Proof. By Lemma 2.4, we obtain the following (split) short exact sequence

(2.1) \begin{equation} \{1\} \rightarrow N_{f_{\kappa }} \overset {\varepsilon }{\rightarrow } F_{\kappa } \overset {p}{\rightarrow } F_{\kappa -1} \rightarrow \{1\}, \end{equation}

where $p$ is the epimorphism from $F_{\kappa }$ onto $F_{\kappa -1}$ satisfying the conditions $p(f_{i}) = f_{i}$ , $i \in \{1, \ldots, \kappa -1\}$ , and $p(f_{\kappa }) = 1$ and $\varepsilon$ is the inclusion map from $N_{f_{\kappa }}$ into $F_{\kappa }$ . Let $Y_{\kappa }$ be the subgroup of $F_{\kappa }$ generated by $f_{\kappa }$ . Since $N_{f_{\kappa }} = Y_{\kappa } [Y_{\kappa },F_{\kappa }]$ and $[Y_{\kappa },F_{\kappa }] \subseteq F^{\prime }_{\kappa }$ , we have $\textrm {I}_{1}(N_{f_{\kappa }}) = \langle y_{\kappa } \rangle$ . Clearly, $[Y_{\kappa }, \,_{(c-1)}F_{\kappa }] \subseteq N_{f_{\kappa },c} \subseteq N_{f_{\kappa }}$ for all $c \geq 2$ . Since $N_{f_{\kappa }} \cap F_{\kappa -1} = \{1\}$ , we get $[Y_{\kappa }, \,_{(c-1)}F_{\kappa }] \cap \gamma _{c}(F_{\kappa -1}) = \{1\}$ for all $c \geq 2$ . For $r \geq 1$ , let $G_{1} = F_{\kappa }$ and, for $r \geq 2$ , $G_{r} = [Y_{\kappa }, \,_{(r-1)}F_{\kappa }] \rtimes \gamma _{r}(F_{\kappa -1})$ . Notice that $G_{1} \supseteq G_{2} \supseteq \cdots \supseteq G_{r} \supseteq \cdots$ . Using the group commutator identity $[ab,cd] = [a,d]^{b} [b,d] [a,c]^{bd}[b,c]^{d}$ and Lemma 2.3, we obtain $(G_{r})_{r \geq 1}$ is a strongly central series of $F_{\kappa }$ and so $G_{r} \supseteq \gamma _{r}(F_{\kappa })$ for all $r$ . Obviously, $G_{r} = \gamma _{r}(F_{\kappa })$ for all $r$ . Since $[Y_{\kappa }, \,_{(c-1)}F_{\kappa }] \subseteq N_{f_{\kappa }}$ for all $c \geq 2$ , we have, by the modular law, $N_{f_{\kappa }} \cap G_{c} = [Y_{\kappa }, \,_{(c-1)}F_{\kappa }](N_{f_{\kappa }} \cap \gamma _{c}(F_{\kappa -1}))$ . Since $N_{f_{\kappa }} \cap \gamma _{c}(F_{\kappa -1}) = \{1\}$ and $G_{c} = \gamma _{c}(F_{\kappa })$ for all $c$ , we get $[Y_{\kappa }, \,_{(c-1)}F_{\kappa }] = N_{f_{\kappa },c}$ for all $c \geq 2$ . Hence, $\textrm { I}_{d}(N_{f_{\kappa }}) = [Y_{\kappa }, \,_{(c-1)}F_{\kappa }]\gamma _{c+1}(F_{n+1})/\gamma _{c+1}(F_{n+1})$ for all $c \geq 2$ .

As before, ${\mathcal {L}}_{\textrm {gr}}(F_{\kappa -1}) \cong \textrm {gr}(F_{\kappa -1})$ as Lie algebras in a natural way. By definition of ${\mathcal {L}}_{\textrm {gr}}(F_{\kappa -1})$ and since $L(\overline {\mathcal{F}}_{\kappa -1})$ is a free Lie algebra on $\overline {\mathcal{F}}_{\kappa -1}$ , we obtain ${\mathcal {L}}_{\textrm {gr}}(F_{\kappa -1}) = L(\overline {\mathcal{F}}_{\kappa -1})$ . Let $p_{L}$ be the Lie algebra epimorphism from ${\mathcal{L}}_{\textrm {gr}}(F_{\kappa })$ onto ${\mathcal {L}}_{\textrm { gr}}(F_{\kappa -1})$ satisfying the conditions $p_{L}(y_{i}) = y_{i}$ (for $i \in \{1, \ldots, \kappa -1\}$ ) and $p_{L}(y_{\kappa }) = 0$ . Thus, $\textrm {Ker}p_{L} = L(\{y_{\kappa }\} \wr (\overline {\mathcal{F}}_{\kappa -1})^{*})$ . Hence, we obtain the following short exact sequence of Lie algebras

(2.2) \begin{equation} \{0\} \rightarrow L(\{y_{\kappa }\} \wr (\overline {\mathcal{F}}_{\kappa -1})^{*}) \overset {\varepsilon _{L}}{\longrightarrow } {\mathcal {L}}_{\textrm {gr}}(F_{\kappa }) \overset {p_{L}}{\longrightarrow } {\mathcal {L}}_{\textrm {gr}}(F_{\kappa -1}) \rightarrow \{0\}, \end{equation}

where $\varepsilon _{L}$ is the inclusion map from $L(\{y_{\kappa }\} \wr (\overline {\mathcal{F}}_{\kappa -1})^{*})$ into ${\mathcal {L}}_{\textrm {gr}}(F_{\kappa })$ . Clearly, for all $c \geq 1$ , we get by Eq.(2.2) the following short exact sequence of $\mathbb{Z}$ -modules

(2.3) \begin{equation} \{0\} \rightarrow L^{c}_{\textrm {grad}}(\{y_{\kappa }\} \wr (\overline {\mathcal{F}}_{\kappa -1})^{*}) \overset {\varepsilon _{L,c}}{\longrightarrow } {\mathcal{L}}_{\textrm {gr},c}(F_{\kappa }) \overset {p_{L,c}}{\longrightarrow } {\mathcal{L}}_{\textrm {gr},c}(F_{\kappa -1}) \rightarrow \{0\}, \end{equation}

where $p_{L,c}$ and $\varepsilon _{L,c}$ are the induced group homomorphisms by $p_{L}$ and $\varepsilon _{L}$ , respectively.

For $c \geq 1$ , let $p_{c}$ be the group epimorphism from ${\mathcal{L}}_{\textrm {gr},c}(F_{\kappa })$ onto ${\mathcal{L}}_{\textrm { gr},c}(F_{\kappa -1})$ induced by $p$ (Eq. (2.1)). Further we write $\varepsilon _{c}$ for the group monomorphism from $\textrm {I}_{c}(N_{f_{\kappa }})$ into ${\mathcal{L}}_{\textrm {gr},c}(F_{\kappa })$ induced by $\varepsilon$ (Eq. (2.1)). By definition of $(N_{f_{\kappa },c})_{c \geq 1}$ , we have $\textrm {Im}\varepsilon _{c} = \textrm {Ker}p_{c} = \textrm { I}_{c}(N_{f_{\kappa }})$ for all $c$ , that is, we obtain the following short exact sequence of $\mathbb{Z}$ -modules for all $c$

(2.4) \begin{equation} \{0\} \rightarrow \textrm {I}_{c}(N_{f_{\kappa }}) \overset {\varepsilon _{c}}{\longrightarrow } {\mathcal{L}}_{\textrm { gr},c}(F_{\kappa }) \overset {p_{c}}{\longrightarrow } {\mathcal{L}}_{\textrm {gr},c}(F_{\kappa -1}) \rightarrow \{0\}. \end{equation}

Since $L^{c}_{\textrm {grad}}(\{y_{\kappa }\} \wr (\overline {\mathcal{F}}_{\kappa -1})^{*}) \subseteq \textrm {I}_{c}(N_{f_{\kappa }})$ , we have by Eqs (2.4) and (2.3) that $\textrm {I}_{c}(N_{f_{\kappa }}) = L^{c}_{\textrm { grad}}(\{y_{\kappa }\} \wr (\overline {\mathcal{F}}_{\kappa -1})^{*})$ for all $c$ . Therefore, ${\mathcal {L}}(N_{f_{\kappa }}) = L(\{y_{\kappa }\} \wr (\overline {\mathcal{F}}_{\kappa -1})^{*})$ . By Theorem2.1 and since ${\mathcal {L}}_{\textrm {gr}}(F_{\kappa -1}) = L(\overline {\mathcal{F}}_{\kappa -1})$ , we get ${\mathcal L}_{\textrm {gr}}(F_{\kappa }) = {\mathcal L}_{\textrm {gr}}(F_{\kappa -1}) \oplus {\mathcal L}(N_{f_{\kappa }})$ for all $\kappa \in \{2, \ldots, n+1\}$ .

