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.

The map $\phi:x\in G\mapsto x^{-1}\in G$ is an automorphism of $G$ so, in particular, it induces a bijection $G\setminus\{e\}\to G\setminus\{e\}$. It maps $b=a_1\cdots a_n$ to itself, so that $b=b^{-1}$ and, therefore, $b^2=e$.
