All Questions
Tagged with gr.group-theory semigroups-and-monoids
120 questions
0
votes
0
answers
61
views
Defining rank of an abelian subgroup using the second centralizer
I recently posted this on MSE, but didn't receive any feedback; so I'm posting it on MO.
I recently came across this article which explored the maximal abelian subgroups of the symmetric group $S_n$. ...
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 ...
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,...
4
votes
0
answers
174
views
Centers and conjugacy classes of groups relative to a pair of group homomorphisms
$\newcommand{\defeq}{\mathbin{\overset{\mathrm{def}}{=}}}$Given a group $G$, its center $\mathrm{Z}(G)$ and set of conjugacy classes $\mathrm{Cl}(G)$ are defined by
\begin{align*}
\mathrm{Z}(G) &\...
6
votes
0
answers
151
views
On dual notions of morphisms of algebraic structures obtained by replacing equaliser with coequalisers
This question is based on this discussion from the Category Theory Zulip. See also the earlier question Natural cotransformations and "dual" co/limits.
Let $G$ and $H$ be groups. We define ...
2
votes
1
answer
423
views
Conjecture about semigroups
Let $G$ be a finite semigroup with order $n$ odd. Let $S_i \in G, i=1,\ldots,\binom{n}{(n+1)/2}$ be all the subsets of $G$ of size $(n+1)/2$.
Let $E(S_i)$ be the set obtained "expanding" $...
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 ...
0
votes
1
answer
327
views
Can we generalise groupoids to monoid-oids? [closed]
Groups correspond to one object categories where every morphism is an isomorphism. Monoids correspond to one object categories.
Groupoids correspond to small categories where every morphism is an ...
3
votes
2
answers
257
views
Cancelable commutative monoids with finite maximal subgroups
Suppose $\mathcal{M} = (M, +, 0)$ is a cancelable commutative monoid. Let $G$ be the maximal subgroup of $M$, i.e.
$$G = \{ a \in M \colon (\exists b \in M)\, a + b = 0 \}.$$
For $a, b \in M$ say $a \...
6
votes
0
answers
183
views
Examples of groups with a positive homogeneous presentation without the Haagerup property or not of type $F_\infty$
I am looking for groups with a certain presentation that do not have the Haagerup property or are finitely presented but not of type $F_\infty$ (meaninig that for some $n\geq 3$ we cannot find any ...
1
vote
0
answers
70
views
Another matrices for a semigroup with intermediate growth
Nathanson showed that the Okninski's semigroup $S$ of $2×2$ matrices which is generated by the set $H=\{A,B\}$, where
$
A=\begin{bmatrix}
1&1\\
0&1\\
\end{bmatrix}
,
B=\begin{bmatrix}
1&0\\...
1
vote
0
answers
274
views
Functional equation $f(x*y) = f(f(x)*f(y))$
Find all endo-functions $f$ on a commutative semigroup $(\mathbb{S},*)$ such that
$f(x*y) = f(f(x)*f(y))$.
Typical case of interest are $(\mathbb{N},+)$ or $(\mathbb{Z}/k\mathbb{Z},+)$ or $(\mathbb{Z}/...
4
votes
0
answers
74
views
Is each TS-topologizable group TG-topologizable?
Definition 1. A topology $\tau$ on a group $X$ is called
$\bullet$ a semigroup topology if the multiplication $X\times X\to X$, $(x,y)\mapsto xy$, is continuous in the topology $\tau$;
$\bullet$ a ...
7
votes
0
answers
295
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 ...
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\...
0
votes
0
answers
41
views
Polyextremal groups
A polynomial of a semigroup $X$ is a function $f:X\to X$ of the form
$f(x)=a_0xa_1\cdots xa_n$, where $a_0,a_1,\dots,a_n$ some elements of the semigroup $X^1=X\cup\{1\}$, called the coefficients of ...
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 $x\in X$. ...
7
votes
0
answers
138
views
The smallest cardinality of a cover of a group by algebraic sets
$\DeclareMathOperator\cov{cov}$A subset $A$ of a semigroup $X$ is called algebraic if $$A=\{x\in X: a_0xa_1x...xa_n=b\}$$ for some $b\in X$ and $a_0,a_1,...,a_n \in X^1=X\cup \{1\}$. The smallest ...
5
votes
0
answers
138
views
Can we define partial group actions on (finite) sets via generators and relators?
Let $G = \langle Y | R \rangle$ be a finitely presented group. A partial group action on a set $X$ is a premorphism into the inverse semigroup
$$
\mathcal I (X) = \{ f: A \to B : A, B \subseteq X, f\...
4
votes
1
answer
417
views
Is a solvable group satisfying a semigroup law?
Let $S$ be the free semigroup on the set $\{x_1,\ldots ,x_n\}$, where $n$ is a positive integer. Suppose that $\mu=\mu (x_1,\ldots ,x_n)$ and $\nu = \nu (x_1,\ldots ,x_n)$ are two elements in $S$. We ...
0
votes
1
answer
139
views
Computationally intractable orbit of a monoid action on a finite set
Suppose for each integer $n\geq 1$ we have a submonoid $M_n\subset \mathrm{Self}(\{1, \dots, n\})$ of self-maps of $\{1, \dots, n\}$.
A characterization of $M_n$ is an algorithm that takes an integer $...
