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...

If finite abelian group $ G$ has an elt $\, j\, $ of order $\,2\,$ then $\,g \to\ j g\,$ pairs its elts so $ G$ has even order.
This is a special case of the often useful fact that the cardinalities of a finite set and its fixed-point set under an involution have equal parity, since the non-fixed points are paired by the involution. Hence, as above, when there are no fixed points $\,( j\ne 1\, \Rightarrow\, j\,g\ne g)\,$ the set has even cardinality.
Such simple symmetries often lie at the heart of elegant proofs, e.g. the famous Heath-Brown-Zagier proof that every prime $\,\equiv 1\pmod{\!4}\, $ is a sum of two squares.
