Discrete subgroups of $S^{1}$ are finite cyclic

Question

Suppose $G$ is a discrete subgroup of the unit circle. I have to prove that it is finite cyclic. Due to the discreteness of $G$, there exists $r$ in [0,1) such that $e^{2i\pi r}$ has the smallest argument. Now, if $r$ is rational, we are done. If it is irrational, then I think there should exist some contradiction to our choice of $r$. I don’t know how to show that $r$ cannot be irrational.

Answer 1

If $r$ is irrational, then $\{ e^{2\pi inr} \}_{n\in \mathbb{Z}}\subseteq G$ is dense in $S^1$. This would contradict the discreteness of $G$.

To show that the above set is dense, you can use Dirichlet's approximation theorem.