Group isomorphism between two groups . Let $SO_2(\mathbb{R})$ be the group of rotations of the circle under the operation of composition. And consider $\mathbb{R}$ as an additive group.
Prove that $SO_2(\mathbb{R}) \cong \begin{Bmatrix} A\in GL_2(\mathbb{R}) : A^tA = I, detA = 1 \end{Bmatrix}$
Hint: $SO_2(\mathbb{R}) \cong \mathbb{R}/2\pi\mathbb{Z}$
Attempt: 
I have verified that $A $ has the form of a rotational matrix, $\begin{bmatrix}
    \cos\theta & -\sin\theta \\
    \sin\theta & \cos\theta
  \end{bmatrix}$, where $0 \leq \theta < 2\pi $, and that the rotational matrices are a subgroup of $GL_2(\mathbb{R})$.
Now I want to show $SO_2(\mathbb{R}) \cong \begin{Bmatrix} A\in GL_2(\mathbb{R}) : A^tA = I, detA = 1 \end{Bmatrix}$, 
Let $\phi : SO_2(\mathbb{R} → \begin{Bmatrix} A\in GL_2(\mathbb{R}) : A^tA = I, detA = 1 \end{Bmatrix}$.
Let $\phi(\theta) = \begin{bmatrix}
    \cos\theta & -\sin\theta \\
    \sin\theta & \cos\theta
  \end{bmatrix}$
Then I have verified it's a homomorphism by $\phi(\theta_1 + \theta_2) = \phi(\theta_1)\phi(\theta_2)$.
Now I will show it's a bijective and conclude it's isomorphic to $\begin{Bmatrix} A\in GL_2(\mathbb{R}) : A^tA = I, detA = 1 \end{Bmatrix}$
Is this approach correct? Can someone please verify the define homomorphism makes sense . I don't know how the hint might help.
Any better approach or feedback would really help!
 A: This problem is a little odd in that I would consider
$$SO_2(\mathbb{R}) = \{A \in GL_2(\mathbb{R}) \;|\; A^tA=I \mbox{ and } \det(A)=1 \}$$
to be the definition of the special orthogonal group of $2 \times 2$ matrices!
Anyway, I think you're on the right track. You would not only need to check that your map is operation preserving and bijective but first of all you need to make sure that it actually makes sense (which it does). What I mean by this is that you need to verify that $\phi(\theta)^t\phi(\theta)=I$, $\det(\phi(\theta))=1$ (good so far), and also $\phi(\theta)=\phi(\theta+2\pi k)$ for $k \in \mathbb{Z}$. The last condition is obviously true (since sine and cosine are periodic) but should be stated to clarify that your mapping does not depend on the representative you chose for your coset in $\mathbb{R}/2\pi\mathbb{Z}$.
I might try a slightly different approach and use the first isomorphism theorem.
Define $\varphi:\mathbb{R} \to GL_2(\mathbb{R})$ by $\varphi(t) = \begin{bmatrix} \cos(t) & -\sin(t) \\ \sin(t) & \cos(t) \end{bmatrix}$.
Notice that $\det(\varphi(t)) = \cos^2(t)+\sin^2(t)=1 \not=0$ so $\varphi$ really does map into $GL_2(\mathbb{R})$.
Next, $\varphi(s)\varphi(t)=\begin{bmatrix} \cos(s) & -\sin(s) \\ \sin(s) & \cos(s) \end{bmatrix}\begin{bmatrix} \cos(t) & -\sin(t) \\ \sin(t) & \cos(t) \end{bmatrix}$ $=\begin{bmatrix} \cos(s)\cos(t)-\sin(s)\sin(t) & -(\sin(s)\cos(t)+\cos(s)\sin(t)) \\ \sin(s)\cos(t)+\cos(s)\sin(t) & \cos(s)\cos(t)-\sin(s)\sin(t) \end{bmatrix}$ $= \begin{bmatrix} \cos(s+t) & -\sin(s+t) \\ \sin(s+t) & \cos(s+t) \end{bmatrix}=\varphi(s+t)$
(using angle addition formulas).
