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18 votes
1 answer
783 views

Are there any "simple" monoids with intermediate growth?

The discovery of the Grigorchuk group which has intermediate growth caused a number of other such groups to be found, but they are all fairly complicated, and as far as I know none of them are ...
saolof's user avatar
  • 1,947
10 votes
1 answer
274 views

A flatness result of Fiedorwicz for amalgamated free products of monoids in connection with classifying spaces of monoids

In Lemma 5.2(a) of Z. Fiedorowicz, Classifying Spaces of Topological Monoids and Categories American Journal of Mathematics Vol. 106, No. 2 (Apr., 1984), pp. 301-350 the author proves the following. ...
Benjamin Steinberg's user avatar
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,...
Salvo Tringali's user avatar
8 votes
1 answer
321 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 ...
Salvo Tringali's user avatar
7 votes
0 answers
294 views

A minimal semigroup generating subset of the additive reals

I asked this on MSE, but I was told to ask it here because it is a difficult question. Consider the additive magma of the real numbers, $(\mathbb{R};+)$. Does there exist a subset $S$ of the reals ...
user107952's user avatar
  • 2,013
6 votes
1 answer
903 views

Monoids and groups of fractions

Let $G$ be a group containing a monoid $M$ that spans $G$ as a group. Is it possible to have a proper quotient $\varphi \colon G \to Q$ of $G$ such that the restriction of $\varphi$ to $M$ is ...
Stefan Witzel's user avatar
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 ...
goblin GONE's user avatar
  • 3,793
5 votes
2 answers
364 views

Have semigroups with actions on themselves that have a dual to the compatibility axiom ever been studied?

For a semigroup $G$ with a left action on itself, the axiom for compatibility becomes: $$ \forall f,g,h\in G:hg(f)=h(g(f)) $$ Now suppose there is additional axiom, or constraint if you prefer, ...
James Smith's user avatar
4 votes
1 answer
434 views

Regarding a new algebraic structure

By "left semigroup-joined-semigroup" I mean an algebraic structures $(S,\cdot,*)$ such that both $\cdot,*$ are associative, and the following property holds (see this ) $$ x*(y\cdot z)=x*y*z\;\; ; \;...
M.H.Hooshmand's user avatar
4 votes
1 answer
154 views

Right reversibility of submonoids of nilpotent groups

Let $G$ be a finitely generated group (optionally torsion-free). Let $N$ be a submonoid of $G$ (that is, a subsemigroup with $1$). A (cancellative) monoid/semigroup $S$ is right reversible if for ...
David Handelman's user avatar
4 votes
0 answers
158 views

Is the group ring of an amenable group, viewed as multiplicative monoid, amenable?

Motivated by this question, it seems natural to ask the following: Question 1: Is there a [finitely generated discrete] torsion-free virtually Abelian (but not Abelian) group $G$ so that the ...
ARG's user avatar
  • 4,432
3 votes
2 answers
1k views

Semiring naturally associated to any monoid?

For any monoid $M$, we can naturally construct a semiring $S$ as follows: Let the additive monoid of $S$ be the free commutative monoid on $M$ Let the multiplicative monoid of $S$ be $M$ Then, if ...
Mike Battaglia's user avatar
2 votes
1 answer
229 views

Has the "semidirect monoid of a semiring" been considered anywhere?

Given a semiring $S$, we get a monoid $M(S)$ as follows: The underlying set of $S$ is $S^2$ The identity element is $(0,1)$ The law of composition is given by $$(a,A)(b,B) = (Ba+b,AB),$$ where $a,A,b$...
goblin GONE's user avatar
  • 3,793
0 votes
0 answers
250 views

Has this theorem on cancellative monoid actions been discovered and published?

Does a statement equivalent to Theorem 3 below appear in the literature? If it does, what is the earliest published reference? Theorem 1. Let $W$ be a non-trivial cancellative invertible-free [1] ...
David Pokorny's user avatar