normaliser of dihedral group of order 4 Let $D_{2n}$ be the dihedral group of order $2n$. I would like to show that if $4$ divides $n$, then the normaliser $N_{D_{2n}}(D_{4})=D_{8}$. I know that 
$$|D_{2n}:N_{D_{2n}}(D_{4})|=|\mathcal{C}|,$$
where $\mathcal{C}$ is the conjugacy class of $D_{4}$ in $D_{2n}$. I am not sure how I would find the size of $\mathcal{C}$ and whether that would even help answer the question.
Thanks for any help in advance.

Let $G=D_{2n} = \langle a,b \mid a^n=b^2=(ab)^2=1 \rangle$. Then we can take the $D_4$ subgroup to be $\langle c,b \rangle$, where $c=a^{n/2}$. The conjugacy class of $b$ in $G$ is $\{ ba^{2k} : \ 1 \le k \le n/2 \}$ of size $n/2$, and this  class includes $bc$. Since $c$ is central in $G$ and hence in every conjugate of $D_4$, each conjugate of $D_4$ contains $c$ and two conjugates of $b$, so there are a total of $n/4$ such conjugates, which gives you the required order for the normalizer of $D_4$.
