I'm reading Aluffi's "Algebra: Chapter 0", and have some difficulties with exercise II.1.8:
Let G be a finite group with exactly one element f with order 2. Prove that $П_{g\in G}(g)=f$
I've proved it for abelian groups, is it really true for all groups?
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 $\wp(G)=\{g_{\sigma(1)} \cdot g_{\sigma(2)} \cdot \cdot \cdot g_{\sigma(n-1)}\cdot g_{\sigma(n)}: g_i \in G, i=1,... ,n$ and $\sigma \in S_n\}$, in other words $\wp(G)$ 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 $S$ be a Sylow $2$-subgroup of $G$.
If $S$ is non-cyclic or $\{1\}$ (that is $|G|$ is odd), then $\wp(G)=G'$.
If $S$ is cyclic, then $\wp(G)=xG'$, where $x$ is the unique element of order $2$ of $S$.