# Is it necessary for this abelian group to have $4n+2$?

$$\DeclareMathOperator{\ord}{ord}$$
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.

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.

• Thank you for the counterexample ! Could you also tell me which part of it is wrong? Commented Jul 13, 2019 at 16:28
• @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 $p=e$. Commented Jul 13, 2019 at 16:30
• This is pretty subtle, I got a bit carried away making that assumption. Thank you for your help ! Commented Jul 13, 2019 at 16:32

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 is $$4n+2=2(2n+1)$$, 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.

• 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. Commented Jul 13, 2019 at 16:28
• No need to refer to an offsite statement of the theorem when onsite we have proofs, e.g. here. Commented Jul 13, 2019 at 17:47