This question is quite direct, but I guess I might be falling in redundancy.
Prove that there is only 1 homeomorphism $h: D_{2n}\rightarrow GL_2(\mathbb{R})$, such that:
$h(r)=$ rotation matrix with angle $\theta=\frac{2\pi}{n}$
$h(s)=$ permutation matrix
My idea is to assume there are two homeomorphism, but when I conclude that both are equal it seems that I'm saying something that I assumed previously. What should I add, or how would you aproach the problem?
In case my comment is correct, let $\varphi$ be one such homomorphism. Suppose $\psi$ is also a homomorphism with the required property.
Since $D_{2n}$ is generated by $r, s$ - i.e. every element is of the form $s^lr^k$, where $l \in \{0, 1\}$ and $k \in \{0, …, n - 1\}$ — and $\psi(s^lr^k) = \psi(s^l) \psi(r^k) = \psi(s)^l \psi(r)^k$, which in turn equals $ \varphi(s)^l \varphi(r)^k = \varphi(s^lr^k)$, we conclude that $\psi = \varphi$, since they coincide in every element of $D_{2n}$.
Notice how, in this proof, I assumed there are two and proved they are equal. I haven’t used any of the thesis in my proof - if you did something similar, you’re probably good to go.
Also notice, as pointed out in the comments, that I used as a hypothesis that there is one. So you would also have to show that the function $\varphi: D_{2n} \to GL_2(\mathbb{R})$ that maps $$r \mapsto \begin{bmatrix} \cos(\frac{2 \pi}{n}) & -\sin(\frac{2 \pi}{n}) \\ \sin(\frac{2 \pi}{n}) & \cos(\frac{2 \pi}{n}) \end{bmatrix}$$ $$s \mapsto \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$$ is a homomorphism, when extended to the entirety of $D_{2n}$, to show one exists.
Does that make sense?