A semigroup is an algebraic structure consisting of a set together with an associative binary operation. A semigroup with an identity element is called monoid.

learn more… | top users | synonyms (1)

5
votes
0answers
41 views

Semigroups and solutions of equation

It is easy to prove: in a finite semigroup if for all $a$ and $b$, $ax=b$ and $ya=b$ has unique solution. then it is group. But if in a finite semigroup, if for all $a$ and $b$, $ax=b$ and $ya=b$ has ...
4
votes
8answers
403 views

Does a non-abelian semigroup without identity exist?

I was introduced to semigroups today and had a question. So all the examples of semigroups I was given were either monoids or groups. So I was curious, does there exist a semi-group which is not ...
4
votes
0answers
32 views

Semigroup satisfying the cancellation property which cannot be embedded in a group

There is a basic result about commutative semigroups which says Any commutative semigroup satisfying the cancellation property can be embedded in a group. However, I read that Not every ...
1
vote
0answers
48 views

Schrödinger Semigroup is compact if potential goes to infinity

Several papers (e.g. this one: arXiv:0810.3275v1 [math.SP] 17 Oct 2008 ) claim that if $H=-\Delta+V$ and $V(x)\to\infty$ if $|x|\to\infty$, then the semigroup $e^{-tH}$ is eventually compact. Does ...
6
votes
1answer
88 views

Is there a name for a regular semigroup with zero in which the product of any two different idempotents is zero?

As the title says, my question is: is there a name for a regular semigroup with zero in which the product of any two different idempotents is zero? Note that any such semigroup is necessarily an ...
9
votes
4answers
312 views

Partition of $\mathbb{R}^+$ into two semigroups

Can the semigroup $(\mathbb R^+ ,+)$ be partitioned into two semigroups? I have been trying but haven't found anything, please help.
11
votes
6answers
463 views

Just How Strong is Associativity?

A friend of mine is using a lot of algebra that is not associative for an advanced Chemistry project. We were discussing it recently and I found it rather amusing how often she said things like ...
-5
votes
1answer
41 views

Question on Semi-groups [closed]

Let (A, *) be a semigroup, Furthermore, for every a and b in A, if a!=b, then a*b != b*a. (a) Show that for every a in A a*a =a (b) Show that for every a, b in A a*b*a =a (c) Show that for every a, ...
1
vote
1answer
47 views

How to show that S/J is a chain where S is a semigroup with no proper generating sets

I'm Struggling with exercise 4.7.17 of Howies Fundamentals of Semigroup theory Let $S$ be a non-trivial semigroup with the property that no proper subset of $S$ generates $S$. Show that each ...
-1
votes
0answers
20 views

Inverse transversal is perfect if and only if $li=i^0lil^0$ for $i \in{I}$, $l \in{\Lambda}$

I am attempting show that an inverse transversal is perfect if and only if $li=i^0lil^0$ for $i \in{I}$, $l \in{\Lambda}$. An inverse transversal is perfect if $li(li)^0=(li)^0li$. I have shown the ...
1
vote
2answers
55 views

Does Green's $\mathcal{J}$-relation define a total order on the equivalence classes?

In a semigroup $S$ we define $a\le_\mathcal{J} b$ iff $a=xby$ for some $x,y\in S^1$. Defining $a\equiv_\mathcal{J} b$ by $a\le_\mathcal{J} b$ and $b\le_\mathcal{J} a$ gives a partial order on ...
2
votes
1answer
62 views

Do the idempotents in an inverse semigroup commute?

I have been looking at this for hours now. Why is it true that idempotents of an inverse semigroup commute? It seems like this should be straightforward but I just can't get it. Any help is greatly ...
13
votes
1answer
763 views

How many associative binary operations there are on a finite set?

I am studying Scott's book Group Theory. In the Exercise $1.1.17$ he asks us to show that if $S$ is a set and $|S|=n$, then there are $n^{\frac{n^{2}+n}{2}}$ commutative binary operations on $S$. But ...
6
votes
1answer
119 views

About translating subsets of $\Bbb Z.$

This is a continuation of About translating subsets of R2. Is it possible to find a pair of sets $A,B\subseteq\Bbb Z$ such that A is a union of translated (only translations are allowed) copies of ...
1
vote
1answer
50 views

If $\lambda\subseteq\mathscr{L}, \rho\subseteq\mathscr{R}$, then $\lambda\circ\rho=\rho\circ\lambda$.

This is (the first part of) Exercise 2.4 of Howie's Fundamentals of Semigroup Theory. The Details. Quoting Howie (on page 22 of my copy): Definition: A relation $R\subseteq S\times S$ on a set ...
2
votes
2answers
47 views

Proving that the binary operator of a group is commutative if the relation on the group is a partial order

What's probably worst is that this is a repost, as I was stuck on 3. before. The question is this: Throughout this question, we shall denote by $\neg$ the relation on a semigroup $(A, * )$ defined ...
6
votes
1answer
86 views

Proving that if the semigroup (A, *) is a group, then the relation is an equivalence relation.

