[proofplan] This is a clean application of the [orbit–stabiliser theorem](/theorems/796) to the conjugation action of $G$ on its set of subgroups. We define the action $G \curvearrowright \mathcal{S}(G)$ by $g \cdot K := gKg^{-1}$, identify the orbit of $H$ with the set of conjugates of $H$, identify the stabiliser of $H$ with the normaliser $N_G(H)$ by definition, and read off the conclusion from orbit–stabiliser. The argument is self-contained modulo the orbit–stabiliser theorem, so the work is verifying that the conjugation action is well-defined, that orbits and stabilisers are the named objects, and that the cardinality of the orbit equals the index of the stabiliser even when $G$ is infinite. [/proofplan] [step:Define the conjugation action of $G$ on the set of subgroups] Let $\mathcal{S}(G) := \{K : K \leq G\}$ denote the set of all subgroups of $G$. Define \begin{align*} \alpha: G \times \mathcal{S}(G) &\to \mathcal{S}(G) \\ (g, K) &\mapsto gKg^{-1} := \{gkg^{-1} : k \in K\}. \end{align*} We verify that $\alpha$ is a well-defined group action. **$gKg^{-1}$ is a subgroup.** For any $g \in G$ and $K \leq G$, the conjugation map $c_g: G \to G$, $x \mapsto gxg^{-1}$ is a group automorphism (its inverse is $c_{g^{-1}}$, and it preserves products: $c_g(xy) = gxyg^{-1} = (gxg^{-1})(gyg^{-1}) = c_g(x) c_g(y)$). Therefore $gKg^{-1} = c_g(K)$ is the image of a subgroup under a homomorphism, hence a subgroup of $G$. **Identity acts as identity.** $\alpha(e, K) = eKe^{-1} = K$. **Compatibility.** For $g, h \in G$, \begin{align*} \alpha(g, \alpha(h, K)) = g(hKh^{-1})g^{-1} = (gh)K(gh)^{-1} = \alpha(gh, K). \end{align*} So $\alpha$ is a left action of $G$ on $\mathcal{S}(G)$. [guided] We use the [orbit–stabiliser theorem](/theorems/796), which requires a group action. The natural action here is **conjugation on subgroups**: the group $G$ acts on the set of all subgroups $\mathcal{S}(G) := \{K : K \leq G\}$ by sending $K$ to its conjugate $gKg^{-1}$. Why is this even an action? We need to check three things: 1. **Well-definedness**: $gKg^{-1}$ must again be a subgroup, so that $\alpha$ lands in $\mathcal{S}(G)$. This is true because conjugation by $g$ defines an inner automorphism $c_g \in \operatorname{Aut}(G)$, $c_g(x) = gxg^{-1}$, which sends subgroups to subgroups. (To see $c_g$ is an automorphism: it's a homomorphism since $c_g(xy) = gxyg^{-1} = gxg^{-1} \cdot gyg^{-1} = c_g(x)c_g(y)$, and its inverse is $c_{g^{-1}}$.) 2. **Identity acts as the identity**: $eKe^{-1} = K$. 3. **Compatibility (left-action axiom)**: $g \cdot (h \cdot K) = (gh) \cdot K$. We compute: \begin{align*} g \cdot (h \cdot K) = g(hKh^{-1})g^{-1} = (gh)K(h^{-1}g^{-1}) = (gh)K(gh)^{-1} = (gh) \cdot K. \end{align*} All three axioms hold, so $\alpha$ is a left action of $G$ on $\mathcal{S}(G)$. [/guided] [/step] [step:Identify the orbit of $H$ with the set of conjugates of $H$] The orbit of $H$ under $\alpha$ is by definition \begin{align*} G \cdot H = \{g \cdot H : g \in G\} = \{gHg^{-1} : g \in G\}. \end{align*} This is exactly the set whose cardinality appears on the left-hand side of the theorem statement. [guided] The orbit of $H$ under the action $\alpha$ is, by definition, \begin{align*} G \cdot H = \{\alpha(g, H) : g \in G\} = \{gHg^{-1} : g \in G\}. \end{align*} This is *literally* the set of conjugates of $H$ — the very set whose cardinality appears in the statement. So the LHS of our equation is $|G \cdot H|$, the cardinality of the orbit. This is the place where the choice of action pays off: the set we want