# One-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 $G$ if and only if:

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

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

$\Box$

### 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 $G$.

$\blacksquare$

## Also presented as

This result can also be presented as:

$H$ is a subgroup of $G$ if and only if:

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

and the same argument applies.

## Linguistic Note

The One-Step Subgroup Test is so called despite the fact that, on the face of it, there are two steps to the test.

This is because the fact that the subset must be non-empty is frequently assumed as one of the "givens", and is then not specifically included as one of the tests to be made.