Intersection of subgroups is subgroup - Revision history

Vipul at 19:39, 31 March 2022

2022-03-31T19:39:33Z

← Older revision		Revision as of 19:39, 31 March 2022
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	==Proof==		==Proof==

	~~<center>{{#widget:YouTube\|id=kYAk-olv3B0}}</center>~~

	''Given'': Let <math>H_i\!</math> be an arbitrary collection of subgroups of a [[group]] <math>G\!</math> indexed by <math>i \in I.</math> Let us denote <math>H = \textstyle\bigcap_{i \in I} H_i.</math> Here, <math>e\!</math> denotes the identity element of <math>G.\!</math>		''Given'': Let <math>H_i\!</math> be an arbitrary collection of subgroups of a [[group]] <math>G\!</math> indexed by <math>i \in I.</math> Let us denote <math>H = \textstyle\bigcap_{i \in I} H_i.</math> Here, <math>e\!</math> denotes the identity element of <math>G.\!</math>

Vipul: /* Related facts */

2013-07-03T00:01:38Z

Related facts

← Older revision		Revision as of 00:01, 3 July 2013
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	* [[Invariance implies strongly intersection-closed]]		* [[Invariance implies strongly intersection-closed]]

			===Intersections of subsets other than subgroups===

			{\| class="sortable" border="1"
			! Subset property !! Proof that it is closed under arbitrary intersections
			\|-
			\| [[twisted subgroup]] \|\| [[intersection of twisted subgroups is twisted subgroup]]
			\|-
			\| [[1-closed subset]] \|\| [[intersection of 1-closed subsets is 1-closed subset]]
			\|-
			\| [[symmetric subset]] \|\| [[intersection of symmetric subsets is symmetric subset]]
			\|}
	===Analogues in other algebraic structures===		===Analogues in other algebraic structures===

	For any [[variety of algebras]], an intersection of subalgebras is a subalgebra. The proof is exactly the same as that for groups. In fact, the result holds in a slightly greater generality than varieties of algebras. For instance, an intersection of subfields is a subfield, although fields do not form a variety of algebras.		For any [[variety of algebras]], an intersection of subalgebras is a subalgebra. The proof is exactly the same as that for groups. In fact, the result holds in a slightly greater generality than varieties of algebras. For instance, an intersection of subfields is a subfield, although fields do not form a variety of algebras.
	<section begin=beginner/>		<section begin=beginner/>

	==Proof==		==Proof==

Vipul: /* Proof */

2013-04-04T21:32:18Z

Proof

← Older revision		Revision as of 21:32, 4 April 2013
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	<section begin=beginner/>		<section begin=beginner/>
	==Proof==		==Proof==

			<center>{{#widget:YouTube\|id=kYAk-olv3B0}}</center>

	''Given'': Let <math>H_i\!</math> be an arbitrary collection of subgroups of a [[group]] <math>G\!</math> indexed by <math>i \in I.</math> Let us denote <math>H = \textstyle\bigcap_{i \in I} H_i.</math> Here, <math>e\!</math> denotes the identity element of <math>G.\!</math>		''Given'': Let <math>H_i\!</math> be an arbitrary collection of subgroups of a [[group]] <math>G\!</math> indexed by <math>i \in I.</math> Let us denote <math>H = \textstyle\bigcap_{i \in I} H_i.</math> Here, <math>e\!</math> denotes the identity element of <math>G.\!</math>

Vipul at 17:56, 24 February 2011

2011-02-24T17:56:59Z

← Older revision		Revision as of 17:56, 24 February 2011
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	{{basic fact}}		{{basic fact}}
			[[difficulty level::1\| ]]
	<section begin=beginner/>		<section begin=beginner/>
	==Statement==		==Statement==

Jon Awbrey: /* Proof */ png to normalize character size + punctuation conventions

