1. (1) This is straightforward.

2. (2) Let $[x,y^{m}] \in [N,G]$ for some $m \in \mathbb{N}$. By Lemma 1 (1) (for $\kappa = m$), we have $[x,y^{m}] = [x,y]^{m}w$, with $w \in [N,G]$, and so $[x,y]^{m} \in [N,G]$.

3. (3) Let $[x^{m},y] \in [N,G]$. By Lemma 1 (1) (for $\kappa = m$), we have $[x^{m},y] = [x,y]^{m}w_{1}$, with $w_{1} \in [N,G]$, and so $[x,y]^{m} \in [N,G]$.

Proof. Since a finite p-group is nilpotent, we have every residually finite p-group is also residually nilpotent. Hence, $G/(Np)_{\omega}$ is residually nilpotent and so $\gamma_{\omega}(G)\le (Np)_{\omega}$. Since G is finitely generated and $G/\gamma_{\omega}(G)$ is residually nilpotent, we have $G/\gamma_{\omega}(G)$ is residually finite. We claim that $N_{\omega}\le \gamma_{\omega}(G)$. Let $g \in N_{\omega}$ and $g \notin \gamma_{\omega}(G)$. Since $G/\gamma_{\omega}(G)$ is residually finite, there exists a normal subgroup N_g of G such that $g \notin N_{g}$, $\gamma_{\omega}(G) \subseteq N_{g}$ and $G/N_{g}$ is finite, which is a contradiction since $g \in N_{\omega}$.

Write $B = \bigcap\limits_{p\ {\rm prime}}(Np)_{\omega}$. To get a contradiction, we assume that $g \in B$ and $g \notin \gamma_{\omega}(G)$. In the next few lines, let $\widetilde{G}=G/\gamma_{\omega}(G)$. Since G is finitely generated and $\widetilde{G}$ is residually nilpotent, there exists an epimorphism ϕ from $\widetilde{G}$ onto a finitely generated nilpotent group H with $\phi(g\gamma_{\omega}(G)) \neq 1$. Since H is polycyclic, we have H is residually finite. Thus, there exists a finite nilpotent group $\hat{H}$ and an epimorphism $\hat{\phi}: \widetilde{G} \rightarrow \hat{H}$ with $\hat{\phi}(g \gamma_{\omega}(G)) \neq 1$. Since $\hat{H}$ is the direct product of its Sylow p-subgroups, there exist a prime number p and a Sylow p-subgroup S_p of $\hat{H}$ such that $\hat{\phi}(g \gamma_{\omega}(G)) \in S_{p} \setminus \{1\}$. Since S_p is a finite p-group, we have $g \notin (Np)_{\omega}$, and so $g\not\in B$, which is a contradiction.

Proof. The elements x_i have orders $p_i^{r_i}$, and so the orders of x_i and x_j are coprime for every i ≠ j. We write $\widetilde{G} = G/\gamma_{\omega}(G)$. By the definition of residual nilpotence, for every non-trivial element $\tilde{g}\in\tilde{G},$ there exist a finite nilpotent group $\hat{H}$ and an epimorphism $\hat{\phi}: \widetilde{G}\rightarrow\hat{H}$ such that $\hat{\phi}(\tilde{g})\neq 1$. On the other hand, $\hat{H}$ is the direct product of finite Sylow p-subgroups S_p. But since the orders of $t^{-k}x_it^k$ and x_j are also coprime, $t^{-k}x_it^k$ and x_j belong to different direct factors S_p of $\hat{H}$. So if $[t^{-k}x_it^{k},x_j]$ are non-trivial in $\tilde{G},$ they always vanish under any $\hat{\phi}$, a contradiction. Therefore, $[t^{-k}x_{i}t^{k},x_{j}] \in \gamma_{\omega}(G)$ for all $k \in \mathbb{Z}$ and $i, j \in \{1, \ldots, n\}$, with i ≠ j.

For the converse, we will show that $G/N$ is residually nilpotent, where N is the normal closure of the set $\{[t^{-k}x_it^{k},x_j]: i,j\in\{1,\ldots, n\}; i\neq j; k\in\mathbb Z\}$ in G. Let $g\in G/N$ with g ≠ 1. If the exponent sum of t in g is non-zero, then we can take the homomorphism $\phi : G/N \rightarrow \mathbb Z$ with $x_i\mapsto 0$ and $t\mapsto 1$. Then $\phi(g)\neq 1,$ and since $\mathbb Z$ is residually nilpotent, the result follows. On the other hand, assume that the exponent sum of t in g is zero. Notice that the relations $[t^{-k}x_it^{k},x_j]$ are equivalent to $[t^{-s}x_it^{s},t^lx_jt^{-l}]$ for every $s,l\in\mathbb Z$. Using these relations, g can be written as

\begin{equation*} g=(t^{-s_1}x_1^{w_1}t^{s_1}\cdots t^{-s_{m_1}}x_1^{w_{m_1}}t^{s_{m_1}}) \cdots (t^{-q_1}x_k^{z_1}t^{q_1}\cdots t^{-q_{m_k}}x_k^{z_{m_k}}t^{q_{m_k}}), \end{equation*}

where the words

\begin{equation*} (t^{-s_1}x_1^{w_1}t^{s_1}\cdots t^{-s_{m_1}}x_1^{w_{m_1}}t^{s_{m_1}}), \ldots, (t^{-q_1}x_k^{z_1}t^{q_1}\cdots t^{-q_{m_k}}x_k^{z_{m_k}}t^{q_{m_k}}) \end{equation*}

are reduced. Since $g\neq 1,$ at least one of the $w_1,\ldots, w_{m_1},\ldots, z_1,\ldots, z_{m_k}\neq 0$. Then we take the homomorphism $\phi : G/N\rightarrow \mathbb Z*\mathbb Z_{p_i^{r_i}}$ with $\phi(\langle t\rangle )=\mathbb Z$, $\phi(\langle x_i\rangle )=\mathbb Z_{p_i^{r_i}}$ and $\phi(x_j)=0$ for all j ≠ i. Since $\mathbb Z*\mathbb Z_{p_i^{r_i}}$ are residually nilpotent (see [Reference Gruenberg4, Theorem 4.1]), the result follows.

Proof. Let $m=p_1^{r_1}\cdots p_n^{r_n}$, with $n\ge 2$ be the prime number decomposition of m. For $i \in \{1, \ldots, n\}$, let $q_{i} = \frac{m}{p^{r_{i}}_{i}}$. Since ${\rm gcd}(q_{1}, \ldots, q_{n}) = 1$, there are $d_{i} \in \mathbb{Z}$ such that $q_{1}d_{1} + \cdots + q_{n}d_{n} = 1$. For $i \in \{1, \ldots, n\}$, let $u_{i} = a^{q_{i}d_{i}}$. Then, for any $i \in \{1, \ldots, n\}$, the order of u_i is $p^{r_{i}}_{i}$ and $a = u_{1}u_{2} \cdots u_{n}$. Since $G\cong \mathbb Z*(\mathbb Z_{p_1^{r_1}}\times\cdots\times\mathbb Z_{p_n^{r_n}})$, G admits a presentation as in Lemma 3, and the isomorphism between the two presentations implies that each x_i maps to u_i.

