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.

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    $\begingroup$ $SO(3)$ embeds into $SO_3(\mathbb C)$. Using choice, fix an isomorphism between $\mathbb C$ and $\overline{\mathbb Q_p}$, giving an isomorphism between $SO_3(\mathbb C)$ and $SO_3(\overline{\mathbb Q_p})$. Then doesn't $SO_3(\overline{\mathbb Q_p})$ act faithfully on $SO_3(\overline{\mathbb Q_p})/SO_3(\overline{\mathbb Z_p})$, which is countable because it is the union over all finite extensions $K_p$ of $\mathbb Q_p$ with ring of integers $\mathcal O_{K_p}$ of the countable $SO_3(K_p)/ SO_3(\mathcal O_{K_p})$? $\endgroup$
    – Will Sawin
    Sep 6, 2021 at 0:15
  • $\begingroup$ Related: mathoverflow.net/q/269008/89334 $\endgroup$
    – Uri Bader
    Sep 10, 2021 at 13:32

2 Answers 2


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.]

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.)


This is possible (filling in some details in the argument from my comment).

We use choice to fix an isomorphism $\mathbb C \cong \overline{\mathbb Q_p}$. Using this, we embed $\mathrm{SO}(3) \subset \mathrm{SO}_3(\mathbb C) \cong \mathrm{SO}_3(\overline{\mathbb Q_p})$.

We let $\mathrm{SO}_3 (\overline{\mathbb Q_p})$ act on the set $\mathrm{SO}_3 (\overline{\mathbb Q_p})/\mathrm{SO}_3( \overline{\mathbb Z_p})$, where $\overline{\mathbb Z_p}$ consists of the elements of $\overline{\mathbb Q_p}$ that are integral over $\mathbb Z_p$.

This action is faithful since an element acts trivially if and only if all its conjugates lie in $\mathrm{SO}_3( \overline{\mathbb Z_p})$, and conjugating by an element of a split maximal torus, we can see that this forces the element to lie in that torus, hence to lie in every split torus, thus lie in the center, which is trivial.

(If, instead of $\mathrm{SO}(3)$, we had a group like $\mathrm{SU}(2)$ with center, we could mod out by the integral elements congruent to $1$ mod $p$)

The set $\mathrm{SO}_3 (\overline{\mathbb Q_p})/\mathrm{SO}_3( \overline{\mathbb Z_p})$ is countable. Indeed, every class in it is represented by some element in $\mathrm{SO}_3 (\overline{\mathbb Q_p})$, which must lie in $\mathrm{SO}_3(K_p)$ for some finite extension $K_p$ of $\mathbb Q_p$, and thus arises from a class in $\mathrm{SO}_3( K_p)/ \mathrm{SO}_3(\mathcal O_{K_p})$ where $\mathcal O_{K_p}$ is the ring of integers.

The set $\mathrm{SO}_3( K_p)/ \mathrm{SO}_3(\mathcal O_{K_p})$ is countable by a standard argument - it is a countable union of the compact sets of matrices with bounded negative valuations, and because $\mathrm{SO}_3(\mathcal O_{K_p})$ is open, the topology is discrete.

Because there are countably many extensions $K_p$, the set is countable in total.

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    $\begingroup$ Very nice. In the fourth paragraph, did you mean that an element acts trivially if and only if all its conjugates lie in $\mathrm{SO}_3( \overline{\mathbb Z_p})$? $\endgroup$
    – GH from MO
    Sep 6, 2021 at 3:02
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    $\begingroup$ @GHfromMO yes, exactly. $\endgroup$
    – Will Sawin
    Sep 6, 2021 at 3:02
  • $\begingroup$ this feels like it should work for $\mathrm{Gl}_n(\mathbb{C})$ as well. that and the structure theory of lie groups might get one up to connected lie groups. $\endgroup$ Sep 6, 2021 at 3:25
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    $\begingroup$ @MarkSapir One group that the argument doesn't obviously work for is the universal cover of $SL_2(\mathbb R)$. I'm not sure if it can be modified to handle that case. $\endgroup$
    – Will Sawin
    Sep 6, 2021 at 3:34
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    $\begingroup$ This very argument was, to my knowledge, first used by Simon Thomas about 20 years ago (he told me later he wasn't aware of Ulam's question). The argument actually extends to the case when $\mathrm{SO}(3,\mathbf{R})$ is replaced with an arbitrary subgroup $G$ of $\mathrm{GL}_n(K)$ whenever $K$ is a field of cardinal $\le 2^{\aleph_0}$. [Oops, I see this latter fact was already pointed out in the comments. Anyway it's known.] $\endgroup$
    – YCor
    Sep 6, 2021 at 8:41

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