In this topic, I will use the word uncountable group referring to groups whose cardinality is $\leq|\mathbb R|$.

Notation: $R$ is the hyperfinite $II_1$-factor, $\omega$ is a free ultrafilter on the natural numbers, $R^\omega$ is the tracial ultrapower, $\tau$ is the unique normalized trace on $R^\omega$, $U(R^\omega)$ is the unitary group of $R^\omega$.

Definition: A group $G$ is called hyperlinear if there is a group monomorphism $\theta:G\rightarrow U(R^\omega)$ such that $\tau(\theta(g))=0$ for all $g\neq1$.

Question 1: Does there exist an uncountable non-hyperlinear group?

A bit of background: the same question for countable groups is known as Connes' embedding problem for groups and it's still unsolved. When I began my PhD (Nov. 2008), my former advisor Florin Radulescu told L. Paunescu and myself that he would have liked to have a better understanding of the problem in higher cardinality. In particular, he asked us to see whether the free group on uncountable many generators, here denoted by $\mathbb F_c$, and the circle group $S_1$ were hyperlinear. Maybe he expected that one of them was not, but we came up with the general result that every subrgroup of $U(R^\omega)$ is hyperlinear (this is basically proved in http://arxiv.org/abs/0911.4978); in particular $S_1$ is hyperlinear and, with an additional 10-line argument, also $\mathbb F_c$ turns out to be hyperlinear. Moreover, the result is quite general and excludes many possible example a priori, making Question 1, in my modest opinion, non-trivial and interesting. At some point, maybe also talking with someone else (but I don't remember exactly who), I got quite convinced that Question 1 has the same level of difficulty of Connes' problem and that they may be actually equivalent. In this case, I would like to find a formal way to express that. In this view, a positive answer to the following question would not completely solve the problem, but would be a nice starting point.

Question 2: Let $G$ be an uncountable group. Do there always exist countable groups $G_1,G_2,\ldots$ and a free ultrafilter $\omega$ such that $G$ embeds into the algebraic ultraproduct of the $G_i$'s?

Update: as shown my Simon Thomas below, the answer is positive assuming CH. On the other hand, Joel's answer shows that without CH we have some weaker result. For instance, Question 2 has affirmative answer if we allow the sequence $G_i$ to be indexed by a possibly non-countable set $I$.

Thanks in advance,


  • $\begingroup$ Don't you mean that uncountable groups have cardinality $\geq |\mathbb{R}|$ (instead of $\leq |\mathbb{R}|$) ? $\endgroup$
    – Max Horn
    Nov 6, 2011 at 16:10
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    $\begingroup$ No, it's right as I wrote. If were $\geq|\mathbb R|$, the answer would have been trivial: take a big enough group. Indeed I am using this non-standard terminology just to avoid easy counterexamples. $\endgroup$ Nov 6, 2011 at 16:21
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    $\begingroup$ Since any group is a metric group with the trivial distance $d(g,h)=1-\delta_{gh}$, it seems to me that your question 2 asks whether every ``uncountable'' group embeds into the algebraic ultraproduct of countable groups. I'm sure that model-theorists know about that... $\endgroup$ Nov 6, 2011 at 21:30
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    $\begingroup$ Alain, at the beginning I wrote basically what you have now written, but (fault of my lack of knowledge) it was not clear to me what is a purely algebraic ultraproduct of countable groups. What is the equivalence relation we mod out with? Maybe it's just: two sequence in the cartesian product are equal if they coincide on an element of $\omega$.. $\endgroup$ Nov 6, 2011 at 21:50
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    $\begingroup$ @Valerio: what else? (:-) $\endgroup$ Nov 6, 2011 at 21:54

2 Answers 2


The general situation, where CH fails, may be informed by the Keisler-Shelah isomorphism theorem, which asserts that two first-order structures have isomorphic ultrapowers if and only if they have the same first-order theory.

In particular, for any infinite group $G$ at all, of any size, we may take a countable elementary subgroup $H$, meaning in particular that they have the same first-order theory, and so there is a nonprincipal ultrafilter $U$ on an index set $I$ such that the ultrapowers $G^I/U\cong H^I/U$ are isomorphic. Since every first-order structure maps elementarily into its ultrapowers, this means in particular that $G$ maps elementarily (and hence monomorphically) into an ultrapower of $H$, a countable group.

