Set Equivalence of Regular Representations

Theorem
If $S$ is a finite subset of a group $G$, then:


 * $\left|{a \circ S}\right| = \left|{S}\right| = \left|{S \circ a}\right|$

That is, $a \circ S$, $S$ and $S \circ a$ are equivalent: $a \circ S \sim S \sim S \circ a$.

Proof
Follows immediately from the fact that both the left and right regular representation are permutations, and therefore bijections.