Let $G = D_{2n}$ and $p$ an odd prime dividing $2n$, i.e. $2n = 2m\cdot p^{\alpha}$ with $p \nmid 2m$. I need to show that $P \in \operatorname{Syl}_p(G)$ is normal in $G$ and cyclic.

As far as normal goes, I know that to show $P$ is normal, I should show that $n_p = 1$. So as such, I know $n_p \equiv_p 1$ and thus $n_p = 1+kp$ for some $k \in \mathbb{Z}^{\ge 0}$. Further, I know $n_p \mid 2m$, and so $a\cdot n_p = 2m$, and substitution gives $a\cdot (1+kp) = 2m \Rightarrow a + akp = 2m$. I'm not sure how to continue with this identity however.

I'm looking for a gentle hint in the right direction. Should I try more divisibility tricks with $n_p$ or is there a nicer way to go about the whole deal?

For the record, this problem is Dummit and Foote, 6.5.5.


Hint: Note that $D_{2n}$ has, as a subgroup, a cyclic group of order $n$. As $p$ is odd we get that $p \ | \ 2n$ implies $p \ | \ n$. So any Sylow $p$-subgroup of $D_{2n}$ is also a Sylow $p$-subgroup of it's cyclic subgroup of order $n$.

  • $\begingroup$ Nice! Hardly a "gentle hint" in my opinion, but very informative nonetheless. $\endgroup$ – walkar Mar 12 '15 at 0:59

As per @Jim's hint:

Let $D_{2n} = \left\langle r,s \mid r^n = s^2 = e, rs = sr^{-1}\right\rangle$ be the dihedral group of order $2n$. If $p$ is an odd prime dividing $n$, i.e. $n = m\cdot p^{\alpha}$ for $p \nmid m$, then $\left\langle r^m \right\rangle = P$ is the required subgroup. It has order $p^\alpha$ (as $|r^m| = p^\alpha$), it is cyclic (as it has a single generator), and normal as $s r^{km} s = r^{-km} \in P$ and $r^j r^{km} r^{-j} = r^{j + km - j} = r^{km} \in P$, which suffices to show normality due to the structure of $D_{2n}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.