Let $K$ be a subgroup of $\mathbb{Z}_n$. How do I prove that there exists $\bar{m} \in K$ such that $m\mid n$?

Question

I'm trying to use this in a different proof (describing all the subgroups of $\mathbb{Z}_n$), but everything I've thought of breaks at some point.

Answer 1

Hint Pick the smallest positive integer $k \in \mathbb Z$ such that $\bar{k} \in K$.

By long division you have $$n=q \cdot k +r$$

Use this relation to conclude that $\bar{r} \in K$.

Answer 2

The statement is not true. Consider the subgroup $K$ of $\mathbb{Z}_4$ generated by $\bar{2}$. Then $1\mid 4$ but $\bar{1}\notin K$. Also, you cannot say that $m\in \mathbb{Z}_k$ if $m$ is just a number! Elements of $\mathbb{Z}_k$ are equivalences classes. People tend to get more sloppy with notation as they understand more things but don't be sloppy at the beginning, it might slow down your learning process.