# Cancelable commutative monoids with finite maximal subgroups

Suppose $$\mathcal{M} = (M, +, 0)$$ is a cancelable commutative monoid. Let $$G$$ be the maximal subgroup of $$M$$, i.e.

$$G = \{ a \in M \colon (\exists b \in M)\, a + b = 0 \}.$$

For $$a, b \in M$$ say $$a \preceq b$$ if $$(\exists c \in M)\, a + c = b$$ and say $$a \equiv b$$ if $$a \preceq b$$ and $$b \preceq a$$. Note for all $$a$$, $$\{b \in M \colon a \equiv b\} = \{a + g \colon g \in G\}$$. In particular, if $$a \equiv b$$ and $$c \equiv d$$ then $$a + b \equiv c + d$$ and so there is a monoid $$\mathcal{M}/\equiv$$.

Is it always the case that $$\mathcal{M}\cong (\mathcal{M}/ \equiv) \times G$$ as monoids?

What if $$G$$ is finite?

What if for all $$a \in M$$, $$\{b \in M\colon b\preceq a\}$$ is finite?

• If $G$ is not a direct factor in the group envelope of $M$ this is probably not the case. For instance take $M$ the additive monoid of sequences $\mathbf{N}\to\mathbf{Z}$ that are eventually non-negative.
– YCor
Mar 24 at 8:49
• This shouldn't be true. I don't think much about cancellative commutative monoids but I don't see why they should have a submonoid transversal to Green's H-relation. Mar 24 at 10:58
• @YCor doesnt the eventually nonnegative sequences have trivial group of units and hence what the OP wants is true because the congruence is trivial? Mar 24 at 12:33
• @BenjaminSteinberg the group of "units" is then the group of eventually zero sentences.
– YCor
Mar 24 at 13:36
• @Ycor, as yes I forgot 0. Sorry Mar 24 at 17:05

Here is a general construction that encompasses @R. van Dobben de Bruyn's example but the idea is taken from his answer. I'll write commutative monoids additively. I'll use $$K(M)$$ for the Grothendieck group of a commutative monoid $$M$$ and $$M^\times$$ for the group of units.
Let $$0\to A\to B\xrightarrow{f} C\to 0$$ be any exact sequence of finitely generated abelian groups with $$C\ncong A\times B$$, e.g., $$0\to2\mathbb Z/4\mathbb Z\to \mathbb Z/4\mathbb Z\to \mathbb Z/2\mathbb Z\to 0$$. Let $$Y$$ be a finite generating set for $$B$$.
Let $$N$$ be any nontrivial finitely generated cancellative monoid with trivial group of units, say with finite set of generators $$X$$ with $$0\notin X$$. Let $$M$$ be the submonoid of $$N\times B$$ generated by $$N\times \{0\}$$, $$\{0\}\times A$$ and $$X\times Y$$. Notice that $$K(M)\cong K(N)\times B$$ by construction since $$(0,y)=(x,y)-(x,0)$$ for $$x\in X$$ and $$y\in Y$$. By construction, $$M^\times = \{0\}\times A$$ and $$M/M^{\times}$$ is isomorphic to the submonoid of $$N\times C$$ generated by $$N\times \{0\}$$ and $$X\times f(Y)$$. In particular $$K(M/M^\times)\cong K(N)\times C$$ by the same argument as before. Thus if $$M\cong (M/M^\times) \times M^\times$$, then $$K(N)\times C\cong K(N)\times A\times B$$. But then $$C\cong A\times B$$ by the structure theorem for finitely generated abelian groups, a contradiction.
Here is a counterexample that is finitely generated and has finite unit group $$M^*$$, so in particular the down-sets are finite. (I'm using the notation from Ogus's Lectures on logarithmic algebraic geometry.)
Example. Let $$A = \mathbf Z \oplus (\mathbf Z/4\mathbf Z)$$, and let $$M \subseteq A$$ be the submonoid generated by $$(1,0)$$, $$(1,1)$$, and $$(0,2)$$. Then $$M^* = 0 \oplus (2\mathbf Z/4\mathbf Z)$$ and $$M/M^*$$ is isomorphic to the submonoid of $$\mathbf Z \oplus (\mathbf Z/2\mathbf Z)$$ generated by $$(1,0)$$ and $$(1,1)$$. Now $$M$$ cannot be isomorphic to $$M^* \times M/M^*$$ since $$M^{\operatorname{gp}}$$ has an element of order $$4$$ but $$(M^* \times M/M^*)^{\operatorname{gp}}$$ does not.