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2 answers
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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 $x\in X$. ...
Taras Banakh's user avatar
  • 40.2k
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\...
cha21's user avatar
  • 328
5 votes
1 answer
517 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 ...
David Pokorny's user avatar
7 votes
2 answers
492 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
3 votes
1 answer
188 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^...
YCor's user avatar
  • 59.4k
1 vote
0 answers
242 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 $\{a,b\} \notin E(G)$...
GA316's user avatar
  • 1,209
49 votes
3 answers
2k 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 ...
Taras Banakh's user avatar
  • 40.2k
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 ...
Benjamin Steinberg's user avatar
23 votes
3 answers
2k 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{...
Martin Brandenburg's user avatar
12 votes
1 answer
705 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
11 votes
2 answers
536 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 ...
Pace Nielsen's user avatar
  • 17.6k
10 votes
1 answer
399 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 ...
Pierre-Yves Gaillard's user avatar
10 votes
1 answer
261 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
7 votes
1 answer
718 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
6 votes
0 answers
587 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
6 votes
0 answers
188 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\...
Taras Banakh's user avatar
  • 40.2k
6 votes
1 answer
358 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 $\mathbb B = (B, +)$ ...
Salvo Tringali's user avatar
5 votes
2 answers
355 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
430 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 ...
Salvo Tringali's user avatar
4 votes
0 answers
154 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
  • 4,342
2 votes
1 answer
221 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 ...
Salvo Tringali's user avatar
1 vote
1 answer
228 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 ...
ABIM's user avatar
  • 5,001
-7 votes
1 answer
335 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 ...
firn's user avatar
  • 25