Proof. Let $J_{1}$ and $J_{2}$ be the ideals of $\textrm {gr}(F_{n+1})$ generated by $ {\mathcal {J}}_{1} = \{[y_{a},y_{b}]\,:\, a \gt b;\, (a,b) \in \theta \}$ and $ {\mathcal {J}}_{2} = \{[\overline {t}, y_{i}, y_{j}], [[\overline {t},y_{a}],[\overline {t},y_{b}]]\,:\, i,j \in \{1,\ldots, n\};\, a \gt b;\, (a,b) \in \theta \}$ , respectively. Clearly, $J = J_{1} + J_{2}$ . Since $L(\{\overline {t}\} \wr (\overline {\mathcal{F}}_{n})^{*})$ is an ideal in $\textrm {gr}(F_{n+1})$ and ${\mathcal {J}}_{2} \subseteq L(\{\overline {t}\} \wr (\overline {\mathcal{F}}_{n})^{*})$ , we conclude that $J_{2} \subseteq L(\{\overline {t}\} \wr (\overline {\mathcal{F}}_{n})^{*})$ and so $J_{2} \subseteq J \cap L(\{\overline {t}\} \wr (\overline {\mathcal{F}}_{n})^{*})$ . But

\begin{equation*} \begin{array}{rll} J \cap L(\{\overline {t}\} \wr (\overline {\mathcal{F}}_{n})^{*}) & = & (J_{1} + J_{2}) \cap L(\{\overline {t}\} \wr (\overline {\mathcal{F}}_{n})^{*}) \\ (\textrm {modular law})& = & (J_{1} \cap L(\{\overline {t}\} \wr (\overline {\mathcal{F}}_{n})^{*})) + J_{2}. \end{array} \end{equation*}

In order to show that $J \cap L(\{\overline {t}\} \wr (\overline {\mathcal{F}}_{n})^{*}) \subseteq J_{2}$ , it is enough to prove that $J_{1} \cap L(\{\overline {t}\} \wr (\overline {\mathcal{F}}_{n})^{*}) \subseteq J_{2}$ . By Lemma 2.2 (and Eq. (2.5)), we have $ J_{1} = (J_{1} \cap \mathcal{L}_{\textrm {gr}}(F_{n})) \oplus (J_{1} \cap L(\{\overline {t}\} \wr (\overline {\mathcal{F}}_{n})^{*})$ . Clearly, $J_{1} \cap L(\{\overline {t}\} \wr (\overline {\mathcal{F}}_{n})^{*})$ is an ideal of $\textrm {gr}(F_{n+1})$ . By a result of Duchamp and Krob [Reference Duchamp and Krob6, Theorem II.7], we have $J_{1} \cap \mathcal{L}_{\textrm {gr}}(F_{n})$ is the ideal of $\mathcal{L}_{\textrm {gr}}(F_{n})$ generated by ${\mathcal {J}}_{1}$ and $J_{1} \cap L(\{\overline {t}\} \wr (\overline {\mathcal{F}}_{n})^{*})$ is the ideal of $L(\{\overline {t}\} \wr (\overline {\mathcal{F}}_{n})^{*})$ generated by ${\mathcal {T}} = \{[\overline {t},z_{1}, \ldots, z_{\mu },[y_{a},y_{b}],z^{\prime }_{1}, \ldots, z^{\prime }_{\nu }]\,:\, z_{1} \cdots z_{\mu }, z^{\prime }_{1} \cdots z^{\prime }_{\nu } \in (\overline {\mathcal{F}}_{n})^{*}, (a,b) \in \theta \}$ . (In the statement of Theorem II.7 of [Reference Duchamp and Krob6], the set $T_{1} = \emptyset$ , since in our case $(n+1,i), (i,n+1) \notin \theta$ for all $i \in \{1, \ldots, n\}$ .) Using the Jacobi identity in the form $[x,[y,z]] = [x,y,z] - [x,z,y]$ , we get ${\mathcal {T}} \subseteq J_{2}$ . Since $J_{2}$ is an ideal in $\textrm {gr}(F_{n+1})$ , we get $J_{1} \cap L(\{\overline {t}\} \wr (\overline {\mathcal{F}}_{n})^{*}) \subseteq J_{2}$ and so $J \cap L(\{\overline {t}\} \wr (\overline {\mathcal{F}}_{n})^{*}) = J_{2}$ . That is, $J \cap L(\{\overline {t}\} \wr (\overline {\mathcal{F}}_{n})^{*})$ is the ideal in $\textrm {gr}(F_{n+1})$ generated by ${\mathcal J}_{2}$ .

Proof.

1. 1. By Eq. (2.5), the definition of $J$ and Lemma 2.2, we have

   (2.9) \begin{equation} J = (J \cap \mathcal{L}_{\textrm {gr}}(F_{n})) \oplus (J \cap L(\{\overline {t}\} \wr (\overline {\mathcal{F}}_{n})^{*})). \end{equation}
   Using Eq. (2.7), we obtain the required decomposition of $J$ .

2. 2. Since $J_{1} \subseteq J$ , we have $J_{1} \cap {\mathcal{L}}_{\textrm {gr}}(F_{n}) \subseteq J \cap {\mathcal{L}}_{\textrm { gr}}(F_{n})$ . By Eq. (2.9), Lemma 2.6 and the definition of $J$ , we get $J \cap {\mathcal{L}}_{\textrm {gr}}(F_{n}) \subseteq J_{1} \cap {\mathcal{L}}_{\textrm {gr}}(F_{n})$ and so $J_{1} \cap {\mathcal{L}}_{\textrm {gr}}(F_{n}) = J \cap {\mathcal{L}}_{\textrm { gr}}(F_{n})$ . By the proof of Lemma 2.6, we obtain the desired result.

3. 3. By Lemma 2.6, we have $J \cap L(\{\overline {t}\} \wr (\overline {\mathcal{F}}_{n})^{*}) \subseteq \bigoplus _{m \geq 3}\textrm { gr}_{m}(F_{n+1})$ . Let $\phi$ be the natural map from $L(\{\overline {t}\} \wr (\overline {\mathcal{F}}_{n})^{*})$ onto $L(\Omega )$ . Thus $\phi (\overline {t}) = \overline {t}$ , $\phi ([\overline {t},y_{i}]) = [\overline {t},y_{i}]$ , $i \in \{1,\ldots, n\}$ , and $\phi ([\overline {t},b_{1}, \ldots, b_{k}]) = 0$ for all $b_{i} \in \overline {{\mathcal F}}_n=\{y_1,\ldots, y_n\}$ , with $k \geq 2$ . Since $\phi$ is a Lie algebra epimorphism, we have $\phi (J \cap L(\{\overline {t}\} \wr (\overline {\mathcal{F}}_{n})^{*}))$ is the ideal of $L(\Omega )$ generated by the set $\{[[\overline {t},y_{a}],[\overline {t},y_{b}]]\,:\, a\gt b;\, (a,b) \in \theta \}$ . It is clear enough that $\phi (J \cap L(\{\overline {t}\} \wr (\overline {\mathcal{F}}_{n})^{*})) \subseteq J \cap L(\Omega )$ . Let $y \in J \cap L(\Omega )$ . There exists $x \in L(\{\overline {t}\} \wr (\overline {\mathcal{F}}_{n})^{*})$ such that $\phi (x) = y$ . By Eq. (2.6), we get $x = u + v$ , where $u \in L(\Omega )$ and $v \in L({\mathcal E} \wr \Omega ^{*})$ . Since $\phi ({\mathcal E} \wr \Omega ^{*}) = 0$ and $L({\mathcal E} \wr \Omega ^{*})$ is a free Lie algebra on ${\mathcal E} \wr \Omega ^{*}$ , we have $\phi (x) = \phi (u) + \phi (v) = u = y$ and so $u \in J \cap L(\Omega )$ . By Eq. (2.7), we conclude that $x \in J \cap L(\{\overline {t}\} \wr (\overline {\mathcal{F}}_{n})^{*})$ . Hence, $y \in \phi (L(\{\overline {t}\} \wr (\overline {\mathcal{F}}_{n})^{*}))$ and so $ \phi (J \cap L(\{\overline {t}\} \wr (\overline {\mathcal{F}}_{n})^{*})) = J \cap L(\Omega )$ . Therefore, the ideal $J \cap L(\Omega )$ of $L(\Omega )$ is generated by the set $\{[[\overline {t},y_{a}],[\overline {t},y_{b}]]\,:\, a \gt b;\, (a,b) \in \theta \}$ .

