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 .
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. If we take a = x and b = x, then ab−1 = xx−1 = e, where e is the identity element. Therefore e is in H.
- Let x be an element in H and we have just shown the identity element, e, is in H. Then let a = e and b = x, it follows that ab−1 = 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.