TITLE: Commutativity of Ordinal Multiplication
QUESTION [12 upvotes]: I was recently asked whether the following two statements about ordinal multiplication were true: 

$\alpha \beta = \beta \alpha \Rightarrow \alpha ^2 \beta^2 = \beta^2 \alpha^2$
$\alpha^2 \beta^2 = \beta^2 \alpha^2 \Rightarrow \alpha \beta = \beta \alpha$

For 1): 
$\alpha \alpha \beta \beta = \alpha \beta \alpha \beta = \beta \alpha \beta \alpha = \beta \beta \alpha \alpha$, so 1 is true. 
However, I can't quite seem to conclude anything for 2). I am given the hint to assume associativity, but I can't quite see how that is helpful. I can't even decide if I think it's true or not. I can't think of a counter example but I'm so used to the asymmetry of ordinal arithmetic that I wouldn't expect this to be true.

REPLY [6 votes]: I might have overlooked something, but it seems that one only needs four simple properties of ordinals ($\alpha$, $\beta$ and $\gamma$ will denote ordinals):

Comparison trichotomy: $(\alpha<\beta)$ or $(\alpha=\beta)$ or $(\alpha>\beta)$.
Multiplication associativity: $\alpha(\beta\gamma)=(\alpha\beta)\gamma$
If $\alpha\leq\beta$, we have $\alpha\gamma \leq \beta\gamma$ and $\gamma\alpha\leq \gamma\beta$ for any $\gamma$.
If $\alpha<\beta$ and $0<\gamma$, we have $\gamma\alpha<\gamma\beta$.

Let's start by assuming $\alpha\beta<\beta\alpha$ and see what we can conclude about relation between $\alpha^2\beta^2$ and $\beta^2\alpha^2$.
Multiplying the assumed inequality $\alpha\beta<\beta\alpha$ by $\alpha$ on the left turns it into a non-strict one and a subsequent multiplication by $\beta$ on the right keeps it so. Therefore, we have
$$\alpha^2\beta^2=\alpha\alpha\beta\beta=\left(\alpha(\alpha\beta)\right)\beta \leq \left(\alpha(\beta\alpha)\right)\beta=\alpha\beta\alpha\beta$$
Another application of the third property yields
$$\alpha\beta\alpha\beta=(\alpha\beta)(\alpha\beta)\leq (\beta\alpha)(\alpha\beta)$$
Now we can use the fourth property to get
$$(\beta\alpha)(\alpha\beta)<(\beta\alpha)(\beta\alpha)=\beta\alpha\beta\alpha$$
(note that $\beta\alpha$ cannot be zero due to being strictly greater than $\alpha\beta$)
The last step follows the same reasoning as the first one and gives us
$$\beta\alpha\beta\alpha=(\beta(\alpha\beta))\alpha\leq (\beta(\beta\alpha))\alpha=\beta\beta\alpha\alpha=\beta^2\alpha^2$$
Putting it all together, we have
$$\alpha\beta <\beta\alpha \implies \alpha^2\beta^2<\beta^2\alpha^2$$
Since the same kind of reasoning applies if we swap $\alpha$ and $\beta$, we can use the first property to finish the proof: $\alpha^2\beta^2=\beta^2\alpha^2$ can only be true if $\alpha\beta=\beta\alpha$.