Symmetric group on infinite coinfinite subset is not conjugacy-closed
This article gives the statement, and possibly proof, of a particular subgroup or type of subgroup not satisfying a particular subgroup property (namely, Conjugacy-closed subgroup (?)) in a particular group or type of group (namely, Symmetric group (?)).
Suppose are sets such that both and are infinite. Let denote the symmetric groups on and respectively, with embedding naturally in : any permutation on is extended to a permutation on by fixing all elements of .
With this embedding, is not a conjugacy-closed subgroup of .
(This proof assumes the axiom of choice).
Given: such that both and are finite.
To prove: The symmetric group is not conjugacy-closed in . In other words, there exist elements such that and are not conjugate in but are conjugate in .
- Let be a countably infinite subset of such that have the same size (this can be done if is infinite). Also note that there is a bijection between and . Combining these two, we get a bijection such that and .
- Suppose is a permutation of that moves every element of . Such a permutation exists: for instance, we can use the axiom of choice to partition into subsets of size two, and then select a permutation that interchanges the two elements in any subset.
- Consider the permutations and . is a permutation that moves every element of and fixes all elements of , while moves every element of and fixes every element of . We note that:
- and are not conjugate in , because has fixed points but doesn't have fixed points.
- and are conjugate in by .