1
vote
1answer
47 views

Every subgroup of a cyclic group is characteristic (using lattice theory).

I want to show for $n\in\Bbb N$, which is not square-free, that every subgroup of $Z_n=\langle x\;|\;x^n\rangle$ is characteristic. But I want to show it in a convoluted way. Every automorphism ...
0
votes
1answer
22 views

Finite Linearly Ordered Abelian Monoids

This question concerns the proper definition of the phrase "finite linearly ordered abelian monoids". The sequence A030453 of OEIS counts the number of "finite linearly ordered abelian monoids". The ...
4
votes
1answer
108 views

Who first studied semilattices?

Historically, who first studied semilattices, as opposed to lattices or Boolean algebras? (With or without identity, I do not mind.)
1
vote
1answer
48 views

Distributive lattices and Birkhoff theorem

I am trying to prove the teorem (Birkhoff) $L$ is a nondistributive lattice iff $M_5$ or $N_5$ can be embedded into $L$ The only part of the proof which I can't understand is this (I am copying from ...
0
votes
1answer
27 views

Lattice inside a finite dimensional vector space

I have an integral domain $R$ and its field of fractions $K$. Let $V$ be a finite dimensional $K$ vector space. Let $M$ be a finitely generated $R$-module contained in $V$. Why is $K\cdot M=V$ ...
3
votes
2answers
37 views

All tree orders are lattice orders?

Say that a set is tree ordered if the downset $\downarrow a =\{b:b\leq a\}$ is linearly ordered for each $a$. In a comment, Keinstein says that such sets are also semi-lattices, provided they are ...
4
votes
4answers
68 views

Number of join-irreducible elements of a lattice: is it monotonic?

Let $\mathcal L$ be a sub-lattice of $\mathcal P(X)$, where $X$ is a finite set. Denote by $\mathcal I(\mathcal L)$ the set of union-irriducible elements of $\mathcal L$ (i.e. $A\in \mathcal ...
0
votes
1answer
121 views

Is every sub-lattice of $\mathcal P(X)$ isomorphic to a sub-lattice of $\mathcal P(X')$ containing singleton sets?

Let $X$ be a finite set. $\mathcal{P}(X)$ denotes the set of all subsets of $X$. Let $\Gamma$ be a sub-lattice of $\mathcal{P}(X)$, i.e. $\Gamma$ is a collection of subsets of $X$ closed under union ...
1
vote
1answer
13 views

extension of an increasing function over a lattice

Let $X$ be a finite set. $\mathcal{P}(X)$ denotes the set of all subsets of $X$. Let $\Gamma$ be a sub-lattice of $\mathcal{P}(X)$, i.e. $\Gamma$ is a collection of subsets of $X$ closed under union ...
0
votes
0answers
50 views

Exercise 1.17 from Bell & Slomson's Models and Ultraproducts

I'm attempting to prove the following theorem left as an exercise from Bell & Slomson's Models & Ultraproducts (1969). I'd like to know whether my attempted proof is correct, and if not, I'd ...
3
votes
2answers
84 views

Examples of Stone algebras which are not Boolean algebras

Grätzer, in his Lattice Theory: Foundation, describes a Stone algebra as a distributive lattice with pseudocomplementation $L$ which satisfies the Stone identity: for every $a \in L$, $\neg a \vee ...
0
votes
0answers
26 views

Operation table of Hasse diagram

Consider the following Hasse diagram: My book gives the following join and meet operation tables for this diagram: $$\begin{array}{|c || c | c|} \hline Subset & x \wedge y & x \vee y \\ ...
0
votes
1answer
35 views

How to show that complement of prime filter is ideal? [closed]

How to show that in any lattice L, F is a prime filter if an only if its complement L\F is an ideal?
1
vote
0answers
33 views

Characterization of subgroup dual to Frattini Subgroup

Let $G$ be a group and let $\mathcal{L}(G)$ denote the complete lattice of subgroups of $G$. We have that every automorphism of $G$ induces a lattice-automorphism on $\mathcal{L}(G)$. From here we see ...
0
votes
0answers
62 views

Ascending chain condition holds in a lattice implies every ideal is principal

Proof: Suppose for contradiction that ACC holds for a lattice L, but there exists an ideal which is not principal. Thus $\exists$ I $\subset$ L s.t. ~$\exists$x: x $\le$a for a$\in$L. Thus $\exists$I ...
0
votes
1answer
54 views

Definition of finite direct decomposition of elements and indecomposable elements at arbitrary lattice

How can i define finite direct decomposition of elements and indecomposable elements at arbitrary lattice . I think i can say an element of lattice is finite direct decomposition of elements if it be ...
1
vote
2answers
67 views

Name for this axiom $(X \bullet R) \sqcup (Y \bullet S) \equiv (X \sqcup Y) \bullet (R \sqcup S)$

I am trying to give a name to this axiom in a definition: $(X \bullet R) \sqcup (Y \bullet S) \equiv (X \sqcup Y) \bullet (R \sqcup S)$ (for all $X, Y, R, S$) where $\sqcup$ is the join of a ...
3
votes
0answers
38 views

Proof of lattice distributivity by a grading function

Looking for an algorithm for recognizing finite distributive lattices, I came across Linear Time Recognition Algorithm for Distributive Lattices by Michel Habib and Lhouari Nourine. Just before the ...
2
votes
1answer
41 views

Looking for an algebraic structure

I'm looking for the name of algebraic structures (in which the elements are partially ordered) with the following properties: Top element defined, bottom optional; Join defined for all elements, ...
0
votes
1answer
78 views

Condition for the boolean algebra of clopen sets to be extremely disconnected.

