Conjugate of Set by Group Product

Theorem
Let $\left({G, \circ}\right)$ be a group.

Let $S \subseteq G$.

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

Then:
 * $\left({S^a}\right)^b = S^{b \circ a}$

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:
 * $S^a := \left\{{y \in G: \exists x \in S: y = a^{-1} \circ x \circ a}\right\} = a^{-1} \circ S \circ a$

Then:
 * $\left({S^a}\right)^b = S^{a \circ b}$

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:


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

Thus we can express $\left({S^a}\right)^b$ as $b \circ \left({a \circ S \circ a^{-1}}\right) \circ b^{-1}$, 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:
 * $\left({S^a}\right)^b = S^{a \circ b}$