# Is the monoid of all cancellative finitely generated commutative monoids cancellative?

$$\DeclareMathOperator\Mon{Mon}\DeclareMathOperator\Grp{Grp}$$Let $$\Mon'$$ be the set of isomorphism classes of (small) commutative, unital, cancellative ($$a + t = b + t$$ implies $$a = b$$) monoids. It is a commutative unital monoid under product $$(M, N) \mapsto M \times N$$. The monoid $$\Mon'$$ is not cancellative because of the Eilenberg swindle'' $$\prod^\infty \mathbb N \times \mathbb N \simeq \prod^\infty \mathbb N$$. For similar reasons, $$\mathbb R^2 \simeq \mathbb R$$ gives another counterexample.

Restrict to finitely generated monoids $$\Mon \subseteq \Mon'$$.

Q: Is $$\Mon$$ a cancellative monoid?

There is a similar monoid of finitely generated abelian groups under cartesian product $$\Grp$$ which is cancellative by the fundamental theorem of finitely generated abelian groups. The map $$\Mon \to \Grp$$ which sends a monoid $$M$$ to its associated group $$M^{\mathrm{gp}}$$ is monoidal and surjective, so $$\Mon$$ is not the zero monoid as in the non-finitely generated case. But this is hardly enough to conclude that $$\Mon$$ is cancellative.

If the answer to the question above is no, there are weaker versions given by demanding that the monoids be saturated'' or that the associated group $$M^{\mathrm{gp}}$$ is torsion free. There are also relative variants for maps either to or from a fixed base monoid $$P$$, where the multiplication is coproduct/product. Can properties of this monoid tell us about the properties of $$P$$?

• (Small nitpick: $Mon'$ is not a set because there are monoids satisfying the given condition of arbitrarily large cardinality.) Commented Oct 12, 2022 at 17:12
• There's another morphism from Mon to a group that doesn't factor through the map to Grp -- take the number of irreducible elements. So the associated group of Mon is at least strictly larger than Grp. Commented Oct 12, 2022 at 18:05
• @QiaochuYuan I took the parenthetical "(small)" to mean that $Mon'$ was only collecting those isomorphism classes inside a given universe, so that it would be a set in the bigger universe. Was that wrong? Commented Oct 12, 2022 at 18:26
• This seems related to the Zarski-Cancellation problem. If we have $A\times \mathbb N\cong B\times \mathbb N$ then $KA[x]\cong KB[x]$. If $A,B$ are normal then $KA\cong KB$ implies $A\cong B$ Commented Oct 12, 2022 at 21:25
• The number of irreducible relations among irreducible elements gives another nontrivial homomorphism from Mon to Grp that doesn't factor through either taking the associated group or the above map. Commented Oct 13, 2022 at 11:52