Number of Distinct Conjugate Subsets is Index of Normalizer

Theorem
Let $G$ be a group.

Let $S$ be a subset of $G$.

Let $\map {N_G} S$ be the normalizer of $S$ in $G$.

Let $\index G {\map {N_G} S}$ be the index of $\map {N_G} S$ in $G$.

The number of distinct subsets of $G$ which are conjugates of $S \subseteq G$ is $\index G {\map {N_G} S}$.

Proof
We have that:
 * $S^a = S^b \iff S^{a b^{-1}} = S$ (reference to be determined).

That is:
 * $S^a = S^b \iff a b^{-1} \in \map {N_G} S)$

which is equivalent to:
 * $a^{-1} \equiv b^{-1} \pmod {\map {N_G} S}$

Thus we have a bijection between:
 * the conjugacy class $\conjclass S$ of subsets of $G$ conjugate to $S$

and: the left coset space $G / \map {N_G} S$

given by $S^a \to a^{-1} \map {N_G} S$.

Since $G / \map {N_G} S$ has $\index G {\map {N_G} S}$ elements, the result follows.