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 $\displaystyle G$ such that for every two non-empty, finite subsets $\displaystyle A,\, B \subseteq G$, the product set $\displaystyle AB$ contains at least one element $\displaystyle ab$, with $\displaystyle a \in A$ and $\displaystyle b \in B$, such that for any $\displaystyle a'\in A$ and $\displaystyle b'\in B,$ if $\displaystyle a'b'=ab,$ then $\displaystyle a'=a$ and $\displaystyle b'=b.$

At.u.p.-groupis a group $\displaystyle G$ such that for every two non-empty, finite subsets $\displaystyle A,\, B \subseteq G$, such that $\displaystyle |A|+|B|>2$, the product set $\displaystyle AB$ contains at least two elements $\displaystyle ab$ and $\displaystyle cd$, with $\displaystyle a,c \in A$ and $\displaystyle b,d \in B$, which enjoy the same uniqueness property as in the previous definition.

Show that a group $\displaystyle G$ 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 $\displaystyle K$ be an arbitrary field. If $\displaystyle G$ is a u.p.-group, then $\displaystyle K[G]$ has no proper divisors of zero. If $\displaystyle G$ is a t.u.p.-group, then $\displaystyle K[G]$ 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