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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.

user1729
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Chris Z
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    @onurcanbektas That is not true: the free group on two generators has a subgroup that is not finitely generated. – Angina Seng Oct 27 '18 at 13:41
  • @LordSharktheUnknown Can you provide that example ? – Our Oct 27 '18 at 14:09
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    @onurcanbektas Take the subgroup of $F_2$ generated by ${ab^na \mid n \geqslant 0}$ – J.-E. Pin Oct 27 '18 at 14:27
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    @onurcanbektas Say free group on $x,y$ then the group generated by $y^nxy^{-n}$ for $n>0$ is an example(which is pretty easy to verify). In fact if you have a surjective homomorphism to the integers from a free group of rank more than 1 then it is not finitely generated(all infinite index normal subgroups are not finitely generated) –  Oct 27 '18 at 14:40
  • @PaulPlummer I did not know that. Thanks for pointing out. – Our Oct 27 '18 at 15:52

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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.

Lee Mosher
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YCor
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  • This seems to be a much more difficult result than I expected considering it is an exercise in a fairly elementary textbook. – Chris Z Oct 31 '18 at 17:10
  • Actually the initial proof I once saw (I don't know where) was probably more "by hand". Actually none of the steps of the proof I sketched is really difficult, but these are standard facts/constructions, and useful for other purposes. Anyway it was certainly not an easy exercise. – YCor Oct 31 '18 at 18:07