Questions tagged [ordered-groups]

An ordered group is a group with a (partial) order which the group operation preserves.

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Union of powers of a well-ordered set is well-ordered.

While studing a certain type of rings I was trying to solve the following exercise (which is what I need to prove that this ring is well defined): Let $S$ be a well-ordered subset of an ordered group $...
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Relation of the triple ordered pair

I do not understand the problem. This is my homework which requires me to prove the triple ordered pair theorem and ask me the equal notation "=" from that theorem is reflexive or symmetric ...
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The composite valuation

I am currently studying valuation theory and came across the concept of composite valuation. I think, I proved that the composite valuation is always given by a lexicographic product of the two ...
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If a field has unique ordering, must its subfields also have unique ordering?

Let $L$ be a field and suppose that there exists a unique total order $\leq$ on $L$ with respect to which $(L,\leq)$ is an ordered field. Now let $K$ be a subfield of $L$. Clearly $K$ is an ordered ...
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What is an isolated subgroup? [closed]

" As in the case of finite numbers, the infinitesimal numbers form an isolated subgroup  of $R_{inf}$ of $^*R$"page 152 What does this sentence mean? What is an isolated subgroup?
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Generalization of order of the product of elements in an abelian group as follows.

Let $(G,*)$ be a group. Let $a_i \in G, |a_i|=n_i, 1 \le i \le m$. Suppose $\gcd(n_i,n_j)=1$ and $a_ia_j=a_ja_i$, for all $i$ and $j$. Let $x=a_1*a_2*\ldots*a_m$. Show that $|x| = n_1n_2\ldots n_m$. ...
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Let $G$ be a group with $|a_i|=n_i$ for all $1\le i \le m$. Show that $|x|=n_1n_2\ldots n_m$ where $x$ defined as follows

Let $(G,*)$ be a group. Let $a_i \in G, |a_i|=n_i, 1 \le i \le m$. Suppose $\gcd(n_i,n_j)=1$ and $a_ia_j=a_ja_i$, for all $i$ and $j$. Let $x=a_1*a_2*\ldots*a_m$. Show that $|x| = n_1n_2\ldots n_m$. ...
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Let $(G,*)$ be a group and $a \in G$. Suppose that $|a|=n$ and $n=mk$ for some positive integers $m$ and $k$. What is $|a^k|$?

Let $(G,*)$ be a group and $a \in G$. Suppose that $|a|=n$ and $n=mk$ for some positive integers $m$ and $k$. What is $|a^k|$? attempt: Let $a \in G$ such that $|a|=n$. Then, \begin{equation*} a^n = a^...
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Order of an element in a group problem as follows

Let $(G,*)$ be a group, $a \in G$, and $|a|=p$, where $p$ is a prime. Prove that $|a^k|=p$ for all $1 \le k \lt p$. Prove that for all $m \in \Bbb N$, either $a^m = e_G$ or $|a^m|=p$, where $e_G$ is ...
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Let $a$ and $n$ be integers with $n>0$. Show that the additive order of $a$ modulo $n$ is $\frac{n}{\gcd(a,n)}$.

Let $a$ and $n$ be integers with $n>0$. Show that the additive order of $a$ modulo $n$ is $\frac{n}{\gcd(a,n)}$. attempt: Let $a$ and $n$ be integers with $n>0$. Let $m=\frac{n}{\gcd(a,n)}$. ...
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Divisible totally ordered additive abelian groups [closed]

Let $(G,+,\leq)$ be a divisible totally ordered additive abelian group and $g_{1},g_{2}\in G$. If for every integer $n>1$, $g_{1}\geq (1-\frac{1}{n})g_{2}$, Can we have $g_{1}\geq g_{2}$? Thanks ...
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Understanding totally ordered abelian groups

Let $(\Lambda, \leq ,+)$ be a totally ordered abelian group. Say such an abelian group is simple if it has no nontrivial quotients (the only quotients are $0$ and itself). One might wish to understand ...
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Question about ordered group and its order ideal.

I am reading a book entitled An introduction to the classification of amenable C*-algebras. It reads, Definition 3.3.1 An ordered group $(G,G_+)$ is an abelian group $G$ with a distinguished ...
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Ordered Groups: Left Multiplication vs Right Multiplication

Given that $G$ is a linearly ordered group (bi-ordered). I want to try and understand the difference between the “size” of left multiplication vs right multiplication (which I have written below using ...
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Decomposition of $2^{[n]}$ into $\binom{n}{n/2}$ chains

I am struggling with this problem: Let n be an even number, and denote $[n]=\{1,2,...,n\}$. A sequence of sets $S_1 , S_2 , \cdots , S_m \subseteq [n]$ is considered graceful if: $m$ is odd. $S_1 \...
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ordered topological groups

Is there a partially order relation on $S^1$ whose corresponding order topology coincides to the standard topology of circle?Is there a totally order relation with such property? what is a theory ...
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On convex subgroups of totally ordered groups.

