Quotient groups of IA-automorphisms of a free group of rank 3

TL;DR

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

Quotients of the McCool group are isomorphic to two copies of free group quotients.

Provides a necessary and sufficient condition for embedding the graded Lie algebra into the Johnson Lie algebra.

Enhances understanding of the algebraic structure of IA-automorphisms of free groups.

Abstract

We prove that, for any positive integer $c$, the quotient group $\gamma_{c}(M_{3})/\gamma_{c+1}(M_{3})$ of the lower central series of the McCool group $M_{3}$ is isomorphic to two copies of the quotient group $\gamma_{c}(F_{3})/\gamma_{c+1}(F_{3})$ of the lower central series of a free group $F_{3}$ of rank $3$ as $\mathbb{Z}$-modules. Furthermore, we give a necessary and sufficient condition whether the associated graded Lie algebra ${\rm gr}(M_{3})$ of $M_3$ is naturally embedded into the Johnson Lie algebra ${\cal L}({\rm IA}(F_{3}))$ of the IA-automorphisms of $F_{3}$.

Equations

$$\mathcal{L}(\mathrm{IA}(F_n))=\underset{r\ge 2}{⨁}{\mathcal{L}}^r(\mathrm{I}\mathrm{A}(F_n)).$$

$$[\varphi {\mathrm{I}}_{j+1}\mathrm{A}(F_n),\psi {\mathrm{I}}_{\kappa +1}\mathrm{A}(F_n)]=({\varphi }^{-1}{\psi }^{-1}\varphi \psi ){\mathrm{I}}_{j+\kappa }\mathrm{A}(F_n),$$

$${\gamma }_c(M_3)/{\gamma }_{c+1}(M_3)\cong {\gamma }_c(F_3)/{\gamma }_{c+1}(F_3)\oplus {\gamma }_c(F_3)/{\gamma }_{c+1}(F_3)$$

$$\frac{1}{c}\sum _{d|c}\mu (d)n^{c/d}\le \mathrm{rank}({\mathcal{L}}^{c+1}(\mathrm{IA}(F_n)))$$

$$\frac{1}{c}\sum _{d|c}\mu (d)2^{c/d}+\frac{1}{c}\sum _{d|c}\mu (d)3^{c/d}\le \mathrm{rank}({\mathcal{L}}^{c+1}(\mathrm{IA}(F_3)))$$

$$\begin{array}[]{ll}U={\mathbb{Z}}{\mbox{-}}{\rm span}\{x_{1},x_{3},x_{5}\}~{}~{}{\rm and{}}&V={\mathbb{Z}}\mbox{-}{\rm span}\{v_{2},v_{4},v_{6}\}.\end{array}$$

$$\begin{array}[]{ll}{\cal J}=&\{[v_{2},x_{1}],[v_{4},x_{3}],[v_{6},x_{5}],[v_{4},v_{2}]-[v_{4},x_{1}],[v_{2},v_{4}]-[v_{2},x_{3}],[v_{4},v_{6}]-[v_{4},x_{5}],\\ &\\ &[v_{6},x_{1}],[v_{6},x_{3}],[v_{2},x_{5}]\}.\end{array}$$

$$\begin{array}[]{lll}L&=&L(U\oplus V)\\ &=&L(U)\oplus L(V\wr U)\\ &=&L(U)\oplus L(V)\oplus L(W),\end{array}$$

$$W_m=\underset{a+b=m-2}{⨁}[V,U,{}_{a}U,{}_{b}V].$$

$$[\mathcal{V},\mathcal{U}]=\{[v_2,x_1],[v_4,x_3],[v_6,x_5],[v_4,x_1],[v_2,x_3],[v_4,x_5],[v_6,x_1],[v_6,x_3],[v_2,x_5]\}$$

$${\mathcal{X}}_{V,U}=\mathcal{V}\cup (\bigcup _{a\ge 1}[\mathcal{V},{}_a\mathcal{U}]),$$

$${\psi }_2([v_4,x_1])=[v_4,v_2]-[v_4,x_1],{\psi }_2([v_2,x_3])=[v_2,v_4]-[v_2,x_3],{\psi }_2([v_4,x_5])=[v_4,v_6]-[v_4,x_5]$$

$$\Psi :{\mathcal{X}}_{V,U}\to L(V\wr U)$$

$$\Psi (L(W))=L(\Psi (W)),$$

$$\begin{array}[]{lll}L(V\wr U)&=&\Psi(L(V\wr U))\\ ({\rm By~{}Corollary}~{}\ref{c1})&=&\Psi(L(V)\oplus L(W))\\ (\Psi~{}{\rm automorphism})&=&\Psi(L(V))\oplus\Psi(L(W))\\ &=&L(V)\oplus L(\Psi(W))\end{array}$$

$$\begin{array}[]{lll}L&=&L(U)\oplus L(V)\oplus L(\Psi(W)).\end{array}$$

$$\mathcal{C}=\Psi ([\mathcal{V},\mathcal{U}])\cup (\bigcup _{\genfrac{}{}{0ex}{}{a+b\ge 1}{a,b\ge 0}}[\Psi ([\mathcal{V},\mathcal{U}]),{}_a\mathcal{U},{}_b\mathcal{V}]).$$

$$\begin{array}[]{lll}[w,x_{1}]&=&[\psi_{2}([v_{4},x_{1}]),x_{3},v_{2},v_{4},x_{1}]\\ &=&[\psi_{2}([v_{4},x_{1}]),x_{3},v_{2},x_{1},v_{4}]+[\psi_{2}([v_{4},x_{1}]),x_{3},v_{2},[v_{4},x_{1}]]\\ &=&[\psi_{2}([v_{4},x_{1}]),x_{3},v_{2},x_{1},v_{4}]+\\ &&[\psi_{2}([v_{4},x_{1}]),x_{3},v_{2},[v_{4},v_{2}]]-[\psi_{2}([v_{4},x_{1}]),x_{3},v_{2},\psi_{2}([v_{4},x_{1}])]\\ &=&[\psi_{2}([v_{4},x_{1}]),x_{3},x_{1},v_{2},v_{4}]+[\psi_{2}([v_{4},x_{1}]),x_{3},[v_{2},x_{1}],v_{4}]+\\ &&[\psi_{2}([v_{4},x_{1}]),x_{3},v_{2},[v_{4},v_{2}]]-[\psi_{2}([v_{4},x_{1}]),x_{3},v_{2},\psi_{2}([v_{4},x_{1}])]\in L(\Psi(W)).\end{array}$$

$${\gamma }_c(M_3)/{\gamma }_{c+1}(M_3)\cong {\gamma }_c(F_3)/{\gamma }_{c+1}(F_3)\oplus {\gamma }_c(F_3)/{\gamma }_{c+1}(F_3)$$

$$\begin{array}[]{lll}r_{1}=(x_{1},x_{2}),&r_{2}=(x_{3},x_{4}),&r_{3}=(x_{5},x_{6})\\ r_{4}=(x_{1}x_{2},x_{5}),&r_{5}=(x_{3}x_{4},x_{6}),&r_{6}=(x_{1}x_{2},x_{4})\\ r_{7}=(x_{3}x_{4},x_{2}),&r_{8}=(x_{5}x_{6},x_{3}),&r_{9}=(x_{5}x_{6},x_{1}),\end{array}$$

$$\mathcal{I}(N)=\underset{d\ge 2}{⨁}{N}_d{\gamma }_{d+1}(F)/{\gamma }_{d+1}(F)=\underset{d\ge 2}{⨁}{\mathcal{I}}_d(N).$$

$${\gamma }_c(M_3)/{\gamma }_{c+1}(M_3)\cong {\gamma }_c(F){\gamma }_{c+1}(F)N/{\gamma }_{c+1}(F)N\cong {\gamma }_c(F)/({\gamma }_c(F)\cap {\gamma }_{c+1}(F)N).$$

$${\gamma }_c(F)/({\gamma }_c(F)\cap {\gamma }_{c+1}(F)N)={\gamma }_c(F)/{\gamma }_{c+1}(F)N_c.$$

$${\gamma }_c(F)/{\gamma }_{c+1}(F)N_c\cong ({\gamma }_c(F)/{\gamma }_{c+1}(F))/{\mathcal{I}}_c(N)\cong L^c/J^c.$$

$${\gamma }_c(M_3)/{\gamma }_{c+1}(M_3)\cong L^c/J^c\cong (L^c{)}^{\ast },$$

$${\gamma }_c(M_3)/{\gamma }_{c+1}(M_3)\cong {\gamma }_c(F_3)/{\gamma }_{c+1}(F_3)\oplus {\gamma }_c(F_3)/{\gamma }_{c+1}(F_3)$$

$$\begin{array}[]{lll}{\cal L}^{c}_{1}(E_{n})&\cong&\gamma_{c}(E_{n})/(\gamma_{c}(E_{n})\cap({\rm I}_{c+2}{\rm A}(F_{n})))\\ &=&\gamma_{c}(E_{n})/(E_{n}\cap{\rm I}_{c+2}{\rm A}(F_{n})))\\ &=&{\rm gr}_{c}(E_{n})\end{array}$$

$$\frac{1}{c}\sum _{d|c}\mu (d)n^{c/d}\le \mathrm{rank}({\mathcal{L}}^{c+1}(\mathrm{IA}(F_n)))$$

$$\mathrm{rank}({\mathcal{L}}_1^c(H))+\frac{1}{c}\sum _{d|c}\mu (d)3^{c/d}\le \mathrm{rank}({\mathcal{L}}^{c+1}(\mathrm{IA}(F_3))).$$

