This paper investigates the structure of quotient groups derived from the McCool group of rank 3, revealing their relation to free groups and conditions for embedding into Johnson Lie algebras.
Contribution
It establishes an isomorphism between certain quotients of the McCool group and free groups, and characterizes when the associated graded Lie algebra embeds into the Johnson Lie algebra.
Findings
01
Quotients of the McCool group are isomorphic to two copies of free group quotients.
02
Provides a necessary and sufficient condition for embedding the graded Lie algebra into the Johnson Lie algebra.
03
Enhances understanding of the algebraic structure of IA-automorphisms of free groups.
Abstract
We prove that, for any positive integer c, the quotient group γc(M3)/γc+1(M3) of the lower central series of the McCool group M3 is isomorphic to two copies of the quotient group γc(F3)/γc+1(F3) of the lower central series of a free group F3 of rank 3 as Z-modules. Furthermore, we give a necessary and sufficient condition whether the associated graded Lie algebra gr(M3) of M3 is naturally embedded into the Johnson Lie algebra L(IA(F3)) of the IA-automorphisms of F3.
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TopicsGeometric and Algebraic Topology · Homotopy and Cohomology in Algebraic Topology · Finite Group Theory Research
Full text
Quotient groups of IA-automorphisms of a free group of rank 3
V. Metaftsis, A.I. Papistas and H. Sevaslidou
Abstract
We prove that, for any positive integer c, the quotient group γc(M3)/γc+1(M3) of the lower central series of the McCool group M3 is isomorphic to two copies of the quotient group γc(F3)/γc+1(F3) of the lower central series of a free group F3 of rank 3 as Z-modules. Furthermore, we give a necessary and sufficient condition whether the associated graded Lie algebra gr(M3) of M3 is naturally embedded into the Johnson Lie algebra L(IA(F3)) of the IA-automorphisms of F3.
1 Introduction and Notation
Let G be a group. For a positive integer c, let γc(G) be the c-th term of
the lower central series of G. We point out that γ2(G)=G′; that is, the derived group of G. We write IA(G) for the kernel of the
natural group homomorphism from Aut(G) into Aut(G/G′) and we call its elements IA-automorphisms of G. For a
positive integer c≥2, the natural group epimorphism from
G onto G/γc(G) induces a group homomorphism πc from
Aut(G) of G into
Aut(G/γc(G)) of G/γc(G). Write IcA(G)=Kerπc. Note that I2A(G)=IA(G). It is proved by
Andreadakis [2, Theorem 1.2], that if G is
residually nilpotent (that is, ⋂c≥1γc(G)={1}), then ⋂c≥2IcA(G)={1}.
Throughout this paper, by “Lie algebra”, we mean Lie
algebra over the ring of integers Z. Let G be a
group. Write grc(G)=γc(G)/γc+1(G) for c≥1 and
denote by (a,b) the commutator (a,b)=a−1b−1ab with a,b∈G. The (restricted) direct sum of
the quotients grc(G) is the associated graded
Lie algebra of G, gr(G)=⨁c≥1grc(G). The Lie bracket multiplication in gr(G) is [aγc+1(G),bγd+1(G)]=(a,b)γc+d+1(G),
where aγc+1(G) and bγd+1(G) are the
images of the elements a∈γc(G) and b∈γd(G) in the quotient groups grc(G) and
grd(G), respectively, and (a,b)γc+d+1(G)
is the image of the group commutator (a,b) in the quotient
group grc+d(G). Multiplication is then extended to
gr(G) by linearity.
For a
positive integer n, with n≥2, we write Fn for a free
group of rank n with a free generating set {x1,…,xn}. For c≥2, we
write Fn,c−1=Fn/γc(Fn). Thus, Fn,c−1 is the free nilpotent group of rank n and class c−1. The natural group epimorphism from Fn onto Fn,c−1
induces a group homomorphism πn,c−1:Aut(Fn)⟶Aut(Fn,c−1).
We write IcA(Fn)=Kerπn,c−1. It is well known that, for t,s≥2, (ItA(Fn),IsA(Fn))⊆It+s−1A(Fn).
Since Fn is residually nilpotent, we have ⋂c≥2IcA(Fn)={1}. Since Fn,c/grc(Fn)≅Fn,c−1
and grc(Fn) is a fully invariant subgroup of Fn,c, the natural group epimorphism from
Fn,c onto Fn,c−1 induces a group homomorphism ψc,c−1:Aut(Fn,c)⟶Aut(Fn,c−1).
It is well known that ψc,c−1 is onto. For c≥2, we define Ac∗(Fn)=Imπn,c∩Kerψc,c−1.
For t∈{2,…,c}, the natural group epimorphism from Fn,c onto Fn,c/γt(Fn,c) induces a group homomorphism σc,t:Aut(Fn,c)→Aut(Fn,c/γt(Fn,c)). Write ItA(Fn,c)=Kerσc,t,
and, for t=2, IA(Fn,c)=I2A(Fn,c). We note that, for c≥2, Fn,c/γc(Fn,c)≅Fn,c−1. Thus, for c≥2,
Ac∗(Fn)=Imπn,c∩IcA(Fn,c). It is easily proved that Ac∗(Fn)≅IcA(Fn)/Ic+1A(Fn) as free abelian groups (see, also, [2, Section 4, p. 246]). Furthermore, for n,c≥2, rank(Ac∗(Fn))≤nrank(grc(Fn))=cn∑d∣cμ(d)nc/d, where μ is the Möbius function. We point out that rank(grc(Fn))=c1∑d∣cμ(d)nc/d for all n,c≥2 (see, for example, [9]).
