# Cancellation property for commutative monoid

Let $$(M,+,e)$$ be a commutative monoid with unit $$e$$. An element $$a\in M$$ is called cancellative element if

for any $$b,c \in M$$ such that $$a+b=a+c$$ implies that $$b=c$$.

Let $$(\mathbf{N},+,0)$$ the commutative monoid of natural numbers.

suppose that

1. we have two morphisms of monoids $$f:(\mathbf{N},+,0)\rightarrow (M,+,e)$$ and $$g:(M,+,e)\rightarrow (\mathbf{N},+,0)$$ such that $$g\circ f= id$$.
2. The monoid $$(M,+,e)$$ is torsion-free.

My question is the following: is the element $$a=f(1)$$ automatically a cancellative element in $$(M,+,e)$$ ?

Edit: By torsion-free I do mean that there does not exist a natural number $$n>0$$ and some element $$x\in M-\{e\}$$ such that $$n x=e$$.

• By torsion free you mean that there does not exist a natural number $n$ and some element $x\in M$ such that $n\cdot x=e$? Jan 2, 2021 at 1:14
• The usual definition of torsion free for commutative monoids is na=nb implies a=b. Jan 2, 2021 at 1:30

The answer is no. Let $$U=\{0,1\}$$ under multiplication. Let $$P$$ be the semigroup of positive integers under $$+$$. Consider $$S=P\times U$$, the direct product and let $$M=S\cup \{I\}$$ where $$I$$ is an adjoined identity. Then $$M$$ is torsion-free, there is a homomorphism $$f\colon \mathbf N\to M$$ given by $$f(0)=I$$ and $$f(n)=(n,0)$$ for $$n>0$$ and $$g\colon M\to \mathbf N$$ with $$g(I)=0$$ and $$g(n,x)=n$$ for $$n>0$$ and $$x\in \{0,1\}$$ and clearly $$gf=1$$ but $$(1,0)$$ is not cancellable.
Consider the monoid $$M=\mathbb{N}\times \{0,1\}$$ where $$(n,a)*(m,b):=(n+m, a\cdot b).$$ The unit element is $$e:=(0,1)$$. Note that this monoid is torsion free. Now consider the maps $$g:(M,*,e)\rightarrow (\mathbb{N}, +,0), g(n,a)=n$$ and $$f: (\mathbb{N}, +,0) \rightarrow (M,*,e)$$ such that $$f(0)=e$$ and $$f(n)=(n,0)$$. Then we have $$g\circ f=id$$, but $$f(1)=(1,0)$$ is not cancellative as $$(1,0)*(0,0)=(1,0)=(1,0)*(0,1).$$
• You answered with essentially the same construction as me while I was typing it. I just removed $(0,0)$ to make it more torsion-free in that there are no elements generating a finite subsemigroup except the identity. Jan 2, 2021 at 1:40