2009-02-25T05:10:55Z

Proof: png to normalize character size + punctuation conventions

← Older revision		Revision as of 05:10, 25 February 2009
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	<section begin=beginner/>		<section begin=beginner/>
	==Proof==		==Proof==
	''Given'': Let <math>H_i\!</math> be an arbitrary collection of subgroups of a [[group]] <math>G\!</math> indexed by <math>i \in I</math>. Let us denote <math>H = \textstyle\bigcap_{i \in I} H_i</math>. Here, <math>e\!</math> denotes the identity element of <math>G\!</math>.		''Given'': Let <math>H_i\!</math> be an arbitrary collection of subgroups of a [[group]] <math>G\!</math> indexed by <math>i \in I.</math> Let us denote <math>H = \textstyle\bigcap_{i \in I} H_i.</math> Here, <math>e\!</math> denotes the identity element of <math>G.\!</math>

	''To prove'': We need to show that <math>H</math> is a subgroup. In other words, we need to show the following:		''To prove'': We need to show that <math>H</math> is a subgroup. In other words, we need to show the following:
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	# <math>e \in H</math>		# <math>e \in H</math>
	# If <math>g \in H</math> then <math>g^{-1} \in H</math>		# If <math>g \in H</math> then <math>g^{-1} \in H</math>
	# If <math>g,h \in H</math> then <math>gh \in H</math>		# If <math>g, h \in H</math> then <math>gh \in H</math>

	''Proof'': Let's prove these one by one:		''Proof'': Let's prove these one by one:

	# Since <math>e \in H_i</math> for every <math>i</math>, <math>e \in H</math>		# Since <math>e \in H_i</math> for every <math>i,\!</math> <math>e \in H.</math>
	# Take <math>g \in H</math>. Then <math>g \in H_i</math> for every <math>i \in I</math>. Since each <math>H_i</math> is a subgroup, <math>g^{-1} \in H_i</math> for each <math>i \in I</math>. Thus, <math>g^{-1} \in H</math>.		# Take <math>g \in H</math>. Then <math>g \in H_i</math> for every <math>i \in I.</math> Since each <math>H_i\!</math> is a subgroup, <math>g^{-1} \in H_i</math> for each <math>i \in I.</math> Thus, <math>g^{-1} \in H.</math>
	# Take <math>g,h \in H</math>. Then <math>g,h \in H_i</math> for every <math>i</math>, so <math>gh \in H_i</math> for every <math>i \in I</math>. Thus <math>gh \in H</math>.		# Take <math>g, h \in H.</math> Then <math>g, h \in H_i</math> for every <math>i,\!</math> so <math>gh \in H_i</math> for every <math>i \in I.</math> Thus <math>gh \in H.</math>
	<section end=beginner/>		<section end=beginner/>

Jon Awbrey: /* Proof */ png + \textstyle

2009-02-25T05:03:03Z

Proof: png + \textstyle

← Older revision		Revision as of 05:03, 25 February 2009
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	<section begin=beginner/>		<section begin=beginner/>
	==Proof==		==Proof==
	''Given'': Let <math>H_i</math> be an arbitrary collection of subgroups of a [[group]] <math>G</math> indexed by <math>i \in I</math>. Let us denote <math>H = \bigcap_{i \in I} H_i</math>. Here, <math>e</math> denotes the identity element of <math>G</math>.		''Given'': Let <math>H_i\!</math> be an arbitrary collection of subgroups of a [[group]] <math>G\!</math> indexed by <math>i \in I</math>. Let us denote <math>H = \textstyle\bigcap_{i \in I} H_i</math>. Here, <math>e\!</math> denotes the identity element of <math>G\!</math>.

	''To prove'': We need to show that <math>H</math> is a subgroup. In other words, we need to show the following:		''To prove'': We need to show that <math>H</math> is a subgroup. In other words, we need to show the following:

Jon Awbrey: /* Symbolic statement */

2009-02-25T05:01:04Z

Symbolic statement

← Older revision		Revision as of 05:01, 25 February 2009
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	Let <math>H_i\!</math> be an arbitrary collection of subgroups of a [[group]] <math>G\!</math> indexed by <math>i \in I</math>. Then, <math>\textstyle\bigcap_{i \in I} H_i</math> is again a subgroup of <math>G\!</math>.		Let <math>H_i\!</math> be an arbitrary collection of subgroups of a [[group]] <math>G\!</math> indexed by <math>i \in I</math>. Then, <math>\textstyle\bigcap_{i \in I} H_i</math> is again a subgroup of <math>G\!</math>.

