Normalizer modulo centralizer acts faithfully by conjugation on the subgroup.Suggest a slogan
Let \(G\) be a group and \(H\leq G\) a subgroup, with normalizer \(N_G(H)\) and centralizer \(C_G(H)\). The map \(c\colon N_G(H)\to\Aut(H)\), \(c(g)(h)=ghg^{-1}\), is a group homomorphism with kernel \(C_G(H)\). Hence \(C_G(H)\normal N_G(H)\) and \(c\) induces an injective homomorphism \[ N_G(H)/C_G(H)\hookrightarrow\Aut(H). \]
For \(g\in N_G(H)\) we have \(gHg^{-1}=H\), so \(c(g)\) is a bijection \(H\to H\), and \(c(g)(hh')=ghg^{-1}gh'g^{-1}=c(g)(h)\,c(g)(h')\); thus \(c(g)\in\Aut(H)\). Since \(c(gg')(h)=gg'h(gg')^{-1}=c(g)(c(g')(h))\), the map \(c\) is a homomorphism. Its kernel is \(\theset{g\in N_G(H)\suchthat ghg^{-1}=h\ \forall h\in H}=C_G(H)\). The first isomorphism theorem gives \(N_G(H)/C_G(H)\cong c(N_G(H))\leq\Aut(H)\).