2
votes
0
answers
203
views
Can the relation between count of commuting pairs and conjugacy classes for finite groups be generalized to semigroups?
It is well-known that number of pairs of commuting elements in finite group G is equal to number of conjugacy classes multiplied by cardinality of G.
More generally here (MO275769) Qiaochu Yuan ...
-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 ...
2
votes
0
answers
74
views
Terminology and notation for generated subgroups
I would like to think about formation of the smallest subgroup (or monoid, or whatever) $H$ of $G$ containing two given subgroups $A$ and $B$ as an operation on subgroups, and I wonder if there is a ...
4
votes
0
answers
72
views
When is the submonoid preserving a subspace finitely generated?
Let $T$ be a topological space with at least one open set whose closure is not open.
Let $G$ be a finitely generated group acting by homeomorphisms on $T$. Let $S\subset T$ be a subspace.
Under what ...
12
votes
2
answers
785
views
Is the Petersen graph a "Cayley graph" of some more general group-like structure?
The Petersen graph is the smallest vertex-transitive graph which is not a Cayley graph. Is it the "Cayley graph" of some slightly more general group-like structure?
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 ...
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 ...
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 ...
5
votes
1
answer
208
views
Conjugacy classes of monoids II: Abelianising a monoid, wrongly
$\newcommand{\unsim}{\mathord{\sim}}$Let $G$ be a group. What is
$$
G/\left(ab\sim ba\ \middle|\ a,b\in G\right)?
$$
Answer: not $G^{\mathrm{ab}}$, but the set of conjugacy classes of $G$.
When ...
0
votes
0
answers
250
views
Has this theorem on cancellative monoid actions been discovered and published?
Does a statement equivalent to Theorem 3 below appear in the literature? If it does, what is the earliest published reference?
Theorem 1. Let $W$ be a non-trivial cancellative invertible-free [1] ...
5
votes
2
answers
804
views
Cancellation property for commutative monoid
Let $(M,+,e)$ be a commutative monoid with unit $e$. An element $a\in M$ is called cancellative element if
for any $b,c \in M$ such that $a+b=a+c$ implies that $b=c$.
Let $(\mathbf{N},+,0)$ the ...
10
votes
1
answer
422
views
Generalized cancelation properties ensuring a monoid embeds into a group
Context: an obvious necessary condition for a monoid to embed into a group (as submonoid) is to satisfy the left and right cancelation rules:
$$xy=xz \quad\Longrightarrow y=z;$$
$$yx=zx \quad\...
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 ...
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 ...
6
votes
1
answer
202
views
Identities of finite inverse semigroups
An inverse semigroup is an algebra with two operations: binary $\cdot$ and unary $^{-1}$ such that $\cdot$ is associative and $xx^{-1}x=x, xx^{-1}yy^{-1}=yy^{-1}xx^{-1}$. The Brandt semigroup with 1, $...
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^...
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\...
3
votes
1
answer
125
views
Quasi-isometries and E-unitary inverse semigroups
Let $S = \langle K\rangle$ be a finitely generated inverse semigroup, where $K \subset S$ is a fixed, finite and symmetric set of generators.
Preliminaries: Recall that we say that $s, t \in S$ are $\...
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 ...
2
votes
0
answers
91
views
Is the natural action of the monoid of endomorphisms is a complete invariant for group?
Let $\alpha$ and $\beta$ be actions of semigroups $A$ and $B$ on sets $X$ and $Y$ respectively. Recall that $\alpha$ and $\beta$ are called isomorphic if there exists an isomorphism $\phi$ between ...
14
votes
1
answer
792
views
Associativity may fail by little?
It is a well-known result on group theory that if a group has many pairs of commuting elements then it is abelian.
This motivated the following pseudo-conjecture.
If a (possibly infinite) set $S$ ...
1
vote
1
answer
326
views
Closed submonoid of $(\mathbb{C}^*)^n$
The answer of this question might be known but I was not able to find any answer. Let $n\geq 1$ and $S$ be a closed submonoid of $(\mathbb{C}^*)^n$, that is, a closed and stable by product subset of $(...
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 ...
18
votes
2
answers
1k
views
Monoids of endomorphisms of nonisomorphic groups
Can monoids of endomorphisms of nonisomorphic groups be isomorphic ?
7
votes
0
answers
260
views
Generating the monoid of injective endomorphisms of the free group
Let $F$ be the free group of rank $2$ (or any finite rank if this does not matter). The set of injective group endomorphisms $F\to F$ forms a monoid $M$ by compositions. Is there a simple looking set ...
8
votes
2
answers
585
views
Is the equational theory of groups axiomatized by the associative law?
Consider the class of groups in the signature {*}. Is the equational theory of that class axiomatized by the associative law? I asked this on math stack exchange but I didn't receive a satisfactory ...
4
votes
1
answer
307
views
Characterization of Archimedean linearly ordered monoids
In this question, it is shown that all Archimedean ordered groups are isomorphic to an ordered subgroup of $\mathbb R$. Additionally, it is shown that if such a group is complete, then it is ...
-3
votes
1
answer
234
views
A common name for a functorial construction of Commutative Algebra?
I am interested whether the following construction naturally appearing in Commutative Algebra has some know and acceped name.
Given a commutative monoid $(M,+)$ and a set $X$, consider the family $F(...
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 $a,A,b$...