Proof that subgroup of circle group has infinite order

Question

Let $z=\cos{\theta}+i\sin{\theta}$ be in the subgroup of the circle group where $\theta \in \mathbb{Q}$. As we already know, the circle group has infinite order. It's also intuitive for me that this subgroup also has infinite order since there are infinite rational numbers between $0$ and $2\pi$ and thus there are infinite complex numbers $z=\cos{\theta}+i\sin{\theta}$. So, is there anything tricky here?

Answer 1

There are infinitely many rationals in $[0,2\pi]$, so infinitely many points $e^{i\theta}$ in the sugroup (you have already established that it's a subgroup, apparently). The points are on the unit circle at angle $\theta $ with the $x$-axis...