1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $D_4$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.