Let $G$ be an abelian group with even order and $M:=\{g\in G:g^2=e\}$.
It is easy to show that $M$ is a sugroup of $G$ and the number of elements of $M$ must be a power of $2$.
I want to prove that the product of the elements of $G$ (which in this case is equal to the product of the elements of $M$) is not the identity element $e$, if and only if #$M=2$ (this means that there is only one element with order $2$). I found this claim when I studied the Wilson criterion for the primality of a number.
The case #$M=4$ is easy. If $a,b\in M$, we can show that $ab$ is not $a,b,e$, hence must be some other element $c$ and we get $abc=c^2=e$. But what about #$M=8$? If we have the elements $e,a,b,c,d,f,g,h$ and $ab=c$ and $df=g$, the product would be $h$ which is impossible considering my claim.
Who can complete my proof ?