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Questions tagged [semigroups]

A semigroup is an algebraic structure consisting of a set together with a single 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/universal algebra. Please use the more specific tag (semigroup-of-operators) whenever appropriate.

1 vote
1 answer
56 views

This old question asks for an example of a semigroup $(G, \ast)$ for which (a) $G$ has a left identity, i.e., an element $e \in G$ such that $e \ast g = g$ for all $g$, and (b) every element of $G$ ...
Travis Willse's user avatar
0 votes
1 answer
82 views

According to a German textbook there are many semigroups $(G,\odot)$ with the following properties: (1) $|G|>2$. (2) There is a left identity element $e\in G$, that is $e\odot g=g$ for all $g\in G$....
BW M's user avatar
  • 9
1 vote
0 answers
47 views

I have been looking into undecidable groups, and was shown this presentation: {\begin{array}{lllll}\langle &a,b,c,d,e,p,q,r,t,k&|&&\\&p^{10}a=ap,&pacqr=rpcaq,&ra=ar,&\\&...
Steven Morrell's user avatar
0 votes
0 answers
32 views

I am working with parametrized polynomial plane curves and I have several related questions about Milnor numbers and singularities. I would be grateful for references, precise statements (theorem ...
Mousa Hamieh's user avatar
0 votes
0 answers
79 views

I have already solved this problem: Suppose that a minimal left ideal $L$ of a semigroup is commutative. Prove that $L$ is a group. My question is how to solve this harder problem: Let $S$ be a ...
testaccount's user avatar
6 votes
3 answers
172 views

I'm trying to understand the proof that idempotent elements commute in an inverse semigroup, but first I need to understand this lemma about the inverse of a product, which is apparently the same as ...
zlaaemi's user avatar
  • 1,725
2 votes
0 answers
62 views

Recently, I've been studying non-autonomous parabolic problems and, as a consquence, the non-autonomous version of semigroups, evolution operators, has appeared. I'm very interested in how resolvent ...
Michelangelo's user avatar
8 votes
1 answer
245 views

Suppose $f$ and $g$ are operators. If for all $x$, we have $f(g(x))=f(x)$ and $g(f(x))=g(x)$ (an example: $f$ is relative interior, $g$ is closure, and $x$ is convex set), then is there a term to ...
Ypbor's user avatar
  • 1,226
4 votes
1 answer
96 views

Given a family of Markov processes $ \{ ( X_{ t} ^{ x})_{ t \geqslant 0}\colon x \in E \}$, where $ X_{ t} ^{ x}$ denotes a time-homogeneous Markov process starting at $ x \in E$, I'm trying to show ...
Cadlag's user avatar
  • 41
3 votes
1 answer
127 views

Are there examples of commutative and associative binary operations over $\mathbb R$ that cannot be expressed as $x \oplus y = f^{-1}(f(x)+f(y))$ where $f$ is an invertible function $f:\mathbb R\...
Fabius Wiesner's user avatar
4 votes
2 answers
202 views

I am exploring the following question: Does there exist an associative binary operation (i.e., a semigroup) on a set ( G ) that has at least a right identity and such that every element has at least ...
remarque's user avatar
  • 189
0 votes
1 answer
40 views

Let $X$ be an infinite abelian semigroup and fix a finite subset $S\subseteq X$. Question. Is it true that there exist $x,y \in X$ such that $\{x,x+y,x+2y\}\cap S=\emptyset$? Ps. It would be enough ...
Paolo Leonetti's user avatar
0 votes
0 answers
24 views

I’m reading nonlinear evolution equations by Song-Mu Zheng and I have some questions regarding the following statement THEOREM 2.4.1 ; COROLLARY 2.4.1 ; COROLLARY 2.4.2 ; COROLLARY 2.4.3 All are ...
Alucard-o Ming's user avatar
0 votes
0 answers
44 views

Let $G$ be a finitely-generated subgroup of $\mathbb{Z}^n$ and $A$ a finitely-generated sub-semigroup of $\mathbb{N}^n$ (which contains $(0,\ldots,0)$). Say I have fixed generating sets $g_1,\ldots,...
walkar's user avatar
  • 4,442
1 vote
0 answers
148 views

I am looking at the textbook by Bakry, Gentil, and Ledoux "Analysis and Geometry of Markov Diffusion Operators". This discussion is from section 3.1.6 which does not show computations and I ...
SparklyCape290's user avatar

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