Prove that $(a_1a_2\cdots a_n)^{2} = e$ in a finite Abelian group Let $G$ be a finite abelian group, $G = \{e, a_{1}, a_{2}, ..., a_{n} \}$. Prove that $(a_{1}a_{2}\cdot \cdot \cdot a_{n})^{2} = e$.
I've been stuck on this problem for quite some time. Could someone give me a hint?
Thanks in advance.

Here is a hint: for any given $a_i\in G$, there are two possibilities: either 


*

*$a_i$ is its own inverse, or 

*$a_i$ is not its own inverse, but rather $a_j=a_i^{-1}$ for some $j\neq i$.

