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This result is not an exercise, but more so a hunch that I am trying to prove for myself. Let's say that $G$ is a finite group with the property that it is both abelian and every non-identity element has order $2$. By Lagrange's theorem, the group order must be even. My hunch is that $G \cong \mathbb{Z}/2 \times \ldots \mathbb{Z}/2$ ($n$ copies). I'm not 100% sure that this is the only group up to isomorphism, but it seems to work for some smaller cases.

The proof I attempted is by induction on $n$, letting $G$ be a finite abelian group with each non-identity element having order $2$ and $|G| = 2n$. For $n = 1$, $|G| = 2$, and the only group of order $2$ is $\mathbb{Z}/2$. For $n = 4$, there are two groups of order $4$: the cyclic group of order $4$ and $\mathbb{Z}/2 \times \mathbb{Z}/2$. $G$ lacks an element of order $4$, so it must be $\mathbb{Z}/2 \times \mathbb{Z}/2$.

The induction step is the tough part. Assume that the assertion holds for $n$, and let $G$ have $2(n+1)$ elements. Each of the non-identity elements has order $2$, so if I remove two such elements, $a$ and $b$, then $H = G - \{a,b\}$ is still a group: these elements were their own inverses, after all. $H$ has $2n$ elements, so I believe I can say the following. $$ (\mathbb{Z}/2)^{n+1} \cong (\mathbb{Z}/2)^n \times \mathbb{Z}/2 \cong H \times \mathbb{Z}/2. $$ My goal is now to show that $H \times \mathbb{Z}/2 \cong G$, but $\mathbb{Z}/2$ only has one non-identity element of order $2$, so I don't see a way to "put back" $a$ and $b$. This leads me to believe that this approach doesn't work.

I'd greatly appreciate any help with proving (or disproving, if it's false) this result.

Shaun
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Brad G.
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    Yes, this is true, but your approach doesn't work. Your claim about removing two elements from $G$ to get a group is not true and violates Lagrange's theorem. And note that $(\mathbb{Z}/2)^n$ has $2^n$ elements, not $2n$. The desired result follows from the structure theorem for finite abelian groups or from a little linear algebra over the finite field $\mathbb{F}_2$. – Qiaochu Yuan Sep 19 '23 at 01:36

2 Answers2

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There is an invalid step in the argument: "if I remove two such elements, $a$ and $b$, then $H = G - \{a,b\}$ is still a group" is incorrect. The reason is that there are plenty of remaining pairs of elements $c,d\in H$ with $cd=a$, and so $H$ is not closed under the group operation.

The ultimate result is actually true, and one doesn't even need to assume that $G$ is abelian—it follows from the fact that every nonidentity element has order $2$: $$ e=(xy)^2=xyxy \implies xy = xey = x(xyxy)y = x^2yxy^2 = yx. $$ Proofs of the result can be found in various places, for instance here.

Greg Martin
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An abelian group $G$ in which every non-identity element has order $2$ is a vector space over $\mathbb{F}_2$. Picking a basis for this vector space, $G$ is isomorphic to $\bigoplus_{i\in I} \mathbb{F}_2$. If $G$ is finite, then $G$ is isomorphic to a finite product of copies of $\mathbb{Z}/2\mathbb{Z}$.

Alex Kruckman
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