R}_{2}}\cap N_{t}=\\{1\\}$. In the next few lines, we write ${\cal
Y}_{x}=\\{y_{x,1},\ldots,y_{x,n}\\}$, where $y_{x,j}=x_{j}F^{\prime}_{n}$,
$j\in[n]$. So, ${\rm gr}(F_{n})$ is a free Lie algebra on ${\cal Y}_{x}$. We
denote by $J_{{\cal Y}_{x}}$ the ideal of ${\rm gr}(F_{n})$ generated the set
${\cal R}_{x,L,2}=\\{[y_{x,a},y_{x,b}]:(a,b)\in\theta_{n,\delta}\\}$. Since
$J_{{\cal Y}_{x}}$ is a homogeneous ideal, we have $J_{{\cal
Y}_{x}}=\bigoplus_{d\geq 2}J^{d}_{{\cal Y}_{x}}$, where $J^{d}_{{\cal
Y}_{x}}=J_{{\cal Y}_{x}}\cap{\rm gr}_{d}(F_{n})$. Let ${\cal L}_{{\cal
Y}_{x}}(N_{{\cal R}_{2}})=\bigoplus_{d\geq 2}I_{x,d}(N_{{\cal R}_{2}})$, where
$I_{x,d}(N_{{\cal R}_{2}})=(N_{{\cal
R}_{2},x,d}\gamma_{d+1}(F_{n}))/\gamma_{d+1}(F_{n})$ and, for any positive
integer $d\geq 2$, $N_{{\cal R}_{2},x,d}=N_{{\cal
R}_{2}}\cap\gamma_{d}(F_{n})$. Since $N_{{\cal R}_{2}}$ is normal in $F_{n}$,
we have ${\cal L}_{{\cal Y}_{x}}(N_{{\cal R}_{2}})$ is an ideal in ${\rm
gr}(F_{n})$. Since ${\cal R}_{x,L,2}\subset I_{x,2}(N_{{\cal R}_{2}})$ and
${\cal L}_{{\cal Y}_{x}}(N_{{\cal R}_{2}})$ is an ideal of ${\rm gr}(F_{n})$,
we have $J_{{\cal Y}_{x}}\subseteq{\cal L}_{{\cal Y}_{x}}(N_{{\cal R}_{2}})$.
Moreover, for any positive integer $d$, with $d\geq 2$, $J^{d}_{{\cal
Y}_{x}}\subseteq I_{x,d}(N_{{\cal R}_{2}})$.

By Corollary 1, $J_{{\cal Y}_{x}}$ is a direct summand of ${\rm gr}(F_{n})$
and so, ${\rm gr}(F_{n})/J_{{\cal Y}_{x}}$ is a free $\mathbb{Z}$-module. By
Theorem 4, the $\mathbb{Z}$-module $J_{{\cal Y}_{x}}/[J_{{\cal
Y}_{x}},J_{{\cal Y}_{x}}]$ is a free (right) $U({\rm gr}(F_{n})/J_{{\cal
Y}_{x}})$-module with a free generating set ${\cal R}_{x,L,2}+[J_{{\cal
Y}_{x}},J_{{\cal Y}_{x}}]$. By a result of Labute [13, Theorem 1], ${\rm
gr}(F_{n}/N_{{\cal R}_{2}})\cong{\rm gr}(F_{n})/J_{{\cal Y}_{x}}$ as Lie
algebras. (Note that in the proof of Theorem 1 in [13], the left action of the
universal enveloping algebra $U(L/I)$ on the $\mathbb{Z}$-module $I/[I,I]$ it
is used.) In the proof of Theorem 1 in [13], since $J_{{\cal Y}_{x}}/[J_{{\cal
Y}_{x}},J_{{\cal Y}_{x}}]$ is a free (right) $U({\rm gr}(F_{n})/J_{{\cal
Y}_{x}})$-module, the induced homomorphism $\xi$ (Lemma 6 for $N=N_{{\cal
R}_{2}}$ and $I=J_{{\cal Y}_{x}}$) is onto and so, ${\cal L}_{{\cal
Y}_{x}}(N_{{\cal R}_{2}})=J_{{\cal Y}_{x}}$.

Let ${\cal L}_{{\cal Y}}(N_{{\cal R}_{2}})=\bigoplus_{d\geq 2}I_{d}(N_{{\cal
R}_{2}})$, where $I_{d}(N_{{\cal R}_{2}})=(N_{{\cal
R}_{2},d}\gamma_{d+1}(F_{n+1}))/\gamma_{d+1}(F_{n+1})$ and, for any positive
integer $d\geq 2$, $N_{{\cal R}_{2},d}=N_{{\cal
R}_{2}}\cap\gamma_{d}(F_{n+1})$. Since $N_{{\cal R}_{2}}\cap N_{t}=\\{1\\}$,
we have $N_{{\cal R}_{2}}\cap\gamma_{d}(F_{n+1})=N_{{\cal
R}_{2}}\cap\gamma_{d}(F_{n})$. Hence, for any positive integer $d$, with
$d\geq 2$, $N_{{\cal R}_{2},x,d}=N_{{\cal R}_{2},d}$. It is clear enough that
${\cal L}_{\cal Y}(N_{{\cal R}_{2}})$ is an ideal in ${\cal
L}_{1}(F_{n})=L({\cal Y})$. The Lie algebra ${\rm gr}(F_{n})$ is isomorphic as
a Lie algebra to ${\cal L}_{1}(F_{n})=L({\cal Y})$ by a Lie algebra
isomorphism $\chi$ defined as $\chi(y_{x,j})=y_{j}$ for all $j\in[n]$.
Clearly, $\chi({\cal L}_{{\cal Y}_{x}}(N_{{\cal R}_{2}}))={\cal L}_{{\cal
Y}}(N_{{\cal R}_{2}})$ and $\chi(J_{{\cal Y}_{x}})=J_{\cal Y}$. Since ${\cal
L}_{{\cal Y}_{x}}(N_{{\cal R}_{2}})=J_{{\cal Y}_{x}}$ and $\chi$ is a Lie
algebra isomorphism, we obtain ${\cal L}_{\cal Y}(N_{{\cal R}_{2}})=J_{\cal
Y}$. So, for any $d\geq 2$, $I_{d}(N_{{\cal R}_{2}})=J^{d}_{\cal Y}$. For
$d=2$, $I_{2}(N_{{\cal R}_{2}})=J^{2}_{\cal Y}=\langle{\cal R}_{2,L}\rangle$.
Thus, we assume that $d\geq 3$. By $(4.2)$, the definition of each ${\cal
P}^{\prime}_{{\cal Y}_{\kappa},1,2}\wr{\cal P}^{\prime\prime}_{{\cal
Y}_{\kappa}}$, Lemma 5 (1) and the Jacobi identity in the form
$[x,[y,z]]=[x,y,z]-[x,z,y]$ (where it is needed), we have $I_{d}(N_{{\cal
R}_{2}})(=J^{d}_{\cal Y})\subseteq([F_{n}]_{X_{{\cal
R}_{2},d-2}}\gamma_{d+1}(F_{n+1}))/\gamma_{d+1}(F_{n+1})$. Since
$[F_{n}]_{X_{{\cal R}_{2},d-2}}\subseteq N_{{\cal R}_{2},d}=N_{{\cal
R}_{2}}\cap\gamma_{d}(F_{n})$, we get $([F_{n}]_{X_{{\cal
R}_{2},d-2}}\gamma_{d+1}(F_{n+1}))/\gamma_{d+1}(F_{n+1})\subseteq
I_{d}(N_{{\cal R}_{2}})$ and so, we have $([F_{n}]_{X_{{\cal
R}_{2},d-2}}\gamma_{d+1}(F_{n+1}))/\gamma_{d+1}(F_{n+1})=I_{d}(N_{{\cal
R}_{2}})=J^{d}_{\cal Y}$. We summarize the above observations in the following
result.

###### Proposition 5

With the previous notations, ${\rm gr}(F_{n}/N_{{\cal R}_{2}})\cong{\cal
L}_{1}(F_{n})/{\cal L}_{\cal Y}(N_{{\cal R}_{2}})$ as Lie algebras in a
natural way, with ${\cal L}_{1}(F_{n})=L({\cal Y})$ and ${\cal L}_{\cal
Y}(N_{{\cal R}_{2}})=J_{\cal Y}$. Moreover, for any $d\geq 3$,

$([F_{n}]_{X_{{\cal
R}_{2},d-2}}\gamma_{d+1}(F_{n+1}))/\gamma_{d+1}(F_{n+1})=I_{d}(N_{{\cal
R}_{2}})=J^{d}_{\cal Y}.$

###### Remark 1

The group $F_{n}/N_{{\cal R}_{2}}$ is the raag $H$. A description of ${\cal
L}_{1}(F_{n})/{\cal L}_{\cal Y}(N_{{\cal R}_{2}})$ has been also given in [8]
(see, also, [9]). We point out that in [9], it has been shown that
$F_{n}/N_{{\cal R}_{2}}$ is a Magnus group.

###  6.2 Recognition of $L({\cal E}\wr\Omega)$ in ${\rm gr}(F_{n+1})$

In this section, we describe the ideal $L({\cal E}\wr\Omega)$ in ${\rm
gr}(F_{n+1})$.

###### Proposition 6

With the previous notations,

  1. 1.

