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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!

Shaun
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    Take $G$ to be the free group of rank $2$ generated by $x$ and $y$, and let $H$ be the normal closure of $y$. Is this finitely generated? – Arturo Magidin Sep 09 '19 at 16:25
  • In general, if $G$ is free of rank at least two, and $H$ is a normal subgroup such that $G/H$ is infinite then $H$ is not finitely generated. I wrote out a proof of this in an old answer here (the proof is at the very end). Obviously this is applicable here, as is being hinted at in the above comment as well as the given answer! – user1729 Sep 10 '19 at 09:38

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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 $(\Bbb Z, +)$.

Shaun
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