*Geometry &Topology* *GGGG*
*GG*

*GGG GGGGGG*
*T T TTTTTTT*
*TT*

*TT*
*TT*
Volume 8 (2004) 1281–1300

Published: 28 September 2004

**The proof of Birman’s conjecture on** **singular braid monoids**

Luis Paris

*Institut de Math´**ematiques de Bourgogne, Universit´**e de Bourgogne*
*UMR 5584 du CNRS, BP 47870, 21078 Dijon cedex, France*

Email: lparis@u-bourgogne.fr

URL: http://math.u-bourgogne.fr/topo/paris/index.html

**Abstract**

Let*B** _{n}* be the Artin braid group on

*n*strings with standard generators

*σ*

_{1}

*, . . . ,*

*σ*

*n*

*−*1, and let

*SB*

*n*be the singular braid monoid with generators

*σ*

^{±}_{1}

^{1}

*, . . . , σ*

_{n}

^{±}

_{−}^{1}

_{1}

*,*

*τ*

_{1}

*, . . . , τ*

_{n}

_{−}_{1}. The desingularization map is the multiplicative homomorphism

*η*:

*SB*

_{n}*→*Z[B

*] defined by*

_{n}*η(σ*

_{i}

^{±}^{1}) =

*σ*

_{i}

^{±}^{1}and

*η(τ*

*) =*

_{i}*σ*

_{i}*−σ*

^{−}

_{i}^{1}, for 1

*≤*

*i≤n−*1. The purpose of the present paper is to prove Birman’s conjecture, namely, that the desingularization map

*η*is injective.

**AMS Classification numbers** Primary: 20F36
Secondary: 57M25. 57M27

**Keywords:** Singular braids, desingularization, Birman’s conjecture

Proposed: Joan Birman Received: 6 January 2004

Seconded: Robion Kirby, Cameron Gordon Revised: 21 September 2004

**1** **Introduction**

Define an *n–braid* to be a collection *b*= (b_{1}*, . . . , b** _{n}*) of disjoint smooth paths
in C

*×*[0,1], called the

*strings*of

*b, such that the*

*k-th string*

*b*

*k*runs mono- tonically in

*t*

*∈*[0,1] from the point (k,0) to some point (ζ(k),1), where

*ζ*is a permutation of

*{*1,2, . . . , n

*}*. An

*isotopy*in this context is a deformation through braids which fixes the ends. Multiplication of braids is defined by concatenation. The isotopy classes of braids with this multiplication form a group, called

*braid group on*

*n*

*strings, and denoted by*

*B*

*. This group has a well-known presentation with generators*

_{n}*σ*1

*, . . . , σ*

*n*

*−*1 and relations

*σ**j**σ** _{k}*=

*σ*

_{k}*σ*

*j*if

*|j−k|>*1

*,*

*σ*

_{j}*σ*

_{k}*σ*

*=*

_{j}*σ*

_{k}*σ*

_{j}*σ*

*if*

_{k}*|j−k|*= 1

*.*

The group *B**n* has other equivalent descriptions as a group of automorphisms
of a free group, as the fundamental group of a configuration space, or as the
mapping class group of the *n*–punctured disk, and plays a prominent rˆole in
many disciplines. We refer to [4] for a general exposition on the subject.

The Artin braid group *B**n* has been extended to the *singular braid monoid*
*SB** _{n}* by Birman [5] and Baez [1] in order to study Vassiliev invariants. The
strings of a singular braid are allowed to intersect transversely in finitely many
double points, called

*singular points. As with braids, isotopy is a deformation*through singular braids which fixes the ends, and multiplication is by concate- nation. Note that the isotopy classes of singular braids form a monoid and not a group. It is shown in [5] that

*SB*

*n*has a monoid presentation with generators

*σ*

_{1}

^{±}^{1}

*, . . . , σ*

_{n}

^{±}

_{−}^{1}

_{1},

*τ*

_{1}

*, . . . , τ*

_{n}

_{−}_{1}, and relations

*σ**i**σ*_{i}^{−}^{1}=*σ*_{i}^{−}^{1}*σ**i* = 1*,* *σ**i**τ**i* =*τ**i**σ**i* if 1*≤i≤n−*1*,*
*σ*_{i}*σ** _{j}* =

*σ*

_{j}*σ*

_{i}*,*

*σ*

_{i}*τ*

*=*

_{j}*τ*

_{j}*σ*

_{i}*,*

*τ*

_{i}*τ*

*=*

_{j}*τ*

_{j}*τ*

_{i}*,*if

*|i−j|>*1

*,*

*σ*_{i}*σ*_{j}*σ** _{i}*=

*σ*

_{j}*σ*

_{i}*σ*

_{j}*,*

*σ*

_{i}*σ*

_{j}*τ*

*=*

_{i}*τ*

_{j}*σ*

_{i}*σ*

_{j}*,*if

*|i−j|*= 1

*.*

Consider the braid group ring Z[B*n*]. The natural embedding *B**n* *→* Z[B*n*]
can be extended to a multiplicative homomorphism *η*: *SB*_{n}*→* Z[B* _{n}*], called

*desingularization map, and defined by*

*η(σ*_{i}^{±}^{1}) =*σ*_{i}^{±}^{1}*,* *η(τ**i*) =*σ**i**−σ*^{−}_{i}^{1}*,* if 1*≤i≤n−*1*.*

This homomorphism is one of the main ingredients of the definition of Vassiliev invariants for braids. It has been also used by Birman [5] to establish a relation between Vassiliev knot invariants and quantum groups.

One of the most popular problems in the subject, known as “Birman’s conjec-
ture”, is to determine whether *η* is an embedding (see [5]). At the time of this

writing, the only known partial answer to this question is that *η* is injective on
singular braids with up to three singularities (see [17]), and on singular braids
with up to three strings (see [13]).

The aim of the present paper is to solve this problem, namely, we prove the following.

**Theorem 1.1** *The desingularization map* *η*: *SB*_{n}*→*Z[B* _{n}*]

*is injective.*

Let *S*_{d}*B**n* denote the set of isotopy classes of singular braids with *d* singular
points. Recall that a *Vassiliev invariant of type* *d* is defined to be a homo-
morphism *v*: Z[B* _{n}*]

*→A*of Z–modules which vanishes on

*η(S*

_{d+1}*B*

*). One of the main results on Vassiliev braid invariants is that they separate braids (see [3], [15], [16]). Whether Vassiliev knot invariants separate knots remains an important open question. Now, it has been shown by Zhu [17] that this sepa- rating property extends to singular braids if*

_{n}*η*is injective. So, a consequence of Theorem 1.1 is the following.