From now on, for r≥2, we write Lr(IA(Fn))=IrA(Fn)/Ir+1A(Fn). Form the (restricted) direct sum of the free
abelian groups Lr(IA(Fn)), and
denoted by
[TABLE]
It has the structure of a graded Lie algebra with Lr(IA(Fn)) as component of degree r−1 in the grading and Lie multiplication given by
[TABLE]
for all ϕ∈IjA(Fn), ψ∈IκA(Fn)(j,κ≥2). Multiplication is then extended to
L(IA(Fn)) by linearity. The above Lie algebra is usually called the
Johnson Lie algebra ofIA(Fn). We point out that, for a positive integer
c, γc(IA(Fn))⊆Ic+1A(Fn).
Let H be a finitely generated subgroup of IA(Fn) with H/H′ torsion-free. For
a positive integer q, let L1q(H)=γq(H)(Iq+2A(Fn))/Iq+2A(Fn). Form the
(restricted) direct sum of abelian groups L1(H)=⨁q≥1L1q(H).
It is easily verified that L1(H) is a Lie subalgebra of L(IA(Fn)). Furthermore, if {y1H′,…,ymH′} is a Z-basis for H/H′, then L1(H) is generated as Lie algebra by the set {y1(I3A(Fn)),…,ym(I3A(Fn))}. By a natural embedding of gr(H) into L(IA(Fn)), we mean that there exists a Lie algebra isomorphism ϕ from gr(H) onto L1(H) satisfying the conditions ϕ(yiH′)=yi(I3A(Fn)), i=1,…,m. In this case, we also say that gr(H) is naturally isomorphic to L1(H).
For n≥2, it was shown by Magnus [8], using work of
Nielsen [13], that IA(Fn) has a finite generating
set {χij,χijk:1≤i,j,k≤n;i=j,k;j<k}, where χij maps xi↦xi(xi,xj) and
χijk maps xi↦xi(xj−1,xk−1), with
both χij and χijk fixing the remaining basis elements. Let Mn be the subgroup of IA(Fn) generated by the subset
S={χij:1≤i,j≤n;i=j}. Then, Mn is called
the McCool group or the basis conjugating automorphisms
group. It is easily verified that the following relations are
satisfied by the elements of S, provided that, in each case, the
subscripts i,j,k,q occurring are distinct:
(χij,χkj)=(χij,χkq)=(χijχkj,χik)=1. It has been proved in [10] that Mn has a presentation
⟨S∣Z⟩, where Z is the set of all possible
relations of the above forms. Since γc(Mn)⊆γc(IA(Fn))⊆Ic+1A(Fn)
for all c≥1, and since Fn is residually nilpotent, we
have ⋂c≥1γc(Mn)={1} and so,
Mn is residually nilpotent.
In the present paper, we show the following result.
Theorem 1
For a positive integer c,
[TABLE]
as free abelian groups.
2. 2.
Let H be the subgroup of M3 generated by χ21,χ12,χ23. Then, L1(M3) is additively equal to the direct sum of the Lie subalgebras L1(H) and L1(Inn(F3)), where Inn(F3) denotes the group of inner automorphisms of F3. Furthermore, gr(M3) is naturally isomorphic to L1(M3) as Lie algebras if and only if gr(H) is naturally isomorphic to L1(H) as Lie algebras.
In [11, Theorem 1], it is shown that M3 is a Magnus group. The proof of it was long and tedious. In Section 2, we present a rather simple proof avoiding many of the technical results. The new approach gives us the description of each quotient group γc(M3)/γc+1(M3) as in Theorem 1 (1). By a result of Sjogren [15] (see, also, [6, Corollary 1.9]), M3 satisfies the Subgroup Dimension Problem. That is, each γc(M3) is equal to the c-th dimension subgroup of M3. Furthermore, the new approach helps us to give a necessary and sufficient condition for a natural embedding of gr(M3) into L(IA(F3)). For n≥2, let Inn(Fn) denote the subgroup of IA(Fn) consisting of all inner automorphisms of Fn. In Section 3, by using an observation of Andreadakis [2, Section 6, p. 249], we show that gr(Inn(Fn)) is naturally embedded into L(IA(Fn)). Hence, γc(Inn(Fn))/γc+1(Inn(Fn)) is isomorphic to a subgroup of Lc+1(IA(Fn)) for all c≥1. Since Inn(Fn)≅Fn, we obtain
[TABLE]
for all n,c, with n≥2. For c=1 and n≥2, we have rank(L2(IA(Fn)))=2n2(n−1) (see, [2, Theorem 5.1]). For n=2 we have IA(F2)=Inn(F2), by a result of Nielsen [12] and by a result of Andreadakis [2, Theorem 6.1], we have rank(Lc+1(IA(F2)))=c1∑d∣cμ(d)2c/d. For n=3, by Theorem 1(2), we may give a lower bound of rank(Lc+1(IA(F3))) in terms of the rank of L1c(H) for all c. In fact, we observe that
[TABLE]
(see, Remark 2 below). For n≥4 and c≥2, Satoh [16, Corollary 3.3] provides a lower bound for
rank(Lc+1(IA(Fn))).
