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

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

Theorem for Alternative Definition
The concept of set conjugate can be defined in a different way:

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

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 RHS refers to subset products.

From Subset Product of Associative 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:
 * $$S^a \ \stackrel {\mathbf {def}} {=\!=} \ a \circ S \circ a^{-1}$$

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