According to Wikipedia the Dihedral group $D_n \cong \; \langle r,s \mid r^n = 1, s^2 = 1, s^{-1}rs = r^{-1}\rangle$. But why does this apply?
As far as I understand the group presentation means that I take $\langle r,s\rangle$ and remove all elements that do not satisfy the provided relations. The Dihedral group $D_n$ consists of $n$ rotations and $n$ reflections. If considering $s$ as reflection, $s^2 = 1$ means that repeating a reflection leads to the identity. But what about the other relations, and why does this apply in general?
Well, you can take $r = \left(\begin{array}{clcr} \cos(\frac{2 \pi }{n}) & -\sin(\frac{2 \pi }{n})\\ \sin(\frac{2 \pi }{n}) & \cos(\frac{2 \pi }{n}) \end{array} \right) $ and $s = \left( \begin{array}{clcr} 0 & 1\\ 1 & 0 \end{array} \right) $ to see that $D_{n}$ satisfies the relations in that presentation. This means that $D_{n}$ is a homomorphic image of the presented group. On the other hand, notice that every element of the presented group is expressible in the form $r^{j} s^{i}$ with $ 0 \leq j \leq n$ and $0 \leq i \leq 1.$ Hence the presented group has order $2n,$ so the homomorphism onto $D_{n}$ must be injective, hence an isomorphism.