# 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 (?)).

## Statement

### For the symmetric group, plain

Let $A$ be a set with at least three elements. Let $\operatorname{Sym}(A)$ denote the symmetric group on $A$ (the group of all permutations on $A$). Then, $\operatorname{Sym}(A)$ is trivial.

### For the symmetric group and finitary symmetric group

Let $A$ be a set with at least three elements. Let $\operatorname{Sym}(A)$ denote the symmetric group on $A$ (the group of all permutations on $A$, and $\operatorname{FSym}(A)$ denote the subgroup comprising all finitary permutations -- permutations moving only finitely many elements. Then the centralizer of $\operatorname{FSym}(A)$ is $\operatorname{Sym}(A)$ 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.

## Proof

Given: A set $A$ with at least three elements. $G = \operatorname{Sym}(A)$ is the group of all permutations of $A$, and $H = \operatorname{FSym}(A)$ is the subgroup comprising finitary permutations.

To prove: $C_G(H)$ is the trivial group.

Proof: Suppose $\sigma \in C_G(H)$ and $a \in A$. Our goal is to show that $\sigma(a) = a$.

Since $A$ has at least three elements, we can find a $b \in A$ such that $b \ne a$ and $b \ne \sigma(a)$. Let $\tau$ be the transposition $(a,b)$. Then, we have:

$\sigma\tau\sigma^{-1} = \tau$.

On the other hand, we know that:

$\sigma\tau\sigma^{-1} = (\sigma(a),\sigma(b))$.

Thus, the transposition $(a,b)$ and the transposition $(\sigma(a),\sigma(b))$ are equal. This forces $\{a,b\} = \{ \sigma(a),\sigma(b)\}$, so either $a = \sigma(a)$ or $b = \sigma(a)$. The latter case was ruled out by choice of $b$, so $a = \sigma(a)$, completing the proof.