# Difference between revisions of "Characteristic not implies elementarily characteristic"

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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>. | + | 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== | ||

− | {{ | + | 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>). |

+ | |||

+ | * <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>. | ||

+ | * 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) neednotsatisfy 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 of a group that is not an elementarily characteristic subgroup of , i.e., there exists a subgroup of such that the first-order theories of the group-subgroup pairs and are elementarily equivalent.

## Proof

Suppose are two infinite sets with |T_1| > |T_2|</math>. Let be the direct product where is the finitary alternating group on and is the finitary alternating group on . Let be the subgroup of that is the first direct factor () and be the subgroup of that is the second direct factor ().

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