Z_m module as Z_n module I think every Z_n module can be viewed as Z_m module. I would like to confirm that we don't require any restriction on n and m, such as m divides n. 
 A: No, that is not true. You do in fact need the condition "$m$ divides $n$". Easy counterexample: $\mathbb{Z}_3$ cannot be viewed as a $\mathbb{Z}_2$-module. This is because $\mathbb{Z}_2$ is a field, so every module is free and therefore has a cardinality which is a power of $2$; but $3 = |\mathbb{Z}_3|$ is not a power of $2$.
More generally, if what you said were true, you could view $\mathbb{Z}_m$ as a $\mathbb{Z}_n$-module, which would mean that $n \cdot x = 0$ for all $x \in \mathbb{Z}_m$; that only happens when $m$ divides $n$.
Reciprocally if $m$ divides $n$, you it's immediate to check that the obvious formula works. With more details: let $M$ be a $\mathbb{Z}_m$-module, then define $[k] \cdot m = k \cdot m$ where $k \in \mathbb{Z}_m$ and $[k]$ is the reduction modulo $n$. Then this is well-defined, because if $k \equiv k' \pmod{m}$, then $k \equiv k' \pmod{n}$ (because $m$ divides $n$), and then this defines a module structure.
Note that in the above, it's defining the map at all that's a problem when $m$ does not divides $n$; once you've defined it, checking the module axioms is immediate. I suspect that this is what you overlooked when you asked this question.
