All Questions
25 questions
25
votes
2
answers
1k
views
The number of polynomials on a finite group
A function $f:X\to X$ on a group $X$ is called a polynomial if there exist $n\in\mathbb N=\{1,2,3,\dots\}$ and elements $a_0,a_1,\dots,a_n\in X$ such that $f(x)=a_0xa_1x\cdots xa_n$ for all $x\in X$. ...
- 41.9k
13
votes
1
answer
1k
views
For what sets $X$ do there exist a pair of functions from $X$ to $X$ with the identity being the only function that commutes with both?
It is not too difficult to show that if $X$ is an infinite set, then there exists a two-element subset of the group $\operatorname{Sym}(X)$ with trivial centralizer iff $\lvert X\rvert \leq \lvert\...
- 328
5
votes
1
answer
597
views
Can every cancellative invertible-free monoid be embedded in a group?
A monoid is invertible-free if $xy=1$ implies $x=y=1$ for all $x,y$.
Question: Can every cancellative invertible-free monoid be embedded in a group?
I'm fairly sure that a quotient of the free product ...
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 ...
- 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
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 ...
- 10.5k
3
votes
1
answer
203
views
Centralizer of a single element in the monoid of self-maps of a set
This is a follow-up to this question: For what sets $X$ do there exist a pair of functions from $X$ to $X$ with the identity being the only function that commutes with both?
Let $X$ be a set, and $X^...
- 63.9k
1
vote
0
answers
256
views
partially commutative monoid [closed]
Let $G$ be a simple graph with vertex $I$ and edge set $E$. I am defining $M(G)$ to be the quotient of the free monoid $I^*$ on $I$ by the relations $ab=ba$ and $c^2 = 1$ whenever $\{a,b\} \notin E(G)$...
- 1,269
51
votes
3
answers
3k
views
Is each squared finite group trivial?
A semigroup $S$ is defined to be squared if there exists a subset $A\subseteq S$ such that the function $A\times A\to S$, $(x,y)\mapsto xy$, is bijective.
Problem: Is each squared finite group ...
- 41.9k
41
votes
4
answers
2k
views
What is the probability two random maps on n symbols commute?
It is well known that two randomly chosen permutations of $n$ symbols commute with probability $p_n/n!$ where $p_n$ is the number of partitions of $n$. This is a special case of the fact that in a ...
- 38.6k
24
votes
3
answers
3k
views
Non-abelian Grothendieck group
By general nonsense the forgetful functor from groups to monoids has a left adjoint. It maps a monoid $(X,\cdot,1)$ to the free group on $\{\underline{x} : x \in X\}$ modulo the relations $\underline{...
- 63.1k
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 ...
- 38.6k
11
votes
2
answers
574
views
Identifying a group without 2-torsion
Suppose we have a finitely presented group $G$ with solvable word problem. (For instance, the command RWSGroup in Magma terminates giving us a finite [but possibly gigantic] rewrite system.) Is there ...
- 18.7k
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
10
votes
1
answer
409
views
Does every set have a rigid self-map?
The question was asked on Mathematics Stackexchange
but has remained unanswered so far.
A self-map is a map $f:X\to X$ from a set $X$ to itself. There is an obvious notion of morphism, and thus of ...
- 4,416
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?
- 3,110
6
votes
0
answers
190
views
The highest degree of a polynomial on a finite group
This question is motivated by the comments and the answer to this MO-question.
First let us recall some definitions.
A function $f:X\to X$ on a group $X$ is called a polynomial if there exists $n\in\...
- 41.9k
6
votes
1
answer
371
views
Embedding a cancellative monoid into another in such a way that a prescribed element becomes left-invertible
Let $\mathbb A = (A, +_A)$ be a cancellative, but possibly non-commutative, monoid with identity $0$, and fix an element $x \in A$. Does there always exist a cancellative monoid $\mathbb B = (B, +)$ ...
- 10.5k
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 ...
- 38.6k
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
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
4
votes
1
answer
446
views
What is a "cusp" ("кусок") in relation to Guba's embedding theorem?
I'm confused by the definition of a "cusp" as found in
V.S. Guba, Conditions for the embeddability of semigroups into groups, Math. Notes 56 (1994), Nos. 1-2, 763-769 (link).
In the words of Mark ...
- 10.5k
2
votes
1
answer
271
views
Apropos of two groups being globally isomorphic iff they are isomorphic
Denote by $\mathcal P(S)$ the semigroup obtained by endowing the non-empty subsets of a "ground semigroup" $S$ (written multiplicatively) with the operation of setwise multiplication induced ...
- 10.5k
1
vote
1
answer
231
views
Continuous semigroup homomorphism of composition to additive structure
Let $G$ be the topological semigroup whose underlying space is $C(\mathbb{R}^d,\mathbb{R}^d)$ equipped with composition as semigroup operation and let $H$ be the topological group whose underlying ...
- 5,405
-8
votes
1
answer
351
views
Are there overwhelmingly more finite monoids than finite spaces? [closed]
A function $f:\mathbb{Z}_{\geq 1}\to\mathbb{Z}_{\geq 1}$ overwhelms $g:\mathbb{Z}_{\geq 1}\to\mathbb{Z}_{\geq 1}$ if for any $k\in \mathbb{Z}_{\geq 1}$ the inequality $f(n)\leq g(n+k)$ holds only for ...
- 23