4. 4. Throughout the proof, we write $S = L({\mathcal E} \wr \Omega ^{*})$ . To show that $S$ is an ideal of $\textrm {gr}(F_{n+1})$ , it is enough to show that $[u,v] \in S$ for all $u \in S$ and $v \in \textrm {gr}(F_{n+1})$ . By Eq. (2.5) and since $S$ is an ideal of $L(\{\overline {t}\} \wr (\overline {\mathcal{F}}_{n})^{*})$ , it is enough to show that $[u,v] \in S$ for all $u \in S$ and $v \in \mathcal{L}_{\textrm {gr}}(F_{n})$ . Since $\mathcal{L}_{\textrm {gr}}(F_{n})$ is generated by the set $\overline {\mathcal{F}}_{n}$ , every element $v$ of $\mathcal{L}_{\textrm {gr}}(F_{n})$ is a $\mathbb{Z}$ -linear combination of Lie commutators of the form $[y_{i_{1}}, \ldots, y_{i_{\kappa }}]$ , $\kappa \geq 1$ , $y_{i_{1}}, \ldots, y_{i_{\kappa }} \in \overline {\mathcal{F}}_{n}$ . Using the Jacobi identity in the form $ [x, [y,z]] = [x,y,z] - [x,z,y]$ , we write each Lie commutator of the form $[u,[y_{i_{1},1}, \ldots, y_{i_{k_{1}},1}], \ldots, [y_{i_{1},\mu }, \ldots, y_{i_{k_{\mu }},\mu }]]$ as a $\mathbb{Z}$ -linear combination of Lie commutators of the form $[u,y_{j_{1},1}, \ldots, y_{j_{k_{1}},1}, \ldots, y_{j_{1},\mu }, \ldots, y_{j_{k_{\mu }},\mu }]$ . By the multi-linearity of Lie commutators, it is enough to show that $[u, y_{i_{1}}, \ldots, y_{i_{\kappa }}] \in S$ for all $u \in S$ and $y_{i_{1}}, \ldots, y_{i_{\kappa }} \in \overline {{\mathcal F}}_n$ , $\kappa \geq 1$ . For a positive integer $\alpha$ , let $[{\mathcal {E}},\,_{\alpha }\Omega ] = [{\mathcal E}, \Omega, \ldots, \Omega ] = \{[v, z_{1}, \ldots, z_{\alpha }]\,:\, v \in {\mathcal {E}}, z_{1}, \ldots, z_{\alpha } \in \Omega \}$ . Since $ {\mathcal E} \wr \Omega ^{*} = {\mathcal E} \cup (\bigcup _{\alpha \geq 1}[{\mathcal {E}},\,_{\alpha }\Omega ])$ and $S$ is a free Lie algebra on ${\mathcal E} \wr \Omega ^{*}$ , it is enough to show that $[w, y_{i_{1}}, \ldots, y_{i_{\kappa }}] \in S$ for all $w \in {\mathcal E} \wr \Omega ^{*}$ and $y_{i_{1}}, \ldots, y_{i_{\kappa }} \in \overline {\mathcal{F}}_{n}$ , $\kappa \geq 1$ . Since ${\mathcal E} \wr \Omega ^{*}$ is a free generating set of $S$ , we may assume that $w \in [{\mathcal {E}},\,_{\alpha }\Omega ]$ with $\alpha \geq 1$ . Since $S$ is an ideal of $L(\{\overline {t}\} \wr (\overline {\mathcal{F}}_{n})^{*})$ and using the Jacobi identity in the form $ [x,y,z] = [x,z,y] + [x,[y,z]]$ on the elements $[v, z_{1}, \ldots, z_{\alpha }]$ with $v \in {\mathcal {E}}$ and $z_{1}, \ldots, z_{\alpha } \in \Omega$ , we may take $w$ to have the form $ w = [u, \,_{m}\overline {t}, \,_{\mu _{1}}[\overline {t},y_{1}], \ldots, \,_{\mu _{n}}[\overline {t},y_{n}]]$ , where $u \in {\mathcal E}$ , $m, \mu _{1}, \ldots, \mu _{n} \geq 0$ and $m + \mu _{1} + \cdots + \mu _{n} \geq 1$ . Using the aforementioned Jacobi identity and since $S$ is an ideal of $L(\{\overline {t}\} \wr (\overline {\mathcal{F}}_{n})^{*})$ , we get $ [w, y_{i_{1}}, \ldots, y_{i_{\kappa }}] = [u, y_{i_{1}}, \ldots, y_{i_{\kappa }}, {}_{m}{\overline {t}}, {}_{\mu _{1}}{[\overline {t},y_{1}]}, \ldots, {}_{\mu _{n}}[{\overline {t},y_{n}]]} + v^{\prime }$ where $v^{\prime } \in S$ . Since $ [u, y_{i_{1}}, \ldots, y_{i_{\kappa }}, {}_{m}{\overline {t}}, {}_{\mu _{1}}{[\overline {t},y_{1}]}, \ldots, {}_{\mu _{n}}{[\overline {t},y_{n}]]} \in S$ , we conclude that $[w, y_{i_{1}}, \ldots, y_{i_{\kappa }}] \in S$ . Therefore, $L({\mathcal E} \wr \Omega ^{*})$ is an ideal of $\textrm { gr}(F_{n+1})$ .

Proof. Since $ {\mathcal E} \wr \Omega ^{*} = \{[\overline {t}, y_{i_{1}}, \ldots, y_{i_{\kappa }}, b_{j_{1}}, \ldots, b_{j_{\lambda }}]\,:\, \kappa \geq 2;\, \lambda \geq 0;\, y_{i_{1}} \cdots y_{i_{\kappa }} \in (\overline {{\mathcal F}}_{n})^{*};\, b_{j_{1}} \cdots b_{j_{\lambda }} \in \Omega ^{*}\}$ , we have the $\mathbb{Z}$ -module $L({\mathcal E} \wr \Omega ^{*})/\gamma _{2}(L({\mathcal E} \wr \Omega ^{*}))$ is free on ${\mathcal E} \wr \Omega ^{*}$ modulo $\gamma _{2}(L({\mathcal E} \wr \Omega ^{*}))$ . For the next few lines, let $e_{ij} = [\overline {t}, y_{i}, y_{j}]$ for all $i, j \in \{1,\ldots, n\}$ and $\widehat {J} = L({\mathcal E} \wr \Omega ^{*})$ . We point out that $e_{ij} = [\omega _{i},y_{j}]$ with $\omega _{i} = [\overline {t},y_{i}]$ for all $i, j \in \{1,\ldots, n\}$ . Let $ {\mathcal{L}}_{\textrm {gr},\widehat {J}}(F_{n}) = \frac {\mathcal{L}_{\textrm {gr}}(F_{n}) + \widehat {J}}{\widehat {J}}$ and $L_{\widehat {J}}(\Omega ) = \frac {L(\Omega ) + \widehat {J}}{\widehat {J}}$ . By Eq. (2.8), we obtain $ \textrm {gr}(F_{n+1})/\widehat {J} = {\mathcal{L}}_{\textrm {gr},\widehat {J}}(F_{n}) \oplus L_{\widehat {J}}(\Omega )$ . Thus, the free $\mathbb{Z}$ -module $\textrm {gr}(F_{n+1})/\widehat {J}$ is a direct sum, as $\mathbb{Z}$ -modules, of its Lie subalgebras ${\mathcal{L}}_{\textrm {gr},\widehat {J}}(F_{n})$ and $L_{\widehat {J}}(\Omega )$ . We point out that ${\mathcal{L}}_{\textrm {gr},\widehat {J}}(F_{n}) \cong \mathcal{L}_{\textrm {gr}}(F_{n})$ and $L_{\widehat {J}}(\Omega ) \cong L(\Omega )$ as Lie algebras in a natural way. For $i \in \{1,\ldots, r\}$ , let $\overline {w}_i = w_i + \widehat {J} \in U(\textrm {gr}(F_{n+1})/\widehat {J}) \setminus \{0\}$ and $w_i \in \textrm {gr}(F_{n+1})$ . Write each $w_{i} = u_{i} + v_{i} + t_{i}$ with $u_{i} \in \mathcal{L}_{\textrm {gr}}(F_{n})$ , $v_{i} \in L(\Omega )$ and $t_{i} \in \widehat {J}$ . Furthermore, we assume that each $u_{i}$ , $v_{i}$ is a simple Lie commutator of degree $\geq 1$ in terms of the elements of the sets $\overline {\mathcal{F}}_{n}$ and $\Omega$ , respectively. Then, for all $i, j \in \{1,\ldots, n\}$ ,

