Conjugate of Set by Group Product

Theorem
Let $\struct {G, \circ}$ be a group.

Let $S \subseteq G$.

Let $S^a$ denote the $G$-conjugate of $S$ by $a$ as:
 * $S^a := \set {y \in G: \exists x \in S: y = a \circ x \circ a^{-1} } = a \circ S \circ a^{-1}$

Then:
 * $\paren {S^a}^b = S^{b \circ a}$

Also defined as
The concept of set conjugate can be defined in a different way:

Let $S^a$ denote the $G$-conjugate of $S$ by $a$ as:
 * $S^a := \set {y \in G: \exists x \in S: y = a^{-1} \circ x \circ a} = a^{-1} \circ S \circ a$

Then:
 * $\paren {S^a}^b = S^{a \circ b}$

Proof
$S^a$ is defined as $a \circ S \circ a^{-1}$ from the definition of the conjugate of a set.

From the definition of subset product with a singleton, this can be seen to be the same thing as:


 * $S^a = \set a \circ S \circ \set {a^{-1} }$.

Thus we can express $\paren {S^a}^b$ as $b \circ \paren {a \circ S \circ a^{-1} } \circ b^{-1}$, and understand that the refers to subset products.

From Subset Product within Semigroup is Associative (which applies because $\circ$ is associative), it then follows directly that:

Proof for Alternative Definition
Using the same preliminary argument as above, we then follow:

Comment
This is not always correct in the literature.

For example, defines set conjugate as:
 * $S^a := a \circ S \circ a^{-1}$

but then states (without proof) the assertion:
 * $\left({S^a}\right)^b = S^{a \circ b}$