Finitary symmetric group is centralizer-free in symmetric group
This article gives the statement, and possibly proof, of a particular subgroup or type of subgroup (namely, Finitary symmetric group (?)) satisfying a particular subgroup property (namely, Centralizer-free subgroup (?)) in a particular group or type of group (namely, Symmetric group (?)).
For the symmetric group, plain
Let be a set with at least three elements. Let denote the symmetric group on (the group of all permutations on ). Then, is trivial.
For the symmetric group and finitary symmetric group
Let be a set with at least three elements. Let denote the symmetric group on (the group of all permutations on , and denote the subgroup comprising all finitary permutations -- permutations moving only finitely many elements. Then the centralizer of is is the trivial group. In other words, no non-identity permutation commutes with every finitary permutation.
Relation between the two formulations
The second formulation is stronger than the first. For finite sets, the two formulations are precisely the same.
Given: A set with at least three elements. is the group of all permutations of , and is the subgroup comprising finitary permutations.
To prove: is the trivial group.
Proof: Suppose and . Our goal is to show that .
Since has at least three elements, we can find a such that and . Let be the transposition . Then, we have:
On the other hand, we know that:
Thus, the transposition and the transposition are equal. This forces , so either or . The latter case was ruled out by choice of , so , completing the proof.