I am trying to find an example of a totally ordered abelian group $(\Gamma,+)$ such that the set of its convex subgroups is $\textbf{not}$ totally ordered. I think there should be such an example, ...
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Does the Unique Product Property mean that there are only two elements in the set that can create any other in the set, given an operation?

I was under the impression that the way it's defined in the title was the correct way to interpret the UPP(as that was what I was searching for) from the definition here: https://mathoverflow.net/...
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Coproduct in the Category of Totally Ordered Groups

This is a question that arised from a problem in valuation theory of commutative rings. There I need to construct a totally ordered Abelian group "containing" every member of a given family of totally ...
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Property of ordered groups.

A totally ordered abelian group is an abelian group (G,+) with a total order $\leq$ such that for all $a,b,c \in G$ if $a \leq b$ then $a+c \leq b+c$. We will say that an ordered abelian group is ...
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Is there a better term for "non-negative"?

Real numbers $x$ satisfying $x \geq 0$ are said to be non-negative. Some alternative phrases include positive or zero and at least zero. However, I find these phrases to be unsatisfactory, for a ...
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an ordered abelian group has no order units

An element $e$ in $G^{+}$ is called an ordered unit in an ordered abelian group $(G,G^{+})$ if for any $g\in G$,there exits a positive integer such that $-ne\leq g \leq ne$. In Rordam's book,there ...
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isomorphism in ordered monoids

I read that a morphism $\gamma : S \rightarrow T$ is an isomorphism if there exists a morphism $\Psi : T \rightarrow S$ such that $\gamma \circ \Psi = I(T)$ and $\Psi \circ \gamma = I(S)$, where $I$ ...
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Order preserving injection $f$ from set of rationals $Q$ into $R$ with discrete image.

How to construct an order-preserving injection $f:Q\rightarrow R$ , such that the image of $f$ is discrete subspace of $R$ (set of reals).
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Looking at Automorphisms of Subgroups of $(\mathbb R,+)$ With Positive Slope

Let $(R,+)$ be a non-trivial subgroup of $(\mathbb R,+)$. We say that an automorphism $\phi$ of $R$ has positive slope if $\tag 1 \phi(R \cap \mathbb R^{\gt 0}) \subset \mathbb R^{\gt 0}$ What ...
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Continuity of absolute value in topological ordered abelian groups

Let $(G,+,0)$ be an abelian topological ordered group, that is, $G$ is endowed with a total order $\leq$ such that, for any $a,b,c\in G$, we have that $a\leq b$ implies $a+c\leq b+c$. Moreover, $G$ is ...
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Totally ordered abelian group with a unique " isolated subgroup"

Let $(G,+,<)$ be a totally ordered abelian group i.e. $(G,+)$ is an abelian group with partial order $<$ such that for every $a,b\in G$, exactly one of $a=b$ or $a<b$ or $b<a$ holds; and ...
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Ordered groups: do irreducible elements always commute?

The title is the main question. Explaining the notation, an ordered group is a group $X$ with a partial order such that for every $x,y,z\in X$, if $x\leq y$, then $zx\leq zy$ and $xz\leq yz$. An ...
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Definition of totally ordered monoid?

Suppose I have the monoid $(\mathbb{N},\times)$. It is my understanding that for the relation $\le$ to form a total order on the above monoid, the following must be true: $$a\le b\iff (\forall c\in\...
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On totally ordered abelian groups having exactly one convex subgroup

Let $(G, <)$ be a totally ordered abelian group. Let us call a proper subgroup $H$ of $G$ to be convex if for every $a \in H$, $[a,-a]:=\{x\in G : -a \le x \le a\} \subseteq H$. If $G$ has exactly ...
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Does every ordered divisible abelian group admit an expansion (and how many) to an ordered field?

Let $(G,+,<)$ be an ordered divisible abelian group. $1)$ Is it always the case that there exists a binary function $*:G\times G \rightarrow G$ such that $(G,+,*,<)$ is an ordered field? $2)$ ...
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Uniqueness of total orders on abelian groups

Suppose $G$ is an abelian group totally bi-ordered by $\leq$ and by $\leq'$. Does it follow that $\leq'$ is either equal to or the converse of $\leq$?
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On the definition of the totally ordered abelian group $\Gamma/\Delta$

If $\Gamma$ is a totally ordered abelian group, and $\Delta\subset\Gamma$ is a convex subgroup (meaning if $\delta,\delta'\in\Delta$ and $\delta\leq\gamma\leq\delta'$ then $\gamma\in\Delta$), then we ...
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Suppose a ring R is a PID. What does this say about possible orders on the set of R?

