Necessary and Sufficient Conditions for Subgroup Inclusions 
Let $\overline{a} , \overline{b} \in \mathbb{Z}/n\mathbb{Z}$. Then $\langle \overline{a} \rangle \le \langle \overline{b} \rangle$ if and only if $(b,n) \mid (a,n)$, where $(x,y)$ denotes the greatest common divisor of $x$ and $y$. 

This problem has plagued me for about a week or so now, and I have reached my limit. I have pages and pages of attempts. I could use some help. 

Recall the fourth isomorphism theorem, that says that the subgroups of a quotient are the same as subgroups of the original group which contain the kernel of the homomorphism.
So then we have the canonical projection:
$$\pi_n:\Bbb Z\to \Bbb Z/n\Bbb Z$$
We know the kernel is exactly $\langle n\rangle$ and that all subgroups containing the kernel are exactly $\langle d\rangle$ where $d|n$. The lift of a given $\bar{x}$ is just $\gcd(x,n)$ so we see

$$\langle\bar{a}\rangle\le\langle\bar{b}\rangle\iff\langle a,n\rangle\le\langle b,n\rangle$$

for the lifted subgroups which are the ones generated by $\gcd(a,n)$ and $\gcd(b,n)$. And this completes the proof.

Following the op's indication he has yet to learn the isomorphism theorems, here is an ad hoc approach:
Note that $\langle \bar a\rangle$ is made up of things in $\Bbb Z$ which are linear combinations of $n$ and $a$, since these generate the kernel. But then $\langle a,n \rangle = \{ax+ny : x,y\in\Bbb Z\}=\langle \gcd(a,n)\rangle$ because $\langle\bar a\rangle = \{\bar a\bar x : \bar x\in\Bbb Z/n\}$. But then as subsets of $\Bbb Z$ we note that $\langle x\rangle\subseteq\langle y\rangle$ iff all multiples of $x$ are also multiples of $y$, i.e. iff $y|x$. Hence $\langle\bar a\rangle \le \langle\bar b\rangle$ iff $\gcd(b,n) | \gcd(a,n)$.
