A weak cancellation property for monoids Suppose $M$ is a (commutative) monoid.
Typically the cancellation property is defined as $a + c = b + c \Rightarrow a = b$ for all $a,b,c \in M$.
Recently I was working on a problem where I thought I needed cancellation, but it turned out that the weaker version $a + c = c \Rightarrow a = 0$ for all $a,c \in M$ would already be sufficient.
My questions are:

*

*Is this actually a weaker property than cancellation?
It is implied by cancellation by choosing $b = 0$, but despite trying some things out myself I am not yet 100% convinced that it is not just cancellation in disguise.

*If it is actually a weaker version of cancellation, is there some reading or other material on it anywhere or does it even have a name?

Note: Commutativity is not really needed, but it was where I stumbled upon this so I just kept it for the sake of simplicity.
 A: An example. Consider the set $\{n\mid n\gt 0\}\cup \{u_n \mid n\gt 0\}\cup\{0\}$ with operation + which is the usual + on natural numbers, $n+u_k=u_k+n=(n+k), u_k+u_n=u_{k+n}$, $0+x=x+0=x$ for every $x$. It is a commutative non-cancelative monoid satisfying your condition. I do not think this class of monoids has a name.
There are books on commutative semigroups (Redei, Grillet,...).
A: Last week, I had formulated the exact same definition for commutative monoids, and even called it weak cancellability.  I was wondering whether others found it useful.
Define divisibility in a commutative monoid $M$ with identity $e$ in the usual way:  for all $a,b\in M,$ we have $a\mid b$ if there exists $c$ with $ac=b.$ I was investigating when divisibility is a partial order (as it is on the positive integers). It is easy to see that the divisibility relation is reflexive and transitive, but it is not necessarily antisymmetric.
I found the following result:  Given a commutative monoid $M$ with identity $e,$ if the weak cancellation property holds and $e$ is the only unit, then the divisibility relation is antisymmetric.
The proof is straightforward.  Suppose $a\mid b$ and $b\mid a.$  Then there are $p,q\in M$ with $ap=b$ and $bq=a.$  Thus, $$a=bq=(ap)q=(pq)a.$$ Since the weak cancellation property holds, we have $pq=e.$  Since there is only unit in $M,$ then necessarily $p=q=e,$ and thus $a=b.$  So divisibility is antisymmetric.
Of course this is implied by the cancellation property. The simplest example I found of a monoid with the weak canellation property which does not possess the cancellation property is essentially a matrix version of what markvs answered, but I have not found a simpler one yet.
It is easy to prove that if divisibilty is antisymmetric, then $e$ is the only unit in $M.$  But the divisibility relation on the set $\{0,1\}$ with multiplication is antisymmetric, but does not possess the weak cancellation property since $0\cdot0=0,$ while $0\ne1.$ Thus, weak cancellation is a distinct concept.
A: For a monoid M, there exists identity e such that a° e = a, for all a∈M, what have you given it is none other than the property for existence of identity element.(since, if b not equal to e then a°b fails to give a again(for some a in M), that's the law what you given),.
And so, there is no connection between that property and the cancellation law, we can take examples ," set of all n×n matrices with multiplication operation".
