When does the product of all elements in a finite abelian group equal $1$?

Question

Clearly for $G \approx \Bbb{Z}_2$ this is not true since $G = \{1, a\}$ and so the product equals $a$. Was wondering what sufficient conditions are such that the product of all the group elements amounts to $1$.

For $G \approx \Bbb{Z}_3$ it's true since $G = \{1, a, b\}$ with $ab = 1$. I'm a little lost as to how to proceed.

I think if each element pairs with an inverse distinct from it, then it's true that the product equals $1$. But is there another way to state that, and is that a neccessary condition as well?

In addition, is it possible to cover all finite abelian groups by taking the product of the squares of all elements?

Answer 1

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.

Answer 2

If the order of $G$ is odd and $G$ is abelian. Then no element in $G$ except identity is self inverse. So, in this case the product of all elements of $G$ turn out to be $1$.

Answer 3

Regarding the question about taking the product of the squares of all elements: Yes, this product is 1 for every finite abelian group, because as per lhf's answer, if the product of all elements is not 1 then it is $x$ for the only element $x$ of order 2, whose square is $1$.