# Permutation groups generated by finitely many point stabilisers

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.

• I do not quite understand: the question is about discrete groups or not? – Mark Sapir Nov 28 '19 at 0:34
• For a counterexample, can't we take $G = A \rtimes \langle t \rangle$ with $t^2=1$, $A$ infinite abelian of odd finite exponent, and $t^{-1}at=a^{-1}$ for all $a \in A$, with action on the set $X$ of cosets of $\langle t \rangle$ in $G$. Then $G$ is transitive on $X$, all point stabilizers have order $2$, and we need $|A|$ point stabilizers to generate $G$? – Derek Holt Nov 28 '19 at 9:10
• @DerekHolt, yes, I think that's a valid (and interesting) counterexample. If you want to elaborate on it in an answer, I will accept and close this question. Thank you! – Jens Bossaert Dec 5 '19 at 11:51
• @MarkSapir the question is very clear, and is indeed about abstract groups. Then the last paragraph is a motivation, and addresses some group topology on the given group. – YCor Dec 5 '19 at 12:49

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 $$G$$.