Show that
$$\begin{align} \langle c, d: c^n , d^2, (cd)^2\rangle &\to D_n,\\ c&\mapsto r,\\ d&\mapsto s\end{align}$$ is an isomorphism, where $D_n$ is the dihedral group, $r$ represents the rotation about the origin by $\frac{2\pi}n$ radians, and $s$ represents reflection about the $x$-axis.
I know that $\langle c, d: c^n , d^2, (cd)^2\rangle$ is defined to be $F(c,d)/N(c^n, d^2, (cd)^2),$ where $F(c,d)$ is the free group on $\{c,d\}$ and $N(c^n, d^2, (cd)^2)$ is the smallest normal subgroup of $F(c,d)$ containing $\{c^n, d^2, (cd)^2\}.$ Let $F := F(c,d)$ and $N := N(c^n,d^2, (cd)^2).$ Also, isomorphisms are bijective homomorphisms. I know that every element of $D_n$ can be expressed as $r^i s^j$ for some $0\leq i< n, 0\leq j < 2,$ so since $c^ir^j\mapsto s^i r^j,$ this implies the map is surjective. However, I'm not sure how to show that the map is injective. Also, I'm not really sure how to show it's a homomorphism.
Let $G = \langle c,d : c^n, d^2,(cd)^2\rangle$. $c\mapsto r$ and $d\mapsto s$ defines a surjective morphism $F(c,d)\to D_n$. Since $r^n = s^2=(rs)^2 = 1$, this induces a surjective morphism $\phi : G\to D_n$.
Now, in $G$ we have $cd = d^{-1}c^{-1} = dc^{n-1}$. Therefore, any $x\in G$ can be written $x = d^ic^j$ with $i\in\{0,1\}$ and $j\in\{0,\ldots,n-1\}$. Therefore, we have $|G|\leq 2n$. Since $|D_n | = 2n$, we have $|G|=2n$ and $\phi$ is an isomorphism.