# Symmetric group on infinite coinfinite subset is not conjugacy-closed

From Groupprops

This article gives the statement, and possibly proof, of a particular subgroup or type of subgroupnotsatisfying a particular subgroup property (namely, Conjugacy-closed subgroup (?)) in a particular group or type of group (namely, Symmetric group (?)).

## Statement

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 .

## Proof

(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 .

**Proof**:

- 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 .