Assume that $G\leq\operatorname{Sym}(X)$ is a permutation group generated by all its point stabilisers, i.e. $G=\langle G\_x \mid x\in X\rangle$. There is no cardinality restriction on $X$. Furthermore, assume that $G$ has *finitely many* orbits, and that $G$ is subdegree-finite, i.e. all point stabilisers have only *finite* orbits.


Is $G$ then necessarily generated by *finitely many* point stabilisers?


The question arises in the context of the permutation topology. When we endow $G$ with the permutation topology, $G$ is a totally disconnected locally compact group when all $G\_x$ are compact, and I want to know if having finitely many orbits is then sufficient for $G$ to be compactly generated.



Permutation groups generated by finitely many point stabilisers
Let $A$ be an infinite abelian group of odd finite exponent. For example we could take $A$ to be the direct product of infinitely many copies of a cyclic group $C\_n$, with $n>1$ odd.


Let $\langle t \rangle$ be a cyclic group of order $2$, define $\phi:\langle t \rangle \to {\rm Aut}(A)$ by $\phi(t): a \mapsto a^{-1}\ (a \in A)$, and let $G = A \rtimes\_\phi \langle t \rangle$ be the associated semidirect product (so $tat^{-1}= a^{-1}$ for all $a \in A$).


Now consider the action of $G$ by left multiplication on the left cosetsof $\langle t \rangle$ in $G$. These have the form $a\langle t \rangle$ for $a \in A$, and the stabilizer in the action of this coset is the subgroup $a\langle t \rangle a^{-1}$


Now, for a subset $B$ of $A$, the subgroup of $G$ generated by the stabilizers of the cosets $\langle b \langle t \rangle$ is contained in the subgroup $\langle B,t \rangle$ of $G$ which is equal to $\langle B \rangle \langle t \rangle$ and has order $2\langle B \rangle$.


So, if $B$ is finite then the subgroup generated by these stabilizers is also finite, and hence $G$ cannot be generated by finitely many stabilizers. In fact we need $|A|$ stabilizers to generate $G$.


