Prove that the normaliser of a Sylow $2$-subgroups of $D_{2n}$ is itself.

Question

Let $2n = 2^a k$ for $k$ odd. Prove that the normaliser of a Sylow $2$-subgroups of $D_{2n}$ is itself.

Here $|N|\mid |G|$ where $N$ is the normaliser of a sylow $2$ subgroup $H$. But $N$ can properly contain $H$. How to rule out this possibility?

Again I have observed if $a$ is greater than $2$, then $H$ can't contain $r$ to the power $i$ where $i$ is odd.

Answer 1

The statement holds for $k=1$ so we may assume that $k\ne 1$.

We know that $D_{2n}\cong C_{n}\rtimes \langle x\rangle$, where $\langle x\rangle \cong C_{2}\le G$ and acts Frobeniusly on $C_{n}$. Now assume that $p \mid |N_{G}(Q)|$, where $p$ is an odd prime and $Q$ is a Sylow $2$-subgroup of $D_{2n}$. Since any $p$-subgroup of $D_{2n}$ is normal, we get that $PQ=P\rtimes Q$, where $P\ne 1$ is the Sylow $p$-subgroup of $N_{G}(Q)$. Obviously $Q\triangleleft N_{G}(Q)$, so $PQ=P\times Q$. Hence $P\le C_{G}(x)$, a contradiction.

Answer 2

A visual way of seeing this is to think of a regular $n$-gon and its symmetries (to me this group is called $G=D_n$, but I'm aware that at least Dummit & Foote denote it $D_{2n}$). If $n=2^ak$, $k$ odd, then within a regular $n$-gon we see exactly $k$ regular $2^a$-gons, gotten from each other by rotating the picture by $\pi/n$.

Each of those $2^a$-gons has a copy of $D_{2^a}$ as its group of symmetries. As those symmetries are also symmetries of the $n$-gon, they form a Sylow $2$-subgroup $P$. Because there are exactly $k$ such subgroups, and the number of Sylow $2$-subgroups is equal to $[G:N_G(P)]$, we can conclude that $|N_G(P)|=|G|/k=|P|$. Hence $N_G(P)=P$.

Below please find the example picture of a regular $20$-gon together with the $5$ circumscribed squares. In the action of $D_{20}$ each of the five squares is stabilized by its own copy of a subgroup isomorphic to $D_4$. That's five distinct Sylow $2$-subgroups right there! Hence their normalizers must have order $|D_{20}|/5=8=|D_4|$.