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0 votes
1 answer
58 views

Is there a characterization of monoids that distribute over each other?

Let $(M, e_1, \times_1, e_2, \times_2)$ be an algebraic structure such that $(M, e_1, \times_1)$ and $(M, e_2, \times_2)$ are monoids $x \times_1 (y \times_2 z) = (x \times_1 y) \times_2 (x \times_1 ...
11 votes
0 answers
427 views

Is there a theory of completions of semirings similar to $I$-adic completions of rings?

Let $L = \text{Con } (\mathbb{N}, 0, +) \setminus \Delta$ be the lattice of monoid congruences on the naturals, excluding the trivial congruence. As it happens, every $\theta \in L$ is the meet of ...
8 votes
1 answer
437 views

Function $\phi$ such that $f(\phi(x,y)) = f(x) + f(y)$

I have a continuous function $f:\mathbb{R}^n\to\mathbb{R}$, and I am looking for a continuous (or at least measurable) function $\phi:\mathbb{R}^{2n}\to\mathbb{R}^n$ such that $f(\phi(x,y))=f(x)+f(y)$....
2 votes
0 answers
92 views

Geometric interpretation of flags and the role of the rook monoid and Kazhdan–Lusztig theory in $M_n(\mathbb{C})$

Let $G = GL_n(\mathbb{C})$, $B$ be its Borel subgroup, and $P$ a parabolic subgroup. The space $G/B$ corresponds to complete flags in $ \mathbb{C}^n$, and $G/P$ corresponds to partial flags. The ...
4 votes
1 answer
364 views

Values attained by the coheight of $(H \setminus H^\times)^k$ as a function of $H$ and $k$

Edit (Apr 24, 2017). I'm updating this post in the light of the latest developments of a related thread. Let $H$ be a multiplicatively written, commutative monoid, and set $M := H \setminus H^\times$,...
0 votes
0 answers
61 views

Defining rank of an abelian subgroup using the second centralizer

I recently posted this on MSE, but didn't receive any feedback; so I'm posting it on MO. I recently came across this article which explored the maximal abelian subgroups of the symmetric group $S_n$. ...
4 votes
1 answer
239 views

True or false? Every left or right cancellative, duo semigroup is cancellative

A semigroup $S$ is duo if $aS = Sa$ for all $a \in S$, where $aS := \{ax: x \in S\}$ and similarly for $Sa$; for instance, every commutative semigroup is duo, and so is every group. On the other hand, ...
2 votes
1 answer
404 views

Reference request: a cousin to the log semiring

Let $f$ be strictly increasing on $\mathbb{R}$. Then $x \oplus y := f^{-1}(f(x)+f(y))$ gives rise to a strict symmetric monoidal ($\Rightarrow$ commutative monoid) structure on $(\mathbb{R},\ge)$ with ...
6 votes
3 answers
551 views

Conjecture about commutative semigroups

Conjecture: given any commutative semigroup $S$ of order $n \ge 4$, there exist $a, b \in S$ with $a \ne b$, an integer $m \ge \lfloor (n-1)/2 \rfloor$, and two $m$-element subsets $X = \{x_1, \ldots, ...
8 votes
1 answer
322 views

Does every cancellative duo semigroup embed into a group?

Prompted by the comments to a recent answer by YCor to a related question (here), I'd like to ask the following: Q. Does every cancellative duo semigroup embed into a group? A (multiplicatively ...
7 votes
2 answers
488 views

Is every cancellative semigroup a subdirect product of subdirectly irreducible cancellative semigroups?

By a classical result of Birkhoff (that is, Theorem 2 in [G. Birkhoff, Subdirect unions in universal algebra, Bull. AMS, 1944]) and the trivial fact that the class of semigroups is closed under the ...
8 votes
2 answers
596 views

If a semigroup embeds into a group, then is it a subdirect product of groups?

