Product of all elements in an odd finite abelian group is 1 This should be an easy exercise: Given a finite odd abelian group $G$, prove that $\prod_{g\in G}g=e$. Indeed, using Lagrange's theorem this is trivial: There is no element of order 2 (since the order must divide the order of $G$, but it is odd), and so every element except $e$ has a unique inverse which is different from it. Hence both the element and its inverse participate in the product and cancel each other.
My problem is simple - I need to solve this without Lagrange's theorem. So either there's a smart way to prove the nonexistance of an element of order 2 in an odd abelian group, or I'm missing something even more basic...

Let $G$ be a finite abelian group of odd order. Then we have 
$$
\prod_{g \in G}g \prod_{g \in G} g = 
\prod_{g \in G} g \prod_{g \in G} g^{-1} = \prod_{g \in G} gg^{-1} = e.
$$
Hence $x:= \prod_{g \in G}\;g$ is its own inverse. Now you just need to show that the only element that's its own inverse in $G$ is $e$.
