Characteristic not implies injective endomorphism-invariant

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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., injective endomorphism-invariant 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 injective endomorphism-invariant subgroup
EXPLORE EXAMPLES YOURSELF: View examples of subgroups satisfying property characteristic subgroup but not injective endomorphism-invariant subgroup|View examples of subgroups satisfying property characteristic subgroup and injective endomorphism-invariant subgroup


Statement with symbols

It is possible to have a group G with a characteristic subgroup H that is not injective endomorphism-invariant: in other words, every automorphism of G sends H to itself, but every injective endomorphism of G does not send H to itself.

Facts used

  1. Finitary symmetric group is characteristic in symmetric group
  2. Finitary symmetric group is not injective endomorphism-invariant in symmetric group
  3. Center is characteristic
  4. Center not is injective endomorphism-invariant


Example of the finitary symmetric group

Let S be an infinite set. Let G = \operatorname{Sym}(S) be the symmetric group on S, and let H = \operatorname{FSym}(S) be the subgroup comprising finitary permutations. Then, H is characteristic in G (fact (1)) but is not I-characteristic in G (fact (2)).

Example of the center

Facts (3) and (4) give another kind of example: the center of a group is always characteristic, but need not be injective endomorphism-invariant.