If $p>2$ is a prime number that divides $n \in \mathbb N$, then $D_n = \langle r,s \rangle$ (with $r^n=s^2=id, sr=r^{-1}s$) has a unique $p$-Sylow subgroup $H \leq D_n$.
I'm having trouble with this question, specifically with showing rigorously a subgroup of $D_n$ is normal. This is my attempt:
Since $p>2$ and $|D_n|=2n$, $p$ divides $n$ but does not divide $2$. So there exists $k \in \mathbb N$ and $m \in \mathbb N$ such that $2n = 2mp^k$ $\implies$ $n=mp^k$ with $p$ not dividing $m$. We know $\langle r \rangle$ is a cyclic subgroup of $D_n$ with order $n$. $\langle r^m \rangle$ is normal in $\langle r \rangle$ (since it is cyclic hence abelian) and $|\langle r^m \rangle|=p^k$ (because $r^{mp^k}=r^n$) so it is a normal $p$-Sylow subgroup of $\langle r \rangle$, hence unique in $\langle r \rangle$. But now I'm stuck - I'm not sure if this implies it is unique in $D_n$. Does it have something to do with the structure of $D_n$?
From the fact that $D_{n}$ has a normal abelian subgroup of order $n$, we call it $N$. We get that first of all, Syl$_{p}(G)$$=$Syl$_{p}(N)$. Now since $N$ is abelian, it has a normal(characteristic) Sylow $p$-subgroup. Hence, $G$ has a normal Sylow $p$-subgroup. That's it, we know that a normal Sylow $p$-subgroup is unique.