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

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...