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Conditions (Z1), (Z2), (Z3)

Now we are ready to check that the presentationZ(M,Λ,ra)satisfies conditions (Z1), (Z2), (Z3) from Section 3.

Here we are going to use notation from Section 3 again, as in Section 8.

Let R0=S0 be the set Z(M,ra) of all relations in Z(M,Λ,ra) except for the hub, and let S1/2 = {Λ(0)}, so Z(M,Λ,ra) = R1/2. We let Y = K, so all letters from RA are 0-letters.

Condition (Z1.1) holds by an easy inspection. Condition (Z1.2) holds by Propo-sition 8.2.

The length of Λ is (2n+ 3)N > n, so (Z2.1) holds. The condition (Z2.2) is obviously true.

The next lemma gives us (Z2.3).

Lemma 9.8. — Assume thatv1w1 andv2w2 are cyclic permutations of Λ(0)±1 and |v1| ≥ ε|Λ(0)|. Then if u1v1 =v2u2 holds modulo S0 for some 0-words u1, u2, then we have u2w1 = w2u1 modulo S0.

Proof. — Suppose that u1v1 = v2u2 modulo S0. Then there exists a van Kam-pen diagram Γ over S0 with boundary p1q1p21q21 such that φ(p1)≡u1, φ(p2)≡u2, φ(q1)≡v1, φ(q2)≡v2 (see Figure 6). TheK-bands ofΓ starting on q1 end onq2 since φ(q1) is a linear word and u1,u2 do not contain letters from KK1. The labels of the start and end edges of each of these bands are equal. Hencev2φ(q2)φ(q1)≡ v1 ≡ v. Since |v| > 0, and Λ(0) and Λ(0)1 do not have common letters, v cannot be a subword of both Λ(0) and Λ(0)1. Without loss of generality assume that v is a subword of Λ(0).

Recall that for every j = 1, ...,N, we denote Λj(0) to be the subword of Λ(0) starting with λ(j) and ending with λ(j+1)( as usual “N+1” here is 1).

By Lemma 9.6 there exists a word h over M such that v·h =v, u1 =L(h,z1), u2 = R(h,z2) modulo Z(M,ra) for some z1,z2. Hence we can assume that u1 = L(h,z1), u2 =R(h,z2) in the free group and that Γ= T(v,h). Since T(v,h) is a dia-gram overZ(M,a), Γ does not contain Burnside ra-cells.

Since v ≡ φ(q1)φ(q2) is a subword of Λ(0), we can attach to Γ two hubs Π1 and Π2 along the paths q1 and q2, and consider the resulting diagram ∆ with boundary label w2u1w11u21. We need to show that ∆ can be replaced by a diagram over S0 with the same boundary label.

Since |v1| ≥ ε|Λ(0)| and N > n > 3, we conclude that v1 contains Λj(0) for some j > 1.

Let q1 be the subpath of q1 such that φ(q1) ≡ Λj(0). Since Λ is linear, every K-band starting on q1 ends on q2. Let T1 be the λ(j)-band and T2 be the λ(j+1) -band starting on q1. Then T1 ends on the edge of q2 labeled by λ(j,0) and T2 ends on the edge of q2 labeled by λ(j+1,0).

The bands T1, T2, the path q1 and the path q2 bound a subdiagram Γ of Γ which is a trapezium. By Lemma 7.5, Γ = Tj(0),h) and Λj(0)·h = Λj(0). Let t be the subpath of ∂(Γ)which is a side of T1.

Consider the subdiagram ∆ of ∆ consisting of Π1, Π2 and t. Since ∆\∆ con-sists of 0-cells, it suffices to show that the boundary label of ∆ is equal to 1 mod-ulo S0.

Thus we need to show that φ(t)=L(h,λ(j)) commutes with Λj(0)λ(j+1,0)1Λj+1(0)λ(j+2,0)1...Λj1(0)λ(j,0)1. (26)

By Proposition 6.16, for every i =1, ...,N we have Λi(0)·h=Λi(0). The trapezium Ti(0),h) gives us the equality

Λi(0)1L(h,λ(i))Λi(0)=R(h,λ(i+1))

=λ(i+1,0)−1(L(h,λ(i+1))λ(i+1,0), i =1, ...,N modulo S0. Hence

λ(i+1,0i(0)1L(h,λ(i))Λi(0)λ(i+1,0)1=L(h,λ(i+1)), i=1, ...,N.

Hence if we conjugate L(h,λ(i)) by the word (26), we get L(h,λ(i)) as required.

