Does every minimally transitive subgroup of the symmetric group on a countably infinite set have finite point stabiliser? Let $X$ be a countably infinite set with $x \in X$. Does every minimally transitive subgroup $H \leq$ Sym($X$) have $\vert \mbox{Stab}_{H}(x)\vert < \infty$?
A subgroup $H \leq$ Sym($X$) is transitive if for all $x,y \in X$ there exists $h \in H$ such that $h(x) = y$. A transitive subgroup $H \leq \mbox{Sym}(X)$ is minimally transitive if $K \leq H$ with $K$ transitive implies $K = H$.
For a subgroup $H \leq \mbox{Sym}(X)$ we define the stabiliser $\mbox{Stab}_{H}(x) = \{h \in H \vert h(x) = x\}$.
Trivial minimally transitive subgroups $H \leq \mbox{Sym}(X)$ are the regular subgroups, these are ones such that for all $x, y \in X$ there exists a unique element $h \in H$ such that $h(x) = y$. These are all realised by countable groups, i.e. if $G$ is a countable group then by Cayleys theorem there is a natural map $G \rightarrow Sym(G)$ given by the right regular action. These all have trivial stabilisers. However I was wondering if this extends to minimally transitive subgroups, the idea being that if you had an infinite stabiliser you should be able to drop to a smaller subgroup contained in that transitive subgroup cutting out some of the stabiliser.
I thought to construct a counter example by making a transitive subgroup with stabiliser $\mathbb{Z}$ however whenever I try to do this, I generate some junk in my stabiliser. E.g, $\langle (1,2), (2,3,4, \ldots) \rangle$. 
 A: I think there are examples of minimally transitive groups with infinite point stabilizers.
If $G$ is a countably infinite group and $x \in G$ has infinite order, then we could consider the permutation action of $G$ on the right cosets of the subgroup $\langle x \rangle$. A proper transitive subgroup would be a proper subgroup $H$ with $\langle x \rangle H = G$. If you chose $G$ to have no proper subgroups of finite index, then we would have $\langle x \rangle \cap H = 1$.
I am sure that there must be more elementary examples, but when you can't think of anything better, you can always try Tarski Monsters. There are such monsters in which all proper nontrivial subgroups are infinite cyclic (see discussion here). In such a group $H$ would be infinite cyclic.
Now Ito proved that, if a group $G$ is the product of two abelian subgroups, then $G$ is metabelian (i.e. solvable of derived length at most $2$). The proof of this result is clever but elementary. The original reference (in German for some reason) is N. Ito, Über das Produkt von zwei abelschen Gruppen, Math. Z. 62 (1955), 400-401.
Now it easy to see that noncyclic solvable groups have proper subgroups that are not infinite cyclic, so we have a contradiction. Hence $H$ cannot exist and $G$ is minimally transitive.
