Marginal implies unconditionally closed: Difference between revisions
No edit summary |
|||
| Line 1: | Line 1: | ||
{{subgroup property implication| | |||
stronger = marginal subgroup| | |||
weaker = unconditionally closed subgroup}} | |||
==Statement== | ==Statement== | ||
Suppose <math>G</math> is a [[T0 | 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>. | ||
In particular, the result applies to the cases that <math>G</math> is a | In particular, the result applies to the cases that <math>G</math> is a [[Lie group]]. | ||
==Related facts== | ==Related facts== | ||
| Line 9: | Line 13: | ||
===Applications=== | ===Applications=== | ||
* [[Center is closed in T0 | * [[Center is closed in T0 topological group]] | ||
Revision as of 18:27, 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)
View all subgroup property implications | View all subgroup property non-implications
Get more facts about marginal subgroup|Get more facts about unconditionally closed subgroup
Statement
Suppose is a T0 topological group (i.e., a qtopological group whose underlying set is a T0 space) and is a marginal subgroup of as an abstract group. Then, is a closed subgroup of (i.e., it is a closed subset in the topological sense). In fact, is a closed normal subgroup of .
In particular, the result applies to the cases that is a Lie group.