Let $G$ be a finitely generated group. The following is a part 1 of proposition 6.3.10 in Clara Löh's Geometric Group Theory and is left as an exercise, but I can`t figure it out.
Let $G$ be a finitely generated group with polynomial growth that admits a surjective homomorphism $\pi: G \to \mathbb{Z}$. Show that the kernel of $\pi$ is finitely generated.
Best Answer
Let $N$ be the kernel of $\pi$ and fix $t\in\pi^{-1}(1)$. Since $G$ is finitely generated and $\mathbf{Z}$ is finitely presented, the kernel is finitely normally generated: there exists a finitely generated subgroup $M$ of $N$ such that $N$ is generated by $\bigcup_{n\in\mathbf{Z}}t^nMt^{-n}$.
If $M$ is both contained in the subgroup generated by $\bigcup_{n\ge 1}t^nMt^{-n}$ and in the one generated by $\bigcup_{n\le -1}t^nMt^{-n}$, then for some $N$ it is both contained in the subgroup generated by $\bigcup_{1\le n\le N}tMt^{-1}$ and in the one generated by $\bigcup_{-N\le n\le -1}t^nMt^{-n}$, and it is easy to deduce that $N$ is finitely generated.
Otherwise, up to change $t$ to $t^{-1}$, $M$ is not contained in the subgroup $N_1=\bigcup_{n\ge 1}t^nMt^{-n}$. Hence, denoting $N_k$ as the subgroup generated by $\bigcup_{n\ge k}t^nMt^{-n}$, we have $N_1=tN_0t^{-1}$ strictly contained in $N_0$, and $N=\bigcup_{k\in\mathbf{Z}} N_k$, so $G$ is a strictly ascending HNN-extension of $N_0$. In particular, by standard arguments, $G$ has exponential growth and contains a free subsemigroup on 2 generators.