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$\DeclareMathOperator{\ord}{ord}$
Let $G$ be a finite abelian group with $|G|=4n+2$, where $n\in \mathbb{N}$. Prove that the product of all of $G$'s elements is different from $e$.
I have two solutions to this problem. I am sure that the first one is correct, but I am not sure whether the second one also works.
In both of them I will be using the fact that $$\prod_{x\in G}x=\prod_{\substack{x\in G \\ \ord x\le2 }}(*)$$
in any finite abelian group.
Solution 1: We will prove that $G$ has only one element of order $2$.
From Cauchy's theorem $\exists a \in G$ such that $\ord(a)=2$. Suppose $\exists b \in G$ such that $\ord(b)=2$.
Since $G$ is an abelian group we have $$(ab)^2=a^2b^2=e,$$so $\ord(ab)=2$.
Consider the set $H=\{e,a,b,ab\}\subset G$,$|H|=4$.
It is easy to see that $H$ is a subgroup of $G$ and from Lagrange's theorem we have that $$\ord(H) | \ord(G) \iff 4|(4n+2),$$ which is obviously a contradiction, so $a$ is the unique element of order $2$ in $G$.
Hence, using $(*)$,$\prod\limits_{x\in G}x=a \neq e$ since $\ord(a)=2$.
Solution 2: I want to prove the following stronger statement :
Let $G$ be a finite abelian group with an even number of elements.Then the product of all of $G$'s elements is different from $e$.
Again from Cauchy's Theorem the group has at least an element of order $2$.
Let $a_1,a_2,...,a_n \in G$ such that $\ord(a_1)=\ord(a_2)=...=\ord(a_n)=2$.(Note: $n$ is an odd number since $|G|$ is even).
We know from $(*)$ that $$\prod_{x\in G}x=\prod_{i=1}^n a_i.$$
Since $G$ is abelian we have $$(a_1 \cdot a_2 \cdot... a_n)^2=a_1^2 \cdot a_2^2 \cdot ... a_n^2=e,$$ so $\ord(a_1 \cdot a_2 \cdot... a_n)=2$.
From here it follows that $\ord\left(\prod\limits_{x\in G}x \right)=2$, so $\prod\limits_{x\in G}x \neq e$.
Since $4n+2$ is even the result follows.
To me the general statement I proved in my second solution seems true, I can't spot any flaws in my proof. I would be grateful if you could look over it and give me some feedback.

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The second proof is wrong. Take the four-group which has one element of order $1$ and three elements of order $2$. The product of all the elements is the identity.

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  • $\begingroup$ Thank you for the counterexample ! Could you also tell me which part of it is wrong? $\endgroup$ – Alexdanut Jul 13 at 16:28
  • $\begingroup$ @Alexdanut You assume that if $p^2=e$ then $p$ has order $2$ ($p$ being the product). However it is also possible for the order of $p$ to be $1$ ie $p=e$. $\endgroup$ – Mark Bennet Jul 13 at 16:30
  • $\begingroup$ This is pretty subtle, I got a bit carried away making that assumption. Thank you for your help ! $\endgroup$ – Alexdanut Jul 13 at 16:32
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More generally, we have Wilson's theorem for finite Abelian groups:

The product of all elements in a finite Abelian group is either $1$ or the element of order $2$ if there is only one such element.

In your case, since the number of elements if $4n+2=2(2n+1)$, there is only one element of order $2$ (otherwise, the order of the group would be divisible by $4$). This is essentially what Solution 1 argues.

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  • $\begingroup$ Indeed, since the elements of order $2$ form an elementary abelian subgroup of order $2^m$ for some $m$, the product of elements of order $2$ is seen to be $1$ for $m\gt 1$, and the other elements of the group are either $1$ or can be paired with their inverses. $\endgroup$ – Mark Bennet Jul 13 at 16:28
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    $\begingroup$ No need to refer to an offsite statement of the theorem when onsite we have proofs, e.g. here. $\endgroup$ – Bill Dubuque Jul 13 at 17:47

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