Number of Distinct Conjugate Subsets is Index of Normalizer/Proof 1

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.

$\blacksquare$

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Sources

• 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 49$