# if $G$ is a finite group of an odd order, then the product of all the elements of $G$ is in $G'$

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?

• You might want to point out that $G/G'$ is of odd order. Commented Jun 21, 2012 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. Commented Jun 21, 2012 at 9:58
• What is meant by $G'$? Commented Jun 21, 2012 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. Commented Jun 21, 2012 at 12:31

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/

• 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. Commented Jun 21, 2012 at 12:36
• Of course, @Geoff. I Just messed between the abelian group thing ( was thinking additively ) and my multiplicative writing. Thanks Commented Jun 21, 2012 at 12:38
• Yes, I thought that was probably what had happened. Commented Jun 21, 2012 at 12:50
• @DonAntonio: As usual, neat and solid proof. Commented Jun 21, 2012 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. Commented Jun 21, 2012 at 17:20