One-Step Subgroup Test using Subset Product

Theorem
Let $G$ be a group.

Let $\varnothing \subset H \subseteq G$ be a non-empty subset of $G$.

Then $H$ is a subgroup of $G$ iff $H H^{-1} \subseteq H$.

Proof
This is a reformulation of the One-Step Subgroup Test in terms of subset product.

Let $H$ is a subgroup of $G$.

Let $x, y \in H$. Then $y \in H^{-1}$.

Then $x y^{-1} \in H$ by the group axioms.

That is:
 * $\forall x, y \in H: x y^{-1} \in H$

and so by definition of subset product:
 * $H H^{-1} \subseteq H$

Now suppose that $H H^{-1} \subseteq H$.

From the definition of subset product:
 * $\forall x, y \in H: x y^{-1} \in H$

So by the One-Step Subgroup Test, $H$ is a subgroup of $G$.