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 $\mathrm{S}\mathrm{y}\mathrm{m}(A)$ denote the symmetric group on $A$ (the group of all permutations on $A$). Then, $\mathrm{S}\mathrm{y}\mathrm{m}(A)$ is trivial.

For the symmetric group and finitary symmetric group

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

Related facts

Stronger facts

• Finitary alternating group is centralizer-free in symmetric group

Corollaries

• Finitary symmetric group is automorphism-faithful in symmetric group

Proof

Given: A set $A$ with at least three elements. $G=\mathrm{S}\mathrm{y}\mathrm{m}(A)$ is the group of all permutations of $A$, and $H=\mathrm{F}\mathrm{S}\mathrm{y}\mathrm{m}(A)$ 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.