I'm aware that posting exam questions is probably frowned upon, but this isn't homework, I think I'm genuinely misunderstanding some part of the algebra. The question is this: Throughout this ...
2
votes
1answer
34 views

About a semigroup equipped by zero

I didn't find a useful related link in which someone could equip a semigroup with a zero while using GAP via web. Theoretically, we can make the semigroup $S$ fat to be $S^{0}=S\cup\{0\}$ with ...
1
vote
2answers
31 views

Non-unital commutative semigroups $S$ such that for all $x \in S$, the function $S \rightarrow S$ given by $y \mapsto x+y$ is a bijection?

Does there exist a commutative semigroup $S$ with the following (additively denoted) properties? For all $x \in S$, the function $S \rightarrow S$ given by $y \mapsto x+y$ is a bijection. $S$ has no ...
3
votes
1answer
58 views

Can a smooth function on the reals form a non-commutative semigroup?

Let $f\colon \mathbb{R}^2 \to\mathbb{R} $ be a smooth function. Can there exist an algebraic structure $(\mathbb{R}, \cdot)$ such that for $x,y \in \mathbb{R}$, $x \cdot y = f(x,y)$ that is a ...
1
vote
1answer
66 views

Subsets of a monoid closed under left-multiplication by elements of a submonoid

Let $M, T$ be monoids (or, semigroups) with $M \subset T$. Then we can consider subsets $S$ of $T$ that are closed under left-multiplication by something in $M$, i.e. $$ a \in S, m \in M \implies ma ...
1
vote
3answers
88 views

Example of an associative binary operation, without identities or inverses.

In essence, I am looking for an example of a semigroup or a semicategory (closure is not that important, but it is useful) that is NOT a monoid or category. Hopefully, there is a neat and ...
5
votes
1answer
156 views

Must a pseudoinverse of a von Neumann regular element be regular?

Background Let $R$ be a ring or a semigroup. We say that $x\in R$ is a von Neumann regular element of $R$ if there exists $y\in R$ such that $$xyx=x.$$ Any $y\in R$ satisfying the above equation is ...
0
votes
1answer
20 views

Can we define unitary representations on semigroups

A representation on a semigroup $S$ is a pair $(\pi,H_\pi)$ where $\pi$ is a homomorphism from $S$ into $B(H_\pi)$ and $H_\pi$ is a Hilbert space. In the group case, a representation $\pi$ of a group ...
1
vote
0answers
87 views

About the multiplicity of a semigroup ring.

Let $A=K[X^{n_1}, \dots, X^{n_s}]$, $S=\langle n_1, \dots, n_s \rangle$ ($n_1 < \cdots < n_s$) a numerical semigroup and let $\mathfrak m$ be the maximal ideal $(X^{n_1}, \dots, X^{n_s})$. We ...
0
votes
1answer
46 views

Is this a*b = a is semigroup and commutative?

Let S be any nonempty set with the operation a * b = a. Is (S,*) a semigroup? Is it commutative? I dont know what to do if a * b = merely a. Usually if a * b = anything that have operation i know how ...
2
votes
1answer
36 views

$A_1,A_2,A_3$ forms a partition of $\mathbb N_{>0}$ and $a,b,c \in A_i \implies a+b+c \in A_i$ then at-least one of $A_i$ is closed under addition?

If $A_1,A_2,A_3$ forms a partition of $\mathbb N_{>0}$ such that for any $i=1,2,3$ ; $a,b,c \in A_i \implies a+b+c \in A_i$ then is it true that at-least one of $A_i$ is closed under addition ? I ...
0
votes
1answer
41 views

What do the operator semi-groups have to do with PDE's?

Can anybody please help me to understand what does the semi-group do with partial differential equations? We started this subject very recently and we are now in the proof of Hille-Yosida Theorem, ...
1
vote
2answers
35 views

How to prove the generator for strongly continous contraction semigroup?

I have a self adjoint operator, which is also negative semi-definite, I have also a Hilbert space. How to show that the given operator with given X Hilbert space is the generator for strongly ...
9
votes
4answers
289 views

give a counterexample of monoid

$G$ is a monoid,$e$ is its identity,if $ab=e$ and $ac=e$, can you give me a counterexample such that $b\neq c$. if not,please prove $b=c$ thanks a lot.
0
votes
1answer
34 views

What's a semidirect product of semigroups?

I see many references to the notion in the title on the internet, but I can't find a definition. Could you please give one? A short introduction to the theory of such products (especially one relating ...
1
vote
0answers
43 views

Reference request on numerical semigroups

I watched some talks about numeric semigroups, and thei relation whti algebraic geometry (such as Weierstrass semigroup of a curve), and I'm interested in take a deeper look in this topic, can anyone ...
8
votes
1answer
279 views

Describing the Wreath product categorically.

The Details: Let's have a recap of some definitions (taken from "Nine Chapters in the Semigroup Art" (pdf), by A. J. Cain). Definition 1: Let $P$ be a semigroup. The left action of $P$ on a set ...
3
votes
1answer
49 views

How to construct a semi-group over a non-numerical set?

