Let $m$ and $n$ be fixed natural numbers, both at least 2. Let $X \in Z_m$ and $Y \in Z_n$. Prove that $X \subseteq Y$ if and only if $n | m$.
[ Hint: $X$ and $Y$ are subsets of $Z$ since they are elements in two (different) partitions of $Z$. In fact, since any two elements of $X$ differ by a multiple of $m$, we have, for some fixed $x \in X$,
$X = \{x + mk | k \in Z\}$. Obtain a similar representation for $Y$ (being sure to use a different dummy variable in the set-builder language for $Y$ than the ‘$k$’ used for $X$). Now, interpret $X \subset Y$ in this context. What happens for $k = 0$ and $k = 1$ help you characterize $m$? ]
For ideals in $\mathbb{Z}$, "to contain is to divide", i.e., $\mathbb{Z}n\supseteq \mathbb{Z}m$ if and only if $n\mid m$. This holds more generally for PIDs, and for Dedekind rings.