For questions about finite semigroups, finite sets equipped with an associative binary operation.
Questions tagged [finite-semigroups]
39 questions
35
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
- 17,616
- 6
- 49
- 121
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…
liczman
- 507
- 2
- 9
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
- 2,276
- 1
- 12
- 21
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
- 7,169
- 3
- 23
- 62
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
- 587
- 5
- 21
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
- 6
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
- 6,321
- 1
- 16
- 32
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
- 1,789
- 7
- 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
- 17,616
- 6
- 49
- 121
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
- 1,000
- 2
- 7
- 16
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
- 119
- 6
2
votes
1 answer
Let $(S,*)$ be a finite semigroup with identity. Prove that $S$ is a group iff $S$ has only one element $x$ such that $x^2=x$.
Let $(S,*)$ be a finite semigroup with identity. Prove that $S$ is a group iff $S$ has only one element $x$ such that $x^2=x$.
Attempt: Does this approach true?
$(\Rightarrow)$
Let $S$ be a group. Denote $e_S$ be an identity element in $S$. Clearly,…
lap lapan
- 1
- 2
- 13
2
votes
2 answers
Is the inverse image of a group also a group for semigroup homomorphisms
If $\varphi : S \to T$ is a surjective semigroup homomorphism between semigroups and $G \subseteq T$ is a group, then is $\varphi^{-1}(G)$ also a group?
I know that this result holds if $S$ and $T$ are finite, as then I can find an idempotent…
StefanH
- 17,616
- 6
- 49
- 121
2
votes
2 answers
A question on Finite Semigroup
Why is it so, that a finite semigroup, say $(S, \circ)$ has $a^m=a^n$ for positive integers $m$ and $n$ with $m>n$ for $a\in S$?
Does it imply some sort of periodicity in the binary composition?
Subhasis Biswas
- 2,570
- 1
- 11
- 27