Normalizer of a Sylow 2-subgroups of dihedral groups I can't solve the following exercise which is the last exercise in page 146 of Dummi & Foote's Abstract Algebra:

Let $2n=2^ak$ where $k$ is odd. Prove that the number of Sylow 2-subgroups of $D_{2n}$ is $k$. [Prove that if $P\in Syl_2(D_{2n})$ then $N_{D_{2n}}(P)=P.$]

Here $D_{2n}=<r,s\mid r^n=s^2=1,\,rs=sr^{-1}>$ is the dihedral group of order $2n$ and $Syl_2(D_{2n})$ is the set of all Sylow 2-subgroups of $D_{2n}$.
It is easy to solve the exercise if we prove that $N_{D_{2n}}(P)=P$ and that's what I can't prove. Could you give me some hints?

My attempt
I got the following results. They might be useless, but who knows?
1) If $a=1$ then Sylow $2$-subgroups of $D_{2n}$ have order $2$ and so their number is the number of elements of $D_{2n}$ of order $2$ which is $n$ because the set of elements of order $2$ of $D_{2n}$ is $\{sr^i:i\in\{0,\,\dots,\,n-1\}\}$. So we can assume that $a>1$.
2) Let $P\in Syl_2(D_{2n})$ and suppose $a>1$ (from 1). Then $P$ must have an element of order $2$ in $\{sr^i:i\in\{0,\,\dots,\,n-1\}\}$ . In fact, we know that all cyclic subgroups of order larger than $2$ of a dihedral group are generated by a power of $r$, so if $P$ is cyclic then $\exists m\in\mathbb{Z}^+,\,P=<r^m>$ and $m$ is a divisor of $n$. Then we have:$|r^m|=\frac{2^{a-1}k}{m}=2^a$ and so $k=2m$ which is false because $k$ is odd. Thus $P$ isn't cyclic and so $P$ is generated by at least two elements ($P$ is finite so it has a finite number of generators and let $S$ be a set of generators of $P$ so that $P=<S>$). If all elements of $S$ have order larger than $2$ they must be powers of $r$ and so $S\subset <R>$ thus $P\le <r>$ which is impossible, otherwise $P$ would be cyclic. Thus at least one element in $S$ has order $2$. But we know that there exists a unique element of $<r>$ which order is $2$. Thus we have two possible cases:


*

*Case 1: There is an element of order larger than $2$ in $S$.


This element must be a power of $r$. If all other elements of $S$ are powers of $r$, then $P\le <r>$ and we get the same contradiction as before.


*

*Case 2: All elements of $S$ have order $2$.


There are at least two distinct ones otherwise $P$ would be cyclic. They can't both be powers of $r$ since only one power of $r$ has order $2$ and so one isn't a power of $2$.


*

*Conclusion: In both cases, there's at least one element of $S$ which order is $2$ and isn't a power of $r$, so $\{sr^i:i\in\{0,\,\dots,\,n-1\}\}\cap P\neq\emptyset$



So here's all I the informations I could get (that might be useless and the exercise might not need any of them ^^). Could you please give me some hints?
 A: There are $n=2^{a−1}k$ elements of $D_{2n}$ lying outside of the cyclic subgroup $\langle r \rangle$, all of order 2. (These are usually called reflections.) Each such reflection is contained in at least one Sylow $2$-subgroup. A Sylow $2$-subgroup has order $2^a$ and contains $2^{a-1}$ reflections. So there must be at least $k$ Sylow $2$2-subgroups.
But a Sylow $2$-subgroup has index $k$, so there at most $k$ Sylow $2$-subgroups. Hence there are exactly $k$, and each reflection is contained in exactly one Sylow $2$-subgroup.
A: Your step (1) looks good; after that, I would look at a factor group:
Assume that $a>1$ and let $N:=\langle r\rangle$ and $P\in Syl_2(D_{2n})$. Then $N$ is cyclic and therefore abelian, and $Q:=P\cap N$ has index $2$ in $P$, so $Q$ is normalized by $\langle P, N\rangle = D_{2n}$. By looking at the relations in the definition of $D_{2n}$, it's easy to see that $D_{2n}/Q\simeq D_{2k}$, and so the claim follows from (1).
A: Just note that each $P$ can separate into two part $A,B$, which contains elements of the form $r^i$ and $sr^j$ respectively.
Soppose $r^i\in P$ has the least power, then all element of the form $r^j\in P$ must have $i\vert j$. As your step (2), $i=sk$ where $2\nmid s$, which implies that all $P\in Syl_2(D_{2n})$ has the same part $A=<r^k>$. 
So we may find $k$ different $B$ to make $P$ be a subgroup. It's easy to check that $P_i=\{A,B_i\}$ where $B_i=\{sr^{k+i},sr^{2k+i},\cdots,sr^{2^{\alpha-1}k+i}\}$ satisfies this property.
Since $n_2|k$ and we have already find $k$ Sylow $2$-subgroups, we have $n_2=k$.
