Suppose $G=\{a_1,...,a_n\}$ is a finite abelian group, and let $x=a_1a_2\dotsm a_n$. Prove that if there is more than one element of order $2$ then $x=e$.
What I've done so far: (#1 is just for illustration, There should be at least $3$ such elements)
$G$ has an even number of elements with order $>2$ (They are paired so: $(a,a^{-1})$ ). And then there are $e$ and $\{b|o(b)=2\}$. Also, if $G$ has at least one element of order 2 then $|G|$ is even (Lagrange). Thus $|\{b|o(b)=2\}|$ must be odd.
If there is exactly one element $b$ of order $2$ then $x=b$. That's because $G$ is abelian, and we can write: $x=e\cdot b\cdot(a_1 a_{1}^{-1}\dotsm a_n a_{n}^{-1})=b\cdot(e\dotsm e)=b$.
If there are exactly $3$ such elements $a,b,c\in G$, then, as shown above, $x=abc$ and thus $x=e$ (That's because $(abc)^2=a^2b^2c^2=e$ which means $abc\in \{a,b,c,e\}$. If $abc=a$ then $b=c^{-1}=c$ which is false, same goes for $abc=b,abc=c$.)
I haven't managed to find a reason why the claim must hold for $5, 7, ...$ (i.e. All other odd integers). Does anyone have an idea?
Thanks in advance. Please excuse my English.
edit: Not a canonical answer but a simple one (if such even exists).