1. ## Non- Abelian Group

Let G be a non-Abelian group of order 689

Show that, in the action of G by conjugation on the set of its Sylow 13-supbgroups, each such subgroup is it's own stabilizer.

I don't understand any part of this quesiton, the wording alone gives me a headache, lol. Can anyone out there help me on this?

Bex

2. Originally Posted by bex23
Let G be a non-Abelian group of order 689

Show that, in the action of G by conjugation on the set of its Sylow 13-supbgroups, each such subgroup is it's own stabilizer.

I don't understand any part of this quesiton, the wording alone gives me a headache, lol. Can anyone out there help me on this?
Since $13|689$ there exists Sylow subgroups. Let $S$ be this set. Define the function $*:G\times S \to S$ by $g*P = gPg^{-1}$. It is easy to check that this is a group action on $S$. A stabilizer of $P\in S$ is the set $\{ g\in G : g*P = P\} = \{p\in G: gPp^{-1} = P\}$. But this is precisely what the normalizer is. Thus, you need to show $P = N(P)$ i.e. each Sylow 13-subgroup is equal to its normalizer.

To show this we use Sylow's third theorem. The number of Sylow $13$-subgroups is congruent to $1$ mod $13$ and divides $53$ since $|G| = 13\cdot 53$. Note that $53$ is prime. This means the number of Sylow 13-subgroups is $1$ or $53$. If the number of Sylow $13$-subgroups was $1$ then $G$ be cyclic (I assume you know this result) and therefore abelian. Thus, the number of Sylow $13$-subgroups is $53$.

We will use a well-known result: If $G$ is a finite group and $H$ is a subgroup then the number of subgroups conjugate to $H$ is $[G:N(H)]$ (the index of the normalizer in the group). However, by Sylow's 2nd theorem the number of conjugates to $P$ (Sylow $13$-subgroup) are in fact all the Sylow $13$-subgroups. Thus, $|S| = [G:N(P)]$. We know that $|S|=53$ by above. Thus, $|S| = |G|/|N(P)| \implies |N(P)| = |G|/|S| = 689/53=13$. But $P\subseteq N(P)$ with $|P| = |N(P)| = 13$ thus $P=N(P)$.

3. Originally Posted by ThePerfectHacker

We will use a well-known result: If $G$ is a finite group and $H$ is a subgroup then the number of subgroups conjugate to $H$ is $[G:N(H)]$ (the index of the normalizer in the group). However, by Sylow's 2nd theorem the number of conjugates to $P$ (Sylow $13$-subgroup) are in fact all the Sylow $13$-subgroups. Thus, $|S| = [G:N(P)]$. We know that $|S|=53$ by above. Thus, $|S| = |G|/|N(P)| \implies |N(P)| = |G|/|S| = 689/53=13$. But $P\subseteq N(P)$ with $|P| = |N(P)| = 13$ thus $P=N(P)$.
this is easier: since $P \leq N(P) < G,$ by Lagrange's theorem we must have $|N(P)|=13,$ i.e. $N(P)=P. \ \ \ \square$