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5.40 (1976)

Solved

Let $G$ be a countable group acting on a set $\Omega$. Suppose that $G$ is $k$-fold transitive for every finite $k$, and $G$ contains no non-trivial permutations of finite support. Is it true that $\Omega$ can be identified with the rational line $\mathbb{Q}$ in such a way that $G$ becomes a group of autohomeomorphisms?

Contributor: P. M. Neumann

Progress

Not always (A. H. Mekler, J. London Math. Soc., 33 (1986), 49–58).

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