0
votes
1answer
40 views

How to show $c-b\lt b-a$

The question: Let $G$ be an Arf semigroup and $a\lt b\lt c$ be three consecutive elements in $G$. How to show that $c-b\lt b-a$ and how to show that this is not necessarily the case for every ...
0
votes
1answer
51 views

Frobenius number and Arf ring of a semigroup [on hold]

Let $$G=\{5m+7n \mid m, n\in \Bbb N\}.$$ Firstly, I want to find the complement of $G$ in $\Bbb N $ is finite. Secondly, how do I find the Frobenius number of $G$ (I guess, the larger ...
0
votes
0answers
31 views

What are $\Gamma$-semigroups?

I have some problems with $\Gamma$-Semigroups, the definition that I've found is A $\Gamma$-Semigroup is a pair $(M,\Gamma)$ defined as follow If $x,y$ and $z$ are in $M$ and $\alpha$ and ...
0
votes
1answer
32 views

Isomorphism of direct product of semigroups

I would appreciate some help with the following problem. Consider four semigroups $A,B,C,D$. I was able to prove that $A\cong C\wedge B\cong D$ implies $A\times B\cong C\times D$. But does also ...
1
vote
1answer
33 views

positive integer polynomial under the usual polynomial multiplication

consider the set of polynomials with positive integer coefficients together with the operation, usual multiplication of polynomials. now my first question is does this set together with the ...
1
vote
1answer
45 views

Generalising cover maps from monoids to semigroups

Let $T,S$ be monoids. A partial surjective mapping $\psi : T \to S$ is called a cover map if for each $s \in S$ there exists some $\hat{s} \in T$ called a cover of $s$ such that for each $t \in ...
1
vote
1answer
61 views

What does $\mathcal{J}$ stand for in Green relations?

Following this book The Algebraic Theory of Semigroups, Volume I , we see that: $a\mathcal{L}b$ means $a$ and $b$ generate the same principle left ideal of the semigroup $S$. $a\mathcal{R}b$ means ...
2
votes
1answer
44 views

Testing for the cancellation laws

When I'm given a presentation of a finitely generated semigroup/monoid, are there any tricks I could use to check if it is cancellative on both sides? I'm not asking for a general algorithm, as I ...
4
votes
1answer
87 views

Naturally Ordered Semigroup: Why does axioms imply order of group is infinite countable ? Why are every group equal up to isomorphism?

Naturally Ordered Semigroup: http://www.proofwiki.org/wiki/Definition:Naturally_Ordered_Semigroup I know $\mathbb N$ is a naturally ordered semigroup by definition. Also I know that the naturally ...
6
votes
0answers
47 views

What can we learn about a magma by studying these monoids?

Given a magma $(X,*)$, we get three monoids in the following way. First, define a pair of functions $L,R : X \rightarrow (X \rightarrow X).$ $$(Lx)(y) = x*y,\quad (Rx)(y) = y*x$$ Then each of the ...
1
vote
1answer
42 views

Does Cartesian Product and Collection of all Sets Perform a Semigroup?

We know that the Cartesian Product is a binary operation. Also it is an associative operation. We know that Cartesian Product of two set is again set, there is even closure axiom. So I need to know ...
0
votes
0answers
57 views

Principal ideal in a semigroup ring.

Let $S= \langle(0,1),(3,2),(5,2) \rangle \subset \mathbb{N}^2 $ be a semigroup and consider the semigroup ring $K[S]$, with $K$ a field. We can consider the principal ideal generated by ...
1
vote
2answers
50 views

If a semigroup satisfies these identities, is it necessarily commutative?

Suppose a semigroup $X$ satisfies the following identities. $$xya\equiv yxa,\quad axy \equiv ayx$$ Without assuming anything further, can we deduce that $X$ satisfies $xy \equiv yx$? In particular, ...
0
votes
1answer
23 views

Do cosets of this semigroup partition this ring?

