The group $\mathbb Z_n$ is isomorphic to a subgroup of $GL_2(\mathbb R)$.

Question

I need to prove the following:

The group $\mathbb Z_n$ is isomorphic to a subgroup of $GL_2(\mathbb R)$.

How can I prove this?

$\mathbb Z_n$ is of order $n$ so it is isomorphic to a subgroup of $GL_n(\mathbb R)$.

I know this is true but here it is given $GL_2(\mathbb R)$.

Answer 1

Consider the group of those matrices of the form\begin{pmatrix}\cos\left(\frac{2k\pi}n\right)&-\sin\left(\frac{2k\pi}n\right)\\\sin\left(\frac{2k\pi}n\right)&\cos\left(\frac{2k\pi}n\right)\end{pmatrix}($k\in\{0,1,\ldots,n-1\}$). In other words, it is the subgroup of $GL_2(\mathbb{R})$ generated by the matrix$$\begin{pmatrix}\cos\left(\frac{2\pi}n\right)&-\sin\left(\frac{2\pi}n\right)\\\sin\left(\frac{2\pi}n\right)&\cos\left(\frac{2\pi}n\right)\end{pmatrix},$$which has order $n$.