How to build the character table of $D_{2n}$ where $n$ is even? So I'm supposed to build the character table of $D_{2n}$ for even $n$. 
$G=D_{2n}= \langle x,y| x^n=y^2=1,\,  yxy^{-1}=x^{-1}\rangle$, and the commutator$[y,x]=x^{n-2}$, where $n$ is even, so $G'=\langle x^{2(\frac{n}{2}-1)} \rangle \cong C_n$, thus the Abelization, $G/ G'\cong C_2$. And I suppose, at this point I should somehow apply Frobenius reciprocity, but I don't think I really understand it at this point. 
Can you help me, please? 
 A: $G=D_{2n}\geq H=⟨x^{n-2}⟩$, where $H$ is the commutator subgroup. Now if we factorize G by H, we get $G/H \cong ⟨x,y| x^2=y^2=1, xy=yx⟩\cong C_2\times C_2$. The order of this group is 4, so obviously the number of 1 dimensional representations must be 4:
$ \chi_{id}$, the trivial representation,$ \chi_{2}$ takes value $1$ on $id \in G$ and $x^i, i =1,2,...,\frac{n-1}{2}$ and $-1$ on $y$ and $yx$, $\chi_3$ and $\chi_4$ are $(-1)^i$ for $ i=0,1,...,\frac{n-1}{2}$ and $\chi_3 (y)=1$, $\chi_4 (y)=-1$, $\chi_3(xy)=-1$, $\chi_4(xy)=1$.
Now we can construct the further representations with induction. We will use the fact, that
$$\chi\uparrow_H^G = \sum_{[t\in G/H], t^{-1}gt\in H}\chi (t^{-1}gt)$$
Let $\xi_n$ be a primitive $n$th root of unity. Then
for $ 1\leq s \leq \frac{n}{2}-1$, $\chi_{\xi_n ^s}(id)=2$ (these are 2 dimensional representations) and on $x^i, i=1,...,\frac{n-1}{2} $ it is $\xi_n ^i+\xi_n ^{-i}$, $\chi_{\xi_n ^s}(y)=\chi_{\xi_n ^s}(xy)=0$. 
We can use Frobenius reciprocity to check this character table. 
