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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

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1 Answer 1

up vote 3 down vote accepted

I apologize for the dumb question, it is obvious,

$G_{k-1}$ is a normal subgroup of $G_{k-2}$ so $a^{-1}b^{-1}a \in G_{k-1}$, then , $$a^{-1}b^{-1}ab \in G_{k-1}.$$

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