Let $\mathcal{M}=\{[t^{-k}u_it^k,u_j] ; i,j\in \{1,\ldots,n\}; i\neq j; k\in\mathbb Z\}$ be the generating set of $\gamma_{\omega}(G)$ described in Lemma 3 and $\mathcal{K} = \{[t^{-k}a^{\mu}t^{k},a^{\nu}]: k \in \mathbb{Z}; \mu, \nu \in \mathbb{N}; {\rm gcd}(\mu, \nu) = 1; \mu \nu = m\}$. Write K for the normal closure of the set $\mathcal{K}$ in G. We claim that $\gamma_{\omega}(G) = K$. We first show that $\gamma_{\omega}(G) \subseteq K$. Let $[t^{-k}u_{i}t^{k},u_{j}] \in {\mathcal{M}}$, and without loss of generality, we assume that i < j. Write $s_{1} = p^{r_{i+1}}_{i+1} \cdots p^{r_{j-1}}_{j-1}p^{r_{j+1}}_{j+1} \cdots p^{r_{n}}_{n}$ and $s_{2} = p^{r_{1}}_{1} \cdots p^{r_{i-1}}_{i-1}$. Then

\begin{align*} \left[t^{-k}u_{i}t^{k},u_{j}\right] & = \left[t^{-k}a^{q_{i}d_{i}}t^{k},a^{q_{j}d_{j}}\right] = \left[t^{-k}\left(a^{p_1^{r_1}\ldots p_{i-1}^{r_{i-1}}p_j^{r_j}}\right)^{d_i s_1}t^{k},\right.\nonumber \\ & \qquad \left.\left(a^{p_i^{r_i}\cdots p_{j-1}^{r_{j-1}}p_{j+1}^{r_{j+1}}\cdots p_n^{r_n}}\right)^{d_j s_2}\right]. \end{align*}

Working in $G/K$,

\begin{equation*} \left[t^{-k}\left(a^{p_1^{r_1}\cdots p_{i-1}^{r_{i-1}}p_j^{r_j}}\right)t^{k}, \left(a^{p_i^{r_i}\cdots p_{j-1}^{r_{j-1}}p_{j+1}^{r_{j+1}}\cdots p_n^{r_n}}\right)\right]=1 \end{equation*}

for all $k \in \mathbb{Z}$. By using the commutator identities $[xy,z]=[x,z]^y[y,z]$, $[x,yz]=[x,z][x,y]^z$, $[x^{-1},y] = ([x,y]^{-1})^{x^{-1}}$ and $[x,y^{-1}] = ([x,y]^{-1})^{y^{-1}}$ repeatedly, we get

\begin{equation*} \left[t^{-k}\left(a^{p_1^{r_1}\cdots p_{i-1}^{r_{i-1}}p_j^{r_j}}\right)^{d_i s_1}t^{k}, \left(a^{p_i^{r_i}\cdots p_{j-1}^{r_{j-1}}p_{j+1}^{r_{j+1}}\ldots p_n^{r_n}}\right)^{d_j s_2}\right]=1 \end{equation*}

for all $k \in \mathbb{Z}$. Hence, $[t^{-k}u_{i}t^{k},u_{j}] \in K$ for i < j. Applying similar arguments as above, we have, for

i > j, $[t^{-k}u_{i}t^{k},u_{j}] \in K$. Consequently, $\gamma_{\omega}(G)\subseteq K$.

For the converse, since $a = u_{1}u_{2} \cdots u_{n}$, ${\rm gcd}(\mu,\nu) = 1$ and $\mu \nu = m$, the elements of $\mathcal{K}$ are

\begin{equation*} [t^{-k}a^{\mu}t^{k},a^{\nu}] = [t^{-k}(u_1\cdots u_n)^{\mu}t^{k},(u_1\cdots u_n)^{\nu}]=[t^{-k} (u_{i_1}\cdots u_{i_{l}})^{\mu}t^{k}, (u_{j_1}\cdots u_{j_{n-l}})^{\nu}], \end{equation*}

with $\{u_{i_1},\ldots, u_{i_l}\}\bigcup \{u_{j_1},\ldots, u_{j_{n-l}}\}=\{u_1,\ldots, u_n\}$ and $\{u_{i_1},\ldots, u_{i_{l}}\}\bigcap \{u_{j_1},\ldots, u_{j_{n-l}}\}=\emptyset.$ Now one can easily show by using the commutator identities $[xy,z]=[x,z]^y[y,z]$ and $[x,yz]=[x,z][x,y]^z$ repeatedly that the elements of K belong to $\gamma_{\omega}(G)$. Therefore, $K \subseteq \gamma_{\omega}(G)$ and so $\gamma_{\omega}(G) = K$.

1. (1) Assume that there is a prime number p such that $n\equiv m \pmod p$. Since ${\rm gcd}(m,n)=1$, we have p divides neither m nor n and so m and n satisfy the conditions of Proposition 3 (2). Therefore, we have $(Np)_{w}$ is the normal closure in G of $[t^{-k}at^k,a]$. On the other hand, by Proposition 2, we have N_ω is the normal closure in G of the set $\{[t^{-k}at^{k},a]: k\in\mathbb Z\}$. Hence, the description of $\gamma_{\omega}(G)$ is an immediate consequence of Lemma 2.

2. (2) Assume that $m\not\equiv n \pmod p$ for every prime number p. Since $\gcd(m,n)=1$, we have by Proposition 3 (1), $(Np)_{\omega}$ is the normal closure of a in G. So $a \in \bigcap\limits_{p\ {\rm prime}}(Np)_{w}$. By Lemma 2, we obtain the required result.

Proof. By Corollary 1, we have $[t^{-k}a^{d}t^{k}, a] \in \gamma_{\omega}(G)$. Hence, in the case d is a power of a prime number, that is, µ = 1 or ν = 1, the required result follows. Thus, in what follows, we may assume that $\mu, \nu \gt 1$. Fix some $k \in \mathbb{Z}$ and let us denote $u=t^{-k}a^{\mu}t^{k}$. Assume that $[u, a^{\nu}] \in \gamma_{i}(G)$ for some $i \geq 2$. Since $[u^{\nu}, a^{\nu}] = [t^{-k}a^{\mu\nu}t^{k}, a^{\nu}]=[t^{-k}a^dt^{k}, a^{\nu}]$, it follows from Corollary 1 that $[u^{\nu}, a^{\nu}] \in\gamma_{\omega}(G)$ and hence, $[u^{\nu}, a^{\nu}] \in \gamma_j(G)$ for all $j \in {\mathbb N}$. In particular, we have $[u^{\nu}, a^{\nu}] \in \gamma_{i+1}(G)$. Since $[u, a^{\nu}] \in \gamma_{i}(G)$ and $[u^{\nu}, a^{\nu}] \in \gamma_{i+1}(G) = [\gamma_{i}(G),G]$, we get, by Lemma 1 (3) (for $N = \gamma_{i}(G)$), $[u, a^{\nu}]^{\nu} \in \gamma_{i+1}(G)$. Similarly, since $[u, a^{\mu\nu}] = [u, a^{d}] = [a^{\mu}, t^{k}a^{d}t^{-k}]^{t^{k}}$ and $\gamma_{\omega}(G)$ is normal in G, it follows from Corollary 1 that $[u, a^{\mu\nu}] \in \gamma_{\omega}(G)$. In particular, we have $[u, a^{\mu\nu}] \in\gamma_{i+1}(G)$. As before, by Lemma 1 (2), we get $[u, a^{\nu}]^{\mu} \in \gamma_{i+1}(G)$. Thus, $[u, a^{\nu}]^{\nu}, [u, a^{\nu}]^{\mu} \in \gamma_{i+1}(G)$. Since ${\rm gcd}(\mu, \nu) = 1$, we have $[u, a^{\nu}] \in \gamma_{i+1}(G)$. We carry on this process, and we obtain the required result.

