- Subgroup test
In
Abstract Algebra , the one-step subgroup test is a theorem that states that for any group, asubset 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 G be a group and let H be a nonempty subset of G. If for all a and b in H, ab-1 is in H, then H is a subgroup of G.
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. Letting a = x and b = x, we have that the identity e = xx-1 = ab-1 which is in H, so e is in H.
* 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.
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