Full text

1 Introduction and Notation

Let $G$ be a group. For a positive integer $c$, let $\gamma_{c}(G)$ be the $c$-th term of the lower central series of $G$. We point out that $\gamma_{2}(G)=G^{\prime}$; that is, the derived group of $G$. We write ${\rm IA}(G)$ for the kernel of the natural group homomorphism from ${\rm Aut}(G)$ into ${\rm Aut}(G/G^{\prime})$ and we call its elements IA-automorphisms of $G$. For a positive integer $c\geq 2$, the natural group epimorphism from $G$ onto $G/\gamma_{c}(G)$ induces a group homomorphism $\pi_{c}$ from ${{\rm Aut}}(G)$ of $G$ into ${{\rm Aut}}(G/\gamma_{c}(G))$ of $G/\gamma_{c}(G)$. Write ${\rm I}_{c}{\rm A}(G)={\rm Ker}\pi_{c}$. Note that ${\rm I}_{2}{\rm A}(G)={\rm IA}(G)$. It is proved by Andreadakis [2, Theorem 1.2], that if $G$ is residually nilpotent (that is, $\bigcap_{c\geq 1}\gamma_{c}(G)=\{1\}$), then $\bigcap_{c\geq 2}{\rm I}_{c}{\rm A}(G)=\{1\}$.

Throughout this paper, by “Lie algebra”, we mean Lie algebra over the ring of integers $\mathbb{Z}$. Let $G$ be a group. Write ${\rm gr}_{c}(G)=\gamma_{c}(G)/\gamma_{c+1}(G)$ for $c\geq 1$ and denote by $(a,b)$ the commutator $(a,b)=a^{-1}b^{-1}ab$ with $a,b\in G$. The (restricted) direct sum of the quotients ${\rm gr}_{c}(G)$ is the associated graded Lie algebra of $G$, ${\rm gr}(G)=\bigoplus_{c\geq 1}{\rm gr}_{c}(G)$. The Lie bracket multiplication in ${\rm gr}(G)$ is $[a\gamma_{c+1}(G),b\gamma_{d+1}(G)]=(a,b)\gamma_{c+d+1}(G)$, where $a\gamma_{c+1}(G)$ and $b\gamma_{d+1}(G)$ are the images of the elements $a\in\gamma_{c}(G)$ and $b\in\gamma_{d}(G)$ in the quotient groups ${\rm gr}_{c}(G)$ and ${\rm gr}_{d}(G)$, respectively, and $(a,b)\gamma_{c+d+1}(G)$ is the image of the group commutator $(a,b)$ in the quotient group ${\rm gr}_{c+d}(G)$. Multiplication is then extended to ${\rm gr}(G)$ by linearity.

For a positive integer $n$, with $n\geq 2$, we write $F_{n}$ for a free group of rank $n$ with a free generating set $\{x_{1},\ldots,x_{n}\}$. For $c\geq 2$, we write $F_{n,c-1}=F_{n}/\gamma_{c}(F_{n})$. Thus, $F_{n,c-1}$ is the free nilpotent group of rank $n$ and class $c-1$. The natural group epimorphism from $F_{n}$ onto $F_{n,c-1}$ induces a group homomorphism $\pi_{n,c-1}:{\rm Aut}(F_{n})\longrightarrow{\rm Aut}(F_{n,c-1})$. We write ${\rm I}_{c}{\rm A}(F_{n})={\rm Ker}\pi_{n,c-1}$. It is well known that, for $t,s\geq 2$, $({\rm I}_{t}{\rm A}(F_{n}),{\rm I}_{s}{\rm A}(F_{n}))\subseteq{\rm I}_{t+s-1}{\rm A}(F_{n})$. Since $F_{n}$ is residually nilpotent, we have $\bigcap_{c\geq 2}{\rm I}_{c}{\rm A}(F_{n})=\{1\}$. Since $F_{n,c}/{\rm gr}_{c}(F_{n})\cong F_{n,c-1}$ and ${\rm gr}_{c}(F_{n})$ is a fully invariant subgroup of $F_{n,c}$, the natural group epimorphism from $F_{n,c}$ onto $F_{n,c-1}$ induces a group homomorphism $\psi_{c,c-1}:{\rm Aut}(F_{n,c})\longrightarrow{\rm Aut}(F_{n,c-1})$. It is well known that $\psi_{c,c-1}$ is onto. For $c\geq 2$, we define $A^{*}_{c}(F_{n})={\rm Im}\pi_{n,c}\cap{\rm Ker}\psi_{c,c-1}$. For $t\in\{2,\ldots,c\}$, the natural group epimorphism from $F_{n,c}$ onto $F_{n,c}/\gamma_{t}(F_{n,c})$ induces a group homomorphism $\sigma_{c,t}:{\rm Aut}(F_{n,c})\rightarrow{\rm Aut}(F_{n,c}/\gamma_{t}(F_{n,c}))$. Write ${\rm I}_{t}{\rm A}(F_{n,c})={\rm Ker}\sigma_{c,t}$, and, for $t=2$, ${\rm IA}(F_{n,c})={\rm I}_{2}{\rm A}(F_{n,c})$. We note that, for $c\geq 2$, $F_{n,c}/\gamma_{c}(F_{n,c})\cong F_{n,c-1}$. Thus, for $c\geq 2$, $A^{*}_{c}(F_{n})={\rm Im}\pi_{n,c}\cap{\rm I}_{c}{\rm A}(F_{n,c})$. It is easily proved that $A^{*}_{c}(F_{n})\cong{\rm I}_{c}{\rm A}(F_{n})/{\rm I}_{c+1}{\rm A}(F_{n})$ as free abelian groups (see, also, [2, Section 4, p. 246]). Furthermore, for $n,c\geq 2$, ${\rm rank}(A^{*}_{c}(F_{n}))\leq n~{}{\rm rank}({\rm gr}_{c}(F_{n}))=\frac{n}{c}\sum_{d|c}\mu(d)n^{c/d}$, where $\mu$ is the Möbius function. We point out that ${\rm rank}({\rm gr}_{c}(F_{n}))=\frac{1}{c}\sum_{d|c}\mu(d)n^{c/d}$ for all $n,c\geq 2$ (see, for example, [9]).

From now on, for $r\geq 2$, we write ${\cal L}^{r}({\rm I}{\rm A}(F_{n}))={\rm I}_{r}{\rm A}(F_{n})/{\rm I}_{r+1}{\rm A}(F_{n})$. Form the (restricted) direct sum of the free abelian groups ${\cal L}^{r}({\rm I}{\rm A}(F_{n}))$, and denoted by

[TABLE]

It has the structure of a graded Lie algebra with ${\cal L}^{r}({\rm IA}(F_{n}))$ as component of degree $r-1$ in the grading and Lie multiplication given by

[TABLE]

for all $\phi\in{\rm I}_{j}{\rm A}(F_{n})$, $\psi\in{\rm I}_{\kappa}{\rm A}(F_{n})$ $(j,\kappa\geq 2)$. Multiplication is then extended to ${\cal L}({\rm IA}(F_{n}))$ by linearity. The above Lie algebra is usually called the Johnson Lie algebra of ${\rm IA}(F_{n})$. We point out that, for a positive integer $c$, $\gamma_{c}({\rm IA}(F_{n}))\subseteq{\rm I}_{c+1}{\rm A}(F_{n})$.

Let $H$ be a finitely generated subgroup of ${\rm IA}(F_{n})$ with $H/H^{\prime}$ torsion-free. For a positive integer $q$, let ${\cal L}^{q}_{1}(H)=\gamma_{q}(H)({\rm I}_{q+2}{\rm A}(F_{n}))/{\rm I}_{q+2}{\rm A}(F_{n})$. Form the (restricted) direct sum of abelian groups ${\cal L}_{1}(H)=\bigoplus_{q\geq 1}{\cal L}^{q}_{1}(H)$. It is easily verified that ${\cal L}_{1}(H)$ is a Lie subalgebra of ${\cal L}({\rm IA}(F_{n}))$. Furthermore, if $\{y_{1}H^{\prime},\ldots,y_{m}H^{\prime}\}$ is a $\mathbb{Z}$-basis for $H/H^{\prime}$, then ${\cal L}_{1}(H)$ is generated as Lie algebra by the set $\{y_{1}({\rm I}_{3}{\rm A}(F_{n})),\ldots,y_{m}({\rm I}_{3}{\rm A}(F_{n}))\}$. By a natural embedding of ${\rm gr}(H)$ into ${\cal L}({\rm IA}(F_{n}))$, we mean that there exists a Lie algebra isomorphism $\phi$ from ${\rm gr}(H)$ onto ${\cal L}_{1}(H)$ satisfying the conditions $\phi(y_{i}H^{\prime})=y_{i}({\rm I}_{3}{\rm A}(F_{n}))$, $i=1,\ldots,m$. In this case, we also say that ${\rm gr}(H)$ is naturally isomorphic to ${\cal L}_{1}(H)$.

