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.



Does $\mathrm{SO}(3)$ act faithfully on a countable set?
From MathSciNet:


Thomas, Simon Infinite products of finite simple groups. II.
J. Group Theory 2 (1999), no. 4, 401–434.


Summary: "(...) In the course of our classification proof, we also show that if $K$ is a field of cardinality $2^\omega$ and $G$ is a non-trivial linear group over $K$, then there exists a subgroup $H$ of $G$ such that $1<[G:H] \le \omega$."


[This is almost the same since enlarging $G$, we can always assume $G$ is simple, say $G=\mathrm{PSL}\_m(K)$, and then the induced homomorphism $G\to \mathrm{Sym}(G/H)$ is injective. The proof uses countable valuations as well.]




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Ershov, Yu. L. ; Churkin, V. A. On a problem of Ulam. (Russian)
Dokl. Akad. Nauk 399 (2004), no. 3, 307-309.


Theorem 1. Every linear group over a field with continuum cardinality (in particular over the field $\mathbf{R}$ of all real numbers or over the field $\mathbf{C}$ of all complex numbers) can be embedded (as an abstract group) in the permutation group of a countable set.


(Unlike S. Thomas, they were aware of Ulam's question. On the other hand they were visibly not aware of S. Thomas paper.)


