Existence of countable subgroup How can one show that there exists a countable subgroup $G$ of $SO(2)$?

(where $SO(2) =$ {$ M \in \mathbb R^{2 \times 2} | M^TM = Id, det(M) = 1$})
Consider a matrix in ${\rm SO}(2)$ it can be written (uniquely for some $0\leq \theta &lt; 2\pi$) as  

$$\begin{bmatrix} \cos\theta &amp; \sin\theta\\ -\sin\theta &amp; \cos\theta\end{bmatrix}.$$
This is of finite order if and only if $\theta$ is rational multiple of $2\pi$. 

(1) You can choose $\theta$ which is not rational multiple of $2\pi$; order of above matrix will be infinite, so subgroup generated by it is one countable (infinite) subgroup.

(2) Or, you can collect all matrices of above type for all $\theta$ that are rational multiples of $2\pi$; this is certainly a countable subgroup.
Consider a matrix in ${\rm SO}(2)$ it can be written (uniquely for some $0\leq \theta &lt; 2\pi$) as  

$$\begin{bmatrix} \cos\theta &amp; \sin\theta\\ -\sin\theta &amp; \cos\theta\end{bmatrix}.$$
This is of finite order if and only if $\theta$ is rational multiple of $2\pi$. 

(1) You can choose $\theta$ which is not rational multiple of $2\pi$; order of above matrix will be infinite, so subgroup generated by it is one countable (infinite) subgroup.

(2) Or, you can collect all matrices of above type for all $\theta$ that are rational multiples of $2\pi$; this is certainly a countable subgroup.