**Corollary 1.2** *Vassiliev braid invariants classify singular braids.*

Let Γ be a graph (with no loop and no multiple edge), let *X* be the set of
vertices, and let *E* =*E(Γ) be the set of edges of Γ. Define the* *graph monoid*
of Γ to be the monoid *M(Γ) given by the monoid presentation*

*M(Γ) =hX* *|xy* =*yx* if *{x, y} ∈E(Γ)i*^{+}*.*

Graph monoids are also known as *free partially commutative monoids* or as
*right-angled Artin monoids. They were first introduced by Cartier and Foata [7]*

to study combinatorial problems on rearrangements of words, and, since then, have been extensively studied by both computer scientists and mathematicians.

The key point of the proof of Theorem 1.1 consists in understanding the struc-
ture of the multiplicative submonoid of Z[B* _{n}*] generated by the set

*{ασ*

_{i}^{2}

*α*

^{−}^{1}

*−*1;

*α∈B*

*and 1*

_{n}*≤i≤n−*1

*}*. More precisely, we prove the following.

**Theorem 1.3** *Let* Ω *be the graph defined as follows.*

*•* Υ =*{ασ*^{2}_{i}*α*^{−}^{1}; *α∈B*_{n}*and* 1*≤i≤n−*1*}* *is the set of vertices of* Ω;

*• {u, v}* *is an edge of* Ω *if and only if we have* *uv* =*vu* *in* *B*_{n}*.*

*Let* *ν*: *M*(Ω)*→*Z[B* _{n}*]

*be the homomorphism defined by*

*ν(u) =u−*1, for all

*u∈*Υ. Then

*ν*

*is injective.*

The proof of the implication Theorem 1.3 *⇒* Theorem 1.1 is based on the
observation that *SB** _{n}* is isomorphic to the semi-direct product of

*M*(Ω) with the braid group

*B*

*n*, and that

*ν*:

*M*(Ω)

*→*Z[B

*n*] is the restriction to

*M*(Ω) of the desingularization map. The proof of this implication is the subject of Section 2. Let

*A*

*, 1*

_{i j}*≤i < j*

*≤n, be the standard generators of the pure braid*group

*P B*

*n*. In Section 3, we show that Υ is the disjoint union of the conjugacy classes of the

*A*

*’s in*

_{i j}*P B*

*. Using homological arguments, we then show that we can restrict the study to the submonoid of*

_{n}*M*(Ω) generated by the conjugacy classes of two given generators,

*A*

*i j*and

*A*

*r s*. If

*{i, j} ∩ {r, s} 6*=

*∅*, then the subgroup generated by the conjugacy classes of

*A*

*and*

_{i j}*A*

*is a free group, and we prove the injectivity using a sort of Magnus expansion (see Section 4).*

_{r s}The case *{i, j} ∩ {r, s}*=*∅* is handled using the previous case together with a
technical result on automorphisms of free groups (Proposition 5.1).

**Acknowledgement** My first proof of Proposition 5.1 was awful, hence I asked
some experts whether they know another proof or a reference for the result. The
proof given here is a variant of a proof indicated to me by Warren Dicks. So, I
would like to thank him for his help.

**2** **Theorem 1.3 implies Theorem 1.1**

We assume throughout this section that the result of Theorem 1.3 holds, and we prove Theorem 1.1.

Let *δ**i* = *σ**i**τ**i* for 1 *≤* *i* *≤* *n−*1. Then *SB**n* is generated as a monoid by
*σ*_{1}^{±}^{1}*, . . . , σ*_{n}^{±}_{−}^{1}_{1}, *δ*_{1}*, . . . , δ*_{n}_{−}_{1}, and has a monoid presentation with relations

*σ*_{i}*σ*^{−}_{i}^{1} =*σ*_{i}^{−}^{1}*σ** _{i}* = 1

*,*

*σ*

_{i}*δ*

*=*

_{i}*δ*

_{i}*σ*

_{i}*,*if 1

*≤i≤n−*1

*,*

*σ*

_{i}*σ*

*=*

_{j}*σ*

_{j}*σ*

_{i}*,*

*σ*

_{i}*δ*

*=*

_{j}*δ*

_{j}*σ*

_{i}*,*

*δ*

_{i}*δ*

*=*

_{j}*δ*

_{j}*δ*

_{i}*,*if

*|i−j|>*1

*,*

*σ**i**σ**j**σ**i*=*σ**j**σ**i**σ**j**,* *σ**i**σ**j**δ**i*=*δ**j**σ**i**σ**j**,* if *|i−j|*= 1*.*
Moreover, the desingularization map *η*: *SB*_{n}*→*Z[B* _{n}*] is determined by

*η(σ*_{i}^{±}^{1}) =*σ*_{i}^{±}^{1}*,* *η(δ**i*) =*σ*_{i}^{2}*−*1*,* if 1*≤i≤n−*1*.*
The following lemma is a particular case of [12], Theorem 7.1.

**Lemma 2.1** *Let* *i, j* *∈ {*1, . . . , n*−*1*}, and let* *β* *∈* *SB*_{n}*. Then the following*
*are equivalent:*

(1) *βσ*^{2}* _{i}* =

*σ*

^{2}

_{j}*β;*(2)

*βδ*

*i*=

*δ*

*j*

*β.*

This lemma shows the following.

**Lemma 2.2** *Let* Ωˆ *be the graph defined as follows.*

*•* Υ =ˆ *{αδ*_{i}*α*^{−}^{1}; *α∈B*_{n}*and* 1*≤i≤n−*1*}* *is the set of vertices of* Ω;ˆ

*• {u,*ˆ *v*ˆ*}* *is an edge of* Ωˆ *if and only if we have* *uˆ*ˆ*v* = ˆ*vˆu* *in* *SB*_{n}*.*

*Then there exists an isomorphismϕ:* *M*( ˆΩ)*→ M*(Ω) *which sendsαδ*_{i}*α*^{−}^{1} *∈*Υˆ
*to* *ασ*^{2}_{i}*α*^{−}^{1} *∈*Υ *for all* *α∈B**n* *and* 1*≤i≤n−*1.

