( J ) . Suppose w ~ G \ R ( J ) . I f w is of non-zero exponent sum in t, w 6 N and since N 3___M, it suffices to consider the case where w is of zero exponent sum in t. Thus {w} ~ M C N; say {w} w M C N_,,,. Since M0 is solvably separable in D~ -- b y induction this holds since ~ (P) < ~ (B) -- repeated applications of Lemma 6 and 7 lead to the conclusion that M0 is solvably separable in 2V_,,,. Thus there exists 1 such that w 6M0d,(N_,.d. Suppose that for some ] > i, w eMod~(N_~,~). Then w = mob, m0 e M0 and ~ e d~(N_~,~). Hence e N 4 , , ~ 5~ (N_~-,~) = ~ (N-t,,) by Lemma 6. But this contradicts w 6 M0dt (N-~,d. Hence w 6 Md~+l (G).

Case (ii) 0 e J. We note that by interchanging the roles of x and y we could embed G in the group G~ which has generators the same as G except for y which is replaced b y ~, and G is embedded in this group by identifying y with ~ . I f we denote the corresponding free product of the groups H(*) and the enlarged Ht0) by H2, we let Me = H~ (J).

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We let H I denote the corresponding product in which x was identified with ~ . Thus in the notation above, H I = / ~ . We put Mz = H I (J) = / ~ (J). Both G1 and G2 are of course naturally embedded in G3 which has generators ~ and ~ instead of x and y. I f w e Mz n M2, in view of the fact t h a t M3 = / / 3 (J) is a free product of direct products of free groups, consideration of the normal form leads to the conclusion t h a t w e M. Hence if w ~ M, w does not belong to both M1 and M2. So assume without loss of generality t h a t w ~ M1. We now abandon the notation we have just introduced and use G again instead of Gz, and ~ instead of Mz. We must show t h a t w ~ ~ ~z (G) for some 1. Without loss of generality we m a y assume w ~ iV, since for each integer r, w ~ ~ ((~) if and only ff wt r ~ -~z (~). Thus it suffices to show t h a t w ~ / ~ (~ (N) for some 1. I f ] ~ J , then ~ -----gp(t) V, where V = E . H ~ ) ( i e J \ O ) , since t normalises V. ~ o w V i s a Magnus subgroup of D* and hence by the induction hypothesis, it is solvably separable in D*. Hence V is solvably separable in hr~,t b y repeated applications of L e m m a 7. Suppose w e N-i,~. Then there exists l such t h a t w ~ V~(N_~,~), so t h a t if for some m > i, w = v ~ with ~ e ~ (N-m, m) and v e V, then

which is a contradiction. Suppose now that/r e J . Then there is at least one ]r ~ J and we can use ]r instead of k in the procedure described at the beginning of this section.

9. EachH (iJ) is cyclic. Suppose now t h a t each H(LS) is cyclic. Case (i). Suppose for each i t h a t there is only one direct factor H (~J). Then in fact G is a positive one-relator group and the result either follows from [4] or else we can alternatively use the argument of 8 b y linking two cycles H(o, H(~') to form a free factor of r a n k 2. Case (ii). I~or some i there exist at least two direct components, say H (~ and H (~ This case follows almost exactly the arguments of 8. T h a t is, suppose H(~ is generated b y x and H (~ is generated b y y and t h a t x appears with exponent sum and y with exponent sum fl in ~ . We replace H (05) b y the infinite cycle generated b y ~ and identify x with ~ . We will need to define an M1 and an M2 as in 8, and so will need to adjoin an ~7 with ~ = y and [~, 7] = 1. Again it will be easy to establish t h a t if an element w ~ M, then w ~ Mz f~ M2. The argument can then be completed as in 8. References

[1] B. B~U~SLAGand S. J. PSLU~,An extension of the Freiheitssatz. Math. Proe. Cambridge Phil. Soc. 89, 35--41 (1981). [2] K. MURASUGI,The center of a group with a single defining relation. Math. Ann. 155, 246--251 (1964). [3] W. M_~G~-us, A. K~RASS and D. SoLrr~, Elements of finite order in groups with a single defining relation. Comm. Pure Appl. Math. XIII, 57--66 (1960). [4] G. BAv~s~G, Positive one-relator groups. Trans. Amer. Math. Soc. 156, 165--183 (1971).

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[5] K. W. GRUENBERG,Residual properties of infinite soluble groups. Proc. Lond. Math. Soc. 3rd Series voL VII no. 25, 29--62 (1957). [6] F. Lvvr~, Solutions of equations over groups. Bull. Amer. Math. Soe. 68, 603--604 (1962). [7] R. C. LYlVDO~and P. E. SCHUPP, Combinatorial Group Theory. Berlin-Heidelberg-New York 1977. Eingegangen am 18.2. 1981 Anschrift des Autors: Benjamin Baumslag Department of Mathematics Imperial College 180 Queen's Gate London SW 7 2 B 2 England