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?

  • 3
    $\begingroup$ 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. $\endgroup$
    – YCor
    Mar 24 at 8:49
  • $\begingroup$ 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. $\endgroup$ Mar 24 at 10:58
  • $\begingroup$ @YCor doesnt the eventually nonnegative sequences have trivial group of units and hence what the OP wants is true because the congruence is trivial? $\endgroup$ Mar 24 at 12:33
  • 1
    $\begingroup$ @BenjaminSteinberg the group of "units" is then the group of eventually zero sentences. $\endgroup$
    – YCor
    Mar 24 at 13:36
  • $\begingroup$ @Ycor, as yes I forgot 0. Sorry $\endgroup$ Mar 24 at 17:05

2 Answers 2


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.


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