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

**ymar** · December 24th 2011, 03:48 PM

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

**ymar** · January 2nd 2012, 04:29 PM

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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