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I need to show that the groups $S^1, SO_2, G, H$ are isomorphic, where

$S^1 = \{z\in \mathbb C\mid |z| = 1\}$

$SO_2 = \{A \in GL_2(\mathbb R)\mid AA^T = \text{Id,} \det(A) = 1\}$

$G = \{T:\mathbb R^2 \to \mathbb R^2\mid T$ is a rotation with center at origin$\}$

$H = \mathbb R/\mathbb Z$

So, I know how to do

$H = \mathbb R/\mathbb Z\cong S^1$

by using the isomorphism theorem and the homomorphism

$\varphi:\Bbb R\to S^1,\quad\varphi(x)=e^{2\pi i x}$

because $\mbox{ker}(\varphi) = \mathbb Z$

For the group $SO_2$ I don't know how to define a function from $SO_2$ to another group. Which is the better way to do it? Using the isomorphism theorem?

Also, for $G$, maybe I can relate it to $S^1$, since $\varphi(x) e^{2\pi i x}$ is connected to rotations.

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Define $\phi:S^1\to SO_2$ as

$\phi (e^{i\theta})=\begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}$

And show that this is an isomorphism

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    $\begingroup$ This is part of your other answer, you should merge them to create a single, complete, answer. $\endgroup$ – hjhjhj57 Jul 15 '15 at 4:52
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Define $\psi:G \to S^1$ as follows

If $T\in G$ then suppose $T$ is rotation by an angle $\theta$ then

$T(x,y)=(x\cos \theta+y\sin \theta ~,~-x\sin \theta+y\cos\theta)$

then $$\psi (T)=e^{i\theta}$$

And show that this is an isomorphism.

Since being isomorphism is an equivalence relation so $G, S^1~\&~SO_2$ are isomorphic

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    $\begingroup$ This is part of your other answer, you should merge them to create a single, complete, answer. $\endgroup$ – hjhjhj57 Jul 15 '15 at 4:52
  • $\begingroup$ but $e^{i\theta}$ isn't bijective, its periodic :( $\endgroup$ – Guerlando OCs Jul 15 '15 at 5:07
  • $\begingroup$ we do not need $e^{i\theta}$ to be bijective, we need $\psi$ to be bijective. This is 100% correct you check $\endgroup$ – brinkle martyn Jul 15 '15 at 5:33

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