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 $\struct {G, \circ}$ :


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

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:

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

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.

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 $H$.

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 $\struct {G, \circ}$.

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

Also see

 * One-Step Subgroup Test using Subset Product
 * Two-Step Subgroup Test