Conjugate of Subgroup is Subgroup/Proof 1

Proof
Let $H \le G$.

First, we show that $x, y \in H^a \implies x \circ y \in H^a$:

Next, we show that $x \in H^a \implies x^{-1} \in H^a$:

Thus by the Two-Step Subgroup Test, $H^a \le G$.