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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Artin-Rees Lemma for Semigroups [on hold]

(Artin-Rees Lemma) Let $S$ be a Noetherian semigroup and $A,B$ be ideals of $S$. A∩B^i=(A∩B^N)B^(i-N) for each İ≥N where N is a natural number. Does anyone know its proof? Hi, I use İ and N as a ...
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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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Can we extend the definition of a homomorphism to binary relations?

This is going to be quite a long post. The actual questions will be at the end of it in section "Questions." INTRODUCTION After receiving an answer to this question about extending the definition of ...
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A rather abstract strongly continuous semigroup

Define $X$ as the Hilbert space $L^{2}(0,\infty)$ and let the operators $T(t):X\to X$, $t\ge 0$ be defined by $(T(t)f)(\zeta):=f(t+\zeta)$ I want to show that $(T(t))_{t\ge 0}$ is a ...
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Is there a logarithmic size generating set for some classes of finite semigroups?

Following my question Why is the minimum size of a generating set for a finite group at most $\log_2 n$?, we know that finite groups have generating sets of size at most $\log_2 n$, and a similar ...
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rectangular groups are completely simple and orthodox

Let $S$ be a rectangular group. i.e $S$ is isomorphic to the direct product of a group and a rectangular band.
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Why do the interesting antihomomorphisms tend to be involutions?

Given a semigroup $S$, define that an antihomomorphism on $S$ is a function $$* :S \rightarrow S$$ satisfying $(xy)^* = y^*x^*.$ Examples abound. Consider: Transposition, where $S$ equals the set ...
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Does the “equality semigroup” have an accepted name?

Given a set $G$, we get a semigroup on $G \cup \{0\}$ as follows: Define $x^2 = x$ for all $x \in G \cup \{0\}$. Define $xy = 0$ for all distinct $x,y \in G \cup \{0\}$. Question 0. Does this ...
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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 $y = z$. ...
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Find all homomorphisms from a semigroup $(\mathbb{Z},+) \to (\mathbb{Q},+)$

In the book Karpfinger & Meyberg (2013) "Algebra" I encountered this problem. So far I have figured out $\tau(x) = qx$ wehre $q \in \mathbb{Q}$. Are there other homomorphisms? Is there a general ...
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equivalent definitions of a perfect inverse transversal [closed]

I am trying to show that the following two conditions are equivalent; $S^0$ is a perfect transversal ( $(li)(li)^0=(li)^0(li) $ where $i \in{I}, l \in{\Lambda}$ ) for all $i \in{I}, l \in{\Lambda}$ ...
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Perturbations of Positive Semigroups with Applications

P is an operator, if P is p-admissible for A then P is p-admissible for A + P. the proof of this theorem is based on the theorem Miyadira-Voiyt. but if we will demonstrate this result in general I ...
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20 views

Proof involving strong continuous semigroups

Let $T(t)$ be a $C_{0}$ semigroup on the Hilbert space $X$ with infinitesimal generator $A$ and let $\rho\in(0,1)$. I want to prove that $\displaystyle \sup_{t\ge 0}||T(t)-I||\le \rho$ is equivalent ...
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1answer
25 views

Showing that A is NOT an infinitesimal generator

As a state space, choose $X=L^{2}(0,1)$. Let $A$ be defined as $\displaystyle Af=\frac{df}{d\zeta}$ with domain $D(A)=\{f\in L^{2}(0,1)|f$ is absolutely continuous and $\frac{df}{d\zeta}\in ...
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Finding the infinitesimal generation of a strongly continuous semigroup

Let $X$ be a Hilbert space, $A\in\mathcal{L}(X)$ and $\displaystyle T(t)=e^{At}=\sum_{n=0}^{\infty}\frac{(At)^{n}}{n!}$. I have already shown that $T(t)$ defines a $C_{0}$ semigroup. But now I need ...
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190 views

L,R,H,D,J relations on a completely simple semi group represented my a rees matrix

I am trying to tackle the following semigroup question. I can't see why my answer is wrong but I haven't used the fact the semigroup is COMPLETELY simple anywhere so I think there must be an error ...
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Showing that an operator generates a unitary group

Consider the following operator on $X=L^{2}(0,1)$: $\displaystyle Af=\frac{df}{d\zeta}$ with domain: $D(A)=\{f\in L^{2}(0,1)|f$ is absolutely continuous, $\frac{df}{d\zeta}\in L^{2}(0,1)$ and ...
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1answer
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Proving one of the properties of a strongly continuous semigroup