1. (1) Let T be the normal closure of the set $\{a^{d}, [t^{-k}a^{\mu}t^{k},a^{\nu}]: k\in\mathbb Z, \mu,\nu \in \mathbb{N}, {\rm gcd}(\mu, \nu) = 1 ~{\rm and}~\mu \nu = d\}$ in G. We first show that $T \subseteq \gamma_{\omega}(G)$. By Lemma 5, it is enough to show that $a^{d} \in \gamma_{\omega}(G)$. Let $d = p^{r_{1}}_{1} \cdots p^{r_{\kappa}}_{\kappa}$ be the prime number decomposition of d. For $i \in \{1, \ldots, \kappa\}$, we write $m = p^{r_{i}}_{i} m^{\prime}_{1i}$ and $n = p^{r_{i}}_{i} n^{\prime}_{1i}$ where $m^{\prime}_{1i} = \frac{d}{p^{r_{i}}_{i}}m_{1}$ and $n^{\prime}_{1i} = \frac{d}{p^{r_{i}}_{i}}n_{1}$. Now, $n^{\prime}_{1i} - m^{\prime}_{1i} = \frac{d}{p^{r_{i}}_{i}}(n_{1} - m_{1})$. Since $p_{i} \nmid (n_{1}-m_{1})$, we obtain $p_{i} \nmid (n^{\prime}_{1i}-m^{\prime}_{1i})$. By Proposition 3 (1), we have $a^{p^{r_{i}}_{i}} \in (Np_{i})_{\omega}$ for every $i \in \{1, \ldots, \kappa\}$ and so $a^{d} \in (Np_{i})_{\omega}$ for any $i \in \{1, \ldots, \kappa\}$. By Proposition 3 (1), for every $q \notin \{p_{1}, \ldots, p_{\kappa}\}$, we have $a \in (Nq)_{\omega}$, which again implies that $a^{d} \in (Nq)_{\omega}$. Therefore, $a^{d} \in \bigcap\limits_{p\ {\rm prime}}(Np)_{w}$. By Lemma 2, we have $a^{d} \in \gamma_{\omega}(G)$ and so $T \subseteq \gamma_{\omega}(G)$. On the other hand, by Lemma 4, $G/T\cong (\mathbb Z*\mathbb Z_d)/\gamma_{\omega}(\mathbb Z*\mathbb Z_d),$ which is residually nilpotent. Hence, $\gamma_{\omega}(G)\subseteq T$. Therefore, $\gamma_{\omega}(G)=T,$ and we obtain the required result.

2. (2) Assume that there is some prime p such that $n_1\equiv m_1 \pmod p$ and let M be the normal closure of the set $\{[t^{-k}a^{\mu}t^{k},a^{\nu}]: k\in\mathbb Z, \mu,\nu \in \mathbb{N},\ {\rm gcd}(\mu, \nu) = 1 ~{\rm and}~\mu \nu = d\}$ in G. By Lemma 5, we have $M \subseteq \gamma_{\omega}(G)$. We claim that $G/M$ is residually nilpotent. Notice that $G/M$ has a presentation of the form

   \begin{equation*}G/M=\langle a,t\mid t^{-1}a^mt=a^n, [t^{-k}a^{\mu}t^{k},a^{\nu}]=1, \ k\in\mathbb Z\rangle \end{equation*}
   with $ \mu,\nu \in \mathbb{N},\ {\rm gcd}(\mu, \nu) = 1 ~{\rm and}~\mu \nu = d.$ Let g be an element in $G/M$. Then g is a word in $a,t$.

   Assume first that the exponent sum of t in g is non-zero. Then there is a map $\phi:G/M\rightarrow \mathbb Z$ such that $a\mapsto 0$ and $t\mapsto 1$. It can easily be seen that ϕ is a homomorphism and that $\phi(g)\neq 1$. Since $\mathbb Z$ is nilpotent, the result follows.

   Assume now that the exponent sum of t in g is zero. Then g can be written in reduced form

   (3.1)\begin{align} g & = a^{\rho_0}\left(t^{\varepsilon_{1}}a^{\rho_1}t^{-\varepsilon_{1}}\right)\left(t^{\varepsilon_{1}+\varepsilon_{2}}a^{\rho_2}t^{-(\varepsilon_{1}+\varepsilon_{2})}\right) \cdots\nonumber \\ & \qquad \left(t^{\varepsilon_{1}+ \cdots +\varepsilon_{\kappa-1}}a^{\rho_{\kappa-1}}t^{-(\varepsilon_{1}+ \cdots + \varepsilon_{\kappa-1})}\right)a^{\rho_{\kappa}}, \end{align}
   with $|\rho_i| \lt m,$ if $\varepsilon_1+\cdots+\varepsilon_i\le-1$ and $|\rho_i| \lt n$ if $\varepsilon_1+\cdots+\varepsilon_i\ge1$. Write each $a^{\rho_i} = a^{d \lambda_i} a^{r_i}$, with $r_i \in \{0, \ldots, d-1\}$, $t^{-k} a^{\rho_i}t^{k} = (t^{-k}(a^{d})^{\lambda_i}t^{k})(t^{-k}a^{r_i}t^{k})$. Using the identity $t^{-k}a^{\zeta}t^{k} = a^{\zeta}[a^{\zeta},t^{k}]$, we rewrite all the above and replace them in the expression (3.1). Then, using the identity $ab=ba[a,b]$ as many times as needed and the identities
   (3.2)\begin{equation} [ab,c] = [a,c]~[[a,c],b]~[b,c], \end{equation}

   (3.3)\begin{equation} [a,bc] = [a,c]~[a,b]~[[a,b],c], \end{equation}
   g has an expression of the form
   \begin{equation*} g=a^{\lambda}~[(a^{d})^{\lambda_{1}},t^{k_1}] \cdots [(a^{d})^{\lambda_{s}},t^{k_s}]\cdot w, \end{equation*}
   where $\lambda_{1}, \ldots, \lambda_{s} \in \mathbb{N}$ and w is a product of group commutators of the form $[h_{1}, \ldots, h_{r}]$, with $r \geq 2$ and $h_{1}, \ldots, h_{r} \in \{a, \ldots, a^{d-1}\} \cup \{t^{k}: k \in \mathbb{Z} \setminus \{0\}\}$. Note that $d\lambda_{1}, \ldots, d\lambda_{s} \lt m_{1}, n_{1}$. Next, we separate two cases.

   1. (a) Let w = 1. For the next few lines, let $G_{1} = {\rm BS}(m_{1},n_{1}) = \langle \bar{t}, \bar{a}: (\bar{t})^{-1}(\bar{a})^{m_{1}}\bar{t} = (\bar{a})^{n_{1}} \rangle$. Since ${\rm gcd}(m_{1}, n_{1}) = 1$ and $n_{1} \equiv m_{1} \pmod p$ for a prime integer p, we have, by Proposition 4 (1) and (3.2), $\gamma_{\omega}(G_{1})$ is the normal closure of the set $\{[\bar{t}^{-k}\bar{a}\bar{t}^{k}, \bar{a}]: k \in \mathbb{Z}\}$ in G ₁. Then there is a natural homomorphism $\psi: G/M\rightarrow G_1/\gamma_{\omega}(G_1)$ with $a\mapsto \bar{a}$ and $t\mapsto\bar{t}$ such that $\psi(g)\neq 1$. Since $G_1/\gamma_{\omega}(G_1)$ is residually nilpotent, the result follows.

