# 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 1

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.

$\blacksquare$

## Proof 2

Let $G$ act on its power set $\powerset G$ by the rule:

- $\forall g \in G, S \in \powerset G: g * S := S^{g^{-1} } = \set {x \in G: g^{-1} x g \in S}$

That is, the conjugacy action on subsets.

From Conjugacy Action on Subsets is Group Action, $*$ is a group action.

The orbit of $S \in \powerset G$ is the conjugacy class of $S$.

That is, $\Orb S$ is the set of distinct subsets of $G$ which are conjugate to $S$

The stabilizer of $S$ is its normalizer $\map {N_G} S$.

From the Orbit-Stabilizer Theorem:

- $\order {\Orb x} = \index G {\Stab x} = \dfrac {\order G} {\order {\Stab x} }$

from which the result follows directly.

$\blacksquare$

## Sources

- 1966: Richard A. Dean:
*Elements of Abstract Algebra*... (previous) ... (next): $\S 1.9$: Exercise $5.16 \ \text{(ii)}$