Mathematics

Questions tagged [finite-semigroups]

For questions about finite semigroups, finite sets equipped with an associative binary operation.

46 questions
39
votes
7 answers

Is there an idempotent element in a finite semigroup?

Let $(G,\cdot)$ be a non-empty 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? Theorem 2.2.1. [R. Ellis] Let $S$ be a…
user59671
6
votes
1 answer

Inclusion relations between equationally defined classes of finite semigroups

Let $S, T$ be two semigroups. In the following all semigroups are supposed to be finite. We write $S \prec T$ if there exists a surjective semigroup morphism from a subsemigroup of $T$ onto $S$. A class of finite semigroups $\mathcal V$ is defined…
StefanH
  • 18,786
5
votes
1 answer

Implicit operations in finite semigroups.

what are some examples of implicit operations in finite semigroups other than expressions involving $\omega$ ? Like $x^\omega y^\omega$ or $x^{\omega+1}$ . By Reiterman's theorem, pseudovarieties of finite semigroups are given by a set of…
theDumbGeometer
  • 617
4
votes
1 answer

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 is, how are $r$ and $n$ related? More…
E W H Lee
  • 2,366
4
votes
1 answer

If $G$ is a finite semi-group and $\forall x,y,z \in G: xy=yz \implies x=z$ then $G$ is an Abelian group

If $G$ is a finite semi-group and $\forall x,y,z \in G: xy=yz \implies x=z$ then $G$ is an Abelian group. I have no idea where to start. I'm stuck! I can't prove even the existence of the identity element :|
user66733
  • 7,469
4
votes
1 answer

Is a finite semigroup with a unique idempotent $e$ always a group with unit $e$?

Given a finite semigroup $S$ containing a unique idempotent $e$ , I can show that every element $s\in S$ has an ''inverse" $s^{-1}\in S$ in the sense that $s s^{-1} = s^{-1}s = e$ : Since $S$ is finite, every element $s$ has an idempotent power…
Lemma 5
  • 71
4
votes
1 answer

Show that $H_x$ is a group for all $x.$

$H_{x}$ denotes the class of $x$ for the Green relation $\mathcal{H}$ . Let $S$ be a finite semigroup where all elements can be written as a product of idempotents, that is, $x=e_1 e_2\dots e_n,$ for idempotents $e_1,e_2,\dots, e_n \in S$ for any $x…
King Ghidorah
  • 617
3
votes
1 answer

A finite semigroup has only trivial subgroups iff $\mathcal{H}$ is the identity relation.

I am struggling with the rightward implication. Here is some of my working: $(\Leftarrow)$ Every subgroup is contained within a maximal subgroup that is the $\mathcal{H}$ class of that subgroup's idempotent. As each $\mathcal{H}$ class is trivial by…
Levent Michael Dasar
  • 58
3
votes
1 answer

Is any finite semigroup of this type a left monoid?

Let $(S, \cdot, e)$ be a semigroup $(S, \cdot)$ with binary operation $e$ in which the identities $e(x, y)\cdot x\approx x$ and $e(x, y)\approx e(y, x)$ hold. In this question I asked if any such semigroup is necessarily a left monoid. Example given…
Jakobian
  • 16,581
3
votes
2 answers

How many non-isomorphic semigroups are there of orders $2$ and $3$?

For order $2$, I have found 5. There are 16 maps from $\{a, b\} \times \{a, b\} \to\{a, b\}$. They form $10$ equivalence classes of non-isomorphic binary operations, $5$ of which are associative. Of these $2$ are monoids and $1$ is a group.They are…
Stephen Meskin
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3
votes
0 answers

The Krull-Schmidt-Remak Theorem for Semigroups and Monoids

For finite groups, the Krull-Schmidt-Remak-Theorem holds, i.e. if $$ H_1 \times H_2 \times \ldots \times H_k \cong G_1 \times G_2 \times \ldots \times G_l $$ where the $H_i, G_i$ could not be further decomposed, then $k = l$ and there exists a…
StefanH
  • 18,786
2
votes
1 answer

Subsemigroup generated by an element contains unique idempotent

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 $$ i+k = 2i+2k \pmod p .$$ I'm stuck here (this is my first modulo…
Tegiri Nenashi
  • 1,050
2
votes
0 answers

Semigroups with the property $S=S^1(S\setminus S^2)$

We have encountered semigroups $S$ which have the property $S=S^1(S\setminus S^2)$ (see Equality of two specific classes of subsets of a group ), or equivalently $S\subseteq S^1(S\setminus S^2)$ . Are there any characterizations (or classifications)…
M.H.Hooshmand
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2
votes
1 answer

On the rank of finite semigroups

For any semigroup $S$ , let $A$ be a non-empty subset of $S$ . Then the subsemigroup generated by $A$ that is, the smallest subsemigroup of $S$ containing $A$ , is denoted by $\langle A\rangle$ . If there exists a finite subset $A$ of $S$ such that…
1ENİGMA1
  • 1,257
2
votes
1 answer

When do $2\times2$ matrices generate a finite semigroup?

Let $A_i$ , $i=1,\ldots, k$ , be $2\times2$ real-valued matrices with determinant 1 or -1. Under what circumstances is the semigroup generated by these matrices finite? I can see that this will be the case if there exists $P \in GL(2,\mathbb{R})$…
Tom Rush
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