# Set closed under under multiplication

• Jun 20th 2009, 07:12 PM
Bruno J.
Set closed under under multiplication
Hello, this is a nice problem.

Suppose $\displaystyle S$ is a set of real numbers, closed under multiplication.
Suppose $\displaystyle S$ is partitioned into $\displaystyle A, B$, such that both $\displaystyle A,B$ are closed under multiplication of three elements (so that $\displaystyle a_1,a_2,a_3 \in A$ implies $\displaystyle a_1a_2a_3 \in A$, and similarily for $\displaystyle B$).

Show that either $\displaystyle A$ or $\displaystyle B$ is closed under multiplication.
• Jun 20th 2009, 10:51 PM
NonCommAlg
Quote:

Originally Posted by Bruno J.
Hello, this is a nice problem.

Suppose $\displaystyle S$ is a set of real numbers, closed under multiplication.
Suppose $\displaystyle S$ is partitioned into $\displaystyle A, B$, such that both $\displaystyle A,B$ are closed under multiplication of three elements (so that $\displaystyle a_1,a_2,a_3 \in A$ implies $\displaystyle a_1a_2a_3 \in A$, and similarily for $\displaystyle B$).

Show that either $\displaystyle A$ or $\displaystyle B$ is closed under multiplication.

nice but pretty easy! (Nod) $\displaystyle S$ doesn't have to be a subset of real numbers. the result is true for any semigroup $\displaystyle S$:

suppose neither $\displaystyle A$ nor $\displaystyle B$ is closed under multiplication. then, since $\displaystyle S$ is multiplicatively closed, there must exist $\displaystyle a_1,a_2 \in A, \ b_1,b_2 \in B$ such that $\displaystyle a_1a_2 \in B, \ b_1b_2 \in A.$ but then we would get

something impossible: $\displaystyle a_1a_2b_1b_2=(a_1a_2)b_1b_2=a_1a_2(b_1b_2) \in A \cap B = \emptyset. \ \ \Box$
• Jun 21st 2009, 07:43 AM
Bruno J.
Yup that is perfect! (Cool)