All Questions
7 questions
1
vote
0
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74
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Pure (ordered) subgroups
Let $H,G$ be abelian groups with $H \leq G$. We say that $H$ is a pure subgroup of $G$ if for every $n \in \mathbb N$ and $h \in H$ the following holds: If $h$ is $n$-divisible in $G$, then $h$ is $n$-...
- 111
3
votes
1
answer
125
views
Uncountable divisible groups and the existence of order-preserving isomorphisms of their subsets
Let $(G,+,0,<)$ be an ordered divisible group of uncountable dimension. Consider the subset $G^{<0}$ of $G$.
Question: Are $G$ and $G^{<0}$ isomorphic as ordered sets? Does there exist an ...
- 33
3
votes
1
answer
171
views
Compatible total orderings of the group $\mathbb{Z}^\mathbb{N}$
Given the additive group of the module $\mathbb{Z}^\mathbb{N}$ and a total ordering of the group that is compatible with addition and where $\chi_{\{n\}} > 0$ for all $n \in \mathbb{N}$, can we say ...
2
votes
1
answer
236
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Example involving partially ordered Abelian groups
Definition 1:
Let $(G,\leq)$ be a nonzero partially ordered Abelian group with order unit $u$. (Recall that $u\in G$ is a order unit if, for every $g\in G$, there exists $N\in\mathbb N$ such that $-Nu\...
user34424
2
votes
2
answers
474
views
Non-archimedean group over the reals
I have a totally ordered group $(\mathbb{R};\leq,\oplus,0,-)$ with the reals as base set satisfying monotonicity, i.e.
for all $x,y,z$ we have that if $y\leq z$ then $x\oplus y \leq x\oplus z$, and I ...
- 43
2
votes
1
answer
327
views
Is $\mathbb{Z}^2$ endowed with the square of the strict order, a lattice-ordered group?
I was looking some lattice-ordered group structure. I have kind of difficulty to figure out about the group $\mathbb{Z}^{2}$ with positive cone is $\mathbb{N}_{>0} \times \mathbb{N}_{>0} \cup \{(...
- 33
7
votes
1
answer
266
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Positive cone of a subgroup of $\mathbb{Z}^n$
This question sounds like it should be very well known, but for some reason I failed to find a decent answer anywhere. Let $G\subset\mathbb{Z}^n$ be a subgroup, and $G_+=G\cap\mathbb{Z}_{\ge0}^n$ be a ...
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