Subgroup test: Difference between revisions

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In [[abstract algebra]], the one-step '''subgroup test''' is a theorem that states that for any group, a nonempty [[subset]] of that [[Group (mathematics)|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 <math>G

</math> be a group and let <math>H</math> be a nonempty subset of <math>G</math>. If for all <math>a</math> and <math>b</math> in <math>H</math>, <math>a b ^{-1}</math> is in <math>H</math>, then <math>H</math> is a subgroup of <math>G</math>.

===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. 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==