	Note that if the collection <math>I</math> is ''empty'', the intersection is defined to be the whole group. In this case, the intersection is clearly a subgroup. It should be noted that the case of an empty intersection is covered in the language of the general proof.		Note that if the collection <math>I\!</math> is ''empty'', the intersection is defined to be the whole group. In this case, the intersection is clearly a subgroup. It should be noted that the case of an empty intersection is covered in the language of the general proof.
	<section end=beginner/>		<section end=beginner/>

Jon Awbrey: /* Symbolic statement */ testing PNG and \textstyle

2009-02-25T05:00:16Z

Symbolic statement: testing PNG and \textstyle

← Older revision		Revision as of 05:00, 25 February 2009
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	===Symbolic statement===		===Symbolic statement===

	Let <math>H_i</math> be an arbitrary collection of subgroups of a [[group]] <math>G</math> indexed by <math>i \in I</math>. Then, <math>\bigcap_{i \in I} H_i</math> is again a subgroup of <math>G</math>.		Let <math>H_i\!</math> be an arbitrary collection of subgroups of a [[group]] <math>G\!</math> indexed by <math>i \in I</math>. Then, <math>\textstyle\bigcap_{i \in I} H_i</math> is again a subgroup of <math>G\!</math>.

	Note that if the collection <math>I</math> is ''empty'', the intersection is defined to be the whole group. In this case, the intersection is clearly a subgroup. It should be noted that the case of an empty intersection is covered in the language of the general proof.		Note that if the collection <math>I</math> is ''empty'', the intersection is defined to be the whole group. In this case, the intersection is clearly a subgroup. It should be noted that the case of an empty intersection is covered in the language of the general proof.

Vipul at 12:50, 2 November 2008

2008-11-02T12:50:32Z

← Older revision		Revision as of 12:50, 2 November 2008
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	===Other facts about combining subgroups in different ways===		===Other facts about combining subgroups in different ways===

	* [[Union of two subgroups is not a subgroup]], ~~unless one~~ of ~~them~~ is contained in the other.		* [[Union of two subgroups is not a subgroup unless they are comparable]]: If we have two subgroups of a group, neither of which is contained in the other ,their union is not a subgroup.
	* [[Directed union of subgroups is subgroup]]: In particular, the union of an ascending chain of subgroups of a group is again a subgroup.		* [[Directed union of subgroups is subgroup]]: In particular, the union of an ascending chain of subgroups of a group is again a subgroup.

Vipul at 23:23, 1 November 2008

2008-11-01T23:23:27Z

@@ Line 1: / Line 1: @@
 {{basic fact}}
-<section begin=main/>
+<section begin=beginner/>
 ==Statement==
@@ Line 12: / Line 12: @@
 Let <math>H_i</math> be an arbitrary collection of subgroups of a [[group]] <math>G</math> indexed by <math>i \in I</math>. Then, <math>\bigcap_{i \in I} H_i</math> is again a subgroup of <math>G</math>.
 ==Proof==
-''Given'': Let <math>H_i</math> be an arbitrary collection of subgroups of a [[group]] <math>G</math> indexed by <math>i \in I</math>. Let us denote <math>H = \bigcap_{i \in I} H_i</math>.
+''Given'': Let <math>H_i</math> be an arbitrary collection of subgroups of a [[group]] <math>G</math> indexed by <math>i \in I</math>. Let us denote <math>H = \bigcap_{i \in I} H_i</math>. Here, <math>e</math> denotes the identity element of <math>G</math>.
 ''To prove'': We need to show that <math>H</math> is a subgroup. In other words, we need to show the following:
@@ Line 26: / Line 65: @@
 # Take <math>g \in H</math>. Then <math>g \in H_i</math> for every <math>i \in I</math>. Since each <math>H_i</math> is a subgroup, <math>g^{-1} \in H_i</math> for each <math>i \in I</math>. Thus, <math>g^{-1} \in H</math>.
 # Take <math>g,h \in H</math>. Then <math>g,h \in H_i</math> for every <math>i</math>, so <math>gh \in H_i</math> for every <math>i \in I</math>. Thus <math>gh \in H</math>.
-<section end=main/>
+<section end=beginner/>
 ==References==