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Say that a Positive Group is a poset $(G,\geq)$ with an operator $\cdot$ s.t.

(Closure) For all $a, b \in G$ we have $a \cdot b \in G$

(Associativity) For all $a, b, c \in G$ we have $a \cdot (b \cdot c) = (a \cdot b) \cdot c$

(Identity element) $\exists e \in G$ s.t. $\forall a \in G$ we have $a \cdot e = e \cdot a = a$

(Divisibility) For all $a, b \in G$ s.t. $b \geq a$, $\exists c$ s.t. $a \cdot c = b$

First question: I made up the name "Positive group". Is there a more standard name for this?

Second question: We can add the following axiom to get something we might term Strictly Positive Group:

(Non-invertability) For any $a\neq e \in G$, $\nexists b$ s.t. $a \cdot b = e$

Is there a standard name for what I am calling a "Strictly positive group".

The basic idea here is that we want to modify the definition of a group so that combining two elements always leads to a larger elements.

$(\mathbb{Z}, +)$ is a group. $(\mathbb{N},+)$ is a strictly positive group.

$(\mathbb{R_+}, *)$ is a group. $( \{ x\in\mathbb{R}|x\geq1 \},* )$ is a strictly positive group.

Third question: Suppose we have a set $G$ with an operator $\cdot$. Let $X$ be some subset of $G$. Let $-X$ denote the set of all inverses of the elements of $X$.

Conjecture: If $X \cup -X$ is a group, then $X$ is a positive group and $X \setminus -X$ is a strictly positive group.

Is this a known result?

exk
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  • You need to modify your conjecture, since at the moment you haven't ruled out just taking $X=G$. – Greg Martin Feb 13 '14 at 21:40
  • @GregMartin Why would I need to rule that out? If $X$ is itself a group then $X \cup -X = X$ is a positive group and $X \setminus -X$ is a strictly positive group. No? – exk Feb 14 '14 at 01:22
  • Well, I guess $X\setminus-X=\emptyset$ wouldn't satisfy the Identity Element axiom (in fact, $X\setminus-X$ will never contain the identity element). But even more seriously: as a subset of $\Bbb Z$, take $X$ to be ${0,1}$ and all the primes and all the negatives of composite numbers. Then $X\cup-X=\Bbb Z$ but neither $X$ nor $-X$ is closed under addition, hence note a positive group. – Greg Martin Feb 14 '14 at 01:52
  • @GregMartin I see what you mean. My notation was mistaken. What I had in mind is that for each pair $(b, b')$ s.t. $b \cdot b' = e$ you take out one of the two elements. – exk Feb 14 '14 at 15:00

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You should probably take a look at ordered groups and ordered semigroups and keep an eye out for the word "cone," if you haven't already.

The two examples you gave are canonical examples of ordered semigroups, and they are the cones of elements greater than the identity in their respective ordered groups.

If you take any ordered group, I believe the cone of elements equal and greater to the identity is an example of your "strictly positive group."

I'm can't immediately see why these definitions would be totally subsumed by this class of examples, so I wouldn't be surprised if you found an example other than what I've described.

rschwieb
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  • Thanks! Those were very helpful search terms. What I was calling a strictly positive group seems closely related to the notion of a positive cone of an unperforated ordered group. I suspect that the divisibility axiom is related (or equivalent) to the group being unperforated. – exk Feb 14 '14 at 16:07