Show that \(\Inn(G)\normal\Aut(G)\).
Solution
For \(g\in G\), let \(c_g\in\Inn(G)\) denote conjugation by \(g\): \[ c_g(x)=gxg^{-1}. \] Take \(\phi\in\Aut(G)\). Then for every \(x\in G\), \[ (\phi c_g\phi^{-1})(x) =\phi\bigl(g\phi^{-1}(x)g^{-1}\bigr) =\phi(g)x\phi(g)^{-1} =c_{\phi(g)}(x). \] Thus \[ \phi c_g\phi^{-1}=c_{\phi(g)}\in\Inn(G). \] Hence \(\phi\Inn(G)\phi^{-1}\subseteq\Inn(G)\) for every \(\phi\in\Aut(G)\). Applying the same inclusion to \(\phi^{-1}\) gives equality, so \(\Inn(G)\normal\Aut(G)\).