Talk:Partially ordered group

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It seems easy to verify that if an ordered group is finite, the order has to be equality; i.e., apart from this trivial case an ordered group is infinite. If we have a source we should add this.--Patrick 07:52, 18 May 2007 (UTC)

Moreover, every element in the positive cone apart from the identity element has infinite order.--Patrick 08:04, 18 May 2007 (UTC)

+ for multiplication?[edit]

Why is + being used for multiplication here? As far as I know, orderability has nothing to do with abelian groups, so this notation is just...wrong.

Unless someone can come up with a good reason I will change this notation to be multiplication, as is standard.

Farpov (talk) 12:57, 5 August 2010 (UTC)

I would like to point out that the if and only if statement in the introduction of the article assumes G to be commutative as well. The correct statement for non-commutative G is: for all a,b in G: a<=b iff (-a*b is in G or b*-a is in G). Not assuming this results in the statement being false for all groups with non trivial centralizor!

I thus completely agree with Farpov. — Preceding unsigned comment added by (talk) 20:29, 23 February 2012 (UTC)

Abelian case[edit]

There's quite a lot of information behind partially ordered abelian groups that I would like to add. Should I add it to this page or create a new page? It may be distracting to have the abelian case here since the nonabelian case is quite different. Minimalrho (talk) 06:42, 7 January 2016 (UTC)

Upon further inspection of this article, I'm concerned about the extent to which the abelian and non-abelian cases are confused. Is perforation a useful concept in the non-abelian case? Do the definitions of Riesz groups and Riesz interpolation property require being abelian or is it standard terminology for non-abelian groups? Minimalrho (talk) 07:06, 12 January 2016 (UTC)