2 The associated Lie algebra of M3
2.1 Lazard elimination
For a free Z-module A, let L(A) be the free
Lie algebra on A, that is, the free Lie algebra on A, where
A is an arbitrary Z-basis of A. Thus, we may
write L(A)=L(A). For a positive integer c, let
Lc(A) denote the cth homogeneous component of L(A). It
is well-known that L(A)=⨁c≥1Lc(A).
For Z-submodules A and B of any Lie algebra, let [A,B] be the Z-submodule spanned by
[a,b] where a∈A and b∈B. Furthermore, B≀A
denotes the Z-submodule defined by B≀A=B+[B,A]+[B,A,A]+⋯.
Throughout this paper, we use the left-normed convention for Lie commutators. The following
result is a version of Lazard’s ”Elimination Theorem” (see
[3, Chapter 2, Section 2.9, Proposition 10]). In the form
written here it is a special case of [4, Lemma 2.2] (see, also, [11, Section 2.2]).
Lemma 1
Let U and V be free Z-modules, and consider the
free Lie algebra L(U⊕V). Then, U and V≀U freely
generate Lie subalgebras L(U) and L(V≀U), and there is a
Z-module decomposition L(U⊕V)=L(U)⊕L(V≀U). Furthermore, V≀U=V⊕[V,U]⊕[V,U,U]⊕⋯
and, for each n≥0, there is a Z-module isomorphism [V,nU,…,U]≅V⊗nU⊗⋯⊗U under which [v,u1,…,un] corresponds to v⊗u1⊗⋯⊗un for all v∈V and all u1,…un∈U.
As a consequence of Lemma 1, we have the following result.
Corollary 1
For free Z-modules U and V, we write L(U⊕V) for the free
Lie algebra on U⊕V. Then, there is a
Z-module decomposition L(U⊕V)=L(U)⊕L(V)⊕L(W),
where W=W2⊕W3⊕⋯ such that, for all m≥2, Wm is the direct sum of submodules [V,U,aU,…,U,bV,…,V] with a+b=m−2 and a,b≥0. Each [V,U,aU,…,U,bV,…,V] is isomorphic to V⊗U⊗aU⊗⋯⊗U⊗bV⊗⋯⊗V as
Z-module. Furthermore, L(W) is the ideal of L(U⊕V) generated by the module
[V,U].
2.2 A decomposition of a free Lie algebra
Let X be the free Z-module with a Z-basis {x1,…,x6} and L=L(X) the free Lie algebra on X.
For i=1,2,3, let v2i=x2i−1+x2i. Furthermore, we write
[TABLE]
Since X=U⊕V, we have L=L(U⊕V) and so, L is free on X={x1,x3,x5,v2,v4,v6}. Let J be the subset of L,
[TABLE]
The aim of this section is to show the following result.
Proposition 1
With the above notation, let L=L(U⊕V) be the free Lie algebra on U⊕V. Let J be the ideal of L generated by the set J. Then, L=L(U)⊕L(V)⊕J. Moreover, J is a free Lie algebra.
For non negative integers a and b, we write [V,U,aU,bV] for [V,U,aU,…,U,bV,…,V]. By Lemma 1 and Corollary 1, we have
[TABLE]
where W=W2⊕W3⊕⋯ such that, for all m≥2,
[TABLE]
Furthermore, L(V≀U) and L(W) are the ideals in L generated by the modules V≀U and [V,U], respectively. In particular, L(W) is the ideal in L generated by the natural Z-basis
[TABLE]
of [V,U]. Let XV,U be the natural Z-basis of V≀U. That is,
[TABLE]
where [V,aU] is the natural Z-basis of the module [V,aU]. Let ψ2 be the Z-linear mapping from [V,U] into L(V≀U) with
[TABLE]
and ψ2 fixes the remaining elements of [V,U]. It is clear enough that ψ2 is a Z-linear monomorphism of [V,U] into L(V≀U). For a≥3, let ψa be the mapping from [V,U,(a−2)U] into L(V≀U) satisfying the conditions ψa([v,u,u3,…,ua])=[ψ2([v,u]),u3,…,ua] for all v∈V and u,u3,…,ua∈U. We define a map
[TABLE]
by Ψ(v)=v for all v∈V and, for a≥2, Ψ(v)=ψa(v) for all v∈[V,U,(a−2)U]. Since L(V≀U) is free on XV,U, we obtain Ψ is a Lie algebra homomorphism. By applying Lemma 2.1 in [5], we see that Ψ is a Lie algebra automorphism of L(V≀U). Since L(W) is a free Lie subalgebra of L(V≀U) and Ψ is an automorphism, we have Ψ(L(W)) is a free Lie subalgebra of L(V≀U). In fact,
[TABLE]
that is, Ψ(L(W)) is a free Lie algebra on Ψ(W).
