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A finitely generated group $G$ algebraically fibers if there is an epimorphism $G\to\mathbb{Z}$ with finitely generated kernel. Since this kernel is finitely generated, we can ask whether *it* algebraically fibers. If it does, then we can keep going, etc etc etc. In any examples I can think of, this process always terminates, either by hitting a group that does not map onto $\mathbb{Z}$ at all (for example a perfect group), or that does but never with finitely generated kernel (for example a free group). But I don't see any reason in general that it should necessarily terminate, and I'm curious to find an example where it doesn't.

So, here is the concrete question: Does there exist an infinite sequence $G_1,G_2,G_3,\dots$ of finitely generated groups such that for each $i$ there exists a short exact sequence $1\to G_{i+1} \to G_i \to \mathbb{Z} \to 1$? It would suffice for example to find a residually solvable, non-solvable group whose derived series consists entirely of finitely generated groups. Another thing that would suffice is finding a finitely generated group $G$ and an automorphism $\phi$ of $G$ such that $G\cong G\rtimes_\phi \mathbb{Z}$.

One reason to be interested in such an example would be a sort of domino effect of BNS-invariants: in such a sequence we would conclude that each $\Sigma^1(G_i)$ is non-empty, as "revealed" by $G_{i+1}$.

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  • $\begingroup$ (Oh, in the residually solvable, non-solvable thing, the derived subgroups would also have to all have infinite abelianization.) $\endgroup$
    – Matt Zaremsky
    Commented Oct 2, 2022 at 13:54

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Let $G_1$ be the wreath product of $\mathbb{Z}$ with $\mathbb{Z}$. There is a surjective homomorphism $G_1\rightarrow \mathbb{Z}$ whose kernel $G_2$ is isomorphic to $G_1$. This gives a sequence of groups in which each $G_i$ is isomorphic to $G_1$.
If we write $G_1:=\langle a,b: [a,a^{b^i}]\,\forall i\rangle$, the homomorphism is defined by $a\mapsto 1$ and $b\mapsto 0$. The elements $a':=aba^{-1}b^{-1}$ and $b$ generate the kernel and satisfy the same relations as $a$ and $b$.

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    $\begingroup$ Oh, neat! It never occurred to me that that kernel is isomorphic to $\mathbb{Z}\wr\mathbb{Z}$ again. Is it obvious that that the map $a\mapsto a'$, $b\mapsto b$ is injective? I.e., that there aren't any "extra" relations that $a'$ and $b$ satisfy? $\endgroup$
    – Matt Zaremsky
    Commented Oct 3, 2022 at 12:35
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    $\begingroup$ I guess it is obvious. The conjugates of $a'$ by powers of $b$ clearly generate a copy of $\mathbb{Z}^\infty$, and any relation between $a'$ and $b$ must have net exponent of $b$ equal to 0 and hence be equivalent to a product of conjugates of $a'$ by powers of $b$. So the "free" in free abelian does all the work. That's great that such a straightforward group yielded an example! $\endgroup$
    – Matt Zaremsky
    Commented Oct 3, 2022 at 13:58

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