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(1) I am looking for an example of a u.p (unique product) group which is not right orderable (RO).

Almost any group I pick up (obviously torsion-free, as u.p. group cannot have nontrivial torsion elements) turns out to be RO, whether it be easy groups like integers, modulo $n$ etc, or groups on matrices, or free groups, etc.
An example of a torsion-free non-RO group can be found at this question.

A group $G$ is said to be u.p., if given any two finite nonempty subsets $A$ and $B$ of $G$, there exist at least one element $x$ which has a unique representation $x=ab$ where $a \in A$ and $b \in B$.

Similarly, a two unique product group (t.u.p. group) is a group $G$ such that if given any two finite nonempty subsets $A$ and $B$ of $G$ with $|A|+|B| > 2$, there exist at least two elements $x$ and $y$ which have unique representations $x=ab$ and $y=cd$ where $a,c \in A$ and $b,d \in B$.

(2) Is there a u.p. group which is not t.u.p.? (clearly, every t.u.p. group is a u.p. group).

As every RO-group is a t.u.p. group, which in turn is u.p. group, (1) was a natural question that came to my mind.

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    $\begingroup$ The problem is open. See e.g. Carter, William. New examples of torsion-free non-unique product groups. J. Group Theory 17 (2014), no. 3, 445--464. $\endgroup$
    – Alireza Abdollahi
    Commented May 30, 2015 at 8:25
  • $\begingroup$ What is some example of torsion free non RO group. That must be there somewhere. $\endgroup$
    – Bhaskar Vashishth
    Commented May 30, 2015 at 8:28
  • $\begingroup$ See the answer of Ian Agol to a question of mine. mathoverflow.net/questions/200080/… $\endgroup$
    – Alireza Abdollahi
    Commented May 30, 2015 at 8:30
  • $\begingroup$ Examples of torsion-free non-RO groups appeared as an answer to this older question. $\endgroup$
    – YCor
    Commented Feb 15, 2020 at 21:26
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    $\begingroup$ This post has two numbered emphasized questions: (1) which is a duplicate, and (2) which makes one answer (reduced to a reference). Finally the third question is the one suggested by the title, and then is hidden after the second emphasized question, and has an answer which is the accepted answer. The post should be rewritten so as to emphasize this latter question (and remove the first question, replacing it with a reference, maybe to the older question). $\endgroup$
    – YCor
    Commented Feb 15, 2020 at 21:30

2 Answers 2

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Such a group has been found by N. Dunfield, see the appendix to the paper

  • Steffen Kionke, Jean Raimbault, Nathan Dunfield, On geometric aspects of diffuse groups, Documenta Mathematica, Vol. 21 (2016), 873-915, journal, arXiv:1411.6449.

The group is the fundamental group of a compact hyperbolic three--manifold which has injectivity radius large enough so that it is known to have unique products (and a little more) by a result of Delzant--Bowditch, but Nathan checked "by hand" that it is not left-orderable (by the same method as in his Inventiones paper with D. Calegari,

which you should check out if you want more examples of non-left/right-orderable groups).

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Every U.P. group is a t.u.p group. See

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  • $\begingroup$ I cannot access it right now as I am at home, but I have asked a friend to send it to me, but I have strong belief that it cant be. Otherwise why distinction b\w u.p and t.u.p. t.u.p must be strictly stronger than u.p $\endgroup$
    – Bhaskar Vashishth
    Commented May 30, 2015 at 9:09
  • $\begingroup$ Also for t.u.p groups, KG always have trivial units for any field K, but for u.p., only when char$K=0$ $\endgroup$
    – Bhaskar Vashishth
    Commented May 30, 2015 at 9:11
  • $\begingroup$ Thanks. Now I believe it. It came in 1980, I am reading Passman, which was written before that, so may be thats why. Now we need not distinguish them. $\endgroup$
    – Bhaskar Vashishth
    Commented May 30, 2015 at 9:55
  • $\begingroup$ Sorry to ask but is there anyway to access the artcile of Andrzej Strojnowski for free ? I would really like to read this proof. $\endgroup$
    – Pierre21
    Commented Mar 11, 2019 at 15:51

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