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 $G$ $H H^{-1} \subseteq H$.

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

Necessary Condition
Let $H$ be a subgroup of $G$.

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

Then by definition of subset product:
 * $x \in H, y^{-1} \in H^{-1}$

By definition of inverse of $H^{-1}$:
 * $\left({y^{-1} }\right)^{-1} \in \left({H^{-1} }\right)^{-1}$

From Inverse of Inverse of Subset of Group:
 * $\left({H^{-1} }\right)^{-1} = H$

and from Inverse of Inverse:
 * $\left({y^{-1} }\right)^{-1} = y$

Thus:
 * $y \in H$

As $H$ is a group, by group axiom $G\,3$:
 * $y^{-1} \in H$

and so by group axiom $G\,0$:
 * $x y^{-1} \in H$

So by definition of subset:
 * $H H^{-1} \subseteq H$

Sufficient Condition
Let:
 * $H H^{-1} \subseteq H$

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 $G$.