Exercise 4.5.5 (If $p$ is an odd prime, a Sylow $p$-subgroup of $D_{2n}$ is cyclic and normal)

Proof. Let $p$ be an odd prime, and let $P$ be a $p$-subgroup of $D_{2n}$, where $2n=p^{\alpha }m,p\nmid m$, so that $|P|=p^{\alpha }$.

Then $|P|$ is odd, and by Lagrange’s Theorem, $P$ does not contain any element of order $2$. This shows that

Since the subgroup $⟨r⟩$ is cyclic, by Theorem 7 p. 58, $⟨r⟩$ has only one subgroup of order $p^{\alpha }$, which is cyclic, i.e.,

(This is true even if $p\nmid n$, in which case $\alpha =0$, and $P=\{1\}=⟨r^n⟩$.)

So there is only one subgroup $P$ of $D_{2n}$ of order $p^{\alpha }$. By Corollary 20 p. 142, $P$ is normal in $D_{2n}$.

Every Sylow $p$-subgroup of $D_{2n}$ is cyclic and normal for every odd prime $p$. □