Prove that, considering $\mathbb{R}$ as an additive group, we have $SO_2(\mathbb{R})\cong \mathbb{R}/2\pi\mathbb{R}$ . Let $SO_2(\mathbb{R})$ be the group of rotations of the circle under the operation of composition.
Prove that, considering $\mathbb{R}$ as an additive group, we have
$SO_2(\mathbb{R})\cong \mathbb{R}/2\pi\mathbb{R}$ .
attempt: Suppose $\phi: \mathbb{R} → SO_2(\mathbb{R})$ . Then we need to show $\phi $ is an homomorphism.
And that $ker\phi = 2\pi\mathbb{R}$, and $Im(\phi) = SO_2(\mathbb{R})$. Then conclude by first isomorphism theorem $SO_2(\mathbb{R})\cong \mathbb{R}/2\pi\mathbb{R}$ .
Define $\phi$ by $\phi(\theta + 2\pi\mathbb{R})= \begin{pmatrix}  a & b
\\ c & d\end{pmatrix}  $ , where the determinant of the matrix is not zero.
$\phi$ is homomomrphism. Take $r_1 = \theta_1 + 2\pi \mathbb{R}, r_2 = \theta_2 + 2\pi \mathbb{R} \in \mathbb{R}/ 2\pi \mathbb{R}$,
and $\phi(r_1 r_2)  =\phi[(\theta_1 + 2\pi \mathbb{R})(\theta_2 + 2\pi \mathbb{R})] = \phi(\theta_1\theta_2 +2\pi \mathbb{R} ) = $.
Can someone please help me? I dont know if I have define $\phi$ well. 
I am confuse about elements in $SO_2(\mathbb{R})$. 
For showing that $2\pi \mathbb{R}$ is the kernel of some homomorphism, can I just say that given any $\theta$ , $\theta + 2\pi \mathbb{R}$ will only return to the same spot. Like a revolution, so it's kind of like clock arithmetic. Thus, it must be the kernel.
Thank you very much!.

You ought to say what $\phi:\mathbb{R}\to SO_2(\mathbb{R})$ is.  Define $\phi(\theta)=\begin{pmatrix}\cos\theta&-\sin\theta\\ \sin\theta & \cos\theta\end{pmatrix}$ and show that $\phi(\alpha+\beta)=\phi(\alpha)\phi(\beta)$ (so then it is a homomorphism).  It is surjective by definition of $SO_2(\mathbb{R})$.  Since $\phi(\theta)$ is the identity matrix if and only if $\theta$ is a multiple of $2\pi$, then the kernel is $2\pi\mathbb{Z}$, and so we have an isomorphism (by the first isomorphism theorem) $\mathbb{R}/2\pi\mathbb{Z}\cong SO_2(\mathbb{R})$.
Note that the kernel is not $\mathbb{R}/2\pi\mathbb{R}$, which, since $2\pi\mathbb{R}=\mathbb{R}$, is trivial.
