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Sylow Subgroups of Symmetric Groups

  1. May 13, 2009 #1
    1. The problem statement, all variables and given/known data
    Find a set of generators for a p-Sylow subgroup K of Sp2
    .
    Find the order of K and determine whether it is normal in Sp2 and if it is abelian.

    2. Relevant equations



    3. The attempt at a solution
    So far I have that the order of Sp2 is p2!. So p2 is the highest power of p that divides the order of the group. Thus the Sylow p-subgroup has order p2 and because of that, has to be abelian. The other parts I'm not so sure on.
     
  2. jcsd
  3. May 13, 2009 #2

    matt grime

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    Are you sure you're right so far? Take p=2, S_4 has order 24, so its Sylow 2 subgroup has order 8. You seem to have assumed that p^2!=p!^2.
     
  4. May 13, 2009 #3
    Sorry, it's supposed to be restricted to just odd primes, in which case I think that still holds, correct?
     
  5. May 13, 2009 #4

    matt grime

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    If you believe your conjecture, then it is the matter of no time at all to check if it's true for p=3 (and it obviously isn't). If you can't prove something in general try an example or 2.
     
  6. May 13, 2009 #5
    Okay, that's true. Sorry.
    I guess I really just have no idea how to find generators for this.
    I was trying with p=3 for my example and found that 3^4 is the highest power of 3 that divides 9! since there are 4 factors of 3 in 9*8*7*6*5*4*3*2*1. So the Sylow p-subgroup would have 81 elements. So in general for odd prime p the order of the Sylow subgroup would be p^(p+1)?

    I'm working on generators right now...
     
    Last edited: May 13, 2009
  7. May 13, 2009 #6
    Here is my idea although it may be way off the mark.
    Still looking at p=3, the subgroup could contain powers of 10 different 9-cycles that way you'd have 80 different elements plus the identity. I think that works.

    Although I guess we can't be sure that it's closed.

    Well, I'm basically out of ideas now...
     
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