Exist any kind of intriguing semigroups that aren't monoids?

Exist any kind of intriguing and also all-natural instances of semigroups that are not monoids (that is, they do not have an identification component)?

To be a little bit extra specific, I presume I need to ask if there any kind of intriguing instances of semigroups $(X, \ast)$ for which there is not a monoid $(X, \ast, e)$ where $e$ remains in $X$. I do not take into consideration an instance like the set of actual numbers more than 10 (taken into consideration under enhancement) to be a completely 'all-natural' semigroup for my objectives; if the domain name can be expanded in a noticeable means to include an identification component then that's not what I'm after.

0
2019-05-04 17:41:52
Source Share
Answers: 6

Here is an economical solution which takes the position that semigroups are naturally intriguing Consider the adhering to definition of a team :

$\textbf{Definition}$ A semigroup $S$ is claimed to be a team if the adhering to hold :

  1. There is an $e \in S$ such that $ea=a$ for all $a\in S$
  2. For each $a \in S$ there is a component $a^{-1} \in S$ with $a^{-1}a=e$

At one factor in my life, it appeared all-natural to ask what takes place if we change axiom 2 with the really comparable axiom

2$^\prime$. For each and every $a \in S$ there is a component $a^{-1} \in S$ with $aa^{-1} = e$.

It is an enjoyable workout to exercise several of the effects that arise from this. Below are a couple of realities concerning a semigroup $S$ which pleases $1$ and also $2^\prime$ :

  • If $e$ is the one-of-a-kind component of $S$ pleasing axiom 1, after that $S$ is a team
  • If $S$ has an identification (in the common feeling) after that $S$ is a team
  • The principal left excellent $Sa = \{sa \mid s \in S\}$ is a team for all $a \in S$, and also actually all major left perfects of $S$ are isomorphic as teams.

It is uncomplicated to locate instances of such semigroups that are not teams. As an example, take into consideration the adhering to set of $2\times 2$ matrices (with matrix reproduction as the procedure) : $$\left\{\begin{pmatrix} a & b \\ 0 & 0\end{pmatrix} \mid a,b \in \mathbb{R}, a \neq 0\right\}$$ Or, an instance that looks like workout 30 in area 4 of Fraleigh is abstract algebra message : the nonzero actual numbers under the procedure $\ast$ specified by $a\ast b = |a|b$.

Absolutely it is open to question whether semigroups pleasing axioms 1 and also 2$^\prime$ are "interesting" or "natural". Yet I presume I assume they are. And also, I am not the just one (or the first one, by a slim chance!) to assume this. See Mann, On certain systems which are almost groups (MR). (A little googling will certainly show up even more outcomes if you are interested.)

0
2019-12-03 05:13:23
Source

Excerpts from Grillets Semigroups Chapter 1, Page 1,

There are 1160 distinctive semigroups of order 5 ; 15793semigroups of order 6 ; 836,021 semigroups of order 7

Granted, most of those semigroups are not that intriguing. Nonetheless, makeover semigroups (not monoids unless you count the identification map in) are intriguing due to the fact that the majority of makeovers (features) are not bijective. Hence, makeover semigroups are extra all-natural than permutation teams.

0
2019-06-01 12:01:27
Source

Convolution of functions/distributions serves in a selection of areas, and also the identification component, the dirac delta, is not purely a function.

0
2019-05-08 07:41:17
Source

Let G be the set of (continual) operates f : R - > R where f (x) often tends to 0 as x often tends to infinity : $lim_{x\to \infty}f(x) = 0$. The driver is the common factor - sensible reproduction of features.

G is shut under * given that lim (f (x)) = 0 and also lim (g (x)) = 0 indicate lim (f (x) *g (x)) = 0. G is a subgroup of f : R - > R , so the identification has to coincide - the function which is frequently 1. Yet this identification is not in G.

EDIT : As Harry appropriately mentions, f : R - > R is not a team. Consequently the adhering to improvement is required : take into consideration just features such that $f(x) \neq 0$ almost everywhere.

0
2019-05-08 07:15:38
Source

Finite collections of matrices of differing measurements, where the item A *B = P in A & Q in B & dim (Q) = codim (P) , and also dim & codim are the measurements of the resource & target rooms of a matrix.

The boundless instance has a noticeable device.

0
2019-05-08 07:11:04
Source

One resource of monoids is offered by taking rings with identification, and also forgeting enhancement. So in a similar way, one resource of semigroups that are not monoids is taking rings without identification, and also forgeting enhancement. With this in mind, allow me clarify one standard resource of rings without identification.

A standard resource of rings is offered by taking features pleasing some practical problem on a room, as an example continual actual or intricate valued features on a room, with pointwise enhancement and also reproduction. Certainly, the constant function 1 is continual, therefore this offers a ring with identification.

Yet intend since we enforce some problem, such as "all features that are continual, and also which in addition disappear at some defined factor". This throws away the constant function 1, therefore offers a ring without identification. Currently you can normally object that this is fabricated ( based on the need in the inquiry that there not be a noticeable expansion to a monoid),. so allow me add extra description regarding why it need not be.

One instance of an indicate take into consideration is "the factor at infinity", i.e. we can consider all features which disappear at infinity, i.e. which on the enhance of bigger and also bigger portable collections, expand smaller sized and also smaller sized. This is an all-natural problem to enforce in several analytic contexts, therefore offers an all-natural instance. (The factor that this sort of development. problem is all-natural in evaluation is that, on a non-compact room, as an example the actual line, an arbitrary continual function might not be integrable (equally as an instance), and also enforcing some degeneration at infinity (probably of the kind I defined, or probably something extra quantitive) comes to be a means to save the scenario.) (Note additionally that the instance that Tomer Vromen offers is specifically of this kind.)

Ultimately, note that if your semigroup does not have an identification, after that you can constantly officially join one, simply by including an added component e and also proclaiming that ex-spouse = x for all x.

One can do a comparable point for rings without identification. If A is a C-algebra (say) without an identification, after that one can create A + C e (the straight amount), and also proclaim that e works as a multiplicative identification. This is a frequently-used strategy in the concept of rings-without-identity.

P.S. I do not recognize much literary works concerning semigroups without identification, but also for rings without identification, the most effective literary works I recognize of remains in useful evaluation books ; as an example Naimark's timeless Normed Rings usually deals with the instance of Banach algebras (and so on) without identification along with the instance when they do have identification, specifically so regarding have the ability to take care of instances such as the ring of continual features on an in your area portable room that disappear at infinity.

0
2019-05-08 04:13:45
Source