Let $X$ be a Hilbert space and $A\in\mathcal{L}(X)$ and $\displaystyle T(t)=e^{At}=\sum_{n=0}^{\infty}\frac{(At)^{n}}{n!}$. I want to show that $T(t+\tau)=T(t)T(\tau)$. So we have that ...
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19 views

semigroups defined on banach algebra

Let $(X,\|.\|)$ be a Banach space and $Z:=\{Z(t)\}_{t\geq 0}$ is strongly continuous semigroup defined on it. If $X$ turns out to be a Banach Algebra, i.e. for $x,y\in X$, $xy\in X$. Is $Z$ still ...
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Find a first order sentence satisfied in every finite semigroup but not in every compact semigroup

Several properties that hold in nonempty finite semigroups also hold in nonempty compact semigroups. Furthermore, many of these properties can be formulated by a first order sentence. For instance, ...
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30 views

Does this show that it is a bounded linear operator?

Let $X$ be a Hilbert space and $A\in\mathcal{L}(X)$. I want to show that $\displaystyle e^{At}:=\sum_{n=0}^{\infty}\frac{(At)^{n}}{n!}=T(t)$ defines a strongly continuous semigroup (i.e. a ...
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Is there an idempotent element in a finite semigroup?

Let $(G,\cdot)$ be a finite semigroup. Is there any $a\in G$ such that: $$a^2=a$$ It seems to be true in view of theorem 2.2.1 page 97 of this book (I'm not sure). But is there an elementary proof? ...
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Subsemigroup generated by an element contains unique idempotent [duplicate]

Possible Duplicate: A cyclic subsemigroup of a semigroup S that is a group My homework: An element $s^{i+k}$ on the cycle is idempotent iff $$ s^{i+k} = s^{2i+2k} ,$$ or equivalently ...
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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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If S is finite and $x/\rho$ is an idempotent of $S/\rho$ then $x/ \rho$ contains an idempotent

Currently working on the following problem, need a little help with the solution to the last part, any hints? Q: Let S be a semigroup, and let ρ be a congruence on S. Prove that if e ∈ S is an ...
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49 views

Drawing a directed graph in Maple

I have been playing around with some Maple today for a semigroup/ graph theory style project. I want to draw a left cayley graph with vertices $\{1,....,12\}$ and edges $[5, 5]$, $\{[6, 2].[7, 1], [1, ...
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154 views

Proof of Lallement’s Lemma

i am a bit weak with my congruence manipulation when it comes to semigroup theory. Could you possibly check my proof and give me constructive criticism on it? Lemma: Let $S$ be a regular semigroup ...
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Is the unit ball of $H^\infty(\mathbb{D})$ a metrizable topological semigroup under multiplication?

The space $H^\infty(\mathbb{D})$ of all bounded holomorphic functions on the open unit disc carries many different topologies. One such topology is given by uniform convergence on compact subsets; ...
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Prove that $\{S(t)\}_{t \ge 0}$ is not a contraction semigroup on $L^\infty(\mathbb{R}^n)$

Define for $t > 0$ $$[S(t)g](x) = \int_{\mathbb{R}^n} \Phi(x-y,t)g(y) \, dy \quad (x \in \mathbb{R}^n),$$ where $g : \mathbb{R}^n \to \mathbb{R}$ and $\Phi$ is the fundamental solution of the ...
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Contraction semigroup on $X$ with generator $A$

Let $\{S(t)\}_{t \ge 0}$ be a contraction semigroup on $X$, with generator $A$. Inductively define $D(A^k):=\{u\in D(A^{k-1}) \mid A^{k-1}u \in D(A)\}$ ($k=2,\ldots$). Show that if $u \in D(A^k)$ ...
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“Universal solution” to factorization of arithmetic mean through a semigroup homomorphism.