   2. (b) Let w ≠ 1. Then it suffices to map $G/M$ to $(G/M)/\langle a^d\rangle \cong (\mathbb Z*\mathbb Z_d)/\gamma_{\omega}(\mathbb Z*\mathbb Z_d)$. The image of g is $w\neq 1,$ which is reduced. The result follows from that fact that $(G/H)/\langle a^d\rangle \cong (\mathbb Z*\mathbb Z_d)/\gamma_{\omega}(\mathbb Z*\mathbb Z_d)$ is residually nilpotent.

Proof. Throughout the proof, we write $u_k = t^{-k}a^dt^k$, with $k \in \mathbb{Z}$. By Corollary 1, $[u_k, a^{d}] \in \gamma_{\omega}(G)$ for all $k \in \mathbb{Z}$. Furthermore, we write $m = dm_{1}$ and $n = dn_{1}$, where $\gcd(m_{1}, n_{1}) = 1$.

1. (1) Since $[u_k ,a^d]=([u_{-k}, a^d]^{-1})^{t^k}$ and $[\gamma_{\omega}(G),G]$ is normal in G, it suffices to show that $[u_k, a^{d}] \in [\gamma_{\omega}(G), G]$ for all $k \in \mathbb{N}$. We use induction on k. Assume at first that k = 1. Since, as aforementioned, $[u_1, a^{d}] \in \gamma_{\omega}(G)$ and since $[u_{1}^{m_{1}}, a^{d}] = [t^{-1}a^{m}t, a^{d}] = [a^{n},a^{d}] = 1$, it follows from Lemma 1 (3) (for $N=\gamma_{\omega}(G)$) that $[u_1, a^{d}]^{m_{1}} \in [\gamma_{\omega}(G), G]$. Since $[u_1, a^{dn_{1}}] = [a^{d}, ta^{n}t^{-1}]^{t} = [a^{d}, a^{m}]^{t} = 1\ in\ [\gamma_{\omega}(G),G]$, by Lemma 1 (2) (for $N=\gamma_{\omega}(G)$), we get $[u_1, a^{d}]^{n_{1}} \in [\gamma_{\omega}(G), G]$. But $\gcd(m_{1}, n_{1}) = 1$, and so the result follows for k = 1.

   Assume that $[u_k, a^{d}] \in [\gamma_{\omega}(G), G]$ for some $k \in \mathbb{N}$. Then, by Lemma 1 (1) (for $N=[\gamma_{\omega}(G),G]$), we have $[u_k^x,a^d]\in [\gamma_{\omega}(G),G]$ for any $x\in\mathbb N$. Hence,

   \begin{equation*} \left[u_{k+1}^{m_{1}}, a^{d}\right] = \left[t^{-(k+1)}a^mt^{k+1},a^d\right] = \left[t^{-k}(t^{-1}a^{m}t)t^{k}, a^{d}\right] = \left[u_k^{n_{1}},a^d\right] \in [\gamma_{\omega}(G),G]. \end{equation*}

   Since $[u_{k+1},a^d]\in\gamma_{\omega}(G)$, it follows from Lemma 1 (2) (for $N=\gamma_{\omega}(G)$) that $[u_{k+1}, a^d]^{m_1}\in [\gamma_{\omega}(G),G]$.

   As above, since $[u_k,a^d]\in[\gamma_{\omega}(G),G]$, we have, by Lemma 1 (1) (for $N=[\gamma_{\omega}(G),G])$ that $[u_k,a^{dm_1}]\in [\gamma_{\omega}(G),G]$ and therefore

   \begin{align*} \left[u_{k+1}, a^{dn_{1}}\right]^{t^{-1}} & = \left[t^{-(k+1)}a^{d}t^{k+1},a^n\right]^{t^{-1}} = [u_{k}, ta^{n}t^{-1}] = [u_{k}, a^{m}]\nonumber \\ & \qquad = [u_k, a^{dm_{1}}] \in [\gamma_{\omega}(G),G]. \end{align*}

   But $[\gamma_{\omega}(G),G]$ is normal in G, so $[u_{k+1}, a^{dn_{1}}] \in [\gamma_{\omega}(G),G]$. Again, by Lemma 1 (2) (for $N=\gamma_{\omega}(G)$), $[u_{k+1}, a^{d}] ^{n_1}\in [\gamma_{\omega}(G),G]$. Since $\gcd(m_{1}, n_{1}) = 1,$ we obtain the required result.

2. (2) By Corollary 1, $[u_k, a] \in \gamma_{\omega}(G)$, and by Lemma 6 (1), $[u_k, a^d] \in [\gamma_{\omega}(G), G]$ for all $k \in \mathbb{Z}$, the result follows from Lemma 1 (2) (for $N=\gamma_{\omega}(G)$.

3. (3) By Corollary 1, $[t^{-k}at^{k}, a^{d}] \in \gamma_{\omega}(G)$. Since $[t^{-k}at^{k}, a^{d}] = ([t^{k}a^{d}t^{-k}, a]^{-1})^{t^{k}}$, the result follows from Lemma 6 (2).

Proof. Let $d = p^{\mu}$, with $\mu \geq 1$. Thus, we may write $m = p^{r}m_{1}$ and $n = p^{s}n_{1}$, where $\mu = \min\{r, s\}$ and $\gcd(p, m_{1}) = \gcd(p, n_{1}) = 1$. Throughout the proof, we write $u_k = t^{-k}a^{d}t^k$ and $v_{k} = t^{-k}at^{k}$. By Corollary 1, we have $[u_{k}, a], [v_k, a^d] \in \gamma_{\omega}(G)$ for all $k \in \mathbb{N}$. We separate several cases. In the following, we repeatedly use the fact that $[\gamma_{\omega}(G),G]$ is normal in G.

1. (1) Let r = s. In this case, we have $d = p^{r}$. At first, we show that $[u_k, a] \in [\gamma_{\omega}(G), G]$ for all $k \in \mathbb{N}$. We use induction on k. Let k = 1. Since

   \begin{equation*}[u_{1}^{m_{1}}, a] = [t^{-1}a^{m}t, a] = [a^{n},a] = 1 \in [\gamma_{\omega}(G),G]\end{equation*}
   and $[u_{1},a] \in \gamma_{\omega}(G)$, we have from Lemma 1 (3) (for $N=\gamma_{\omega}(G))$ that $[u_1, a]^{m_{1}} \in [\gamma_{\omega}(G), G]$. Furthermore, by Lemma 6 (2), $[u_1, a]^{p^{r}} \in [\gamma_{\omega}(G), G]$. But $\gcd(m_{1}, p^{r}) = 1$ and so we have $[u_1, a] \in [\gamma_{\omega}(G), G]$. Thus, our claim is valid for k = 1.

   Assume that $[u_k, a] \in [\gamma_{\omega}(G), G]$ for some $k \in \mathbb{N}$. Using Equation (3.2) as many times as needed and since $[u_k, a] \in [\gamma_{\omega}(G), G]$, we get $[u_{k}^{n_{1}}, a] \in [\gamma_{\omega}(G), G]$. Since

   \begin{equation*}\left[u_{k+1}^{m_{1}}, a\right] = \left[t^{-(k+1)}a^{m}t^{k+1}, a\right] = \left[t^{-k}a^{n}t^{k}, a\right] = \left[u_{k}^{n_{1}}, a\right] \in \left[\gamma_{\omega}(G), G\right]\end{equation*}
   and $[u_{k+1},a] \in \gamma_{\omega}(G)$, it follows from Lemma 1 (3) (for $N=\gamma_{\omega}(G))$ that $[u_{k+1}, a]^{m_{1}} \in [\gamma_{\omega}(G), G]$. Furthermore, by Lemma 6 (2), $[u_{k+1}, a]^{p^{r}} \in [\gamma_{\omega}(G), G]$. But $\gcd(m_{1}, p^{r}) = 1$ and so we have $[u_{k+1}, a] \in [\gamma_{\omega}(G), G]$. Therefore, $[u_{k}, a] \in [\gamma_{\omega}(G),G]$ for all $k \in \mathbb{N}$. By Equation (3.2), we get
   \begin{equation*}[u_{k}, a] = \left[a^{p^{r}}, t^{k}, a\right] \in [\gamma_{\omega}(G),G]\end{equation*}
   for all $k \in \mathbb{N}$.

