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 $$G$$.