Call a monoid group-like if it embeds into its group completion. There exists an obvious tension between group-like and idempotent monoids in that a group cannot contain non-trivial idempotent elements, so any idempotent elements of a monoid have to be trivialized by its group completion. Furthermore, almost all important examples of monoids are either one or the other. My question is, has this been formalized anywhere? Is there maybe a decomposition theorem for monoids into their grouplike and idempotent parts?
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asked Jun 25, 2022 at 14:23
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7$\begingroup$ What do you do about cancellative monoids that don't embed in a group? They have problems not caused by idempotents. $\endgroup$– Benjamin SteinbergCommented Jun 25, 2022 at 15:26
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1$\begingroup$ Also what do you mean by grouplike part? Do you mean submonoids or maybe subsemigroups on which the group completion is injective? $\endgroup$– Benjamin SteinbergCommented Jun 25, 2022 at 15:28
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$\begingroup$ Malcev gave a necessary and sufficient condition for a cancellation monoid to embed in a group. Finding such conditions for semigroups has a long and rich history. The monoid with generators $a,b,c,d$ and two relations $ab=cd$ and $aeb=ced$ is cancellative but not group-embeddable, and of course this has no non-trivial idempotent. $\endgroup$– Carl-Fredrik Nyberg BroddaCommented Jun 25, 2022 at 15:31
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$\begingroup$ For inverse monoids the idempotents are responsible for all identifications but usually very little of the monoid embeds with the exception of the inverse hull of a group embeddable cancellative monoid $\endgroup$– Benjamin SteinbergCommented Jun 25, 2022 at 15:44
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1$\begingroup$ I asked a related question about rigs. The answer seems to be no. $\endgroup$– Zhen LinCommented Nov 23, 2022 at 23:09
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2 Answers
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For finite semigroups (and thus monoids), the Krohn–Rhodes theorem gives a decomposition into (simple) groups and aperiodic semigroups (subsemigroups of the flip-flop, which is idempotent). However, the decomposition is more complicated than a separation into a “group-like part” and an “idempotent part”; it is in terms of iterated wreath products (or alternatively, iterated semidirect products, I guess).
answered Jul 21, 2023 at 11:59
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Not exactly what you are looking for, but there may be some connection. The following situation is frequent: a monoid $M$, an equivalence relation $R$ on $M$ which is compatible with the monoid operator, and $M/R$ is an idempotent monoid. Examples:
- Polynomials with a positive highest-degree coefficient, as an additive cancellative monoid:
$P \mathbin R Q$ is $\operatorname{deg}(P) = \operatorname{deg}(Q)$. - Functions $\mathbb{R}^+ \rightarrow \mathbb{R}^+$ as an additive cancellative monoid, with Big O equivalence relation on some point $a$ (where $a \in \mathbb{R}$ or $a=+\infty$ or $a=-\infty$):
$f \mathbin R g$ is $f(x) = \mathcal{O}(g(x))$ as $x \to a$. - Sets with infinite cardinal, operation is union:
$A \mathbin R B$ is $\operatorname{card}(A) = \operatorname{card}(B)$. - Geometric or topological sets, with union:
$\mathbin R$ is having the same dimension (for some dimensions, e.g. Hausdorff).
All these examples share the same pattern: the resulting idempotent monoid is a totally ordered set, and the operator is $\max$. One could build more complex examples with the same ingredients, e.g. multiple-variable polynomials, with $\operatorname{deg}$ being the set of highest-degree terms, but that looks artificial.
One possible question: given an idempotent monoid, can it always be considered as the quotient of a cancellative monoid?
answered Jun 25, 2022 at 22:07
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3$\begingroup$ Free monoid are cancellative so all monoids are quotients of cancellative monoids. Also every monoid has a universal quotient that is idempotent $\endgroup$ Commented Jun 25, 2022 at 22:19
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$\begingroup$ What does "with their equivalence relation on $+\infty$" mean? Does it mean that the difference of the functions tends to $0$, that their quotient is well defined near $+\infty$ and that its limit is $1$, or something else? $\endgroup$– LSpiceCommented Jul 21, 2023 at 3:21
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1$\begingroup$ @LSpice Thanks for your edits. This was one of my first posts (in MO or MSE), so I was not fully aware of typographic rules. As for your question, the equivalence relation I had in mind is the "Big O" one. I shall state that in the text. $\endgroup$ Commented Jul 21, 2023 at 15:19
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1$\begingroup$ Re, my pleasure, and I hope you've been enjoying the experience here! Just so you know, while it's always good to be semantically correct, $R$ vs. $\mathbin R$
$R$ vs. $\mathbin R$makes no difference in rendering; the only difference comes when a binary op. is surrounded by things on which to operate: $A R B$ vs. $A \mathbin R B$$A R B$ vs. $A \mathbin R B$. Lots of things TeX already knows are binary operators, likeA + B, but you have to tell it about $R$ specifically. You can also suppress the built-in behaviour: $A{+}B$A{+}B. $\endgroup$– LSpiceCommented Jul 21, 2023 at 18:00