Closed subgroup of finite index implies open: Difference between revisions

Latest revision as of 23:32, 23 June 2012

Statement

Statement for left-topological, right-topological, or semitopological groups

In a left-topological group or right-topological group, any closed subgroup of finite index (i.e., a closed subgroup that is also a subgroup of finite index) must be an open subgroup.

Note that a semitopological group is both a left-topological group and a right-topological group, so the result applies to semitopological groups.

Statement for topological groups

In a topological group, any closed subgroup of finite index (i.e., a closed subgroup that is also a subgroup of finite index) must be an open subgroup.

Note that topological groups are semitopological groups, so the result applies to these.

Related facts

Proof

Proof for left-topological groups

Given: A right-topological group $G$ , a closed subgroup $H$ of finite index in $G$ .

To prove: $H$ is an open subgroup of $G$

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	For all $g\in G$ , the map $G\to G$ given by $x\mapsto gx$ is a self-homeomorphism of $G$ .	Definition of left-topological group	$G$ is a left-topological group.		[SHOW MORE] From the definition of left-topological group, the multiplication map $G\times G\to G$ is continuous in its second input. The map $x\mapsto gx$ is a continuous map. Further, its inverse is the map $x\mapsto g^{-1}x$ . Thus, $x\mapsto gx$ is a homeomorphism for any $g\in G$ .
2	Every left coset of $H$ in $G$ is a closed subset of $G$ .	Homeomorphisms take closed subsets to closed subsets		Step (1)	By Step (1), $x\mapsto gx$ is a self-homeomorphism of $G$ , so it takes the closed subset $H$ to the closed subset $gH$ . Thus, for any $g\in G$ , $gH$ is closed in $G$ .
3	The union of all the left cosets of $H$ other than $H$ itself is closed in $G$	Union of finitely many closed subsets is closed	$H$ has finite index in $G$	Step (2)	Step-fact combination direct.
4	$H$ is open in $G$	A subset is open iff its set-theoretic complement is closed.		Step (3)	The set-theoretic complement of $H$ in $G$ is precisely the union of all the left cosets other than $H$ itself, and by Step (3), this is closed. Hence, $H$ is open.

Proof for right-topological groups

The proof is analogous to the proof for left-topological groups, except that we use right cosets instead of left cosets.

@@ Line 1: / Line 1: @@
 ==Statement==
-In a [[topological group]], any [[fact about::closed subgroup of finite index;1| ]][[uses property satisfaction of::closed subgroup of finite index]] (i.e., a [[fact about::closed subgroup;1| ]][[uses property satisfaction of::closed subgroup]] that is also a [[fact about::subgroup of finite index;2| ]][[uses property satisfaction of::subgroup of finite index]]) must be an [[fact about::open subgroup;1| ]][[uses property satisfaction of::open subgroup]].
+===Statement for left-topological, right-topological, or semitopological groups===
+In a [[left-topological group]] ''or'' [[right-topological group]], any [[fact about::closed subgroup of finite index;1| ]][[uses property satisfaction of::closed subgroup of finite index]] (i.e., a [[fact about::closed subgroup;1| ]][[uses property satisfaction of::closed subgroup]] that is also a [[fact about::subgroup of finite index;2| ]][[uses property satisfaction of::subgroup of finite index]]) must be an [[fact about::open subgroup;1| ]][[proves property satisfaction of::open subgroup]].
+Note that a semitopological group is both a left-topological group and a right-topological group, so the result applies to semitopological groups.
+===Statement for topological groups===
+In a [[topological group]], any [[fact about::closed subgroup of finite index;1| ]][[uses property satisfaction of::closed subgroup of finite index]] (i.e., a [[fact about::closed subgroup;1| ]][[uses property satisfaction of::closed subgroup]] that is also a [[fact about::subgroup of finite index;2| ]][[uses property satisfaction of::subgroup of finite index]]) must be an [[fact about::open subgroup;1| ]][[proves property satisfaction of::open subgroup]].
+Note that topological groups are semitopological groups, so the result applies to these.
 ==Related facts==
@@ Line 11: / Line 20: @@
 ==Proof==
-===Proof details===
+===Proof for left-topological groups===
-'''Given''': A [[topological group]] <math>G</math>, a [[closed subgroup]] <math>H</math> of finite index in <math>G</math>.
+'''Given''': A right-topological group <math>G</math>, a [[closed subgroup]] <math>H</math> of finite index in <math>G</math>.
 '''To prove''': <math>H</math> is an open subgroup of <math>G</math>
@@ Line 21: / Line 31: @@
 ! Step no. !! Assertion/construction !! Facts used !! Given data used !! Previous steps used !! Explanation
 |-
-| 1 || For all <math>g \in G</math>, the map <math>G \to G</math> given by <math>x \mapsto gx</math> is a self-homeomorphism of <math>G</math>. || Definition of topological group || <math>G</math> is a topological group. || || <toggledisplay>From the definition of topological group, the multiplication map <math>G \times G \to G</math> is continuous. The map <math>x \mapsto gx</math> is a composite of the fiber inclusion <math>x \mapsto (g,x)</math> and the multiplication map, hence it is a continuous map. Further, its inverse is the map <math>x \mapsto g^{-1}x</math>. Thus, <math>x \mapsto gx</math> is a homeomorphism for any <math>g \in G</math>.</toggledisplay>
+| 1 || For all <math>g \in G</math>, the map <math>G \to G</math> given by <math>x \mapsto gx</math> is a self-homeomorphism of <math>G</math>. || Definition of left-topological group || <math>G</math> is a left-topological group. || || <toggledisplay>From the definition of left-topological group, the multiplication map <math>G \times G \to G</math> is continuous in its second input. The map <math>x \mapsto gx</math> is a continuous map. Further, its inverse is the map <math>x \mapsto g^{-1}x</math>. Thus, <math>x \mapsto gx</math> is a homeomorphism for any <math>g \in G</math>.</toggledisplay>
 |-
 | 2 || Every [[left coset]] of <math>H</math> in <math>G</math> is a closed subset of <math>G</math>. || Homeomorphisms take closed subsets to closed subsets || || Step (1) || By Step (1), <math>x \mapsto gx</math> is a self-homeomorphism of <math>G</math>, so it takes the closed subset <math>H</math> to the closed subset <math>gH</math>. Thus, for any <math>g \in G</math>, <math>gH</math> is closed in <math>G</math>.
@@ Line 29: / Line 39: @@
 | 4 || <math>H</math> is open in <math>G</math> || A subset is open iff its set-theoretic complement is closed.|| || Step (3) || The set-theoretic complement of <math>H</math> in <math>G</math> is precisely the union of all the left cosets other than <math>H</math> itself, and by Step (3), this is closed. Hence, <math>H</math> is open.
 |}
+===Proof for right-topological groups===
+The proof is analogous to the proof for left-topological groups, except that we use right cosets instead of left cosets.