\begin{equation*} \begin{array}{rll} (e_{ij} + \gamma _{2}(\widehat {J})) \overline {w}_{1} \cdots \overline {w}_{r} & = & e_{ij}(\textrm {ad}w_{1}) \cdots (\textrm { ad}w_{r}) + \gamma _{2}(\widehat {J}) \\ & = & [e_{ij}, w_{1}, \ldots, w_{r}] + \gamma _{2}(\widehat {J}) \\ & = & [e_{ij}, u_{1} + v_{1} + t_{1}, \ldots, u_{r} + v_{r} + t_{r}] + \gamma _{2}(\widehat {J}). \end{array} \end{equation*}

By the multi-linearity of the Lie commutator, $e_{ij}, t_{1}, \ldots, t_{r} \in \widehat {J}$ and by using the Jacobi identity in the form $[a,[b,c]] = [a,b,c] - [a,c,b]$ , we can write the element $(e_{ij}+ \gamma _{2}(\widehat {J})) \overline {w}_{1} \cdots \overline {w}_{r}$ as a $\mathbb{Z}$ -linear combination of elements of the form $[e_{ij}, z_{1}, \ldots, z_{\mu }] + \gamma _{2}(\widehat {J})$ , where $z_{1}, \ldots, z_{\mu } \in \overline {\mathcal{F}}_{n} \cup \Omega$ . For a moment, we consider the product $z_{1} \cdots z_{\mu }$ as a $\mu$ -tuple $(z_{1}, \ldots, z_{\mu })$ and assume that, for some $j \in \{1,\ldots, n\}$ , $y_{j}$ occurs in the $k$ -th position ( $k \geq 2$ ) of $(z_{1}, \ldots, z_{\mu })$ . If in the $(k-1)$ -th position of $(z_{1}, \ldots, z_{\mu })$ is $\omega _{\tau }$ , for some $\tau \in \{1,\ldots, n\}$ , then

(4.1) \begin{equation} [e_{ij},z_{1}, \ldots, \omega _{\tau },y_{j}, \ldots, z_{\mu }] + \gamma _{2}(\widehat {J}) = [e_{ij},z_{1}, \ldots, y_{j}, \omega _{\tau }, \ldots, z_{\mu }] + \gamma _{2}(\widehat {J}), \end{equation}

since both $e_{ij}$ and $[\omega _{\tau },y_{j}] = e_{\tau j} \in \widehat {J}$ . If in the $(k-1)$ -position of $(z_{1}, \ldots, z_{\mu })$ there is a $\overline {t}$ , then

(4.2) \begin{equation} [e_{ij},z_{1}, \ldots, \overline {t},y_{j}, \ldots, z_{\mu }] - ([e_{ij},z_{1}, \ldots, y_{j}, \overline {t}, \ldots, z_{\mu }] + [e_{ij},z_{1}, \ldots, \omega _{j}, \ldots, z_{\mu }]) \in \gamma _{2}(\widehat {J}). \end{equation}

Having in mind (4.1) and (4.2), we may write each element $[e_{ij}, z_{1}, \ldots, z_{\mu }] + \gamma _{2}(\widehat {J})$ , where $z_{1}, \ldots, z_{\mu } \in \overline {\mathcal{F}}_{n} \cup \Omega$ , as a $\mathbb{Z}$ -linear combination of elements of the form

\begin{equation*} e_{ij}(\textrm {ad}y_{i_{1}}) \cdots (\textrm {ad}y_{i_{\kappa }})(\textrm {ad}b_{j_{1}}) \cdots (\textrm {ad} b_{j_{\lambda }}) + \gamma _{2}(\widehat {J}) = (e_{ij} + \gamma _{2}(\widehat {J})) \overline {y}_{i_{1}} \cdots \overline {y}_{i_{\kappa }} \overline {b}_{j_{1}} \cdots \overline {b}_{j_{\lambda }}, \end{equation*}

where $y_{i_{1}} \cdots y_{i_{\kappa }} \in (\overline {\mathcal{F}}_{n})^{*}$ and $ b_{j_{1}} \cdots b_{j_{\lambda }} \in \Omega ^{*}$ . Hence, since the set ${\mathcal E} \wr \Omega ^{*}$ is a free generating set for $\widehat {J}$ , we have the required result.

Proof. By Proposition 2.7 (4), $L({\mathcal E} \wr \Omega ^{*})$ is the ideal of $\textrm {gr}(F_{n+1})$ generated by ${\mathcal{R}}_{3,L} = \{[\overline {t},y_{i},y_{j}]\,:\, i, j \in \{1, \ldots, n\}\}$ . Since ${\mathcal R}_{3,L} \subset \textrm {I}_{3}(M_{{\mathcal R}_{3}})$ and ${\mathcal L}(M_{{\mathcal R}_{3}})$ is an ideal of $\textrm { gr}(F_{n+1})$ , we have $L({\mathcal E} \wr \Omega ^{*}) \subseteq {\mathcal L}(M_{{\mathcal R}_{3}})$ . By Eq. (2.8), $L({\mathcal E} \wr \Omega ^{*})$ is a direct summand of $\textrm {gr}(F_{n+1})$ and so $\textrm {gr}(F_{n+1})/L({\mathcal E} \wr \Omega ^{*})$ is a free $\mathbb{Z}$ -module. By Lemma 4.1, $L({\mathcal E} \wr \Omega ^{*})/\gamma _{2}(L({\mathcal E} \wr \Omega ^{*}))$ is a free (right) $U(\textrm { gr}(F_{n+1})/L({\mathcal E} \wr \Omega ^{*}))$ -module with a free generating set ${\mathcal R}_{3,L}+\gamma _{2}(L({\mathcal E} \wr \Omega ^{*}))$ . By a result of Labute [Reference Labute13, Theorem 1], $ \textrm {gr}(F_{n+1}/M_{{\mathcal R}_{3}}) \cong \textrm {gr}(F_{n+1})/L({\mathcal E} \wr \Omega ^{*})$ as Lie algebras. Since $L({\mathcal E} \wr \Omega ^{*}) = \bigoplus _{c \geq 1}(L({\mathcal E} \wr \Omega ^{*}) \cap \textrm {gr}_{c}(F_{n+1}))$ and $\textrm { gr}(F_{n+1}) = \bigoplus _{c \geq 1}\textrm {gr}_{c}(F_{n+1})$ , we have $\textrm {gr}(F_{n+1})/L({\mathcal E} \wr \Omega ^{*}) = \bigoplus _{c \geq 1}(\frac {\textrm {gr}_{c}(F_{n+1})+L({\mathcal E} \wr \Omega ^{*})}{L({\mathcal E} \wr \Omega ^{*})})$ . Since $\textrm {gr}(F_{n+1})/L({\mathcal E} \wr \Omega ^{*}) \cong \textrm {gr}(F_{n+1}/M_{{\mathcal{R}_{3}}})$ , we get, for $c \geq 1$ , $\frac {\textrm {gr}_{c}(F_{n+1})+L({\mathcal E} \wr \Omega ^{*})}{L({\mathcal E} \wr \Omega ^{*})} \cong \textrm { gr}_{c}(F_{n+1}/M_{\mathcal{R}_{3}})$ as $\mathbb{Z}$ -modules. By Eq. (2.8), $\textrm {gr}(F_{n+1})/L({\mathcal E} \wr \Omega ^{*})$ is a free $\mathbb{Z}$ -module and so, for $c \geq 1$ , $\textrm { gr}_{c}(F_{n+1}/M_{\mathcal{R}_{3}})$ is a free abelian group of finite rank. As in the proof of Proposition 3.3, for $c \geq 1$ , $ \textrm {gr}_{c}(F_{n+1}/M_{\mathcal{R}_{3}}) \cong \textrm {gr}_{c}(F_{n+1})/\textrm {I}_{c}(M_{\mathcal{R}_{3}})$ in a natural way. For $c \geq 1$ ,

