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?

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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.