Conjugate of Set by Group Product

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

Let $$S \subseteq G$$.

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

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:

Comment
This proves that $$\left({S^a}\right)^b = S^{b \circ a}$$, whereas, section 45, has $$\left({S^a}\right)^b = S^{a \circ b}$$. This may need to be resolved.