Given ordinal $\alpha \geq \omega_1$ does $\alpha \omega_1 = \omega_1 \alpha$

Question

As in the question, I am asked whether or not ordinal multiplication is commutative for an ordinal greater that $\omega_1$ and $\omega_1$.

My reasoning was that it shouldn't:

Take $\alpha = \omega_1 ^{\omega}$, then we have $\omega_1 \alpha = \omega_1 ^{1+\omega} = \omega_1 ^{\omega} = \alpha$

But $\alpha\omega_1 = \omega_1 ^{\omega+1}$

However, I'm not very familiar with how $\omega_1$ might behave. Is it really true here that $\omega_1^{\omega} < \omega_1^{\omega + 1}$ or is there some property about the uncountability of $\omega_1$ that actually allows for equality here?

Answer 1

We have $\omega_1^{\omega + 1} = \omega_1^\omega \omega_1$ by definition of exponentiation, and $\omega_1 > 1$ clearly. Multiplication (and addition) on the right preserves strict equality, so $\omega_1^{\omega} \omega_1 > \omega_1^\omega$.

Answer 2

As B. Mehta observes, you have indeed found a counterexample to the claim. However, there is an arguably simpler counterexample: $$\omega_1(\omega_1+1)=\omega_1^2+\omega_1\quad\mbox{but}\quad (\omega_1+1)\omega_1=\omega_1^2.$$