Subgroup test: Difference between revisions

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===Proof===

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

* Since the operation of <math>H</math> is the same as the operation of <math>G</math>, the operation is associative since <math>G</math> is a group.

* Since <math>H</math> is not empty there exists an element <math>x</math> in <math>H</math>. If we take <math>a = x</math> and <math>b = x</math>, then ab<sup>−1</supmath>ab^{-1} = xx⁻¹x x^{-1} = e</math>, where <math>e</math> is the identity element. Therefore <math>e</math> is in <math>H</math>.

* Let <math>x</math> be an element in <math>H</math> and we have just shown the identity element, <math>e</math>, is in <math>H</math>. Then let <math>a = e</math> and <math>b = x</math>, it follows that ab<sup>−1</supmath>ab^{-1} = ex⁻¹^{-1} = x<sup>−1^{-1}</supmath> in <math>H</math>. So the inverse of an element in <math>H</math> is in <math>H</math>.

* Finally, let <math>x</math> and <math>y</math> be elements in <math>H</math>, then since <math>y</math> is in <math>H</math> it follows that y<supmath>−1y^{-1}</supmath> is in <math>H</math>. Hence <math>x (y⁻¹^{-1})⁻¹^{-1} = xy</math> is in <math>H</math> and so <math>H</math> is closed under the operation.

Thus <math>H</math> is a subgroup of <math>G</math>.

==Two-step subgroup test==