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Group of order 765 is abelian

  1. Sep 28, 2009 #1
    1. The problem statement, all variables and given/known data
    Show that the group of order 765 is abelian (Hint: let G act by conjugation on a normal Sylow p subgroup)


    2. Relevant equations
    Sylow theorems


    3. The attempt at a solution
    By using Sylow`s third theorem, I have calculated that the number of Sylow-3 subgroups and Sylow 17 subgroups is both 1; so both of them are normal subgroups of G. I just don't really understand how to proceed from here; by G acting on either one of those subgroups; I get a group homomorphism G->S_Q (where Q is either the Sylow 17 subgroup or Sylow 3 subgroup). I believe that eventual goal is to show that G is cyclic.

    Some random thoughts in my head: Aut(Q)=C_16 (Q being the 17-Sylow subgroup). Much thanks and any help is appreciated
     
  2. jcsd
  3. Sep 29, 2009 #2

    MxM

    User Avatar

    First do a counting argument to show that the number of p-Sylow subgroups for all three values of p must be equal to 1. This will show the p-Sylow subgroups exhaust G. Hence their direct product is isomorphic to G. But each p-Sylow subgroup is abelian (because each order is either prime or a square of a prime) and so all of G is abelian. Done diddly done! You're going to UBC aren't you? :)
     
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