As in Section 3, for every cyclically reduced essential element g ∈ Hkra we de-fine 0(g) as the maximal subgroup of Hra which contains g in its normalizer. For an arbitrary essential element g of the form vuv1 where u is cyclically reduced, we de-fine 0(g) = v0(u)v1. By Lemma 3.7, 0(g) is well defined (does not depend on the presentationg =vuv1).

The following lemma will imply both (Z3.1) and (Z3.2).

Lemma 9.9. — Let w be an essential element represented by a cyclically minimal in rank 1/2 word W, let x = 1 be a 0-element in Hkra such that w4xw4 is a 0-element. Then the following two statements hold.

(i) There exists a 0-element u such that wu1 commutes with every element of 0(w). (ii)x ∈0(w).

Proof. — We divide the proof into several steps.

1. Let P1 be a 0-word representing x. By Lemma 3.4 every cyclic shift of W is a cyclically K-reduced word. Therefore we can assume that W starts with a letter z∈KK1 (by Lemma 3.7).

By Lemma 3.6, (Wz)1P1Wz is equal in Hkra to a 0-word P. By Lemma 9.5 Wz=CBzD in Hkra where B does not contain R-letters and starts with a K-letter z such that Cz = zC, zD = Dz modulo Z(M,a) where D ≡ L(f,z), C ≡L(f,z) for some words f,f over M.

In particular since Wz = CBzD =CBDz, we have W= CBD in Hkra. Since DCz =zDC, passing if necessary to a cyclic shift BDC of W, we can assume that C is empty, so z =z, and replace W by BD where Dz=zD modulo Z(M,a). The wordB is equal to zB1Uwhere B1 is either empty or ends with a letter zKK1, Uis a word over A.

2. By Lemma 3.6, we have the following equalities in Hkra: (zB1UDzB1Uz)1P1(zB1UDzB1Uz)=P2

(27)

(zB1UDzB1UDzB1Uz)1P1(zB1UDzB1UDzB1Uz)=P3

(28)

whereP2 and P3 are 0-words.

By Lemma 9.7 (which can be applied because zB1UDzB1Uz is reduced since this word is equal to a subword of the same length of the reduced word W3) we can assume that equality (27) holds moduloZ(M,a)(replacing P1 and P2 by equal in Hkra

words if necessary), Bz is a reduced admissible word. By Lemma 3.6, since Dz =zD, we have that(zB1Uz)1P1(zB1Uz)is a0-word moduloZ(M,a). Hence by Lemma 9.6 P1=L(h,z) modulo Z(M,a) and

Bz·h=Bz. (29)

3. Now consider a diagram∆ overZ(M,a)corresponding to equality (27),∂(∆)

= p1q1p21q21 where φ(p1) ≡ P1, φ(p2) ≡ P2, φ(q1) ≡ BDBz ≡ φ(q2). Then q1 = s1t1s1, where φ(s1)≡B, φ(t1) ≡D, φ(s1)≡Bz. Consider the maximal z-band T1 in

∆starting on the first edge ofs1. As usual, we can assume that p2 is a side of az-band T2. The bands T1 and T2 bound a subdiagram ∆1 of ∆ which is a trapezium. By Lemma 7.5, ∆1=T(Bz,h1) where h1 is the history of T1.

Consider the natural homomorphism ψ from Hkra to the free group freely gen-erated by M. This homomorphism takes K and A to 1, and takes each letter from R to the corresponding rule from M. It is clear that this homomorphism is correctly defined because killing all letters from KA in every relation from Z(M,a) and re-placing letters from R by the corresponding rules from M gives us a trivial relation 1= 1. Since D = L(f,z), ψ(D)= f. Applying this homomorphism to the bound-ary label of the subdiagram of∆ bounded by p1 and a side of T1, we get h1=f−1hf. Hence we have

Bz·f1hf =Bz. (30)

4. Now let ∆1 be a van Kampen diagram corresponding to equality (28). Ap-plying Lemma 9.6 to the subdiagram∆2 of∆1 bounded by two z-bands T1, T2 cor-responding to the two last distinguished occurrences of z in zB1UDzB1UDzB1Uz, we obtain zB1Uz ·h2 = zB1Uz, i.e. Bz ·h2 = Bz where h2 is the history of the z-band T1. Applying the homomorphism ψ to the boundary label of the subdiagram

of∆2=∆1\∆2 we obtain the equality ψ(D)2hψ(D)2h21=f2hf2h21 modulo Burn-side relations. Hence f2hf2h21 =1 modulo Burnside relations. We conclude that

Bz·h2 =Bz (31)

whereh2=f2hf2 modulo Burnside relations.

