Show that a Sylow p-subgroup of the dihedral group is cyclic and normal for every odd prime.
Let $\displaystyle p$ be an odd prime, $\displaystyle D_{p}$ the dihedral group with $\displaystyle 2p$ elements and $\displaystyle e$ its identity element.
A $\displaystyle p$-Sylow in $\displaystyle D_{p}$ has an order of $\displaystyle p$, so it is cyclic: indeed, the order of an element of such a $\displaystyle p$-Sylow subgroups is either $\displaystyle 1$ or $\displaystyle p$ (because it divides $\displaystyle p$) so if the element is different than $\displaystyle e$, then it's a generator, and since $\displaystyle p$ is different from $\displaystyle 1$, there are generators.
Furthermore, $\displaystyle D_{p}$ has $\displaystyle 2p$ elements. The number $\displaystyle N_{p}$ of $\displaystyle p$-Sylow and $\displaystyle 1$ are congruent modulo $\displaystyle p$, and $\displaystyle N_{p}$ divides $\displaystyle \frac{|Dp|}{p}$, so divides $\displaystyle 2$. So $\displaystyle N_{p}=1$. But all $\displaystyle p$-Sylow subgroups are conjugate, so when there is only one $\displaystyle p$-Sylow $\displaystyle H$, it is normal.
Indeed, $\displaystyle \forall x \in D_{p},\ xHx^{-1}$ has the same order than $\displaystyle H$. Thus it's a $\displaystyle p$-Sylow, and $\displaystyle xHx^{-1}=H$.
clic-clac, the dihedral group is not necessarily in the form $\displaystyle D_p.$ it could be $\displaystyle D_n$ with $\displaystyle p \mid n.$ i'll prove that in fact every sugroup of odd order in $\displaystyle D_n$ is normal and cyclic. to see this, let
$\displaystyle D_n=<a,b: \ a^2=b^n=1, \ aba=b^{-1}>$ and suppose $\displaystyle H \subset D_n$ is a subgroup with odd order. then $\displaystyle H \subseteq <b>,$ because $\displaystyle ab^j, \ 0 \leq j < n,$ have order 2 and hence they cannot be in
H. so H is cyclic. besides a cyclic group has only one subgroup of any order, i.e. H is normal. Q.E.D.