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?
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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?
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**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,


Valerio



Connes' embedding conjecture for uncountable groups
The general situation, where CH fails, may be informed by
the [Keisler-Shelah isomorphism
theorem](https://encyclopediaofmath.org/wiki/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.


