Show that SO(2) is isomorphic with the complex circle group

Question

For my math study I have to prove that $SO(2)$ is isomorphic with the complex circle group. Some steps in this prove are a bit difficult to me, so I hope you could help me.

With $SO(2)$ I mean the group of all rotations $\rho_{x}$ in $\mathbb{R^{2}}$ with $x$ the angle of rotation around the origin. With $U_{1}$ I mean the complex circle group, so $U_{1}$ = {$z \in \mathbb{C}: |z| = 1$}.

This is what I've done so far:

I defined $f: SO(2) \to U_{1}$ with $f(\rho_{x}) = e^{2\pi ix}$ and proved that it is an homomorphism and that it is surjective. So the only things I have to do is proving that $f$ is well-defined and injective.For injectivity, I have come so far:

Assume $f(\rho_{x}) = f(\rho_{y})$ $\Rightarrow$ $e^{2\pi ix} = e^{2\pi iy}$ $\Rightarrow$ $x - y \in \mathbb{Z}$

How do I have to complete the injectivity prove and prove that $f$ is well-defined?

Thanks in advance!

Answer 1

Alternately, one can show this, while avoiding checking well-definedness manually, using the First Isomorphism Theorem.

First observe for concreteness (i.e., making easier writing down certain maps later) that:

1. The element in $SO(2)$ that corresponds to (anticlockwise) rotation by $x$ revolutions (equivalently, $2 \pi x$ radians) can be written as the rotation matrix $$\begin{pmatrix}\cos 2 \pi x & -\sin 2\pi x\\ \sin 2 \pi x& \cos 2 \pi x\end{pmatrix};$$ in particular, the representation has the feature that group multiplication in $SO(2)$ is just (the restriction of) matrix multiplication.
2. Any element in $U_1 \subset \mathbb{C}$ can be written as $$e^{2 \pi i x} = \cos 2 \pi x + i \sin 2\pi x.$$

This suggests a correspondence $$\begin{pmatrix}\cos 2 \pi x & -\sin 2\pi x\\ \sin 2 \pi x& \cos 2 \pi x\end{pmatrix} \leftrightarrow e^{2 \pi i x} = \cos 2 \pi x + i \sin 2\pi x.$$

We now formalize this idea: Define a map $F: \mathbb{R} \to SO(2)$ by $$F : x \mapsto \begin{pmatrix}\cos 2 \pi x & -\sin 2\pi x\\ \sin 2 \pi x& \cos 2 \pi x\end{pmatrix};$$ one can readily show that it is surjective, and using the usual angle sum identities shows that it is a homomorphism. Now, $\ker F = \mathbb{Z}$, so by the F.I.T., $F$ descends to an isomorphism $$\tilde{F}: \mathbb{R} / \mathbb{Z} \stackrel{\cong}{\to} \text{im } F = SO(2).$$ Similarly, one can induce an isomorphism $\tilde{G}: \mathbb{R} / \mathbb{Z} \stackrel{\cong}{\to} U_1$.