$L({\cal E}\wr\Omega)$ is the ideal of ${\rm gr}(F_{n+1})$ generated by the
set ${\cal R}_{3,L}=\\{[s,y_{i},y_{j}]:i,j\in[n]\\}$.

  2. 2.

Let $N_{{\cal R}_{3}}={\cal R}_{3}^{F_{n+1}}$ be the normal closure of ${\cal
R}_{3}$ in $F_{n+1}$. Then, ${\rm gr}(F_{n+1}/N_{{\cal R}_{3}})\cong{\rm
gr}(F_{n+1})/{\cal L}(N_{{\cal R}_{3}})$ as Lie algebras in a natural way,
with ${\cal L}(N_{{\cal R}_{3}})=L({\cal E}\wr\Omega)$. For any $d\geq 3$,
$L^{d}_{\rm grad}({\cal E}\wr\Omega)=I_{d}(N_{{\cal
R}_{3}})=([F_{n+1}]_{X_{{\cal
R}_{3},d-3}}\gamma_{d+1}(F_{n+1}))/\gamma_{d+1}(F_{n+1})$, where $X_{{\cal
R}_{3}}$ is the subgroup of $F_{n+1}$ generated by the set ${\cal R}_{3}$.

Proof.

  1. 1.

Write $I_{{\cal R}_{3,L}}$ for the ideal of ${\rm gr}(F_{n+1})$ generated by
the set ${\cal R}_{3,L}$. Since ${\cal E}\wr\Omega\subset I_{{\cal R}_{3,L}}$,
we have $L({\cal E}\wr\Omega)\subseteq I_{{\cal R}_{3,L}}$. On the other hand,
since ${\cal R}_{3,L}\subset L({\cal E}\wr\Omega)$ and, by Proposition 3 (3),
$L({\cal E}\wr\Omega)$ is an ideal of ${\rm gr}(F_{n+1})$, we have $I_{{\cal
R}_{3,L}}\subseteq L({\cal E}\wr\Omega)$ and so, we get the required result.

  2. 2.

Recall that ${\cal L}(N_{{\cal R}_{3}})=\bigoplus_{d\geq 3}I_{d}(N_{{\cal
R}_{3}})$, where $I_{d}(N_{{\cal R}_{3}})=(N_{{\cal
R}_{3},d}\gamma_{d+1}(F_{n+1}))/\gamma_{d+1}(F_{n+1})$ and $N_{{\cal
R}_{3},d}=N_{{\cal R}_{3}}\cap\gamma_{d}(F_{n+1})$. Since ${\cal
R}_{3,L}\subset I_{3}(N_{{\cal R}_{3}})=L^{3}_{\rm grad}({\cal E}\wr\Omega)$
and ${\cal L}(N_{{\cal R}_{3}})$ is an ideal of ${\rm gr}(F_{n+1})$, we have,
by Proposition 6 (1), $L({\cal E}\wr\Omega)\subseteq{\cal L}(N_{{\cal
R}_{3}})$. For any positive integer $d$, with $d\geq 3$, $L^{d}_{\rm
grad}({\cal E}\wr\Omega)\subseteq I_{d}(N_{{\cal R}_{3}})$. By $(6.3)$,
$L({\cal E}\wr\Omega)$ is a direct summand of ${\rm gr}(F_{n+1})$ and so,
${\rm gr}(F_{n+1})/L({\cal E}\wr\Omega)$ is a free $\mathbb{Z}$-module. By
Theorem 6, the $\mathbb{Z}$-module $L({\cal E}\wr\Omega)/[L({\cal
E}\wr\Omega),L({\cal E}\wr\Omega)]$ is a free (right) $U({\rm
gr}(F_{n+1})/L({\cal E}\wr\Omega))$-module with a free generating set ${\cal
R}_{3,L}+[L({\cal E}\wr\Omega),L({\cal E}\wr\Omega)]$. Using similar arguments
as before, we have ${\cal L}(N_{{\cal R}_{3}})=L({\cal E}\wr\Omega)$. Hence,
for any $d\geq 3$, $L^{d}_{\rm grad}({\cal E}\wr\Omega)=I_{d}(N_{{\cal
R}_{3}})$. For non-negative integers $\kappa$ and $\lambda$, with $\kappa\geq
2$, let $V_{\kappa+\lambda}$ be the free $\mathbb{Z}$-module spanned by the
set
$\\{[s,y_{i_{1}},\ldots,y_{i_{\kappa}},z_{j_{1}},\ldots,z_{j_{\lambda}}]:i_{1},\ldots,i_{\kappa}\in[n];z_{j_{1}},\ldots,z_{j_{\lambda}}\in\Omega\\}$.
Clearly,

$L({\cal E}\wr\Omega)=L\left(\bigoplus_{\kappa\geq 2,\lambda\geq
0\atop\kappa+\lambda\geq 2}V_{\kappa+\lambda}\right).$

Then, for all $d\geq 3$, $L^{d}_{\rm grad}({\cal E}\wr\Omega)$ is the
$\mathbb{Z}$-module of $L({\cal E}\wr\Omega)$ consisting of all Lie
commutators of the form $[v_{1},\ldots,v_{\nu}]$ where $v_{i}\in
V_{\kappa_{i}+\lambda_{i}}$, $i\in[\nu]$ and
$(\kappa_{1}+\lambda_{1})+\cdots+(\kappa_{\nu}+\lambda_{\nu})=d-\nu$. For
$d\geq 3$, write $\overline{I}_{d}(N_{{\cal R}_{3}})=[F_{n+1}]_{X_{{\cal
R}_{3},d-3}}\gamma_{d+1}(F_{n+1})/\gamma_{d+1}(F_{n+1})$. Since
$[F_{n+1}]_{X_{{\cal R}_{3},d-3}}\subseteq N_{{\cal R}_{3},d}=N_{{\cal
R}_{3}}\cap\gamma_{d}(F_{n+1})$ for all $d\geq 3$, we have
$\overline{I}_{d}(N_{{\cal R}_{3}})\subseteq I_{d}(N_{{\cal
R}_{3}})=L^{d}_{\rm grad}({\cal E}\wr\Omega)$. On the other hand, since any
Lie commutator $[v_{1},\ldots,v_{\nu}]$ of the above form belongs to
$\overline{I}_{d}(N_{{\cal R}_{3}})$, we get $L^{d}_{\rm grad}({\cal
E}\wr\Omega)\subseteq\overline{I}_{d}(N_{{\cal R}_{3}})$ and so, we obtain the
required result. $\Box$

###### Remark 2

The group $F_{n+1}/N_{{\cal R}_{3}}$ is a Formanek-Procesi group $FP(F_{n})$
with $F_{n}$ a free group of rank $n\geq 2$. That is, $FP(F_{n})=\langle
t,g_{1},\ldots,g_{n}:[t,g_{i},g_{j}]=1,i,j\in[n]\rangle$. A presentation of
${\rm gr}(F_{n+1}/N_{{\cal R}_{3}})$ has been also given in [17, Theorem 1
(1)]. Furthermore, it has been shown that $FP(F_{n})$ is a Magnus group [17,
Theorem 1 (3)].

###  6.3 Recognition of $I_{\Omega}$ in ${\rm gr}(F_{n+1})$

For $i\in[n]$, we write $q_{i}=[t,x_{i}]$,
$\omega_{i}=q_{i}\gamma_{3}(F_{n+1})$ and
$\Omega_{0}=\\{\omega_{1},\ldots,\omega_{n}\\}$. Let $F_{\Omega_{0}}$ be the
free group on the set $\\{q_{1},\ldots,q_{n}\\}$. Note that

$F_{\Omega_{0}}\subseteq F^{\prime}_{n+1}~{}~{}{\rm
and}~{}~{}F_{\Omega_{0}}\gamma_{3}(F_{n+1})/\gamma_{3}(F_{n+1})=[X,F_{n}]\gamma_{3}(F_{n+1})/\gamma_{3}(F_{n+1}),$

where $X$ is the subgroup of $F_{n+1}$ generated by $t$. For a positive
integer $\kappa$, let ${\cal L}_{{\rm
od},\kappa}(F_{\Omega_{0}})=\gamma_{\kappa}(F_{\Omega_{0}})\gamma_{2\kappa+1}(F_{n+1})/\gamma_{2\kappa+1}(F_{n+1})$
and form the (restricted) direct sum ${\cal L}_{{\rm od}}(F_{\Omega_{0}})$ of
the abelian groups ${\cal L}_{{\rm od},k}(F_{\Omega_{0}})$. Since
$[\gamma_{\kappa}(F_{\Omega_{0}}),\gamma_{\lambda}(F_{\Omega_{0}})]\subseteq\gamma_{\kappa+\lambda}(F_{\Omega_{0}})$
for all $\kappa,\lambda\in\mathbb{N}$, we have ${\cal L}_{\rm
od}(F_{\Omega_{0}})$ is a Lie subalgebra of ${\rm gr}(F_{n+1})$. In fact,
${\cal L}_{\rm od}(F_{\Omega_{0}})=L({\Omega_{0}})=L(I_{2}(N_{t}))$. By
$(6.2)$, ${\cal L}(N_{t})=L(\Omega)\oplus L({\cal E}\wr\Omega)$. We point out
that $L(\Omega)=L(I_{2}(N_{t})\oplus I_{1}(N_{t}))$. For a positive integer
$d$, $L^{d}_{\rm grad}(\Omega)$ is the $\mathbb{Z}$-submodule of $L(\Omega)$
spanned by all Lie commutators $[v_{1},\ldots,v_{\mu}]$, $\mu\geq 1$,
$v_{j}\in I_{i(j)}(N_{t})$, $i(j)\in[2]$, $j\in[\mu]$ and
$i(1)+\cdots+i(\mu)=d$. Thus, by $(6.2)$ and Proposition 6 (2), for $d\geq 1$,

