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.

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Semigroups and solutions of equation

It is easy to prove: in a finite semigroup if for all $a$ and $b$, $ax=b$ and $ay=b$ has unique solution. then it is group. But if in a finite semigroup, if for all $a$ and $b$, $ax=b$ and $ay=b$ has ...
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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 ...
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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 ...
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8answers
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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 ...
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1answer
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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, ...
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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 ...
6
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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 ...
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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 ...
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1answer
61 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 ...
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2answers
53 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
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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
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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
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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 ...
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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 ...
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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 ...
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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 ...
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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 ...
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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 ...
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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 ...
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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 ...
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1answer
40 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, ...
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34 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 ...
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462 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 ...
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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.
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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 ...
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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 ...
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1answer
31 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.
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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
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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 ...
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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 ...
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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 ...
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1answer
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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 ...
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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 ...
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1answer
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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 ...
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1answer
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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 ...
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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 ...
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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 ...
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1answer
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finite semigroup on one generator,cycle, tail,group,zero element

Suppose we have a finite semigroup on one generator. It has a tail of length r and cycle of length c.The cycle is a group, but what can be chosen as a neutral element of it?Why is not ANY element ...
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Solution of nonhomogeneous problem using semigroup of linear operators

Let $X$ be a complex Hilbert space; and let $A:D(A)\subset X \to X$ be a $\mathbb C-$linear operator. Assume that $A$ is self-adjoint(so that $D(A)$ is a dense subset of $X$) and that $A\leq 0$ (i.e., ...
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How to prove that: If two binary operations are anti-isomorphic and one of them is associative then the second one also will be associative?

We know what is called an anti-isomorphic operation on a set S. it is just a one two one $ g $ function mapping from $S$ to $S$. $ g: S \rightarrow S$. and it satisfy this condition $ g(xy)= ...
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Haar measure on locally compact semigroups

I'm reading on Haar measure and we know that every locally compact group admits a Haar measure, is the same true for semigroups? if not, is there a class of semigroups that admits a Haar measure? ...
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1answer
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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 $S$ ...
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1answer
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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 ...
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1answer
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An estimate involving gaps in a subsemigroup of $(\mathbb N,+)$

I think this question can be solved by a high school student, maybe there is some trick on it or I'm forgetting something. Before my question, some background is required: Definition: A ...
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1answer
76 views

Period of semigroup

Let $S$ be a finite semigroup of order $n$. Suppose that $S$ has index $m$ and period $r$, i.e. $S$ satisfies the identity $x^{m+r} = x^m$. Then it is quite easy to show that $m \leq n$. My question ...
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About automorphisms of commutative semigroups

Suppose that $M$ is a commutative monoid and that the product $P$ of $M$ and the nonnegative integers $\mathbb{N}$ with addition has no nontrivial automorphisms. The set $S$ of pairs $(m,n)$ in $P$ ...
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Finite Linearly Ordered Abelian Monoids

This question concerns the proper definition of the phrase "finite linearly ordered abelian monoids". The sequence A030453 of OEIS counts the number of "finite linearly ordered abelian monoids". The ...
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Terminology for idempotents that commute with every other idempotent

Given a semigroup $S$, is there terminology for those $x \in S$ such that the following hold? $x$ is idempotent Given any idempotent $y \in S$, we have $xy=yx$. Comments. Let $E$ denote the set ...
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1answer
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Help in this characterization of the gaps of the symmetric numerical semigroups

Before my question, some background: Definition 1: A numerical semigroup is a subsemigroup $N$ of the additive semigroup $\mathbb N$ of the non-negative integers such that $\mathbb N-N$ is ...
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117 views

Who first studied semilattices?

Historically, who first studied semilattices, as opposed to lattices or Boolean algebras? (With or without identity, I do not mind.)