Given a subgroup \(H\leq G\), let \[ \mathcal C(H)=\ts{gHg^{-1}:g\in G} \] be the set of subgroups conjugate to \(H\). Show that \[ \abs{\mathcal C(H)}=[G:N_G(H)]. \]
Solution
1. Let \(G\) act on the set of its subgroups by conjugation:\[ g \cdot K = g K g^{-1} \quad \text{for } g \in G, \; K \in \mathcal{S}. \]Here \(\mathcal S\) denotes the set of subgroups of \(G\).
2. The orbit of \(H\) is exactly the set of its conjugate subgroups:\[ \operatorname{Orb}_G(H) = \{g \cdot H \mid g \in G\} = \{g H g^{-1} \mid g \in G\} = \mathcal{C}(H). \]
3. The stabilizer of \(H\) is its normalizer:\[ \operatorname{Stab}_G(H) = \{g \in G \mid g \cdot H = H\} = \{g \in G \mid g H g^{-1} = H\} = N_G(H), \]
4. By orbit–stabilizer,\[ |\mathcal{C}(H)| = |\operatorname{Orb}_G(H)| = |G / N_G(H)| = [G : N_G(H)]. \]