For $n\geq 2$, it was shown by Magnus [8], using work of Nielsen [13], that IA$(F_{n})$ has a finite generating set $\{\chi_{ij},\chi_{ijk}:1\leq i,j,k\leq n;i\neq j,k;j<k\}$, where $\chi_{ij}$ maps $x_{i}\mapsto x_{i}(x_{i},x_{j})$ and $\chi_{ijk}$ maps $x_{i}\mapsto x_{i}(x^{-1}_{j},x^{-1}_{k})$, with both $\chi_{ij}$ and $\chi_{ijk}$ fixing the remaining basis elements. Let $M_{n}$ be the subgroup of IA$(F_{n})$ generated by the subset $S=\{\chi_{ij}:1\leq i,j\leq n;\ i\neq j\}$. Then, $M_{n}$ 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: $(\chi_{ij},\chi_{kj})=(\chi_{ij},\chi_{kq})=(\chi_{ij}\chi_{kj},\chi_{ik})=1$. It has been proved in [10] that $M_{n}$ has a presentation $\langle S\mid Z\rangle$, where $Z$ is the set of all possible relations of the above forms. Since $\gamma_{c}(M_{n})\subseteq\gamma_{c}({\rm IA}(F_{n}))\subseteq{\rm I}_{c+1}{\rm A}(F_{n})$ for all $c\geq 1$, and since $F_{n}$ is residually nilpotent, we have $\bigcap_{c\geq 1}\gamma_{c}(M_{n})=\{1\}$ and so, $M_{n}$ 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 $M_{3}$ generated by $\chi_{21},\chi_{12},\chi_{23}$. Then, ${\cal L}_{1}(M_{3})$ is additively equal to the direct sum of the Lie subalgebras ${\cal L}_{1}(H)$ and ${\cal L}_{1}({\rm Inn}(F_{3}))$, where ${\rm Inn}(F_{3})$ denotes the group of inner automorphisms of $F_{3}$. Furthermore, ${\rm gr}(M_{3})$ is naturally isomorphic to ${\cal L}_{1}(M_{3})$ as Lie algebras if and only if ${\rm gr}(H)$ is naturally isomorphic to ${\cal L}_{1}(H)$ as Lie algebras.

In [11, Theorem 1], it is shown that $M_{3}$ 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 $\gamma_{c}(M_{3})/\gamma_{c+1}(M_{3})$ as in Theorem 1 (1). By a result of Sjogren [15] (see, also, [6, Corollary 1.9]), $M_{3}$ satisfies the Subgroup Dimension Problem. That is, each $\gamma_{c}(M_{3})$ is equal to the $c$-th dimension subgroup of $M_{3}$. Furthermore, the new approach helps us to give a necessary and sufficient condition for a natural embedding of ${\rm gr}(M_{3})$ into ${\cal L}({\rm IA}(F_{3}))$. For $n\geq 2$, let ${\rm Inn}(F_{n})$ denote the subgroup of ${\rm IA}(F_{n})$ consisting of all inner automorphisms of $F_{n}$. In Section 3, by using an observation of Andreadakis [2, Section 6, p. 249], we show that ${\rm gr}({\rm Inn}(F_{n}))$ is naturally embedded into ${\cal L}({\rm IA}(F_{n}))$. Hence, $\gamma_{c}({\rm Inn}(F_{n}))/\gamma_{c+1}({\rm Inn}(F_{n}))$ is isomorphic to a subgroup of ${\cal L}^{c+1}({\rm IA}(F_{n}))$ for all $c\geq 1$. Since ${\rm Inn}(F_{n})\cong F_{n}$, we obtain

[TABLE]

for all $n,c$, with $n\geq 2$. For $c=1$ and $n\geq 2$, we have ${\rm rank}({\cal L}^{2}({\rm IA}(F_{n})))=\frac{n^{2}(n-1)}{2}$ (see, [2, Theorem 5.1]). For $n=2$ we have ${\rm IA}(F_{2})={\rm Inn}(F_{2})$, by a result of Nielsen [12] and by a result of Andreadakis [2, Theorem 6.1], we have ${\rm rank}({\cal L}^{c+1}({\rm IA}(F_{2})))=\frac{1}{c}\sum_{d|c}\mu(d)2^{c/d}$. For $n=3$, by Theorem 1(2), we may give a lower bound of ${\rm rank}({\cal L}^{c+1}({\rm IA}(F_{3})))$ in terms of the rank of ${\cal L}^{c}_{1}(H)$ for all $c$. In fact, we observe that

[TABLE]

(see, Remark 2 below). For $n\geq 4$ and $c\geq 2$, Satoh [16, Corollary 3.3] provides a lower bound for ${\rm rank}({\cal L}^{c+1}({\rm IA}(F_{n}))).$

2 The associated Lie algebra of $M_{3}$

2.1 Lazard elimination

For a free $\mathbb{Z}$-module $A$, let $L(A)$ be the free Lie algebra on $A$, that is, the free Lie algebra on $\cal A$, where $\cal A$ is an arbitrary $\mathbb{Z}$-basis of $A$. Thus, we may write $L(A)=L({\cal A})$. For a positive integer $c$, let $L^{c}(A)$ denote the $c$th homogeneous component of $L(A)$. It is well-known that $L(A)=\bigoplus_{c\geq 1}L^{c}(A)$. For $\mathbb{Z}$-submodules $A$ and $B$ of any Lie algebra, let $[A,B]$ be the $\mathbb{Z}$-submodule spanned by $[a,b]$ where $a\in A$ and $b\in B$. Furthermore, $B\wr A$ denotes the $\mathbb{Z}$-submodule defined by $B\wr A=B+[B,A]+[B,A,A]+\cdots$.

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 $\mathbb{Z}$-modules, and consider the free Lie algebra $L(U\oplus V)$. Then, $U$ and $V\wr U$ freely generate Lie subalgebras $L(U)$ and $L(V\wr U)$, and there is a $\mathbb{Z}$-module decomposition $L(U\oplus V)=L(U)\oplus L(V\wr U)$. Furthermore, $V\wr U=V\oplus[V,U]\oplus[V,U,U]\oplus\cdots$ and, for each $n\geq 0$, there is a $\mathbb{Z}$-module isomorphism $[V,\underbrace{U,\ldots,U}_{n}]\cong V\otimes\underbrace{U\otimes\cdots\otimes U}_{n}$ under which $[v,u_{1},\ldots,u_{n}]$ corresponds to $v\otimes u_{1}\otimes\cdots\otimes u_{n}$ for all $v\in V$ and all $u_{1},\ldots u_{n}\in U$.

As a consequence of Lemma 1, we have the following result.

Corollary 1

For free $\mathbb{Z}$-modules $U$ and $V$, we write $L(U\oplus V)$ for the free Lie algebra on $U\oplus V$. Then, there is a $\mathbb{Z}$-module decomposition $L(U\oplus V)=L(U)\oplus L(V)\oplus L(W)$, where $W=W_{2}\oplus W_{3}\oplus\cdots$ such that, for all $m\geq 2$, $W_{m}$ is the direct sum of submodules $[V,U,\underbrace{U,\ldots,U}_{a},\underbrace{V,\ldots,V}_{b}]$ with $a+b=m-2$ and $a,b\geq 0$. Each $[V,U,\underbrace{U,\ldots,U}_{a},\underbrace{V,\ldots,V}_{b}]$ is isomorphic to $V\otimes U\otimes\underbrace{U\otimes\cdots\otimes U}_{a}\otimes\underbrace{V\otimes\cdots\otimes V}_{b}$ as $\mathbb{Z}$-module. Furthermore, $L(W)$ is the ideal of $L(U\oplus V)$ generated by the module $[V,U]$.

2.2 A decomposition of a free Lie algebra

Let $X$ be the free $\mathbb{Z}$-module with a $\mathbb{Z}$-basis $\{x_{1},\ldots,x_{6}\}$ and $L=L(X)$ the free Lie algebra on $X$. For $i=1,2,3$, let $v_{2i}=x_{2i-1}+x_{2i}$. Furthermore, we write

[TABLE]

Since $X=U\oplus V$, we have $L=L(U\oplus V)$ and so, $L$ is free on ${\cal X}=\{x_{1},x_{3},x_{5},v_{2},v_{4},v_{6}\}$. Let ${\cal J}$ be the subset of $L$,

[TABLE]

The aim of this section is to show the following result.