Let W1=Bz, h1=f−1hf, g =f, b=h2. ThenW1·h1=W1 by (30),W1·gh1g−1 = W1 by (29) and W1·b=W1 by (31) where b=g1h1g modulo Burnside relations.

Suppose that Bz contains an accepted subwordB. ThenB·h is a cyclic shift C ofΛ(0)for some word h overM. Then for some admissible subwordB ofBandB, B·h =C where CC ≡C, |C| ≤ 1. By Lemma 7.5 B =UCV in Hkra where U andVare 0-words. HenceB =U(C)1Vin rank 1/2. Since |U(C)1V|<|UCV|, and B is a subword of W, we get a contradiction with the assumption that W is cyclically minimal in rank 1/2.

Thus all conditions of Proposition 6.17 hold. By that proposition and by Re-mark 6.18, we can conclude that for every integers there exists an word bs+1 over M which is equal to gsh1gs =fs1hfs+1 modulo Burnside relations and such that

Bz·bs+1=Bz.

5. For arbitrary integer s consider the trapezia Ts = T(Bz,bs). Let ∂(Ts) = p1q1p−12 q−12 , where φ(p1)=L(bs,z), φ(p2)=R(bs,z), φ(qi)=Bz, i=1,2.

Considering the subdiagram of Ts obtained by removing the last z-band (whose side is p2), we get the equality

L(bs,z)BL(bs,z)1=B

moduloZ(M,a). ThusBcommutes with L(bs,z)modulo Z(M,a). Notice thatb0=h.

HenceBcommutes withL(f,z)sL(h,z)L(f,z)s moduloZ(M,ra). Recall thatL(f,z)

≡ D, L(h,z) ≡ P1 ≡ L(b0,z). Hence (D)−sx(D)s commutes with B for every inte-ger s.

Therefore for every non-negative integer s, we have the following equalities in Hkra:

WsP1Ws=

(BD)sP1(BD)s=(D)1B1...(D)1B1(D)1(B1P1B)DBD...BD= (D)1B1...(D)1B1((D)1P1D)BD...BD=

(D)1B1...B1((D)2P1(D)2)B...BD=...

(D)sP1(D)s (32)

(we underline the parts of the words which are changing). Similar equalities hold for negatives. Therefore for every integers, wsxws is a0-element inHkra. Hencex∈0(w)

(since the subgroup generated by wsxws, s = 0,±1, ..., consists of 0-elements and is normalized by w).

6. Let us prove that Bcommutes with every elementy∈0(w)modulo Z(M,ra). Let Py be a word in RA representing y. Since y ∈ 0(w), W2PyW2 is equal mod-ulo Z(M,ra) to a 0-word. By Lemma 3.6, (BDz)1Py(BDz) is a 0-word (modulo Z(M,ra)). Since Dz = zD modulo Z(M,ra), we have that (BzD)1Py(BzD) is a 0-word moduloZ(M,ra). By Lemma 3.6,(Bz)1Py(Bz)is a0-wordPy moduloZ(M,ra). Considering a van Kampen diagram for this equality, we deduce (by a usual argu-ment) that Py = L(hy,z), Py = R(hy,z) for some words hy,hy over M. Applying the homomorphism ψ to the equality

(Bz)1L(hy,z)(Bz)=R(hy,z)

we get that hy =hy modulo Burnside relations. The last equality implies B1PyB=B1L(hy,z)B=zR(hy,z)z1 =L(hy,z)=Py

(33)

modulo Z(M,ra) (by Lemma 7.1). Hence B commutes with Py modulo Z(M,ra), as required.

7. Now let y ∈0(w) be represented by a word Py. Let us prove by induction on l that (D)lPy(D)l belongs to 0(w) for every l ≥ 0. This is clearly true for l = 0.

Suppose that it is true for some l ≥0. Then

(D)l+1Py(D)l1=(D)(D)lPy(D)l(D)1

=B1(BD)(D)lPy(D)l(BD)1B. (34)

Since(D)lPy(D)l belongs to0(w)=0(BD), we have that(BD)(D)lPy(D)l(BD)1 belongs to 0(w) (by the definition of 0(w)). Since B commutes with every element of 0(w), (34) implies that

(D)l+1Py(D)l1=(BD)(D)lPy(D)l(BD)10(w).

Similarly one can prove that (D)lPy(D)l belongs to 0(w)for all l < 0.

8. Let u be the element ofHkra represented by D. Then clearlyu is a0-element.

Let w be the element represented by W(D)1. We need to prove that w commutes with y, that is, W(D)−1 commutes withPy modulo Z(M,ra), or, equivalently,

(BD)1Py(BD)=(D)1Py(D)

modulo Z(M,ra). This has been proved already in (33).