Headnote A: my apologies for the long winded exposition; I couldn't find a more compact way to convey what I'm looking for, why I'm looking for it, and what have I tried myself. Suggestions for ...
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Justifying an equality involving a closed operator $A$

Justify the equality $$A \int_0^\infty e^{-\lambda t} S(t) u \, dt = \int_0^\infty e^{-\lambda t} AS(t) u \, dt$$ used in (16) of §7.4.1. (Hint: Approximate the integral by a Riemann sum and recall ...
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A finite pseudo-ring such that $ab^2=b$ for some $(a,b)\in A$

I have this exercise which I found very difficult: $(A,+,.)$ is a finite pseudo-ring (a structure satisfying all the axioms of a ring except for the existence of a multiplicative identity), and ...
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Prove that $\{S(t)\}_{t \ge 0}$ is a contration semigroup on $L^2(\mathbb{R}^n)$

Define for $t > 0$ $$[S(t)g](x) = \int_{\mathbb{R}^n} \Phi(x-y,t)g(y) \, dy \quad (x \in \mathbb{R}^n),$$ where $g : \mathbb{R}^n \to \mathbb{R}$ and $\Phi$ is the fundamental solution of the ...
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Properties of resolvent operators

I am asked to prove the identities of $(12)$ and $(13)$, which are given on page 438 of the textbook PDE Evans, 2nd edition as follows: THEOREM 3 (Properties of resolvent operators). (i) If ...
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Doubt about semigroups in this article (anyone can help).

I need help in this article. My doubt is very arithmetical and I think follows directly from the definitions. So I think anyone could help me. The author defines what is a semigroup, gaps and ...
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1answer
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Invertibility of operators related to Markov processes in Ethier-Kurtz

Lemma 2.3 of the book by Ethier and Kurtz (first edition, I believe) defines $$ g_n := (\lambda - A)(\lambda_n - A)^{-1}g $$ for some fixed $ g $ but I see no guarantee that $(\lambda_n - A)^{-1} g ...
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Semi group presentation $<a, b | a^{2} = b^{2} = 0, aba = a, bab = b>$

Another semi group question here, trying to get my head around the topic. Consider the semi group $S=\left<a, b | a^{2} = b^{2} = 0, aba = a, bab = b\right>$ I need to prove that $S$ has order ...
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How is the following expresson be obtained and the meaning of the expression in blue box?

Let me introduce the term {$E_\lambda:\lambda \geq0$} is the spectral resolution of identity of a self adjoint densely defined, positive and closed operator $A:D(A)\subset X\rightarrow X$ , Where X ...
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Let $S$ be a completely regular semigroup, expressed as a semilattice $Y$ of completely simple semigroups $S_{\alpha}$ ($\alpha \in{Y})$

Show that, if $L$ is a left ideal of $S_{\alpha}$, then $$L \cup [\cup \{S_{\beta} : \beta < \alpha\}]$$ is a left ideal of $S$. Suppose now that $S= ...
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A semigroup with identity having exactly one idempotent is a group

Let $G$ be a finite semi-group with identity such that it has only one idempotent.Is $G$ a group? It only remains to show that for any $a\in G$ $\exists b\in G$ such that $ab=ba=e$ where $e$ is the ...
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Isomorphism of a semigroup S and (Aᴬ, ⋅).

I would like to ask you for help with proving the following theorem from our textbook: Any semigroup is isomorphic to a subsemigroup of ($A^A, \cdot$) for a suitable set A. The theorem is then ...
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Associativity in category theory [closed]

In Category Theory, the composition of morphisms must be associative. What would happen if we give up this associative law? For example Lie algebras are vector spaces with non-associative binary ...
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Eilenberg's rational hierarchy of nonrational automata & languages

In the preface to his very influential books Automata, Languages and Machines (Volumes A, B), Samuel Eilenberg tantalizingly promised a Volume C dealing with "a hierarchy (called the rational ...
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give a counterexample of monoid

If $G$ 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.
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Closed linear operator

I am having some trouble in showing the following map is closed: For $f\in L^2(\mathbb{R^2})$ with $(x+iy)f(x,y)\in L^2(\mathbb{R^2})$, $M(f)(x,y)=(x+iy)f(x,y)$. I am also asked to find the ...
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Shift operators and c0 semigroups

I am asked where the bounded continuous functions on $\mathbb{R}$ (with sup norm) with the right shift operators ( $T_t(f)(x)=f(x-t)$ ) form a c0 semigroup. I believe the answer is no, but I am ...
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Quotients of finitely generated semigroups are finitely generated

Is it true that the quotient of a finitely generated semigroup by a normal subsemigroup is finitely generated? If so, could you suggest a place where the proof is written down?
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Example of a semigroup with unique idempotent which is not a monoid

I am searching for an example of a semigroup, with unique idempotent element, that is not a monoid. Please help.
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Group identities and inverses

A set is a set. A magma is a set with a binary operator. A semigroup is a magma with an associative binary operator. A monoid has a two-sided identity. And a group has two-sided inverses. I am ...