1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Sylow theorems, simple groups

  1. May 8, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove that no group of order 96 is simple.

    2. Relevant equations
    The sylow theorems

    3. The attempt at a solution
    96 = 2^5*3. Using the third Sylow theorem, I know that n_2 = 1 or 3 and n_3 = 1 or 16. I need to show that either n_2 = 1 or n_3 = 1, but I am unsure how to do this.
     
  2. jcsd
  3. May 8, 2012 #2

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Let H be a 2-Sylow subgroup. Then the index of [itex]N_G(H)[/itex] is 3 by the Sylow theorems.

    Can you prove that there exists a homomorphism [itex]G\rightarrow S_3[/itex]??

    Hint: define an action by left multiplication on [itex]N_G(H)[/itex].

    What can you infer from the kernel being normal?
     
  4. May 8, 2012 #3
    I'm not too good at group actions. What would the action be? And where would the homomorphism come from?

    If the kernel is normal, then it must be trivial, otherwise there would be a non-trivial normal subgroup of G.
     
  5. May 8, 2012 #4

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    I gave you the action: left multiplication on [itex]N_G(H)[/itex].
     
  6. May 8, 2012 #5
    So does that action look like (g, n) |-> gn where n is in N_G(H)?

    Ok, so how about the homomorphism? Do you map an element of g to the action of left multiplication by g on N_G(H)?
     
  7. May 8, 2012 #6

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    The action is

    [tex](g,aN_G(H))\rightarrow gaN_G(H)[/tex]
     
  8. May 8, 2012 #7
    So we define the map phi : G -> S_3 by phi(g) = (left multiplication by g).

    I can see why this is a homomorphism. (Since left multiplication by g_1g_2 is the same as left multiplication by g_2, then left multiplication by g_1).

    Is this right?
     
  9. May 9, 2012 #8

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Yes, that is right.

    The kernel of this map is a normal subgroup, so what do we get from G being simple?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Sylow theorems, simple groups
  1. Sylow Theorem (Replies: 0)

Loading...