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Homework Help: Abstract-Sylow Theorems

  1. Dec 10, 2009 #1
    1. The problem statement, all variables and given/known data
    In the permutations group S4, let H be the cyclic group the is generated by the cycle
    (1 2 3 4).
    A. Prove that the centralizer C(H) of H is excatly H ( C(H)={g in G|gh=hg for every h in H} )
    B. Prove that the normalizer of H is a 2-sylow group of S4.

    2. Relevant equations

    3. The attempt at a solution

    Well, there's a quick calculation of H:
    (13)(24)(1234)=(1 4 3 2)
    Hence: H={(1234),(13)(24),(1432),1}

    The order of the group is 4. Hence [S4:H]=6...
    Since it's a cyclic group, it's abelian. Which means H is a sub group of C(H)...
    How can I prove that C(H) is in H? Or how excatly can I prove the 1st statement?

    About 2, I actually have no clue.... My abilities in sylow&homomorphzm are realy lame and I need to send a lot of excercices 'till next week so I realy need your help :(

    TNX to all the helpers!
  2. jcsd
  3. Dec 10, 2009 #2
    I haven't really thought a lot about it, but a fairly straightforward way I see to solve it (without the Sylow theorems; don't know if that's a problem) would be to use the theorem that for any subset H of a group G, the number of conjugates of H equals the index of the normalizer of H in G. So count the number of conjugates of H (there should be 3) and you will get the order of the normalizer of H in G, which will prove (B). As for (A) use the special case that says that for some element x of a group G, the number of conjugates of x equals the index of the centralizer of x in G. So count the number of conjugates of (1 2 3 4) (there should be 6 in total; all 4-cycles) and then you get the order of the centralizer of (1 2 3 4) which will convince you that you have found all elements of the centralizer of H.

    I assume you algebra book has these theorems (in Dummit-Foote they are proposition 6 in section 4.3).
  4. Dec 11, 2009 #3
    Hey there, I know these theorems of course, but the way you've suggested is very "not elegant" ...There might be an easier way?

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