Show that $\Bbb{Q}/\Bbb{Z}$ as a $\Bbb{Z}$-module has no free non-trivial submodules.
I already proved that $\Bbb{Q}/\Bbb{Z}$ is not finitely generated as a $\Bbb{Z}$ module. I want to prove the aforementioned statement. I tried looking at all $\phi:\Bbb{Z}\rightarrow\Bbb{Q}/\Bbb{Z}$, and concluded that if $\phi(1)=\frac{m}{n}$ then $\operatorname{Ker}{\phi}=n\Bbb{Z}$ and $\operatorname{Im}(\phi)=\{\frac{am}{n}:1\leq a\leq n\}$, but this didn't get me too far. Any help would be appreciated.
You're on the right track! Remember a (nontrivial) free $\mathbb{Z}$ module contains a copy of $\mathbb{Z}$, and so if one existed you would be able to find an injective map $\phi : \mathbb{Z} \to \mathbb{Q} / \mathbb{Z}$.
But you're super close to showing that $\ker(\phi) \neq 0$ for any such $\phi$. Once you do this, you'll have shown that no injective map $\mathbb{Z} \to \mathbb{Q}/\mathbb{Z}$ exists!
Do you see how to take what you've done and conclude that every $\phi : \mathbb{Z} \to \mathbb{Q}/\mathbb{Z}$ has nontrivial kernel? As a commenter notes, it's helpful to remember that every element of $\mathbb{Q} / \mathbb{Z}$ has finite order.
As one quick critique, $\text{Ker} (\phi) = n \mathbb{Z}$ not $\mathbb{Z} / n \mathbb{Z}$. I assume this was a typo, though. But it does show how close you are -- is it possible for $n \mathbb{Z}$ to equal $0$?
I hope this helps ^_^