Lemme check this: Under ZFC, regardless of CH, there is a sub-field of the reals with cardinality $\aleph_1$, right?

If CH holds, $\beth_1 = \aleph_1$, so the reals themselves are an example. If ￢CH holds, using Axiom of Choice, we choose $\aleph_1$ elements from the $\beth_1$ elements that extend $\mathbb{Q}$.

@DannyuNDos sure. Choose aleph_1 distinct elements of R and the field generated by them has size aleph_1

I've just noticed something peculiar about group completion. For consider the ordinal $\omega^2$ as an additive monoid. Since $1 + \omega = \omega$, the group completion will cancel the $\omega$ out to yield $1 = 0$.

A following conclusion is that, group completion is not necessarily injective.

yeah, you need some kind of cancellation condition to hold in the original thing for there to be any embedding in a group

otherwise, the collapsing can be very dramatic