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Math Help - u.p. = t.u.p. for groups

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    u.p. = t.u.p. for groups

    It says in the sticky that I need to send a 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.

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

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

    Show that a group 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 K be an arbitrary field. If G is a u.p.-group, then K[G] has no proper divisors of zero. If G is a t.u.p.-group, then 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
    Last edited by CaptainBlack; December 24th 2011 at 11:14 PM.
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  2. #2
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    Re: u.p. = t.u.p. for groups

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

    Suppose first G is t.u.p. Let's take two non-empty finite subsets A,B\subseteq G. If |A|+|B|>2 we have more than we need. If |A|+|B|=1, then A=B=\{x\}, so AB=\{x^2\} and the only element of this set decomposes uniquely. If |A|+|B|=2, then it must be that A=\{x\},\,B=\{y\}, with x\neq y. Therefore AB=\{xy\} and the only element of this decomposes uniquely.

    To prove the opposite implication, suppose that G is u.p. and not t.u.p. Then there are two non-empty, finite subsets A,B\subseteq G, such that |A|+|B|>2 and there exists exactly one element ab\in AB,\,a\in A,\,b\in B, such that the product representation is unique.

    Let C=a^{-1}A and D=Bb^{-1}. We will show that in CD only 1\cdot 1 has a unique representation.

    1=a^{-1}a\in C and 1=bb^{-1}\in D. Now, if xy=1,\, x\in C,\,y\in D, then x=a^{-1}a_1 for some a_1\in A and y=b_1b^{-1} for some b_1\in B, so we have

    a^{-1}a_1b_1b^{-1}=1, that is
    a_1b_1=ab,

    and therefore a_1=a and b_1=b, whence x=1=y. Therefore 1\cdot 1 is represented uniquely as a product of elements of C and D. It is the only such element in CD. Indeed, suppose we have c=a^{-1}a_1\in C and d=b_1b^{-1}\in D for some a_1\in A and b_1\in B, such that cd=a^{-1}a_1b_1b^{-1} is represented uniquely as a product of elements of C and D. Suppose

    a_1b_1=a_2b_2,

    for some a_2\in A and b_2\in B. Then cd=a^{-1}a_2b_2b^{-1}, and so we have, from the uniqueness of c and d,

    a^{-1}a_1 = a^{-1}a_2 and b_1b^{-1}=b_2b^{-1}, that is
    a_1=a_2 and b_1=b_2.

    From this we see that the product a_1b_1 is uniquely represented, so we must have

    a_1=a and b_1=b.

    Therefore, c=1=d as we wanted. We have proven that 1\cdot 1 is the only element in CD with a unique representation.

    Now, let E=D^{-1}C and F=DC^{-1}. E and F are obviously non-empty finite sets. We will show that EF has no uniquely represented elements, which will be a contradiction with the assumption that G is u.p. The contradiction will end the proof.



    Let (d_1^{-1}c_1)(d_2c_2^{-1})\in EF, for some d_1,d_2\in D and c_1,c_2\in C. We will consider three cases.

    Case 1. If c_1\neq 1 or d_2\neq 1, then c_1d_2 is not uniquely represented in CD. Therefore also (d_1^{-1}c_1)(d_2c_2^{-1}) is not uniquely represented in EF.

    Case 2. If c_2\neq 1 or  d_1\neq 1, then  c_2d_1 isn't uniquely represented in CD. Therefore, we have elements c_3\in C and d_3\in D such that c_2\neq c_3 and c_2d_1=c_3d_3. Then we have d_1^{-1}c_2^{-1}=d_3^{-1}c_3^{-1}, and so the element (d_1^{-1}\cdot 1)(1\cdot c_2^{-1}) has another representation (d_3^{-1}\cdot 1)(1\cdot c_3^{-1}).

    Case 3. If c_1=c_2=d_1=d_2=1, then we can take an element 1\neq c. We will find such an element in C or in D, because |C|+|D|=|A|+|B|>2. Say c\in C. Then we have (1\cdot 1)(1\cdot 1) = (1\cdot c)(1\cdot c^{-1}), another representation in EF.

    This ends the proof.
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