Conjugate of Subgroup is Subgroup

Theorem
For any subgroup $$H \le G$$, the conjugate of $$H$$ by $$a$$ is a subgroup of $$G$$:

$$\forall H \le G, a \in G: H^a \le G$$

Proof
Let $$H \le G$$.

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


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

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