Let G be an abelian group and let $\displaystyle g,h \in G$.

If $\displaystyle |g| = m$ is finite then prove that, for $\displaystyle n \in \mathbb {Z}$, $\displaystyle ng=0$ iff $\displaystyle m|n$.

Now the obvious cases are if $\displaystyle n=0$, $\displaystyle ng=0$ and $\displaystyle m|n$ trivially. Same for $\displaystyle n=m$.

The rest all seems rather straightforward, however I'm having trouble with finding a structure for my proof, especially since n is an integer, not just a natural number.

For the case where $\displaystyle 0<n<m$ I have if $\displaystyle g+...+g (m times) =0$ then $\displaystyle [g+...+g (m times)] + g+ ... +g (ntimes)= g+...+g (n times)$ which, by definition of the order of an element cannot be equal to zero. I find it hard to prove this for the similar case of $\displaystyle m<n<0$.

Am I making this too difficult or what is it that I'm missing?