TITLE: Abelian groups: proving $\prod\limits_{g\in G}g=\prod\limits_{\substack{g\in G\\g^2=1}}g$
QUESTION [9 upvotes]: Let $G$ be a finite abelian group. Why is $$\prod_{g\in G}g=\prod_{\substack{g\in G\\g^2=1}} g \ ?$$
I've tried to consider an element with its inverse but I don't get why you can combine the elements - which are different of its inverse - and they annul themselves in the product. Any help?

REPLY [6 votes]: We have $$\prod_{g\in G}g = \prod_{\substack{g\in G\\g^2=1}}g \cdot\prod_{\substack{g\in G\\g^2\neq1}}g $$ In the product $\displaystyle\prod_{\substack{g\in G\\g^2\neq1}}g$ every element cancels with its inverse.