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#REDIRECT [[Subgroup#Subgroup tests]]
In [[Abstract Algebra]], the one-step '''subgroup test''' is a theorem that states that for any group, a nonempty [[subset]] of that [[Group_%28mathematics%29|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.
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==One-step subgroup test==
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Let G be a group and let H be a nonempty subset of G. If for all a and b in H, ab<sup>-1</sup> is in H, then H is a subgroup of G.
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===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<sup>-1</sup> 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<sup>-1</sup> = ab<sup>-1</sup> 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<sup>-1</sup> = x<sup>-1</sup> 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<sup>-1</sup> is in H. Hence x(y<sup>-1</sup>)<sup>-1</sup> = 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 [[Closure (mathematics)|closed]] under the operation as well as under the taking of inverses.
[[Category:Mathematical theorems]]
[[Category:Articles containing proofs]]
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