Sylow p-subgroups

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November 10th 2008, 08:12 AM
grad444

Sylow p-subgroups

Show that a Sylow p-subgroup of the dihedral group is cyclic and normal for every odd prime.
November 10th 2008, 08:56 AM
clic-clac

Let $p$ be an odd prime, $D_{p}$ the dihedral group with $2p$ elements and $e$ its identity element.

A $p$ -Sylow in $D_{p}$ has an order of $p$ , so it is cyclic: indeed, the order of an element of such a $p$ -Sylow subgroups is either $1$ or $p$ (because it divides $p$ ) so if the element is different than $e$ , then it's a generator, and since $p$ is different from $1$ , there are generators.
Furthermore, $D_{p}$ has $2p$ elements. The number $N_{p}$ of $p$ -Sylow and $1$ are congruent modulo $p$ , and $N_{p}$ divides $\frac{|Dp|}{p}$ , so divides $2$ . So $N_{p}=1$ . But all $p$ -Sylow subgroups are conjugate, so when there is only one $p$ -Sylow $H$ , it is normal.
Indeed, $\forall x \in D_{p},\ xHx^{-1}$ has the same order than $H$ . Thus it's a $p$ -Sylow, and $xHx^{-1}=H$ .
November 10th 2008, 09:55 AM
NonCommAlg

clic-clac, the dihedral group is not necessarily in the form $D_p.$ it could be $D_n$ with $p \mid n.$ i'll prove that in fact every sugroup of odd order in $D_n$ is normal and cyclic. to see this, let

$D_n=<a,b: \ a^2=b^n=1, \ aba=b^{-1}>$ and suppose $H \subset D_n$ is a subgroup with odd order. then $H \subseteq <b>,$ because $ab^j, \ 0 \leq j < n,$ 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.