Question Show that a Sylow $p$-subgroup of dihedral group $D_{2n}$ is cyclic and normal when $p$ is an odd prime.
$$
D_{2n}:=\left\{ \langle x,y\rangle|x^{n}=1,y^2=1,xy=yx^{-1}=yx^{n-1} \right\} .
$$
My thought
Assume that $x^i y\in S$ where $S$ is a Sylow $p$-subgroup,
$$
\left( x^iy \right) ^2=x^iy(x^iy)=x^iy(yx^{n-i})=x^i(yy)x^{n-i}=x^ix^{n-i}=1,
$$
implying $\text{order}(x^i y)=2$($i=0,1,\cdots,n-1$), so it can't be a member of $S$ as $|S|=p^r$ is odd, contradiction! Then we know that $S$ is a subgroup of $\langle x\rangle$, cyclic and normal indeed.
However, this deduction has nothing to do with the prime property of $p$, only using the fact that $p$ is odd. Therefore, I wonder if there is anything wrong with the idea or it is actually the case. Thank you in advance!
Best Answer
Your argument is correct! Indeed, any odd order subgroup of a dihedral group is both cyclic and normal, and this follows from the argument you suggest.
(If you think about the dihedral group geometrically, then half of the group consists of rotations generating a cyclic group, and the other half consists of reflections which all have order two --- this is just another way of saying what you computed geometrically)