Proposition 1

With the above notation, let $L=L(U\oplus V)$ be the free Lie algebra on $U\oplus V$. Let $J$ be the ideal of $L$ generated by the set ${\cal J}$. Then, $L=L(U)\oplus L(V)\oplus J$. Moreover, $J$ is a free Lie algebra.

For non negative integers $a$ and $b$, we write $[V,U,~{}_{a}U,~{}_{b}V]$ for $[V,U,\underbrace{U,\ldots,U}_{a},\underbrace{V,\ldots,V}_{b}]$. By Lemma 1 and Corollary 1, we have

[TABLE]

where $W=W_{2}\oplus W_{3}\oplus\cdots$ such that, for all $m\geq 2$,

[TABLE]

Furthermore, $L(V\wr U)$ and $L(W)$ are the ideals in $L$ generated by the modules $V\wr U$ and $[V,U]$, respectively. In particular, $L(W)$ is the ideal in $L$ generated by the natural $\mathbb{Z}$-basis

[TABLE]

of $[V,U]$. Let ${\cal X}_{V,U}$ be the natural $\mathbb{Z}$-basis of $V\wr U$. That is,

[TABLE]

where $[{\cal V},~{}_{a}{\cal U}]$ is the natural $\mathbb{Z}$-basis of the module $[V,~{}_{a}U]$. Let $\psi_{2}$ be the $\mathbb{Z}$-linear mapping from $[V,U]$ into $L(V\wr U)$ with

[TABLE]

and $\psi_{2}$ fixes the remaining elements of $[{\cal V},{\cal U}]$. It is clear enough that $\psi_{2}$ is a $\mathbb{Z}$-linear monomorphism of $[V,U]$ into $L(V\wr U)$. For $a\geq 3$, let $\psi_{a}$ be the mapping from $[V,U,~{}_{(a-2)}U]$ into $L(V\wr U)$ satisfying the conditions $\psi_{a}([v,u,u_{3},\ldots,u_{a}])=[\psi_{2}([v,u]),u_{3},\ldots,u_{a}]$ for all $v\in{\cal V}$ and $u,u_{3},\ldots,u_{a}\in{\cal U}$. We define a map

[TABLE]

by $\Psi(v)=v$ for all $v\in{\cal V}$ and, for $a\geq 2$, $\Psi(v)=\psi_{a}(v)$ for all $v\in[{\cal V},{\cal U},~{}_{(a-2)}{\cal U}]$. Since $L(V\wr U)$ is free on ${\cal X}_{V,U}$, we obtain $\Psi$ is a Lie algebra homomorphism. By applying Lemma 2.1 in [5], we see that $\Psi$ is a Lie algebra automorphism of $L(V\wr U)$. Since $L(W)$ is a free Lie subalgebra of $L(V\wr U)$ and $\Psi$ is an automorphism, we have $\Psi(L(W))$ is a free Lie subalgebra of $L(V\wr U)$. In fact,

[TABLE]

that is, $\Psi(L(W))$ is a free Lie algebra on $\Psi(W)$.

Lemma 2

With the above notation, $L(\Psi(W))$ is an ideal in $L$.

*Proof. *Since $\Psi$ is an automorphism of $L(V\wr U)$, we obtain $\Psi(L(W))=L(\Psi(W))$ is an ideal in $L(V\wr U)$. We point out that

[TABLE]

and so,

[TABLE]

To show that $L(\Psi(W))$ is an ideal in $L$, it is enough to show that $[w,u]\in L(\Psi(W))$ for all $w\in L(\Psi(W))$ and $u\in L$. Since $L(\Psi(W))$ is an ideal in $L(V\wr U)$ and Lemma 1, it is enough to show that $[w,u]\in L(\Psi(W))$ for all $w\in L(\Psi(W))$ and $u\in L(U)$. Furthermore, we may show that $[w,x_{i_{1}},\ldots,x_{i_{k}}]\in L(\Psi(W))$ for all $w\in L(\Psi(W))$ and $x_{i_{1}},\ldots,x_{i_{k}}\in\{x_{1},x_{3},x_{5}\}$. Write

[TABLE]

Since $\cal C$ is a $\mathbb{Z}$-basis for $\Psi(W)$, we have $L(\Psi(W))=L({\cal C})$. Thus, we need to show that $[w,x_{i_{1}},\ldots,x_{i_{k}}]\in L(\Psi(W))$ for all $w\in{\cal C}$ and $x_{i_{1}},\ldots,x_{i_{k}}\in\{x_{1},x_{3},x_{5}\}$. Since $\cal C$ is a free generating set of $L(\Psi(W))$, the equation (1) and the linearity of the Lie bracket, we may assume that $w\in[\Psi([{\cal V},{\cal U}]),~{}_{a}{\cal U},~{}_{b}{\cal V}]$ with $a+b\geq 1$. Clearly, we may assume that $b\geq 1$. Since $L(\Psi(W))$ is ideal in $L(V\wr U)$ and, by the Jacobi identity, we may further assume that $w$ has a form $[v,y_{j_{1}},\ldots,y_{j_{a}},~{}_{\mu}v_{2},~{}_{\nu}v_{4},~{}_{\rho}v_{6}]$ with $y_{j_{1}},\ldots,y_{j_{a}}\in{\cal U}$, $\mu,\nu,\rho\geq 0$ and $\mu+\nu+\rho\geq 1$. By using the Jacobi identity in the expression $[w,x_{i_{1}},\ldots,x_{i_{k}}]$, and replacing $[v_{4},x_{1}],[v_{2},x_{3}]$ and $[v_{4},x_{5}]$ by $[v_{4},v_{2}]-\psi_{2}([v_{4},x_{1}])$, $[v_{2},v_{4}]-\psi_{2}([v_{2},x_{3}])$ and $[v_{4},v_{6}]-\psi_{2}([v_{4},x_{5}])$, respectively, as many times as it is needed and since $L(\Psi(W))$ is an ideal in $L(V\wr U)$, we may show that $[w,x_{i_{1}},\ldots,x_{i_{k}}]\in L(\Psi(W))$. Therefore, $L(\Psi(W))$ is an ideal in $L$. $\Box$

Example 1

In the present example, we explain the procedure described in the above proof. Let $w=[\psi_{2}([v_{4},x_{1}]),x_{3},v_{2},v_{4}]$. Then,

[TABLE]

Proof of Proposition 1. Since $\psi_{2}([V,U])\subseteq J$ and $J$ is ideal in $L$, we get $L(\Psi(W))\subseteq J$. Since ${\cal J}\subseteq L(\Psi(W))$, by Lemma 2, we obtain $J\subseteq L(\Psi(W))$. Therefore, $J=L(\Psi(W))$. That is, $J$ is a free Lie algebra. Furthermore, $L=L(U)\oplus L(V)\oplus J$. $\Box$

For $c\geq 2$, let $J^{c}=J\cap L^{c}$. Since $J$ is homogeneous, we get $J=\bigoplus_{c\geq 2}J^{c}$. From the above proof, we have the following result.

Corollary 2

With the above notation, let $\Psi$ be the Lie algebra automorphism of $L(V\wr U)$ defined naturally on $V\wr U$ by means of $\psi_{2}$. Then, $J=L(\Psi(W))$. Furthermore, for $c\geq 2$, $L^{c}=L^{c}(U)\oplus L^{c}(V)\oplus J^{c}$.

2.3 A description of ${\rm gr}(M_{3})$

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 $M_{3}$ be the McCool subgroup of the IA-automorphisms ${\rm IA}(F_{3})$ of $F_{3}$. Then, ${\rm gr}(M_{3})\cong L/J$ as Lie algebras. In particular, ${\rm gr}(M_{3})$ 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 $\{x_{1},\ldots,x_{6}\}$. It is well known that ${\rm gr}(F)$ is a free Lie algebra of rank $6$; freely generated by the set $\{x_{i}F^{\prime}:i=1,\ldots,6\}$. The free Lie algebras $L$ and ${\rm gr}(F)$ are isomorphic and from now on we identify the two Lie algebras. Furthermore, $L^{c}=\gamma_{c}(F)/\gamma_{c+1}(F)$ for all $c\geq 1$. Define

[TABLE]

and ${\cal R}=\{r_{1},\ldots,r_{9}\}$.

Let $N={\cal R}^{F}$ be the normal closure of $\cal R$ in $F$. For a positive integer $d$, let $N_{d}=N\cap\gamma_{d}(F)$. We point out that for $d\leq 2$, $N_{d}=N$. Further, for $d\geq 2$, $N_{d+1}=N_{d}\cap\gamma_{d+1}(F)$. Define ${\cal I}_{d}(N)=N_{d}\gamma_{d+1}(F)/\gamma_{d+1}(F)$. It is easily verified that ${\cal I}_{d}(N)\cong N_{d}/N_{d+1}$ as $\mathbb{Z}$-modules. The following result was shown in [11, Section 6].

