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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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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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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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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2answers
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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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167 views

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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1answer
42 views

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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1answer
38 views

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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1answer
88 views

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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33 views

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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18 views

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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1answer
98 views

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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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 ...
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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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52 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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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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50 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 ...
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1answer
92 views

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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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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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. Show that a semigroup is completely simple and orthodox if and only if it is a ...
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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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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 ...
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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 ...
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90 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 ...
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1answer
50 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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37 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
76 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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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
23 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
80 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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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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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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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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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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340 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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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
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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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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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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
91 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
54 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 ...
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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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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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54 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 ...
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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 ...