Characteristic not implies elementarily characteristic: Difference between revisions

Revision as of 19:21, 10 January 2010

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 of a group $G$ that is not an elementarily characteristic subgroup of $G$ .

Proof

PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]

@@ Line 9: / Line 9: @@
 ==Proof==
-There are various examples involving symmetric groups on infinite sets. The key idea in all these examples is the [[Baer-Schreier-Ulam theorem]], which states that the characteristic subgroups are precisely the trivial subgroup, the whole group, the [[finitary alternating group]], the [[finitary symmetric group]], and the <math>\kappa</math>-ary symmetric groups for various cardinals <math>\kappa</math>.
+{{fillin}}
-The idea is that, for distinct cardinals smaller than the cardinality of the whole set but both infinite, we cannot distinguish between the corresponding <math>\kappa</math>-ary symmetric groups. ''Note: This still needs a rigorous proof! See [http://mathoverflow.net/questions/10919/elementary-equivalence-of-infinitary-symmetric-groups this page for a discussion].