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Math Help - Group Theory - Sylow and Conjugation

  1. #1
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    Group Theory - Sylow and Conjugation

    A couple of questions I had hard time solving (and not succeeded).

    1. We showed that for every p-sylow group P in G, the following is true:
     N_G(N_G(P)) = N_G(P) . Show that equation isn't true for some subgroup H (not sylow) of  S_4 (symmetric group), meaning  N_G(N_G(H)) \neq N_G(H) .
    I don't fully understand how to do this without doing it brute-force, by applying all elements of G on the chosen H.

    2. H is a subgroup of  S_4 , created by the cycle (1234).
    (a) Show that  C_G(H)=H for  C_G(H)= \{g \in G \| gh=hg \forall h \in H \}
    (b) Show that  N_G(H) is a 2-sylow subgroup of G.

    I have some difficulty by showing that too.

    I will very appreciate any help.
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  2. #2
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    Quote Originally Posted by pavelr View Post
    1. We showed that for every p-sylow group P in G, the following is true:
     N_G(N_G(P)) = N_G(P) . Show that equation isn't true for some subgroup H (not sylow) of  S_4 (symmetric group), meaning  N_G(N_G(H)) \neq N_G(H) .
    I don't fully understand how to do this without doing it brute-force, by applying all elements of G on the chosen H.
    I did not actually work out the details for this problem, but I had an idea that you might want to try. Let H=S_4 = \{ \sigma\in S_4 | \sigma (4) = 4\} i.e. all permutations that fix 4. Then it is seem that |H| = 6 (in fact H is naturally isomorphic to S_3). Now since H is not normal it means N(H) \not = S_4. If N(H) \not = H i.e. if there is \sigma \in S_4 - H so that \sigma H \sigma^{-1} = H then it means H is propely contained in N(H). By Lagrange's theorem it would mean N(H) = A_4 since |N(H) | = 12. But N(A_4) = S_4. Therefore, N(N(H)) \not = N(H). I cannot gaurentte you this would work because I was kinda lazy to work out the details but this would be the first thing I would try to do.
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