For questions about finite semigroups, finite sets equipped with an associative binary operation.
Questions tagged [finite-semigroups]
47 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
- 18,586
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
- 597
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
- 7,449
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…
King Ghidorah
- 617
- 5
- 22
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…
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
- 15,280
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
- 1,889
- 9
- 17
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
- 18,586
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…
Tegiri Nenashi
- 1,030
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
- 2,529
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
- 1,247
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 $P \in GL(2,\mathbb{R})$ such…
Tom Rush
- 128