Show that \[ \operatorname{Inn}(G)\cong G/Z(G). \]
Solution
Define \[ \Phi:G\longrightarrow\operatorname{Aut}(G), \qquad \Phi(g)(x)=gxg^{-1}. \] Then \(\Phi\) is a homomorphism and \[ \operatorname{im}\Phi=\operatorname{Inn}(G). \] 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 \[ G/Z(G)\cong\operatorname{Inn}(G). \]