Lemma 2
With the above notation, L(Ψ(W)) is an ideal in L.
*Proof. *Since Ψ is an automorphism of L(V≀U), we obtain Ψ(L(W))=L(Ψ(W)) is an ideal in L(V≀U). We point out that
[TABLE]
and so,
[TABLE]
To show that L(Ψ(W)) is an ideal in L, it is enough to show that [w,u]∈L(Ψ(W)) for all w∈L(Ψ(W)) and u∈L. Since L(Ψ(W)) is an ideal in L(V≀U) and Lemma 1, it is enough to show that [w,u]∈L(Ψ(W)) for all w∈L(Ψ(W)) and u∈L(U). Furthermore, we may show that [w,xi1,…,xik]∈L(Ψ(W)) for all w∈L(Ψ(W)) and xi1,…,xik∈{x1,x3,x5}. Write
[TABLE]
Since C is a Z-basis for Ψ(W), we have L(Ψ(W))=L(C). Thus, we need to show that [w,xi1,…,xik]∈L(Ψ(W)) for all w∈C and xi1,…,xik∈{x1,x3,x5}. Since C is a free generating set of L(Ψ(W)), the equation (1) and the linearity of the Lie bracket, we may assume that w∈[Ψ([V,U]),aU,bV] with a+b≥1. Clearly, we may assume that b≥1. Since L(Ψ(W)) is ideal in L(V≀U) and, by the Jacobi identity, we may further assume that w has a form [v,yj1,…,yja,μv2,νv4,ρv6] with yj1,…,yja∈U, μ,ν,ρ≥0 and μ+ν+ρ≥1. By using the Jacobi identity in the expression [w,xi1,…,xik], and replacing [v4,x1],[v2,x3] and [v4,x5] by [v4,v2]−ψ2([v4,x1]), [v2,v4]−ψ2([v2,x3]) and [v4,v6]−ψ2([v4,x5]), respectively, as many times as it is needed and since L(Ψ(W)) is an ideal in L(V≀U), we may show that [w,xi1,…,xik]∈L(Ψ(W)). Therefore, L(Ψ(W)) is an ideal in L. □
Example 1
In the present example, we explain the procedure described in the above proof. Let w=[ψ2([v4,x1]),x3,v2,v4]. Then,
[TABLE]
Proof of Proposition 1. Since ψ2([V,U])⊆J and J is ideal in L,
we get L(Ψ(W))⊆J.
Since J⊆L(Ψ(W)), by Lemma 2, we obtain J⊆L(Ψ(W)).
Therefore, J=L(Ψ(W)). That is, J is a free Lie algebra.
Furthermore, L=L(U)⊕L(V)⊕J. □
For c≥2, let Jc=J∩Lc. Since J is homogeneous, we get J=⨁c≥2Jc. From the above proof, we have the following result.
Corollary 2
With the above notation, let Ψ be the Lie algebra automorphism of L(V≀U) defined naturally on V≀U by means of ψ2. Then, J=L(Ψ(W)). Furthermore, for c≥2, Lc=Lc(U)⊕Lc(V)⊕Jc.
2.3 A description of gr(M3)
Our aim in this section is to show the following result. For its proof, we use similar arguments as in [11, Section 6].
Theorem 2
Let M3 be the McCool subgroup of the IA-automorphisms IA(F3) of F3. Then, gr(M3)≅L/J as Lie algebras. In particular, gr(M3) is isomorphic as a Lie algebra to an (external) direct sum of two copies of a free Lie algebra of rank 3. Furthermore, for each c,
[TABLE]
as free abelian groups.
Following the notation of the previous subsection, let us denote by F the free group generated by {x1,…,x6}. It is well
known that gr(F) is a free Lie algebra of rank 6; freely
generated by the set {xiF′:i=1,…,6}. The
free Lie algebras L and gr(F) are isomorphic and from now on we identify the two Lie algebras. Furthermore, Lc=γc(F)/γc+1(F) for all c≥1.
Define
[TABLE]
and R={r1,…,r9}.
Let N=RF be the normal closure of R in F. For a positive integer d, let Nd=N∩γd(F). We point out that for d≤2, Nd=N. Further, for d≥2, Nd+1=Nd∩γd+1(F). Define Id(N)=Ndγd+1(F)/γd+1(F). It is easily verified that Id(N)≅Nd/Nd+1 as Z-modules. The following result was shown in [11, Section 6].
Proposition 2
For a positive integer c, Nc+2 is generated by the set
{(r±1,g1,…,gs):r∈R,s≥c,g1,…,gs∈F∖{1}}. Furthermore, Ic+2(N)=Jc+2 for all c≥1.
Since F is residually nilpotent, we have ⋂d≥2Nd={1}. Also N⊆F′, we get I1(N)=0. Moreover, Id(N)≅Nd/Nd+1 as
Z-modules for all d≥2, and by Proposition
2 we have, Nd=Nd+1 for all d≥2. Define
[TABLE]
Since N is a normal subgroup of F, we have I(N) is an
ideal of L (see [7]).