   Next we show that $[v_k, a^{p^{r}}] \in [\gamma_{\omega}(G), G]$ for all $k \in \mathbb N$. As before, we use induction on k. Let k = 1. Since

   \begin{equation*} [v_1, a^{p^{r}n_{1}}] = [v_1, a^{n}] = [a,ta^{n}t^{-1}]^{t} = [a,a^{m}]^{t} =1 \in [\gamma_{\omega}(G),G]\end{equation*}
   and $[v_{1}, a^{p^{r}}] \in \gamma_{\omega}(G)$, it follows from Lemma 1 (2) (for $N=\gamma_{\omega}(G))$ that $[v_1, a^{p^{r}}]^{n_{1}} \in [\gamma_{\omega}(G), G]$. Furthermore, by Lemma 6 (3), $[v_1, a^{p^{r}}]^{p^{r}} \in [\gamma_{\omega}(G), G]$. But $\gcd(p^{r}, n_{1}) = 1,$ and so we have $[v_1, a^{p^{r}}] \in [\gamma_{\omega}(G), G]$. Thus, our claim is true for k = 1.

   Assume that $[v_k, a^{p^{r}}] \in [\gamma_{\omega}(G), G]$ for some $k \in \mathbb{N}$. Using Equation (3.3) as many times as needed and since $[v_{k},a] \in [\gamma_{\omega}(G),G]$, we get $[v_{k},(a^{p^{r}})^{m_{1}}] = [v_{k}, a^{m}] \in [\gamma_{\omega}(G), G]$. Since

   \begin{equation*}[t^{-(k+1)}at^{k+1}, a^{n}] = [t^{-k}at^{k}, a^{m}]^{t} = [v_{k},a^{m}]^{t}\end{equation*}
   and $[\gamma_{\omega}(G),G]$ is normal in G, we get $[v_{k+1}, a^{n}] = [v_{k+1},(a^{p^{r}})^{n_{1}}] \in [\gamma_{\omega}(G), G]$. Since $[v_{k+1},a^{p^{r}}] \in \gamma_{\omega}(G)$, we have from Lemma 1 (1) (for $N=\gamma_{\omega}(G))$ that $[v_{k+1}, a^{p^{r}}]^{n_{1}} \in [\gamma_{\omega}(G), G]$. Furthermore, by Lemma 6 (3), $[v_{k+1}, a^{p^{r}}]^{p^{r}} \in [\gamma_{\omega}(G), G]$. But $\gcd(n_{1}, p^{r}) = 1$ and so we have $[v_{k+1}, a^{p^{r}}] \in [\gamma_{\omega}(G), G]$. Therefore $[v_{k},a^{p^{r}}] \in [\gamma_{\omega}(G),G]$ for all $k \in \mathbb{N}$. By (3.2), we get
   \begin{equation*}[v_{k},a^{p^{r}}] = [a, t^{k}, a^{p^{r}}] \in [\gamma_{\omega}(G), G]\end{equation*}
   for all $k \in \mathbb{N}$.

2. (2) Let r < s. In this case, we have $d = p^{r}$. By similar arguments as in case (1), we get $[u_k, a] \in [\gamma_{\omega}(G), G]$ for all $k \in \mathbb{N}$. Thus, it remains to show that $[v_k, a^{p^{r}}] \in [\gamma_{\omega}(G), G]$ for all $k \in \mathbb{N}$. By Corollary 1 (for $y = x = 1$ and $d = p^{r}$), $[v_{k}, a^{p^{r}}] \in \gamma_{\omega}(G)$ for all $k \in \mathbb{N}$. By Lemma 6 (3) (for $d = p^{r}$), $[v_{k}, a^{p^{r}}]^{p^{r}} \in [\gamma_{\omega}(G),G]$ for all $k \in \mathbb{N}$. We separate two cases.

1. (a) Let $2r \leq s$ and fix a positive integer $k \geq 1$. By Lemma 1 (1) (for $N = \gamma_{\omega}(G)$, $x = v_{k+1}$, $y = a^{p^{r}}$, $\kappa = p^{r} p^{s-2r}n_{1}$), we have

   (4.1)\begin{equation} \left[v_{k+1}, \left(a^{p^{r}}\right)^{p^{r}p^{s-2r}n_{1}}\right] \equiv \left[v_{k+1},a^{p^{r}}\right]^{p^{r} p^{s-2r}n_{1}}~({\rm mod}~[\gamma_{\omega}(G),G]). \end{equation}

   Since $[v_{k+1},a^{p^{r}}]^{p^{r}} \in [\gamma_{\omega}(G),G]$, we get $[v_{k+1},a^{p^{r}}]^{p^{r} p^{s-2r}n_{1}} \in [\gamma_{\omega}(G),G],$ and so by Equation (4.1), we have $[v_{k+1},a^{p^{s}n_{1}}] \in [\gamma_{\omega}(G),G]$. But $[v_{k+1},a^{p^{s}n_{1}}] = [v_{k},ta^{p^{s}n_{1}}t^{-1}]^{t} = [v_{k},a^{p^{r}m_{1}}]^{t}$. Since $[\gamma_{\omega}(G),G]$ is normal in G, we get

   (4.2)\begin{equation} \left[v_{k},a^{p^{r}m_{1}}\right] \in [\gamma_{\omega}(G),G]. \end{equation}

   Since $[v_{k}, a^{p^{r}}] \in \gamma_{\omega}(G)$, it follows from Lemma 1 (1) (for $N = \gamma_{\omega}(G)$, $x = v_{k}$, $y = a^{p^{r}}$, $\kappa = m_{1}$) that

   \begin{equation*} \left[v_{k}, a^{p^{r}m_{1}}\right] \equiv \left[v_{k},a^{p^{r}}\right]^{m_{1}} ~({\rm mod}~[\gamma_{\omega}(G),G]). \end{equation*}

   By Equation (4.2), we obtain $[v_{k},a^{p^{r}}]^{m_{1}} \in [\gamma_{\omega}(G),G]$. Since $[v_{k},a^{p^{r}}]^{p^{r}} \in [\gamma_{\omega}(G),G]$ and ${\rm gcd}(m_{1}, p^{r}) = 1$, we have $[v_{k},a^{p^{r}}] = [t^{-k}at^{k},a^{d}] \in [\gamma_{\omega}(G),G]$.