Let $R = \Bbb{Z}^3$ be the usual direct product ring, and let $S = \{ (k^a, k^b, k^c) : k \in \Bbb{Z} - \{0\}\}$ be a subsemigroup of $R$. Then do the cosets $aS$ partition $R$? Let $R^* = S^{-1}R$ ...
1
vote
1answer
46 views

Unique idempotents in semigroups

In a finite semigroups every element has a unique idempotent power, just take $s^{n!}$ where $n = |S|, s \in S$. In an infinite semigroup there are clearly elements without idempotents, just take ...
3
votes
2answers
70 views

On the powerset of a monoid, does the second operation play nice with the first? Indeed, is it even useful?

Let $X$ denote a monoid. Then we can make $Y = \mathcal{P}(X)$ into a monoid, too. Define $$AB = \{ab \mid a \in A, b \in B\}$$ for all $A,B \in Y.$ We see immediately that $1$ (shorthand for $\{1\}$) ...
2
votes
1answer
49 views

A semigroup $S$ that $|S| > 2$ is regular if and only if $x^2=x$ for any $x \in S$ where $S-\{x\}$ is a group

A semi-group $S$ is called regular if for any $y \in S$ there exists $a \in S$ such that $yay=y$. Let $S$ be a semi-group with more than two elements and $x \in S$ be such an element that $S - ...
1
vote
1answer
37 views

Does there exist a commutative magma such that $\mid$ is transitive, which is not a semigroup?

Let $M$ denote a commutative magma, and write $x \mid y$ iff $xa=y$ for some $a \in M$. If $M$ is a semigroup, then $\mid$ is transitive. Does there exist a commutative magma such that $\mid$ is ...
2
votes
1answer
64 views

If $S$ is a right group, then every $\mathcal L$-class is a $\mathcal H$-class and also a subgroup.

A semigroup $S$ is called right simple if it contains no proper right ideal. A semigroup that is right simple and left cancellative is called a right group. This is equivalent to saying that, for any ...
4
votes
1answer
73 views

definition of (semi)group (co)homology

I'm puzzled why "group cohomology" contains terms 'group' (instead of 'semigroup') and 'cohomology' (instead oh 'homology and cohomology'). I'm new to the subject. Please inform me of any claims ...
4
votes
0answers
63 views

Can we say anything about the structure of the semigroup of non-coprime pairs after this?

Let $S = \{(a,b) : \ a, b \in \Bbb{Z} \wedge \gcd(a,b) \neq 1 \}$. Then it forms a semigroup under componentwise multiplication and if we add an exception, that even though $\gcd(1,1) = 1$, we ...
1
vote
0answers
47 views

What operations is this set closed under? The set of all $\{(a,b) : (a,b) \in \Bbb{Z}^2, \gcd(a,b) \neq 1\}$.

Let $S = \{(a,b): a,b \in \Bbb{Z}, \gcd(a,b) \neq 1 \}$. Under what binary operations is $S$ closed (that we can come up with)? I came up with the following: $$ \begin{align*} (a,b)(c,d) = \\ ...
0
votes
0answers
25 views

Proper term for a monotonic magma

Let $M$ be the multiplication table for a finite magma. Entries are labeled $0,\ldots,n-1$. M has the property that $M(i,j)\ge i, M(i,j) \ge j$. What is the proper term for this kind of magma and ...
1
vote
3answers
92 views

Semigroups isomorphism [duplicate]

Does there exist an isomorphism between the semigroups $S(4)$ and ‎‎‎‎‎‎$\mathbf Z_{256‎‎‎‎‎‎‎}$.‎ $S(4)$ is the set of all maps from the set $X$ to itself and $X = \{1, 2, 3, 4\}$. $S(4)$ is a ...
1
vote
1answer
68 views

Semigroup isomorphism

Does there exist an isomorphism between the semigroups $S(4)$ and $\mathbb{‎‎‎‎‎Z}_{‎256}$?‎ $S(4)$ is the set of all maps from the set $X$ to itself and $X =\{1, 2, 3, 4\}$, $S(4)$ is a ...
3
votes
1answer
168 views

G is a group if and only if for all $a,b∈G$, $ax=b$ has solution; true or false? Why?

