# Enlarging a subdegree-finite “almost transitive” permutation group to a transitive one? (follow-up)

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.

• 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. – YCor Sep 28 '18 at 16:09
• 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 $H,K$). – YCor Sep 29 '18 at 10:49

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.

• 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. – YCor Oct 30 '18 at 13:24
• @YCor: That's true. I omitted these trivial cases, as they cannpt glue to anything else. – Ilya Bogdanov Oct 30 '18 at 13:39