More explicitly, is there always some partial or total order we can apply to the set of R?
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a property of a morphism of ordered groups

An ordered group $(G,G^+)$ is an abelian group $G$ together with a subsemigroup $G^+$ containing the identity $0$, having these properties: $G^+-G^+=G$. $G^+\cap (-G^+)=\{0\}$. We have for $x,y\in ...
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Subgroups of the rationals is either generated by one element or infinitely generated

The following situation is given: Consider $(\mathbb{Q},+)$ as an ordered abelian group: $x,y\in \mathbb{Q}, \; x\le y\; :\Leftrightarrow y-x\ge 0$. Let $\mathbb{Q}_+=\{c\in\mathbb{Q}\mid c\ge 0\}.$ ...
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When does an abelian "linearly ordered group " has the property that any non-empty subset of the set of "non-negative" elements has a "least" element

By a theorem of F.W. Levi , we know that an abelian group can be equipped with a linear order (https://en.wikipedia.org/wiki/Linearly_ordered_group) iff the group is torsion free . So let $(G,\le )$ ...
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Existence of a certain near-metric map on an ordered divisible abelian group

Let $\mathcal{M}=(M,0,+,<)$ be a linearly ordered divisible abelian group. Let's define an $\mathcal{M}$-metric on $M$ to be a map $d:M\times M \rightarrow M$ such that (1) $\forall x,y\in M,\, d(...
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If ordered ring $A$ has upper or lower bound, then, $A = \{0\}$ (alternative proof)

Knowing that any ordered ring $A \neq \{0\}$ is infinite, can I say that if an ordered ring is finite, it must be $\{0\}$ ? By this result - If an ordered ring $A$ has an upper bound, does it have ...
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1 vote
1 answer
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Counting permutations without fixpoint

Let $n$ be a positive integer. Consider an ordered set $S_n = [1,2,3,...,n]$ where the $j$ th element from the left equals $j$. Now consider a function defined on $S_n$ as a permutation of that set. ...
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2 votes
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Theory with non archimedean models

Is there a consistent theory in the language of ordered groups (or ordered rings) whose models are non archimedean ordered groups (or rings or fields)? (note: I am not asking for the existence of an ...
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Density definition in totally ordered Abelian group

In my analysis textbook, there's a section about totally ordered Abelian groups, and there's a definition of density, which goes as follows: A subset $A \subset G$ of a totally ordered Abelian group $...
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Existence of a sequence converging to $0$

Let $(G,+,\le)$ be a partially ordered group (with identity $0$) and suppose that for each positive $g \in G$ there exists $g^\prime \in G$ such that $0<g^\prime<g$. Is it true that there ...
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Is there any order isomorphic function between $\mathbb N \times \mathbb Z$ to $\mathbb Z \times \mathbb N$?

Is there any order isomorphic function between $\mathbb N \times \mathbb Z$ to $\mathbb Z \times \mathbb N$? (With lexicographical order) Thanks
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Ordered ring and mutiplicative ordinal

If $(E,<)$ is a linear order, let $s(E)$ denote the supremum of the set of ordinals which (order-)embed in $(E,<)$. $s(E)$ is also the set of ordinals which embed in $E$ with a non cofinal range....
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Probability to find at least one alphabetically ordered subset of K elements in a set of N elements

I would like to know how to calculate the probability of finding an alphabetically ordered subset of at least K elements in a set of N alphabetical elements. For example, for a set of N letters from ...
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3 votes
1 answer
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Intervals in divisible ordered groups

Is it true that if $(G,+,0,<)$ is a divisible ordered abelian group with at least two elements, then for $a,b >0 \in G$, there is an injective order preserving map from $[0;a)$ to $[0;b)$? It ...
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On ordered abelian groups containing $\mathbb{Z}$

Let $\Delta$ be an ordered abelian group containing $\mathbb{Z}$ as a subgroup of index $e$. I need to show that for any positive element $\delta \in \Delta$, we have $e\delta \geq 1$. I have no ...
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5 votes
1 answer
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Can the order on an ordered, cancellative monoid be extended to its Grothendieck group?

Suppose we have an ordered, cancellative monoid and we wish to apply the Grothendieck group construction to it. Can the total order be extended to the larger group? Example: consider the ordered ...
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2 votes
2 answers
368 views

Partial order relation on the group of integers

We know that the usual $\leq$ is a partial order relation on the group of integers $\mathbb Z$ and $\mathbb Z$ is a totally ordered with this partial order relation. Is there any other partially order ...
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