Let \(G\) be a group and \(H\le G\).
- Prove that \(C_G(H)\) and \(N_G(H)\) are subgroups and that \[ C_G(H)\subseteq N_G(H)\le G. \]
- Prove that \(C_G(H)\trianglelefteq N_G(H)\) and state the \(N/C\) theorem.
Solution
The centralizer and normalizer are \[ C_G(H)=\{g\in G:gh=hg\text{ for all }h\in H\}, \] \[ N_G(H)=\{g\in G:gHg^{-1}=H\}. \] Both contain the identity and are closed under products and inverses, hence are subgroups. If \(g\in C_G(H)\), then \(ghg^{-1}=h\) for every \(h\in H\), so \(gHg^{-1}=H\). Thus \[ C_G(H)\le N_G(H)\le G. \]
Conjugation gives a homomorphism \[ \Phi:N_G(H)\longrightarrow\operatorname{Aut}(H), \qquad \Phi(g)(h)=ghg^{-1}. \] Its kernel is exactly \(C_G(H)\). Therefore \[ C_G(H)\trianglelefteq N_G(H), \] and the first isomorphism theorem gives the \(N/C\) theorem: \[ N_G(H)/C_G(H)\cong\operatorname{im}\Phi\le\operatorname{Aut}(H). \]