Checking for isomorphism.

Question

What is given is the group $R$.

G(R)

Prove that $R$ is isomorphic to $G$ by first giving an formula for a map $ϕ:R→G$, and checking that your map.

Answer 1

Let matrices of $G$ be denoted as $M(\theta)$, in which we just plug in $\theta$ for each entry in the rotation matrix. Just define $\phi : R \to G$ as $\phi(x):=M(x)$. First let's check this is a group homomorphism:

$$\phi(x+y)=M(x+y)=M(x)\cdot M(y)=\phi(x)\cdot\phi(y)$$

which follows from the fact that $M(\theta)$ is a rotation matrix. Note that the identity in $G$ is given by $$\begin{bmatrix}1 & 0\\ 0& 1\end{bmatrix}=\begin{bmatrix}\cos(0) & -\sin(0)\\ \sin(0)& \cos(0)\end{bmatrix}=\begin{bmatrix}\cos(2\pi n) & -\sin(2 \pi n)\\ \sin(2 \pi n)& \cos(2 \pi n)\end{bmatrix}$$ for any $n \in \mathbb{Z}$. Hence, the kernel of $\phi$ is given as $$Ker(\phi):=\{2\pi n : n \in \mathbb{Z}\}$$ We want to use the first isomorphism, but first we have to show $\phi$ is a surjection. $\phi$ is surjective if for all $M(\theta) \in G$, there exists some $x \in R$ such that $M(\theta)=\phi(x)$. This is trivial; since $\theta$ is a real number just choose $x=\theta$, then for all $M(\theta)$ we have that there exists an $x$ so that $\phi(x)=\phi(\theta)=M(\theta)$. Since $\phi$ is a surjection we can now use the first isomorphism theorem to assert
$$R/Ker(\phi)=R/H\approxeq G$$