- Show that \(C_G(H) \subseteq N_G(H) \leq G\).
Solution
Recall \[ C_G(H)=\{g\in G:gh=hg\text{ for all }h\in H\} \] and \[ N_G(H)=\{g\in G:gHg^{-1}=H\}. \]
1. One has\[ C_G(H)\subseteq N_G(H). \]
If \(g\in C_G(H)\), then for every \(h\in H\), \[ ghg^{-1}=h. \] Thus conjugation by \(g\) fixes every element of \(H\), so \(gHg^{-1}=H\) and \(g\in N_G(H)\).
2. One has\[ N_G(H)\le G. \]
The identity normalizes \(H\). If \(g_1,g_2\in N_G(H)\), then \[ (g_1g_2)H(g_1g_2)^{-1} =g_1(g_2Hg_2^{-1})g_1^{-1} =g_1Hg_1^{-1} =H, \] so \(g_1g_2\in N_G(H)\). If \(g\in N_G(H)\), then \(gHg^{-1}=H\), and conjugating this equality by \(g^{-1}\) gives \(g^{-1}Hg=H\), so \(g^{-1}\in N_G(H)\).
Therefore \[ C_G(H)\subseteq N_G(H)\le G. \]