Corollary 3
I(N)=J.
*Proof. *Since J=⨁d≥2Jd and I2(N)=J2, we have from Proposition 2 that I(N)=J. □
Proof of Theorem 2. Since M3/M3′≅F/NF′=F/F′,
we have gr(M3) is generated as a Lie algebra by the set
{αi:i=1,…,6} with αi=xiM3′. Since L is a free Lie algebra of rank 6
with a free generating set {x1,…,x6}, the map
ζ from L into gr(M3) satisfying the conditions
ζ(xi)=αi, i=1,…,6, extends uniquely
to a Lie algebra homomorphism. Since gr(M3) is generated as a
Lie algebra by the set {αi:i=1,…,6}, we have
ζ is onto. Hence, L/Kerζ≅gr(M3) as Lie
algebras. By definition, J⊆Kerζ, and so
ζ induces a Lie algebra epimorphism ζ from
L/J onto gr(M3). In particular, ζ(xi+J)=αi, i=1,…,6. Note that ζ
induces ζc, say, a Z-linear mapping
from (Lc+J)/J onto γc(M3)/γc+1(M3).
For c≥2,
[TABLE]
Since γc+1(F)⊆γc(F), we have by the
modular law,
by Corollary 2, where (Lc)∗=Lc(U)⊕Lc(V). Since both L(U) and L(V) are free Lie algebras of rank 3, we have L(U)≅L(V)≅gr(F3) in a natural way and so, for c≥1, Lc(U)≅Lc(V)≅γc(F3)/γc+1(F3) as free abelian groups. Hence, for c≥1,
[TABLE]
as free abelian groups and so, rank(γc(M3)/γc+1(M3))=rank(Lc)∗. Since J=⨁c≥2Jc, we have
(Lc+J)/J≅Lc/(Lc∩J)=Lc/Jc≅(Lc)∗ and so, we obtain Kerζc is torsion-free. Since rank(γc(M3)/γc+1(M3))=rank(Lc)∗, we have Kerζc={1}
and so, ζc is isomorphism. Since
ζ is epimorphism and each ζc is
isomorphism, we have ζ is isomorphism. Hence, L/J≅gr(M3) as Lie algebras. □
3 Embeddings
In this section, we shall give a criterion for the natural embedding of gr(M3) into L1(IA(F3)). We shall prove the following result.
Lemma 3
Let H be a finitely generated subgroup of IA(Fn), n≥2, with H/H′ torsion-free, and let {y1H′,…,ymH′} be a Z-basis for H/H′. Then, gr(H) is naturally isomorphic to L1(H) if and only if γc(H)=H∩(Ic+1A(Fn)) for all c.
*Proof. *We assume that γc(H)=H∩(Ic+1A(Fn)) for all c. For c≥1, let ψc be the natural Z-module epimorphism from grc(H) onto L1c(H). Since γc(H)∩(Ic+2A(Fn))=H∩(Ic+2A(Fn))=γc+1(H) for all c, we get ψc is isomorphism for all c≥1. Since gr(H) is the (restricted) direct sum of the quotients grc(H), there exists a group homomorphism ψ from gr(H) into L1(IA(Fn)) such that each ψc is the restriction of ψ on grc(H). It is easily shown that ψ is a Lie algebra homomorphism. Since ψ(yiH′)=yi(I3A(Fn)), i=1,…,n, we get ψ is a Lie algebra epimorphism. Furthermore, since each ψc is a Z-module isomorphism, we obtain ψ is a Lie algebra isomorphism. Conversely, let ϕ be a Lie algebra isomorphism from gr(H) onto L1(H) satisfying the conditions ϕ(yiH′)=yi(I3A(Fn)), i=1,…,m. Then, ϕ induces a Z-module isomorphism ϕc from grc(H) onto L1c(H) for all c. In particular, ϕc((yi1,…,yic)γc+1(H))=(yi1,…,yic)(Ic+2A(Fn)) for all i1,…,ic∈{1,…,m}. Furthermore, grc(H) is Z-module isomorphic to γc(H)/(γc(H)∩(Ic+2A(Fn))). Since grc(H) is polycyclic and so, it is a hopfian group, we have γc+1(H)=γc(H)∩(Ic+2A(Fn)) for all c. We claim that γc(H)=H∩(Ic+1A(Fn)) for all c. Since γc(H)⊆Ic+1A(Fn), it is enough to show that H∩(Ic+1A(Fn))⊆γc(H). To get a contradiction, let α∈H∩(Ic+1A(Fn)) and α∈/γc(H). Since γc(H)⊆Ic+1A(Fn) for all c and ⋂t≥2ItA(Fn)={1}, we get H is residually nilpotent. Thus, there exists a unique d∈N such that α∈γd(H)∖γd+1(H). Therefore, α∈/H∩(Id+2A(Fn)). Since α∈γd(H)∖γd+1(H) and α∈/γc(H), we have γc(H)⊆γd+1(H) and so, d+1≤c. Let k be a non-negative integer such that c=d+1+k. Since H∩(Ic+1A(Fn))=H∩(Id+2+kA(Fn)), we have α∈Id+2+kA(Fn). But Id+2+kA(Fn)⊆Id+2A(Fn) and so, α∈Id+2A(Fn), which is a contradiction. Therefore, γc(H)=H∩(Ic+1A(Fn)) for all c. □
Remark 1
It is known that IA(Fn)/γ2(IA(Fn)), with n≥2, is torsion-free and its rank is 2n2(n−1). Furthermore, γ2(IA(Fn))=I3A(Fn) (see, for example, [14]). It was conjectured by Andreadakis [2] that γc(IA(Fn))=Ic+1A(Fn) for all c. By Lemma 3, gr(IA(Fn)) is naturally isomorphic to L1(IA(Fn)) if and only if Andreadakis conjecture is valid. Now, if H is a finitely generated subgroup of IA(Fn) with H/H′ torsion free, then the statement gr(H) is naturally isomorphic to L1(H) seems to be an “Andreadakis Conjecture” for H.
