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.

  • $\begingroup$ I do not quite understand: the question is about discrete groups or not? $\endgroup$ – Mark Sapir Nov 28 '19 at 0:34
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    $\begingroup$ 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 $G$? $\endgroup$ – Derek Holt Nov 28 '19 at 9:10
  • $\begingroup$ @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! $\endgroup$ – Jens Bossaert Dec 5 '19 at 11:51
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    $\begingroup$ @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. $\endgroup$ – 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 $G$.


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