Proposition 2

For a positive integer $c$, $N_{c+2}$ is generated by the set $\{(r^{\pm 1},g_{1},$ $\ldots,g_{s}):r\in{\cal R},s\geq c,g_{1},\ldots,g_{s}\in F\setminus\{1\}\}$. Furthermore, ${\cal I}_{c+2}(N)=J^{c+2}$ for all $c\geq 1$.

Since $F$ is residually nilpotent, we have $\bigcap_{d\geq 2}N_{d}=\{1\}$. Also $N\subseteq F^{\prime}$, we get ${\cal I}_{1}(N)=0$. Moreover, ${\cal I}_{d}(N)\cong N_{d}/N_{d+1}$ as $\mathbb{Z}$-modules for all $d\geq 2$, and by Proposition 2 we have, $N_{d}\neq N_{d+1}$ for all $d\geq 2$. Define

[TABLE]

Since $N$ is a normal subgroup of $F$, we have ${\cal I}(N)$ is an ideal of $L$ (see [7]).

Corollary 3

${\cal I}(N)=J$.

*Proof. *Since $J=\bigoplus_{d\geq 2}J^{d}$ and ${\cal I}_{2}(N)=J^{2}$, we have from Proposition 2 that ${\cal I}(N)=J$. $\Box$

Proof of Theorem 2. Since $M_{3}/M^{\prime}_{3}\cong F/NF^{\prime}=F/F^{\prime}$, we have ${\rm gr}(M_{3})$ is generated as a Lie algebra by the set $\{\alpha_{i}:i=1,\ldots,6\}$ with $\alpha_{i}=x_{i}M^{\prime}_{3}$. Since $L$ is a free Lie algebra of rank $6$ with a free generating set $\{x_{1},\ldots,x_{6}\}$, the map $\zeta$ from $L$ into ${\rm gr}(M_{3})$ satisfying the conditions $\zeta(x_{i})=\alpha_{i}$, $i=1,\ldots,6$, extends uniquely to a Lie algebra homomorphism. Since ${\rm gr}(M_{3})$ is generated as a Lie algebra by the set $\{\alpha_{i}:i=1,\ldots,6\}$, we have $\zeta$ is onto. Hence, $L/{\rm Ker}\zeta\cong{\rm gr}(M_{3})$ as Lie algebras. By definition, $J\subseteq{\rm Ker}\zeta$, and so $\zeta$ induces a Lie algebra epimorphism $\overline{\zeta}$ from $L/J$ onto ${\rm gr}(M_{3})$. In particular, $\overline{\zeta}(x_{i}+J)=\alpha_{i}$, $i=1,\ldots,6$. Note that $\overline{\zeta}$ induces $\overline{\zeta}_{c}$, say, a $\mathbb{Z}$-linear mapping from $(L^{c}+J)/J$ onto $\gamma_{c}(M_{3})/\gamma_{c+1}(M_{3})$. For $c\geq 2$,

[TABLE]

Since $\gamma_{c+1}(F)\subseteq\gamma_{c}(F)$, we have by the modular law,

[TABLE]

But, by Proposition 2, for $c\geq 3$,

[TABLE]

Since ${\cal I}_{2}(N)=J^{2}$, we obtain, for $c\geq 2$,

[TABLE]

Therefore, for $c\geq 1$,

[TABLE]

by Corollary 2, where $(L^{c})^{*}=L^{c}(U)\oplus L^{c}(V)$. Since both $L(U)$ and $L(V)$ are free Lie algebras of rank $3$, we have $L(U)\cong L(V)\cong{\rm gr}(F_{3})$ in a natural way and so, for $c\geq 1$, $L^{c}(U)\cong L^{c}(V)\cong\gamma_{c}(F_{3})/\gamma_{c+1}(F_{3})$ as free abelian groups. Hence, for $c\geq 1$,

[TABLE]

as free abelian groups and so, ${\rm rank}(\gamma_{c}(M_{3})/\gamma_{c+1}(M_{3}))={\rm rank}(L^{c})^{*}$. Since $J=\bigoplus_{c\geq 2}J^{c}$, we have $(L^{c}+J)/J\cong L^{c}/(L^{c}\cap J)=L^{c}/J^{c}\cong(L^{c})^{*}$ and so, we obtain ${\rm Ker}\overline{\zeta}_{c}$ is torsion-free. Since ${\rm rank}(\gamma_{c}(M_{3})/\gamma_{c+1}(M_{3}))={\rm rank}(L^{c})^{*}$, we have ${\rm Ker}\overline{\zeta}_{c}=\{1\}$ and so, $\overline{\zeta}_{c}$ is isomorphism. Since $\overline{\zeta}$ is epimorphism and each $\overline{\zeta}_{c}$ is isomorphism, we have $\overline{\zeta}$ is isomorphism. Hence, $L/J\cong{\rm gr}(M_{3})$ as Lie algebras. $\Box$

3 Embeddings

In this section, we shall give a criterion for the natural embedding of ${\rm gr}(M_{3})$ into ${\cal L}_{1}({\rm IA}(F_{3}))$. We shall prove the following result.

Lemma 3

Let $H$ be a finitely generated subgroup of ${\rm IA}(F_{n})$, $n\geq 2$, with $H/H^{\prime}$ torsion-free, and let $\{y_{1}H^{\prime},\ldots,y_{m}H^{\prime}\}$ be a $\mathbb{Z}$-basis for $H/H^{\prime}$. Then, ${\rm gr}(H)$ is naturally isomorphic to ${\cal L}_{1}(H)$ if and only if $\gamma_{c}(H)=H\cap({\rm I}_{c+1}{\rm A}(F_{n}))$ for all $c$.

*Proof. *We assume that $\gamma_{c}(H)=H\cap({\rm I}_{c+1}{\rm A}(F_{n}))$ for all $c$. For $c\geq 1$, let $\psi_{c}$ be the natural $\mathbb{Z}$-module epimorphism from ${\rm gr}_{c}(H)$ onto ${\cal L}_{1}^{c}(H)$. Since $\gamma_{c}(H)\cap({\rm I}_{c+2}{\rm A}(F_{n}))=H\cap({\rm I}_{c+2}{\rm A}(F_{n}))=\gamma_{c+1}(H)$ for all $c$, we get $\psi_{c}$ is isomorphism for all $c\geq 1$. Since ${\rm gr}(H)$ is the (restricted) direct sum of the quotients ${\rm gr}_{c}(H)$, there exists a group homomorphism $\psi$ from ${\rm gr}(H)$ into ${\cal L}_{1}({\rm IA}(F_{n}))$ such that each $\psi_{c}$ is the restriction of $\psi$ on ${\rm gr}_{c}(H)$. It is easily shown that $\psi$ is a Lie algebra homomorphism. Since $\psi(y_{i}H^{\prime})=y_{i}({\rm I}_{3}{\rm A}(F_{n}))$, $i=1,\ldots,n$, we get $\psi$ is a Lie algebra epimorphism. Furthermore, since each $\psi_{c}$ is a $\mathbb{Z}$-module isomorphism, we obtain $\psi$ is a Lie algebra isomorphism. Conversely, let $\phi$ be a Lie algebra isomorphism from ${\rm gr}(H)$ onto ${\cal L}_{1}(H)$ satisfying the conditions $\phi(y_{i}H^{\prime})=y_{i}({\rm I}_{3}{\rm A}(F_{n}))$, $i=1,\ldots,m$. Then, $\phi$ induces a $\mathbb{Z}$-module isomorphism $\phi_{c}$ from ${\rm gr}_{c}(H)$ onto ${\cal L}^{c}_{1}(H)$ for all $c$. In particular, $\phi_{c}((y_{i_{1}},\ldots,y_{i_{c}})\gamma_{c+1}(H))=(y_{i_{1}},\ldots,y_{i_{c}})({\rm I}_{c+2}{\rm A}(F_{n}))$ for all $i_{1},\ldots,i_{c}\in\{1,\ldots,m\}$. Furthermore, ${\rm gr}_{c}(H)$ is $\mathbb{Z}$-module isomorphic to $\gamma_{c}(H)/(\gamma_{c}(H)\cap({\rm I}_{c+2}{\rm A}(F_{n})))$. Since ${\rm gr}_{c}(H)$ is polycyclic and so, it is a hopfian group, we have $\gamma_{c+1}(H)=\gamma_{c}(H)\cap({\rm I}_{c+2}{\rm A}(F_{n}))$ for all $c$. We claim that $\gamma_{c}(H)=H\cap({\rm I}_{c+1}{\rm A}(F_{n}))$ for all $c$. Since $\gamma_{c}(H)\subseteq{\rm I}_{c+1}{\rm A}(F_{n})$, it is enough to show that $H\cap({\rm I}_{c+1}{\rm A}(F_{n}))\subseteq\gamma_{c}(H)$. To get a contradiction, let $\alpha\in H\cap({\rm I}_{c+1}{\rm A}(F_{n}))$ and $\alpha\notin\gamma_{c}(H)$. Since $\gamma_{c}(H)\subseteq{\rm I}_{c+1}{\rm A}(F_{n})$ for all $c$ and $\bigcap_{t\geq 2}{\rm I}_{t}{\rm A}(F_{n})=\{1\}$, we get $H$ is residually nilpotent. Thus, there exists a unique $d\in{\mathbb{N}}$ such that $\alpha\in\gamma_{d}(H)\setminus\gamma_{d+1}(H)$. Therefore, $\alpha\notin H\cap({\rm I}_{d+2}{\rm A}(F_{n}))$. Since $\alpha\in\gamma_{d}(H)\setminus\gamma_{d+1}(H)$ and $\alpha\notin\gamma_{c}(H)$, we have $\gamma_{c}(H)\subseteq\gamma_{d+1}(H)$ and so, $d+1\leq c$. Let $k$ be a non-negative integer such that $c=d+1+k$. Since $H\cap({\rm I}_{c+1}{\rm A}(F_{n}))=H\cap({\rm I}_{d+2+k}{\rm A}(F_{n}))$, we have $\alpha\in{\rm I}_{d+2+k}{\rm A}(F_{n})$. But ${\rm I}_{d+2+k}{\rm A}(F_{n})\subseteq{\rm I}_{d+2}{\rm A}(F_{n})$ and so, $\alpha\in{\rm I}_{d+2}{\rm A}(F_{n})$, which is a contradiction. Therefore, $\gamma_{c}(H)=H\cap({\rm I}_{c+1}{\rm A}(F_{n}))$ for all $c$. $\Box$