Let $X$ be a topological space and let $\Gamma \mathcal O(X)$ be it's boolean algebra of clopen subsets. For compact totally disconnected space, show that $\Gamma \mathcal O(X)$ is complete (as a ...
0
votes
1answer
30 views

lowering of a semilattice

In these Lecture Notes the notion of lowering a semilattice is introduced, there it is stated: Sometimes "broken" elements need to be looked at and computed with. Now any semilattice can have an ...
0
votes
1answer
29 views

Are ideals of an sup semilattice always non-empty?

I am trying to do an exercise from the book A Compendium of Continuous Lattices. Exercise: Let $L$ be a set with a transitive relation, and let $A,B$ be ideals of $L$. (i) $A\cap B$ is an ideal of ...
0
votes
1answer
135 views

Showing that a group is lattice-ordered

Say I have a set $S$ with a group operation $\cdot$ and lattice ordering $\leq$. Suppose further that: $x\leq 1\implies xy\leq y$ $x\geq 1\implies xy\geq y$ For all $x,y$. Does it follow that ...
1
vote
2answers
55 views

Notation for “incommensurate” elements?

Say that $x\wedge y\not=x,y$, that is neither $x\leq y$ nor $y\leq x$. Is there a way to denote this? I've been saying $x<>y$ but that's completely made up.
0
votes
1answer
237 views

What is a subdirect product?

I'm having trouble understanding what a subdirect product is. Say $G$ is a subdirect product of $H=\prod H_i$ - this means that the homomorphisms $f_i:G\to H_i$ are surjective, which can be ...
1
vote
2answers
107 views

The kernel of the kernel.

From Wikipedia-Entry on Equivalence Relatin:Lattices The possible equivalence relations on any set X, when ordered by set inclusion, form a complete lattice, called Con X by convention. The ...
2
votes
2answers
195 views

Smallest Congruence Relation generated by a set

$\newcommand{\cl}{\operatorname{cl}}$ Let $R \subset S \times S$ be a binary relation, the smallest i) reflexive relation containing it is $$ \cl_\mathrm{ref} = R \cup \{ (x,x) : x \in S \} $$ ii) ...
1
vote
1answer
63 views

Proof that the lattice of fully invariant congruences is a sublattice of the lattice of all congruences

Let $\mathfrak U$ be an algebra (i.e. a set, called universe, together with several $n$-ary operations) in the sense of universal algebra. Denote by $\operatorname{Con} \mathfrak U$ the set of all ...
5
votes
2answers
98 views

Are filters in lattices exactly the homomorphic preimages $\varphi^{-1}(1)$ of top elements?

Say I got a lattice L, a bounded lattice K with top-element $1$ and a homomorphism $\varphi : L \to K$, then $\varphi^{-1}(1)$ is a filter in L. I wondered whether I can represent every filter $F ...
3
votes
0answers
52 views

How can we define “trivially orthogonal” groups?

In any lattice-ordered group, we say that two elements are orthogonal if their meet is 1. I've been thinking of groups who have only "trivial" orthogonal relations, i.e. $x\perp y\implies x=1$ or ...
1
vote
1answer
75 views

Show that the stabilizer is a prime subgroup

We define a subgroup $H$ as being convex if $g\in H\implies h\in H$ for all $1\leq h\leq g$. A convex subgroup $P$ is prime if for any two convex subgroups $X,Y$: $X\cap Y \subseteq P$ implies that ...
1
vote
0answers
64 views

Intuition behind prime subgroups

In any lattice ordered group, we say that a convex subgroup $P$ is prime if for any two convex subgroups $X,Y$: $X\cap Y \subseteq P$ implies that $X\subseteq P$ or $Y\subseteq P$. This is analogous ...
1
vote
1answer
22 views

Show $g^{-1} \wedge h^{-1}=(g\vee h)^{-1}$

Glass' Partially Ordered GroupsLemma 2.3.2 says: Let G be a p.o. group and $g,h\in G$. If $g\vee h$ exists, then $g^{-1} \wedge h^{-1}=(g\vee h)^{-1}$ Proof: If $f\leq g^{-1},h^{-1}$ then ...
0
votes
1answer
27 views

Show that $\langle G^+\rangle=G$ in a directed group

Lemma 2.1.8 of Glass' Partially Ordered Groups states: $G$ is a directed group if and only if $\langle G^+\rangle=G$ (where $G^+=\{x:x\geq1\}$) This doesn't make any sense to me. For example, ...
3
votes
0answers
58 views

Can all non-archimedean groups be written as a product of archimedean groups?