**Proof** Let *α, β* *∈B**n* and *i, j∈ {*1, . . . , n*−*1*}*. Then, by Lemma 2.1,
*ασ*^{2}_{i}*α*^{−}^{1} =*βσ*_{j}^{2}*β*^{−}^{1} *⇔* (β^{−}^{1}*α)σ*^{2}* _{i}* =

*σ*

_{j}^{2}(β

^{−}^{1}

*α)*

*⇔* (β^{−}^{1}*α)δ**i*=*δ**j*(β^{−}^{1}*α)* *⇔* *αδ**i**α*^{−}^{1} =*βδ**j**β*^{−}^{1}*.*

This shows that there exists a bijection *ϕ: ˆ*Υ *→* Υ which sends *αδ*_{i}*α*^{−}^{1} *∈* Υˆ
to *ασ*^{2}_{i}*α*^{−}^{1} *∈* Υ for all *α* *∈* *B**n* and 1 *≤* *i* *≤* *n−*1. Let *α, β* *∈* *B**n* and
*i, j∈ {*1, . . . , n*−*1*}*. Again, by Lemma 2.1,

(ασ^{2}_{i}*α*^{−}^{1})(βσ_{j}^{2}*β*^{−}^{1}) = (βσ_{j}^{2}*β*^{−}^{1})(ασ_{i}^{2}*α*^{−}^{1})

*⇔* *σ*^{2}* _{i}*(α

^{−}^{1}

*βσ*

^{2}

_{j}*β*

^{−}^{1}

*α) = (α*

^{−}^{1}

*βσ*

^{2}

_{j}*β*

^{−}^{1}

*α)σ*

_{i}^{2}

*⇔* *δ** _{i}*(α

^{−}^{1}

*βσ*

^{2}

_{j}*β*

^{−}^{1}

*α) = (α*

^{−}^{1}

*βσ*

^{2}

_{j}*β*

^{−}^{1}

*α)δ*

_{i}*⇔* (β^{−}^{1}*αδ*_{i}*α*^{−}^{1}*β)σ*^{2}* _{j}* =

*σ*

^{2}

*(β*

_{j}

^{−}^{1}

*αδ*

_{i}*α*

^{−}^{1}

*β)*

*⇔* (β^{−}^{1}*αδ*_{i}*α*^{−}^{1}*β)δ** _{j}* =

*δ*

*(β*

_{j}

^{−}^{1}

*αδ*

_{i}*α*

^{−}^{1}

*β)*

*⇔* (αδ*i**α*^{−}^{1})(βδ*j**β*^{−}^{1}) = (βδ*j**β*^{−}^{1})(αδ*i**α*^{−}^{1})

This shows that the bijection *ϕ: ˆ*Υ*→*Υ extends to an isomorphism *ϕ:* *M*( ˆΩ)

*→ M*(Ω).

Now, we have the following decomposition for *SB** _{n}*.

**Lemma 2.3**

*SB*

*=*

_{n}*M*( ˆΩ)o

*B*

_{n}*.*

**Proof** Clearly, there exists a homomorphism *f*: *M*( ˆΩ)o*B*_{n}*→* *SB** _{n}* which
sends

*β*to

*β∈SB*

*n*for all

*β*

*∈B*

*n*, and sends ˆ

*u*to ˆ

*u∈SB*

*n*for all ˆ

*u∈*Υ. Onˆ the other hand, one can easily verify using the presentation of

*SB*

*n*that there exists a homomorphism

*g*:

*SB*

_{n}*→ M*( ˆΩ)o

*B*

*such that*

_{n}*g(σ*

_{i}

^{±}^{1}) =

*σ*

_{i}

^{±}^{1}

*∈B*

*for all*

_{n}*i∈ {*1, . . . , n

*−*1

*}*, and

*g(δ*

*) =*

_{i}*δ*

_{i}*∈*Υ for allˆ

*i∈ {*1, . . . , n

*−*1

*}*. Obviously,

*f◦g*= Id and

*g◦f*= Id.

**Remarks** (1) Let *G( ˆ*Ω) be the group given by the presentation
*G( ˆ*Ω) =*h*Υˆ *|uˆ*ˆ*v*= ˆ*vu*ˆ if *{ˆu,v} ∈*ˆ *E( ˆ*Ω)i*.*

It is well-known that *M*( ˆΩ) embeds in *G( ˆ*Ω) (see [9], [10]), thus *SB** _{n}* =

*M*( ˆΩ)o

*B*

*embeds in*

_{n}*G( ˆ*Ω)o

*B*

*. This furnishes one more proof of the fact that*

_{n}*SB*

*n*embeds in a group (see [11], [2], [14]).

(2) The decomposition *SB** _{n}*=

*M*( ˆΩ)o

*B*

*together with Lemma 2.2 can be used to solve the word problem in*

_{n}*SB*

*n*. The proof of this fact is left to the reader. Another solution to the word problem for

*SB*

*can be found in [8].*

_{n}**Proof of Theorem 1.1** Consider the homomorphism deg : *B**n* *→* Z defined
by deg(σ*i*) = 1 for 1*≤i≤n−*1. For *k∈*Z, let *B**n*^{(k)}=*{β* *∈B**n*; deg(β) =*k}*.
We have the decomposition

Z[B* _{n}*] =M

*k**∈Z*

Z[B_{n}^{(k)}]*,*

where Z[B*n*^{(k)}] denotes the free abelian group freely generated by *B**n*^{(k)}. Let
*P* *∈*Z[B* _{n}*]. We write

*P*=P

*k**∈Z**P** _{k}*, where

*P*

_{k}*∈*Z[B

*n*

^{(k)}] for all

*k∈*Z. Then

*P*

*is called the*

_{k}*k-th component*of

*P*.

Let*γ, γ*^{0}*∈SB** _{n}* such that

*η(γ*) =

*η(γ*

*). We write*

^{0}*γ*=

*αβ*and

*γ*

*=*

^{0}*α*

^{0}*β*

*where*

^{0}*α, α*

^{0}*∈ M*( ˆΩ) and

*β, β*

^{0}*∈*

*B*

*(see Lemma 2.3). Let*

_{n}*d*= deg(β). We observe that the

*d-th component of*

*η(γ*) is

*±β*, and, for

*k < d*, the

*k-th component*of

*η(γ*) is 0. In particular,

*η(γ*) completely determines

*β*. Since

*η(γ) =η(γ*

*), it follows that*

^{0}*β*=

*β*

*.*

^{0}So, multiplying*γ* and *γ** ^{0}* on the right by

*β*

^{−}^{1}if necessary, we may assume that

*γ*=

*α∈ M*( ˆΩ) and

*γ*

*=*

^{0}*α*

^{0}*∈ M*( ˆΩ). Observe that

(ν*◦ϕ)(γ*) =*η(γ) =η(γ** ^{0}*) = (ν

*◦ϕ)(γ*

*)*

^{0}*.*

Since *ν* is injective (Theorem 1.3) and *ϕ* is an isomorphism (Lemma 2.2), we
conclude that *γ* =*γ** ^{0}*.

