The question is from Dummit and Foote pg146, Exercise 12.
Let $2n=2^ak$ where $k$ is odd. Prove that the number of Sylow $2$-subgroups of $D_{2n}$ is $k$.
This is my attempt.
My Attempt: The order of the Sylow $2$-subgroup must be $2^a$, and obviously $P \leq N_{D_{2n}}(P)$. So, by Sylow's theorem, $$n_2=[G:N_{D_{2n}}(P)]\equiv 1\pmod{2}$$
Since $[G:N_{D_{2n}}(P)]\geq [G: P]=k$, and $n_2$ is odd, we must conclude that $n_2=k$
However, the hint in the question suggests me to prove that $N_{D_{2n}}(P)=P$. I can get that result after I get the answer, not the other way around as the question suggests. How would I get the result $N_{D_{2n}}(P)=P$ first?
Best Answer
Let me be more explicit about what I meant in the comments. If you label your vertices from $1$ to $2^{a-1}k$, you can get a regular $2^{a-1}$-gon by taking the vertices $i, i+k, i+2k,..., i+2^{a-2}k$ for $i\in \{1,..., k-1\}$. This gives us $k$ distinct $2^{a-1}$-gon each of which has symmetry group $D_{2^{a}}$. These are indeed the desired Sylow subgroups. So, we have that $n_2=k$.