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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
    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
    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
    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
    Yes, that is right.

    The kernel of this map is a normal subgroup, so what do we get from G being simple?
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