$I_{d}(N_{t})=L^{d}_{\rm grad}(\Omega)\oplus I_{d}(N_{{\cal R}_{3}}).$ $None$

So, $I_{1}(N_{t})=L^{1}_{\rm grad}(\Omega)=\langle s\rangle$ and
$I_{2}(N_{t})=L^{2}_{\rm grad}(\Omega)=\langle[s,y_{j}]:j\in[n]\rangle$. For
$d\geq 3$, by Proposition 4, it is easily verified that
$I_{d}(N_{t})=([F_{n+1}]_{F_{\Omega_{0}},d-2})\gamma_{d+1}(F_{n+1})/\gamma_{d+1}(F_{n+1})$.
By Proposition 6, $I_{d}(N_{{\cal
R}_{3}})=([F_{n+1}]_{[F_{\Omega_{0}},F_{n}],d-3})\gamma_{d+1}(F_{n+1})/\gamma_{d+1}(F_{n+1})$.
We denote by $L^{d}_{\rm grad,1}(\Omega)$ the $\mathbb{Z}$-submodule of
$L^{d}_{\rm grad}(\Omega)$ spanned by all Lie commutators
$[v_{1},\ldots,v_{\mu}]$, with $v_{j}\in I_{2}(N_{t})$, $j\in[\mu]$ and
$2\mu=d$. Thus, $L^{d}_{\rm grad,1}(\Omega)=\\{0\\}$ if $d$ is odd, and
$L^{2\mu}_{\rm grad,1}(\Omega)={\cal L}_{{\rm od},\mu}(F_{\Omega_{0}})$. We
write $L^{d}_{\rm grad,2}(\Omega)$ for the $\mathbb{Z}$-submodule of
$L^{d}_{\rm grad}(\Omega)$ spanned by all Lie commutators
$[v_{1},\ldots,v_{\mu}]$, with $v_{j}\in I_{i(j)}(N_{t})$, $j\in[\mu]$,
$i(1),\ldots,i(\mu)\in[2]$, at least one of $i(1),\ldots,i(\mu)$ is equal to
$1$ and $i(1)+\cdots+i(\mu)=d$. By Theorem 2, $L(\Omega)=L(I_{2}(N_{t}))\oplus
L(I_{1}(N_{t})\wr I_{2}(N_{t}))$. Thus, for all $d\in\mathbb{N}$, we have
$L^{d}_{\rm grad}(\Omega)=L^{d}_{\rm grad}(I_{2}(N_{t}))\oplus L^{d}_{\rm
grad}(I_{1}(N_{t})\wr I_{2}(N_{t}))$. By the definitions of $L^{d}_{\rm
grad,1}(\Omega)$ and $L^{d}_{\rm grad,2}(\Omega)$, we have, for all
$d\in\mathbb{N}$,

$L^{d}_{\rm grad}(\Omega)=L^{d}_{\rm grad,1}(\Omega)\oplus L^{d}_{\rm
grad,2}(\Omega)$ $None$

and so, by $(6.7)$, for all $d\in\mathbb{N}$,

$L^{d}_{\rm grad}(I_{1}(N_{t})\wr I_{2}(N_{t}))=L^{d}_{{\rm grad},2}(\Omega).$
$None$

For $\mu\geq 1$ and non-negative integers $j(1),\ldots,j(\mu)$, with
$j(1)+\cdots+j(\mu)=d$ and $d\geq 1$, let
$X_{(j(1),\ldots,j(\mu))}=[X_{j(1)},\ldots,X_{j(\mu)}]$, where
$X_{j(i)}\in\\{X,F_{\Omega_{0}}\\}$, $i\in[\mu]$, $j(1),\ldots,j(\mu)\in[2]$,
and at least one of $X_{j(1)},\ldots,X_{j(\mu)}$ equals $X$. Let $X_{{\rm
grad},d}=\prod_{j(1)+\cdots+j(\mu)=d}X_{(j(1),\ldots,j(\mu))}$ and
$M_{2,d}=X_{{\rm grad},d}\gamma_{d+1}(F_{n+1})/\gamma_{d+1}(F_{n+1})$. By the
definitions of $M_{2,d}$ and $L^{d}_{{\rm grad},2}(\Omega)$, we have
$M_{2,d}=L^{d}_{{\rm grad},2}(\Omega)$ for all $d\geq 1$. Form the
(restricted) direct sum ${\cal L}_{M}=\bigoplus_{d\geq 1}M_{2,d}$ of the
abelian groups $M_{2,d}$. By using the Jacobi identity in the form
$[x,[y,z]]=[x,y,z]+[x,z,y]$, we may show that ${\cal L}_{M}$ is a Lie
subalgebra of ${\rm gr}(F_{n+1})$. By $(6.8)$, we get ${\cal
L}_{M}=L(I_{1}(N_{t})\wr I_{2}(N_{t}))$. By $(6.7)$, we have

$L(\Omega)={\cal L}_{\rm od}(F_{\Omega_{0}})\oplus{\cal L}_{M}.$ $None$

For a positive integer $\kappa$, let
$F_{{\Omega},\kappa}=F_{\Omega}\cap\gamma_{\kappa}(F_{n+1})$. Form the
(restricted) direct sum ${\cal L}(F_{\Omega})$ of the abelian groups
$I_{\kappa}(F_{\Omega})=F_{{\Omega},\kappa}\gamma_{\kappa+1}(F_{n+1})/\gamma_{\kappa+1}(F_{n+1})$.
It is clear enough that ${\cal L}(F_{\Omega})$ is a Lie subalgebra of ${\rm
gr}(F_{n+1})$.

###### Proposition 7

With the previous notations, ${\cal L}(F_{\Omega})=L(\Omega)$. In particular,
${\cal L}(F_{\Omega})$ is a free Lie algebra on $I_{1}(F_{\Omega})\oplus
I_{2}(F_{\Omega})$. For any positive integer $d$,
$I_{d}(F_{\Omega})=L^{d}_{\rm grad}(\Omega)$.

Proof. Since $F_{{\Omega_{0}}}\subseteq F^{\prime}_{n+1}$, we have
$I_{1}(F_{\Omega})=I_{1}(N_{t})$ and $I_{2}(F_{\Omega})=I_{2}(N_{t})$. Since
${\cal L}(F_{\Omega})$ is a Lie subalgebra of ${\rm gr}(F_{n+1})$ and
$L(\Omega)$ is free on $I_{1}(N_{t})\oplus I_{2}(N_{t})$, we have
$L(\Omega)\subseteq{\cal L}(F_{\Omega})$. Clearly, ${\cal
L}(F_{\Omega})\subseteq{\cal L}(N_{t})$. By $(6.2)$ and the modular law, we
get

${\cal L}(F_{\Omega})=L(\Omega)\oplus(L({\cal E}\wr\Omega)\cap{\cal
L}(F_{\Omega})).$ $None$

We claim that $L({\cal E}\wr\Omega)\cap{\cal L}(F_{\Omega})=\\{0\\}$. Let
$\overline{u}\in L({\cal E}\wr\Omega)\cap{\cal L}(F_{\Omega})$. By Proposition
6 (2), ${\cal L}(N_{{\cal R}_{3}})=L({\cal E}\wr\Omega)$. Since both ${\cal
L}(N_{{\cal R}_{3}})$ and ${\cal L}(F_{\Omega})$ are graded Lie algebras, we
may assume that $\overline{u}=u\gamma_{c+1}(F_{n+1})\in I_{c}(N_{{\cal
R}_{3}})\cap I_{c}(F_{\Omega})$ with $u\in\gamma_{c}(F_{n+1})$ and some
$c\in\mathbb{N}$. It is clear enough that $c\geq 3$. Then,
$\overline{u}=u_{1}\gamma_{c+1}(F_{n+1})=u_{2}\gamma_{c+1}(F_{n+1})$, where
$u_{1}\in[F_{n+1}]_{X_{{\cal R}_{3}},c-3}$ and $u_{2}\in
F_{\Omega,c}=F_{\Omega}\cap\gamma_{c}(F_{n+1})$. To get a contradiction, we
assume that $u\notin\gamma_{c+1}(F_{n+1})$. Hence,
$u_{1},u_{2}\in\gamma_{c}(F_{n+1})\setminus\gamma_{c+1}(F_{n+1})$. Since
$F_{\Omega}$ is a free group on the set ${\cal
F}_{\Omega}=\\{t,q_{1},\ldots,q_{n}\\}$, $u_{2}$ is (uniquely) written as
$u_{2}=z^{\mu_{1}}_{i_{1}}\cdots z^{\mu_{\kappa}}_{i_{\kappa}}$, with
$z_{i_{1}},\ldots,z_{i_{\kappa}}\in{\cal F}_{\Omega}$, $i_{\lambda}\neq
i_{\lambda+1}$, $\lambda\in[\kappa-1]$ and
$\mu_{1},\ldots,\mu_{\kappa}\in{\mathbb{Z}}\setminus\\{0\\}$. In the next few
lines, we write
$\overline{t}=t\gamma_{c+1}(F_{n+1}),\overline{x}_{i}=x_{i}\gamma_{c+1}(F_{n+1})$,
$i\in[n]$ and $F_{n+1,c}=F_{n+1}/\gamma_{c+1}(F_{n+1})$. Thus, $F_{n+1,c}$ is
a free nilpotent group of rank $n+1$ and class $c$ with a free generating set
$\\{\overline{t},\overline{x}_{1},\ldots,\overline{x}_{n}\\}$. Note that, for
any $j\in[n]$,
$\overline{q}_{j}=q_{j}\gamma_{c+1}(F_{n+1})=[\overline{t},\overline{x}_{j}]\in
F^{\prime}_{n+1,c}$. Using the identity $ab=ba[a,b]$ as many times as needed
in the above expression of $u_{2}$ and working in $F_{n+1,c}$, the element
$\overline{u}_{2}=u_{2}\gamma_{c+1}(F_{n+1})$ is written in $F_{n+1,c}$ as

