# $G'$ and the product of all the elements in an odd order group

Yesterday, through working on a question Groups with only one element of order 2, Don antonio brought out a nice question within the comments:

The product of all the elements in an odd order group $G$ is always contained in the group's derived subgroup $G'$.

Honestly, I tried to link some facts for proving that but, they didn't work. Thanks for any hint for that.

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Let $\,G\,$ be a group, $\,G'=[G,G]=\,$ its derived or commutator subgroup, then we have the following:

(1) $\,G/G'\,$ is abelian, and thus

(2) Any product in $\,G/G'\,$ can be arranged at will by (1), and finally

(3) The product of all the elements in a group with an odd number of elements is in $\,G'\,$ since the product of their images in $\,G/G'\,$ is trivial.

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Thanks so much Don. (-: – Babak S. Nov 16 '12 at 19:55

By considering the quotient $G/G^\prime$, it suffices to show that the product of all the elements in an abelian group of odd order is trivial. This is true because for every $x$ that appears in the product, $x^{-1}$ also appears, and $x\ne x^{-1}$.

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Thanks Sean for the answer. – Babak S. Nov 16 '12 at 19:54

@Babak, there is a very neat but deep result here that goes a step further.

Let $G$ be a finite group of order $n$, say $G=\{g_1, g_2, ..., g_{n-1}, g_n\}$ and define $P(G)=\{g_{\sigma(1)} \cdot g_{\sigma(2)} \cdot \cdot \cdot g_{\sigma(n-1)}\cdot g_{\sigma(n)}: g_i \in G, i=1,... ,n$ and $\sigma \in S_n\}$, in other words $P(G)$ is the set of all possible products of $n$ different elements of $G$. (Of course the result of such product depends on the order of the elements, $G$ does not have to be abelian here!). Let $P$ be a Sylow $2$-subgroup of $G$.

If $P$ is non-cyclic or $\{1\}$ (that is $|G|$ is odd), then $P(G)=G'$.

If $P$ is cyclic, then $P(G)=xG'$, where $x$ is the unique element of order $2$ of $P$.

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