Results 1 to 3 of 3

Math Help - Non- Abelian Group

  1. #1
    Junior Member
    Joined
    Mar 2008
    Posts
    33

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    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 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).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by ThePerfectHacker View Post

    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Abelian Group
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: November 5th 2011, 01:37 PM
  2. Abelian group
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: August 8th 2011, 03:27 PM
  3. Group, abelian
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: October 14th 2010, 05:05 AM
  4. Abelian group
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 5th 2010, 02:46 PM
  5. Is the subgroup of an abelian group always abelian?
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: December 6th 2009, 11:38 PM

Search Tags


/mathhelpforum @mathhelpforum