$\overline{u}_{2}=\overline{t}^{\alpha}\overline{q}^{\alpha_{1}}_{1}\cdots\overline{q}^{\alpha_{n}}_{n}w_{3}(\overline{t},\overline{q}_{1},\ldots,\overline{q}_{n})\cdots
w_{c}(\overline{t},\overline{q}_{1},\ldots,\overline{q}_{n}),$ $None$

where $\alpha,\alpha_{1},\ldots,\alpha_{n}\in\mathbb{Z}$,
$w_{3}(\overline{t},\overline{q}_{1},\ldots,\overline{q}_{n}),\ldots,w_{c}(\overline{t},\overline{q}_{1},\ldots,\overline{q}_{n})$
are products of group commutators in $\overline{t},$
$\overline{q}_{1},\ldots,\overline{q}_{n}$ and have weights $3,4,\ldots,c$ in
$F_{n+1}$, respectively. Thus, for $j\in[c]\setminus\\{1,2\\}$,
$w_{j}(t,q_{1},\ldots,q_{n})\in
F_{\Omega,j}=F_{\Omega}\cap\gamma_{j}(F_{n+1})$.

Form the (restricted) direct sum ${\rm gr}(F_{n+1,c})$ of the abelian groups
$\gamma_{r}(F_{n+1,c})/\gamma_{r+1}(F_{n+1,c})$. It is well known that ${\rm
gr}(F_{n+1,c})$ is a free nilpotent Lie algebra of rank $n+1$ and class $c$
with a free generating set
$\\{\overline{t}F^{\prime}_{n+1,c},\overline{x}_{1}F^{\prime}_{n+1,c},\ldots,$
$\overline{x}_{n}F^{\prime}_{n+1,c}\\}$. By $(6.11)$, $\overline{u}_{2}$ is
written in ${\rm gr}(F_{n+1,c})$ as

$\begin{array}[]{rll}\overline{u}_{2}&=&\overline{t}^{\alpha}F^{\prime}_{n+1,c}+\overline{q}^{\alpha_{1}}_{1}\cdots\overline{q}^{\alpha_{n}}_{n}\gamma_{3}(F_{n+1,c})+w_{3}(\overline{t},\overline{q}_{1},\ldots,\overline{q}_{n})\gamma_{4}(F_{n+1,c})+\cdots+\\\
&&w_{c-1}(\overline{t},\overline{q}_{1},\ldots,\overline{q}_{n})\gamma_{c}(F_{n+1,c})+w_{c}(\overline{t},\overline{q}_{1},\ldots,\overline{q}_{n}).\end{array}$

Since $u_{2}\in\gamma_{c}(F_{n+1})\setminus\gamma_{c+1}(F_{n+1})$, that is,
$\overline{u}_{2}\in\gamma_{c}(F_{n+1,c})$ (a non-trivial element in
$F_{n+1,c}$), we get
$\overline{u}_{2}=w_{c}(\overline{t},\overline{q}_{1},\ldots,\overline{q}_{n})$.
Therefore, $u_{2}\gamma_{c+1}(F_{n+1})\in M_{2,c}$. By $(6.9)$, $(6.8)$ and
$(6.7)$, we get $u_{2}\gamma_{c+1}(F_{n+1})$ $\in L^{c}_{{\rm
grad},2}(\Omega)$. On the other hand, $u_{1}\gamma_{c+1}(F_{n+1})\in
I_{c}(N_{{\cal R}_{3}})$. By $(6.7)$ and $(6.6)$, we obtain the required
contradiction. Therefore, $L({\cal E}\wr\Omega)\cap{\cal
L}(F_{\Omega})=\\{0\\}$ and so, by $(6.10)$, ${\cal L}(F_{\Omega})=L(\Omega)$.
Since $L(\Omega)$ is free on $\Omega$ and
$\langle\Omega\rangle=I_{1}(F_{\Omega})\oplus I_{2}(F_{\Omega})$, we have
${\cal L}(F_{\Omega})$ is a free Lie algebra on $I_{1}(F_{\Omega})\oplus
I_{2}(F_{\Omega})$. Since both ${\cal L}(F_{\Omega})$ and $L(\Omega)$ are
graded Lie algebras, we have $I_{d}(F_{\Omega})=L^{d}_{\rm grad}(\Omega)$ for
all $d$. $\Box$

Let $N_{{\cal R}_{4}}={\cal R}_{4}^{F_{\Omega}}$ be the normal closure of
${\cal R}_{4}$ in $F_{\Omega}$ and $X_{{\cal R}_{4}}$ be the subgroup of
$F_{\Omega}$ generated by the set ${\cal R}_{4}$. In fact, $X_{{\cal
R}_{4}}\subseteq F^{\prime}_{\Omega_{0}}$. By Lemma 5 (2), $N_{{\cal
R}_{4}}=X_{{\cal R}_{4}}[X_{{\cal R}_{4}},F_{\Omega}]$. Thus, $N_{{\cal
R}_{4}}\subseteq F^{\prime}_{\Omega}\cap\gamma_{4}(F_{n+1})$. For a positive
integer $c$, let $N_{{\cal R}_{4},c}=N_{{\cal R}_{4}}\cap\gamma_{c}(F_{n+1})$.
For $c\in[4]$, $N_{{\cal R}_{4},c}=N_{{\cal R}_{4}}$. For $c\geq 4$, we write
$I_{c}(N_{{\cal R}_{4}})=(N_{{\cal
R}_{4},c}\gamma_{c+1}(F_{n+1}))/\gamma_{c+1}(F_{n+1})$. Form the (restricted)
direct sum ${\cal L}(N_{{\cal R}_{4}})$ of the abelian groups $I_{c}(N_{{\cal
R}_{4}})$. Since $N_{{\cal R}_{4},c}\subseteq F_{{\Omega},c}$ for all $c$, we
have ${\cal L}(N_{{\cal R}_{4}})$ is a Lie subalgebra of ${\cal
L}(F_{\Omega})$. Since $N_{{\cal R}_{4}}$ is a normal subgroup of
$F_{\Omega}$, we obtain ${\cal L}(N_{{\cal R}_{4}})$ is an ideal of ${\cal
L}(F_{\Omega})$. Let $I_{\Omega}$ be the ideal of ${\cal
L}(F_{\Omega})=L(\Omega)$ generated by the set ${\cal
R}_{4,L}=\\{[\omega_{a},\omega_{b}]:(a,b)\in\theta_{n,\delta}\\}$. For
$\mu\geq 1$ and non-negative integers $j(1),\ldots,j(\mu)$, with
$j(1)+\cdots+j(\mu)=c\geq 1$, let $Y_{{\cal R}_{4},(j(1),\ldots,j(\mu))}=$
$[X_{{\cal R}_{4}},Y_{j(1)},\ldots,Y_{j(\mu)}]$, with
$Y_{j(i)}\in\\{X,F_{\Omega_{0}}\\}$, $i\in[\mu]$, $j(1),\ldots,j(\mu)\in[2]$,
and at least one of $Y_{j(1)},\ldots,Y_{j(\mu)}$ equals $X$. Moreover, let
$Y_{{\cal R}_{4},{\rm grad},c}=\prod_{j(1)+\cdots+j(\mu)=c}Y_{{\cal
R}_{4},(j(1),\ldots,j(\mu))}$. We point out that $F_{\Omega}/N_{{\cal R}_{4}}$
is raag.

###### Proposition 8

With the previous notations, ${\rm gr}(F_{\Omega}/N_{{\cal R}_{4}})\cong{\cal
L}(F_{\Omega})/{\cal L}(N_{{\cal R}_{4}})$ as Lie algebras in a natural way
with ${\cal L}(N_{{\cal R}_{4}})=I_{\Omega}$. For all non-negative integers
$c$, $I_{c+4}(N_{{\cal R}_{4}})=I^{c+4}_{\Omega}$. If $c=2\nu+1$, then
$I_{2\nu+5}(N_{{\cal R}_{4}})=Y_{{\cal R}_{4},{\rm
grad},2\nu+1}\gamma_{2\nu+6}(F_{n+1})/\gamma_{2\nu+6}(F_{n+1})$ and, if
$c=2\nu$, then $I_{2\nu+4}(N_{{\cal R}_{4}})=([F_{\Omega_{0}}]_{X_{{\cal
R}_{4}},\nu}~{}Y_{{\cal R}_{4},{\rm
grad},2\nu})\gamma_{2\nu+5}(F_{n+1})/\gamma_{2\nu+5}(F_{n+1})$.

