Difference between revisions of "Characteristic not implies elementarily characteristic"

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(Proof)
 
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==Statement==
 
==Statement==
  
It is possible to have a [[characteristic subgroup]] of a group <math>G</math> that is not an [[elementarily characteristic subgroup]] of <math>G</math>.
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It is possible to have a [[characteristic subgroup]] <math>H</math> of a group <math>G</math> that is not an [[elementarily characteristic subgroup]] of <math>G</math>, i.e., there exists a subgroup <math>K</math> of <math>G</math> such that the first-order theories of the group-subgroup pairs <math>(G,H)</math> and <math>(G,K)</math> are elementarily equivalent.
  
 
==Proof==
 
==Proof==
  
{{fillin}}
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Suppose <math>T_1, T_2</math> are two infinite sets with |T_1| > |T_2|</math>. Let <math>G</math> be the direct product <math>G_1 \times G_2</math> where <math>G_1</math> is the finitary alternating group on <math>T_1</math> and <math>G_2</math> is the finitary alternating group on <math>T_2</math>. Let <math>H</math> be the subgroup of <math>G</math> that is the first direct factor (<math>G_1 \times \{ e \}</math>) and <math>K</math> be the subgroup of <math>G</math> that is the second direct factor (<math>\{ e \} \times G_2</math>).
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* <math>H</math> is characteristic (in fact, homomorph-containing) in <math>G</math>: By cardinality considerations and the fact that <math>H</math> is simple, any homomorphic image of <math>H</math> in <math>G_2</math> is trivial. Therefore, the image of <math>H</math> under any homomorphism of <math>G</math> is contained in <math>H</math>. Therefore, <math>H</math> is characteristic in <math>G</math>.
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* The first-order theories of the group-subgroup pairs <math>(G,H)</math> and <math>(G,K)</math> are elementarily equivalent, so <math>H</math> is not elementarily characteristic in <math>G</math>: This can be seen from the idea that first-order statements reference only finite subsets, and as far as finite subsets are concerned, <math>G_1</math> and <math>G_2</math> look the same.

Latest revision as of 16:27, 1 June 2020

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., characteristic subgroup) need not satisfy the second subgroup property (i.e., elementarily characteristic subgroup)
View a complete list of subgroup property non-implications | View a complete list of subgroup property implications
Get more facts about characteristic subgroup|Get more facts about elementarily characteristic subgroup
EXPLORE EXAMPLES YOURSELF: View examples of subgroups satisfying property characteristic subgroup but not elementarily characteristic subgroup|View examples of subgroups satisfying property characteristic subgroup and elementarily characteristic subgroup

Statement

It is possible to have a characteristic subgroup H of a group G that is not an elementarily characteristic subgroup of G, i.e., there exists a subgroup K of G such that the first-order theories of the group-subgroup pairs (G,H) and (G,K) are elementarily equivalent.

Proof

Suppose T_1, T_2 are two infinite sets with |T_1| > |T_2|</math>. Let G be the direct product G_1 \times G_2 where G_1 is the finitary alternating group on T_1 and G_2 is the finitary alternating group on T_2. Let H be the subgroup of G that is the first direct factor (G_1 \times \{ e \}) and K be the subgroup of G that is the second direct factor (\{ e \} \times G_2).

  • H is characteristic (in fact, homomorph-containing) in G: By cardinality considerations and the fact that H is simple, any homomorphic image of H in G_2 is trivial. Therefore, the image of H under any homomorphism of G is contained in H. Therefore, H is characteristic in G.
  • The first-order theories of the group-subgroup pairs (G,H) and (G,K) are elementarily equivalent, so H is not elementarily characteristic in G: This can be seen from the idea that first-order statements reference only finite subsets, and as far as finite subsets are concerned, G_1 and G_2 look the same.