en>Rjwilmsi: Citation parameter fixes, using AWB (9461)

2013-09-02T13:19:25Z

Citation parameter fixes, using AWB (9461)

← Older revision		Revision as of 15:19, 2 September 2013
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	~~Hi there~~. ~~My title~~ is ~~Sophia Meagher although it is not~~ the ~~title on my beginning certification~~. ~~Credit authorising~~ is ~~where my primary income arrives from~~. ~~I've always loved residing~~ in ~~Alaska~~. To ~~play domino~~ is ~~something I really appreciate performing~~.<br><br>~~Look at my homepage psychic phone readings ([http:~~//~~brazil~~.~~amor~~-~~amore~~.~~com~~/~~irboothe check~~ this ~~site out~~])		In [[abstract algebra]], the one-step '''subgroup test''' is a theorem that states that for any group, a nonempty [[subset]] of that [[Group_%28mathematics%29\|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<sup>-1</sup> 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<sup>-1</sup> 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<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.

			Thus H is a subgroup of G.

			==Two-step subgroup test==

			A corollary of this theorem is the two-step subgroup test which states that a nonempty subset of a group is itself a group if the subset is [[Closure (mathematics)\|closed]] under the operation as well as under the taking of inverses.

			[[Category:Theorems in group theory]]
			[[Category:Articles containing proofs]]

137.82.118.160: /* Examples */

2012-08-30T20:20:55Z

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Hi there. My title is Sophia Meagher although it is not the title on my beginning certification. Credit authorising is where my primary income arrives from. I've always loved residing in Alaska. To play domino is something I really appreciate performing.<br><br>Look at my homepage psychic phone readings ([http://brazil.amor-amore.com/irboothe check this site out])

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en>Rjwilmsi: Citation parameter fixes, using AWB (9461)

137.82.118.160: /* Examples */