Proof. Since ${\cal R}_{4,L}\subset I_{4}(N_{{\cal R}_{4}})$ and ${\cal
L}(N_{{\cal R}_{4}})$ is an ideal of ${\cal L}(F_{\Omega})$, we have
$I_{\Omega}\subseteq{\cal L}(N_{{\cal R}_{4}})$. Thus, for any positive
integer $d$, with $d\geq 4$, $I^{d}_{\Omega}\subseteq I_{d}(N_{{\cal
R}_{4}})$. As in Section 4.2 (taking $I_{\Omega}$ for $J\cap L(\Omega)$ and
having in mind Proposition 3 (4)), we may show that $I_{\Omega}$ is a direct
summand of ${\cal L}(F_{\Omega})=L(\Omega)$ (by using $(4.13)$, $(4.14)$,
$(4.16)$ and $(4.18)$) and so, ${\cal L}(F_{\Omega})/I_{\Omega}$ is a free
$\mathbb{Z}$-module. By Theorem 4 (for $I_{m}=I_{\Omega}$ and ${\cal
Y}_{m}=\\{s\\}\cup\Omega_{0}$), the $\mathbb{Z}$-module
$I_{\Omega}/[I_{\Omega},I_{\Omega}]$ is a free (right) $U({\cal
L}(F_{\Omega})/I_{\Omega})$-module with a free generating set ${\cal
R}_{4,L}+[I_{\Omega},I_{\Omega}]$. By a result of Labute [13, Theorem 1],
${\rm gr}(F_{\Omega}/N_{{\cal R}_{4}})\cong{\cal L}(F_{\Omega})/I_{\Omega}$ as
Lie algebras. In the proof of Theorem 1 in [13], since
$I_{\Omega}/[I_{\Omega},I_{\Omega}]$ is a free (right) $U({\cal
L}(F_{\Omega})/I_{\Omega})$-module, the induced homomorphism $\xi$ (Lemma 6
for $N=N_{{\cal R}_{4}}$ and $I=I_{\Omega}$) is onto and so, ${\cal
L}(N_{{\cal R}_{4}})=I_{\Omega}$. Hence, for all $d\geq 4$, $I_{d}(N_{{\cal
R}_{4}})=I^{d}_{\Omega}$. For a non-negative integer $c$,
$[F_{\Omega_{0}}]_{X_{{\cal R}_{4},c}}\subseteq N_{{\cal R}_{4},2c+4}=N_{{\cal
R}_{4}}\cap\gamma_{2c+4}(F_{n+1})$ and so,

$([F_{\Omega_{0}}]_{X_{{\cal
R}_{4},c}}\gamma_{2c+5}(F_{n+1}))/\gamma_{2c+5}(F_{n+1})\subseteq
I_{2c+4}(N_{{\cal R}_{4}})=I^{2c+4}_{\Omega}.$ $None$

Since $Y_{{\cal R}_{4},{\rm grad},c}\subseteq N_{{\cal R}_{4},c+4}=N_{{\cal
R}_{4}}\cap\gamma_{c+4}(F_{n+1})$ for all $c\geq 1$, we have

$(Y_{{\cal R}_{4},{\rm
grad},c}\gamma_{c+5}(F_{n+1}))/\gamma_{c+5}(F_{n+1})\subseteq I_{c+4}(N_{{\cal
R}_{4}})=I^{c+4}_{\Omega}.$ $None$

In section 4.2, we gave a decomposition of $I_{\Omega}$ (by $(4.17)$ and
$(4.18)$). In particular, for $d\geq 4$,

$I^{d}_{\Omega}=L^{d}(\Omega^{(1)}_{2})\oplus L^{d}_{\rm grad}({\cal
P}^{\prime}_{{\Omega_{2},1,2}}\wr{\cal
P}^{\prime\prime}_{{\Omega}_{2}})\oplus\cdots\oplus L^{d}_{\rm grad}({\cal
P}^{\prime}_{{\Omega_{n-1},1,2}}\wr{\cal
P}^{\prime\prime}_{{\Omega}_{n-1}})\oplus L^{d}_{\rm grad}({\cal
P}^{\prime}_{{\Omega_{0}}}\wr{\cal P}^{\prime\prime}_{{\Omega}_{0}}),$ $None$

where
$\Omega^{(1)}_{2}=\\{[\omega_{2},\omega_{1},~{}_{a}\omega_{1},~{}_{b}\omega_{2}]:a+b\geq
0\\}$. That is, $\Omega^{(1)}_{2}$ is the standard free generating set for
$\gamma_{2}(L(\Omega_{2}))$. It follows, by $(6.12)$ (for $c=0$) and $(6.14)$
(for $d=4$), that $I_{4}(N_{{\cal R}_{4}})=I^{4}_{\Omega}=\langle{\cal
R}_{4,L}\rangle$. Thus, we may assume that $d\geq 5$. Write $d=c+4$, with
$c\geq 1$. Let $c=2\nu+1$. By $(6.14)$, the definitions of the sets ${\cal
P}^{\prime\prime}_{{\Omega_{2},\omega_{3}}},\ldots,$ ${\cal
P}^{\prime\prime}_{{\Omega_{n-1},\omega_{n}}}$, ${\cal
P}^{\prime}_{{\Omega_{2},1,2}}\wr{\cal
P}^{\prime\prime}_{{\Omega}_{2}},\ldots,$ ${\cal
P}^{\prime}_{{\Omega_{n-1},1,2}}\wr{\cal P}^{\prime\prime}_{{\Omega}_{n-1}}$,
${\cal P}^{\prime}_{{\Omega_{0}}}\wr{\cal P}^{\prime\prime}_{{\Omega}_{0}}$,
Lemma 5 (1) and the Jacobi identity in the form $[x,[y,z]]=[x,y,z]-[x,z,y]$
(where it is needed), we have $I_{2\nu+5}(N_{{\cal R}_{4}})\subseteq Y_{{\cal
R}_{4},{\rm grad},2\nu+1}\gamma_{2\nu+6}(F_{n+1})/\gamma_{2\nu+6}(F_{n+1})$.
By $(6.13)$ (for $c=2\nu+1$), we get $I_{2\nu+5}(N_{{\cal R}_{4}})=Y_{{\cal
R}_{4},{\rm
grad},2\nu+1}~{}\gamma_{2\nu+6}(F_{n+1})/\gamma_{2\nu+6}(F_{n+1})$. Let
$c=2\nu$. By using similar arguments as above and by $(6.12)$ and $(6.13)$, we
obtain $I_{2\nu+4}(N_{{\cal R}_{4}})=([F_{\Omega_{0}}]_{X_{{\cal
R}_{4}},\nu}~{}~{}Y_{{\cal R}_{4},{\rm
grad},2\nu})\gamma_{2\nu+5}(F_{n+1})/\gamma_{2\nu+5}(F_{n+1})$. $\Box$

###  6.4 Proof of Theorem 1

####  6.4.1 A presentation of ${\rm gr}(FP(H))$

For $j\in[4]\setminus\\{1\\}$, let $H_{j}$ be the subgroup of $F_{n+1}$
generated by the set ${\cal R}_{j}$ and $H$ be the subgroup of $F_{n+1}$
generated by the set ${\cal R}$. It follows, by Lemma 5 (for $G=F_{n+1}$,
${\cal G}={\cal F}=\\{x_{1},\ldots,x_{n},t\\}$, ${\cal S}={\cal R}$ and
$N_{\cal S}=N$), that $N=H[H,F_{n+1}]$. For $j\in[4]\setminus\\{1\\}$, let
${\cal H}_{j,n+1}$ be the set
$\\{[r,z^{\varepsilon_{1}}_{1},\ldots,z^{\varepsilon_{\kappa}}_{\kappa}]:r\in{\cal
R}_{j};\kappa\geq 1;\varepsilon_{i}=\pm
1;i\in[\kappa];z_{1},\ldots,z_{\kappa}\in{\cal F}\\}$. By Lemma 5 (2), the set
${\cal R}_{j}\cup{\cal H}_{j,n+1}$ is a generating set for
$H_{j}[H_{j},F_{n+1}]$. In other words, $H_{j}[H_{j},F_{n+1}]$ is the normal
closure of ${\cal R}_{j}$ in $F_{n+1}$.

###### Lemma 7

  1. 1.

Let $N={\cal R}^{F_{n+1}}$ be the normal closure of ${\cal R}$ in $F_{n+1}$.
Then,

$N=H[H_{2},F_{n+1}][H_{3},F_{n+1}][H_{4},F_{n+1}].$

  2. 2.

For $v=[r,z^{\varepsilon_{1}}_{1},\ldots,z^{\varepsilon_{\kappa}}_{\kappa}]$
with $r\in{\cal R}_{2}$, $\kappa\geq 1$, $\varepsilon_{i}=\pm 1$,
$i\in[\kappa]$ and at least one of $z_{1},\ldots,z_{\kappa}$ is $t$, we have
$v\gamma_{\kappa+3}(F_{n+1})\in I_{\kappa+2}(N_{{\cal R}_{3}})$. Otherwise,
$v\gamma_{\kappa+3}(F_{n+1})\in I_{\kappa+2}(N_{{\cal R}_{2}})$.

  3. 3.

