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Math Help - Conjugate subgroups.

  1. #1
    Junior Member platinumpimp68plus1's Avatar
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    Conjugate subgroups.

    Let H be a subgroup of G, with index p (prime). What are the possible numbers of conjugate subgroups of H?

    I'm pretty lost. I think it has something to do with Sylow's Theorems, and the fact that if p=|G|/|H|, so then p divides|G| and there is a p-Sylow subgroup, and Sylow-p's are conjugate to each other? But then I get really confused and forget what I'm trying to do, and the question didn't say G is finite so I don't even know if that method applies. Any help?
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    Quote Originally Posted by platinumpimp68plus1 View Post
    Let H be a subgroup of G, with index p (prime). What are the possible numbers of conjugate subgroups of H?

    I'm pretty lost. I think it has something to do with Sylow's Theorems, and the fact that if p=|G|/|H|, so then p divides|G| and there is a p-Sylow subgroup, and Sylow-p's are conjugate to each other? But then I get really confused and forget what I'm trying to do, and the question didn't say G is finite so I don't even know if that method applies. Any help?

    I don't think there's one unique answer to this question, which is most probably directed to finite groups (read at the beginning of the chapter/section whether all the groups in that chapter/section are assumed to be finite).
    Now, if p was the MINIMAL prime that divides the order of p then we have a very simple and nice answer...

    Tonio
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    Junior Member platinumpimp68plus1's Avatar
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    Quote Originally Posted by tonio View Post
    I don't think there's one unique answer to this question, which is most probably directed to finite groups (read at the beginning of the chapter/section whether all the groups in that chapter/section are assumed to be finite).
    Now, if p was the MINIMAL prime that divides the order of p then we have a very simple and nice answer...

    Tonio
    Hmmm, okay. It didn't say p was the smallest prime but I'm think it is supposed to be assumed, just based on other problems in the chapter. I don't know how to use that though...
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    Quote Originally Posted by platinumpimp68plus1 View Post
    Hmmm, okay. It didn't say p was the smallest prime but I'm think it is supposed to be assumed, just based on other problems in the chapter. I don't know how to use that though...

    Then H is normal in G... . The prove passes through the action of G on the set of left cosets of G in H, and then using the associated homomorphism's kernel and etc.

    Tonio
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    Junior Member platinumpimp68plus1's Avatar
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    Quote Originally Posted by tonio View Post
    Then H is normal in G... . The prove passes through the action of G on the set of left cosets of G in H, and then using the associated homomorphism's kernel and etc.

    Tonio
    Thanks... I'm still really confused though. I'm pretty lost in general when it comes to G-sets and conjugate groups. I dunno how to start/how this tells me how many conjugate subgroups of H I have. Could you explain in some more detail?
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    Quote Originally Posted by platinumpimp68plus1 View Post
    Thanks... I'm still really confused though. I'm pretty lost in general when it comes to G-sets and conjugate groups. I dunno how to start/how this tells me how many conjugate subgroups of H I have. Could you explain in some more detail?

    If p is the minimal prime that divides the order of G, or if simple it is given H is a normal sbgp. of G, then all the conjugates of H are...H itself, of course!

    Tonio
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    Quote Originally Posted by platinumpimp68plus1 View Post
    Let H be a subgroup of G, with index p (prime). What are the possible numbers of conjugate subgroups of H?

    I'm pretty lost. I think it has something to do with Sylow's Theorems, and the fact that if p=|G|/|H|, so then p divides|G| and there is a p-Sylow subgroup, and Sylow-p's are conjugate to each other? But then I get really confused and forget what I'm trying to do, and the question didn't say G is finite so I don't even know if that method applies. Any help?
    If H \leq G, then the number of conjugates of H in G is equal to the index of its normalizer, i.e., [G:N_G(H)].

    Since H is a maximal subgroup of G, N_G(H) is either H or G.
    If N_G(H)=H, then the number of conjugates of H in G is p. If N_G(H)=G, then the number of conjugates of H in G is 1.

    If you need to prove that the number of conjugates of H in G is equal to the index of its normalizer, you need to define a function f:[H] \rightarrow G/N given by f(aHa^{-1})=aN, where [H] denotes the family of conjugates of H and N denotes N_G(H). I'll leave it to you to verify that f is a bijection.
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