How to prove that a quotient group is isomorphic to a group of matrices Consider the group $(\mathbb{R},+)$ and the subgroup
$$H=\{2n \pi |n\in \mathbb{Z}\}$$
Let $G$ denote the group of matrices of the form
\begin{bmatrix}
    \cos(\theta) & -\sin(\theta) \\
    \sin(\theta) & \cos(\theta)
  \end{bmatrix}
Where $\theta$ is any real number and the group operation is matrix multiplication. Prove that $\mathbb{R}/H$ is isomorphic to $G$ by first giving an explicit formula for a map $\phi:\mathbb{R}/H\to\ G$, and then checking that your map is well-formed, bijective, and respects the group operation.
I know that for the map to be well defined, the mapping has to be element to element, and that I need to ensure that its one-to-one and onto. But I am not making any progress with the proof nor able to give a formula for the map.
Can someone please help me out!
 A: Problem asks for an explicit map $\phi: R/H \to G$ but probably, the best approach here is to use the First Isomorphism Theorem. We will try to define a homomorphism $\phi: \mathbb{R} \to G$ such that $\phi$ is surjective and $\text{Ker}(\phi) = H$.

Another thing is if $T:\mathbb{R}^2 \to \mathbb{R}^2$ is a linear transformation that rotates a vector $\theta$ radians around origin clockwise, and $B$ is the ordered standard basis for $\mathbb{R}^2$, then $[T]_B^B = \begin{bmatrix}
    \cos(\theta) & -\sin(\theta) \\
    \sin(\theta) & \cos(\theta)
  \end{bmatrix}$. So, we can consider $G$ as group of rotations in $\mathbb{R}^2$ (I am writing this part just for intuition).

Now, define $\phi: \mathbb{R} \to G$ such that $\phi(\theta) = \begin{bmatrix}
    \cos(\theta) & -\sin(\theta) \\
    \sin(\theta) & \cos(\theta)
  \end{bmatrix}, \forall \theta \in \mathbb{R}$. Then, $\phi$ is well-defined and a homomorphism (I leave this part to you. To prove it is a homomorphism, you can use the intuition part). Also $\phi$ is clearly surjective and,
$$\text{Ker}(\phi) = \bigg\{\theta \in \mathbb{R} \mid \phi(\theta) = \begin{bmatrix}
    1 & 0 \\
    0 & 1
  \end{bmatrix}\bigg\} = \{\theta \in \mathbb{R} \mid \cos(\theta) = 1, \sin(\theta) = 0\} = H$$
(Note that we could also find this by using the intuition part above. Because kernel of this map is simply the angles in radians that when we rotate a vector that much, we get the same vector)
Then, by using the proof of First Isomorphism Theorem, we can easily obtain a map $\psi: \mathbb{R}/H \to G$ such that $\psi(\theta H) = \phi(\theta) = \begin{bmatrix}
    \cos(\theta) & -\sin(\theta) \\
    \sin(\theta) & \cos(\theta)
  \end{bmatrix}$. I am leaving it to you to check that $\psi$ is well-defined, homomorphism and bijective.
A: Hint:
Check that the map
\begin{align}
\mathbf R&\longrightarrow G \\
x &\longmapsto \begin{bmatrix}
    \cos x & -\sin x\\
    \sin x & \phantom{-}\cos x
  \end{bmatrix}
\end{align}
is a homomorphism from the additive group $(\mathbf R, +)$ onto the multiplicative  group $G$, and dtermine its kernel, to apply the $1^\text{st}$ isomorphism theorem.