For $v=[r,z^{\varepsilon_{1}}_{1},\ldots,z^{\varepsilon_{\kappa}}_{\kappa}]$
with $r\in{\cal R}_{4}$, $\kappa\geq 1$, $\varepsilon_{i}=\pm 1$,
$i\in[\kappa]$ and $z_{1},\ldots,z_{\kappa}\in{\cal F}$, we have
$v\gamma_{\kappa+5}(F_{n+1})\in I_{\kappa+4}(N_{{\cal R}_{4}})\oplus
I_{\kappa+4}(N_{{\cal R}_{3}})$.

Proof.

  1. 1.

It is enough to show that
$[H,F_{n+1}]\subseteq[H_{2},F_{n+1}][H_{3},F_{n+1}][H_{4},F_{n+1}]$. By using
the identities $[ab,c]=[a,c][a,c,b][b,c]$, $[a,bc]=[a,c][a,b][a,b,c]$ and
Lemma 5 (1), we have $N=H[H_{2},F_{n+1}][H_{3},F_{n+1}][H_{4},F_{n+1}]$.

  2. 2.

Clearly,
$v\gamma_{\kappa+3}(F_{n+1})=[r,z_{1},\ldots,z_{\kappa}]^{\varepsilon}\gamma_{\kappa+3}(F_{n+1})$,
with $\varepsilon=\pm 1$. Without loss the generality, we may assume
$v=[r,z_{1},\ldots,z_{\kappa}]$, with $r\in{\cal R}_{2}$ and
$z_{1},\ldots,z_{\kappa}\in{\cal F}$. In the next few lines, we consider the
word $z_{1}\cdots z_{\kappa}$ as the $\kappa$-tuple
$(z_{1},\ldots,z_{\kappa})$. Let $1\leq i_{1}<\cdots<i_{\nu}\leq\kappa$ be the
$\nu$ different positions of $t$ in the $\kappa$-tuple
$(z_{1},\ldots,z_{\kappa})$. By our hypothesis, $\nu\geq 1$. Put
$v(0)=[r,z_{1},\ldots,z_{i_{1}-1}]$ with $z_{1},\ldots,z_{i_{1}-1}\in{\cal
X}$. Note that, for $i_{1}=1$, $v(0)=r$. By Lemma 5 (1)$(e)$ and, working
modulo $\gamma_{\kappa+3}(F_{n+1})$,

$\begin{array}[]{rll}v&=&-[[t,v(0)],z_{i_{1}+1},\ldots,z_{\kappa}]\\\ &&\\\
&=&\sum^{i_{1}-1}_{\lambda=0}(\sum_{\sigma\in
S_{i_{1}-1,\lambda}}(-1)^{\lambda+1}[t,z_{\sigma(1)},\ldots,z_{\sigma(i_{1}-1)},z_{i_{1}+1},\ldots,z_{\kappa}]).\end{array}$

We point out that $r$ occurs in the word
$z_{\sigma(1)}\cdots,z_{\sigma(i_{1}-1)}$. By Lemma 5 (1)$(d)$ (Jacobi
identity), each of the above group commutators belongs to $[F_{n+1}]_{X_{{\cal
R}_{3},\kappa-1}}$. By Proposition 7, we have $v\gamma_{\kappa+3}(F_{n+1})\in
I_{\kappa+2}(N_{{\cal R}_{3}})$. Otherwise, by Proposition 5, we have
$v\gamma_{\kappa+3}(F_{n+1})\in I_{\kappa+2}(N_{{\cal R}_{2}})$.

  3. 3.

We proceed as above. Without loss of generality, we may assume
$v=[r,z_{1},\ldots,z_{\kappa}]$, with $r\in{\cal R}_{4}$ and
$z_{1},\ldots,z_{\kappa}\in{\cal F}$. We separate two cases. First, we assume
that $z_{1},\ldots,z_{\kappa}\in{\cal X}$. Since
$[[t,x_{a}],z],[[t,x_{b}],z]\in{\cal R}_{3}$ for all
$r=[[t,x_{a}],[t,x_{b}]]$, with $(a,b)\in\theta_{n,\delta}$, and $z\in{\cal
X}$, we have $v\gamma_{\kappa+5}(F_{n+1})\in I_{\kappa+4}(N_{{\cal R}_{3}})$.
Thus, we assume that at least one of $z_{1},\ldots,z_{\kappa}$ is $t$. Let
$1\leq i_{1}<\cdots<i_{\nu}\leq\kappa$ be the $\nu$ different positions of $t$
in the $\kappa$-tuple $(z_{1},\ldots,z_{\kappa})$. By our hypothesis, $\nu\geq
1$. If $i_{1}\geq 2$, then, $z_{1}\in{\cal X}$ and so, by using the above
argument, we get $v\gamma_{\kappa+5}(F_{n+1})\in I_{\kappa+4}(N_{{\cal
R}_{3}})$. Therefore, $z_{1}=t$. By working modulo $I_{\kappa+4}(N_{{\cal
R}_{3}})$, using the Jacobi identity in the form $[x,y,z]=[x,z,y]+[x,[y,z]]$
as many times as needed and by Proposition 8, we obtain
$v\gamma_{\kappa+5}(F_{n+1})\in I_{\kappa+4}(N_{{\cal R}_{4}})\oplus
I_{\kappa+4}(N_{{\cal R}_{3}})$. Thus, in any case,
$v\gamma_{\kappa+5}(F_{n+1})\in I_{\kappa+4}(N_{{\cal R}_{4}})\oplus
I_{\kappa+4}(N_{{\cal R}_{3}})$.

###### Proposition 9

With the previous notations,

  1. 1.

${\cal L}(N)=J_{N}$.

  2. 2.

$\rm gr(FP(H))={\rm gr}(F_{n+1}/N)\cong{\rm gr}(F_{n+1})/J_{N}$ as Lie
algebras.

Proof.

  1. 1.

Write ${\cal L}={\cal L}_{\cal Y}(N_{{\cal R}_{2}})+{\cal L}(N_{{\cal
R}_{4}})+{\cal L}(N_{{\cal R}_{3}})$. By Propositions 5, 8, 6 (2) and by
$(6.4)$, we have ${\cal L}$ is the additive direct sum of the Lie subalgebras
${\cal L}_{\cal Y}(N_{{\cal R}_{2}})$, ${\cal L}(N_{{\cal R}_{4}})$ and ${\cal
L}(N_{{\cal R}_{3}})$ of ${\rm gr}(F_{n+1})$, and ${\cal L}=J_{N}$. Hence,
${\cal L}$ is the ideal of ${\rm gr}(F_{n+1})$ generated by the set ${\cal
R}_{L}$. Since ${\cal R}_{L}\subset I_{2}(N)\oplus I_{3}(N)\oplus I_{4}(N)$
and ${\cal L}(N)$ is an ideal of ${\rm gr}(F_{n+1})$, we have
$J_{N}\subseteq{\cal L}(N)$. Suppose that there exists $d\in\mathbb{N}$ such
that $J^{d}_{N}\nsubseteqq I_{d}(N)$ and $J^{s}_{N}=I_{s}(N)$ for all $s\leq
d-1$. It is clear enough that $d\geq 5$. Thus, there exists $u\in
N_{d}=N\cap\gamma_{d}(F_{n+1})$ such that $u\gamma_{d+1}(F_{n+1})\notin
J^{d}_{N}$ and so, $u\notin\gamma_{d+1}(F_{n+1})$. It is easily seen that
$[H_{j},F_{n+1}]\subseteq\gamma_{j+1}(F_{n+1})$ for $j\in[4]\setminus\\{1\\}$.
By Lemma 7 (1), every element $u\in N\setminus\\{1\\}$ is written as
$u=hh_{2}h_{3}h_{4}$ with $h\in H$ and $h_{j}\in[H_{j},F_{n+1}]$,
$j\in[4]\setminus\\{1\\}$. Using the identity $ab=ba[a,b]$ as many times as
needed in the above expression of $u$ and working in $F_{n+1,d}$, the element
$\overline{u}=u\gamma_{d+1}(F_{n+1})$ is written in $F_{n+1,d}$ as a product
$\overline{w}_{2}\overline{w}_{3}\cdots\overline{w}_{\nu}$, where each $w_{j}$
is of weight $j$ in $F_{n+1}$ and it is a product of group commutators in
elements of ${\cal R}\cup(\bigcup_{j\in[4]\setminus\\{1\\}}{\cal H}_{j,n+1})$.
Thus, for example, $\overline{w}_{2}$ is a product of $[x_{a},x_{b}]$,
$(a,b)\in\theta_{n,\delta}$. As in the proof of Proposition 7, $\overline{u}$
is written in ${\rm gr}(F_{n+1,d})$ as
$\overline{u}=\overline{w}_{2}\gamma_{3}(F_{n+1,d})+\overline{w}_{3}\gamma_{4}(F_{n+1,d})+\cdots+\overline{w}_{d-1}\gamma_{d}(F_{n+1,d})+w_{d}$.
Since $u\in\gamma_{d}(F_{n+1})\setminus\gamma_{d+1}(F_{n+1})$, that is,
$\overline{u}\in\gamma_{d}(F_{n+1,d})$ (a non-trivial element in $F_{n+1,d}$),
we get $\overline{u}=\overline{w}_{d}$. Therefore, by working modulo
$\gamma_{d+1}(F_{n+1})$, $u=\prod u_{\mu}$ is a product of group commutators
of weight $d$ in $F_{n+1}$ where in each $u_{\mu}$ at least one element of
${\cal R}$ occurs. We claim that
$\overline{u}=\overline{u}_{2}+\overline{u}_{4}+\overline{u}_{3}$, where
$\overline{u}_{j}\in I_{d}(N_{{\cal R}_{j}})$, $j\in[4]\setminus\\{1\\}$. We
point out that if $r\in{\cal R}_{3}\cup{\cal H}_{3,n+1}$ occurs in $u_{\mu}$
for some $\mu$, then, by Proposition 6, $u_{\mu}\in I_{d}(N_{{\cal R}_{3}})$.
For some $\mu$, let $u_{\mu}=w(a_{1},\ldots,a_{\lambda})\in{\rm
gr}_{d}(F_{n+1})$ with $a_{1},\ldots,a_{\lambda}\in{\cal R}_{2}\cup{\cal
R}_{4}\cup{\cal H}_{2,n+1}\cup{\cal H}_{4,n+1}$. By using the Jacobi identity
as many as needed and by Lemma 7 (2)-(3), we get $u_{\mu}\in I_{d}(N_{{\cal
R}_{4}})\oplus I_{d}(N_{{\cal R}_{3}})$. Therefore,
$\overline{u}=\overline{u}_{2}+\overline{u}_{4}+\overline{u}_{3}$, where
$\overline{u}_{j}\in I_{d}(N_{{\cal R}_{j}})$, $j[4]\setminus\\{1\\}$. By
$(6.4)$, Propositions 5, 6 (2) and 8, we obtain $\overline{u}\in J^{d}_{N}$
which is the required contradiction. Thus, for all $d\geq 2$,
$J^{d}_{N}=I_{d}(N)$ and so, ${\cal L}(N)=J_{N}$.