2. (b) Let $2r \gt s$ and fix a positive integer $k \geq 1$. Since $[v_{k},a^{p^{r}}] \in \gamma_{\omega}(G)$, it follows from Lemma 1 (1) (for $x = v_{k}$, $y = a^{p^{r}}$, $\kappa = m_{1}$) that $[v_{k}, a^{p^{r}m_{1}}] \in \gamma_{\omega}(G)$ and

   \begin{equation*} \left[v_{k},a^{p^{r}m_{1}}\right] \equiv \left[v_{k}, a^{p^{r}}\right]^{m_{1}}~({\rm mod}~[\gamma_{\omega}(G),G]). \end{equation*}

   Since $[\gamma_{\omega}(G),G]$ is normal in G,

   (4.3)\begin{equation} \left[v_{k}, a^{p^{r}m_{1}}\right]^{p^{2r-s}} \equiv \left[v_{k},a^{p^{r}}\right]^{p^{2r-s}m_{1}}~({\rm mod}~[\gamma_{\omega}(G),G]). \end{equation}

   Since $[v_{k+1}, a^{p^{s}n_{1}}] = [v_{k}, ta^{p^{s}n_{1}}t^{-1}]^{t} = [v_{k}, a^{p^{r}m_{1}}]^{t}$ and $[v_{k}, a^{p^{r}m_{1}}] \in \gamma_{\omega}(G)$, we have

   \begin{equation*} \left[v_{k+1}, a^{p^{s}n_{1}}\right] \equiv \left[v_{k}, a^{p^{r}m_{1}}\right]~({\rm mod}~[\gamma_{\omega}(G),G]). \end{equation*}

   By Equation (4.3), we get

   (4.4)\begin{equation} \left[v_{k+1}, a^{p^{s}n_{1}}\right]^{p^{2r-s}} \equiv \left[v_{k}, a^{p^{r}}\right]^{p^{2r-s}m_{1}}~({\rm mod}~[\gamma_{\omega}(G),G]). \end{equation}

   Since $[v_{k+1}, a^{p^{r}}] \in \gamma_{\omega}(G)$ and r < s, it follows from Lemma 1 (1) (for $x = v_{k+1}$, $y = a^{p^{r}}$, $\kappa = p^{s-r}n_{1}$) that

   (4.5)\begin{equation} \left[v_{k+1}, a^{p^{s}n_{1}}\right] \equiv \left[v_{k+1}, a^{p^{r}}\right]^{p^{s-r}n_{1}}~({\rm mod}~[\gamma_{\omega}(G),G]). \end{equation}

   By Equations (4.4) and (4.5), we have

   (4.6)\begin{equation} \left[v_{k}, a^{p^{r}}\right]^{p^{2r-s}m_{1}} \equiv \left[v_{k+1}, a^{p^{r}}\right]^{p^{r}}~({\rm mod}~[\gamma_{\omega}(G),G]). \end{equation}

   Since $[v_{k+1},a^{p^{r}}]^{p^{r}} \in [\gamma_{\omega}(G),G]$, we obtain by Equation (4.6) that $[v_{k}, a^{p^{r}}]^{p^{2r-s}m_{1}} \in [\gamma_{\omega}(G),G]$. Since ${\rm gcd}(p^{r}, p^{2r-s}m_{1}) = p^{2r-s}$, we get $[v_{k}, a^{p^{r}}]^{p^{2r-s}} \in [\gamma_{\omega}(G),G]$. If $3r \leq 2s$, then, by applying similar arguments as in case $2r \leq s$, we have $[v_{k}, a^{p^{r}}] \in [\gamma_{\omega}(G),G]$. If $3r \gt 2s$, then, by applying similar arguments as in case $2r \gt s$, we get $[v_{k},a^{p^{r}}]^{p^{3r-2s}} \in [\gamma_{\omega}(G),G]$. Since $2r-s \gt 3r-2s \gt \cdots$ and since there is y such that $(y+1)r \leq ys$ (for $\frac{r}{s-r} \in \mathbb{N}$, let $y = \frac{r}{s-r},$ and for $\frac{r}{s-r} \notin \mathbb{N}$, let y be the integral part of $\frac{r}{s-r}$), by continuing this process, we obtain $[v_{k}, a^{p^{r}}] \in [\gamma_{\omega}(G),G]$.

1. (3) Let s < r. By applying similar arguments as in case (2), we obtain the required result.

By cases (1), (2) and (3), we get the desired result.

Proof. Let $d = \gcd(m, n)$. Since, by Lemma 2, $N_{\omega} \leq \gamma_{\omega}(G),$ and by Proposition 2, N_ω coincides with the normal closure of the set $\{[t^{-k}a^{d}t^{k}, a]: k \in \mathbb{Z}\}$ in G, it suffices to show that $[t^{-k}a^{d}t^{k}, a] \in [\gamma_{\omega}(G), G]$ for all $k \in \mathbb{Z}$. Furthermore, since $[t^{-k}a^xt^k,a^y]=([t^{k}a^yt^{-k},a^x]^{-1})^{t^k}$ and $[\gamma_{\omega}(G),G]$ is normal in G, it is enough to show that $[t^{-k}a^{d}t^{k}, a], [t^{-k}at^{k}, a^{d}] \in [\gamma_{\omega}(G), G]$ for all $k \in \mathbb{N}$. For d = 1, the required result follows from Lemma 6 (1) and so, from now on, we may assume that d > 1. Let $d = p_{1}^{\mu_{1}}\ldots p_{\lambda}^{\mu_{\lambda}}$ be the prime factor decomposition of d. To prove the result, we use induction on λ. For λ = 1, the required result follows from Proposition 5. Assume that the result is true for some $\lambda \geq 1$ and let $d = p_{1}^{\mu_{1}}\ldots p_{\lambda+1}^{\mu_{\lambda+1}}$. Thus, we may write $m = p_{1}^{r_{1}}\ldots p_{\lambda+1}^{r_{\lambda+1}}\mu$ and $n = p_{1}^{s_{1}}\ldots p_{\lambda+1}^{s_{\lambda+1}}\nu$, where $\mu_{i} = \min\{r_{i}, s_{i}\}$ and $\gcd(p_{i}, \mu) = \gcd(p_{i}, \nu) = 1$, with $i = 1, \ldots, \lambda+1$. For $j \in \{1, \ldots, \lambda+1\}$, let $u = a^{p_{j}^{\mu_{j}}}$ and let K_j be the subgroup of G generated by the set $\{u, t\}$. For convenience, we write $m_{j} = m/({p_{j}^{\mu_{j}}})$, $n_{j} = n/({p_{j}^{\mu_{j}}})$ and $d_{j} = d/({p_{j}^{\mu_{j}}})$. We point out that $K_{j} = {\rm BS}(m_{j}, n_{j})$ (see [Reference Levitt7, Lemma 7.10]). Since $\gcd(m_{j}, n_{j}) = p_{1}^{\mu_{1}}\ldots p_{j-1}^{\mu_{j-1}}p_{j+1}^{\mu_{j+1}} \ldots p_{\lambda+1}^{\mu_{\lambda+1}}$, by our inductive argument, we get $[t^{-k}u^{d_{j}}t^{k}, u] \in [\gamma_{\omega}(K_{j}), K_{j}] \subseteq [\gamma_{\omega}(G), G]$, that is, $\left[t^{-k}a^{p_{j}^{\mu_{j}}d_{j}}t^{k}, a^{p_{j}^{\mu_{j}}}\right] \in [\gamma_{\omega}(G), G]$ and so $\left[t^{-k}a^{d}t^{k}, a^{p_{j}^{\mu_{j}}}\right] \in [\gamma_{\omega}(G), G]$. Since $[t^{-k}a^{d}t^{k}, a] \in \gamma_{\omega}(G)$, it follows that $[t^{-k}a^{d}t^{k}, a]^{p_{j}^{\mu_{j}}} \in [\gamma_{\omega}(G), G]$. Therefore, for $j_{1}, j_{2} \in \{1, \ldots, \lambda+1\}$, with $j_{1} \neq j_{2}$, we have $[t^{-k}a^{d}t^{k}, a]^{p_{j_{1}}^{\mu_{j_{1}}}}, [t^{-k}a^{d}t^{k}, a]^{p_{j_{2}}^{\mu_{j_{2}}}} \in [\gamma_{\omega}(G), G]$. Since ${\rm gcd}\left(p_{j_{1}}^{\mu_{j_{1}}}, p_{j_{2}}^{\mu_{j_{2}}}\right) = 1$, we get $[t^{-k}a^{d}t^{k}, a] \in [\gamma_{\omega}(G), G]$ and the result follows.

