Subgroup test: Difference between revisions

Revision as of 22:40, 20 January 2017

Revision as of 22:48, 20 January 2017

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* 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 ofin H. Sinceand 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,. soSo 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.