  2. 2.

Write $G=F_{n+1}/N$. Since $N\subseteq F^{\prime}_{n+1}$, we have
$G/G^{\prime}\cong F_{n+1}/F^{\prime}_{n+1}$ and ${\rm gr}(G)$ is generated as
a Lie algebra by the set ${\cal G}^{\prime}=\\{\alpha_{i},b:i\in[n]\\}$ where
$\alpha_{i}=x_{i}G^{\prime}$, $i\in[n]$, and $b=tG^{\prime}$. Since ${\rm
gr}(F_{n+1})$ is free on the set ${\cal A}=\\{y_{1},\ldots,y_{n},s\\}$, where
$y_{i}=x_{i}F^{\prime}_{n+1}$, $i\in[n]$ and $s=tF^{\prime}_{n+1}$, the map
$\zeta$ from ${\rm gr}(F_{n+1})$ into ${\rm gr}(G)$ satisfying the conditions
$\zeta(y_{i})=\alpha_{i}$, $i\in[n]$, and $\zeta(s)=b$, extends uniquely to a
Lie algebra homomorphism. Since ${\rm gr}(G)$ is generated as a Lie algebra by
the set ${\cal G}^{\prime}$, we have $\zeta$ is onto. Hence, ${\rm
gr}(F_{n+1})/{\rm Ker}\zeta\cong{\rm gr}(G)$ as Lie algebras. By definition,
$J_{N}\subseteq{\rm Ker}\zeta$ and so, $\zeta$ induces a Lie algebra
epimorphism $\overline{\zeta}$ from ${\rm gr}(F_{n+1})/J_{N}$ onto ${\rm
gr}(G)$. In particular, $\overline{\zeta}$ induces $\overline{\zeta}_{d}$,
say, a $\mathbb{Z}$-linear mapping from $({\rm gr}_{d}(F_{n+1})+J_{N})/J_{N}$
onto ${\rm gr}_{d}(G)=\gamma_{d}(G)/\gamma_{d+1}(G)$. For $d\geq 2$,

$\begin{array}[]{rll}{\rm
gr}_{d}(G)&\cong&(\gamma_{d}(F_{n+1})\gamma_{d+1}(F_{n+1})N)/(\gamma_{d+1}(F_{n+1})N)\\\
&&\\\
&\cong&\gamma_{d}(F_{n+1})/(\gamma_{d}(F_{n+1})\cap(\gamma_{d+1}(F_{n+1})N)).\end{array}$

Since $\gamma_{d+1}(F_{n+1})\subseteq\gamma_{d}(F_{n+1})$, we have, by the
modular law,

$\gamma_{d}(F_{n+1})/(\gamma_{d}(F_{n+1})\cap(\gamma_{d+1}(F_{n+1})N))=\gamma_{d}(F_{n+1})/(\gamma_{d+1}(F_{n+1})N_{d})\cong{\rm
gr}_{d}(F_{n+1})/I_{d}(N_{d}).$

For $d\geq 2$, and in the view of Proposition 9 (1), we have
$I_{d}(N)=J^{d}_{N}$ and so, we get ${\rm rank}({\rm gr}_{d}(G))={\rm
rank}({\rm gr}_{d}(F_{n+1})/J^{d}_{N})$. Since $J_{N}=\bigoplus_{d\geq
2}J^{d}_{N}$, we have

$\begin{array}[]{rll}({\rm gr}_{d}(F_{n+1})+J_{N})/J_{N}&\cong&{\rm
gr}_{d}(F_{n+1})/({\rm gr}_{d}(F_{n+1})\cap J_{N})\\\ &&\\\ &=&{\rm
gr}_{d}(F_{n+1})/J^{d}_{N}.\end{array}$

By Theorem 5, we obtain ${\rm Ker}\overline{\zeta}_{d}$ is torsion-free. Since
${\rm rank}({\rm gr}_{d}(G))={\rm rank}({\rm gr}_{d}(F_{n+1})/J^{d}_{N})$, we
have ${\rm Ker}\overline{\zeta}_{d}=\\{1\\}$ and so, $\overline{\zeta}_{d}$ is
an isomorphism. Since $\overline{\zeta}$ is an epimorphism and each
$\overline{\zeta}_{d}$ is isomorphism, we have $\overline{\zeta}$ is a Lie
algebra isomorphism. Hence, we have the required result. $\Box$

###### Corollary 3

With the previous notations,

$\rm gr(FP(H))={\rm gr}(F_{n+1}/N)\cong{\rm gr}(F_{n}/N_{{\cal
R}_{2}})\oplus{\rm gr}(F_{\Omega}/N_{{\cal R}_{4}})$

as $\mathbb{Z}$-modules.

Proof. By Proposition 5, Proposition 8 and by $(6.5)$, we obtain the desired
result. $\Box$

####  6.4.2 $F_{n+1}/N$ is residually nilpotent

Let $\chi_{n}$ be the natural epimorphism from $F_{n+1}$ onto $F_{n}$
satisfying the conditions $\chi_{n}(x_{i})=x_{i}$, $i\in[n]$, and
$\chi_{n}(t)=1$. Note that ${\rm Ker}\chi_{n}=N_{t}$, the normal closure of
$\\{t\\}$ in $F_{n+1}$. By Proposition 4 (1), $F_{n+1}$ is the semidirect
product of $N_{t}$ by $F_{n}$. Let $\pi_{F}$ be the natural epimorphism from
$F_{n}$ onto $F_{n}/N_{{\cal R}_{2}}$. By the definitions of ${\rm
Ker}(\pi_{F}\chi_{n})$ and $N_{t}N_{{\cal R}_{2}}$, we have ${\rm
Ker}(\pi_{F}\chi_{n})=N_{t}N_{{\cal R}_{2}}$. Hence, $F_{n+1}/N_{t}N_{{\cal
R}_{2}}\cong F_{n}/N_{{\cal R}_{2}}$. By a result of Duchamp and Krob [9],
$F_{n}/N_{{\cal R}_{2}}$ is residually nilpotent. For
$j\in[4]\setminus\\{1\\}$, let $N_{{\cal R}_{j},\theta}$ be the normal closure
of ${\cal R}_{j}$ in $F_{n+1}$. By Lemma 7 (1) and Lemma 5 (1)$(a)$, we get
$N=N_{{\cal R}_{2},\theta}N_{{\cal R}_{3},\theta}N_{{\cal R}_{4},\theta}$. We
claim that $N\subseteq N_{t}N_{{\cal R}_{2}}$. Indeed, it is enough to show
that $N_{{\cal R}_{2},\theta}\subseteq N_{t}N_{{\cal R}_{2}}$. Furthermore, it
is enough to show that $w^{-1}[x_{a},x_{b}]w\in N_{t}N_{{\cal R}_{2}}$ for all
$w\in F_{n+1}$ and $(a,b)\in\theta_{n,\delta}$. By writing
$w^{-1}[x_{a},x_{b}]w=[x_{a},x_{b}][[x_{a},x_{b}],w]$ and the definitions of
$N_{t}$ and $N_{{\cal R}_{2}}$, we obtain the required result.

In the next few lines, let $G=F_{n+1}/N$ and $H=N_{t}N_{{\cal R}_{2}}$. Since
$N\subseteq F^{\prime}_{n+1}$, we have $G/G^{\prime}\cong
F_{n+1}/F^{\prime}_{n+1}$. Let $h=wN\in G$ with $w\notin N$. If $w\notin
F^{\prime}_{n+1}$ and since $G/G^{\prime}\cong
F_{n+1}/F^{\prime}_{n+1}\cong\mathbb{Z}^{n+1}$ in a natural way, we get
$h\notin G^{\prime}$ and $G/G^{\prime}$ is nilpotent. Thus, we assume that
$w\in F^{\prime}_{n+1}$ and so, $h\in G^{\prime}$. The element $w$ is written
as a product of group commutators $w(x_{1},\ldots,x_{n},t)$. We separate two
cases.

