Subgroup test: Difference between revisions

Revision as of 02:29, 29 January 2014

Revision as of 22:59, 12 February 2015

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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^-1−1 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. Then the identity is in H since we can write it as e = x x^-1−1 which is in H by the initial assumption.

* Let x be an element of H. Since the identity e is in H it follows that ex^-1−1 = x^-1−1 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^-1−1 is in H. Hence x(y^-1−1)^-1−1 = xy is in H and so H is closed under the operation.

Thus H is a subgroup of G.