All Questions
14 questions
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 ...
- 10.5k
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,...
- 10.5k
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 ...
- 2,013
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 ...
- 1,947
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 ...
- 4,432
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] ...
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 $a,A,b$...
- 3,793
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.
...
- 38.6k
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 )
$$
x*(y\cdot z)=x*y*z\;\; ; \;...
- 995
4
votes
1
answer
154
views
Right reversibility of submonoids of nilpotent groups
Let $G$ be a finitely generated group (optionally torsion-free). Let $N$ be a submonoid of $G$ (that is, a subsemigroup with $1$).
A (cancellative) monoid/semigroup $S$ is right reversible if for ...
- 4,679
5
votes
2
answers
364
views
Have semigroups with actions on themselves that have a dual to the compatibility axiom ever been studied?
For a semigroup $G$ with a left action on itself, the axiom for compatibility becomes:
$$
\forall f,g,h\in G:hg(f)=h(g(f))
$$
Now suppose there is additional axiom, or constraint if you prefer, ...
- 505
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 ...
- 3,793
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 ...
- 2,050
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 ...
- 4,897