**3** **Proof of Theorem 1.3**

We start this section with the following result on graph monoids.

**Lemma 3.1** *Let* Γ *be a graph, let* *X* *be the set of vertices, and let* *E* =*E(Γ)*
*be the set of edges of* Γ*. Let* *x*_{1}*, . . . , x*_{l}*, y*_{1}*, . . . , y*_{l}*∈* *X* *and* *k* *∈ {*1,2, . . . , l*}*
*such that:*

*•* *x*_{1}*x*_{2}*. . . x** _{l}*=

*y*

_{1}

*y*

_{2}

*. . . y*

_{l}*(in*

*M*(Γ));

*•* *y** _{k}*=

*x*

_{1}

*, and*

*y*

_{i}*6*=

*x*

_{1}

*for all*

*i*= 1, . . . , k

*−*1.

*Then* *{y**i**, x*1*} ∈E(Γ)* *for all* *i*= 1,2, . . . , k*−*1.

**Proof** Let *F*^{+}(X) denote the free monoid freely generated by *X*. Let *≡*1 be
the relation on *F*^{+}(X) defined as follows. We set *u≡*1 *v* if there exist *u*1*, u*2*∈*
*F*^{+}(X) and *x, y∈X* such that *u*=*u*_{1}*xyu*_{2}, *v* =*u*_{1}*yxu*_{2}, and *{x, y} ∈* *E(Γ).*

For *p* *∈* N, we define the relation *≡**p* on *F*^{+}(X) by setting *u* *≡**p* *v* if there
exists a sequence *u*0 = *u, u*1*, . . . , u**p* = *v* in *F*^{+}(X) such that *u**i**−*1 *≡*1 *u**i* for
all *i*= 1, . . . , p. Consider the elements *u* = *x*_{1}*x*_{2}*. . . x** _{l}* and

*v*=

*y*

_{1}

*y*

_{2}

*. . . y*

*in*

_{l}*F*

^{+}(X). Obviously, there is some

*p∈*N such that

*u≡*

*p*

*v*. Now, we prove the result of Lemma 3.1 by induction on

*p*.

The case *p*= 0 being obvious, we may assume *p≥*1. There exists a sequence
*u*_{0} =*u, u*_{1}*, . . . , u*_{p}_{−}_{1}*, u** _{p}* =

*v*in

*F*

^{+}(X) such that

*u*

_{i}

_{−}_{1}

*≡*1

*u*

*for all*

_{i}*i*= 1, . . . , p.

By definition of*≡*1, there exists*j∈ {*1,2, . . . , l*−*1*}* such that*{y**j**, y**j+1**} ∈E(Γ)*
and *u*_{p}_{−}_{1} =*y*_{1}*. . . y*_{j}_{−}_{1}*y*_{j+1}*y*_{j}*y*_{j+2}*. . . y** _{l}*. If either

*j < k−*1 or

*j > k*, then, by the inductive hypothesis, we have

*{x*

_{1}

*, y*

_{i}*} ∈*

*E(Γ) for all*

*i*= 1, . . . , k

*−*1. If

*j*=

*k−*1, then, by the inductive hypothesis, we have

*{x*1

*, y*

*i*

*} ∈*

*E(Γ) for all*

*i*= 1, . . . , k

*−*2. Moreover, in this case,

*{y*

_{j}*, y*

_{j+1}*}*=

*{y*

_{k}

_{−}_{1}

*, y*

_{k}*}*=

*{y*

_{k}

_{−}_{1}

*, x*

_{1}

*} ∈*

*E(Γ). If*

*j*=

*k*, then, by the inductive hypothesis, we have

*{y*

_{i}*, x*

_{1}

*} ∈E(Γ) for*all

*i*= 1, . . . , k

*−*1 and

*i*=

*k*+ 1.

Now, consider the standard epimorphism *θ*: *B*_{n}*→* Sym* _{n}* defined by

*θ(σ*

*) = (i, i+ 1) for 1*

_{i}*≤i≤n−*1. The kernel of

*θ*is called the

*pure braid group on*

*n*

*strings, and is denoted by*

*P B*

*. It has a presentation with generators*

_{n}*A** _{i j}* =

*σ*

_{j}

_{−}_{1}

*. . . σ*

_{i+1}*σ*

^{2}

_{i}*σ*

^{−}

_{i+1}^{1}

*. . . σ*

^{−}

_{j−1}^{1}

*,*1

*≤i < j*

*≤n ,*and relations

*A*^{−}_{r s}^{1}*A*_{i j}*A** _{r s}* =

*A*

*if*

_{i j}*r < s < i < j*or

*i < r < s < j ,*

*A*

^{−}

_{r s}^{1}

*A*

*i j*

*A*

*r s*=

*A*

*r j*

*A*

*i j*

*A*

^{−}

_{r j}^{1}if

*s*=

*i ,*

*A*^{−}_{r s}^{1}*A*_{i j}*A** _{r s}*=

*A*

_{i j}*A*

_{s j}*A*

_{i j}*A*

^{−}

_{s j}^{1}

*A*

^{−}

_{i j}^{1}if

*i*=

*r < s < j ,*

*A*^{−}_{r s}^{1}*A*_{i j}*A** _{r s}*=

*A*

_{r j}*A*

_{s j}*A*

^{−}

_{r j}^{1}

*A*

^{−}

_{s j}^{1}

*A*

_{i j}*A*

_{s j}*A*

_{r j}*A*

^{−}

_{s j}^{1}

*A*

^{−}

_{r j}^{1}if

*r < i < s < j .*

(See [4]). We denote by*H*_{1}(P B* _{n}*) the abelianization of

*P B*

*, and, for*

_{n}*β∈P B*

*, we denote by [β] the element of*

_{n}*H*

_{1}(P B

*) represented by*

_{n}*β*. A consequence of the above presentation is that

*H*1(P B

*n*) is a free abelian group freely generated

by *{*[A* _{i j}*]; 1

*≤*

*i < j*

*≤*

*n}*. This last fact shall be of importance in the remainder of the paper.

