I need to prove (or find a counter example) that if $G$ is solvable and $H \leq G$ is a subgroup of finite index, then the commutator subgroup $D(H)$ is also a subgroup of finite index in $D(G)$. This would allow me to say :
If $G_1$ and $G_2$ are commensurable (i.e. there are $H_1 \leq G_1$ and $H_2 \leq G_2$ such that $|G_i : H_i| < \infty$ and $H_1 \cong H_2$) then $D^n(G_1$) and $D^n(G_2)$ are commensurable for all $n$.
It's easy to show that $D(H)$ is a subgroup but I can't even figure out how to start the main part of this proof. Does anyone have a clue ?
EDIT : Thanks to Tsemo Aristide who answered the original question. The fact that it's not always true doesn't help me at all for my research, so here's a follow up question : (not sure that editing is the right way to ask a follow up question though so please let me know if I shouldn't be doing this !)
What if $G$ was nilpotent instead of solvable ? Would it work then ?
Thanks in advance !
The infinite dihedral group $D_{\infty}$ is a solvable group generated by $x,a$ with the relation $x^2=e, xax=a^{-1}$ The subgroup $H$ generated by $a$ is an infinite subgroup of index $2$.
$D(H)=e$ remark that $xax^{-1}a^{-1}=xaxa^{-1}=a^{-2}$ implies that $D(D_{\infty})$ is infinite.