Let be $G$ a solvable group, let
$ G=G_0\supset G_1\supset\cdots\supset G_k=1$
be the derived series for $G.$ Is clear that $G_ {k-1}$ is abelian. Now take $b\in G_{k-1}$ e $a\in G_{k-2}$ my question is how to see that:
$a^{-1}b^{-1}ab \in G_{k-1}~~~?$
this is a crucial step in a demonstration of Riemannian geometry I'm reading, the theorem of Byers