# 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 ...
• 169
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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 ...
• 169
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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 ...
• 4,414
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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? $2)$ ...
• 3,129
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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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### 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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### 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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### 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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