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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:

  1. 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.
  2. 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.

markvs
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Pseudoradius
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3 Answers3

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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,...).

markvs
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  • Nice find, thank you. :)

    I was wondering if there is a simpler example, then I found that for finite monoids my property is equivalent to the monoid being a group, so I doubt there is anything simpler.

     – Pseudoradius Jul 10 '20 at 08:16
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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.

Vince Matsko
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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".

A learner
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  • Your statement and example are incorrect. The set of all $n\times n$ matrices with multiplication does not satisfy OP's property since you have the example of $A^2 = A$ with $A \neq I_n$. For example, take $n = 2$ and $$A = \begin{bmatrix}1 & 0 \ 0 & 0\ \end{bmatrix}.$$ – Aryaman Maithani Jul 07 '20 at 15:16
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    If $a \ne e$, then one can only conclude $\exists b \in M, ab \ne b$, not $\forall b \in M, ab \ne b$. – Geoffrey Trang Jul 07 '20 at 15:30
  • @Aryaman Maithani the property given by op says it is ,, if a+c=c for all a,c then it implies a=0, and this case possible only if we take a identity and c runs over on whole monoid, and your example works for just on A what you defined, but it should work on all the matrice by op's property. – A learner Jul 07 '20 at 15:36
  • @Subhajit: What OP means is that if $a + c = c$ for any $a, c \in M$, then we should have $a = 0$. – Aryaman Maithani Jul 07 '20 at 15:38
  • @Geoffrey Trang yes, what I want to say! – A learner Jul 07 '20 at 15:40
  • @Aryaman Maithani why don't you go for the contrapositive part of the rule! It understands you what I want to say! – A learner Jul 07 '20 at 15:43
  • @Subhajit: without resorting to contrapositive, why don't you see that the result fails for the example you mentioned. We had $A\cdot A = A$ without $A = I_2$! – Aryaman Maithani Jul 07 '20 at 15:45
  • @Aryaman Maithani, exactly AB=A possible only if B=I,and for your conclusion, AA=A possible for only A not for all B such that BA=B – A learner Jul 07 '20 at 15:48
  • @Subhajit: uh, I've written a counterexample to that. – Aryaman Maithani Jul 07 '20 at 15:48