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Math Help - Sylow p-group

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    Sylow p-group

    Will anyone give me a hand on this problem? I am trying to show that if H \vartriangleleft G and H contains P, a Sylow p-subgroup in G. Then n_p(H)=n_p(G). I am not sure that this is true, but I see that since H \vartriangleleft G and P is contained in H, then H must contain all the conjugates of P. By Sylow's theorem, we know all Sylow p-subgroups of G are conjugates. So, n_p(H)=n_p(G). Do I need to construct a 1-1 correspondence here? I think that n_p(H) and n_p(G) are just the number of Sylow p-subgroups, how can I really construct a 1-1 correspondence if they are not sets? I don't claim that H and G have the same set of Sylow p-subgroups.
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  2. #2
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    Quote Originally Posted by namelessguy View Post
    Will anyone give me a hand on this problem? I am trying to show that if H \vartriangleleft G and H contains P, a Sylow p-subgroup in G. Then n_p(H)=n_p(G). I am not sure that this is true, but I see that since H \vartriangleleft G and P is contained in H, then H must contain all the conjugates of P. By Sylow's theorem, we know all Sylow p-subgroups of G are conjugates. So, n_p(H)=n_p(G). Do I need to construct a 1-1 correspondence here? I think that n_p(H) and n_p(G) are just the number of Sylow p-subgroups, how can I really construct a 1-1 correspondence if they are not sets? I don't claim that H and G have the same set of Sylow p-subgroups.
    if Q is a Sylow p-subgroup of G, then Q=gPg^{-1}, for some g \in G. thus Q =gPg^{-1} \subseteq gHg^{-1} = H, i.e. Q is a Sylow p-subgroup of H. the converse is trivial because |P| divides |H|.
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