Let $\varphi:Z_2\rightarrow Aut(Z_n)$ be a homomorphism such that $\varphi(\overline{1})$ is the automorphism by inversion.
Set $Dih_n\triangleq Z_n\rtimes_{\varphi} Z_2$.
How do I prove that $<x,y|x^n=y^2=(xy)^2=1>$ is a presentation for $Dih_n$?
Here's how I tried:
Let $F(x,y)$ be the free group on $\{x,y\}$
Define $p(x)=(\overline{1},\overline{0})$ and $p(y)=(\overline{0},\overline{1})$.
Then, $p$ can be extended to a homomorphism $\Omega:F(x,y)\rightarrow Dih_n$ and $\Omega$ is surjective. Hence, $F(x,y)/\ker(\Omega)\cong Dih_n$.
Moreover, $x^n,y^2,(xy)^2$ are in $\ker(\Omega)$.
Let $N$ be the normal closure of $\{x^n,y^2,(xy)^2\}$. Then $N\subset \ker(\Omega)$. However, how do I prove that $N=\ker(\Omega)$?
You can simplify things by counting.
Let $G_n$ be the group with presentation $\langle r,s:r^n,s^2,rsrs\rangle$. Then $G$ has at most $2n$ elements. Indeed, every element is a word in $r,s$, and by $srs=r^{-1}$ we can arrange any word to be in the form $s^i r^j$ with $i=0,1$ and $j=0,\ldots,n-1$, since $r^n=1$ and $s^2=1$ -- this means we have at most $2n$ distinct words. Note any word of the form $sr^i$ has order a divisor of $2$ ($sr^isr^i=ssr^{-i}r^i=1$), but since $r,s$ don't commute, it cannot be trivial for any $i$ (the case $i=0$ being obvious). With this one checks the elements $1,r,r^2,\ldots,r^{n-1},s,sr,sr^2,\ldots,sr^{n-1}$ are all distinct, hence $G_n$ has order exactly $2n$.
On the other hand, it is clear $D_n =C_n\rtimes C_2$ has exactly $2n$ elements, and as you said we have a group morphism $\eta:G_n \to D_n$ that sends $r$ the the generator of $C_n$ and $s$ to the generator of $C_2$. Since this is surjective and our sets are of equal cardinality, it is injective and hence an isomorphism.