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    $\Inn(G)\cong G/Z(G)$

    E-AMD-FPHQEZZL

    Area
    Algebra
    Topics
    Automorphisms, Centralizers and Normalizers, Isomorphism Theorems
    Status
    draft

    Show that \(\operatorname{Inn}(G) \cong G / Z(G)\).

    Solution

    Define \[ \Phi:G\to\Aut(G),\qquad \Phi(g)(x)=gxg^{-1}. \] Then \[ \Phi(gh)=\Phi(g)\Phi(h), \] so \(\Phi\) is a homomorphism. Its image is, by definition, \(\Inn(G)\).

    Moreover, \[ g\in\ker\Phi \iff gxg^{-1}=x\ \text{for all }x\in G \iff g\in Z(G). \] Thus \(\ker\Phi=Z(G)\). The first isomorphism theorem gives \[ \boxed{G/Z(G)\cong\Inn(G)}. \]

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