# One-Step Subgroup Test

## Theorem

Let $\struct {G, \circ}$ be a group.

Let $H$ be a subset of $G$.

Then $\struct {H, \circ}$ is a subgroup of $\struct {G, \circ}$ if and only if:

$(1): \quad H \ne \O$, that is, $H$ is non-empty
$(2): \quad \forall a, b \in H: a \circ b^{-1} \in H$.

## Proof

### Necessary Condition

Let $H$ be a subset of $G$ that fulfils the conditions given.

It is noted that the fact that $H$ is non-empty is one of the conditions.

It is also noted that the group product of $\struct {H, \circ}$ is the same as that for $\struct {G, \circ}$, that is, $\circ$.

So it remains to show that $\struct {H, \circ}$ is a group.

We check the four group axioms:

G1: Associativity:

From Subset Product within Semigroup is Associative, associativity is inherited by $\struct {H, \circ}$ from $\struct {G, \circ}$.

G2: Identity

Let $e$ be the identity of $\struct {G, \circ}$.

Since $H$ is non-empty, $\exists x \in H$.

If we take $a = x$ and $b = x$, then $a \circ b^{-1} = x \circ x^{-1} = e \in H$, where $e$ is the identity element.

G3: Inverses

If we take $a = e$ and $b = x$, then $a \circ b^{-1} = e \circ x^{-1} = x^{-1} \in H$.

Thus every element of $H$ has an inverse also in $H$.

G0: Closure

Let $x, y \in H$.

Then $y^{-1} \in H$, so we may take $a = x$ and $b = y^{-1}$.

So:

$a \circ b^{-1} = x \circ \paren {y^{-1} }^{-1} = x \circ y \in H$

Thus, $H$ is closed.

Therefore, $\struct {H, \circ}$ satisfies all the group axioms, and is therefore a group.

Therefore $\struct {H, \circ}$ is a subgroup of $\struct {G, \circ}$.

$\Box$

### Sufficient Condition

Now suppose $\struct {H, \circ}$ is a subgroup of $\struct {G, \circ}$.

$(1): \quad H \le G \implies H \ne \O$ from the fact that $H$ is a group and therefore can not be empty.
$(2): \quad$ As $\struct {H, \circ}$ is a group, it is closed and every element has an inverse. So it follows that $\forall a, b \in H: a \circ b^{-1} \in H$.

$\blacksquare$

## Comment

This is called the one-step subgroup test although, on the face of it, there are two steps to the test.

This is because the fact that $H$ 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.