3.1 The associated Lie algebra of Inn(Fn)
In this section, we show that the associated Lie algebra gr(Inn(Fn)) of the inner automorphisms Inn(Fn) of Fn, with n≥2, is naturally embedded into L(IA(Fn)). Throughout this section, we write En=Inn(Fn). Recall that, for g∈Fn, τg(x)=gxg−1 for all x∈Fn. Thus, En={τg:g∈Fn}. The following result has been proved in [2, Section 6].
Lemma 4
For a positive integer c, γc(En)=En∩Ic+1A(Fn).
Using the above we may show the following.
Proposition 3
Let n be positive integer, with n≥2. Then, gr(En)
is naturally embedded into L(IA(Fn)). In particular, for all c, grc(En) is isomorphic to a Z-submodule of Lc+1(IA(Fn)).
*Proof. *Since Fn is centerless, we have Fn≅En in a natural way and so, En is finitely generated. Moreover, gr(En) is a free Lie algebra of rank n. Namely, gr(En) is freely generated by the set {τxiEn′:i=1,…,n}. Let ϕ be the mapping from {τxiEn′:i=1,…,n} to L1(En) satisfying the conditions ϕ(τxiEn′)=τxiI3A(Fn), i=1,…,n. Since gr(En) is free on {τxiEn′:i=1,…,n}, ϕ is extended to a Lie algebra epimorphism. By Lemma 3, it is enough to show that γc(En)=En∩Ic+1A(Fn) for all c, which is valid by Lemma 4.
Therefore, gr(En)
is naturally embedded into L(IA(Fn)). Since
[TABLE]
for all c, we have grc(En) is isomorphic to a Z-submodule of Lc+1(IA(Fn)) for all c. □
Corollary 4
For positive integers n and c, with n≥2,
[TABLE]
where μ is the Möbius function.
*Proof. *Since Fn≅En as groups and gr(Fn) is a free Lie algebra, we get gr(Fn)≅gr(En) as Lie algebras. Hence, for c≥1, grc(Fn)≅grc(En) as Z-modules. Since rank(grc(Fn))=c1∑d∣cμ(d)nc/d, we obtain, by Proposition 3, the required result. □
3.2 The Lie algebra L1(M3)
Write b1=χ21, b2=χ12, b3=χ23, u2=χ31χ21, u4=χ32χ12 and u6=χ23χ13. Then, M3 is generated by the set {b1,b2,b3,u2,u4,u6} and its presentation is given by
We point out that u2=τx1−1, u4=τx2−1 and u6=τx3−1. Let H and E be the subgroups of M3 generated by the sets {b1,b2,b3} and {u2,u4,u6}, respectively. We point out that E=Inn(F3). One can easily see that H is a free group of rank 3. Thus, gr(H)≅gr(E)≅gr(F3) as Lie algebras.
