Powers of Element form Subgroup

Theorem
Let $\left({G, \circ}\right)$ be a group.

Then:
 * $\forall a \in G: H = \left\{{a^n: n \in \Z}\right\} \le G$

That is, the subset of $G$ comprising all elements possible as powers of $a \in G$ is a subgroup of $G$.

Proof
Clearly $a \in H$, so $H \ne \varnothing$.

Let $x, y \in H$.

Thus by the One-step Subgroup Test, $H \le G$.