$\DeclareMathOperator\Mon{Mon}\DeclareMathOperator\Grp{Grp}$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 $P$?

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    $\begingroup$ (Small nitpick: $Mon'$ is not a set because there are monoids satisfying the given condition of arbitrarily large cardinality.) $\endgroup$ Commented Oct 12, 2022 at 17:12
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    $\begingroup$ 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. $\endgroup$
    – Leo Herr
    Commented Oct 12, 2022 at 18:05
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    $\begingroup$ @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? $\endgroup$ Commented Oct 12, 2022 at 18:26
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    $\begingroup$ 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 $A\cong B$ $\endgroup$ Commented Oct 12, 2022 at 21:25
  • $\begingroup$ 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. $\endgroup$
    – Leo Herr
    Commented Oct 13, 2022 at 11:52


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