Re: u.p. = t.u.p. for groups

OK, it wasn't a good idea apparently, but here's the solution.

Suppose first is t.u.p. Let's take two non-empty finite subsets If we have more than we need. If then , so and the only element of this set decomposes uniquely. If then it must be that with Therefore and the only element of this decomposes uniquely.

To prove the opposite implication, suppose that is u.p. and not t.u.p. Then there are two non-empty, finite subsets such that and there exists exactly one element such that the product representation is unique.

Let and We will show that in only has a unique representation.

and Now, if then for some and for some so we have

that is

and therefore and whence Therefore is represented uniquely as a product of elements of and It is the only such element in Indeed, suppose we have and for some and such that is represented uniquely as a product of elements of and Suppose

for some and Then and so we have, from the uniqueness of and

and that is

and

From this we see that the product is uniquely represented, so we must have

and

Therefore, as we wanted. We have proven that is the only element in with a unique representation.

Now, let and and are obviously non-empty finite sets. We will show that has no uniquely represented elements, which will be a contradiction with the assumption that is u.p. The contradiction will end the proof.

Let for some and We will consider three cases.

Case 1. If or then is not uniquely represented in . Therefore also is not uniquely represented in

Case 2. If or then isn't uniquely represented in Therefore, we have elements and such that and Then we have and so the element has another representation

Case 3. If then we can take an element We will find such an element in or in , because Say Then we have another representation in

This ends the proof.