Let $R=Z_n$
I'm trying to show an abelian group H is a R-module if and only if the order of every element in the group divides n.
I was thinking to prove in the forward direction you could write $$h\in H $$ $h+h+...+h $ (n times)$=nh=0$ as we're over $Z_n$
But not sure how to prove in the other direction.
Short direct answer
Let $X$ be an abelian group such that for each $x$ in $X$, $nx=0$. For each $m$, $k$ in $\mathbb{Z}$ note that $(m+kn)x=mx+k(nx)=mx$ and hence it makes sence to define for each $[m]$ in $\mathbb{Z}_n$, $[m]x = mx$. One then checks that all the required identities hold (e.g. $([m]+[n])x=[m]x+[n]x$).
Longer more thorough answer
For an abelian group $X$ let $\text{End}(X)$ be the ring with underlying set the homorphisms from $X$ to itself, and addition defined pointwise (i.e. (f+g)(x)=f(x)+g(x)) and multiplication by composition. An equivalent way of defining an $R$-module is as an abelian group together with a homorphism from $a:R \to \text{End}(X)$ (one then define $r x = a(r)(x)$).
To prove what you want note that every abelian group $X$ can be made into a $\mathbb{Z}$-module, by defining $a: \mathbb{Z} \to \text{End}(X)$ as follows
$a(n)(x) = \left\{ \begin{array}{l l} 0 & \text{if $n=0$}\\ x + a(n-1)(x) & \text{if $n>0$}\\ -x + a(n+1)(x) & \text{if $n<0$} \end{array}\right.$
For some natural number $n$ let $q: \mathbb{Z} \to \mathbb{Z}_n$ be quotient map in rings. Now suppose that for some abelian group $X$ we know that for each $x$ in $X$, $nx=0$. It follows that $a:\mathbb{Z} \to \text{End}(X)$ described above has the property that $a(m+kn)=a(m)$ and hence $a$ factors through $q$ i.e. there is a ring homorphism $a':\mathbb{Z}_n \to \text{End}(X)$ such that $a=a'q$ making $X$ into an $\mathbb{Z}_n$-module.