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?