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,

Valerio