Show $SO_2(\mathbb{R}) \cong\{ A \in GL_2(\mathbb{R}) : A^tA = I, \det A = 1\}$, where $SO_2(\mathbb{R})$ is the group of rotations of the circle under the operation of composition.
Attempt: Suppose $A =\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$.
I have verified that $S = \{ A \in GL_2(\mathbb{R}) : A^tA = I, \det A = 1\}$ is a subgroup of $SO_2(\mathbb{R})$.
I was thinking to show define a function and show it's a bijective homomorphism to conclude the two groups are isomorphic to each other.
Let $\gamma : SO_2(\mathbb{R}) → S$.
define $\gamma(\theta) = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$.
I am confuse about how to define a function from $SO_2(\mathbb{R})$ to $S$.
Can someone please help me? Any better approach would help or suggestion. Thank you!