Index of commutator subgroup in the commutator group 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 !

This MathOverflow question answers your question about nilpotent groups in the positive, by induction on the Hirsch length.
https://mathoverflow.net/questions/107679/index-of-derived-subgroup-in-derived-group
