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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.

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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 $n$.

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  • $\begingroup$ Nice! Hardly a "gentle hint" in my opinion, but very informative nonetheless. $\endgroup$ – walkar Mar 12 '15 at 0:59
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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 $D_{2n}$.

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