We say that a partially ordered group $(G,\cdot, \geq)$ is Archimedean if for any $g,h >1\in G$ there exists some n such that $g^n > h$. All the non-archimedean groups I know of can be written ...
1
vote
1answer
84 views

When is “mod $n$” a congruence relation on the lattice $(\Bbb N,\gcd,\text{lcm})$?

For which $n\in \Bbb N$, $$a\equiv b,a'\equiv b'\quad \text{implies} \quad \gcd(a,a')\equiv \gcd(b,b'), \text{lcm}(a,a')\equiv \text{lcm}(b,b')$$ all mod $n$. For $n=2$ it is true.
5
votes
1answer
174 views

A Question on the Young Lattice and Young Tableaux

Let: $\lambda \vdash n$ be a partition of $n$ $f^\lambda$ - number of standard Young Tableaux of shape $\lambda$ $\succ$ - be the covering in the Young Lattice (that is, $\mu \succ \lambda$ iff ...
8
votes
2answers
225 views

Example of non-abelian partially ordered group

What is a simple example of a non-abelian partially ordered group?
0
votes
1answer
64 views

an infinite queue preserving equality.

Is there any well-ordered set $(A,\leq)$ such that: $(A,\leq^{-1})$ is well-ordered. $A$ is infinite. there's exactly one function $\theta:A\rightarrow \{0,1\}$ such that 1) for each $a < M$, ...
2
votes
1answer
73 views

Composition length of surjective inverse limits

Let $M$ be a left module over some ring $R$ and suppose that $M$ is an inverse limit of a family of modules $M_i$ with $i\in I$. We suppose also that the maps of the inverse limit $\pi_i:M\to M_i$ are ...
5
votes
2answers
2k views

Given the Hasse diagram tell if the structure is a lattice

Let's consider the following Hasse diagram: I need to tell whether this is a lattice. By lattice definition I can prove the above shown structure $M_5$ to be a lattice if and only if $\forall x,y ...
0
votes
2answers
407 views

How to apply the lattice definition and show if a poset is a lattice

Let $S = \{1,2,3\}$ and let the poset $(\wp(S)\setminus\{\emptyset\}, \sim)$ be defined as follows: $$\begin{aligned} X \sim Y \Leftrightarrow X = Y \text{ or } \max(X) < \max(Y) \end{aligned}$$ ...
3
votes
1answer
179 views

How to derive distributivity from Boolean algebra laws

Let $(L,\le,\bot,\top)$ be a bounded lattice and $\neg: L \rightarrow L$ be a map that satisfies the following laws: $a \wedge b = \bot \Leftrightarrow a \le \neg b$ $\neg\neg a =a$ I'd like to ...
1
vote
2answers
313 views

Draw Hasse diagram as two elements have same image

Let $S=\{1,2,3,4,5,6,7,8,9,10\}$, $P=\{y \in \mathbb N : y \text { is a prime number}\}$, consider the map $f$ defined as follows: $$\begin{aligned} f:x\in S \rightarrow f(x) \in \wp (P) ...
5
votes
0answers
134 views

Prove $(\mathbb Z \times \mathbb Z, \Sigma)$ to be a partial order and tell if its subset $T'$ is a lattice

Let $T = (\mathbb Z\times\mathbb Z, \Sigma) $ be defined as follows: $$\begin{aligned} (a,b) \text{ } \Sigma \text { } (c,d) \Leftrightarrow (a,b) = (c,d) \text{ or } a^2b^2<c^2d^2\end{aligned}$$ ...
4
votes
2answers
351 views

partially ordered group, positive cone, quotient (exercise)

Definitions: A partially ordered group or po-group is a po-set $(G,\leq)$, such that $G$ is a group and $\forall x,y,a,b\!\in\!G\!:x\!\leq\!y\Rightarrow axb\!\leq\!ayb$, i.e. a po-set that is a group ...
1
vote
1answer
320 views

MacNeille completion of a totally ordered set: Dedekind cuts

If $X$ is any partially ordered set with $A\!\subseteq\!X$ and $x\!\in\!X$, define $x\!\leq\!A :\Leftrightarrow \forall a\!\in\!A\!: x\!\leq\!a$ and $A\!\leq\!x :\Leftrightarrow \forall a\!\in\!A\!: ...
2
votes
1answer
79 views

Diamonds of ideals, part 3

I'd like to wrap up the line of questioning started first in this question and then continued in this question. The only variant left to try is: "How close can you get to the Diamond lattice ...
4
votes
2answers
199 views

An analogy between subgroups and equivalence relations.

I have noticed a certain analogy between subgroups of a group $G$ and equivalence relations on a set $X$. I would like to know if there's an explanation for this analogy or a common generalization of ...
2
votes
1answer
148 views

correspondence for universal subalgebras of $U/\vartheta$

Let $U$ be a universal algebra of type $T$, and denote $\mathrm{Con}(U)\!=\!\{\text{congruence relations on }U\}$ and $\mathrm{Sub}(U)\!=\!\{\text{subalgebras of }U\}$. Let "$\leq$" mean "subalgebra". ...