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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:


and finally we conclude $ab...n \in G'$

Does my proof make any sense?

Any other solutions?

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You might want to point out that $G/G'$ is of odd order. – Gerry Myerson Jun 21 '12 at 9:55
This looks more or less correct. Notice though that when you multiply all elements of $G$, when you pass to $G/G',$ every coset of $G'$ is showing up $|G'|$ times in the product. – Geoff Robinson Jun 21 '12 at 9:58
Thank you both! – catch22 Jun 21 '12 at 10:25
What is meant by $G'$? – Najib Idrissi Jun 21 '12 at 12:23
@N.I , $\,G'=[G:G]=\langle\, [x,y]:=x^{-1}y^{-1}xy\;\;;\;\;x,y \in G \,\rangle = $ the derived or commutator subgroup of $\,G\,$ , and it is characterized by being the minimal normal subgroup of $\,G\,$ s.t. $\,G/G'\,$ is abelian. This is a rather important subgroup. – DonAntonio Jun 21 '12 at 12:31
up vote 2 down vote accepted

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/

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I think you want to raisie the barred product at the bottom to the power of $|G'|$, rather than multiplying it by $|G'|$. But the answer is the identity to that , which is still the identity. – Geoff Robinson Jun 21 '12 at 12:36
Of course, @Geoff. I Just messed between the abelian group thing ( was thinking additively ) and my multiplicative writing. Thanks – DonAntonio Jun 21 '12 at 12:38
Yes, I thought that was probably what had happened. – Geoff Robinson Jun 21 '12 at 12:50
@DonAntonio: As usual, neat and solid proof. – Babak S. Jun 21 '12 at 14:03
@Babak, thank you but the OP, Geoff and Gerry did most of it. I just summarized and put a little order in it. – DonAntonio Jun 21 '12 at 17:20

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