The title has it all: Q. If a semigroup $S$ embeds into a group, then is $S$ (isomorphic to) a subdirect product of groups? If yes, then $S$ is a subdirect product of subdirectly irreducible groups,...
3 votes
0 answers
250 views

Action (of a graded monoid) required

Reference request: Did the construction below appear anywhere before? Any mentions of it or especially any links to something commonly known would be really helpful. I feel that it might be related to ...
3 votes
0 answers
89 views

Ordering the elements of a semigroup by $a \le b$ iff $a=b$ or $b=ab=ba$

Let $S$ be a semigroup, written multiplicatively. The binary relation $\le$ on (the underlying set of) $S$, whose graph consists of all pairs $(a,b) \in S \times S$ such that $a = b$ or $b = ab = ba$, ...
4 votes
0 answers
147 views

Isomorphism between the reduced C*-algebra of a groupoid and the crossed product of inverse semigroups

In Paterson's book "Groupoids, Inverse Semigroups and their Operator Algebras" he proves that for any r-discrete groupoid $G$ with unit space $G^0$, its full $C^* $-algebra $C^* (G)$ is ...
0 votes
0 answers
114 views

Clarifications sought on the paper on the semigroup associated with a free polynomial by Ali Abbas and Abdallah Assi

I have three questions regarding the proof of Proposition 4 on page 4 of this paper here. For those interested in addressing these questions, please refer to some definitions in the first two or three ...
3 votes
0 answers
161 views

On generators of the multiplicative semigroup $\{r\in\mathbb Q:\ r>1\}$

The set $M=\{r\in\mathbb Q:\ r>1\}$ is a commutative semigroup with respect to the multiplication. For any integers $a>b\ge1$, we clearly have $$\frac ab=\prod_{n=b}^{a-1}\frac{n+1}n.$$ So the ...
3 votes
0 answers
92 views

Reference for the monoidal category structure $X \otimes Y = X + Y + X \times Y$ on a distributive category

Given a distributive category $\mathscr C$ (more generally a rig category), we can define a (semicocartesian) monoidal category structure on $\mathscr C$ with tensor product given by $X \otimes Y := X ...
6 votes
0 answers
632 views

Generating functions in countable commutative monoids

Let $f: \mathbb{N}_0 \rightarrow \mathbb{C}$ be a function. The power series of $f$ can be viewed as the function $\mathscr{P}_f : q \mapsto \sum_{n \in \mathbb{N}_0}^{} f(n)q^n$ where $q \in \mathbb{...
5 votes
0 answers
191 views

Do most semigroups have a zero?

It is widely believed in finite semigroup theory that asymptotically almost all finite semigroups $S$, up to isomorphism, are 3-nilpotent, i.e., they satisfy $\#\{abc\,:\,a,b,c\in S\} = 1$. My ...
5 votes
3 answers
851 views

What are some examples of non-commutative $\mathbb{Q}$-monoids and/or $\mathbb{R}$-monoids?

Definition 0. Let $R$ denote a commutative semiring with $0$ and $1$. By an $R$-monoid, I mean a monoid $M$ equipped with an action $R \times M \rightarrow M$ denoted $r,m \mapsto m^r$, satisfying the ...
1 vote
1 answer
215 views

Reference about cancellation property for semigroups

Have the semigroups with the following cancellation property been studied? Property: Let $S$ be a semigroup and $x,y\in S$ such that $xz=yz,$ for all $z\in S,$ then $x=y$.
2 votes
1 answer
57 views

Are simplicial commutative inverse semigroups fibrant?

Let $X$ be a simplicial object in the category of commutative inverse semigroups (or monoids, if needed). Is the underlying simplicial set of $X$ always a Kan complex? If so, are there some nice ...
9 votes
1 answer
889 views

Combinatorics for the action of Virasoro / Kac–Schwarz operators: partition polynomials of free probability theory

In the background sections below, I establish the relations among characterizations of the action of Virasoro / Kac–Schwarz operators of 2D gravity models presented in terms of Laurent series by ...
10 votes
0 answers
248 views

What is the tiling semigroup for an einstein "hat" tiling?

