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Math Help - Normalizer and Centralizer

  1. #1
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    Normalizer and Centralizer

    H\leq G and |H|=2. Show that N_G(H)=C_G(H). Deduce that if N_G(H)=G, then H\leq Z(G)

    If h\in H, then |h|=2 if h\neq e.
    N_G(H)=\{g\in G:gHg^{-1}=H\}
    C_G(H)=\{{g\in G:ghg^{-1}=h, \ \forall h\in H\}.

    How can I get to the point where they are equal with this information?
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  2. #2
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    Re: Normalizer and Centralizer

    well it's obvious that C_G(H) is a subgroup of N_G(H). so what we need to show is that the normalizer is a subgroup of the centralizer if |H| = 2.

    so let g be in N_G(H). clearly geg^{-1} = e, and we know that ghg^{-1} \in H, where h is the non-identity element of H.

    but if ghg^{-1} = e, then gh = g, and thus h = e, contradiction. hence ghg^{-1} = h \implies gh = hg, so g \in C_G(H)
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  3. #3
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    Re: Normalizer and Centralizer

    Quote Originally Posted by Deveno View Post
    well it's obvious that C_G(H) is a subgroup of N_G(H). so what we need to show is that the normalizer is a subgroup of the centralizer if |H| = 2.

    so let g be in N_G(H). clearly geg^{-1} = e, and we know that ghg^{-1} \in H, where h is the non-identity element of H.

    but if ghg^{-1} = e, then gh = g, and thus h = e, contradiction. hence ghg^{-1} = h \implies gh = hg, so g \in C_G(H)
    I am good thanks. I marked the thread solved awhile ago.
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