Marginal implies unconditionally closed: Difference between revisions

Revision as of 18:28, 27 July 2013

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., marginal subgroup) must also satisfy the second subgroup property (i.e., unconditionally closed subgroup)
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Statement

Suppose $G$ is a T0 topological group (i.e., a topological group whose underlying set is a T0 space) and $H$ is a marginal subgroup of $G$ as an abstract group. Then, $H$ is a closed subgroup of $G$ (i.e., it is a closed subset in the topological sense). In fact, $H$ is a closed normal subgroup of $G$ .

In particular, the result applies to the cases that $G$ is a Lie group.

Related facts

Applications

Center is closed in T0 topological group

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 ==Statement==
-Suppose <math>G</math> is a [[T0 topological group]] (i.e., a [[qtopological group]] whose underlying set is a [[topospaces:T0 space|T0 space]]) and <math>H</math> is a [[fact about::marginal subgroup;1| ]][[marginal subgroup]] of <math>G</math> as an abstract group. Then, <math>H</math> is a [[closed subgroup]] of <math>G</math> (i.e., it is a closed subset in the topological sense). In fact, <math>H</math> is a [[closed normal subgroup]] of <math>G</math>.
+Suppose <math>G</math> is a [[T0 topological group]] (i.e., a [[topological group]] whose underlying set is a [[topospaces:T0 space|T0 space]]) and <math>H</math> is a [[fact about::marginal subgroup;1| ]][[marginal subgroup]] of <math>G</math> as an abstract group. Then, <math>H</math> is a [[closed subgroup]] of <math>G</math> (i.e., it is a closed subset in the topological sense). In fact, <math>H</math> is a [[closed normal subgroup]] of <math>G</math>.
 In particular, the result applies to the cases that <math>G</math> is a [[Lie group]].