# Enlarging a subdegree-finite “almost transitive” permutation group to a transitive one?

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.

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.
• 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? – Jens Bossaert Sep 28 '18 at 11:40
• You mean $\tau\notin G$? or some additional requirement? Also, do you still assume that $G$ has finite stabilizers? – YCor Sep 28 '18 at 12:22
• 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). – Jens Bossaert Sep 28 '18 at 15:29