Let \(G\) be a group and let \(\gamma: G \to \operatorname{Aut}(G)\) be the conjugation homomorphism defined by \(\gamma(g) = c_g\), where \(c_g(h) = g h g^{-1}\) for all \(h \in G\). Show that the kernel of \(\gamma\) is the center of \(G\): \[\ker(\gamma) = Z(G).\]
Solution
For \(g\in G\), \[ g\in\ker\gamma \iff c_g=\operatorname{id}_G \iff ghg^{-1}=h\quad\forall h\in G. \] Multiplying on the right by \(g\) gives \[ gh=hg\quad\forall h\in G, \] which is exactly the condition \(g\in Z(G)\). Hence \[ \boxed{\ker\gamma=Z(G)}. \] By the first isomorphism theorem this also yields \[ \operatorname{Inn}(G)\cong G/Z(G). \]