Subgroup test: Difference between revisions

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Proof: A much clearer understanding of the Identity proof.
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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. ThenIf thewe identitytake isa in= Hx sinceand web can= writex, itthen as eab<sup>−1</sup> = x xxx<sup>−1</sup> which= ise, inwhere He byis the initialidentity assumptionelement. Therefore 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.