  1. 1.

Let $w\notin H$. Since $N_{t}$ is normal in $F_{n+1}$, we have $H$ contains
some group commutator $w(x_{1},\ldots,x_{n},t)$, occurring in the product of
$w$, having the letter $t$. Hence, $w~{}{\rm modulo}~{}H$ is a product of
group commutators $w(x_{1},\ldots,x_{n})$. Set $w_{1}=\prod
w(x_{1},\ldots,x_{n})$ and so, $w=w_{1}h_{t}h_{2}$ with $h_{t}\in N_{t}$ and
$h_{2}\in N_{{\cal R}_{2}}$. Note that $wH=w_{1}H$ and $w_{1}\notin H$. Now,
$\pi_{F}\chi_{n}(w)=\pi_{F}(w_{1}h_{2})=w_{1}N_{{\cal R}_{2}}$. Since
$F_{n}/N_{{\cal R}_{2}}$ is residually nilpotent and $w_{1}\notin N_{{\cal
R}_{2}}$ (otherwise, $w\in H$ which is a contradiction), there exists a normal
subgroup $\Lambda_{w}/N_{{\cal R}_{2}}$, with $\Lambda_{w}$ normal in $F_{n}$,
such that $w_{1}\notin\Lambda_{w}$ and $F_{n}/\Lambda_{w}$ is nilpotent. Let
$K_{w}=(\pi_{F}\chi_{n})^{-1}(\Lambda_{w}/N_{{\cal R}_{2}})$. Since
$\pi_{F}\chi_{n}$ is epimorphism, we have $K_{w}$ is a normal subgroup of
$F_{n+1}$ and $w\notin K_{w}$ (otherwise, $w_{1}N_{{\cal
R}_{2}}\in\Lambda_{w}/N_{{\cal R}_{2}}$. Hence, there exists
$w^{\prime}_{1}\in\Lambda_{w}$ such that $w_{1}N_{{\cal
R}_{2}}=w^{\prime}_{1}N_{{\cal R}_{2}}$. Since $N_{{\cal
R}_{2}}\subseteq\Lambda_{w}$, we get a contradiction). Since $H\subseteq
K_{w}$, we get $F_{n+1}/K_{w}\cong F_{n}/\Lambda_{w}$. Since
$F_{n}/\Lambda_{w}$ is nilpotent, we obtain $(F_{n+1}/H)/(K_{w}/H)$ is
nilpotent.

  2. 2.

Let $w\in H$. Thus, $h=wN=w^{\prime}N$ for some $w^{\prime}\in N_{t}$. So,
without loss of generality, we may regard $w\in N_{t}$. Since
$N_{t,d}=N_{t}\cap\gamma_{d}(F_{n+1})\subseteq\gamma_{d}(F_{n+1})$ for all
$d\in\mathbb{N}$ and $F_{n+1}$ is residually nilpotent, we get $\bigcap_{d\geq
1}N_{t,d}=\\{1\\}$. Therefore, there exists $d\in\mathbb{N}$ such that $w\in
N_{t,d}\setminus N_{t,d+1}$. Clearly, we may assume that $d\geq 2$. Let $d=2$.
By Proposition 4 (2), $w\gamma_{3}(F_{n+1})$ is a non-trivial element of
$I_{2}(N_{t})=F_{\Omega_{0}}\gamma_{3}(F_{n+1})/\gamma_{3}(F_{n+1})$. By
Propositions 6 (2) and 8, there exists $w_{1}\in F_{\Omega_{0}}\setminus
F^{\prime}_{\Omega_{0}}$ and $w_{1}\gamma_{3}(F_{n+1})\notin I_{2}(N)$ (that
is, $w_{1}\gamma_{3}(F_{n+1})\neq v_{1}\gamma_{3}(F_{n+1})$ for all $v_{1}\in
N$) such that $w\gamma_{3}(F_{n+1})=w_{1}\gamma_{3}(F_{n+1})$. Suppose that
$wN\in\gamma_{3}(G)$. Since $\gamma_{3}(G)\cong\gamma_{3}(F_{n+1})/N_{3}$ in a
natural way, we have $wN_{3}=hN_{3}$ for some $h\in\gamma_{3}(F_{n+1})$.
Hence, $w\in\gamma_{3}(F_{n+1})$, which is a contradiction. Thus,
$wN\notin\gamma_{3}(G)$ and $G/\gamma_{3}(G)$ is nilpotent. So, we assume that
$d\geq 3$. Clearly, $w\gamma_{d+1}(F_{n+1})$ is a non-trivial element of
$I_{d}(N_{t})$. By $(6.6)$, $I_{d}(N_{t})=L^{d}_{\rm grad}(\Omega)\oplus
I_{d}(N_{{\cal R}_{3}})$. By $(6.4)$, Proposition 6 (2) and Proposition 9 (1),
$I_{d}(N_{{\cal R}_{3}})\subseteq I_{d}(N)$. We point out that $N_{{\cal
R}_{3}}=N_{{\cal R}_{3},\theta}$. By the analysis of $I_{d}(N_{t})$ given in
Section 6.3, Proposition 7, Proposition 8 and Proposition 6 (2), there exists
$w_{2}\in F_{\Omega,d}$ and $w_{2}\gamma_{d+1}(F_{n+1})\notin I_{d}(N)$ (that
is, $w_{2}\gamma_{d+1}(F_{n+1})\neq v\gamma_{d+1}(F_{n+1})$ for all $v\in
N_{d}$) such that $w\gamma_{d+1}(F_{n+1})=w_{2}\gamma_{d+1}(F_{n+1})$. As
before, let $wN\in\gamma_{d+1}(G)$. Since
$\gamma_{d+1}(G)\cong\gamma_{d+1}(F_{n+1})/N_{d+1}$ in a natural way, we have
$wN_{d+1}=h^{\prime}N_{d+1}$ for some $h^{\prime}\in\gamma_{d+1}(F_{n+1})$,
which is a contradiction. Thus, $wN\in\gamma_{d}(G)\setminus\gamma_{d+1}(G)$.
Since $G/\gamma_{d+1}(G)$ is nilpotent, we get the required result. Therefore,
$F_{n+1}/N$ is residually nilpotent. $\Box$

## References

  * [1]
  * [2] S. Andreadakis, E. Raptis and D. Varsos, A characterization of residually finite HNN-extensions of finitely generated abelian groups. Arch. Math. (Basel) 50 (1988), no. 6, 495–501.
  * [3] Yu.A. Bahturin, Identical Relations in Lie Algebras, Nauka, Moscow, 1985 (in Russian). English translation: VNU Science Press, Utrecht, 1987.
  * [4] N. Bourbaki, Lie Groups and Lie Algebras Part I, Hermann, Paris, 1987 (Chapters 1-3).
  * [5] R.M. Bryant and R. Stöhr, Fixed points of automorphisms of free Lie algebras, Arch. Math., 67 (1996), 281–289.
  * [6] R.M. Bryant, L.G. Kovács and R. Stöhr, Invariant bases for free Lie rings, Q. J. Math. 53 (2002), no. 1, 1–17.
  * [7] R. Charney, An introduction to right angled Artin groups, Geom. Dedicata 125 (2007), 141–158.
  * [8] G. Duchamp and D. Krob, The free partially commutative Lie algebra: bases and ranks, Adv. Math. 95 (1992), 92–126.
  * [9] G. Duchamp and D. Krob, The lower central series of the free partially commutative group, Semigroup Forum 45 (1992), 385–394.
  * [10] E. Formanek and C. Procesi, The automorphism group of a free group is not linear, J. Algebra 149 (1992), 494–499.
  * [11] M. Hall, A basis for free Lie rings and higher commutators in free groups, Proc. Amer. Math. Soc., 1 (1950), 575–581.
  * [12] J.P. Labute, On the descending central series of groups with a single defining relation, J. Algebra, 14 (1970), 16–23.
  * [13] J.P. Labute, The determination of the Lie algebra associated to the lower central series of a group, Trans. Amer. Math. Soc., 288 (1985), no. 1, 51–57.
  * [14] M. Lazard, Sur les groupes nilpotents et les anneaux de Lie, Ann. Sci. Ecole Norm. Sup., 71 (3) (1954), 101–190.
  * [15] W. Magnus, A. Karrass and D. Solitar, Combinatorial Group Theory: presentations of groups in terms of generators and relations, Dover Publications, 1978.
  * [16] V. Metaftsis, E. Raptis and D. Varsos, On the linearity of HNN-extensions with abelian base. J. Pure Appl. Algebra, 216 (2012), no. 5, 997–1003.
  * [17] I. Sevaslidou, The associated Lie algebra of an HNN-extension of a group, J. Gen. Lie Theory Appl., accepted.

V. Metaftsis, Department of Mathematics, University of the Aegean, Karlovassi,
832 00 Samos, Greece. e-mail<EMAIL_ADDRESS>

A.I. Papistas, Department of Mathematics, Aristotle University of
Thessaloniki, 541 24 Thessaloniki, Greece. e-mail<EMAIL_ADDRESS>