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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.

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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.

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  • $\begingroup$ How can you get that subgroup $H$ without Sylow theory (out of curiosity)? $\endgroup$ – Randall Mar 10 at 3:08
  • $\begingroup$ "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? $\endgroup$ – AColoredReptile Mar 10 at 3:17
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    $\begingroup$ @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. $\endgroup$ – Gerry Myerson Mar 10 at 3:22
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    $\begingroup$ @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. $\endgroup$ – Gerry Myerson Mar 10 at 3:24
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    $\begingroup$ @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. $\endgroup$ – Gerry Myerson Mar 10 at 3:26
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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 $$\prod_{g\in G'}g\cdot ga\cdot gb\cdot gab =\prod_{g\in G'}g^4\cdot (ab)^2=1(\text{since $G$ is Abelian and every element is of order $2$}). $$

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    $\begingroup$ 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? $\endgroup$ – Randall Mar 10 at 3:21
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    $\begingroup$ @Randall Yes, you're right. I don't see any problem. I was trying to extend the OP's approach. $\endgroup$ – Thomas Shelby Mar 10 at 3:24
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    $\begingroup$ Yeah, I understand that approach for the answer. Just curious. Been thinking about this a lot for some reason. $\endgroup$ – Randall Mar 10 at 3:26
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    $\begingroup$ Either way, this is slick. $\endgroup$ – Randall Mar 10 at 3:31
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    $\begingroup$ To pick a nit, I'd call $G'$ a set of coset representatives, not the set. But substantively it's fine. $\endgroup$ – Robert Shore Mar 10 at 8:00

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