My undergraduate dissertation was on inverse semigroups and the key text I used for it was Lawson's, "Inverse Semigroups: The Theory of Partial Symmetries". In said book, Lawson describes ...
4 votes
0 answers
174 views

Centers and conjugacy classes of groups relative to a pair of group homomorphisms

$\newcommand{\defeq}{\mathbin{\overset{\mathrm{def}}{=}}}$Given a group $G$, its center $\mathrm{Z}(G)$ and set of conjugacy classes $\mathrm{Cl}(G)$ are defined by \begin{align*} \mathrm{Z}(G) &\...
2 votes
1 answer
423 views

Conjecture about semigroups

Let $G$ be a finite semigroup with order $n$ odd. Let $S_i \in G, i=1,\ldots,\binom{n}{(n+1)/2}$ be all the subsets of $G$ of size $(n+1)/2$. Let $E(S_i)$ be the set obtained "expanding" $...
6 votes
0 answers
151 views

On dual notions of morphisms of algebraic structures obtained by replacing equaliser with coequalisers

This question is based on this discussion from the Category Theory Zulip. See also the earlier question Natural cotransformations and "dual" co/limits. Let $G$ and $H$ be groups. We define ...
4 votes
2 answers
544 views

Membership problem in monoids

What is the simplest example of a monoid with undecidable membership problem? In other words, I'm looking for a concrete monoid $S$ such that there is no algorithm which takes elements $s_1,...,s_n$ ...
5 votes
0 answers
187 views

Isbell duality for monoids and groups

Isbell Duality $\newcommand{\IsbellSpec}{\mathsf{Spec}}\newcommand{\IsbellO}{\mathsf{O}}\newcommand{\Sets}{\mathsf{Sets}}\newcommand{\rmL}{\mathrm{L}}\newcommand{\rmR}{\mathrm{R}}\newcommand{\B}{\...
0 votes
0 answers
95 views

Which algebraic structure characterizes the set of non-trivial qudratic residues in a finite field?

I understand this question may be too naive to ask, but I am unable to figure it out. Suppose, $\mathbb{QR^*}$ denotes the set of all quadratic residues in a finite field except the identity element $...
8 votes
2 answers
483 views

Posets obtained from a semigroup by the definition $x \leq y \iff x \cdot y = x$

A po-groupoid is a groupoid $\langle A,\cdot\rangle $ such that the relation defined by $$ x \leq y \text{ if and only if } x \cdot y = x $$ is a partial order on $A$, the order related to $\langle ...
8 votes
1 answer
236 views

Quiver and relations for a monoid related to Catalan numbers

Let $C_n$ be the monoid consisting of monotone maps $\{1,...,n\} \rightarrow \{1,...,n\}$ with $f(i) \leq i$ for all $i$. The cardinality of $C_n$ is given by the Catalan numbers. Consider $A_n= \...
3 votes
3 answers
345 views

Examples of cancellative normal semigroups

I've got a couple of things to test against normality in cancellative semigroups. A normal semigroup $S$ is one in which for any $x\in S$ we have $xS=Sx.$ This implies the Ore condition $$x,y\in S\...
5 votes
0 answers
107 views

Structure of well-ordered commutative monoids

Let $(M,+)$ be a commutative monoid. Let $<$ be a well-ordering on $M$, where $\forall a\in M,\ 0\leq a$ $\forall a,b,c\in M,\ a<b\Rightarrow a+c<b+c$ The first condition means $M$ will be ...
2 votes
1 answer
176 views

Generating totally ordered free commutative monoids

Let’s say I have a set $A$. I build the free commutative monoid $M$ generated by $A$. When can a well-order on $A$ be extended to $M$, in a way that is compatible with its monoid structure? I am ...
6 votes
1 answer
185 views

A name for semigroups in which left and right principal ideals coincide

Is there any standard name for semigroups $S$ in which $xS=Sx$ for all $x\in S$? Examples of such semigroups are commutative semigroups and Clifford inverse semigroups.
2 votes
0 answers
80 views

An alternative definition for finitely generated (and principal) ideals in a semigroup

Let $S$ be a semigroup. An ideal (of $S$) is a subset $I$ of $S$ such that $SI$ and $IS$ are both contained in $I$. The non-empty ideals constitute a subsemigroup, $\mathfrak I(S)$, of the power ...
5 votes
0 answers
160 views

$S$ and $T$ globally isomorphic semigroups, with $S$ (commutative and) cancellative, iff $S$ is isomorphic to $T$?