to count is, by construction, the orbit of a single chosen element under the action. [/guided] [/step] [step:Identify the stabiliser of $H$ with the normaliser $N_G(H)$] The stabiliser of $H$ under $\alpha$ is \begin{align*} G_H := \{g \in G : g \cdot H = H\} = \{g \in G : gHg^{-1} = H\}. \end{align*} By the definition of the [normaliser](/page/Normalizer), \begin{align*} G_H = N_G(H). \end{align*} In particular $N_G(H)$ is a subgroup of $G$ (stabilisers under any group action are subgroups). [guided] The stabiliser of $H$ under the conjugation action is, by definition, \begin{align*} G_H := \{g \in G : g \cdot H = H\} = \{g \in G : gHg^{-1} = H\}. \end{align*} Compare this to the **definition of the normaliser** of $H$ in $G$: \begin{align*} N_G(H) := \{g \in G : gHg^{-1} = H\}. \end{align*} The two sets are *literally* the same set — the equation $gHg^{-1} = H$ appears verbatim in both. Hence $G_H = N_G(H)$. This is not a calculation, just a recognition: the normaliser is *defined* to be the stabiliser of $H$ under conjugation on subgroups. A useful by-product: stabilisers under any group action are automatically subgroups (closed under products, contains $e$, closed under inverses — all immediate from the action axioms). So $N_G(H) \leq G$, recovering the standard fact for free. [/guided] [/step] [step:Apply orbit–stabiliser to conclude] By the [orbit–stabiliser theorem](/theorems/796), for any group action of $G$ on a set $X$ and any $x \in X$, the map \begin{align*} \Psi: G/G_x &\to G \cdot x \\ gG_x &\mapsto g \cdot x \end{align*} is a well-defined bijection between the set of left cosets of the stabiliser and the orbit. Taking cardinalities, \begin{align*} |G \cdot x| = |G/G_x| = [G : G_x]. \end{align*} This identity holds whether $G$ is finite or infinite, with both sides being cardinal numbers (and infinite if either is infinite). Applying this to our action $\alpha$ at the point $H \in \mathcal{S}(G)$, and using $G_H = N_G(H)$ from the previous step, \begin{align*} |\{gHg^{-1} : g \in G\}| = |G \cdot H| = [G : G_H] = [G : N_G(H)]. \end{align*} This is the desired equality. [guided] The [orbit–stabiliser theorem](/theorems/796) states that for any action $G \curvearrowright X$ and any $x \in X$, the assignment \begin{align*} \Psi: G / G_x &\to G \cdot x \\ gG_x &\mapsto g \cdot x \end{align*} is a well-defined bijection. Let's recall briefly why: - *Well-defined*: if $g_1 G_x = g_2 G_x$, then $g_1^{-1} g_2 \in G_x$, so $(g_1^{-1} g_2) \cdot x = x$, hence $g_2 \cdot x = g_1 \cdot x$. - *Injective*: if $g_1 \cdot x = g_2 \cdot x$, then $(g_1^{-1} g_2) \cdot x = x$, so $g_1^{-1} g_2 \in G_x$, hence $g_1 G_x = g_2 G_x$. - *Surjective*: every element of the orbit is by definition $g \cdot x$ for some $g \in G$. Taking cardinalities (both sides are well-defined cardinals — for infinite groups too, since a bijection of sets preserves cardinality regardless of size), \begin{align*} |G \cdot x| = |G/G_x| = [G : G_x]. \end{align*} Now we apply this with $X = \mathcal{S}(G)$, action $\alpha$, and $x = H$. From the previous two steps: \begin{align*} G \cdot H &= \{gHg^{-1} : g \in G\}, \\ G_H &= N_G(H). \end{align*} Therefore \begin{align*} |\{gHg^{-1} : g \in G\}| = |G \cdot H| = [G : G_H] = [G : N_G(H)], \end{align*} which is exactly the statement we wanted to prove. The argument works whether $G$ is finite (in which case both sides are finite positive integers, related by $[G : N_G(H)] = |G| / |N_G(H)|$) or infinite (in which case both sides are equal cardinal numbers). [/guided] [/step]