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Math Help - Sylow group

  1. #1
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    Sylow group

    I was asked to prove that given a finite group G, a normal group N in G, and a p-sylow supgroup P in G, then
     (N_{G}(P)N)/{N}\cong N_{G/N}(PN/N)<br />
where N_{G} (P) is the normalizer of P in G (same with N_{G/N}(PN/N) )
    I tried to solve it using the homomorphism
    \varphi:N_{G}(P)N\rightarrow G/N defined by \varphi(gn)=gnN
    I can show that Im(\varphi)\subseteq N_{G/N}(PN/N) and also that ker(\varphi)=N .
    all I need now is to show that N_{G/N}(PN/N)\subseteq Im(\varphi) in order to use the first isomorphism theorem, but for some reason I can manage it.

    any help would be appreciated
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  2. #2
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    Quote Originally Posted by Prometheus View Post
    I was asked to prove that given a finite group G, a normal group N in G, and a p-sylow supgroup P in G, then
     (N_{G}(P)N)/{N}\cong N_{G/N}(PN/N)<br />
where N_{G} (P) is the normalizer of P in G (same with N_{G/N}(PN/N) )
    I tried to solve it using the homomorphism
    \varphi:N_{G}(P)N\rightarrow G/N defined by \varphi(gn)=gnN

    note that \color{red}gnN=gN because \color{red}n \in N. so we define \color{red}\varphi(gn)=gN. you also need to show that \color{red}\varphi is well-defined and a homomorphism.

    I can show that Im(\varphi)\subseteq N_{G/N}(PN/N) and also that ker(\varphi)=N .
    all I need now is to show that N_{G/N}(PN/N)\subseteq Im(\varphi) in order to use the first isomorphism theorem, but for some reason I can manage it.
    you need a little trick to show that N_{G/N}(PN/N)\subseteq Im(\varphi). so let gN \in N_{G/N}(PN/N). then we'll get g^{-1}Pg \subseteq PN. thus g^{-1}Pg is a Sylow p-subgroup of PN. but P is a Sylow p-subgroup of

    PN too. thus P and g^{-1}Pg are conjugates in PN, i.e there exists n \in N: \ g^{-1}Pg=n^{-1}Pn. therefore g_0=gn^{-1} \in N_G(P). hence: \varphi(g_0)=g_0N=gN. \ \Box
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  3. #3
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    we can actually prove a better result, which is: (N_G(P)N)/N=N_{G/N}(PN/N). then you even don't need to define an isomorphism. here's how to prove the equality:

    first if gN \in (N_G(P)N)/N, with g \in N_G(P), then g^{-1}Pg=P, which will quickly give us: gN \in N_{G/N}(PN/N).

    conversely, suppose gN \in N_{G/N}(PN/N). then as i showed in my previous post gn^{-1} \in N_G(P), for some n \in N. thus: g \in N_G(P)N and hence: gN \in (N_G(P)N)/N.
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