Thus, this fully answers the version of question 2 in which we allow the ultrafilter to live on a bigger index set:

Theorem. For every group $G$ there is a countable group $H$ and a free ultrafilter $U$ on a set, such that $G$ embeds into the ultrapower $H^I/U$.

If you want to insist that the ultrafilter concentrate on index set $\mathbb{N}$, however, then things become more complicated. If the CH holds, then the Keisler-Shelah theorem shows that any two groups of size at most $2^{\aleph_0}$ and with the same theory have isomorphic ultrapowers by an ultrafilter on $\aleph_0$, and so the desired result is attained. In the non-CH case, however, what we seem to get is that for any cardinal $\lambda$, if $\beta$ is smallest such that $\lambda^\beta\gt\lambda$, then any two groups of size $\beta$ with the same theory have isomorphic utrapowers using an ultrafilter on $\lambda$. Thus, they each map into an ultrapower of the other.

The Keisler-Shelah theorem was proved first by Keisler in the case that GCH holds, using saturation ideas as in Simon's answer. The need for the GCH was later removed by Shelah.


A partial answer to Alain's question ... Suppose that the Continuum Hypothesis $CH$ holds. Let $G$ be any group of size $2^{\aleph_{0}} = \aleph_{1}$. Then there exists a countable subgroup $H$ of $G$ with the same first order theory $T$. Let $\Gamma = \prod_{\mathcal{U}} H_{n}$ be the ultraproduct such that every $H_{n} = H$. Then $CH$ implies that $\Gamma$ is a saturated model of $T$ and hence $G$ embeds into $\Gamma$.

Your question concerning the possibility of embedding groups $G$ with $|G| \leq 2^{\aleph_{0}}$ into ultraproducts $\prod_{\mathcal{U}} H_{n}$ over a countable index set might be interesting in the case when $CH$ fails. There is a similar open problem concerning sofic groups. It is an easily seen that every group is sofic iff $Sym(\mathbb{N})$ embeds in some universal sofic group. However, even if every group is sofic, it is not clear whether or not $Sym(\mathbb{N})$ embeds in a universal sofic group arising from an ultrafilter over $\mathbb{N}$.

If you are interested in embeddings using ultraproducts over larger index sets, then you can make use of an ancient result of Malcev, which says that every group $G$ embeds into a suitable ultraproduct of its finitely generated subgroups. Here the index set $I$ is the set of its finitely generated subgroups, which is uncountable if $G$ is uncountable.

PS: In Shelah's paper on ultraproducts, he mentions that if Martin's Axiom holds, then there exists an ultrafilter $\mathcal{D}$ over $\mathbb{N}$ such that for every countable structure $M$ for a countable language, the corresponding ultraproduct $\prod_{\mathcal{D}}M$ is saturated. In particular, it is consistent with the failure of $CH$ that every group $G$ with $|G| \leq 2^{\aleph_{0}}$ embeds into an ultraproduct $\prod_{\mathcal{U}} H_{n}$ over a countable index set.

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    $\begingroup$ Except I am told that I am now a set theorist. So naturally I hope that your question is independent of ZFC. $\endgroup$ Nov 6, 2011 at 22:06
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    $\begingroup$ @Simon: Oops... But won't you attend the conference at CIRM, two weeks from now? $\endgroup$ Nov 6, 2011 at 22:11
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    $\begingroup$ @Alain: The organizers are very broad minded. @Valerio: The only reason that I know that a countable number of copies works is a "soft" model-theoretic reason. Assuming CH, the ultraproduct of countably many copies of a countable structure is saturated and this implies that any other structure with the same first order theory must embed into it. $\endgroup$ Nov 6, 2011 at 22:49
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    $\begingroup$ Simon, many thanks for your answer + clarification. I have accepted Joel's answer only because he considers also the case when CH may fail. $\endgroup$ Nov 7, 2011 at 10:45
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    $\begingroup$ Nice observation in the P.S., Simon! $\endgroup$ Nov 7, 2011 at 14:32

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