A semigroup is a set, e.g. $X$, with an associative binary operation, e.g. $\star$. That is for all $x,y,z \in X$, $(x \star y) \star z = x \star (y \star z)$. I have seen some abstract algebraic ...
1
vote
0answers
53 views

looking for books on topological semigroup:

I'm looking for several books on topological semigroup: Topological semigroups: history, theory, applications. Karl Heinrich Hofmann Mathematics Research Library, Tulane University. The ...
2
votes
0answers
130 views

The congruence $\{(a^m, a^{m+r})\}^\#$ on $a^+$.

I've spent a bit too long on this exercise. It's time to ask for help. This is Exercise 1.20 of Howie's Fundamentals of Semigroup Theory. Let $\rho_{m, r}$ (for $m, r\ge 1$) be the congruence ...
4
votes
1answer
49 views

One-sided and two-sided cancellability

Is there a semigroup $S$ with right- and left-cancellable elements but no elements cancellable from both sides? $s\in S$ is left-cancellable if for any $a,b\in S$ we have $$sa=sb\implies a=b$$ and ...
-1
votes
1answer
32 views

Semigroup of a group and identiy

If $(G, *)$ is a group and $£$ is restriction of $*$ on subset $S$ of $G$. Is there some semi-group $(S, £)$ such that identity of $(S, £)$ and $(G, *)$ are different.
1
vote
1answer
29 views

Semigroup without left cancellation law and right cancellation law

Can please someone give example of a Semigroup in which neither left cancellation law hold nor right cancellation law
3
votes
2answers
37 views

Existence of right and left identity in minimalistic algebraic structure

Let $(A,\cdot)$ be some algebraic structure in which there exists elements $e_r,e_l$ such that $$e_l\cdot x = x, \forall x\in A$$ $$x\cdot e_r = x, \forall x\in A$$ By definition, if $(A,\cdot)$ is ...
2
votes
1answer
77 views

How would a category theorist describe Green's relations?

In Semigroup Theory, Green's relations are everywhere. Their equivalence classes, for instance, on a given semigroup $S$ can tell one a lot about the structure of $S$. There is some trivial sense in ...
0
votes
1answer
55 views

For semigroups, $S\preccurlyeq T$ iff there exists an injective relational morpism $\mu: S\to T$.

This is Exercise 1.16 of Howie's Fundamentals of Semigroup Theory. The Details. Definition 1: Let $A$ and $B$ be sets. A relation $\rho$ from $A$ to $B$ is a subset of $A\times B$. Define ...
1
vote
2answers
69 views

Involution on inverse semigroups

I'm trying to prove the following for inverse semigroups $\bf Def:$ an inverse semigroup $S$is a semigruop such that for each $x\in S$ the exists a unique $y\in S$ such that $xyx=x$ and $yxy=y$. An ...
1
vote
0answers
75 views

How many associative ternary operations there are on a finite set?

We know that algebraic operation is a function $f:\underbrace{\left ( X\times X\times \cdots \times X\right )}_{t\ \text{times}}\rightarrow{X}$ If $X$ is a set and and cardinality is $|X|=n$ then ...
2
votes
1answer
75 views

About Rees homomorphism

I am came across the notion Rees Congruence for semigroups. J. Howie defines it as $$\rho_I=(I\times I)\cup {1_S}$$ wherein $I$ is an ideal of semigroup $S$ ...
2
votes
1answer
54 views

Showing the full transformation semigroup $\mathscr{T}_n=\langle\zeta, \tau, \pi\rangle$.

I'm sorry if this is a duplicate in any way. Throughout I use cycle notation and write maps $m:X\to Y$ on the right of their arguments (e.g. $xm=y$ for $m(x)=y$). This is Exercise 1.7 of Howie's ...
2
votes
1answer
46 views

Complicated proof in Transformation Semigroups.

Let $X$ be an infinite set. The relative rank of a subset $T_{X}$ over a subset $S$ is either uncountable or at most $2$. I don't understand how to prove this Corollary in case that $T_{X} \setminus ...
3
votes
0answers
71 views

Monoids, Semigroups, and a Reflective Subcategory.

The following is reflective in the category $\mathbf{Sem}$ of semigroups and their homomorphisms: we add a new neutral element to each semigroup, even monoids; then the reflections are identity ...
2
votes
1answer
64 views

Commutative Cancellative Semigroup: When is an irreducible element prime?

Suppose $(S,\cdot)$ is a semigroup with neutral element $e$ (i.e. $xe=ex=x$ for all $x\in S$) and the following properties: S is commutative: $xy=yx$. S is cancellative: $xy = xz$ implies $y = z$. ...
1
vote
0answers
26 views

Every right principal ideal non-emptily intersects the center — what is that?

This is a follow-up to Do Lipschitz/Hurwitz quaternions satisfy the Ore condition? Jyrki Lahtonen answered the question in the positive by noticing that every right principal ideal in either ring has ...