Example of Non-Cancellative Commutative Monoid

Question

Can anyone help to give an example of non-cancellative commutative monoid? Maybe like, $S_n = \{1,2,3...,n\}$ and a map of $S_n \times S_n \to S_n$.

Thanks.

Answer 1

For any commutative ring $R$ that is not $0$, $(R, \times)$ is not cancellatuve. It's even "less" cancellative is $R$ is not an integral domain.

Examples include $\mathbb{Z}/n\mathbb{Z}$ for $n$ a nonprime non integer; but also infinite rings such as the ring of continuous (or smooth) functions $\mathbb{R}\to \mathbb{R}$; or for any commutative ring $R$ and integer $n\geq 2$, $R^n$, and many more.

Answer 2

The set of all the subgroups of a group $G$ is a commutative monoid under set intersection, with $G$ the identity element. Say $\mathcal G$ this monoid. We get:

1. $G$ is the only invertible element of $\mathcal G$;
2. the subset of all the normal subgroups of $G$ is a submonoid of $\mathcal G$.

$\mathcal G$ is non-cancellative, as $H\cap H=H\cap G$ for every $H\in\mathcal G\setminus\{G\}$, but $H\ne G$.

Answer 3

Let $\mathbf S = \langle S, \vee, 0\rangle$ be a semilattice with bottom element.
Thus $\langle S, \vee\rangle$ is a commutative idempotent semigroup and $s\vee 0 = s$ for every $s \in S$.

So $\mathbf S$ is a monoid. But if $S$ is nontrivial, say $s \in S\setminus\{0\}$, then $$s\vee0 = s = s\vee s,$$ so $\mathbf S$ is not cancellative.

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More generally if $a,b \in S$ and $a\neq b$, then, with $c=a\vee b$, we have $$a\vee b = c = a \vee c = c \vee b,$$ so, if $\mathbf S$ were cancellative, then we would have $a=c$ and $b=c$, a contradiction.