# 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.

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.

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.

• How can you get that subgroup $H$ without Sylow theory (out of curiosity)? Mar 10, 2019 at 3:08
• "So the product of all the elements is $a^r$, where $r$, half the order of the group, is even, and we're done" why does this imply that the product of all the non identity elements is the identity? Mar 10, 2019 at 3:17
• @Randall, we're dealing with abelian groiups here, so we don't need Sylow; the structure theorem for finite abelian groups will do. Alternatively, even for nonabelian groups, you can prove without Sylow that a group of order $p^r$ has a subgroup of order $p^{r-1}$, $p$ being prime. Mar 10, 2019 at 3:22
• @Randall, or, you can do it like this: let $a\ne1$, then $a$ generates a subgroup of order 2; let $b$ not be in that subgroup, then $a,b$ generate a subgroup of order 4; let $c$ not be in that subgroup, then $a,b,c$ generate a subgroup of order 8; and so on. Mar 10, 2019 at 3:24
• @Randall, you may be right about order $p^{r-1}$ and Sylow, I may have been thinking about the proof there's a nontrivial center, which is much less than what's needed. But the other approaches work. Mar 10, 2019 at 3:26

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 $$\prod_{g\in G'}g\cdot ga\cdot gb\cdot gab =\prod_{g\in G'}g^4\cdot (ab)^2=1(\text{since G is Abelian and every element is of order 2}).$$

• Can't you make the same argument using $H=\{e,a\}$? To me, this seems more natural (and less work) than using a 4-element subgroup. Is there somewhere the argument breaks down on a 2-element subgroup? Mar 10, 2019 at 3:21
• @Randall Yes, you're right. I don't see any problem. I was trying to extend the OP's approach.
– cqfd
Mar 10, 2019 at 3:24
• Yeah, I understand that approach for the answer. Just curious. Been thinking about this a lot for some reason. Mar 10, 2019 at 3:26
• Either way, this is slick. Mar 10, 2019 at 3:31
• To pick a nit, I'd call $G'$ a set of coset representatives, not the set. But substantively it's fine. Mar 10, 2019 at 8:00