Order of Conjugate of Subgroup

Theorem
Let $G$ be a group.

Let $H$ be a subgroup of $G$ such that $H$ is of finite order.

Then $\left|{H^a}\right| = \left|{H}\right|$.

Proof
From the definition of Conjugate of Group Subet we have $H^a = a H a^{-1}$.

From Set Equivalence of Regular Representations, $\left|{a H a^{-1}}\right| = \left|{a H}\right| = \left|{H}\right|$.