\begin{equation*} \frac {\textrm {gr}_{c}(F_{n+1})+L({\mathcal E} \wr \Omega ^{*})}{L({\mathcal E} \wr \Omega ^{*})} \cong \frac {\textrm {gr}_{c}(F_{n+1})}{\textrm {gr}_{c}(F_{n+1}) \cap L({\mathcal E} \wr \Omega ^{*})} \cong \frac {\textrm { gr}_{c}(F_{n+1})}{\textrm {I}_{c}(M_{\mathcal{R}_{3}})} \end{equation*}

as free $\mathbb{Z}$ -modules. Since $\textrm {gr}_{c}(F_{n+1}) \cap L({\mathcal E} \wr \Omega ^{*}) \subseteq \textrm { I}_{c}(M_{\mathcal{R}_{3}}) \subseteq \textrm {gr}_{c}(F_{n+1})$ for all $c \geq 2$ , we obtain $\textrm {gr}_{c}(F_{n+1}) \cap L({\mathcal E} \wr \Omega ^{*}) = \textrm {I}_{c}(M_{\mathcal{R}_{3}})$ for all $c \geq 2$ . Hence, we obtain ${\mathcal L}(M_{{\mathcal R}_{3}}) = L({\mathcal E} \wr \Omega ^{*})$ .

Proof. Recall that $N_{t}$ is the normal closure of $\{t\}$ in $F_{n+1}$ . Since $N_{t} \cap M_{\mathcal{R}_{2}} = \{1\}$ , we have $M_{\mathcal{R}_{2}} \cap \gamma _{d}(F_{n+1}) = M_{\mathcal{R}_{2}} \cap \gamma _{d}(F_{n})$ for all $d \geq 1$ . Hence, for $d \geq 1$ , $\textrm {I}_{d}(M_{\mathcal{R}_{2}}) = \frac {(M_{\mathcal{R}_{2}} \cap \gamma _{d}(F_{n})) \gamma _{d+1}(F_{n+1})}{\gamma _{d+1}(F_{n+1})} \leq {\mathcal{L}}_{\textrm {gr},d}(F_{n})$ . Since $M_{\mathcal{R}_{2}}$ is normal in $F_{n}$ , we obtain ${\mathcal{L}}(M_{\mathcal{R}_{2}}) = \bigoplus _{d \geq 2}\textrm {I}_{d}(M_{\mathcal{R}_{2}})$ is an ideal in ${\mathcal{L}}_{\textrm {gr}}(F_{n})$ . By Proposition 2.7 (2), $J \cap {\mathcal{L}}_{\textrm {gr}}(F_{n}) \subseteq {\mathcal{L}}(M_{\mathcal{R}_{2}})$ . By Proposition 3.3 (for $N = M_{\mathcal{R}_{2}}$ , $I = J \cap {\mathcal{L}}_{\textrm { gr}}(F_{n})$ and ${\mathcal{L}}_{\textrm {gr}}(F_{n}) = \textrm {gr}(F_{n})$ ), we get ${\mathcal{L}}(M_{\mathcal{R}_{2}}) = J \cap {\mathcal{L}}_{\textrm {gr}}(F_{n})$ .

For $i \in \{1,\ldots, n\}$ , let $q_{i} = [t,f_{i}]$ , $\omega _{i} = q_{i}\gamma _{3}(F_{n+1}) = [\overline {t},y_{i}]$ and $\Omega _{0} = \{\omega _{1}, \ldots, \omega _{n}\}$ . Thus, $\Omega _{0} \subset \textrm {gr}_{2}(F_{n+1})$ and $\Omega = \{\overline {t}\} \cup \Omega _{0}$ . By Theorem2.1, $L(\Omega ) = L(\Omega _{0}) \oplus L(\{\overline {t}\} \wr \Omega ^{*}_{0})$ . We write $F_{Q}$ for the subgroup of $F_{n+1}$ generated by $Q = \{q_{1}, \ldots, q_{n}, t\}$ . It is clear enough that $F_{Q}$ is a free group of rank $n+1$ . In particular $F_{n+1}$ is obviously isomorphic to $F_{Q}$ .

Let $F_{Q_{0}}$ be the subgroup of $F_{Q}$ generated by $Q_{0} = \{q_{1}, \ldots, q_{n}\}$ . Clearly, $F_{Q_{0}}$ is free on $Q_{0}$ . We point out that $F_{Q}$ is the semidirect product of $N_{Q,t}$ by $F_{Q_{0}}$ where $N_{Q,t}$ is the normal closure of $\{t\}$ in $F_{Q}$ . For $c \geq 1$ , let $ {\mathcal L}_{\textrm {gr},c}(F_{Q_{0}}) = \gamma _{c}(F_{Q_{0}})\gamma _{2c+1}(F_{n+1})/\gamma _{2c+1}(F_{n+1})$ and form the (restricted) direct sum ${\mathcal L}_{\textrm {gr}}(F_{Q_{0}})$ of the abelian groups ${\mathcal L}_{\textrm { gr},c}(F_{Q_{0}})$ . Since $[\gamma _{\kappa }(F_{Q_{0}}), \gamma _{\lambda }(F_{Q_{0}})] \subseteq \gamma _{\kappa +\lambda }(F_{Q_{0}})$ for all $\kappa, \lambda$ , we have ${\mathcal L}_{\textrm {gr}}(F_{Q_{0}})$ is a Lie subalgebra of $\textrm {gr}(F_{n+1})$ . Notice that, for $c \geq 1$ , ${\mathcal L}_{\textrm {gr},c}(F_{Q_{0}}) = L^{c}({\Omega _{0}})$ and so ${\mathcal L}_{\textrm {gr}}(F_{Q_{0}}) = L({\Omega _{0}})$ . Therefore, $L(\Omega ) = {\mathcal L}_{\textrm {gr}}(F_{Q_{0}}) \oplus L(\{\overline {t}\} \wr \Omega ^{*}_{0})$ . For $c \geq 1$ , let $F_{Q,c} = F_{Q} \cap \gamma _{c}(F_{n+1})$ . Form the (restricted) direct sum ${\mathcal L}(F_{Q})$ of the abelian groups $\textrm {I}_{c}(F_{Q}) = F_{{Q},c}\gamma _{c+1}(F_{n+1})/\gamma _{c+1}(F_{n+1})$ . Clearly, ${\mathcal L}(F_{Q})$ is a Lie subalgebra of $\textrm {gr}(F_{n+1})$ .

