Subgroup test

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

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

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

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

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

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

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

Thus ${\displaystyle H\,}$  is a subgroup of ${\displaystyle G\,}$ .