# Subgroup test

In abstract algebra, the one-step **subgroup test** is a theorem that states that for any group, a nonempty 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. The two-step subgroup test is a similar theorem which requires the subset to be closed under the operation and taking of inverses.

## One-step subgroup test

Let be a group and let be a nonempty subset of . If for all and in , is in , then is a subgroup of .

### Proof

Let G be a group, let H be a nonempty subset of G and assume that for all a and b in H, ab^{-1} is in H. To prove that H is a subgroup of G we must show that H is associative, has an identity, has an inverse for every element and is closed under the operation. So,

- Since the operation of H is the same as the operation of G, the operation is associative since G is a group.
- Since H is not empty there exists an element x in H. Then the identity is in H since we can write it as e = x x
^{-1}which is in H by the initial assumption. - Let x be an element of H. Since the identity e is in H it follows that ex
^{-1}= x^{-1}in H, so the inverse of an element in H is in H. - Finally, let x and y be elements in H, then since y is in H it follows that y
^{-1}is in H. Hence x(y^{-1})^{-1}= xy is in H and so H is closed under the operation.

Thus H is a subgroup of G.

## Two-step subgroup test

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.