For the $SO(2)$, I want to show that there is precisely one subgroup of order $n$ for every natural $n$. I can see the obvious case of the identity subgroup of order 1, but from here I get stuck. What would a subgroup of order 2 even look like? What other rotation of the circle would be in this subgroup? Is it some arbitrary multiple $2\pi r$ for some $0 < r \leq 1$? I am having trouble visualizing this.
An idea I've had is that these finite order subgroups should be cyclic, i.e. the subgroup of order 2 looks like $G = \{ I, A_{\pi} \}$, where $I$ is the identity and $A_{\pi}$ is the rotation matrix for a rotation of angle $\pi$. Then the subgroup of order $n$ would be that which contains $2\pi$ divided into $n$ rotations, but I'm unsure if this is the correct approach.