Proposition 4
L1(M3)* is additively equal to the direct sum of the Lie subalgebras L1(H) and L1(E).*
*Proof. *By the proof of Proposition 3, gr(E)≅L1(E) and so, L1(E) is a free Lie algebra of rank 3. Since b1∈/I3A(F3), we have L1(H) is a non-trivial subalgebra of L(IA(F3)). In fact, L1(H) is generated by the set {bi(I3A(F3)):i=1,2,3}. Since (τg,ϕ)=τg−1ϕ−1(g) for all ϕ∈Aut(F3) and g∈F3, we have L1(E) is an ideal in L(IA(F3)) and so, L1(H)+L1(E) is a Lie subalgebra of L1(M3). Let w∈L1(H)∩L1(E). Since both L1(H) and L1(E) are graded Lie algebras, we may assume that w∈L1d(H)∩L1d(E) for some d. Thus, there are u∈γd(H) and v∈γd(E) such that w=u(Id+2A(F3))=v(Id+2A(F3)). To get a contradiction, we assume that u,v∈/Id+2A(F3). Therefore, v∈γd(E)∖γd+1(E) and so, there exists ω∈γd(F3)∖γd+1(F3) such that v=τωρ, where ρ∈γd+1(E). Since γd+1(E)⊆Id+2A(F3), we get u−1τω∈Id+2A(F3). Since u−1 fixes x3, we have x3−1(u−1τω(x3))=(x3,u−1(ω−1))∈γd+2(F3). Since u−1(xj)=xjyj, yj∈γd+1(F3), j=1,2, and γd(F3) is a fully invariant subgroup of F3, we have u−1(ω−1)=ω−1ω1, with ω1∈γd+1(F3) and so, (w,x3)∈γd+2(F3). Since gr(F3) is a free Lie algebra of rank 3 with a free generating set {xiF3′:i=1,2,3} and γd(F3)/γd+1(F3) is the d-th homogeneous component of gr(F3), we obtain (w,x3)∈γd+1(F3)∖γd+2(F3), which is a contradiction. Therefore, L1(H)∩L1(E)={0}. By the proof of Theorem 2, we obtain L1(M3)=L1(H)⊕L1(E). □
Remark 2
By Proposition 4, for all c, we obtain L1c(M3)=L1c(H)⊕L1c(E). Since L1c(E)≅γc(E)/γc+1(E)≅γc(F3)/γc+1(F3), we have rank(L1c(E))=c1∑d∣cμ(d)3c/d. Thus, for any c,
[TABLE]
Let H1 be the subgroup of H generated by the set {χ21,χ23}. We point out that H1 is a free group of rank 2. Then, gr(H1) is a free Lie algebra of rank 2. Since both χ21 and χ23 fix x1 and x3, it may be shown that γc(H1)=H1∩Ic+1A(F3) for all c. Therefore, γc(H1)∩Ic+2A(F3)=γc+1(H1) for all c and so, L1c(H1)≅grc(H1) for all c≥1. Since gr(H1) is a free Lie algebra on H1={χ21H1′,χ23H1′}, the mapping ψ from H1 into L1(H1) satisfying the conditions ψ(aH1′)=a(I3A(F3)) with a∈{χ21,χ23} can be extended to a Lie algebra homomorphism ψ. Since L1(H1) is generated as Lie algebra by the set {χ21(I3A(F3)),χ23(I3A(F3))}, we have ψ is onto. By Lemma 3, ψ is a natural Lie algebra isomorphism from gr(H1) onto L1(H1). Since L1c(H1)≅grc(H1) for all c≥1, we have rank(L1c(H1))=c1∑d∣cμ(d)2c/d. Since L1(H1) is a Lie subalgebra of L1(H), we obtain L1c(H1)≤L1c(H) for all c. Therefore,
[TABLE]
for all c and so,
[TABLE]
for all c.
In our next result, we give a necessary and sufficient condition for a natural embedding of gr(M3) into L(IA(F3)).
Proposition 5
Let H be the subgroup of M3 generated by χ21,χ12,χ23. Then, gr(M3) is naturally isomorphic to L1(M3) as Lie algebras if and only if gr(H) is naturally isomorphic to L1(H) as Lie algebras.
*Proof. *Suppose that gr(M3) is naturally isomorphic to L1(M3) as Lie algebras. Thus, for all c, grc(M3)≅L1c(M3) as Z-modules. By Proposition 4, grc(M3)≅L1c(H)⊕L1c(E). Since, for all c, grc(M3)≅grc(F3)⊕grc(F3) as Z-modules, we obtain
[TABLE]
Therefore, L1c(H)≅grc(H) for all c. Hence, γc(H)∩Ic+2A(F3)=γc+1(H) for all c and so, as in the proof of Lemma 3, γc(H)=H∩Ic+1A(F3) for all c. We point out that H/H′ is torsion-free with a free generating set H={χ21H′,χ12H′,χ23H′}. Since gr(H) is a free Lie algebra with a free generating set H, and L1(H) is generated as a Lie algebra by the set {χ21(I3A(F3)),χ12(I3A(F3)),χ23(I3A(F3))}, we have there exists a natural Lie epimorphism from gr(H) onto L1(H). By Lemma 3, gr(H) is naturally isomorphic to L1(H) as Lie algebras.
Conversely, let gr(H) be naturally isomorphic to L1(H) as Lie algebras. By Lemma 3, γc(H)=H∩(Ic+1A(F3)) for all c. Thus,
[TABLE]
and so, L1c(H)≅grc(H) for all c. By Proposition 4,
Let ψ be the Lie algebra epimorphism from L onto L1(M3) satisfying the conditions ψ(x2j−1)=bj(I3A(F3)) and ψ(v2j)=τx2j−1−1(I3A(F3)), j=1,2,3.