Remark 1

It is known that ${\rm IA}(F_{n})/\gamma_{2}({\rm IA}(F_{n}))$, with $n\geq 2$, is torsion-free and its rank is $\frac{n^{2}(n-1)}{2}$. Furthermore, $\gamma_{2}({\rm IA}(F_{n}))={\rm I}_{3}{\rm A}(F_{n})$ (see, for example, [14]). It was conjectured by Andreadakis [2] that $\gamma_{c}({\rm IA}(F_{n}))={\rm I}_{c+1}{\rm A}(F_{n})$ for all $c$. By Lemma 3, ${\rm gr}({\rm IA}(F_{n}))$ is naturally isomorphic to ${\cal L}_{1}({\rm IA}(F_{n}))$ if and only if Andreadakis conjecture is valid. Now, if $H$ is a finitely generated subgroup of ${\rm IA}(F_{n})$ with $H/H^{\prime}$ torsion free, then the statement ${\rm gr}(H)$ is naturally isomorphic to ${\cal L}_{1}(H)$ seems to be an “Andreadakis Conjecture” for $H$.

3.1 The associated Lie algebra of ${\rm Inn}(F_{n})$

In this section, we show that the associated Lie algebra ${\rm gr}({\rm Inn}(F_{n}))$ of the inner automorphisms ${\rm Inn}(F_{n})$ of $F_{n}$, with $n\geq 2$, is naturally embedded into ${\cal L}({\rm IA}(F_{n}))$. Throughout this section, we write $E_{n}={\rm Inn}(F_{n})$. Recall that, for $g\in F_{n}$, $\tau_{g}(x)=gxg^{-1}$ for all $x\in F_{n}$. Thus, $E_{n}=\{\tau_{g}:g\in F_{n}\}$. The following result has been proved in [2, Section 6].

Lemma 4

For a positive integer $c$, $\gamma_{c}(E_{n})=E_{n}\cap{\rm I}_{c+1}{\rm A}(F_{n})$.

Using the above we may show the following.

Proposition 3

Let $n$ be positive integer, with $n\geq 2$. Then, ${\rm gr}(E_{n})$ is naturally embedded into ${\cal L}({\rm IA}(F_{n}))$. In particular, for all $c$, ${\rm gr}_{c}(E_{n})$ is isomorphic to a $\mathbb{Z}$-submodule of ${\cal L}^{c+1}({\rm IA}(F_{n}))$.

*Proof. *Since $F_{n}$ is centerless, we have $F_{n}\cong E_{n}$ in a natural way and so, $E_{n}$ is finitely generated. Moreover, ${\rm gr}(E_{n})$ is a free Lie algebra of rank $n$. Namely, ${\rm gr}(E_{n})$ is freely generated by the set $\{\tau_{x_{i}}E^{\prime}_{n}:i=1,\ldots,n\}$. Let $\phi$ be the mapping from $\{\tau_{x_{i}}E^{\prime}_{n}:i=1,\ldots,n\}$ to ${\cal L}_{1}(E_{n})$ satisfying the conditions $\phi(\tau_{x_{i}}E^{\prime}_{n})=\tau_{x_{i}}{\rm I}_{3}{\rm A}(F_{n})$, $i=1,\ldots,n$. Since ${\rm gr}(E_{n})$ is free on $\{\tau_{x_{i}}E^{\prime}_{n}:i=1,\ldots,n\}$, $\phi$ is extended to a Lie algebra epimorphism. By Lemma 3, it is enough to show that $\gamma_{c}(E_{n})=E_{n}\cap{\rm I}_{c+1}{\rm A}(F_{n})$ for all $c$, which is valid by Lemma 4. Therefore, ${\rm gr}(E_{n})$ is naturally embedded into ${\cal L}({\rm IA}(F_{n}))$. Since

[TABLE]

for all $c$, we have ${\rm gr}_{c}(E_{n})$ is isomorphic to a $\mathbb{Z}$-submodule of ${\cal L}^{c+1}({\rm IA}(F_{n}))$ for all $c$. $\Box$

Corollary 4

For positive integers $n$ and $c$, with $n\geq 2$,

[TABLE]

where $\mu$ is the Möbius function.

*Proof. *Since $F_{n}\cong E_{n}$ as groups and ${\rm gr}(F_{n})$ is a free Lie algebra, we get ${\rm gr}(F_{n})\cong{\rm gr}(E_{n})$ as Lie algebras. Hence, for $c\geq 1$, ${\rm gr}_{c}(F_{n})\cong{\rm gr}_{c}(E_{n})$ as $\mathbb{Z}$-modules. Since ${\rm rank}({\rm gr}_{c}(F_{n}))=\frac{1}{c}\sum_{d|c}\mu(d)n^{c/d}$, we obtain, by Proposition 3, the required result. $\Box$

3.2 The Lie algebra ${\cal L}_{1}(M_{3})$

Write $b_{1}=\chi_{21}$, $b_{2}=\chi_{12}$, $b_{3}=\chi_{23}$, $u_{2}=\chi_{31}\chi_{21}$, $u_{4}=\chi_{32}\chi_{12}$ and $u_{6}=\chi_{23}\chi_{13}$. Then, $M_{3}$ is generated by the set $\{b_{1},b_{2},b_{3},u_{2},u_{4},u_{6}\}$ and its presentation is given by

$M_{3}=\langle b_{1},b_{2},b_{3},u_{2},u_{4},u_{6}:(u_{2},b_{1}),(u_{2},u_{4}b^{-1}_{2}),(u_{2},b_{3}),$

$(u_{4},u_{2}b^{-1}_{1}),(u_{4},b_{2}),(u_{4},b^{-1}_{3}u_{6}),(u_{6},b_{1}),(u_{6},b_{2}),(u_{6},b_{3})\rangle.$

We point out that $u_{2}=\tau_{x^{-1}_{1}}$, $u_{4}=\tau_{x^{-1}_{2}}$ and $u_{6}=\tau_{x^{-1}_{3}}$. Let $H$ and $E$ be the subgroups of $M_{3}$ generated by the sets $\{b_{1},b_{2},b_{3}\}$ and $\{u_{2},u_{4},u_{6}\}$, respectively. We point out that $E={\rm Inn}(F_{3})$. One can easily see that $H$ is a free group of rank $3$. Thus, ${\rm gr}(H)\cong{\rm gr}(E)\cong{\rm gr}(F_{3})$ as Lie algebras.

