If $K\rtimes \mathbb{Z}$ is a finitely generated group but $K$ isn't, can the following abelianization be finite? Suppose that we have a finitely generated group $G = K\rtimes\mathbb{Z}$, but $K$ is not finitely generated.
Let $T$ be a finite subset of $K$ such that $\langle (0,1), (k,0)\mid k \in T \rangle$ is a generating set of $G$. Let $\phi(n)$ be the automorphism on $K$ corresponding to $n \in \mathbb{Z}$, which defines the semi-direct product. Let $H =  \langle T  \rangle$, a finitely generated subgroup of $K$.
Now, suppose in addition, the automorphisms satisfy that
$$\ldots  \phi(-1)(H)  \supsetneqq  \phi(0)(H)   \supsetneqq \phi(1)(H)  \supsetneqq \phi(2)(H)  \ldots \tag{$*$}$$
(roughly speaking, if we keep applying  $\phi$ on $H$, the size of the group gets "smaller", but never reaches the trivial group)
Question: Is it possible that $H$ has finite abelianization, i.e. the quotient group $H \big/ [H,H]$ is finite?
My thoughts so far: Because the automorphisms satisfy $(*)$, we know $|H|$ has to be infinite (e.g. the Lamplight group doesn't satisfy $(*)$ because in that case $H = \mathbb{Z}_2$  is finite). The only semidirect product I know that satisfy $(*)$ are the ones with $K$ being abelian, which means $H$ is infinite abelian, hence can't have finite abelianization. For example, $G = D\rtimes\mathbb{Z}$, where $D = \{j/2^m \mid j \in \mathbb{Z}, m\in\mathbb{N}  \}$, the dyadic rationals, and $n\in \mathbb{Z}$ acts on $D$ by multiplying every number by $2^n$, this group satisfies $(*)$, but $D$ is abelian.
Any example of such a group would be really appreciated.
 A: Every finitely generated, non-cohopfian group $H$ with finite abelianization yields an example.
Indeed, if $H=\langle T\rangle$ is such a group with $T$ finite, $i$ is an injective non-surjective endomorphism of $H$, the corresponding ascending HNN-extension
$$\langle H,t\mid tht^{-1}=i(h):h\in H\rangle$$
works. Here $K$ is the inductive limit of $H\stackrel{i}\to H\stackrel{i}\to H\stackrel{i}\to H\cdots$.
It remains to provide examples of such groups $H$. Derek Holt mentioned one: Thompson's group $V$ (which is isomorphic to the pointwise stabilizer of $[0,1/2[$ in $V$). A simpler example is the infinite dihedral group $\langle u,x\mid u^2=(ux)^2\rangle$: the endomorphism mapping $(u,x)\mapsto (u,x^2)$ is injective and non-surjective (in the latter case the ascending HNN-extension contains the Baumslag-Solitar group $\mathrm{BS}(1,2)$ with index 2).
A: Or to put it simply,
let $x_n$ be a rotation by an angle of $\theta=2^n\alpha\pi$, where $0<\alpha<1$ and $\alpha\not\in\mathbb{Q}$,
and let $u$ be a reflection along the $x$-axis.
Let $K=\langle x_n,u\;|\;n\in\mathbb{Z}\rangle $ and $H=\langle x_0,u\rangle$.
We can have $\phi(1)$ sending $x_n$ to $x_{n+1}$ and sending $u$ to itself.
As you probably know, the infinite dihedral group has abelianisation $\mathbb{Z}_2\oplus\mathbb{Z}_2$.
