Question

A group $\mathit{H}$is said to be finitely generated if there is a finite subset $\mathit{S}$of $\mathit{H}$such that $\mathit{H}\mathbf{=}\mathbf{⟨}\mathbf{S}\mathbf{⟩}$ (see Theorem 7.18). If$\mathit{N}$ is a normal subgroup of a group G such that the groups $\mathit{N}$and$\mathit{G}\mathbf{/}\mathit{N}$ are finitely generated, prove that$\mathit{G}$ is finitely generated.

Short answer

It is proved that,$G$ is finitely generated.

Step-by-step solution

 Step 1: Given that

Consider $G$be a group and $N$be a normal subgroup of $G$.

Consider $N$and $G/N$are finitely generated.

Now, our aim is to show that $G$is finitely generated.

Assume that$N$ and$G/N$ are generated by finite sets$S=\{s_1,\dots ,s_l\}$ and $T=\{t_{1}N,\dots ,t_{k}N\}$, respectively, where$l,k\in \mathbb{N}$ .

Prove the result

For any $g\in G$, there exists $i_1,\dots ,i_k\in \mathbb{Z}$such that:

$\begin{array}{c}gN={\left(t_{1}N\right)}^{i_1}\dots {\left(t_{k}N\right)}^{i_k}\\ =t_1^{i}N\dots t_k^{i_k}N\\ =t_1^i\dots t_k^{i_k}N\end{array}$

This implies that, $g^{-1}{t}_1^i\dots t_k^{i_k}\in N$.

As we know that $N$is generated by $S$, this implies that $g^{-1}{t}_1^i\dots t_k^{i_k}=s_1^{j_1}\dots s_l^{j_i}$, where $j_1,\dots ,j_i\in \mathbb{Z}$.

Therefore, $g=t_1^i\dots t_k^{i_k}{s}_1^{j_1}\dots s_l^{j_i}$.

This implies that,$G$is generated by the set $\{s_1,\dots ,s_l,t_1,\dots ,t_k\}$, as $g$was an arbitrary element.

Thus, it is proved that$G$ is finitely generated.