I think we only need $ax=b$ or $ya=b$ have solutions in $G$, I'll prove it. Proof: (I) $G$ is a group $\implies$ $a^{-1}ax=a^{-1}b$ $\implies$ $x=a^{-1}b$ $\implies$ $ax=b$ has solutions in $G$ ...
0
votes
1answer
39 views

Why $f(x)=x$ for any $x\in f(X)$?

The set $S(X,X)$ of all mappings of a set $X$ to itself with the composition of mappings in the role of multiplication, where $|X|>1$. Why is not it a group? Let $X$ be a nonempty set. Then the ...
1
vote
3answers
40 views

Example for a claim

If $H$ is a subsemigroup of a semigroup $S$, then it may happen that, for some $a$ and $b$ in $S$, the sets $aH$ and $bH$ don't coincide and, nonetheless, are not disjoint. Does there exist an ...
3
votes
3answers
104 views

How could I understand this sentence:

How could I understand this sentence: For any subgroup $H$ of a group $G$ we have $H^2=HH=H$. However, in general, the same is not true for subsemigroups of semigroups. Thanks a lot!
0
votes
2answers
78 views

How could I prove that the equality $a^na^m=a^{n+m}$ in a semigroup with identity?

Let $S$ be a semigroup with a identity. How could I prove that the equality $a^na^m=a^{n+m}$ which holds for all $n,m \in \mathbb Z$. Note that $a \in S$ and $\mathbb Z$ denotes the set of all ...
1
vote
2answers
215 views

Is associative binary operator closed on this subset?

Here is the problem: Suppose that $*$ is an associative binary operation on a set $S$. Let $$H:= \{a \in S\mid a * x = x * a \mbox{ for all }x\in S\}.$$ In other words, $H$ is consisting of all ...
1
vote
1answer
114 views

If in a semigroup $S$, $\forall x \exists ! y:xyx=x$, then $S$ is a group [duplicate]

If for all $x$ in a semigroup $S$, there exists a unique $y$ such that $x y x=x$, then $S$ is a group. (Not to be confused with inverse semigroup, where only $y$ satisfying both $xyx=x$ and $yxy=y$ is ...
-1
votes
2answers
103 views

Building factor semigroup w.r.t. $I_n = \{ w \in A^+ : |w| \ge n \}$.

I don't understand the following factor semigroup. Consider the pseudovariety $N$ of nilpotent semigroups. For any finite alphabet $A$, let $I_n = \{ w \in A^+ : |w| \ge n \}$. Then $A^+ / I_n$ is ...
1
vote
1answer
70 views

Proving $\mathcal{D}$-class of semigroup.

Theorem 2.17 If $a$ and $b$ are elements of a semigroup $S$, then $ab\in R_a\cap L_b$ if and only if $R_b\cap L_a$ contains an idempotent. If this is the case, then $$aH_b = H_a b = H_a H_b ...
4
votes
1answer
30 views

Semigroup which satisfied $(xy)^{\pi} = (xy)^{\pi}x$ and principal right ideals

Let $(S,\cdot)$ be a finite semigroup, then every $s \in S$ has an unique idempotent power, i.e. there exists a smallest $i \in \mathbb N$ such that $s^i$ is idempotent. It is the unique idempotent in ...
7
votes
3answers
351 views

Is a semigroup $G$ with left identity and right inverses a group?

Hungerford's Algebra poses the question: Is it true that a semigroup $G$ that has a left identity element and in which every element has a right inverse is a group? Now, If both the identity and the ...
5
votes
1answer
104 views

Structure theorem for finitely generated commutative $semi$groups.

