Let \(G\) act by conjugation on its set of subgroups. Show that the number of subgroups conjugate to \(H\le G\) is \[ [G:N_G(H)]. \]
Solution
Let \[ \mathcal O(H)=\{gHg^{-1}:g\in G\} \] be the conjugacy orbit of \(H\).
1. The stabilizer of \(H\) is its normalizer.
By definition, \[ \operatorname{Stab}_G(H) =\{g\in G:gHg^{-1}=H\} =N_G(H). \]
2. Apply orbit-stabilizer.
Orbit-stabilizer gives a bijection \[ G/N_G(H)\longrightarrow \mathcal O(H),\qquad gN_G(H)\longmapsto gHg^{-1}. \] Hence \[ |\mathcal O(H)|=[G:N_G(H)]. \] Thus the number of conjugates of \(H\) is the index of its normalizer.