For 1*≤i < j* *≤n, we set*

Υ*i j* =*{βA**i j**β*^{−}^{1} ; *β* *∈P B**n**}.*
**Lemma 3.2** *We have the disjoint union* Υ =F

*i<j*Υ_{i j}*.*
**Proof** It is esily checked that

*σ*_{r}*A*_{i j}*σ*_{r}^{−}^{1} =

*A** _{i j+1}* if

*r*=

*j ,*

*A*_{j}_{−}_{1}_{j}*A*_{i j}_{−}_{1}*A*^{−}_{j}_{−}^{1}_{1}* _{j}* if

*r*=

*j−*1

*> i ,*

*A*

_{i+1}*if*

_{j}*j−*1

*> i*=

*r ,*

*A*

^{−}

_{i j}^{1}

*A*

_{i}

_{−}_{1}

_{j}*A*

*if*

_{i j}*r*=

*i−*1

*,*

*A**i j* otherwise*.*

This implies that the union S

*i<j*Υ* _{i j}* is invariant by the action of

*B*

*by con- jugation. Moreover,*

_{n}*σ*

_{i}^{2}=

*A*

*i i+1*

*∈*Υ

*i i+1*for all

*i*

*∈ {1, . . . , n*

*−*1}, thus Υ

*⊂*S

*i<j*Υ*i j*. On the other hand, *A**i j* is conjugate (by an element of *B**n*) to
*σ*_{i}^{2}, thus Υ_{i j}*⊂*Υ for all *i < j*, therefore S

*i<j*Υ_{i j}*⊂*Υ.

Let *i, j, r, s* *∈ {*1, . . . , n*}* such that *i < j*, *r < s, and* *{i, j} 6*= *{r, s}*. Let
*u∈* Υ*i j* and *v* *∈*Υ*r s*. Then [u] = [A*i j*]*6= [A**r s*] = [v], therefore *u* *6=v*. This
shows that Υ_{i j}*∩*Υ* _{r s}*=

*∅*.

The following lemmas 3.3 and 3.5 will be proved in Sections 4 and 5, respectively.

Let*F(X) be a free group freely generated by some setX*. Let*Y* =*{gxg*^{−}^{1}; *g∈*
*F*(X) and*x∈* *X}*, and let *F*^{+}(Y) be the free monoid freely generated by *Y*.
We prove in Section 4 that the homomorphism *ν*: *F*^{+}(Y)*→*Z[F(X)], defined
by *ν(y) =y−*1 for all *y∈Y*, is injective (Proposition 4.1). The proof of this
result is based on the construction of a sort of Magnus expansion. Proposition
4.1 together with the fact that *P B** _{n}* can be decomposed as

*P B*

*=*

_{n}*F*o

*P B*

_{n}

_{−}_{1}, where

*F*is a free group freely generated by

*{A*

*; 1*

_{i n}*≤i≤n−*1

*}*, are the main ingredients of the proof of Lemma 3.3.

Choose some *x*_{0}*∈X*, consider the decomposition *F*(X) =*hx*_{0}*i ∗F(X\ {x*_{0}*}*),
and let *ρ*: *F*(X) *→* *F(X) be an automorphism which fixes* *x*_{0} and which
leaves *F*(X*\ {x*0*}) invariant. Let* *y*1*, . . . , y**l* *∈ {gx*0*g*^{−}^{1}; *g∈F*(X)}. We prove
in Section 5 that, if *ρ(y*_{1}*. . . y** _{l}*) =

*y*

_{1}

*. . . y*

*, then*

_{l}*ρ(y*

*) =*

_{i}*y*

*for all*

_{i}*i*= 1, . . . , l (Proposition 5.1). The proof of Lemma 3.5 is based on this result together with Corollary 3.4 below.

**Lemma 3.3** *Let* *i, j, r, s* *∈ {*1, . . . , n*}* *such that* *i < j,* *r < s,* *{i, j} 6*=*{r, s},*
*and* *{i, j} ∩ {r, s} 6*=*∅. Let* *M*[i, j, r, s] *be the free monoid freely generated by*
Υ*i j**∪*Υ*r s**, and let* *ν*¯: *M*[i, j, r, s] *→* Z[B*n*] *be the homomorphism defined by*

¯

*ν(u) =u−*1 *for all* *u∈*Υ_{i j}*∪*Υ_{r s}*. Then* *ν*¯ *is injective.*

**Corollary 3.4** *Let* *i, j* *∈ {*1, . . . , n*}* *such that* *i < j. Let* *M*[i, j] *be the free*
*monoid freely generated by* Υ_{i j}*, and let* *ν*¯: *M*[i, j]*→*Z[B* _{n}*]

*be the homomor-*

*phism defined by*

*ν(u) =*¯

*u−*1

*for all*

*u∈*Υ

*i j*

*. Then*¯

*ν*

*is injective.*

**Lemma 3.5** *Let* *i, j, r, s* *∈ {*1, . . . , n*}* *such that* *i < j,* *r < s, and* *{i, j} ∩*
*{r, s}*=*∅. (In particular, we have* *n≥*4.) Let Ω[i, j, r, s]¯ *be the graph defined*
*as follows.*

*•* Υ_{i j}*∪*Υ_{r s}*is the set of vertices of* Ω[i, j, r, s];¯

*• {u, v}* *is an edge of* Ω[i, j, r, s]¯ *if and only if we have* *uv* =*vu* *in* *B**n**.*
*Let* *M*[i, j, r, s] = *M*( ¯Ω[i, j, r, s]), and let *ν*¯: *M*[i, j, r, s] *→* Z[B* _{n}*]

*be the ho-*

*momorphism defined byν(u) =*¯

*u−*1

*for all*

*u∈*Υ

*i j*

*∪*Υ

*r s*

*. Thenν*¯

*is injective.*

**Proof of Theorem 1.3** Recall the decomposition
Z[B*n*] =M

*k**∈Z*

Z[B_{n}^{(k)}] (1)

given in the proof of Theorem 1.1, where *B**n*^{(k)} =*{β* *∈B**n*; deg(β) = *k}*, and
Z[B*n*^{(k)}] is the free abelian group freely generated by*B**n*^{(k)}. Note that deg(u) = 2
for all *u∈*Υ.

Let *α* *∈ M*(Ω). We write *α* = *u*1*u*2*. . . u** _{l}*, where

*u*

*i*

*∈*Υ for all

*i*= 1, . . . , l. Define the

*length*of

*α*to be

*|α|*=

*l*. We denote by ¯

*α*the element of

*B*

*represented by*

_{n}*α*(ie, ¯

*α*=

*u*

_{1}

*u*

_{2}

*. . . u*

*in*

_{l}*B*

*). Let [1, l] =*

_{n}*{*1,2, . . . , l

*}*. Define a

*subindex*of [1, l] to be a sequence

*I*= (i1

*, i*2

*, . . . , i*

*q*) such that

*i*1

*, i*2

*, . . . , i*

*q*

*∈*[1, l], and

*i*

_{1}

*< i*

_{2}

*<*

*· · ·*

*< i*

*. The notation*

_{q}*I*

*≺*[1, l] means that

*I*is a subindex of [1, l]. The

*length*of

*I*is

*|I|*=

*q*. For

*I*= (i

_{1}

*, i*

_{2}

*, . . . , i*

*)*

_{q}*≺*[1, l], we set

*α(I*) =

*u*

*i*1

*u*

*i*2

*. . . u*

*i*

*q*

*∈ M*(Ω) and ¯

*α(I) denotes the corresponding element*of

*B*

_{n}^{(2q)}.

