Let $G=A\ast \mathbb{Z}$ be the free product of a group $A$ and the cyclic group $\mathbb{Z}$ and suppose $K$ is a subgroup of $G$. By Kurosh Subgroup Theorem we know that $K=F\ast (\ast_{i\in I}(K\cap A^{u_i}))$, where $F$ is free group and $u_i$ are some representatives of double cosets $KxA$ in $G$.
Now suppose further that $A$ has ACC on normal subgroups and $K$ is normal.
Is it true that $K$ is finitely generated?
(This will be true if we can show that $|I|$ and $\operatorname{rank}\ F$ are finite.)
Best Answer
Set $A$ equal to $\mathbb{Z}$, which satisfies the ascending chain condition ("ACC", every strictly ascending chain of (normal) subgroups eventually terminates). Then $G=\mathbb{Z}\ast\mathbb{Z}=F_2$ and $F_2$ contains normal subgroups that are not finitely generated.
Examples:
1) The commutator subgroup is normal and not finitely generated.
2) The subgroup generated by $\left\{b^k a b^{-k}\ |\ k\in\mathbb{Z}\right\}$ is normal and not finitely generated.