Suppose that $G$ is a finitely presented group, $H$ is a subgroup and $G/H$ is isomorphic to $\mathbb{Z}$. Can I deduce from here that $H$ is finitely generated?
I have tried to prove it without success, but I'm not able to find a counterexample neither. Can someone give me a hint, please? Thanks in advance!
The normal closure $\langle\langle b\rangle\rangle$ of $b$ in $$\langle a,b\mid \varnothing \rangle\tag{$\mathcal{P}$}$$ is not finitely generated; one of its generating sets is $\{ a^kba^{-k}\mid k\in\Bbb Z\}$, and one can see from this that $\langle\langle b\rangle\rangle$ is not finitely generated.
But $$\langle a,b\mid \varnothing \rangle/\langle\langle b\rangle\rangle\tag{$\mathcal{Q}$}$$ is just killing $b$ in $(\mathcal{P})$, so $(\mathcal{Q})$ is isomorphic to the free group of rank one, known to be isomorphic to $(\Bbb Z, +)$.