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.



Enlarging a subdegree-finite "almost transitive" permutation group to a transitive one?
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.


