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' $
My proof:
(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:
$1=(aG')(bG')…(aG')=ab…nG'$
and finally we conclude $ab…n \in G'$
Does my proof make any sense?
Any other solutions?
Best Answer
I'll summarize the OP's solution, following the points commented by Gerry and Geoff, so that this question won't remain unanswered:
If $\,G\,$ abelian then we can pair any element in the group with its inverse. Since there can't be element that equals its own inverse (why?), we have an even number of such pairs (why?) and $$x=\prod_{a\in G}a=1\cdot (aa^{-1})(bb^{-1})\cdot\ldots =1\cdot\ldots\cdot 1 =1\in \{1\}=G'$$
If $\,G\,$ is not abelian, let us take $\,G/G'\,$ , which is abelian, and let us denote $\,\overline{a}:=aG'\in G/G'\,$ . By Lagrange' theorem , $\,|gG'|=|G'|\,,\,\forall\,g\in G\,$ , so by the first part we get$$\left(\prod_{a\in G}a\right)G'=\left(\prod_{\overline{a}\in G/G'}\overline{a}\right)^{|G'|}\in \left(G/G'\right)'=\overline{\{1\}}\leq G/G'\Longrightarrow \left(\prod_{a\in G}a\right)\in G'$$
Check please whether the above reflects accurately what you people commented/