Proof. Since $F_{Q}$ is the semidirect product of $N_{Q,t}$ by $F_{Q_{0}}$ and $F_{{Q_{0}}} \subseteq F^{\prime }_{n+1}$ , we have $\textrm {I}_{1}(F_{Q}) = \textrm {I}_{1}(N_t)$ and $\textrm {I}_{2}(F_{Q}) = \textrm {I}_{2}(N_t) = {\mathcal{L}}_{\textrm { gr},1}(F_{Q_{0}})$ . Notice that $L(\Omega ) = L(\textrm {I}_{1}(N_t) \oplus \textrm {I}_{2}(N_t))$ . Since ${\mathcal L}(F_{Q})$ is a Lie subalgebra of $\textrm {gr}(F_{n+1})$ and $L(\Omega )$ is free on $\textrm {I}_{1}(N_t) \oplus \textrm {I}_{2}(N_t)$ , we have $L(\Omega ) \subseteq {\mathcal L}(F_{Q})$ . Clearly, ${\mathcal L}(F_{Q}) \subseteq {\mathcal L}(N_t)$ . By Eq. (2.6) and the modular law, we get

(4.3) \begin{equation} {\mathcal L}(F_{Q}) = L(\Omega ) \oplus (L({\mathcal E} \wr \Omega ^{*}) \cap {\mathcal L}(F_{Q})). \end{equation}

We claim that $L({\mathcal E} \wr \Omega ^{*}) \cap {\mathcal L}(F_{Q}) = \{0\}$ . Let $\overline {u} \in L({\mathcal E} \wr \Omega ^{*}) \cap {\mathcal L}(F_{Q})$ . By Proposition 4.2, ${\mathcal L}(M_{{\mathcal R}_{3}}) = L({\mathcal E} \wr \Omega ^{*})$ . Since both ${\mathcal L}(M_{{\mathcal R}_{3}})$ and ${\mathcal L}(F_{Q})$ are graded Lie algebras, we may assume that $\overline {u} = u \gamma _{c+1}(F_{n+1}) \in \textrm {I}_{c}(M_{{\mathcal R}_{3}}) \cap \textrm {I}_{c}(F_{Q})$ with $u \in \gamma _{c}(F_{n+1})$ and some $c \geq 1$ . 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 $u_{1} \in M_{{\mathcal{R}}_{3}} \cap \gamma _{c}(F_{n+1})$ and $u_{2} \in F_{Q,c} = F_{Q}\cap \gamma _c(F_{n+1})$ . 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 $F_{Q}$ is a free group on $Q = \{q_{1}, \ldots, q_{n}, t\}$ , $u_{2}$ is (uniquely) written as $u_{2} = z^{\mu _{1}}_{i_{1}} \cdots z^{\mu _{\kappa }}_{i_{\kappa }}$ , with $z_{i_{1}}, \ldots, z_{i_{\kappa }} \in Q$ , $i_{\lambda } \neq i_{\lambda + 1}$ , $\lambda \in \{1,\ldots, \kappa - 1\}$ and $\mu _{1}, \ldots, \mu _{\kappa } \in {\mathbb{Z}} \setminus \{0\}$ . Notice that the elements of $Q$ are ordered as $t \lt q_{1} \lt \cdots \lt q_{n}$ . In the next few lines, we write $\widehat {f}_{i} = f_{i}\gamma _{c+1}(F_{n+1})$ , $i \in \{1, \ldots, n\}$ and $\widehat {t}=t\gamma _{c+1}(F_{n+1})$ . Thus, $\{\widehat {f}_{1}, \ldots, \widehat {f}_{n},\widehat {t}\}$ is a free generating set of $F_{n+1,c} = F_{n+1}/\gamma _{c+1}(F_{n+1})$ . Observe that, for any $j \in \{1,\ldots, n\}$ , $\widehat {q}_{j} = q_{j}\gamma _{c+1}(F_{n+1}) = [\widehat {t}, \widehat {f}_{j}] \in \gamma _{2}(F_{n+1,c})$ . Using the collection process for arbitrary group elements on $u_{2} = z^{\mu _{1}}_{i_{1}} \cdots z^{\mu _{\kappa }}_{i_{\kappa }}$ (see [Reference Clement, Majewicz and Zyman4, Chapter 3, Theorem 3.1 and Theorem 3.5]), the element $\widehat {u}_{2} = u_{2}\gamma _{c+1}(F_{n+1})$ is written in $F_{n+1,c}$ as

\begin{equation*} \widehat {u}_{2} = (\widehat {t})^{\alpha }(\widehat {q}_{1})^{\alpha _{1}} \cdots (\widehat {q}_{n})^{\alpha _{n}} w_{3}(\widehat {t}, \widehat {q}_{1}, \ldots, \widehat {q}_{n}) \cdots w_{c}(\widehat {t}, \widehat {q}_{1}, \ldots, \widehat {q}_{n}), \end{equation*}

where $\alpha, \alpha _{1}, \ldots, \alpha _{n}$ are integers, $w_{3}(t,q_{1}, \ldots, q_{n}), \ldots, w_{c}(t,q_{1}, \ldots, q_{n})$ are products of basic group commutators in $t$ , $f_{1}, \ldots, f_{n}$ and have weights $3, 4, \ldots, c$ in $F_{n+1}$ , respectively. Thus, for $j \in \{3,\ldots, c\}$ , $w_{j}(t, q_{1}, \ldots, q_{n}) \in F_{Q,j}$ . Since $u_{2} \in \gamma _{c}(F_{n+1}) \setminus \gamma _{c+1}(F_{n+1})$ and the center of $F_{n+1,c}$ equals $\gamma _{c}(F_{n+1,c})$ , we get $\widehat {u}_{2} = w_{c}(\widehat {t}, \widehat {q}_{1}, \ldots, \widehat {q}_{n})$ . Therefore, $u_{2}\gamma _{c+1}(F_{n+1}) \in L^{c}_{\textrm {grad}}(\Omega )$ . On the other hand, $u_{2}\gamma _{c+1}(F_{n+1}) = u_{1}\gamma _{c+1}(F_{n+1}) \in \textrm {I}_{c}(M_{{\mathcal R}_{3}}) = L^{c}_{\textrm { grad}}(\mathcal{E} \wr \Omega ^{*})$ , which contradicts to the fact $L(\Omega ) \cap L(\mathcal{E} \wr \Omega ^{*}) = \{0\}$ (Eq. (2.8)). Therefore, $L({\mathcal E} \wr \Omega ^{*}) \cap {\mathcal L}(F_{Q}) = \{0\}$ and so, by Eq. (4.3), ${\mathcal L}(F_{Q}) = L(\Omega )$ .

Proof. Let $\psi$ be the Lie algebra isomorphism from $\textrm {gr}(F_{n+1})$ onto ${\mathcal{L}}(F_{Q})$ satisfying the conditions $\psi (\overline {t}) = \overline {t}$ and $\psi (y_{i}) = \omega _{i} = [\overline {t},y_{i}]$ , $i \in \{1, \ldots, n\}$ . Let $J_{1}$ be the ideal of $\textrm { gr}(F_{n+1})$ generated by ${\mathcal{J}}_{1} = \{[y_{a},y_{b}]\,:\, a \gt b;\, (a,b) \in \theta \}$ . Since $\psi$ is a Lie algebra isomorphism and Proposition 2.7 (3), we have $\psi (J_{1}) = J \cap L(\Omega )$ . By Corollary 3.2 (for $n+1$ ), we get

\begin{equation*} \textrm {gr}(F_{n+1}) = \langle \overline {\mathcal{F}}_{n+1}\rangle \oplus (\bigoplus _{j=3}^{n+1}L((\mathcal{T}_{3,j} \setminus \{y_{j}\}) \wr \{y_{j}\}^{*})) \oplus J_{1}, \end{equation*}

where $\langle \overline {\mathcal{F}}_{n+1}\rangle$ is the $\mathbb{Z}$ -module spanned by $\overline {\mathcal{F}}_{n+1}$ and $J_{1} = \bigoplus _{j=2}^{n}L((\mathcal{T}_{1,j} \cup (\mathcal{T}_{2,j}) \wr (\mathcal{T}_{3,j}^{*}) \oplus L(\mathcal{T}_{2,n+1} \wr (\mathcal{T}_{3,n+1}^{*})$ . Notice that since $(n+1,i), (i,n+1) \notin \theta$ for $i \in \{1, \ldots, n\}$ , we have $\mathcal{T}_{1,n+1} = \emptyset$ . Since $\psi$ is a Lie algebra isomorphism, we obtain

\begin{equation*} {\mathcal{L}}(F_{Q}) = \langle \overline {\mathcal{Q}} \rangle \oplus (\bigoplus _{j=3}^{n+1}L((\psi (\mathcal{T}_{3,j}) \setminus \{\omega _{j}\}) \wr \{\omega _{j}\}^{*})) \oplus \psi (J_{1}), \end{equation*}

where $\langle \overline {\mathcal{Q}}\rangle$ is the $\mathbb{Z}$ -module spanned by $\overline {\mathcal{Q}} = \{\overline {t}, \omega _{1}, \ldots, \omega _{n}\}$ . Since $\psi (J_{1}) = J \cap L(\Omega )$ , we obtain the desired result.

