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.

  • $\begingroup$ 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, 2018 at 16:09
  • 3
    $\begingroup$ 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, 2018 at 10:49

1 Answer 1


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.

  • $\begingroup$ 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, 2018 at 13:24
  • $\begingroup$ @YCor: That's true. I omitted these trivial cases, as they cannpt glue to anything else. $\endgroup$ Oct 30, 2018 at 13:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.