By τw we denote the inner automorphisms of F3 with τw(x)=w−1xw. Since J⊆Kerψ and gr(M3)≅L/J as Lie algebras, there exists a Lie algebra epimorphism ψ from gr(M3) onto L1(M3) satisfying the conditions ψ(biM3′)=bi(I3A(F3)) and
ψ(τxi−1M3′)=τxi−1(I3A(F3)), i=1,2,3. Since, in addition, M3∩Ic+1A(F3)=γc(M3) for all c, we obtain from Lemma 3 the required result. □
Remark 3
For c=1,2,3, we show that γc(H)=H∩Ic+1A(F3). That is, L1c(H)≅γc(H)/γc+1(H) for c=1,2,3. For c≥4, the method used becomes very cumbersome. For simplicity, let x=χ21, y=χ12 and z=χ23. For c=1, our claim is trivially true. Let c=2. Since γ2(H)/γ3(H) is a free abelian group of rank 3, we have each element h∈γ2(H)∖γ3(H) is uniquely written as
[TABLE]
where v∈γ3(H) and a1,a2,a3 are non-negative integers. We claim that h∈/I4A(F3). By a direct calculation, (z,x)a1(xi)=xi, i=2, and (z,x)a1(x2)=x2(x3,x1,x2)−a1u, where u∈γ4(F3). Working modulo γ3(H), we have
[TABLE]
We point out that, for ϕ∈IA(F3) and g∈F3, we have (τg,ϕ)=τg−1ϕ−1(g). Since (z,τx2−1)=τ(x3−1,x2−1) and (τx2−1,x)=τ(x2−1,x1−1), we get
[TABLE]
Since both (z,χ32)a2 and (χ32,x)a3 fix x1, we obtain
[TABLE]
where v1,v2∈γ4(F3). Thus, (z,y),(y,x)∈I3A(F3)∖I4A(F3). Furthermore, it is easily shown that, for a1+a2+a3=0, h∈γ2(H) and h∈/I4A(F3). Therefore, γ2(H)=H∩I3A(F3).
For c=3, we apply similar arguments as before. We point out that γ3(H)/γ4(H) is a free abelian group of rank 8. Each element g∈γ3(H)∖γ4(H) is uniquely written as g=g1g2g3u, where g1=(z,x,x)b1(z,x,z)b2, g2=(y,x,x)b3(y,x,z)b4(z,x,y)b5(z,y,z)b6, g3=(y,x,y)b7(z,y,y)b8, u∈γ4(H) and b1,…,b8 are non-negative integers. Since g1(xi)=xi for i=1,3, we have, by direct calculations,
[TABLE]
where v12∈γ5(F3). By the equation (*), and working modulo γ4(H), we have
[TABLE]
Since both (χ32,x,x)b3 and (χ32,x,z)b4 fix x1, we get
[TABLE]
where v3,v4∈γ5(F3). Since (z,x)(x2)=x2(x3,x1,x2)−1u, with u∈γ4(F3), and by the equation (*), we get
[TABLE]
and
[TABLE]
Since both (z,x,χ32)b5 and (z,y,χ32)b6 fix x1, we obtain
[TABLE]
where v5,v6∈γ5(F3). Therefore,
[TABLE]
where v∈γ5(F3).
Next, we point out that (χ32,χ21)(xi)=xi for i=1,2 and (χ32,χ21)(x3)=x3(x2,x1,x3)w, where w∈γ4(F3). Since (χ32,χ21,τx2−1)=IdF3, we have
[TABLE]
where ψ∈γ4(M3). Hence,
[TABLE]
and so,
[TABLE]
Since (τ(x2−1,x1−1),χ32−1)b7=IdF3, we get
[TABLE]
Therefore,
[TABLE]
where v7∈γ5(F3). By direct calculations, (z,χ32)(x1)=x1, (z,χ32)(x2)=x2(x3,x2,x2)−1u2 and (z,χ32)(x3)=x3(x3,x2,x3)u3, where u2,u3∈γ4(F3). It is easily verified that
[TABLE]
where ψ1∈γ4(M3). By the equation (*),
[TABLE]
Since
[TABLE]
and
[TABLE]
we have
[TABLE]
Hence,
[TABLE]
Since (z,χ32,χ32) fixes x1, we have
[TABLE]
where v8∈γ5(F3). Therefore,
[TABLE]
where v7,8∈γ5(F3). Since the ”basic” group commutators of length 4 consists of a basis for γ4(F3)/γ5(F3), we obtain from the equations (1), (2) and (3) that g∈γ4(H) and g∈/I6A(F3), that is, γ4(H)=H∩I5A(F3).
Bibliography16
The reference list from the paper itself. Each links out to its DOI / PubMed record.
1[1]
2[2] S. Andreadakis, On the automorphisms of free groups and free nilpotent groups. Proc. London Math. Soc. 15 (1965), 239–268.
3[3] N. Bourbaki, Lie Groups and Lie Algebras Part I, Hermann, Paris, 1987 (Chapters 1-3).
4[4] R.M. Bryant, L.G. Kovács, Ralph Stöhr, Lie powers of modules for groups of prime order, Proc. London Math. Soc. 84 (2002), 343–374.
5[5] R.M. Bryant, L.G. Kovács, Ralph Stöhr, Invariant bases for free Lie rings, Q. J. Math. 53 (2002), no. 1, 1–17.
6[6] N. Gupta, Free Group Rings, American Mathematical Society, 1987.
7[7] M. Lazard, Sur les groupes nilpotents et les anneaux de Lie, Ann. Sci. Ecole Norm. Sup. (3) 71 (1954) 101-190.
8[8] W. Magnus, Über n 𝑛 n -dimensinale Gittertransformationen, Acta Math. , 64 (1935), 353-367.