If $G/H$ and $H$ are finitely generated, then so is $G$

Question

I'm trying to prove that if $H$ is a normal subgroup of a group $G$ such that $H$ and $G/H$ are finitely generated, then G is finitely generated also. I'm trying to find a finite set $X$ such that $G$ is generated by $X$, but I have no ideal how to find this set using the finite generator sets of $H$ and $G/H$.

I need help

Thanks in advance

Answer 1

Hints: we're given

$$H=\langle\,h_1,\ldots,h_k\,\rangle\;,\;\;G/H:=\langle\,g_1H,\ldots,g_nH\,\rangle$$

Remember now that for all $\,x\in G\,$ there exist unique $\,1\le i_x\le n\,$ and unique $\,h_x\in H\,$ s.t. $\,x=g_{i_x}h_x\,$ and, of course, then $\,x\in g_{i_x}H\,$ , so...

Answer 2

Let $S$ be a finite generating set for $H$. By the normality of $H$, we have for all $g \in G$ that $g = h_{1}gh^{-1}$ for some $h,h_{1} \in H$.

Using the generalized associative law for $G$ and reindexing if necessary we have that

$g = gh_{1_S}=g($finite product in $S)$.

Since $G/H$ partitions $G$ we know that $gh_{1_S}$ is in one of the cosets of $G$ in $H$, meaning that

$gh_{1_S} \in ($finite product of elements in $G$ taken from a (finite subset of $G)=G_{1} )H$ using the fact that in our case, $(gH)^{k} = g^{k}H$ as well as a similar technique to the one above to put $H$ at the end.

Then $g = ($finite product in $G_{1})($finite product in $S)$ for all $g \in G$ where $|S|,|G_{1}|$ are finite.

Hence, one can take $S \cup G_{1}$ to be a finite generating set for $G$.