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    $C_G(H)\trianglelefteq N_G(H)$

    E-AMD-F5ZHMJNK

    Area
    Algebra
    Topics
    Centralizers and Normalizers, Normal Subgroups
    Status
    draft

    Show that \(C_G(H) \trianglelefteq N_G(H)\) is a normal subgroup.

    Solution

    Conjugation gives a homomorphism \[ \Phi:N_G(H)\longrightarrow\operatorname{Aut}(H), \qquad \Phi(g)(h)=ghg^{-1}. \] It is well defined because every \(g\in N_G(H)\) satisfies \(gHg^{-1}=H\).

    Its kernel is \[ \ker\Phi =\{g\in N_G(H):ghg^{-1}=h\text{ for all }h\in H\} =C_G(H). \] Therefore \(C_G(H)\) is the kernel of a group homomorphism, hence \[ \boxed{C_G(H)\trianglelefteq N_G(H).} \]

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