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Math Help - Normalizers

  1. #1
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    Normalizers

    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} )-I've managed to prove this one...
    Can't understand how to prove this one:
    B. Prove that the normalizer of H is a 2-sylow group of S4.

    TNX to all the helpers!
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  2. #2
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    Quote Originally Posted by WannaBe View Post
    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} )-I've managed to prove this one...
    Can't understand how to prove this one:
    B. Prove that the normalizer of H is a 2-sylow group of S4.

    TNX to all the helpers!

    Check that (12)(34)\cdot(1234)\cdot(12)(34)=(1432)=(1234)^{-1}\Longrightarrow  (12)(34)\in N_{S_n}(H)\Longrightarrow 4\mid ord(C(H))\mid 24 so...

    Tonio
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  3. #3
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    I didn't completely understand...
    The equality you've wrote is obvious... By the definition, it means that (12)(34) is in N[Sn](H) ofcourse...
    but why it means that 4|o(C(H))? And why from here we get that it's a sylow-p group? It can has an order 12 also...

    TNX
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  4. #4
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    Quote Originally Posted by WannaBe View Post
    I didn't completely understand...
    The equality you've wrote is obvious... By the definition, it means that (12)(34) is in N[Sn](H) ofcourse...
    but why it means that 4|o(C(H))? And why from here we get that it's a sylow-p group? It can has an order 12 also...

    TNX

    That was a typo: it should have been 4\mid o(N_{S_4}) , and it follows at once from Lagrange's theorem and the fact that for any subgroup H of any group G we always have H\le N_G(H).
    Finally, no: N_{S_4} can't be all A_4 (which is the only subgroup of order 12 in S_4)

    Tonio
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  5. #5
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    There is an easy proof for the fact A4 is the only subgroup of order 12 in S4?


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  6. #6
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    Quote Originally Posted by WannaBe View Post
    There is an easy proof for the fact A4 is the only subgroup of order 12 in S4?


    TNX a lot!

    Well, this depends on what you consider "easy", but the following isn't, imo, hard:

    Lemma: If H\le S_n , then either H\le A_n or else exactly half the elements of H are even permutations (and thus, of course, [H:H\capA_n]=2)

    Proof: (Hint) If H\nleq A_n , define a map from the set of odd permutations of H, H_o to H_e=the set of even permutations of H, by h'\mapsto h'x\,,\,\,\forall\,x\in H_o , and check it is 1-1 ...

    From this lemma it follows at once what we want.

    Tonio
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  7. #7
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    Tnx a lot... If I'll have any questions I'll ask in this post...
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