Prove that if $G$ is a finite abelian group with $n$ elements, then $g^n=e$ for all $g\in G$ Assume we have $G$, a finite abelian group where $|G|=n$. We wish to show that $g^n=e$, for all $g \in G$. 
So, we need to do this without Lagrange's theorem. (If we use Lagrange then this is clearly true for all finite groups.)
We know, since $G$ is abelian that $a*b=b*a$, for all $a,b \in G$ and that $o(g)=k\leq n$ and so we have $g^k=e$ where $k \leq n$
I am unsure of where to take this. Any hints would be appreciated!

Consider the map given by multiplication with $g$. This is bijective, and so $\prod_{a \in G} a = g^n \prod_{a \in G}a$, from which we deduce that $g^n=e$.
