Under what choice assumptions is there a monoid structure on every set?

Question

The question arose when discussing possible cardinalities of hom-sets of whether it's any weaker than the axiom of choice that there exists a monoid of every cardinality. It's well known, or at least known, that the axiom of choice holds just if every set $S$ has a group structure. One direction follows from Lowenheim-Skolem, and the other from using the group structure to inject $S$ into $h(S)\times h(S)$, $h$ the Hartogs number.

So, a proof of choice from monoid structures would certainly have to proceed differently. If the monoid is commutative and infinite, then I guess its Grothendieck group will share its cardinality, and we see the existence of commutative monoid structures on every set implies AC. Is the left adjoint of the forgetful functor from all groups to monoids any worse than the Grothendieck group functor? Or does anyone suggest a different proof?

Answer 1

Every non-empty set admits a commutative monoid structure:

1. The 1-element set is a monoid in a unique way.
2. If $X$ has at least two distinct elements, say $0$ and $1$, then we can make $X$ into a commutative monoid as follows: $$x \cdot y = \begin{cases} x & \text{if } y = 1 \\ y & \text{if } x = 1 \\ 0 & \text{otherwise} \end{cases}$$ Of course, $X$ is not a cancellative monoid in this case.