Symmetric group on infinite coinfinite subset is not conjugacy-closed

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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 (?)).

Statement

Suppose S \subset T are sets such that both S and T \setminus S are infinite. Let \operatorname{Sym}(S), \operatorname{Sym}(T) denote the symmetric groups on S and T respectively, with \operatorname{Sym}(S) embedding naturally in \operatorname{Sym}(T): any permutation on S is extended to a permutation on T by fixing all elements of T \setminus S.

With this embedding, \operatorname{Sym}(S) is not a conjugacy-closed subgroup of \operatorname{Sym}(T).

Proof

(This proof assumes the axiom of choice).

Given: S \subset T such that both S and T \setminus S are finite.

To prove: The symmetric group \operatorname{Sym}(S) is not conjugacy-closed in \operatorname{Sym}(T). In other words, there exist elements g,h \in \operatorname{Sym}(S) such that g and h are not conjugate in \operatorname{Sym}(S) but are conjugate in \operatorname{Sym}(T).

Proof:

  1. Let A be a countably infinite subset of S such that S \setminus A have the same size (this can be done if S is infinite). Also note that there is a bijection between T \setminus S and A \cup (T \setminus S). Combining these two, we get a bijection \alpha: S \to S such that \alpha(S) = S \setminus A and \alpha(T \setminus S) = A \cup (T \setminus S).
  2. Suppose f is a permutation of S that moves every element of S. Such a permutation exists: for instance, we can use the axiom of choice to partition S into subsets of size two, and then select a permutation that interchanges the two elements in any subset.
  3. Consider the permutations f and \alpha f \alpha^{-1}. f is a permutation that moves every element of S and fixes all elements of T \setminus S, while \alpha f \alpha^{-1} moves every element of S \setminus A and fixes every element of A \cup (T \setminus S). We note that:
    • f and \alpha f \alpha^{-1} are not conjugate in \operatorname{Sym}(S), because \alpha f \alpha^{-1} has fixed points but f doesn't have fixed points.
    • f and \alpha f \alpha^{-1} are conjugate in \operatorname{Sym}(T) by \alpha.