What can we say about $a, b \in G$ when $na = nb$ with $G$ a $p$-group and $(n, p) = 1$?

Question

Let $a, b \in G$, with $G$ an abelian $p$-group. Define $n$ such that $(n,p)=1$ Assume $na=nb$. Then, I think it follows that $a=b$.

I think if $G$ is cyclic, $na=nb \Rightarrow na \equiv nb \pmod{p} \Rightarrow a=b$. How would I generalize this to non-cyclic groups?