$\newcommand{\ZZ}{\mathbb{Z}}$ It is a classical theorem that a finitely generated commutative group is isomorphic to one of the form: $$ \ZZ^n \oplus \ZZ/{m_1}\ZZ \oplus \ZZ/{m_2}\ZZ \oplus \dots ...
1
vote
2answers
153 views

Whether $L=\{(a^m,a^n)\}^*$ is regular or not?

I am condidering the automatic structure for Baumslag-Solitar semigroups. And I have a question. For any $m,n \in Z$, whether the set $L=\{(a^m,a^n)\}^*$ is regular or not. Here a set is regular means ...
5
votes
1answer
173 views

Help me prove equivalently of regular semigroup and group.

Let $S$ be a semigroup. Prove that the following are equivalent: $\forall a \in S \exists! x \in S$ such that $ax \in E(S)$ where $E(S)$ is the set of all idempotent. $\forall a \in S \exists! x \in ...
1
vote
1answer
55 views

semigroups-terminology question

Let the element $x$ of a commutative semigroup has the property $x=a+b\Rightarrow a=x~$or $b=x$. I call such an element "prime". My question is: what is the right term about such elements. ...
2
votes
1answer
124 views

Is closure of a semigroup again a semigroup?

Let $S$ be a compact left-topological semi-group (meaning, $S$ is both a semi-group and a compact Hausdorff topological space, and the map $x \mapsto x y$ is continuous for any fixed $y$, but the map ...
3
votes
1answer
99 views

Does someone know other examples of $K$-semigroups?

In our work on projective representations we need use the following object: Let $K$ be a field. By a $K$-semigroup we mean a semigroup $S$ with $0$ and a map $K \times S \to S$ such that $\alpha ...
5
votes
5answers
575 views

Is there an idempotent element in a finite semigroup?

Let $(G,.)$ be a 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? ...
2
votes
1answer
71 views

Finitely generated semigroup gets the finitely generated subsemigroup?

From N.RUSKUC's paper "On Large Subsemigroups and Finiteness conditions of Semigroups", there is a theorem, Here large subsemigroup means $S$\ $T$ is finite. In this side "=>" of the proof in the ...
3
votes
2answers
60 views

How to move from a right semigroup action to a left semigroup action?

Let $S$ be a semigroup and $X$ any set. Define a left action of $S$ on $X$ to be a map $\sigma: S \times X \rightarrow X$ with the property that $(st)x = s(tx)$, where we define $gx = \sigma(g,x)$ ...
9
votes
0answers
187 views

How much do idempotent ultrafilters generate in terms of semigroups?

It is known that the set of ultrafilters on, say, the natural numbers $\mathbb{N}$, can naturally be endowed with the structure of a compact topological left semigroup (which fails to be anything ...
3
votes
3answers
114 views

How are the powers being changed

I have a semigroup $S$ including a generator, say $d$, such that $$d^4=d$$ I am trying to guess the general rule of $d$'s powers such that when I want to calculate $d^n, n\in\mathbb N$; I can simplify ...
1
vote
4answers
165 views

Inverse elements in the absence of identities/associativity.

Lets view groups as consisting of a binary operation, a distinguished element $e$, and unary operation $x \mapsto x^{-1}$. Then the group axioms can be stated as follows. $(xy)z=x(yz).$ $xe=ex=x.$ ...
3
votes
0answers
94 views

Representation for idempotent semiring

I have a semi-ring whose multiplication is non-commutative, and addition is idempotent. That is, $ab \neq ba$ and $a + a = a$. The semi-ring is freely generated from a finite set $\Sigma$, the ...
5
votes
3answers
105 views

Proving a binary operation on $\mathbb{Z}$ gives a semigroup

Let $x\circ y= x +y-xy, \quad (x,y) \in \mathbb{Z}$ where $\circ$ is a binary operation on $\mathbb{Z}$, prove that this is a semigroup. My Work To prove we have to check two things: $\mathbb{Z}$ ...