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 $G\_{i+1}$.



Iterated algebraic fibering
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 $b$.