Proposition 4

${\cal L}_{1}(M_{3})$* is additively equal to the direct sum of the Lie subalgebras ${\cal L}_{1}(H)$ and ${\cal L}_{1}(E)$.*

*Proof. *By the proof of Proposition 3, ${\rm gr}(E)\cong{\cal L}_{1}(E)$ and so, ${\cal L}_{1}(E)$ is a free Lie algebra of rank $3$. Since $b_{1}\notin{\rm I}_{3}{\rm A}(F_{3})$, we have ${\cal L}_{1}(H)$ is a non-trivial subalgebra of ${\cal L}({\rm IA}(F_{3}))$. In fact, ${\cal L}_{1}(H)$ is generated by the set $\{b_{i}({\rm I}_{3}{\rm A}(F_{3})):i=1,2,3\}$. Since $(\tau_{g},\phi)=\tau_{g^{-1}\phi^{-1}(g)}$ for all $\phi\in{\rm Aut}(F_{3})$ and $g\in F_{3}$, we have ${\cal L}_{1}(E)$ is an ideal in ${\cal L}({\rm IA}(F_{3}))$ and so, ${\cal L}_{1}(H)+{\cal L}_{1}(E)$ is a Lie subalgebra of ${\cal L}_{1}(M_{3})$. Let $\overline{w}\in{\cal L}_{1}(H)\cap{\cal L}_{1}(E)$. Since both ${\cal L}_{1}(H)$ and ${\cal L}_{1}(E)$ are graded Lie algebras, we may assume that $\overline{w}\in{\cal L}^{d}_{1}(H)\cap{\cal L}^{d}_{1}(E)$ for some $d$. Thus, there are $u\in\gamma_{d}(H)$ and $v\in\gamma_{d}(E)$ such that $\overline{w}=u({\rm I}_{d+2}{\rm A}(F_{3}))=v({\rm I}_{d+2}{\rm A}(F_{3}))$. To get a contradiction, we assume that $u,v\notin{\rm I}_{d+2}{\rm A}(F_{3})$. Therefore, $v\in\gamma_{d}(E)\setminus\gamma_{d+1}(E)$ and so, there exists $\omega\in\gamma_{d}(F_{3})\setminus\gamma_{d+1}(F_{3})$ such that $v=\tau_{\omega}\rho$, where $\rho\in\gamma_{d+1}(E)$. Since $\gamma_{d+1}(E)\subseteq{\rm I}_{d+2}{\rm A}(F_{3})$, we get $u^{-1}\tau_{\omega}\in{\rm I}_{d+2}{\rm A}(F_{3})$. Since $u^{-1}$ fixes $x_{3}$, we have $x^{-1}_{3}(u^{-1}\tau_{\omega}(x_{3}))=(x_{3},u^{-1}(\omega^{-1}))\in\gamma_{d+2}(F_{3})$. Since $u^{-1}(x_{j})=x_{j}y_{j}$, $y_{j}\in\gamma_{d+1}(F_{3})$, $j=1,2$, and $\gamma_{d}(F_{3})$ is a fully invariant subgroup of $F_{3}$, we have $u^{-1}(\omega^{-1})=\omega^{-1}\omega_{1}$, with $\omega_{1}\in\gamma_{d+1}(F_{3})$ and so, $(w,x_{3})\in\gamma_{d+2}(F_{3})$. Since ${\rm gr}(F_{3})$ is a free Lie algebra of rank $3$ with a free generating set $\{x_{i}F^{\prime}_{3}:i=1,2,3\}$ and $\gamma_{d}(F_{3})/\gamma_{d+1}(F_{3})$ is the $d$-th homogeneous component of ${\rm gr}(F_{3})$, we obtain $(w,x_{3})\in\gamma_{d+1}(F_{3})\setminus\gamma_{d+2}(F_{3})$, which is a contradiction. Therefore, ${\cal L}_{1}(H)\cap{\cal L}_{1}(E)=\{0\}$. By the proof of Theorem 2, we obtain ${\cal L}_{1}(M_{3})={\cal L}_{1}(H)\oplus{\cal L}_{1}(E)$. $\Box$

Remark 2

By Proposition 4, for all $c$, we obtain ${\cal L}^{c}_{1}(M_{3})={\cal L}^{c}_{1}(H)\oplus{\cal L}^{c}_{1}(E)$. Since ${\cal L}^{c}_{1}(E)\cong\gamma_{c}(E)/\gamma_{c+1}(E)\cong\gamma_{c}(F_{3})/\gamma_{c+1}(F_{3})$, we have ${\rm rank}({\cal L}^{c}_{1}(E))=\frac{1}{c}\sum_{d|c}\mu(d)3^{c/d}$. Thus, for any $c$,

[TABLE]

Let $H_{1}$ be the subgroup of $H$ generated by the set $\{\chi_{21},\chi_{23}\}$. We point out that $H_{1}$ is a free group of rank $2$. Then, ${\rm gr}(H_{1})$ is a free Lie algebra of rank $2$. Since both $\chi_{21}$ and $\chi_{23}$ fix $x_{1}$ and $x_{3}$, it may be shown that $\gamma_{c}(H_{1})=H_{1}\cap{\rm I}_{c+1}{\rm A}(F_{3})$ for all $c$. Therefore, $\gamma_{c}(H_{1})\cap{\rm I}_{c+2}{\rm A}(F_{3})=\gamma_{c+1}(H_{1})$ for all $c$ and so, ${\cal L}^{c}_{1}(H_{1})\cong{\rm gr}_{c}(H_{1})$ for all $c\geq 1$. Since ${\rm gr}(H_{1})$ is a free Lie algebra on ${\cal H}_{1}=\{\chi_{21}H^{\prime}_{1},\chi_{23}H^{\prime}_{1}\}$, the mapping $\psi$ from ${\cal H}_{1}$ into ${\cal L}_{1}(H_{1})$ satisfying the conditions $\psi(aH^{\prime}_{1})=a({\rm I}_{3}{\rm A}(F_{3}))$ with $a\in\{\chi_{21},\chi_{23}\}$ can be extended to a Lie algebra homomorphism $\overline{\psi}$. Since ${\cal L}_{1}(H_{1})$ is generated as Lie algebra by the set $\{\chi_{21}({\rm I}_{3}{\rm A}(F_{3})),\chi_{23}({\rm I}_{3}{\rm A}(F_{3}))\}$, we have $\overline{\psi}$ is onto. By Lemma 3, $\overline{\psi}$ is a natural Lie algebra isomorphism from ${\rm gr}(H_{1})$ onto ${\cal L}_{1}(H_{1})$. Since ${\cal L}^{c}_{1}(H_{1})\cong{\rm gr}_{c}(H_{1})$ for all $c\geq 1$, we have ${\rm rank}({\cal L}^{c}_{1}(H_{1}))=\frac{1}{c}\sum_{d|c}\mu(d)2^{c/d}$. Since ${\cal L}_{1}(H_{1})$ is a Lie subalgebra of ${\cal L}_{1}(H)$, we obtain ${\cal L}^{c}_{1}(H_{1})\leq{\cal L}^{c}_{1}(H)$ 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 ${\rm gr}(M_{3})$ into ${\cal L}({\rm IA}(F_{3}))$.

Proposition 5

Let $H$ be the subgroup of $M_{3}$ generated by $\chi_{21},\chi_{12},\chi_{23}$. Then, ${\rm gr}(M_{3})$ is naturally isomorphic to ${\cal L}_{1}(M_{3})$ as Lie algebras if and only if ${\rm gr}(H)$ is naturally isomorphic to ${\cal L}_{1}(H)$ as Lie algebras.

*Proof. *Suppose that ${\rm gr}(M_{3})$ is naturally isomorphic to ${\cal L}_{1}(M_{3})$ as Lie algebras. Thus, for all $c$, ${\rm gr}_{c}(M_{3})\cong{\cal L}^{c}_{1}(M_{3})$ as $\mathbb{Z}$-modules. By Proposition 4, ${\rm gr}_{c}(M_{3})\cong{\cal L}^{c}_{1}(H)\oplus{\cal L}^{c}_{1}(E)$. Since, for all $c$, ${\rm gr}_{c}(M_{3})\cong{\rm gr}_{c}(F_{3})\oplus{\rm gr}_{c}(F_{3})$ as $\mathbb{Z}$-modules, we obtain

[TABLE]

Therefore, ${\cal L}^{c}_{1}(H)\cong{\rm gr}_{c}(H)$ for all $c$. Hence, $\gamma_{c}(H)\cap{\rm I}_{c+2}{\rm A}(F_{3})=\gamma_{c+1}(H)$ for all $c$ and so, as in the proof of Lemma 3, $\gamma_{c}(H)=H\cap{\rm I}_{c+1}{\rm A}(F_{3})$ for all $c$. We point out that $H/H^{\prime}$ is torsion-free with a free generating set ${\cal H}=\{\chi_{21}H^{\prime},\chi_{12}H^{\prime},\chi_{23}H^{\prime}\}$. Since ${\rm gr}(H)$ is a free Lie algebra with a free generating set $\cal H$, and ${\cal L}_{1}(H)$ is generated as a Lie algebra by the set $\{\chi_{21}({\rm I}_{3}{\rm A}(F_{3})),\chi_{12}({\rm I}_{3}{\rm A}(F_{3})),\chi_{23}({\rm I}_{3}{\rm A}(F_{3}))\}$, we have there exists a natural Lie epimorphism from ${\rm gr}(H)$ onto ${\cal L}_{1}(H)$. By Lemma 3, ${\rm gr}(H)$ is naturally isomorphic to ${\cal L}_{1}(H)$ as Lie algebras.

