Difference between revisions of "Lazard Lie property is not subgroup-closed"

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(Created page with "{{group metaproperty dissatisfaction| property = Lazard Lie group| metaproperty = subgroup-closed group property}} ==Statement== It is possible to have a [[Lazard Lie group]...")
 
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It is possible to have a [[Lazard Lie group]] <math>G</math> and a [[subgroup]] <math>H</math> of <math>G</math> that is ''not'' a [[Lazard Lie group]].
 
It is possible to have a [[Lazard Lie group]] <math>G</math> and a [[subgroup]] <math>H</math> of <math>G</math> that is ''not'' a [[Lazard Lie group]].
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==Related facts==
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===Similar facts===
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* [[LCS-Lazard Lie property is not subgroup-closed]]
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* [[Baer Lie property is not subgroup-closed]]
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===Opposite facts===
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* [[Powering-invariant subgroup of Lazard Lie group is Lazard Lie group]]
  
 
==Proof==
 
==Proof==

Latest revision as of 03:11, 3 August 2013

This article gives the statement, and possibly proof, of a group property (i.e., Lazard Lie group) not satisfying a group metaproperty (i.e., subgroup-closed group property).
View all group metaproperty dissatisfactions | View all group metaproperty satisfactions|Get help on looking up metaproperty (dis)satisfactions for group properties
Get more facts about Lazard Lie group|Get more facts about subgroup-closed group property|

Statement

It is possible to have a Lazard Lie group G and a subgroup H of G that is not a Lazard Lie group.

Related facts

Similar facts

Opposite facts

Proof

Further information: unitriangular matrix group:UT(3,Q), unitriangular matrix group:UT(3,Z)

Consider the example G = UT(3,\mathbb{Q}) and H = UT(3,\mathbb{Z}). G is a Lazard Lie group (in fact, a Baer Lie group). H is a group of class exactly two that is not 2-powered, hence, it is not a Lazard Lie group.