Observe that the decomposition of *ν(α) with respect to the direct sum (1) is:*

*ν*(α) =
X*l*
*q=0*

(*−*1)^{l}^{−}* ^{q}* X

*I**≺*[1,l],*|**I**|*=q

¯

*α(I*)*,* (2)

and X

*I**≺*[1,l],*|**I**|*=q

¯

*α(I)∈*Z[B_{n}^{(2q)}]*,*
for all *q*= 0,1, . . . , l.

Let*α** ^{0}*=

*u*

^{0}_{1}

*u*

^{0}_{2}

*. . . u*

^{0}

_{k}*∈ M*(Ω) such that

*ν(α) =ν(α*

*). The decomposition given in (2) shows that*

^{0}*k*=

*l*and

X

*I**≺*[1,l],*|**I**|*=q

¯

*α(I) =* X

*I**≺*[1,l],*|**I**|*=q

¯

*α** ^{0}*(I)

*,*(3)

for all *q*= 0,1, . . . , l.

We prove that *α* = *α** ^{0}* by induction on

*l. The cases*

*l*= 0 and

*l*= 1 being obvious, we assume

*l≥*2.

Suppose first that *u*^{0}_{1}=*u*_{1}. We prove
X

*I**≺*[2,l],*|**I**|*=q

¯

*α(I) =* X

*I**≺*[2,l],*|**I**|*=q

¯

*α** ^{0}*(I) (4)

by induction on *q*. The case *q* = 0 being obvious, we assume *q≥*1. Then
X

*I**≺*[2,l],*|**I**|*=q

¯
*α(I)*

= X

*I**≺*[1,l],*|**I**|*=q

¯

*α(I)−u*_{1}*·* X

*I**≺*[2,l],*|**I**|*=q*−*1

¯
*α(I)*

= X

*I**≺*[1,l],*|**I**|*=q

¯

*α** ^{0}*(I)

*−u*1

*·*X

*I**≺*[2,l],*|**I**|*=q*−*1

¯

*α** ^{0}*(I) (by induction and (3))

= X

*I**≺*[2,l],*|**I**|*=q

¯
*α** ^{0}*(I)

*.*

Let *α*_{1}=*u*_{2}*. . . u** _{l}* and

*α*

^{0}_{1}=

*u*

^{0}_{2}

*. . . u*

^{0}*. By (4), we have*

_{l}*ν(α*

_{1}) =

*l**−*1

X

*q=0*

(*−*1)^{l}^{−}^{1}^{−}* ^{q}* X

*I**≺*[2,l],*|**I**|*=q

¯
*α(I)*

=

*l**−*1

X

*q=0*

(*−*1)^{l}^{−}^{1}^{−}* ^{q}* X

*I**≺*[2,l],*|**I**|*=q

¯

*α** ^{0}*(I) =

*ν(α*

^{0}_{1})

thus, by the inductive hypothesis, *α*1 =*α*^{0}_{1}, therefore *α*=*u*1*α*1=*u*1*α*^{0}_{1}=*α** ^{0}*.

Now, we consider the general case. (3) applied to *q*= 1 gives
X*l*

*i=1*

*u**i*=
X*l*
*i=1*

*u*^{0}_{i}*.* (5)

So, there exists *k* *∈ {*1, . . . , l*}* such that *u*^{0}* _{k}* =

*u*1 and

*u*

^{0}

_{i}*6*=

*u*1 for all

*i*= 1, . . . , k

*−*1. We prove that, for 1

*≤*

*i*

*≤*

*k*

*−*1,

*u*

^{0}*and*

_{i}*u*

_{1}=

*u*

^{0}*multi- plicatively commute (in*

_{k}*B*

*n*or, equivalently, in

*M(Ω)). It follows that*

*α*

*=*

^{0}*u*1

*u*

^{0}_{1}

*. . . u*

^{0}

_{k}

_{−}_{1}

*u*

^{0}

_{k+1}*. . . u*

^{0}*, and hence, by the case*

_{l}*u*1 =

*u*

^{0}_{1}considered before,

*α*=

*α*

*.*

^{0}Fix some *t∈ {*1, . . . , k*−*1*}*. Let *i, j, r, s* *∈ {*1, . . . , n*}* such that *i < j*, *r < s,*
*u*_{1} =*u*^{0}_{k}*∈*Υ* _{i j}*, and

*u*

^{0}

_{t}*∈*Υ

*. There are three possible cases that we handle simultaneously:*

_{r s}(1) *{i, j}*=*{r, s}*;

(2) *{i, j} 6*=*{r, s}* and *{i, j} ∩ {r, s} 6*=*∅*;
(3) *{i, j} ∩ {r, s}*=*∅*.

Let ¯Ω[i, j, r, s] be the graph defined as follows.

*•* Υ*i j**∪*Υ*r s* is the set of vertices of ¯Ω[i, j, r, s];

*• {u, v}* is an edge of ¯Ω[i, j, r, s] if and only if we have *uv* =*vu* in *B**n*.
Let*M*[i, j, r, s] =*M*( ¯Ω[i, j, r, s]), and let ¯*ν*: *M*[i, j, r, s]*→*Z[B*n*] be the homo-
morphism defined by ¯*ν(u) =u−*1 for all*u∈*Υ_{i j}*∪*Υ* _{r s}*. Note that, by Corollary
3.4 and Lemma 3.3, ¯Ω[i, j, r, s] has no edge and

*M*[i, j, r, s] is a free monoid in Cases 1 and 2. Moreover, the homomorphism ¯

*ν*is injective by Lemmas 3.3 and 3.5 and by Corollary 3.4.

