Under what choice assumptions is there a monoid structure on every set? 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?
 A: If you're willing to think about semigroups rather than monoids, then some further comments are in order.
First comment. Every set $X$ carries two canonical semigroup structures:


*

*The "left zero" semigroup: $xy=x$ for all $x,y \in X$.

*The "right zero" semigroup: $xy=y$ for all $x,y \in X$.


This gives two non-equivalent functors
$$\mathbf{SemiGrp} \leftarrow \mathbf{Set}.$$
Obviously, both are sections of the forgetful functor
$$\mathbf{SemiGrp} \rightarrow \mathbf{Set}.$$
Second comment. By a pointed semigroup, I mean a semigroup $X$ together with a distinguished absorbing element $\bot$. This can be equivalently defined as a semigroup object in the monoidal category of pointed sets with smash product.
Its interesting to observe that every pointed set $(X,\bot)$ can be made into a pointed semigroup in the following ways:


*

*The "Zhen Lin" semigroup: we declare that $xy=\bot$ for all $x,y \in X$.

*The "equality" semigroup: we declare that $x^2=x$ for all $x \in X,$ and $xy=\bot$ for all distinct $x,y \in X$.


This gives two non-equivalent functors
$$\mathbf{\bot SemiGrp} \leftarrow \mathbf{\bot Set}.$$
Obviously, both are sections of the forgetful functor
$$\mathbf{\bot SemiGrp} \rightarrow \mathbf{\bot Set}.$$
Third comment.
What we're talking about here is really a kind of "categorified idempotency."
Let me explain.
If $x$ is an element of some monoid, then $x$ is said to be idempotent iff $x=x^2$. It makes sense to write $\mathrm{isIdem}(x)$ for the statement $x=x^2$. Okay, but what happens if we're given an object $X$ living in a monoidal category? Well, we could choose to define that $\mathrm{isIdem}(X)$ means "the set of all isomorphisms $X \leftarrow X \otimes X$." But such isomorphisms seem not to arise very often in practice. Another option would be to defined that $\mathrm{isIdem}(X)$ means "the set of all morphisms $X \leftarrow X \otimes X$." So basically, its the set of all ways of making $X$ into a magma. But darn it, magmas are just so uselessly general! So to make the concept more useful, lets define that $\mathrm{isIdem}(X)$ is the set of all associative morphisms $f : X \leftarrow X \otimes X$. In other words, its the collection of all ways of making $X$ into a semigroup.
Under the philosophy of propositions as types and in light of the above paragraph, we can think of the proposition "$X$ is idempotent" as denoting the collection of all ways of making the object $X$ into a semigroup. Hence if $\mathbf{C}$ is a monoidal category, we can think of the proposition "every object of $\mathbf{C}$ is idempotent" as denoting the collection of all sections of the forgetful functor $$\mathbf{C} \leftarrow \mathrm{Semi}(\mathbf{C}),$$ which are thought of as "proofs" for this statement. Hence, the first comment really consisted of defining two non-equivalent "proofs" that "every object of $(\mathbf{Set},\times,1)$ is idempotent," and the second comment really consisted of defining two non-equivalent "proofs" that "every object of $(\bot\mathbf{Set},\otimes,\{0,\bot\})$ is idempotent."
A: Every non-empty set admits a commutative monoid structure:


*

*The 1-element set is a monoid in a unique way.

*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.

