Suppose $G$ is a finite group with $\vert G\vert>2$ and all non-identity elements of $G$ are order $2$, then the product of all the elements of $G$ is the identity.
Since all non identity elements are order 2 I already know that $G$ is abelian as $ab=(ba)^{-1}=ba$.
I was told there is a solution involving cosets, so here is what I have.
I know that there must be a 4 element subgroup of G. Since for 2 elements a,b. I can form a subgroup $H=\{e,a,b,ab\}$. Now this subgroup has the property if I take the product of the non identity elements, $a(b)(ab)=e$ since $H$ is abelian. Taking cosets of $H$ I get $xH=Hx$ since $G$ is abelian. I know the order of $G$ must be even since it's divisible by 2. And for any coset I have $xH=\{x,xa,xb,xab\}$
I also know that if my product is $a_1\cdot a_2\cdot ...\cdot a_n$ that each member of this product can only appear in each coset once as cosets are disjoint.
From here though I have no idea, I'm not even sure what this coset idea is even supposed to give me.