It says in the sticky that I need tosenda PM with the solution to the moderators, and not that I should wait for the answer. I hope it's OK that I'm posting now. I'm very excited and I just can't wait.

It's a challenge problem, not a question. I will post the solution in a week if no one replies.

Au.p.-group(unique-product group) is a group such that for every two non-empty, finite subsets , the product set contains at least one element , with and , such that for any and if then and

At.u.p.-groupis a group such that for every two non-empty, finite subsets , such that , the product set contains at least two elements and , with and , which enjoy the same uniqueness property as in the previous definition.

Show that a group is u.p. iff it's t.u.p.

The motivation of the problem is that there are many results concerning group algebras made of u.p.- or t.u.p.-groups (also semigroup algebras with u.p.- or t.u.p-semigroups, which are defined identically). For example, here's a theorem from Donald Passman's book,Infinite Group Rings(1971):

Let be an arbitrary field. If is a u.p.-group, then has no proper divisors of zero. If is a t.u.p.-group, then has only trivial units.

1. You are supposed to await approval before posting.

2. I'm not sure this really quallifies, but I will wait and see.

CB