We can check that $\varphi(t)^t\varphi(t) = \begin{bmatrix} \cos(t) & -\sin(t) \\ \sin(t) & \cos(t) \end{bmatrix}^t\begin{bmatrix} \cos(t) & -\sin(t) \\ \sin(t) & \cos(t) \end{bmatrix}$ $= \begin{bmatrix} \cos(t) & \sin(t) \\ -\sin(t) & \cos(t) \end{bmatrix} \begin{bmatrix} \cos(t) & -\sin(t) \\ \sin(t) & \cos(t) \end{bmatrix}$ $= \begin{bmatrix} \cos^2(t)+\sin^2(t) & 0 \\ 0 & \cos^2(t)+\sin^2(t) \end{bmatrix} = I$. This means that the image of $\varphi$ is a subset of $\{ A \in GL_2(\mathbb{R}) \;|\; A^tA=I \mbox{ and } \det(A)=1 \}$.
Conversely, suppose that $A^tA=I$ and $\det(A)=1$. Then if $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, then $A^tA=I$ implies that $a^2+c^2=1$, $ab+cd=0$, and $b^2+d^2=1$. This implies that $(a,c)$ lies on the unit circle so there exists some $t \in \mathbb{R}$ such that $a=\cos(t)$ and $c=\sin(t)$. Likewise, $c=\sin(s)$ and $d=\cos(s)$. But $ab+cd=0$ implies that $\cos(t)\sin(s)+\sin(t)\cos(s)=0$. So $\sin(s+t)=0$ (using an angle addition formula). Without loss of generality we either have that $s+t=0$ or $s+t=\pi$ (we can use periodicity to remove $2\pi k$ ($k \in \mathbb{Z}$) from $s+t$). If $s+t=\pi$, $\det(A)=ad-bc=\cos(t)\cos(\pi-t)-\sin(t)\sin(\pi-t)=\cos(t+\pi-t)=\cos(\pi)=-1$ (wrong). So $s+t=0$ [then $\det(A)=ad-bc=\cos(t)\cos(-t)-\sin(t)\sin(-t)=\cos(t-t)=\cos(0)=1$ (correct)]. This then implies that $A = \begin{bmatrix} \cos(t) & \sin(-t) \\ \sin(t) & \cos(-t) \end{bmatrix} = \begin{bmatrix} \cos(t) & -\sin(t) \\ \sin(t) & \cos(t) \end{bmatrix} = \varphi(t)$. So the image of $\varphi$ is exactly $\{A \in GL_2(\mathbb{R}) \;|\; A^tA=I \mbox{ and } \det(A)=1 \}$.
Finally, $\mathrm{Ker}(\varphi) = \{ t \in \mathbb{R} \;|\; \varphi(t)=I \}$. Now $\varphi(t) = \begin{bmatrix} \cos(t) & -\sin(t) \\ \sin(t) & \cos(t) \end{bmatrix} = I$ implies that $\cos(t)=1$ and $\sin(t)=0$. Thus $t=2\pi k$ for some $k \in \mathbb{Z}$ (conversely $\varphi(2\pi k)=I$). Hence $\mathrm{Ker}(\varphi) = 2\pi \mathbb{Z}$. 
Now the first isomorphism theorem gives us that $\mathbb{R}/\mathrm{Ker}(\varphi) \cong \mathrm{Im}(\varphi)$ and so in particular, $SO_2(\mathbb{R})=\mathbb{R}/2\pi \mathbb{Z} \cong \{A \in GL_2(\mathbb{R}) \;|\; A^tA=I \mbox{ and } \det(A)=1 \}$.
If you are trying to prove $G/H \cong K$, it is usually best to look for a homomorphism from $G$ to $K$ whose kernel is $H$. Then apply the isomorphism theorem. Trying to build the isomorphism directly is usually a bit messier. :)