This completes verification of properties (Z1), (Z2), (Z3). By Proposition 3.19, we have the following statement.

Proposition 9.10. — There exists a graded presentation Rkra containing Z(M,ra) of the factor groupHkra(∞)ofHkra over the subgroup generated by alln-th powers of elements ofHkra, such that every relation of Rkra of rank ≥ 1 has the form un =1 for some word u, and every minimal diagram over Rkra satisfies property A from Section 2.4, and all the lemmas from Section 3.

10. Proof of Theorem 1.3

Lemma 10.1. — For every word w(x1, ...,xm), if w(a1(κ(1)), ...,am(κ(1))) = 1 in Hkra(∞), then w(a1, ...,am)=1 in A(κ(1)).

Proof. — Let ∆ be a g-reduced diagram over Rkra with boundary label w(a1(κ(1)), ...,am(κ(1))). Suppose that ∆ contains cells of rank > 0. By Prop-osition 9.10 ∆ is an A-map. By Lemma 2.5 ∆ contains a cell Π of rank > 0 and a contiguity subdiagram of Π to ∂(∆). Since non-0 letters in presentation Rkra are letters from K, by definition of contiguity subdiagrams from Section 2, ∂(∆) contains a K-edge, a contradiction.

Hence ∆ does not contain cells of rank > 0. Therefore ∆ is a diagram over Hkra. So we can assume that ∆ is a g-reduced diagram over Hkra. By Lemma 9.2 we can assume that ∆ does not contain K-annuli. Hence ∆ does not contain K-edges.

Therefore∆ is a diagram over Rra where R is the set of non-0 letters.

Suppose that ∆ contains a cell of rank > 0. Then by Proposition 8.2 and Lem-ma 2.6 ∆ contains a cell Π of rank > 0 and a contiguity subdiagram of Π to ∂(∆). Hence ∂(∆) contains R-edges, a contradiction. Therefore ∆ does not contain cells of rank > 0. Hence ∆ is a diagram over Hra. By Lemma 7.3, part (3), ∆ does not contain R-annuli. Hence ∆ does not contain R-edges. This means that all cells of ∆ are a-cells. Therefore ∆ is a diagram over Ha.

Since A(κ(1)) is a retract of Ha, we obtain that w(a1(κ(1)), ...,am(κ(1))) is equal to 1 modulo R(κ(1))-relations and Burnside relations, as required.

Given the Burnside group B(m,n) generated by a1, ...,am, we have, by Lem-ma 5.2, a homomorphism φ:B(m,n)H such that φ(ai)=ai(κ(1)), i=1, ...,m.

Assume that φ(w(a1, ...,am))Hn. Then w(a1(κ(1)), ...am(κ(1))) = 1 in Hkra(∞)=H(n)=H/Hn by Proposition 9.10. Setting R(κ(1))=∅, we have from Lemma 10.1 that w = 1 in A(κ(1)) = B(a1, ...,am;n). Hence φ is an embedding, and moreover, the intersection of φ(B(m,n))Hn is trivial. This proves Statement (1) of Theorem 1.3.

Let now L be an arbitrary normal subgroup of B(m,n) (the latter can now be identified with its image inH/Hn). To prove Statement (2) of Theorem 1.3, we have to show that L = LH/Hn ∩B(m,n) = L. Let us define R(κ(1)) to be the set of all words in a1(κ(1)), ...,am(κ(1)) representing elements of L. The new group Hkra(∞) (whose construction depends now on the new choice of R(κ(1))) is a homomorphic image of H/Hn by Proposition 9.10. Therefore it suffices to prove that every word w(a1(κ(1)), ...,am(κ(1))) that is equal to 1 in Hkra(∞), is also trivial in B(m,n)/L ∼= A(κ(1)). But again, this claim is a part of Lemma 10.1.

The proof of Theorem 1.3 is complete.

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properties (M1), (M2), (M3) 54

z,z+ 108 Z(M) 114 Z(M,Λ) 114

Z(M,a) 142 Z(M,ra) 151 Z(M,Λ,ra) 151

A. Yu. O.

Department of Mathematics Vanderbilt University

1326 Stevenson Center Nashville, TN 37240 USA

alexander.olshanskiy@vanderbilt.edu http://www.math.vanderbilt.edu/∼olsh and

Department of Higher Algebra, MEHMAT Moscow State University

Moscow, Russia

olshan@shabol.math.msu.su M. V. S.

Department of Mathematics Vanderbilt University 1326 Stevenson Center Nashville, TN 37240 USA

http://www.math.vanderbilt.edu/∼msapir

Manuscrit reçu le 8 février 2001.