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Continuity of multiplicative character

Let $G$ be a discrete group and $\beta (G)$ denote the Stone-Cech compactification of $G$, a right topological semigroup. By a multiplicative character, I mean a mapping that preserves multiplication ...
nick's user avatar
  • 61
3 votes
0 answers
166 views

A question of terminology - Unitizations of semigroups

There are at least two standard ways of unitizing a (small) semigroup $\mathbb A$: (i) We add an identity regardless that $\mathbb A$ is already unital. (ii) We add an identity only if none is ...
Salvo Tringali's user avatar
4 votes
1 answer
215 views

Name for a regular band

Is there a name for regular bands that satisfy $xyx=yx$ for all $x$,$y$?
user avatar
16 votes
2 answers
1k views

The symmetric monoidal category of finite sets

It is well-known that the (augmented) simplex category is the universal monoidal category with a monoid object. What about a commutative analogue? Consider the category $\mathsf{FinSet}$ of finite ...
Martin Brandenburg's user avatar
9 votes
2 answers
2k views

What is the free monoidal category generated by a monoid?

In several places in a segment on cohomology (for example, here (PDF)) in John Baez's online lecture notes for a course in 2007 on quantum gravity, much is made of the fact that the simplex category $...
ziggurism's user avatar
  • 1,446
1 vote
1 answer
216 views

Counting modular squares in an interval

For an integer $m$, let $S^m_{x_0,x_1} = \{ t | x_0 ≤ t ≤ x_1 $ and $t$ is a square modulo $m \}$. Let $S^m_x$ = $S^m_{0,x}$. Determining whether the sets $S^m_x$ are empty is easy (1 is always a ...
Stephan Wehner's user avatar
1 vote
2 answers
375 views

What are the monoids in which every globally idempotent subsemigroup contains the identity element?

A semigroup is called globally idempotent when for any $x\in S$ there are $y,z\in S$ such that $x=yz$. Is there a name for monoids whose every globally idempotent subsemigroup contains the identity ...
Michał Masny's user avatar
0 votes
1 answer
403 views

When does a power semigroup have a zero, and what can the zero be?

Let $S$ be a semigroup. The power semigroup of $S$ is the set $P(S)=2^S\setminus\lbrace\varnothing\rbrace $ with the operation $$AB=\lbrace ab\ |\ a\in A,\ b\in B\rbrace.$$ This operation is ...
Michał Masny's user avatar
32 votes
5 answers
9k views

How many binary operations are associative?

Let $X$ be a finite set of $n$ elements, and consider a binary operation $\odot: X \times X \rightarrow X$. There are $n^{n^2}$ such binary operations, as the $n \times n$ table entries can each be ...
Joseph O'Rourke's user avatar
7 votes
2 answers
544 views

A linearly orderable monoid which does not embed into a linearly orderable group

It is known (after an example of A.I. Mal'cev) that there exist cancellative semigroups which do not embed into a group. On the other hand, it is not difficult to see that every linearly orderable ...
Salvo Tringali's user avatar
2 votes
2 answers
475 views

On the notion of torsion-freeness in semigroup theory

The following seems to be the "official" notion of torsion-freeness in the context of semigroups: TF1. A (multiplicatively written) semigroup $\mathfrak A$ is torsion-free if there do not ...
Salvo Tringali's user avatar
4 votes
1 answer
518 views

Strictly totally ordered semigroups - Looking for references

Let $\mathfrak A = (A, \cdot)$ be a semigroup (written multiplicatively). We say that $\mathfrak A$ is linearly orderable if there exists a total order $\le$ on $A$ such that $ac < bc$ and $ca < ...
Salvo Tringali's user avatar
4 votes
1 answer
297 views

Reference for subsemigroups of $\mathbb{N}^n$

A well known result about the natural numbers $\mathbb{N}$ says that for any finite subset $A \subset \mathbb{N}$ there exists $R \ge 0$ such that if $n$ is in the subgroup of $\mathbb{Z}$ generated ...
Lee Mosher's user avatar
  • 15.4k
3 votes
1 answer
243 views

Embedding Semigroups in Rings

Let $S$ be a finite commutative semigroup with identity. Under what conditions (on the semigroup $S$) it is possible to find a ring $R$ such that the multiplicative structure of $R - \{0\}$ is ...
zacarias's user avatar
  • 801
6 votes
0 answers
618 views

Duality between conjugacy classes and irreducible characters for finite monoids?

Qiaochu's answer to this question suggests that the proper way to view the bijection between conjugacy classes and irreducible complex representations of a finite group is via a duality. My question ...
Benjamin Steinberg's user avatar
4 votes
2 answers
357 views

Minimal right ideals in finite semigroup

Let $E$ be a finite semigroup. According to N. Bourbaki (Algèbre I p. 121 exerc. 14 c), if $M$ and $M'$ are minimal right ideals in $E$, then they are isomorphic. I spent some time browsing through ...
Pnine's user avatar
  • 43
3 votes
1 answer
297 views

Is the universal inverse semigroup of a commutative semigroup an embedding?

