 Commuting squares of ordinals

Two ordinals commute under ordinal multiplication if and only if their squares commute. One implication follows from associativity. For the converse, squaring is strictly increasing: $A<B$ implies $A^2\le BA<B^2$. If $\alpha \beta <\beta \alpha$, monotonicity givesa contradiction to commuting squares. Swap the factors to exclude the reverse inequality; zero factors cause no exception.