Let *a*_{1} = 1, a_{2}*, . . . , a*_{p}*∈* [1, l], *a*_{1} *< a*_{2} *<* *· · ·* *< a** _{p}*, be the indices such that

*u*

_{a}

_{ξ}*∈*Υ

_{i j}*∪*Υ

*for all*

_{r s}*ξ*= 1,2, . . . , p. Let

*I*

_{0}= (a

_{1}

*, a*

_{2}

*, . . . , a*

*), and let*

_{p}*α(I*0) =

*u*

*a*1

*u*

*a*2

*. . . u*

*a*

*p*

*∈ M*[i, j, r, s]. (It is true that

*M*[i, j, r, s] is a submonoid of

*M*(Ω), but this fact is not needed for our purpose. So, we should consider

*α(I*

_{0}) as an element of

*M*[i, j, r, s], and not as an element of

*M*(Ω).) Recall that, for

*β*

*∈*

*P B*

*n*, we denote by [β] the element of

*H*1(P B

*n*) represented by

*β*. Recall also that

*H*

_{1}(P B

*) is a free abelian group freely generated by*

_{n}*{*[A

*]; 1*

_{i j}*≤i < j*

*≤n}*. Observe that

¯

*ν(α(I*_{0})) =
X*p*
*q=0*

(*−*1)^{p}^{−}* ^{q}* X

*I**≺*[1,l],*|**I**|*=q,
[ ¯*α(I)]**∈Z*[A*i j*]+Z[A*r s*]

¯

*α(I*)*.* (6)

Let *b*_{1}*, . . . , b*_{p}*∈* [1, l], *b*_{1} *< b*_{2} *<* *· · ·* *< b** _{p}*, be the indices such that

*u*

^{0}

_{b}*ξ* *∈*

Υ*i j* *∪*Υ*r s* for all *ξ* = 1,2, . . . , p. (Clearly, (5) implies that we have as many
*a** _{ξ}*’s as

*b*

*’s.) Note that*

_{ξ}*t, k*

*∈ {b*

_{1}

*, . . . , b*

_{p}*}*. Let

*I*

_{0}

*= (b*

^{0}_{1}

*, b*

_{2}

*, . . . , b*

*), and let*

_{p}*α*

*(I*

^{0}_{0}

*) =*

^{0}*u*

^{0}

_{b}1*u*^{0}_{b}

2*. . . u*^{0}_{b}

*p* *∈ M*[i, j, r, s]. By (3) we have
X

*I**≺*[1,l],*|**I**|*=q,
[ ¯*α(I)]**∈Z*[A*i j*]+Z[A*r s*]

¯

*α(I) =* X

*I**≺*[1,l],*|**I**|*=q,
[ ¯*α** ^{0}*(I)]

*∈Z*[A

*i j*]+Z[A

*r s*]

¯
*α** ^{0}*(I)

*,*

for all *q* *∈*N, thus, by (6), ¯*ν(α(I*0)) = ¯*ν(α** ^{0}*(I

_{0}

*)). Since ¯*

^{0}*ν*is injective, it follows that

*α(I*

_{0}) =

*α*

*(I*

^{0}_{0}

*), and we conclude by Lemma 3.1 that*

^{0}*u*

^{0}*and*

_{t}*u*

^{0}*=*

_{k}*u*

_{1}commute.

**4** **Proof of Lemma 3.3**

As pointed out in the previous section, the key point of the proof of Lemma 3.3 is the following result.

**Proposition 4.1** *Let* *F(X)* *be a free group freely generated by some set* *X,*
*let* *Y* =*{gxg*^{−}^{1}; *g* *∈F*(X) *and* *x∈X}, let* *F*^{+}(Y) *be the free monoid freely*
*generated by* *Y, and let* *ν*: *F*^{+}(Y)*→*Z[F(X)] *be the homomorphism defined*
*by* *ν(y) =y−*1 *for all* *y* *∈Y. Then* *ν* *is injective.*

First, we shall prove Lemmas 4.2, 4.3, and 4.4 that are preliminary results to the proof of Proposition 4.1.

Let deg : *F(X)* *→* Z be the homomorphism defined by deg(x) = 1 for all
*x∈X*. Write *A*=Z[F(X)]. For *k∈*Z, let*F** _{k}*(X) =

*{g∈F*(X); deg(g)

*≥k}*, and let

*A*

*k*=Z[F

*(X)] be the free Z–module freely generated by*

_{k}*F*

*(X). The family*

_{k}*{A*

*k*

*}*

*k*

*∈Z*is a filtration of

*A*compatible with the multiplication, that is:

*• A**k**⊂ A**l* if *k≥l*;

*• A**p**· A**q**⊂ A**p+q* for all *p, q∈*Z;

*•* 1*∈ A*0.

Moreover, this filtration is a separating filtration, that is:

*• ∩**k**∈Z**A**k*=*{*0*}*.

Let ˜*A* denote the completion of*A* with respect to this filtration. For *k∈*Z, we
write*F*^{(k)}(X) =*{g∈F*(X); deg(g) =*k}*, and we denote by*A*^{(k)}=Z[F^{(k)}(X)]

the free Z–module freely generated by *F*^{(k)}(X). Then any element of ˜*A*can be

uniquely represented by a formal series P_{+}_{∞}

*k=d**P** _{k}*, where

*d∈*Z and

*P*

_{k}*∈ A*

^{(k)}for all

*k≥d.*

We take a copy *G** _{x}* of Z

*×*Z generated by

*{x,x*ˆ

*}*, for all

*x*

*∈*

*X*, and we set

*G*ˆ =

*∗*

*x*

*∈*

*X*

*G*

*x*. Let

*U*( ˜

*A) denote the group of units of ˜A. Then there is a*homomorphism ˆ

*η*: ˆ

*G→ U*( ˜

*A*) defined by

ˆ

*η(x) =x ,* *η(ˆ*ˆ *x) =x−*1*,* for*x∈X .*
Note that

ˆ

*η(ˆx*^{−}^{1}) =*−*

+*∞*

X

*k=0*

*x*^{k}*,* for*x∈X .*

The homomorphism ˆ*η* defined above is a sort of Magnus expansion and the
proof of the following lemma is strongly inspired by the proof of [6], Ch. II,*§*
5, Thm. 1.

**Lemma 4.2** *The homomorphism* *η*ˆ: ˆ*G→ U*( ˜*A*) *is injective.*

**Proof** Let *g* *∈* *G. Define the*ˆ *normal form* of *g* to be the finite sequence
(g_{1}*, g*_{2}*, . . . , g** _{l}*) such that:

*•* for all *i∈ {*1, . . . , l*}*, there exists *x*_{i}*∈X* such that *g*_{i}*∈G*_{x}_{i}*\ {*1*}*;

*•* *x*_{i}*6*=*x** _{i+1}* for all

*i*= 1, . . . , l

*−*1;

*•* *g*=*g*_{1}*g*_{2}*. . . g** _{l}*.

