Let \(G\) be a group and \(H \le G\) a subgroup. Show that the centralizer \(C_G(H)\) is a normal subgroup of the normalizer \(N_G(H)\) (i.e. \(C_G(H) \trianglelefteq N_G(H)\)), and that \(N_G(H)/C_G(H)\) is isomorphic to a subgroup of \(\operatorname{Aut}(H)\) (the \(N/C\) Theorem).
Solution
For \(n\in N_G(H)\), conjugation restricts to an automorphism of \(H\). Hence \[ \Phi:N_G(H)\longrightarrow \operatorname{Aut}(H),\qquad \Phi(n)(h)=nhn^{-1} \] is a homomorphism. Its kernel is precisely \[ \ker\Phi=\{n\in N_G(H):nhn^{-1}=h\ \forall h\in H\}=C_G(H). \] Therefore \(C_G(H)\trianglelefteq N_G(H)\), and the first isomorphism theorem gives \[ N_G(H)/C_G(H)\cong \operatorname{im}\Phi\le \operatorname{Aut}(H). \] This is the \(N/C\) theorem.