Question :
Show that a group G is solvable if and only if there exist sub-groups $G^{(0)}$, $G^{(1)}$, ..., $G^{(m)}$, with m $\ge$ 1, such that :
$G^{(m)}$ = {$e_G$} and the $G^{(i)}$ are defined inductively by $G^{(0)}$ = G, and $G^{(i)}$ = [$G^{(i-1)}$, $G^{(i-1)}$], where
[G, G] = < [x, y] | x, y ∈ G, [x, y] = xy$x^{-1}$$y^{-1}$ >
I'm able to show the forward implication, but not the backward one (i.e. the $G^{(i)}$ exist ==> G solvable). I have tried by induction on the i, but this doesn't seem to go anywhere. More specifically, I believe that once $G^{(i-1)}$/$G^{(i)}$ is abelian for all i, the normality $G^{(i)}$ $\lhd$ $G^{(i-1)}$ should be straightforward.
Is this the right approach? How can I show that $G^{(i-1)}$/$G^{(i)}$ is abelian?