If H is a normal subgroup of a group G such that H and G/H are finitely generated, then so is G.
Proof- Let $H = \langle x_1,...,x_n \rangle$ and $G/H = \langle y_1H,...,y_mH \rangle$. Claim that $G = \langle x_1,...,x_n,y_1,...,y_m \rangle$. Note first that $\langle x_1,...,x_n,y_1,...,y_m \rangle \subset G$ by closure of the group G. Let $g\in G$. Then $gH\in G/H$ so that we can write $gH = (g_1H)^{n_1}\dots (g_kH)^{n_k}$, where each $g_i = y_j$ for some $1\leq j \leq m$ and $n_i \in \mathbb{Z}$. Since $H$ is normal in $G$, we have $G/H$ is a group under multiplication so we can change $$gH = (g_1H)^{n_1}\dots (g_kH)^{n_k} = g_1^{n_1} \dots g_k^{n_k}H$$. By definition of cosets this gives $g^{-1}(g_1^{n_1} \dots g_k^{n_k}) \in H = \langle x_1,...,x_n \rangle$. Thus $g^{-1}(g_1^{n_1} \dots g_k^{n_k}) = h_1^{m_1} \dots h_l^{m_l}$, where each $h_i = x_j$ for some $1 \leq j \leq n$ and $m_i \in \mathbb{Z}$. It then follows that $g = g_1^{n_1} \dots g_k^{n_k}(h_1^{m_1} \dots h_l^{m_l})^{-1} \in \langle x_1,...,x_n,y_1,...,y_m \rangle$
here is the proof , my question is that why does $g_1^{n_1} \dots g_k^{n_k}(h_1^{m_1} \dots h_l^{m_l})^{-1} \in \langle x_1,...,x_n,y_1,...,y_m \rangle$?
my additional question is that, in this line, "Note first that $\langle x_1,...,x_n,y_1,...,y_m \rangle \subset G$ by closure of the group G", why is it by closure property?
I have edited my proof to make it more detailed:
here is my updated proof:
Let $H \vartriangleleft G$ such that $H$ and $G/H$ are finitely generated. Then\
$H = \langle x_1,\ldots,x_n \rangle$ and $G/H = \langle y_1H,\ldots,y_mH \rangle$ ,
where $\{x_1,\ldots,x_n\} \subseteq H$ and $\{y_1H,\ldots,y_mH\} \subseteq G/H$
\textbf{Claim}: $G = \langle x_1,\ldots,x_n,y_1,\ldots,y_m \rangle$.
\noindent Let $g \in G$. Then $gH \in G/H$.
\noindent By Theorem 2.8, we can write it as \begin{align*} gH = (g_1H)^{n_1}\cdots(g_kH)^{n_k} \end{align*}
\noindent where each $g_i = y_j$ for some $1 \leq j \leq m$, $i = 1,\ldots,k$ and $n_i \in \mathbb{Z}$, $k < \infty$.
Since $H$ is normal in $G$, we have $G/H$ is a group under multiplication by Theorem 5.4. So that \begin{align*} gH = g_1^{n_1} \cdots g_k^{n_k}H. \end{align*}
\noindent By Corollary 4.3 (iii), this gives \begin{align*} g^{-1}(g_1^{n_1} \cdots g_k^{n_k}) \in H = \langle x_1,\ldots,x_n \rangle . \end{align*}
Hence \begin{align*} g^{-1}(g_1^{n_1} \cdots g_k^{n_k}) = h_1^{m_1} \cdots h_l^{m_l} \end{align*} where each $h_i = x_j$ for some $1 \leq j \leq n$, $i = 1,\ldots,l$ and $m_i \in \mathbb{Z}$, $l < \infty$.
Since $h_1^{m_1} \cdots h_l^{m_l} \in H$, we have \begin{align*} g^{-1}(g_1^{n_1} \cdots g_k^{n_k}) (h_1^{m_1} \cdots h_l^{m_l})^{-1} &= h_1^{m_1} \cdots h_l^{m_l} (h_1^{m_1} \cdots h_l^{m_l})^{-1}\\ &= e. \end{align*}
Also, $g^{-1} \in G$. Thus \begin{align*} gg^{-1}(g_1^{n_1} \cdots g_k^{n_k}) (h_1^{m_1} \cdots h_l^{m_l})^{-1} &= ge\\ e(g_1^{n_1} \cdots g_k^{n_k}) (h_1^{m_1} \cdots h_l^{m_l})^{-1} &=ge\\ (g_1^{n_1} \cdots g_k^{n_k}) (h_1^{m_1} \cdots h_l^{m_l})^{-1}&=g. \end{align*}
\noindent It then follows that \begin{align*} g = (g_1^{n_1} \cdots g_k^{n_k})(h_1^{m_1} \cdots h_l^{m_l})^{-1} \in \langle x_1,\ldots,x_n,y_1,\ldots,y_m \rangle. \end{align*}
\noindent Thus \begin{align*} G \subseteq \langle x_1,\ldots,x_n,y_1,\ldots,y_m \rangle. \end{align*}.
Moreover, by closure property in G \begin{align*} \langle x_1,\ldots,x_n,y_1,\ldots,y_m \rangle \subseteq G. \end{align*}
Hence \begin{align*} G = \langle x_1,\ldots,x_n,y_1,\ldots,y_m \rangle. \end{align*} \ \noindent Therefore, G is finitely generated. \qed
Is my proof correct? Pleasee help me