On the section about the Hom sets of modules, Hungerford has an exercise that asks to show that $$\operatorname{Hom}(\mathbb{Z}_m, \mathbb{Z}_n) \cong \mathbb{Z}_{(m,n)}$$ and then in the next exercise he has
If $A,B$ are abelian groups and $m,n$ integers such that $mA = 0 = nB$, then every element of $\operatorname{Hom}(A,B)$ has order dividing $(m,n)$.
What does $(m,n)$ mean in these contexts?