Let $G$ be a group acting highly transitively (and faithfully) on a set $S$. Suppose that $G$ is finitely presented, and that every stabilizer in $G$ of a finite subset of $S$ is finitely generated. I think I can prove that $G$ embeds in a finitely presented simple group, which in particular implies $G$ has decidable word problem, but I'd like a better understanding of why such a $G$ should have decidable word problem. Is there a pre-existing (and/or more direct) reason that a group admitting such an action should have decidable word problem?

(Edit: Here an action of a group $G$ on a set $S$ is called *highly transitive* if for all $n\in\mathbb{N}$ the induced action of $G$ on the set of $n$-tuples of distinct elements of $S$ is transitive.)

uniformly. This places some constraints on what you might expect a solution to the word problem to look like. $\endgroup$10more comments