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. This tag is most frequently used for questions related to the concept of semigroups in the context of abstract algebra. Please use the more ...

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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
144 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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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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1answer
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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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1answer
108 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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0answers
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Semigroup satisfying the cancellation property which cannot be embedded in a group [duplicate]

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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0answers
59 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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1answer
51 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 J−...
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1answer
110 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= \mathcal{S}(Y;G_{\alpha};{\phi}_{\alpha,\...
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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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2answers
208 views

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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1answer
152 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
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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 $S/\...
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2answers
89 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 ...
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1answer
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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
54 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
45 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
102 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 non-...
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4answers
572 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 simple-to-...
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1answer
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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
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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
70 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
38 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
201 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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2answers
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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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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
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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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1answer
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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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1answer
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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
99 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
62 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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1answer
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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 $$a\rho=...
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1answer
98 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 ...
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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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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
42 views

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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1answer
376 views

Why is a monoid with right identity and left inverse not necessarily a group? [duplicate]

This problem is from Herstein's 'Topics in Algebra'. I've thought about it a bit but haven't come up with much. Let $G$ be a non-empty set with an associative product which also satisfies: $\exists ...
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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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1answer
158 views

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)= g(y)...
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1answer
130 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 $S$ ...
4
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1answer
60 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 ...
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1answer
81 views

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
88 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 ...