TITLE: Prove that $(a_1a_2\cdots a_n)^{2} = e$ in a finite Abelian group
QUESTION [15 upvotes]: 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.

REPLY [5 votes]: In addition to Pete Clark's answer, there is also a very neat answer to the question What is the set of all different products of all the elements of a finite group $G$? So $G$ not necessarily abelian. Well, if a 2-Sylow subgroup of $G$ is trivial or non-cyclic, then this set equals the commutator subgroup $G'$. If a 2-Sylow subgroup of $G$ is cyclic, then this set is the coset $xG'$ of the commutator subgroup, with $x$ the unique involution of a 2-Sylow subgroup.