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 $\order {H^a} = \order H$.

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

From Set Equivalence of Regular Representations:
 * $\order {a H a^{-1} } = \order {a H} = \order H$