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Consider a permutation group $G$ acting on an infinite set $X$. Assume $G$ has finitely many orbits, and every point stabiliser $G_x$ has finite orbits.

Can we always find a permutation $\tau\in\operatorname{Sym}(X)$ (not necessarily of finite order or finite support) such that $H=\langle G,\tau\rangle$ is transitive, while every point stabiliser $H_x$ still has finite orbits?

As noted in this previous question some obvious choice of $\tau$ will not work in general.

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  • $\begingroup$ Following from my previous answer, the answer is yes if $G$ acts freely on $X$; the argument extends to the case when all $G$-orbits in $X$ are isomorphic as $G$-sets (equivalently, when all point stabilizers are conjugate to each other): choose a subset of $X$ meeting each orbit once, such that all its elements have the same stabilizer, and add a cycle through these points. $\endgroup$ – YCor Sep 28 '18 at 16:09
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    $\begingroup$ It's a great question, at the moment I have little idea whether it's true, even in seemly easy test-cases (e.g., $G$ acting on $(G/H)\sqcup (G/K)$ for two finite non-conjugate subgroups $H,K$). $\endgroup$ – YCor Sep 29 '18 at 10:49
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It seems that a counterexample looks as follows (it is somewhat siimilar with the counterexample to your previous question). Let $k$ be a large integer. Take an infinite tree $T=(V,E)$, where all degrees equal $k+1$. Let $G$ be its group of automorphisms. $G$ acts on $V\cup E$ with two obvious orbits, and all orbits of the stabilizers are finite.

I could find only quite technical proof that this example works; prehaps, there are easier ones.

Define the metric $d(\cdot,\cdot)$ on $V\cup E$ identifying each edge with its midpoint (the length of every edge is $1$). E.g., the distance between a vertex and an incident edge is $1/2$.

Assume that we took a transitive group $H\geqslant G$ with finite stabilizers' orbits. Let $G_v$ and $G_e$ be stabilizers of a vertex $v$ and an edge $e$ in $G$, and $H_v$ and $H_e$ be those in $H$. Then $H_e$ and $H_v$ are conjugates, so there is a cardinality-preserving bijection of their sets of orbits. Say that the radius of $H_v$-orbit is the maximal distance from $v$ to its element; the same for $H_e$-orbits.

$H_v$-orbits are unions of $G_v$-orbits which have cardinalities $(k+1),(k+1),k(k+1),k(k+1),\dots,k^n(k+1),k^n(k+1),\dots$. Similarly, $H_e$-orbits are unions of $G_e$-orbits whose cardinalities are $2,2k,2k,2k^2,\dots,2k^n,2k^n,\dots$ (notice here that an $H_v$-, and hence $H_e$-, orbit cannot have just $2$ elements). So the cardinality of any $H_v$-orbit $\Omega_v$ has the form either $k^n+O(k^{n-1})$ or $2k^n+O(k^{n-1})$, and that of the corresponding $H_e$-orbit $\Omega_e$ may have the form either $2k^n+O(k^{n-1})$ or $4k^n+O(k^{n-1})$. Hence this (common) cardinality is $2k^n+O(k^{n-1})$. This means that the radius of $\Omega_v$ is an integer $r$< and that of $\Omega_e$ is either $r$ or $r+1/2$.

Case 1. $\Omega_v$ and $\Omega_e$ have the same radius $r$. Consider now some $\tau\in H$ mapping $v$ to $e$; every $H_v$-orbit $\Omega_v$ is mapped to some $H_e$-orbit $\Omega_e$ of the same radius $r$. Notice that a dominating part $\partial\Omega_v$ of $\Omega_v$ consists of far vertices $v'$ with $d(v,v')=r$ and far edges $e'$ with $d(v,e')=r-1/2$; similarly, a dominating part $\partial \Omega_e$ of $\Omega_e$ consists of far edges $e'$ with $d(e,e')=r$. So most of such permutations $\tau$ map a good proportion of far vertces and edges (for $v$) to far edges (for $e$). See footnote for the explanation of the term `most of'.

Now consider some orbit' radius $r$. Take a vertex $v$ ans some $\tau$ mapping $v\mapsto e$. Let $\Omega_v$ and $\Omega_e$ be $H_v$- and $H_e$-orbits of radius $r$. Under $\tau$, most of $k^{r-1}(k+1)$ far edges for $v$ map to far edges for $e$.

Let $e'$ be an edge incident to $v$, and let $\Omega_{e'}$ he $H_{e'}$-orbit of radius $r$. Notice that $k^r$ of far edges for $e'$ are also far edges for $v$. So, most of them also map to far edges/vertices for $\tau(e')$, so thare are almost $k^r$ common far edges for $e$ and $\tau(e')$. This may happen only if $\tau(e')$ is a vertex incident to $e$. But $e'$ can be chosen in $k+1$ ways, while $e$ has only two endpoints. A contradiction.

Case 2. Assume that the radius of $\Omega_e$ is $r+1/2$. By symmetry, we may assume that $d(v,e)=r+1/2$, so that $v\in\Omega_e$. Due to the cardinalities, $\Omega_e$ should contain also either all vertices at distance $r-1/2$ from $e$, or all edges $e'$ at distance $r-1$ from $e$. In the former case, we get that an edge at distance $r-1/2$ and an edge at distance $r+1/2$ are equivalent modulo $H_v$, so $e\in\Omega_v$, which is impossible. In the latter case, we have $\sigma\in H$ with $\sigma(e')=v$, $\sigma(e)=e$. But, since $e$ lies in $\Omega_{e'}$ corresponding to $\Omega_e$, we get $e\in\Omega_v$ again.

Footnote. "Most of permutations $\tau$" is regarded in the following sense. If we consider all permutations $\tau$ mapping some $x\mapsto y$, there are only finitely many $t$ into which may a fixed $z$ map. If we are interested in images of a finite set of such $z$'s, there are finitely many tuples of images, and they are `equally distributed' (since all $\tau$ realizing one tuple form a coset of their joint stabilizer). Now we may speak on the probability in this sense.

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  • $\begingroup$ It does not seem to matter, but you also have a $G_v$ orbit of cardinal 1 and a $G_e$-orbit of cardinal 1. $\endgroup$ – YCor Oct 30 '18 at 13:24
  • $\begingroup$ @YCor: That's true. I omitted these trivial cases, as they cannpt glue to anything else. $\endgroup$ – Ilya Bogdanov Oct 30 '18 at 13:39

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