The question of existence of a universal inverse semigroup of an arbitrary semigroup has been answered before (this is a construction similar to the Grothendieck group). Let's refer to the universal ...
Thomas Klimpel's user avatar
2 votes
2 answers
286 views

Idempotent semigroups: Are they all residually finite?

As pointed out by Mark Sapir in his answer to a related question, every residually finite divisible semigroup is idempotent (hence uniquely divisible). On another hand, it is not difficult to prove ...
Salvo Tringali's user avatar
2 votes
1 answer
360 views

On a property of subsemigroups

Let $H$ denote a subsemigroup of a semigroup $G$. I'm interested in the following property: $$\forall g\in G\exists h\in H:gh\in H.$$ This property is weaker than the property that $H$ is an ideal ...
John's user avatar
  • 21
4 votes
1 answer
314 views

A semigroup with the property that $x^n = a$ has at least one solution

Is there a standard name for a (multiplicatively-written) semigroup $(A, \cdot)$ such that, given an arbitrary $a \in A$, the equation $x^n = a$ has at least one solution $x \in A$ for each $n \in \...
Salvo Tringali's user avatar
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 ...
Michał Masny's user avatar
10 votes
2 answers
1k views

Connective spectra versus simplicial abelian groups - very basic question

Hello, I have very general , "introductory" questions (It is quite hard for me to seek for specific things in the algebraic topology literature). I guess that connective spectra have a model ...
Sasha's user avatar
  • 5,562
8 votes
2 answers
427 views

Is there a general result that theorems about finite structures proved in ZFC can be proved in ZF?

The title question is too vague so let me be specific. Much of modern finite semigroup theory uses profinite semigroups and properties of profinite semigroups that depend on the existence of prime ...
Benjamin Steinberg's user avatar
5 votes
3 answers
718 views

Subsets of $\mathbb{R}^+$ closed under addition

No one's answered the question cumulant problem so here's a simpler question: Has anyone described or catalogued all sets of non-negative real numbers that are closed under addition? In particular, ...
Michael Hardy's user avatar
8 votes
2 answers
826 views

Which semigroups can be linearly ordered?

As usual I consider a semigroup to be a structure $(A, +)$ such that $+$ is an associative binary function over the set $A$. The notion of linearly-ordered semigroup corresponds to structures of the ...
boumol's user avatar
  • 768
1 vote
2 answers
442 views

submonoids of Z_n

Anyone knows how to describe explicitly the submonoids of Z_n, regarded as a multiplicative monoid?
Camilo Salazar's user avatar
11 votes
3 answers
942 views

What is the smallest variety of algebras containing all fields?

A field is a ring whose nonzero elements form a commutative group under multiplication. A field is also a commutative inverse semigroup with respect to multiplication. The unique multiplicative ...
Thomas Klimpel's user avatar
4 votes
0 answers
396 views

Is there a homological way to compute quiver presentations?

I have recently been studying with colleagues the representation theory of certain finite monoids that come up in probability theory and combinatorics, see Ken Brown's beautiful survey here. These ...
Benjamin Steinberg's user avatar
22 votes
2 answers
977 views

Mapping from a finite index subgroup onto the whole group

Dear All, here is the question: Does there exist a finitely generated group $G$ with a proper subgroup $H$ of finite index, and an (onto) homomorphism $\phi:G\to G$ such that $\phi(H)=G$? My guess ...
Victor's user avatar
  • 1,437
5 votes
1 answer
304 views

flat maps of monoids which are not localizations

It is well known that a localization $S^{-1}R$ of a commutative ring $R$ is flat as a $R$-module. Rather, I am looking for extensions of rings which share certain properties of localizations, like ...
Ricardo Andrade's user avatar
1 vote
3 answers
585 views

Terminology for certain monoids which are to monoids like fields are to rings

Let $M$ be a commutative monoid with zero. Then the condition $M^* = M \setminus \{0\}$ is very similar to the condition for a commutative ring to be a field. This analogy is also used in the work "...
14 votes
2 answers
1k views

Economical hard word problem

Can anyone give me an example of a very simple word problem, where by "simple" I mean that it has very few generators and relations, that is nevertheless insoluble. To make the question easier, I am ...
gowers's user avatar
  • 29k
22 votes
2 answers
1k views

Toposes (topoi) as classifying toposes of groupoids

A famous theorem of Joyal and Tierney says that each Grothendieck topos is equivalent to the classifying topos of a localic groupoid. I believe that Butz and Moerdijk have shown that if the topos has ...
Benjamin Steinberg's user avatar
4 votes
0 answers
331 views

What is the pro-algebraic completion of the free semigroup on one generator?

This question is motivated by an attempt to understand what is going on in Tom's post from a certain point of view. Let $\mathbb N^+$ be the free semigroup on one generator (so the positive natural ...
Benjamin Steinberg's user avatar
2 votes
1 answer
301 views

Is there existing terminology for this technical condition on semilattices?

