Two-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:
 * $$H H \subseteq H$$
 * $$H^{-1} \subseteq H$$

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

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

Then $$H$$ is closed, and Groupoid Subset Product with Self gives that $$H H \subseteq H$$.

Also, $$g \in H^{-1} \implies \exists h \in H: g = h^{-1} \implies g \in H$$, so $$H^{-1} \subseteq H$$.

Now suppose that:


 * $$H H \subseteq H$$
 * $$H^{-1} \subseteq H$$

From the definition of [Definition:Subset Product|subset product]]:
 * $$\forall x, y \in H: x y \in H$$
 * $$\forall x \in H^{-1}: x^{-1} \in H$$

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