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Let G be a group. We say that G satisfies the unique product (UP) property, if for every pair of non-empty subsets $$A,B \subset G$$, there exists an element $$c \in AB := \{ ab \; | \; a \in A, b \in B \}$$ such that c can be written in a unique way, i.e. for a unique $$a \in A$$ and a unique $$b \in B$$.

Note that the definition does not require every product to be unique, merely that at least one product is unique.

There exists a superficial variation on the UP property: The two unique products (TUP) property. We say that G satisfies the TUP property, if for every pair of subsets $$A,B \subset G$$, one of which contains at least two elements, then there exist two elements $$c_1, c_2 \in AB$$ such that c1 and c2 can be written in unique ways.

Obviously, any group that satisfies the TUP property automatically satisfies the UP property. However, perhaps more surprisingly, the converse is also true: If a group is UP then it is TUP. This result was due to Strojnowski in 1980.

Unique product groups were first considered as an obvious way to attack Kaplansky's Zero Divisor Conjecture. UP groups can easily be seen to satisfy the zero divisor conjecture, although most authors omit a proof or citation of a proof when remarking on this fact. A proof of this folklore result is provided below:

Proposition: Suppose a torsion-free group G satisfies the UP property. Then G satisfies the zero divisor conjecture.

The proof is by contradiction. Suppose that G has the UP property, but that the group ring $$\mathbb{C} G$$ contains a proper zero divisor. Then there exist $$x,y \in \mathbb{C} G$$, with $$x,y \neq 0$$ such that $$xy= 0$$.

Since $$x,y \neq 0$$, we must have that $$|\text{supp} (x)|, |\text{supp}(y)| \neq 0$$, and, in particular, $$\text{supp}(x), \text{supp}(y) \neq \emptyset$$. We now set $$A:= \text{supp}(x)$$ and $$B:= \text{supp}(y)$$. Since G is a UP group, there exists $$c \in AB$$ such that $$c= ab$$ for a unique $$a \in A$$ and $$b \in B$$.

By definition of group ring multiplication, $$c \in \text{supp} (xy) = AB$$. However, $$xy=0$$, so $$\text{supp}(xy) = \emptyset$$. Therefore, the coefficient of c - $$r_c = r_a r_b$$ (where $$r_a, r_b$$ are the coefficients corresponding to a in x and b in y) - must equal zero. Without loss of generality, assume that $$r_a = 0$$. But then $$a \notin \text{supp}(x)$$, and so $$a \notin A$$. This results in the set AB containing no unique product, which is a contradiction.

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