Clearly, such an expression for *g* always exists and is unique. The*length* of *g*
is defined to be lg(g) =*l.*

Let (p, q)*∈*Z*×*Z, (p, q)*6*= (0,0). Write
(t*−*1)^{p}*t** ^{q}*=

+*∞*

X

*k=d*

*c**k p q**t*^{k}*,*

where *d∈*Z and *c*_{k p q}*∈*Z for all *k≥d. We show that there exists* *a≥d* such
that *a* *6*= 0 and *c*_{a p q}*6*= 0. If *q* *6*= 0, then *a* = *q* *6*= 0 and *c** _{q p q}* =

*±*1

*6*= 0. If

*q*= 0, then

*a*= 1

*6= 0 and*

*c*1

*p*0=

*±p6= 0.*

Let *g∈G,*ˆ *g6= 1. Let (ˆx*^{p}_{1}^{1}*x*^{q}_{1}^{1}*, . . . ,x*ˆ^{p}_{l}^{l}*x*^{q}_{l}* ^{l}*) be the normal form of

*g. We have*ˆ

*η(g) = (x*_{1}*−*1)^{p}^{1}*x*^{q}_{1}^{1}(x_{2}*−*1)^{p}^{2}*x*^{q}_{2}^{2}*. . .*(x_{l}*−*1)^{p}^{l}*x*^{q}_{l}^{l}

= X

*k*1*≥**d*1*,...,k**l**≥**d**l*

*c**k*1*p*1*q*1*c**k*2*p*2*q*2*. . . c**k*_{l}*p*_{l}*q*_{l}*·x*^{k}_{1}^{1}*x*^{k}_{2}^{2}*. . . x*^{k}_{l}^{l}*.*

By the above observation, there exist*a*_{1}*, a*_{2}*, . . . , a*_{l}*∈*Z*\{*0*}* such that*c*_{a}_{i}_{p}_{i}_{q}_{i}*6*=
0 for all*i*= 1, . . . , l. Now, we show that*x*^{k}_{1}^{1}*. . . x*^{k}_{l}^{l}*6*=*x*^{a}_{1}^{1}*. . . x*^{a}_{l}* ^{l}* if (k

_{1}

*, . . . , k*

*)*

_{l}*6*= (a1

*, . . . , a*

*l*). This implies that the coefficient of

*x*

^{a}_{1}

^{1}

*. . . x*

^{a}

_{l}*in ˆ*

^{l}*η(g) is*

*c*

_{a}_{1}

_{p}_{1}

_{q}_{1}

*. . . c*

_{a}

_{l}

_{p}

_{l}

_{q}

_{l}*6*= 0, thus ˆ

*η(g)6*= 1.

Since (ˆ*x*^{p}_{1}^{1}*x*^{q}_{1}^{1}*, . . . ,x*ˆ^{p}_{l}^{l}*x*^{q}_{l}* ^{l}*) is the normal form of

*g*, we have

*x*

*i*

*6*=

*x*

*i+1*for all

*i*= 1, . . . , l

*−*1, thus (x

^{a}_{1}

^{1}

*, . . . , x*

^{a}

_{l}*) is the normal form of*

^{l}*x*

^{a}_{1}

^{1}

*. . . x*

^{a}

_{l}*. Suppose*

^{l}*k*

_{i}*6*= 0 for all

*i*= 1, . . . , l. Then (x

^{k}_{1}

^{1}

*, . . . , x*

^{k}

_{l}*) is the normal form of*

^{l}*x*

^{k}_{1}

^{1}

*. . . x*

^{k}

_{l}*, therefore*

^{l}*x*

^{k}_{1}

^{1}

*. . . x*

^{k}

_{l}

^{l}*6*=

*x*

^{a}_{1}

^{1}

*. . . x*

^{a}

_{l}*if (k1*

^{l}*, . . . , k*

*)*

_{l}*6*= (a1

*, . . . , a*

*). Suppose there exists*

_{l}*i∈ {*1, . . . , l

*}*such that

*k*

*= 0. Then*

_{i}lg(x^{k}_{1}^{1}*. . . x*^{k}_{l}* ^{l}*)

*< l*= lg(x

^{a}_{1}

^{1}

*. . . x*

^{a}

_{l}*)*

^{l}*,*thus

*x*

^{k}_{1}

^{1}

*. . . x*

^{k}

_{l}

^{l}*6*=

*x*

^{a}_{1}

^{1}

*. . . x*

^{a}

_{l}*.*

^{l}For each *x* *∈* *X*, we take a copy *SG**x* of Z*×*N generated as a monoid by
*{x, x*^{−}^{1}*,x*ˆ*}*, and we set *SG* = *∗**x**∈**X**SG** _{x}*. Then there is a homomorphism

*η*:

*SG→*Z[F(X)] defined by

*η(x*^{±}^{1}) =*x*^{±}^{1}*,* *η(ˆx) =x−*1*,* for*x∈X .*
**Lemma 4.3** *The homomorphism* *η*: *SG→*Z[F(X)] *is injective.*

**Proof** We have *SG⊂G, and, since*ˆ *{A**k**}**k**∈Z* is a separating filtration, *A* =
Z[F(X)] is a subalgebra of ˜*A*. Now, observe that *η*: *SG* *→* Z[F(X)] is the
restriction of ˆ*η* to *SG, thus, by Lemma 4.2,* *η* is injective.

Let ˆ*Y* = *{gˆxg*^{−}^{1}; *g* *∈* *F*(X) and *x* *∈* *X} ⊂* *SG, and let* *F*^{+}( ˆ*Y*) be the free
monoid freely generated by ˆ*Y*. The proof of the following lemma is left to the
reader. A more general statement can be found in [9].

**Lemma 4.4** *We have* *SG*=*F*^{+}( ˆ*Y*)o*F*(X).

Now, we can prove Proposition 4.1, and, consequently, Lemma 3.3.

**Proof of Proposition 4.1** Let ˆ*ν*: *F*^{+}( ˆ*Y*) *→* Z[F(X)] be the restriction of
*η*: *SG*=*F*^{+}( ˆ*Y*)o*F(X)→*Z[F(X)] to *F*^{+}( ˆ*Y*), and let *ϕ*: *F*^{+}( ˆ*Y*)*→F*^{+}(Y)
be the epimorphism defined by*ϕ(gˆxg*^{−}^{1}) =*gxg*^{−}^{1} for all*g∈F(X) andx∈X*.
(The proof that *ϕ* is well-defined is left to the reader.) The homomorphism
ˆ

*ν* is injective (Lemma 4.3), *ϕ* is a surjection, and ˆ*ν* = *ν* *◦ϕ, thus* *ϕ* is an
isomorphism and *ν* is injective.