# One-Step Subgroup Test

## Contents

## 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 operation 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:

**$\text G 1$: Associativity**:

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

**$\text G 2$: 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.

**$\text G 3$: 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 $H$.

**$\text G 0$: 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.

## Also see

## Sources

- 1965: J.A. Green:
*Sets and Groups*... (previous) ... (next): $\S 5.2$. Subgroups - 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 8$: Theorem $8.4: \ 1^\circ - 3^\circ$ - 1966: Richard A. Dean:
*Elements of Abstract Algebra*... (previous) ... (next): $\S 1.9$: Subgroups: Lemma $8$ - 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: Subgroups and Cosets: $\S 35 \alpha$ - 1974: Thomas W. Hungerford:
*Algebra*... (previous) ... (next): $\S 1.2$ - 1978: John S. Rose:
*A Course on Group Theory*... (previous) ... (next): $0$: Some Conventions and some Basic Facts - 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): $\S 36.4$: Subgroups - 1982: P.M. Cohn:
*Algebra Volume 1*(2nd ed.) ... (previous) ... (next): $\S 3.2$: Groups; the axioms - 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $4$: Subgroups: Proposition $4.2$