Consider a countable group $H=\{a_0,a_1,a_2, ... \}$.
We define the map $f: \mathbb{Z} \to H $ , $f(2^n):=a_n $ and $f(k)=1$ for $k \notin \{2^0 , 2^1, 2^2, ...\}$.
We consider $f$ as an element of the (unrestricted) wreath product $H \wr \mathbb{Z}$ , or, if you prefer, as an element of the semidirect product $ H^{\mathbb{Z}} \rtimes \mathbb{Z} $ where $\mathbb{Z}= \langle t \rangle $ (take $t=1$) acts on $H^{\mathbb{Z}}$ by shifting the domain : $t^{-1}ft(n):=f(n+1)$
Now consider the subgroup $G:= \langle t,f \rangle $ and its intersection with the basis $M:=G \cap H^{\mathbb{Z}} $.
I want to show that every element of the derived group $[M,M]$ has finite support.
The intuition behind this is that the domain of $f$ is "sparse enough" , so $[t^{-2^i}ft^{2^i},t^{-2^j}ft^{2^j}]$ takes only one nontrivial value at 0.
I tried to prove the statement by induction on the word lenght $w(\cdot)$:
Consider $x,y \in [M,M]$ . if $w(x) $ and $w(y)$ are smaller than 1, then $x,y \in \{ f,f^{-1}, 1\}$ so the commutator vanishes.
Now let $w(x) ,w(y) $ smaller or equal than n, then $x=x_{n-1}a$ and $y=y_{n-1}b$ where $x_{n-1}, y_{n-1}$ are words of lenght smaller or equal than $n-1$ and $a,b$ are elements in $\{ f,f^{-1},t,t^{-1}\}$
If $a,b$ are in $\{f,f^{-1}\}$ the stament follows easily...I'm having some trouble with the case where $a,b$ are in $\{t,t^{-1}\}$. I spent some hours trying to prove it with no results, I'm not sure anymore if my approach here is the right one... does someone have an idea?