Given a semilattice $S$, a subset $E$, and a positive integer $n$, let $E^{[n]}$ be the set of all products of $n$-tuples in $E$. Thus $\bigcup_{n\geq 1} E^{[n]}$ is nothing but the subsemigroup of $S$...
Yemon Choi's user avatar
  • 25.8k
15 votes
2 answers
1k views

Exact sequence of monoids

What is the right definition of an exact sequence of monoid homomorphisms? I can't seem to find a consistent in my searches; indeed Balmer (Remark 2.6, http://www.math.ucla.edu/~balmer/Pubfile/...
John Voight's user avatar
  • 3,009
5 votes
0 answers
2k views

Is the radical of a homogeneous ideal homogeneous?

Let $S$ be an $M$-graded $R$-algebra, where $M$ is some monoid, and $I\subset S$ an homogeneous ideal. The original, naïve, question, was: is it true that $\sqrt{I}$ is homogeneous? In this generality,...
quim's user avatar
  • 1,811
12 votes
1 answer
744 views

Is the following construction of the 0-Hecke monoid (well) known?

Let W be a Coxeter group with Coxeter generators S. The corresponding 0-Hecke monoid H(W) has generating set S, the braid relations of W and the relations that each element of S is an idempotent. If ...
Benjamin Steinberg's user avatar
7 votes
1 answer
722 views

How is called a semigroup...

Does anyone know, how is called a semigroup in which every equation $ax=b$ has only a finite set (maybe empty) of solutions?
Boris Novikov's user avatar
3 votes
1 answer
529 views

Study of free monoids of the recursive S. Eilenberg.

Compared to the usual treatises on recursion (eg, Rogers H. "Computability and Undecidability." McGraw-Hill, New York) the book of Samuel Eilenberg & Calvin C. Elgot "Recursiveness" treats such ...
Buschi Sergio's user avatar
3 votes
1 answer
367 views

submonoid of a matrix monoid with a common eigenvector

Hello, I am considering two real invertible $3\times 3$ matrices $A$ and $B$ and a nonzero vector $v\in\mathbb{R}^3$ and i am wondering if the submonoid $E$ of the monoid $(A,B)$ genererated by $A$ ...
lbdl's user avatar
  • 69
15 votes
0 answers
1k views

Is the category of smooth manifolds equivalent to the opposite category of the category of commutative monoids of some additive symmetric monoidal category?

This is a followup to my previous question, which asked whether the category of commutative or noncommutative C*-algebras or von Neumann algebras is equivalent to the category of commutative or ...
Dmitri Pavlov's user avatar
0 votes
0 answers
179 views

semigroup actions of groups on regular rooted trees

If $G$ is a group which has a semigroup action on a regular rooted tree via prefix-preserving, continuous transformations (I give the tree the path metric), what kinds of algebraic restrictions can we ...
dan's user avatar
  • 125
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 \...
David White's user avatar
  • 30.3k
1 vote
1 answer
260 views

The intersection of Block Groups and R-trivial (finite) monoids

Let $\textbf{BG}$ be the pseudovariety of block groups, also known as $\textbf{EJ}, \textbf{PG},\ldots,\text{etc.}$(see [1]), and let $\textbf{R}$ be the pseudovariety of R-trivial monoids, by the ...
Xorwell's user avatar
  • 424
7 votes
1 answer
266 views

Positive cone of a subgroup of $\mathbb{Z}^n$

This question sounds like it should be very well known, but for some reason I failed to find a decent answer anywhere. Let $G\subset\mathbb{Z}^n$ be a subgroup, and $G_+=G\cap\mathbb{Z}_{\ge0}^n$ be a ...
Vladimir Dotsenko's user avatar
11 votes
1 answer
949 views

Magma "actions" (or alternatively, "What is the Yoneda lemma for magmas?")

Arguably the most import thing about groups, semigroups and more generally categories, is that they can act on sets (or even collections of sets in the case of a category). This is the basis for all ...
Mikola's user avatar
  • 2,392
6 votes
1 answer
516 views

Growth zeta-functions of regular languages

Dear All, my following question may be known and ought to be known, so in case it is folklore please could you give me the references. To start, it is obvious that growth of rational languages are ...
Victor's user avatar
  • 1,437
5 votes
2 answers
754 views

Do all finitely generated nilpotent semigroups have polynomial growth?

The notion of nilpotency passes nicely from groups to semigroups. Define $q_1(x,y)=xy$ and $$q_{i+1}(x,y,z_1,\cdots,z_i)=q_i(x,y,z_1,\cdots,z_{i-1})z_iq_i(y,x,z_1,\cdots,z_{i-1})$$ inductively for all ...
Victor's user avatar
  • 1,437
7 votes
1 answer
433 views

Powers of maps on finite sets

Let $X_n$ be a set with $n$ elements. Write $F(X_n,X_n)$ for the set of maps from $X_n$ to itself. It is a monoid under the operation of composition. Let $m$ be a positive integer. How many maps in $...
Steven Spallone's user avatar

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