Let $\mathrm{SO}(3)$ be the group of rotations of $\mathbb{R}^3$ and let $S_\infty$ be the group of all permutations of $\mathbb{N}$. Is $\mathrm{SO}(3)$ isomorphic to a subgroup of $S_\infty$?

This question is due to Ulam. It is discussed in V.2 of Ulam's "A collection of mathematical problems". It is also discussed in de le Harpe's "Topics in Geometric Group Theory", Appendix III.B. According to de la Harpe it was open as of 2003. Is it still open?

Ulam also asks the more general question of whether every Lie group is isomorphic to a subgroup of $S_\infty$. De la Harpe constructs homomorphic embeddings $\mathbb{R} \to S_\infty$ and $\mathbb{R}/\mathbb{Z} \to S_\infty$. There is an obvious embedding $S_\infty \times S_\infty \to S_\infty$, it follows that any connected abelian Lie group is a subgroup of $S_\infty$. So it is natural to look at $\mathrm{SO}(3)$ or other low dimensional non-abelian Lie groups.

It is worth pointing out that we cannot hope to realize $\mathrm{SO}(3)$ as a subgroup of $S_\infty$ without using a lot of choice. Suppose that $G$ is a compact lie group and $f : G \to S_\infty$ is a homomorphism constructed without choice. Then $f$ is Baire measurable, so by the Pettis lemma $f$ is continuous, hence $f(G)$ is compact and connected. Finally, any compact connected subgroup of $S_\infty$ is trivial. (In general closed subgroups of $S_\infty$ are totally disconnected.) So a faithful action of $\mathrm{SO}(3)$ would be rather different than the usual kinds of actions.