Conversely, let ${\rm gr}(H)$ be naturally isomorphic to ${\cal L}_{1}(H)$ as Lie algebras. By Lemma 3, $\gamma_{c}(H)=H\cap({\rm I}_{c+1}{\rm A}(F_{3}))$ for all $c$. Thus,

[TABLE]

and so, ${\cal L}^{c}_{1}(H)\cong{\rm gr}_{c}(H)$ for all $c$. By Proposition 4,

[TABLE]

for all $c$, and by Theorem 2,

[TABLE]

for all $c$. Therefore,

[TABLE]

for all $c$. As in the proof of Lemma 3, for all $c$,

[TABLE]

Let $\psi$ be the Lie algebra epimorphism from $L$ onto ${\cal L}_{1}(M_{3})$ satisfying the conditions $\psi(x_{2j-1})=b_{j}({\rm I}_{3}{\rm A}(F_{3}))$ and $\psi(v_{2j})=\tau_{x^{-1}_{2j-1}}({\rm I}_{3}{\rm A}(F_{3}))$, $j=1,2,3$. By $\tau_{w}$ we denote the inner automorphisms of $F_{3}$ with $\tau_{w}(x)=w^{-1}xw$. Since $J\subseteq{\rm Ker}\psi$ and ${\rm gr}(M_{3})\cong L/J$ as Lie algebras, there exists a Lie algebra epimorphism $\overline{\psi}$ from ${\rm gr}(M_{3})$ onto ${\cal L}_{1}(M_{3})$ satisfying the conditions $\overline{\psi}(b_{i}M^{\prime}_{3})=b_{i}({\rm I}_{3}{\rm A}(F_{3}))$ and $\overline{\psi}(\tau_{x^{-1}_{i}}M^{\prime}_{3})=\tau_{x^{-1}_{i}}({\rm I}_{3}{\rm A}(F_{3}))$, $i=1,2,3$. Since, in addition, $M_{3}\cap{\rm I}_{c+1}{\rm A}(F_{3})=\gamma_{c}(M_{3})$ for all $c$, we obtain from Lemma 3 the required result. $\Box$

Remark 3

For $c=1,2,3$, we show that $\gamma_{c}(H)=H\cap{\rm I}_{c+1}{\rm A}(F_{3})$. That is, ${\cal L}^{c}_{1}(H)\cong\gamma_{c}(H)/\gamma_{c+1}(H)$ for $c=1,2,3$. For $c\geq 4$, the method used becomes very cumbersome. For simplicity, let $x=\chi_{21}$, $y=\chi_{12}$ and $z=\chi_{23}$. For $c=1$, our claim is trivially true. Let $c=2$. Since $\gamma_{2}(H)/\gamma_{3}(H)$ is a free abelian group of rank $3$, we have each element $h\in\gamma_{2}(H)\setminus\gamma_{3}(H)$ is uniquely written as

[TABLE]

where $v\in\gamma_{3}(H)$ and $a_{1},a_{2},a_{3}$ are non-negative integers. We claim that $h\notin{\rm I}_{4}{\rm A}(F_{3})$. By a direct calculation, $(z,x)^{a_{1}}(x_{i})=x_{i}$, $i\neq 2$, and $(z,x)^{a_{1}}(x_{2})=x_{2}(x_{3},x_{1},x_{2})^{-a_{1}}u$, where $u\in\gamma_{4}(F_{3})$. Working modulo $\gamma_{3}(H)$, we have

[TABLE]

We point out that, for $\phi\in{\rm IA}(F_{3})$ and $g\in F_{3}$, we have $(\tau_{g},\phi)=\tau_{g^{-1}\phi^{-1}(g)}$. Since $(z,\tau_{x^{-1}_{2}})=\tau_{(x^{-1}_{3},x^{-1}_{2})}$ and $(\tau_{x^{-1}_{2}},x)=\tau_{(x^{-1}_{2},x^{-1}_{1})}$, we get

[TABLE]

Since both $(z,\chi_{32})^{a_{2}}$ and $(\chi_{32},x)^{a_{3}}$ fix $x_{1}$, we obtain

[TABLE]

where $v_{1},v_{2}\in\gamma_{4}(F_{3})$. Thus, $(z,y),(y,x)\in{\rm I}_{3}{\rm A}(F_{3})\setminus{\rm I}_{4}{\rm A}(F_{3})$. Furthermore, it is easily shown that, for $a_{1}+a_{2}+a_{3}\neq 0$, $h\in\gamma_{2}(H)$ and $h\notin{\rm I}_{4}{\rm A}(F_{3})$. Therefore, $\gamma_{2}(H)=H\cap{\rm I}_{3}{\rm A}(F_{3})$.

For $c=3$, we apply similar arguments as before. We point out that $\gamma_{3}(H)/\gamma_{4}(H)$ is a free abelian group of rank $8$. Each element $g\in\gamma_{3}(H)\setminus\gamma_{4}(H)$ is uniquely written as $g=g_{1}g_{2}g_{3}u$, where $g_{1}=(z,x,x)^{b_{1}}(z,x,z)^{b_{2}}$, $g_{2}=(y,x,x)^{b_{3}}(y,x,z)^{b_{4}}(z,x,y)^{b_{5}}(z,y,z)^{b_{6}}$, $g_{3}=(y,x,y)^{b_{7}}(z,y,y)^{b_{8}}$, $u\in\gamma_{4}(H)$ and $b_{1},\ldots,b_{8}$ are non-negative integers. Since $g_{1}(x_{i})=x_{i}$ for $i\neq 1,3$, we have, by direct calculations,

[TABLE]

where $v_{12}\in\gamma_{5}(F_{3})$. By the equation (*), and working modulo $\gamma_{4}(H)$, we have

[TABLE]

Since both $(\chi_{32},x,x)^{b_{3}}$ and $(\chi_{32},x,z)^{b_{4}}$ fix $x_{1}$, we get

[TABLE]

where $v_{3},v_{4}\in\gamma_{5}(F_{3})$. Since $(z,x)(x_{2})=x_{2}(x_{3},x_{1},x_{2})^{-1}u$, with $u\in\gamma_{4}(F_{3})$, and by the equation (*), we get

[TABLE]

and

[TABLE]

Since both $(z,x,\chi_{32})^{b_{5}}$ and $(z,y,\chi_{32})^{b_{6}}$ fix $x_{1}$, we obtain

[TABLE]

where $v_{5},v_{6}\in\gamma_{5}(F_{3})$. Therefore,

[TABLE]

where $v\in\gamma_{5}(F_{3})$.

Next, we point out that $(\chi_{32},\chi_{21})(x_{i})=x_{i}$ for $i=1,2$ and $(\chi_{32},\chi_{21})(x_{3})=x_{3}(x_{2},x_{1},x_{3})w$, where $w\in\gamma_{4}(F_{3})$. Since $(\chi_{32},\chi_{21},\tau^{-1}_{x_{2}})={\rm Id}_{F_{3}}$, we have

[TABLE]

where $\psi\in\gamma_{4}(M_{3})$. Hence,

[TABLE]

and so,

[TABLE]

Since $(\tau_{(x^{-1}_{2},x^{-1}_{1})},\chi^{-1}_{32})^{b_{7}}={\rm Id}_{F_{3}}$, we get

[TABLE]

Therefore,

[TABLE]

where $v_{7}\in\gamma_{5}(F_{3})$. By direct calculations, $(z,\chi_{32})(x_{1})=x_{1}$, $(z,\chi_{32})(x_{2})=x_{2}(x_{3},x_{2},x_{2})^{-1}u_{2}$ and $(z,\chi_{32})(x_{3})=x_{3}(x_{3},x_{2},x_{3})u_{3}$, where $u_{2},u_{3}\in\gamma_{4}(F_{3})$. It is easily verified that

[TABLE]

where $\psi_{1}\in\gamma_{4}(M_{3})$. By the equation (*),

[TABLE]

Since

[TABLE]

and

[TABLE]

we have

[TABLE]

Hence,

[TABLE]

Since $(z,\chi_{32},\chi_{32})$ fixes $x_{1}$, we have

[TABLE]

where $v_{8}\in\gamma_{5}(F_{3})$. Therefore,

[TABLE]

where $v_{7,8}\in\gamma_{5}(F_{3})$. Since the ”basic” group commutators of length $4$ consists of a basis for $\gamma_{4}(F_{3})/\gamma_{5}(F_{3})$, we obtain from the equations (1), (2) and (3) that $g\in\gamma_{4}(H)$ and $g\notin{\rm I}_{6}{\rm A}(F_{3})$, that is, $\gamma_{4}(H)=H\cap{\rm I}_{5}{\rm A}(F_{3})$.