Proof. Let us assume first that there is some prime number p such that $m\equiv n\pmod p$. Then by Proposition 4 (1), $\gamma_{\omega}(G)=N_{\omega},$ and so by Proposition 6, $\gamma_{\omega}(G)=[\gamma_{\omega}(G),G]$. On the other hand, if $m\not\equiv n\pmod p$ for every prime integer p, then $m-n=\pm 1$ and so the relation $t^{-1}a^mt=a^n$ becomes $t^{-1}a^{n\pm 1}t=a^n$ or equivalently $t^{-1}a^{n\pm 1}ta^{-(n\pm 1)}=a^{\mp 1}$ or $[t,a^{-(n\pm 1)}]=a^{\mp 1}$. By Proposition 4 (2), we have $a\in \gamma_{\omega}(G)$. Since $[t,a^{-(n\pm 1)}]=a^{\mp 1}$, we obtain $a\in[\gamma_{\omega}(G),G]$. By Proposition 4 (2), $\gamma_{\omega}(G) \subseteq [\gamma_{\omega}(G),G]$ and the result follows.

Proof. Fix some $k \in \mathbb{Z}$. By Proposition 2, $[t^{-k}a^{d}t^{k}, a] \in N_{\omega}$ and so by Proposition 6, we have $[t^{-k}a^{d}t^{k}, a] = [t^{-k}a^{\mu \nu}t^{k},a] \in [\gamma_{\omega}(G), G]$. Using Equation (3.3) as many times as needed and since $[\gamma_{\omega}(G),G]$ is a normal subgroup of G, we obtain $[t^{-k}a^{\mu\nu}t^{k}, a^{\nu}] \in [\gamma_{\omega}(G), G]$. By Lemma 5, $[t^{-k}a^{\mu}t^{k}, a^{\nu}] \in \gamma_{\omega}(G)$ and so by Lemma 1 (3) (for $N=\gamma_{\omega}(G))$, $[t^{-k}a^{\mu}t^{k}, a^{\nu}]^{\nu} \in [\gamma_{\omega}(G), G]$. By Proposition 2, Proposition 6 and since N_ω is a normal subgroup of G, we get $[t^{-k}at^{k}, a^{d}] \in [\gamma_{\omega}(G), G]$. Using Equation (3.2) as many times as needed and since $[\gamma_{\omega}(G),G]$ is a normal subgroup of G, we obtain $[(t^{-k}at^{k})^{\mu}, a^{\mu\nu}] = [t^{-k}a^{\mu}t^{k}, a^{\mu \nu}] \in [\gamma_{\omega}(G), G]$. By Lemma 5, $[t^{-k}a^{\mu}t^{k}, a^{\nu}] \in \gamma_{\omega}(G)$ and so by Lemma 1 (2) (for $N=\gamma_{\omega}(G))$, we have $[t^{-k}a^{\mu}t^{k}, a^{\nu}]^{\mu} \in [\gamma_{\omega}(G), G]$. But $\gcd(\mu, \nu) = 1$; therefore, we get the result.

Proof. Let $d={\rm gcd}(m,n)$, with $m=dm_1$, $n=dn_1$ and ${\rm gcd}(m_1,n_1)=1$. Let n = m. By Theorem 1 (1), Proposition 7 and $[\gamma_{\omega}(G),G] \leq \gamma_{\omega}(G)$, we have $[\gamma_{\omega}(G),G] = \gamma_{\omega}(G)$. If there is a prime number p such that $m_1\equiv n_1\pmod p$, then by Theorem 1 (2), Proposition 7 and $[\gamma_{\omega}(G),G] \leq \gamma_{\omega}(G)$, we have the required result. Finally, let $m_1\not\equiv n_1 \pmod p$ for every prime number p. By Proposition 7 and Theorem 1 (2), it is sufficient to show that $a^d\in [\gamma_{\omega}(G),G]$. Since $m_1\not\equiv n_1 \pmod p$ for every prime number p, we have $m_1-n_1=\pm 1$. Hence, the relation $t^{-1}a^mt=a^n$ implies that $t^{-1}a^{d(n_1 \pm 1)}t=a^{dn_1}$ or equivalently $t^{-1}a^{d(n_1\pm 1)}t=a^{d(n_1\pm 1)}a^{\mp d}$ or $[a^{d(n_1\pm 1)}, t]=a^{\mp d}$. By Theorem 1 (2), $a^d\in\gamma_{\omega}(G)$ and so we obtain the required result.

Proof. For all $c \in \mathbb{N}$, $\gamma_{c}(F/N) = \gamma_{c}(F)N/N$. We have the following natural isomorphisms as abelian groups

\begin{equation*} \begin{array}{rll} {\rm gr}_{c}(F/N) & \cong & \frac{\gamma_{c}(F)N}{\gamma_{c+1}(F)N} = \frac{\gamma_{c}(F)(\gamma_{c+1}(F)N)}{\gamma_{c+1}(F)N} \\ & \cong & \frac{\gamma_{c}(F)}{\gamma_{c}(F) \cap (\gamma_{c+1}(F)N)}. \end{array} \end{equation*}

Since $\gamma_{c+1}(F) \subseteq \gamma_{c}(F)$, by the modular law, we have $\gamma_{c}(F) \cap (\gamma_{c+1}(F) N) = \gamma_{c+1}(F)(\gamma_{c}(F) \cap N)$. Therefore, for all $c \in \mathbb{N}$,

\begin{equation*} {\rm gr}_{c}(F/N) \cong \frac{\gamma_{c}(F)}{\gamma_{c+1}(F)(\gamma_{c}(F) \cap N)} \cong \frac{{\rm gr}_{c}(F)}{\frac{\gamma_{c+1}(F)N_{c}}{\gamma_{c+1}(F)}} = \frac{{\rm gr}_{c}(F)}{{\rm I}_{c}(N)} \end{equation*}

as abelian groups in a natural way.

