One-Step Subgroup Test

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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
$(2): \quad \forall a, b \in H: a \circ b^{-1} \in H$.


Corollary

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

Sufficient 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:


Group Axiom $\text G 1$: Associativity

From Restriction of Associative Operation is Associative, associativity is inherited by $\struct {H, \circ}$ from $\struct {G, \circ}$.


Group Axiom $\text G 2$: Existence of Identity Element

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.


Group Axiom $\text G 3$: Existence of Inverse Element

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 $H$.


Group Axiom $\text G 0$: 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 $\struct {G, \circ}$.

$\Box$


Necessary 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 $\forall a, b \in H: a \circ b^{-1} \in H$.

$\blacksquare$


Also see


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.


Sources

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