Two-Step Subgroup Test using Subset Product

Theorem
Let $G$ be a group.

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

Then $H$ is a subgroup of $G$ :
 * $H H \subseteq H$
 * $H^{-1} \subseteq H$

where:
 * $H H$ is the product of $H$ with itself
 * $H^{-1}$ is the inverse of $H$.

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

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

Then $H$ is closed.

It follows from Magma Subset Product with Self:
 * $H H \subseteq H$

Then:
 * $g \in H^{-1} \implies \exists h \in H: g = h^{-1} \implies g \in H$

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

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

From the definition of 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 $G$.

Also see

 * One-Step Subgroup Test using Subset Product