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Consider a finitely generated group $G$ and a subnormal series of $G$: $$1=G_0\trianglelefteq G_1\trianglelefteq\cdots\trianglelefteq G_{n-1}\trianglelefteq G_n=G$$ Now, suppose that $G_1$ fibres, i.e. there is an epimorphism $\varphi:G_1\to\mathbb{Z}$ with $\ker(\varphi)$ finitely generated. Is there any way to lift the fibration onto $G$? i.e. to construct an epimorphism $\varphi':G\to\mathbb{Z}$ with $\ker(\varphi)$ finitely generated.

I know the answer is no in general, but maybe it is true if we assume some extra properties such as $G_i/G_{i-1}$ being $F_\infty$. More generaly, I would like to study this problem when we know that $G_i/G_{i-1}$ fibres insted of the first quotient.

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    $\begingroup$ Consider for example the subgroup $G$ of $\operatorname{\rm SL}(3,\mathbb{Z})$ consisting of upper triangular matrices with ones on the diagonal, with $G_1$ the central copy of $\mathbb{Z}$ in the top right corner. Any extra properties you require had better not be satisfied by this mild looking group. $\endgroup$
    – Dave Benson
    Commented Jan 17 at 12:28
  • $\begingroup$ @DaveBenson Nice example. But what about ''easy groups''? Maybe we can add some reasonable property to $G$ that avoids weird behaviours. $\endgroup$
    – Marcos
    Commented Jan 17 at 12:50
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    $\begingroup$ When you say "lift" do you mean you want the restriction of $\varphi'$ to $G_1$ to equal $\varphi$? Or do you just want $G$ to fiber, period? Either way I think there's not a good way to ensure this, for example it's easy to find groups with finite abelianization that have finite index subgroups that fiber (e.g., various right-angled Coxeter groups). So even in the "very nice" case when $n=1$ and $G/G_1$ is finite this doesn't need to happen. $\endgroup$
    – Matt Zaremsky
    Commented Jan 17 at 15:09

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