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

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

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 $x, y \in H$.

Then by the definition of subset product:


 * $x y^{-1} \in H H^{-1}$

As $H \le G$, from the One-Step Subgroup Test, $x y^{-1} \in H$.

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

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 $G$.

Also presented as
This result can also be presented as:

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

and the same argument applies.