# Set closed under under multiplication

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

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

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

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

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

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

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

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

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