Proof. For $c \geq 1$ , let $M_{{\mathcal R}_{4},c} = M_{{\mathcal R}_{4}} \cap \gamma _{c}(F_{n+1})$ . Clearly, $M_{{\mathcal R}_{4},c} = M_{{\mathcal R}_{4}}$ for $c \in \{1,2,3,4\}$ . For $c \geq 4$ , let $\textrm {I}_{c}(M_{{\mathcal R}_{4}}) = (M_{{\mathcal R}_{4},c} \gamma _{c+1}(F_{n+1}))/\gamma _{c+1}(F_{n+1})$ and ${\mathcal L}(M_{{\mathcal R}_{4}}) = \bigoplus _{c \geq 4}\textrm { I}_{c}(M_{{\mathcal R}_{4}})$ . Since $M_{{\mathcal R}_{4},c} \subseteq F_{{Q},c}$ for all $c$ , we have ${\mathcal L}(M_{{\mathcal R}_{4}})$ is a Lie subalgebra of ${\mathcal L}(F_{Q})$ . By Lemma 4.4, ${\mathcal {L}}(F_{Q}) = L(\Omega )$ . Since $M_{{\mathcal R}_{4}}$ is a normal subgroup of $F_{Q}$ , we obtain ${\mathcal L}(M_{{\mathcal R}_{4}})$ is an ideal of ${\mathcal L}(F_{Q})$ . By Proposition 2.7 (3), $J \cap L(\Omega )$ is the ideal of ${\mathcal L}(F_{Q})$ generated by ${\mathcal R}_{4,L} = \{[\omega _{a},\omega _{b}]\,:\, a \gt b;\, (a,b) \in \theta \}$ and so $J \cap L(\Omega ) = \bigoplus _{d \geq 4} (J \cap {\mathcal L}_{\textrm { gr},c}(F_{Q_{0}}))$ . By Lemma 4.5, ${\mathcal{L}}(F_{Q})/(J \cap L(\Omega ))$ is a free $\mathbb{Z}$ -module. By the proof of Proposition 3.3 (for $\textrm {gr}(F_{n}) = {\mathcal{L}}(F_{Q})$ , $N = M_{{\mathcal {R}}_{4}}$ and $I = J \cap L(\Omega )$ ), we obtain ${\mathcal L}(M_{{\mathcal R}_{4}}) = J \cap L(\Omega )$ .

Proof.

1. 1. Throughout this proof, we write $\widetilde {R} = R [R_{2}, F_{n+1}] [R_{3}, F_{n+1}] [R_{4}, F_{n+1}]$ . Since each $[R_{j}, F_{n+1}]$ is normal in $F_{n+1}$ , we have each $[v, h_{1}, \ldots, h_{\lambda }] \in [R_{j}, F_{n+1}]$ for all $v \in {\mathcal {R}}_{j}$ and $h_{1}, \ldots, h_{\lambda } \in F_{n+1}$ . Since $M = R\,[R,F_{n+1}]$ , it is enough to show that $[R, F_{n+1}] \subseteq \widetilde {R}$ . In particular, it is enough to show that any $[u,f] \in \widetilde {R}$ for $u \in R$ and $f \in F_{n+1} \setminus \{1\}$ . Write $u = u_{i_{1}}^{\varepsilon _{1}} \cdots u_{i_{\kappa }}^{\varepsilon _{\kappa }}$ with $\kappa \geq 1$ , $\varepsilon _{j} = \pm 1$ , $j \in \{1, \ldots, \kappa \}$ and $u_{i_{1}}, \ldots, u_{i_{\kappa }} \in {\mathcal {R}}$ . Using an inductive argument on $\kappa$ and the following identities

   1. (a) $[ab,c] = [a,c][a,c,b][b,c]$ ,

   2. (b) $[a,bc] = [a,c][a,b][a,b,c]$ ,

   3. (c) $[a^{-1},b] = [a,b,a][a,b,a,a^{-1}][a,b]^{-1}$ ,

   4. (d) $[a,b^{-1}] = [a,b,b][a,b,b,b^{-1}][a,b]^{-1}$

   we write $[u,f]$ as a product of elements of the form $[v,h_{1}, \ldots, h_{m}]^{\varepsilon }$ , with $m \geq 1$ , $\varepsilon = \pm 1$ , $v \in {\mathcal {R}}$ and $h_{1}, \ldots, h_{m} \in F_{n+1}$ . Since each $[R_{j}, F_{n+1}]$ is normal in $F_{n+1}$ , we obtain the required claim. Hence, $M = R [R_{2}, F_{n+1}] [R_{3}, F_{n+1} [R_{4}, F_{n+1}]$ .

2. 2. Clearly, $v \gamma _{\kappa + 3}(F_{n+1}) = [r, z_{1}, \ldots, z_{\kappa }]^{\varepsilon }\gamma _{\kappa + 3}(F_{n+1})$ with $\varepsilon = \pm 1$ and $z_{1}, \ldots, z_{\kappa } \in {\mathcal F}_{n+1}$ . Without loss of generality, we may assume $v = [r, z_{1}, \ldots, z_{\kappa }]$ , with $r \in {\mathcal R}_{2}$ and $z_{1}, \ldots, z_{\kappa } \in {\mathcal F}_{n+1}$ . 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} \lt \cdots \lt i_{\nu } \leq \kappa$ be the $\nu$ different appearances of $t$ in the $\kappa$ -tuple $(z_{1}, \ldots, z_{\kappa })$ . By our hypothesis $\nu \geq 1$ . Set $v(0) = [r,z_{1}, \ldots, z_{i_{1}-1}]$ with $z_{1}, \ldots, z_{i_{1}-1} \in {\mathcal F}_{n}$ . For $i_{1} = 1$ , $v(0) = r$ . Write $g_{1} = r$ , $g_{2} = z_{1}, \ldots, $ $g_{i_{1}} = z_{i_{1}-1}$ . By Lemma 4.7 (2) and working modulo $\gamma _{\kappa + 3}(F_{n+1})$

   \begin{equation*} \begin{array}{rll} v & = & -[[t, v(0)], z_{i_{1}+1}, \ldots, z_{\kappa }] \\ & = & \sum ^{i_{1}}_{\lambda = 0}(\sum _{\sigma \in S_{i_{1},\lambda }}(-1)^{\lambda + 1}[t, g_{\sigma (1)}, \ldots, g_{\sigma (i_{1})}, z_{i_{1}+1}, \ldots, z_{\kappa }]). \end{array} \end{equation*}
   We point out that $r$ occurs in the word $g_{\sigma (1)} \cdots g_{\sigma (i_{1})}$ . By Lemma 4.7 (1) (Jacobi identity), each of the above group commutator belongs to $[R_{3}, F_{n+1}, \ldots, F_{n+1}]$ with $\kappa - 1$ factors of $F_{n+1}$ . By Proposition 4.2, $v \gamma _{\kappa + 3}(F_{n+1}) \in \textrm {I}_{\kappa + 2}(M_{{\mathcal R}_{3}})$ . If there are no appearances of $t$ then, by Proposition 4.3, $v\gamma _{\kappa + 3}(F_{n+1}) \in \textrm {I}_{\kappa + 2}(M_{{\mathcal R}_{2}})$ .

