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##I.Motivation from descriptive set theory##

##II.The problem## Let $G\leq S_{\infty}$ be a permutation group acting on $\mathbb{N}$ (starts from 1 for convenience). Define $G_n$ to be the n-point stabilizer, $G_n=\{g\in G|g(i)=i,1\leq i\leq n\}$, and $G_0=G$. We represent an infinite orbit $O$ of $G_n$ by a node $N$ on level $n$ (finite orbits do not count).

##III.Some attempts##

##IV.Possible modifications of the problem (not strict)##

I.Motivation from descriptive set theory

II.The problem

Let $G\leq S_{\infty}$ be a permutation group acting on $\mathbb{N}$ (starts from 1 for convenience). Define $G_n$ to be the n-point stabilizer, $G_n=\{g\in G|g(i)=i,1\leq i\leq n\}$, and $G_0=G$. We represent an infinite orbit $O$ of $G_n$ by a node $N$ on level $n$ (finite orbits do not count).

III.Some attempts

IV.Possible modifications of the problem (not strict)

Assuming $G$ is transitive, this process defines a map from $G$ to a rooted tree $T_G \in\omega^{<\omega}$. We say it is the orbit tree of $G$ and it roughly describes how the orbits of $G$ split when we keep fixing more and more points. The natural inverse question is

For every tree $T\in\omega^{<\omega}$, is there a group $G$ such that the orbit tree $T_G$ is isomorphic to $T$? Can we construct it?

Assuming $G$ is transitive, this process defines a map from $G$ to a rooted tree $T_G \subset\omega^{<\omega}$. We say it is the orbit tree of $G$ and it roughly describes how the orbits of $G$ split when we keep fixing more and more points. The natural inverse question is

For every tree $T\subset\omega^{<\omega}$, is there a group $G$ such that the orbit tree $T_G$ is isomorphic to $T$? Can we construct it?

Blockquote

1.If $G$ is a permutation group on $\Omega$ and $\alpha\in\Omega$ and the orbits of $G_{\alpha}$ are $\Omega_{\lambda}(\lambda\in\Lambda)$ then $G_{\alpha}$ is a subdirect product of the restrictions $H_{\lambda}:=G_{\alpha}^{\Omega_{\lambda}}$. Specifically $G_{\alpha}$ can be embedded in the Cartesian product $\prod_{\lambda}H_{\lambda}$ such that the projections onto each factor is surjective.

2.Now let $G$ be a free group of countably infinite rank. We want to find a corefree subgroup $H$ of $G$ of countable index such that $H$ is free of infinite rank and has a specified number of double cosets $HxH$ for which $|H:x^{-1}Hx|$ is infinite($H$ is going to play the role of $G_{\alpha}$). We know that every subgroup of a free group is free and many(most?) of them have infinite rank when $G$ has infinite rank. I don't think it is that hard to satisfy the condition about the number of cosets. If this is true then we can go from level 0 to level 1.

3.To go to level 2 we know that $H$ is a subdirect product of the $H_{\lambda}$. Suppose that all the infinite $H_{\lambda}$ are free of infinite rank and that the embedding of $H$ induces an isomorphism onto each of these. Now for each infinite $H_{\lambda}$ choose (corefree, infinite rank) $K_{\lambda}$ so that $H_{\lambda}$ and $K_{\lambda}$ play the roles of $G$ and $H$ in 2 to provide the correct number of infinite orbits at the next level. Can we choose the isomorphisms $H\rightarrow H_{\lambda}$ and a subgroup $K$ of $H$ such that $K$ maps onto $K_{\lambda}$ for each $\lambda$?

4.If the answers to these questions are yes, I think that you can then prove what you want.

Blockquote

1. If $G$ is a permutation group on $\Omega$ and $\alpha\in\Omega$ and the orbits of $G_{\alpha}$ are $\Omega_{\lambda}(\lambda\in\Lambda)$ then $G_{\alpha}$ is a subdirect product of the restrictions $H_{\lambda}:=G_{\alpha}^{\Omega_{\lambda}}$. Specifically $G_{\alpha}$ can be embedded in the Cartesian product $\prod_{\lambda}H_{\lambda}$ such that the projections onto each factor are surjective.

2. Now let $G$ be a free group of countably infinite rank. We want to find a corefree subgroup $H$ of $G$ of countable index such that $H$ is free of infinite rank and has a specified number of double cosets $HxH$ for which $|H:x^{-1}Hx|$ is infinite ($H$ is going to play the role of $G_{\alpha}$). We know that every subgroup of a free group is free and many (most of them?) have infinite rank when $G$ has infinite rank. I don't think it is that hard to satisfy the condition about the number of cosets. If this is true then we can go from level 0 to level 1.

3. To go to level 2 we know that $H$ is a subdirect product of the $H_{\lambda}$. Suppose that all the infinite $H_{\lambda}$ are free of infinite rank and that the embedding of $H$ induces an isomorphism onto each of these. Now for each infinite $H_{\lambda}$ choose (corefree, infinite rank) $K_{\lambda}$ so that $H_{\lambda}$ and $K_{\lambda}$ play the roles of $G$ and $H$ in 2 to provide the correct number of infinite orbits at the next level. Can we choose the isomorphisms $H\rightarrow H_{\lambda}$ and a subgroup $K$ of $H$ such that $K$ maps onto $K_{\lambda}$ for each $\lambda$?

4.If the answers to these questions are yes, I think that you can then prove what you want.