# 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!.

• I fail to see where you defined $\phi$. Also, $\Bbb R$ is a group w.r.t. the sum, not the product. – user228113 Feb 2 '16 at 4:35
• So it would look like this? $\phi(r_1 + r_2) = \phi[(\theta_1 + 2\pi R) + (\theta_2 + 2\pi R)] = \phi(\theta 1 + \theta_2 + 2\pi R)$? – Mahidevran Feb 2 '16 at 4:39
• How much is $\phi(-17+\sqrt{23}+2\pi\Bbb R)$ ? – user228113 Feb 2 '16 at 4:40
• Also, there are a lot of matrices with determinant $1$ that are not in $SO_2(\Bbb R)$. – user228113 Feb 2 '16 at 4:41

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.
• Are you defining your matrix with cosine and sine because of the relationship $cos^2\theta + sin^2\theta = 1$? , and the determinant must have a value of 1 – Mahidevran Feb 2 '16 at 4:46
• I did that because you said $SO_2(\mathbb{R})$ is defined to be rotation matrices. If you instead said it's all matrices $A$ such that $A^TA=I$, then I would have proved that the matrix I wrote satisfies that equation, and furthermore, that all such $A$ satisfying $A^TA=I$ can be written that way. – Kyle Miller Feb 2 '16 at 4:50