Proof. Write $\overline{x} = x\gamma_{2}(F)$ and $\overline{y} = y\gamma_{2}(F)$. Let ${\rm gr}_{1,R}(F)$ be the subgroup of ${\rm gr}_{1}(F)$ generated by ${\mathcal{R}}_{\kappa} = \{\overline{x}, \kappa \overline{y}\}$. It is a free abelian group of rank 2 and has a finite index $|\kappa|$ in ${\rm gr}_{1}(F)$. For a positive integer c, let ${\mathcal{L}}_{c, \mathcal{R}_{\kappa}} = {\mathcal{L}}_{\mathcal{R}_{\kappa}} \cap {\rm gr}_{c}(F)$. That is, ${\mathcal{L}}_{c,{\mathcal{R}}_{\kappa}}$ is spanned by all Lie commutators of the form $[y_{i_{1}}, \ldots, y_{i_{c}}]$ with $y_{i_{j}} \in {\mathcal{R}}_{\kappa}$, $j = 1, \ldots, c$. Clearly, ${\mathcal{L}}_{\mathcal{R}_{\kappa}} = \bigoplus_{c \geq 1}{\mathcal{L}}_{c, \mathcal{R}_{\kappa}}$. Let $w(\overline{x}, \overline{y})$ be a non-zero Lie commutator in ${\rm gr}_{c}(F)$ and let r be the number of occurrences of $\overline{y}$ in $w(\overline{x}, \overline{y})$. By the linearity of Lie bracket, we have $\kappa^{r} w(\overline{x}, \overline{y}) = w(\overline{x}, \kappa \overline{y})$. Since ${\rm gr}_{c}(F)$ is an abelian group generated by the basic Lie commutators $v(\overline{x}, \overline{y})$ (of length c) and since for each $v(\overline{x}, \overline{y})$ there exists a non-zero power of κ, say κ ^r, depending on the number of occurrences of $\overline{y}$, such that $\kappa^r v(\overline{x}, \overline{y}) = v(\overline{x}, \kappa \overline{y}) \in {\mathcal{L}}_{c, \mathcal{R}_{\kappa}}$, we get ${\mathcal{L}}_{c, \mathcal{R}_{\kappa}}$ has a finite index in ${\rm gr}_{c}(F)$. So rank$({\mathcal{L}}_{c, \mathcal{R}_{\kappa}}) = {\rm rank}({\rm gr}_{c}(F))$.

Since ${\rm gr}(F)$ is a free Lie algebra on the set $\{\overline{x}, \overline{y}\}$ and ${\mathcal{L}}_{\mathcal{R}_{\kappa}}$ is generated as a Lie algebra by the set $\{\overline{x}, \kappa \overline{y}\}$, the natural Lie homomorphism ψ from ${\rm gr}(F)$ into ${\mathcal{L}}_{\mathcal{R}_{\kappa}}$, with $\psi(\overline{x}) = \overline{x}$ and $\psi(\overline{y}) = \kappa \overline{y}$, is surjective. It is easily verified that, for any positive integer c, ψ induces a $\mathbb{Z}$-linear mapping ψ_c from ${\rm gr}_{c}(F)$ onto ${\mathcal{L}}_{c,{\mathcal{R}}_{\kappa}}$. Namely, $\psi_{c}(u(\overline{x}, \overline{y})) = u(\overline{x}, \kappa \overline{y})$ for all $u(\overline{x}, \overline{y}) \in {\rm gr}_{c}(F)$. Since ${\mathcal{L}}_{c,{\mathcal{R}}_{\kappa}}$ has finite index in ${\rm gr}_{c}(F)$ and ${\rm gr}_{c}(F)$ is a free abelian group of finite rank, we obtain ψ_c is an isomorphism of abelian groups. Let $\overline{w} \in {\rm Ker}\psi$. Since ${\rm gr}(F)$ is graded, without loss of generality, we may assume that $\overline{w} = w\gamma_{c+1}(F) \in {\rm gr}_{c}(F)$ for some $w \in \gamma_{c}(F)$ and $c \in \mathbb{N}$. To get a contradiction, we assume that $w = w(x,y) \in \gamma_{c}(F) \setminus \gamma_{c+1}(F)$. Then

\begin{equation*}0=\psi(\overline{w}) = \psi(w(\overline{x}, \overline{y})) = \psi_{c}(w(\overline{x}, \overline{y})).\end{equation*}

Since ψ_c is an isomorphism, we have $w(\overline{x},\overline{y})=0$ in ${\rm gr}_{c}(F)$. This implies that $w(x,y) \in\gamma_{c+1}(F)$, a contradiction. By the above, ψ is an isomorphism of Lie algebras. Hence, ${\mathcal{L}}_{\mathcal{R}_{\kappa}}$ is a free Lie algebra with a free generating set $\mathcal{R}_{\kappa}$. By the elimination theorem (see, for example, [Reference Bourbaki3, Chapter 2, Section 2.9, Proposition 10]), ${\mathcal{L}}_{\mathcal{R}_{\kappa}} = \langle \overline{x} \rangle \oplus L(\{\kappa \overline{y}\} \wr \{\overline{x}\})$, where the free Lie algebra $L(\{\kappa \overline{y}\} \wr \{\overline{x}\})$ is the ideal in ${\mathcal{L}}_{\mathcal{R}_{\kappa}}$ generated by $\kappa \overline{y}$. Clearly, $L(\{\kappa \overline{y}\} \wr \{\overline{x}\}) = \langle \kappa \overline{y} \rangle \oplus \left(\bigoplus_{c \geq 2} {\mathcal{L}}_{c,{\mathcal{R}}_{\kappa}}\right)$.

Proof. Write $\delta = n-m$, $\overline{x} = x\gamma_{2}(F)$ and $\overline{y} = y \gamma_{2}(F)$. Assume that δ ≠ 0 and let ${\mathcal{R}}_{\delta} = \{\overline{x}, \delta \overline{y}\}$. By Proposition 8, ${\mathcal{L}}_{\mathcal{R}_{\delta}}$ is a free Lie algebra on the set ${\mathcal{R}}_{\delta}$. Note that ${\rm I}_{1}(N) = \langle \delta \overline{y} \rangle$. Since ${\mathcal{L}}(N)$ is an ideal in ${\rm gr}(F)$, we have $L(\{\delta \overline{y}\} \wr \{\overline{x}\}) \subseteq {\mathcal{L}}(N)$. Hence, for all $c \geq 2$, ${\mathcal{L}}_{c,{\mathcal{R}}_{\delta}} \subseteq {\rm I}_{c}(N)$. By Proposition 8, for any $c \geq 2$, ${\mathcal{L}}_{c,{\mathcal{R}}_{\delta}}$ has finite index in ${\rm gr}_{c}(F)$ and so ${\rm I}_{c}(N)$ has finite index in ${\rm gr}_{c}(F)$. By Lemma 7, we obtain, for any $c \geq 2$, ${\rm gr}_{c}(F/N)$ is a finite abelian group.

Thus, we may assume that δ = 0. In this case, let ${\mathcal{R}}_{m} = \{\overline{x}, m \overline{y}\}$. Then ${\rm I}_{1}(N) = \{0\}$. We use similar arguments as before. By Proposition 8, ${\mathcal{L}}_{\mathcal{R}_{m}}$ is a free Lie algebra on the set ${\mathcal{R}}_{\delta}$. Note that ${\rm I}_{2}(N) = \langle [\overline{x}, m \overline{y}] \rangle$. Since ${\mathcal{L}}(N)$ is an ideal in ${\rm gr}(F)$, we have $\bigoplus_{c \geq 2}\mathcal{L}_{c,{\mathcal{R}}_{m}} \subseteq {\mathcal{L}}(N)$. Hence, for all $c \geq 2$, ${\mathcal{L}}_{c,{\mathcal{R}}_{m}} \subseteq {\rm I}_{c}(N)$. By Proposition 8, for any $c \geq 2$, ${\mathcal{L}}_{c,{\mathcal{R}}_{m}}$ has finite index in ${\rm gr}_{c}(F)$ and so ${\rm I}_{c}(N)$ has finite index in ${\rm gr}_{c}(F)$. By Lemma 7, we obtain, for any $c \geq 2$, ${\rm gr}_{c}(F/N)$ is a finite abelian group.

Proof. For $m, n \in \mathbb{Z} \setminus \{0\}$, let N be the normal closure of the element $r = y^{n-m}[x,y^{m}]$ in F. Write t = xN and a = yN. Clearly, the quotient group $F/N$ has a presentation $\langle t, a \mid a^{n-m} [t,a^{m}] = 1\rangle$. So, by Proposition 9, we obtain the desired result.