Finitary symmetric group is not fully invariant in symmetric group
From Groupprops
This article gives the statement, and possibly proof, of a particular subgroup or type of subgroup (namely, Finitary symmetric group (?)) not satisfying a particular subgroup property (namely, Fully characteristic subgroup (?)) in a particular group or type of group (namely, Symmetric group (?)).
Contents
Statement
The Finitary symmetric group (?) on an infinite set is not fully characteristic as a subgroup of the symmetric group.
Related facts
- Finitary symmetric group is characteristic in symmetric group
- Finitary symmetric group is strictly characteristic in symmetric group
- Finitary symmetric group is not I-characteristic in symmetric group
Proof
Proof idea
We use the fact that an infinite set can be put in bijection with a union of countably many copies of that set. Using this bijection, we construct an endomorphism that sends each permutation to a permutation that acts like it on each of the countably many copies. This is an injective endomorphism, and under this endomorphism, every non-identity permutation gets mapped to a permutation that is not finitary. In particular, the finitary symmetric group is sent to a subgroup that it intersects trivially.