In Abstract Algebra, the one-step subgroup test is a theorem that states that for any group, a subset of that group is itself a group if the inverse of any element in the subset multiplied with any other element in the subset is also in the subset.

Or more formally let $G\,$ be a group and let $H\,$ be a nonempty subset of $G\,$ . If $\forall {a,b\in H},ab^{-1}\in H$ then $H\,$ is a subgroup of $G\,$ .

A corollary of this theorem is the two-step subgroup test which states that a nonempty subset of a group is itself a group if the subset is closed under the operation as well as under the taking of inverses.

Proof

To prove that $H\,$ is a subgroup of $G\,$ we must show that $H\,$ is nonempty, associative, has an identity, has an inverse for every element, and is closed under the operation.

Let $G\,$ be a group, let $H\,$ be a nonempty subset of $G\,$ and assume that $\forall {a,b\in H},ab^{-1}\in H$ .

Since the operation of $H\,$ is the same as the operation of $G\,$ , the operation is associative since $G\,$ is a group.

Next we show that the identity, $e\,$ , is in $H\,$ . Since $H\,$ is not empty there exists an $x\in H$ . Letting $a=x\,$ and $b=x\,$ , we have that $e=xx^{-1}=ab^{-1}\in H$ , so $e\in H$ .

We now show that every element in $H\,$ has an inverse in $H\,$ . Let $x\in H$ . Since $e\in H$ it follows that $ex^{-1}=x^{-1}\in H$ , so $x^{-1}\in H$

Finally we show that $H\,$ is closed under the operation. Let $x,y\in H$ , then since $y\in H$ it follows that $y^{-1}\in H$ . Hence $x{(y^{-1})}^{-1}=xy\in H$ and so $H\,$ is closed under the operation.

Thus $H\,$ is a subgroup of $G\,$ .