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 (?)).

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 $\alpha$.