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

  1. #1
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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 $\displaystyle G$ 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 $\displaystyle b'=b.$

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

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

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

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

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

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

    $\displaystyle a_1b_1=a_2b_2,$

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

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

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

    $\displaystyle a_1=a$ and $\displaystyle b_1=b.$

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

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



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

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

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

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

    This ends the proof.
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