3. 3. We proceed as above. Without loss of generality, we may assume $v = [r, z_{1}, \ldots, z_{\kappa }]$ , with $r \in {\mathcal R}_{4}$ and $z_{1}, \ldots, z_{\kappa } \in {\mathcal F}_{n+1}$ . We separate two cases. First, we assume that $z_{1}, \ldots, z_{\kappa } \in {\mathcal F}_{n}$ . Let $r = [[t,f_{a}], [t,f_{b}]] \in {\mathcal {R}}_{4}$ . Notice that $[[t,f_{a}],z]$ and $[[t,f_{b}],z] \in {\mathcal R}_{3}$ for all $z \in {\mathcal F}_{n}$ . Using the Jacobi identity in the form $[x,y,z] = [x,z,y] + [x,[y,z]]$ , Proposition 2.7 (4), and Proposition 4.2, we have $v\gamma _{\kappa + 5}(F_{n+1}) \in \textrm {I}_{\kappa + 4}(M_{{\mathcal R}_{3}})$ . Thus, we assume that at least one of $z_{1}, \ldots, z_{\kappa }$ is $t$ . Let $1 \leq i_{1} \lt \cdots \lt i_{\nu } \leq \kappa$ be the $\nu$ different appearances of $t$ in the $\kappa$ -tuple $(z_{1}, \ldots, z_{\kappa })$ . By our hypothesis $\nu \geq 1$ . If $i_{1} \geq 2$ , then $z_{1} \in {\mathcal F}_{n}$ and so, using the above argument, we get $v\gamma _{\kappa + 5}(F_{n+1}) \in \textrm {I}_{\kappa + 4}(M_{{\mathcal R}_{3}})$ . Thus, we assume that $z_{1} = t$ . If $\kappa = 1$ , then by Proposition 2.7 (3) and Proposition 4.6, we get $v\gamma _{6}(F_{n+1}) \in \textrm {I}_{5}(M_{{\mathcal{R}}_{4}})$ . Thus we assume that $\kappa \geq 2$ . Working modulo $\textrm {I}_{\kappa + 4}(M_{{\mathcal R}_{3}})$ , using the Jacobi identity in the form $[x,y,z] = [x,z,y] + [x,[y,z]]$ and Proposition 4.6, we obtain $v\gamma _{\kappa + 5}(F_{n+1}) \in \textrm { I}_{\kappa + 4}(M_{{\mathcal R}_{4}}) \oplus \textrm {I}_{\kappa + 4}(M_{{\mathcal R}_{3}})$ . Thus, in any case, $v\gamma _{\kappa + 5}(F_{n+1}) \in \textrm {I}_{\kappa + 4}(M_{{\mathcal R}_{4}}) \oplus \textrm {I}_{\kappa + 4}(M_{{\mathcal R}_{3}})$ .

Proof.

1. 1. It follows from a result of P. Hall [Reference Hall12] (see, also, [Reference Clement, Majewicz and Zyman4, Chapter 3, Theorem 3.1]).

2. 2. We induct on $\kappa$ , and let $\kappa = 2$ . Since $[H,F_{n}]$ is generated by the elements $[u,f]$ with $u \in H$ and $f \in F_{n} \setminus \{1\}$ , it is enough to show that each $[u,f]$ is written in the required form. Using the well-known group commutator identities (see, e.g.,, proof of Lemma 4.8 (1)), working modulo $[H,F_{n},F_{n}]$ and since $[H, F_{n}, F_{n}]$ is normal in $F_{n}$ , we obtain the desired result. Thus, we assume that our claim is valid for some $\kappa \geq 2$ , and let $g \in [H,F_{n}]$ . By our inductive argument, $g$ is written as $c^{d_{1}}_{1} \cdots c^{d_{\kappa -1}}_{\kappa -1}w$ with $w \in [H,\,_{\kappa }F_{n}]$ and each $c_{\mu }$ is a product of weighted group commutators of the form $[v, z_{1}, \ldots, z_{\mu }]$ with $v \in {\mathcal {S}}$ and $z_{1}, \ldots, z_{\mu } \in {\mathcal {F}}_{n}$ . Working modulo $[H, \,_{\kappa + 1}F_{n}]$ , we have $w$ is a product of weighted group commutators of the form $[v, z_{1}, \ldots, z_{\kappa }]$ with $v \in {\mathcal {S}}$ and $z_{1}, \ldots, z_{\kappa } \in {\mathcal {F}}_{n}$ . Hence, any element $g \in [H, F_{n}]$ can be expressed in the required form.

Proof. Write ${\mathcal L} = {\mathcal{L}}(M_{{\mathcal{R}}_{2}}) + {\mathcal{L}}(M_{{\mathcal{R}}_{3}}) + {\mathcal{L}}(M_{{\mathcal{R}}_{4}})$ . By Proposition 4.3, Proposition 4.2, Proposition 4.6 and Proposition 2.7 (1), we have $\mathcal L$ is the additive direct sum of the Lie subalgebras ${\mathcal{L}}(M_{{\mathcal{R}}_{2}})$ , ${\mathcal{L}}(M_{{\mathcal{R}}_{3}})$ and ${\mathcal{L}}(M_{{\mathcal{R}}_{4}})$ of $\textrm {gr}(F_{n+1})$ and ${\mathcal L} = J$ . Thus, for $d \geq 2$ , $J \cap \textrm {gr}_{d}(F_{n+1}) = \textrm {I}_{d}(M_{{\mathcal{R}}_{2}}) \oplus \textrm { I}_{d}(M_{{\mathcal{R}}_{3}}) \oplus \textrm {I}_{d}(M_{{\mathcal{R}}_{4}})$ .

We claim that $J \cap \textrm {gr}_{d}(F_{n+1}) = \textrm {I}_{d}(M)$ for all $d \geq 2$ . Since $J \cap \textrm {gr}_{2}(F_{n+1}) = \textrm {I}_{2}(M_{{\mathcal{R}}_{2}}) = \textrm {I}_{2}(M)$ , we may assume that $d \geq 3$ . Let $u \in M \cap \gamma _{d}(F_{n+1})$ such that $u \notin \gamma _{d+1}(F_{n+1})$ . That is, $u\gamma _{d+1}(F_{n+1})$ is a non-trivial element of $\textrm {I}_{d}(M) \leq \textrm {gr}_{d}(F_{n+1})$ . By Lemma 4.8 (1), $u$ is written as $u = h h_{2} h_{3} h_{4}$ with $h \in R$ and $h_{j} \in [R_{j},F_{n+1}]$ , $j \in \{2,3,4\}$ . We point out that, for a positive integer $\kappa$ , $[R_{j}, \,_{\kappa }F_{n+1}] \subseteq \gamma _{j+\kappa }(F_{n+1})$ with $j \in \{2,3,4\}$ . Next, we work in the free nilpotent group $F_{n+1,d} = F_{n+1}/\gamma _{d+1}(F_{n+1})$ . By Lemma 4.9 (for $G = R$ with a generating set $\mathcal {R}$ , $H = R_{j}$ and ${\mathcal {S}} = {\mathcal {R}}_{j}$ ), and using the collection process for arbitrary group elements on $u$ (see [Reference Clement, Majewicz and Zyman4, Chapter 3, Theorem 3.1 and Theorem 3.5]), the element $\overline {u} = u \gamma _{d+1}(F_{n+1})$ is written in $F_{n+1,d}$ as $\overline {w}_{2}\overline {w}_{3} \cdots \overline {w}_{d}$ , where each $w_{i}$ is a product of weighted group commutators each of which has weight $i$ and in each weighted group commutator at least one element of $\mathcal {R}$ occurs. Notice that, for $i \in \{2, \ldots, d\}$ , $w_{i} \in M \cap \gamma _{i}(F_{n+1})$ . Since $u \in \gamma _{d}(F_{n+1}) \setminus \gamma _{d+1}(F_{n+1})$ and since the center of $F_{n+1,d}$ equals $\gamma _{d}(F_{n+1,d}) = \textrm {gr}_{d}(F_{n+1})$ , we have $\overline {u} = \overline {w}_{d}$ . Hence, working modulo $\gamma _{d+1}(F_{n+1})$ , $u = \prod u^{\prime }_{\mu }$ is a product of weighted group commutators of weight $d$ in $F_{n+1}$ where in each $u^{\prime }_{\mu }$ at least one element of $\mathcal R$ occurs. Fix $\mu$ and let $u^{\prime }_{\mu }$ be a weighted group commutator of weight $d$ in $F_{n+1}$ of the form $w(a_{1}, \ldots, a_{\lambda })$ where each $a_{i}$ is a weighted group commutator of weight $\lt d$ . We separate the following cases.