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Similar questions have been asked without great success to answer. I read what I could about the problem, but no idea. I wrote to university professor, no response.

There has been question about examples of associative binary operation which are not closed. No example was given, I found none.

If definition of associativity is standard as we all know it then all not closed operations must be non-associative - and this is in stark contrast with articles on wikipedia like: https://en.wikipedia.org/wiki/Semigroup

Lets have Cayley table defined on $\{a, b, c\}$, operation $*$ which is commutative and: $$aa = a, bb = b, cc = c, ab = b, ac = c, bc = Y$$

If I try to prove associativity, then it fails with:

$$a*(b*c) = a*Y = ???$$

$$(a*b)*c = Y$$

Sides are simply not equal. First part cannot defined on my structure, there is no answer.

And this in general happens whenever there is element outside the set I my humble opinion. If I am mistaken, I would be very happy to understand what am I missing. So far I have found no concrete counterexample.

One solution would be that binary operation must be closed, then there is conflict with table of structures on wikipedia page.

Other solution would be, that these instances where there is undefined operations, are simply left out. Then we would work only with associative triples where both sides are defined.

Thank you all kindly.

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    $\begingroup$ Have you read about Groupoids? By the way, a "closed set" is a topological concept. You mean something else here. $\endgroup$ – Somos Jul 17 '20 at 22:02
  • $\begingroup$ @Somos There is a concept of closure for binary operations: en.wikipedia.org/wiki/Closure_(mathematics) $\endgroup$ – Alonso Delfín Jul 17 '20 at 22:20
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    $\begingroup$ Yes, the binary operator is closed. Not the set. There is a difference. $\endgroup$ – Somos Jul 17 '20 at 22:31
  • $\begingroup$ Thank you, binary operation is closed. My mistake in question, sorry very much. $\endgroup$ – Josef Hlava Jul 17 '20 at 22:32
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    $\begingroup$ By definition, a binary operation is closed. $\endgroup$ – Carlo Jul 18 '20 at 1:26
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You are talking about partial semigroups. These can be defined as sets with partial binary operation such that one of the three conditions holds:

1 $a(bc)=(ab)c$ provided both sides are defined

2 If $a(bc)$ is defined then $ab$ and $(ab)c$: are defined and $a(bc)=(ab)c$.

3 Similar to 2 with left and right sides switched.

So there are three (nonequivalent) definitions of partial semigroups. Here is an old paper about partial semigroups: Hrmová, Renáta Partial groupoids with some associativity conditions. Mat. Časopis Sloven. Akad. Vied 21 (1971), 285–311.

One easy way to obtain a partial semigroup in the sense 1 is to take a (finite) semigroup, and delete a few entries of its multiplication table. There exist results about when we can reconstruct the original multiplication table then.

This is an even older paper about this Kozlov, K. P. Existence of rigid cells in the Cayley table of a semigroup. Leningrad. Gos. Ped. Inst. Učen. Zap. 387 1968 131–133.

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  • $\begingroup$ I am not talking about partial binary operation, that is not my thought. Lets change letter Y in my definition to nicer letter "d". $\endgroup$ – Josef Hlava Jul 18 '20 at 6:58
  • $\begingroup$ I understand, that thinking about 1 foreing element could be as 1 undefined operation and hence partial binary operation. My intent is that there could be 2 foreing elements or more. Then associativity could not be part of a binary operation. One solution which is missing almost everywhere in definition of associativity is your point 1 ...." provided both sides are defined". That would solve all these questions immediately :) I have never seen definition of associativity..."provided both sides are defined" and I don't see how this would be evident. $\endgroup$ – Josef Hlava Jul 18 '20 at 7:02
  • $\begingroup$ ' have never seen definition of associativity..."provided both sides are defined" and I don't see how this would be evident. ' I gave two references. There are many more. Yes, you are talking about partial operations, your "$Y$" is just substitution for empty space. $\endgroup$ – Mark Sapir Jul 18 '20 at 15:31

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