Find all the possible pairs of submodules $M_1$ and $M_2$ of the $\mathbb{Z}$-module $\mathbb{Z}_{18}$ so that $\mathbb{Z}_{18} = M_1 \oplus M_2$.
I started by considering the possible proper subgroups of $\mathbb{Z}_{18}$, which are $\langle\bar{9}\rangle$, $\langle\bar6\rangle$, $\langle\bar3\rangle$ and $\langle\bar2\rangle$, which are also $\mathbb{Z}$-submodules.
Consider the pair $M_1 = \langle\bar9\rangle$ and $M_2 = \langle\bar{2}\rangle$. Since $\bar1 = \bar9 + 5\bar2$, then for $\bar{k} \in \mathbb{Z_{18}}$ I checked by hand that there's unique $p, q \in \mathbb{Z}$ so that $\bar{k} = p\bar9 + q\bar2$. Unless I'm missing something, I think this implies that this is a valid pair.
I imagine the other pair to consider would be $\langle\bar3\rangle$ and $\langle\bar2\rangle$, but I'm not 100% sure I'm doing this the right way. Even if this is right, I imagine there must be a more elegant way of solving this.
Any help would be greatly appreciated!
As you remarked, there are four non-trivial subgroups of $\mathbb Z_{18}$ given by the divisors of $18$ (others than $1$ and $18$). Now it's good to know what's the sum and the intersection of such subgroups.
In general, in $\mathbb Z_n$ we have $\bar k\mathbb Z_n+\bar l\mathbb Z_n=\overline{\gcd(k,l)}\mathbb Z_n$ and $\bar k\mathbb Z_n\cap\bar l\mathbb Z_n=\overline{\operatorname{lcm}(k,l)}\mathbb Z_n$. (Here $k,l$ are divisors of $n$.) If you want $\bar k\mathbb Z_n\oplus\bar l\mathbb Z_n=\mathbb Z_n$, then must have $\operatorname{lcm}(k,l)=n$ and $\gcd(k,l)=1$. In your case this happens only for the pair $(2,9)$.