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Thread: Non- Abelian Group

  1. #1
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    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
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  2. #2
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    Quote Originally Posted by bex23 View Post
    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 $\displaystyle 13|689$ there exists Sylow subgroups. Let $\displaystyle S$ be this set. Define the function $\displaystyle *:G\times S \to S$ by $\displaystyle g*P = gPg^{-1}$. It is easy to check that this is a group action on $\displaystyle S$. A stabilizer of $\displaystyle P\in S$ is the set $\displaystyle \{ g\in G : g*P = P\} = \{p\in G: gPp^{-1} = P\}$. But this is precisely what the normalizer is. Thus, you need to show $\displaystyle P = N(P)$ i.e. each Sylow 13-subgroup is equal to its normalizer.

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

    We will use a well-known result: If $\displaystyle G$ is a finite group and $\displaystyle H$ is a subgroup then the number of subgroups conjugate to $\displaystyle H$ is $\displaystyle [G:N(H)]$ (the index of the normalizer in the group). However, by Sylow's 2nd theorem the number of conjugates to $\displaystyle P$ (Sylow $\displaystyle 13$-subgroup) are in fact all the Sylow $\displaystyle 13$-subgroups. Thus, $\displaystyle |S| = [G:N(P)]$. We know that $\displaystyle |S|=53$ by above. Thus, $\displaystyle |S| = |G|/|N(P)| \implies |N(P)| = |G|/|S| = 689/53=13$. But $\displaystyle P\subseteq N(P)$ with $\displaystyle |P| = |N(P)| = 13$ thus $\displaystyle P=N(P)$.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post

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