# Two-Step Subgroup Test using Subset Product

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## 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:

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

where:

## 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:

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

$\Box$

### 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 $G$.

$\blacksquare$

## Also see

## Linguistic Note

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

This is because the fact that the subset must be non-empty is usually an unspoken assumption, and is not specifically included as one of the tests to be made.

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 9$: Compositions Induced on the Set of All Subsets: Theorem $9.1: \ 2^\circ$ - 1967: George McCarty:
*Topology: An Introduction with Application to Topological Groups*... (previous) ... (next): Chapter $\text{II}$: Groups: Subgroups - 1982: P.M. Cohn:
*Algebra Volume 1*(2nd ed.) ... (previous) ... (next): $\S 3.2$: Groups; the axioms