TITLE: $G'$ and the product of all the elements in an odd order group
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: @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 $P$.