Marc asks:

<<
Let some set be closed under an associative multiplication operator.

If it commutes then the product of distinct squares is also square: aabb =
abab.

Contrarily, can you construct a non-commuting system where this is usually
false?
>>

The simplest example I can imagine is the set of all finite strings of a's and b's with the operation of concatenation, aka the free semigroup on 2 generators.

--Dan.