If $G$ is a finite group and all non-identity elements of $G$ are order 2 then the product of these elements is the identity.

Question

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.

Answer 1

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.

Answer 2

Write $G=\displaystyle\bigcup_{g\in G }gH $.

For $g_1,g_2\in G $, either $g_1H=g_2H $ or $g_1H\cap g_2H=\emptyset$. Let $G'$ be a set of all $g$'s with pairwise disjoint cosets. Then we can write $G=\displaystyle\bigcup_{g\in G' }gH.$

Now the product of all elements can be written as $$\prod_{g\in G'}g\cdot ga\cdot gb\cdot gab =\prod_{g\in G'}g^4\cdot (ab)^2=1(\text{since $G$ is Abelian and every element is of order $2$}). $$