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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. Now consider a permutation $\tau\in\operatorname{Sym}(X)$ of finite order, and let $H=\langle G,\tau\rangle$. Is it necessarily true that every point stabiliser $H_x$ has finite orbits?

The situation I'm particularly interested in is when $\tau$ is a cycle with one element in each orbit of $G$, such that $H$ is transitive.

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No. Consider $G=\mathbf{Z}$ acting on itself by translation. Let $\tau$ be the transposition $(0,1)$. Then $H$ is the group of permutations of $\mathbf{Z}$ coinciding to translations at infinity; in particular it contains all finitely supported permutations; thus the stabilizer $H_0$ acts transitively on the complement of $\{0\}$.

It remains no in your more restricted setting (one cycle meeting each orbit once). Consider $G$ acting on $\mathbf{Z}$, generated by $\alpha:n\mapsto -n$ and $\beta: n\mapsto 2-n$. Thus $G$ is infinite dihedral, and has 2 orbits (odd and even numbers). Let $t$ be the transposition $(0,1)$ and $H=\langle G,t\rangle$. Then $\beta t\beta^{-1}$ is the transposition $(1,2)$ and hence also belongs to $H$. Given that $\beta\alpha$ is the translation $n\mapsto n+2$, we deduce that all transpositions $(n,n+1)$ belong to $H$ and hence $H$ contains all finitely supported permutations, and hence has infinite point stabilizers.

Note that the answer is yes when $G$ acts freely, and $t$ is an $n$-cycle meeting once each orbit. Indeed, this case, the $G$-conjugates of $t$ pairwise have disjoint support, and generate a group preserving the partition by these supports, which contains all point stabilizers.

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  • $\begingroup$ Thank you for the illuminating counterexamples! I was too optimistic thinking any $\tau$ would work. Might the title question still be salvable in the following way: for general $G$ with finitely many orbits and finite suborbits, can one always find an element $\tau$ (not necessarily of finite order) such that $\langle G,\tau\rangle$ is transitive with finite suborbits? $\endgroup$ – Jens Bossaert Sep 28 '18 at 11:40
  • $\begingroup$ You mean $\tau\notin G$? or some additional requirement? Also, do you still assume that $G$ has finite stabilizers? $\endgroup$ – YCor Sep 28 '18 at 12:22
  • $\begingroup$ Yes, any $\tau\in\operatorname{Sym}(X)$ but $\tau\notin G$ (otherwise $\langle G,\tau\rangle=G$ will not be transitive). The only assumptions on $G$ are (1) finitely many orbits and (2) finite suborbits (orbits of point stabilisers). $\endgroup$ – Jens Bossaert Sep 28 '18 at 15:29
  • $\begingroup$ Ok, this is a reasonable question but should be asked as a separate post. $\endgroup$ – YCor Sep 28 '18 at 15:31
  • $\begingroup$ Follow-up: mathoverflow.net/questions/311637 $\endgroup$ – YCor Oct 30 '18 at 15:52

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