Let $G$ be a finite group of an odd order, and let $x$ be the product of all the elements of $G$ in some order. Prove that $x \in G' $
(1) If $G$ is abelian then it is very simple to prove.
(2) If $G$ is not abelian:
$G/G'$ is abelian, therefore by (1), the product of all the elements of $G/G'$ is in $(G/G')'$. But $(G/G')'=1$. So we have:
and finally we conclude $ab...n \in G'$
Does my proof make any sense?
Any other solutions?