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.
Best Answer
As you note, the group is abelian. Since it has no element of order $p$ for any odd prime $p$, the order of the group is divisible only by the prime $2$, so the order is a power of $2$.
The group then has a subgroup $H$ of index $2$, with a coset $aH$. To each element $h$ of $H$ there corresponds an element $ah$ of $aH$. The product of these two elements is $ah^2$, which is $a$. So the product of all the elements is $a^r$, where $r$, half the order of the group, is even, and we're done.