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Questions tagged [ordered-groups]

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

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Ordered groups over the same underlying set and total order, can they be group isomorphic without being order isomorphic?

Given ordered groups $H:=(A,+)$ and $H':=(A, \boxplus)$ as per the same total order $ \leqslant$ of A, can $H$ and $H'$ be group isomorphic without being order isomorphic? I understand that there can ...
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Is an Archimedean topological group of the reals isomorphic to $(\mathbb R,+)$?

A group $H:=(\mathbb R,\boxplus)$, is given to be Ordered as per the canonical order of $\mathbb R$. Archimedean as per order in 1. Topological as per canonical topology of $\mathbb R$. Can it be ...
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Characterization of weighted total orders on $\mathbb{N}^r$ or $\mathbb{Z}^r$

By a weighted total order on $\mathbb{N}^r$ or $\mathbb{Z}^r$ I mean the relation defined by $$ (m_1,\ldots,m_r) \leq (n_1,\ldots,n_r) \quad\Longleftrightarrow\quad \sum_{i=1}^r c_i m_i \leq \sum_{i=1}...
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if f is a homomorphism from L to L', Should the image f(L) be a sublattice of L'?

I'm a beginner in the subject & my question can be meaningless, so I'm sorry from start if that's the case. I just don't understand why all of the image f(L) can be a sublattice of L' when f is a ...
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Quotients of preordered groups

A preordered group is a group along with a preorder that makes addition monotone. As it is well known, a congruence in a group is equivalent to a normal subgroup. Everyone knows that a preorder on a ...
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Is every ordered abelian group the additive group of an ordered ring?

Let $\Lambda$ be an ordered abelian group, (there is a total order on $\Lambda$ which is compatible with addition). Is there a multiplication map on $\Lambda$ that turns it into an ordered ring? I ...
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The quotient of totally ordered additive abelian groups which is torsion (value groups in valuation theory) [closed]

Let $(G,+,\leq)$ be a totally ordered additive abelian group and $(H,+\leq)$ be a subgroup of $(G,+,\leq)$. Let the quotient $\frac{G}{H}$ be torsion and be of finite order $n$ ($|\frac{G}{H}|=n$), ...
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Linearly ordered, non CT-groups

A linearly ordered group is a group $(G,\dot,1)$ together with a linear ordering $<$ on $G$ with $f<g \Longrightarrow (f h<g h$ and $h f < h g)$ for all $f,g,h \in G$. A commutative-...
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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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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 ...
goblin GONE's user avatar
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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 $I$ ...
Link L's user avatar
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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 ...
Matemáticos Chibchas's user avatar
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1 answer
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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\...
Ozaner Hansha's user avatar
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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 ...
user's user avatar
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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 $\leq$?
Jeremy's user avatar
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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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1 vote
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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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3 votes
1 answer
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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 ...
dumb_undergrad's user avatar
1 vote
1 answer
90 views

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. ...
mick's user avatar
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2 votes
2 answers
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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 ...
nombre's user avatar
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1 vote
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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 $...
Matija Sreckovic's user avatar
3 votes
1 answer
81 views

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 ...
Paolo Leonetti's user avatar