Denote by $\mathcal P(S)$ the semigroup obtained by equipping the non-empty subsets of a "ground semigroup" $S$ (written multiplicatively) with the operation of setwise multiplication ...
2 votes
0 answers
91 views

A recursive description of the smallest divisor-closed subsemigroup containing a set

Let $S$ be a semigroup and $\widehat{S}$ be its unitization, i.e., the monoid obtained from $S$ by adjoining an identity element if necessary (so that $\widehat{S} = S$ when $S$ is already a monoid). ...
13 votes
3 answers
978 views

Model Structure/Homotopy Pushouts in topological monoids?

Let $\mathsf C$ be the category of topological monoids, that is, the category of monoids in $(\textsf{Top}, \times)$. Can the model category structure on $\textsf{Top}$ (Serre fibrations, ...
3 votes
1 answer
171 views

Every homomorphism between (rational) Puiseux monoids is multiplication by a non-negative rational

Let a (rational) Puiseux monoid be a non-trivial submonoid of the non-negative rational numbers under (the usual operation of) addition. It is not difficult to show that, if $f \colon H \to K$ is a (...
6 votes
1 answer
173 views

References on semigroup actions

I posted this question on Math Stack Exchange about 10 days ago, but received no answer (https://math.stackexchange.com/q/4843881/1223994). I would like to ask for references on semigroup actions on ...
3 votes
0 answers
75 views

Are the automorphisms of the power semigroup of a cancellative semigroup cardinality-preserving?

Let $S$ be a semigroup (written multiplicatively) and $f$ be an automorphism of the power semigroup $\mathcal P(S)$ of $S$, that is, a bijective function on the family of all non-empty subsets of $S$ ...
3 votes
0 answers
103 views

An isomorphism problem for semigroups of ideals

An ideal of a semigroup $S$ (written multiplicatively) is a set $I \subseteq S$ such that $IS$ and $SI$ are both contained in $I$ (here, $XY$ means, for all $X, Y \subseteq S$, the setwise product of $...
1 vote
1 answer
142 views

Congruences that aren't "finite from above," take 2: semigroups

This is a hopefully less trivial version of this question. Briefly, say that a congruence is parafinite if it is the largest congruence contained in some equivalence relation with finitely many ...
6 votes
1 answer
622 views

When is the cofibrant replacement of a product the product of the cofibrant replacements?

I'm in a situation where I'd like to prove $Q(E\otimes E) \simeq QE \otimes QE$ for a monoid $E$ in a symmetric monoidal model category. I know it's not true in general that $Q(E\otimes F)\simeq QE \...
14 votes
4 answers
742 views

Prove or disprove: $R^{n+1} \supseteq R \cap R^2 \cap \cdots \cap R^n$ for every binary relation $R$ on a set of size $n$

Prove or disprove: $R^{n+1} \supseteq R \cap R^2 \cap \cdots \cap R^n$ for every binary relation $R$ on a set of size $n$. I have verified the statement for $n \leq 4$ with a Mathematica code. I have ...
0 votes
0 answers
92 views

What is the highest $n\in\Bbb N$ for which a complete classification of inverse semigroups of order up to $n$ is known?

What is the highest $n\in\Bbb N$ for which a complete classification of inverse semigroups of order up to $n$ is known? Given that there are $3{,}684{,}030{,}417$ semigroups of order $8$, I guess $n\...
7 votes
3 answers
525 views

Is the class of inverse semigroups globally determined?

This question is a follow-up to this one I asked on math.stackexchange. I've decided to ask here because I believe this is a research-level question. I'm sorry if I'm wrong -- I'm not a researcher ...

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