Commutativity of Ordinal Multiplication

Statement (2) is true, though I don't know any way to prove it without intricate analysis of Cantor normal forms. The rough idea is that it is extremely rare for ordinal multiplication to commute, so $\alpha^2\beta^2=\beta^2\alpha^2$ forces $\alpha$ and $\beta$ to have a certain highly restricted form that implies $\alpha\beta=\beta\alpha$ as well.

Let me start with a version of (2) for addition.

Lemma: Let $\alpha$ and $\beta$ be ordinals such that $\alpha\cdot 2+\beta\cdot 2=\beta\cdot 2+\alpha\cdot 2$. Then $\alpha+\beta=\beta+\alpha$.

Proof: If $\alpha$ or $\beta$ is finite this is easy (a nonzero finite ordinal commutes only with other finite ordinals), so we assume they are both infinite. Write them in Cantor normal form as $$\alpha=\omega^{\alpha_n}\cdot a_n+\cdots+\omega^{\alpha_1}\cdot a_1+a_0$$ and $$\beta=\omega^{\beta_m}\cdot b_m+\cdots+\omega^{\beta_1}\cdot b_1+b_0.$$ We then have Cantor normal forms $$\alpha\cdot 2=\omega^{\alpha_n}\cdot (2a_n)+\cdots+\omega^{\alpha_1}\cdot a_1+a_0$$ and $$\beta\cdot 2=\omega^{\beta_m}\cdot (2b_m)+\cdots+\omega^{\beta_1}\cdot b_1+b_0.$$ By the criterion for commutativity of addition in this answer, $\alpha\cdot 2+\beta\cdot 2=\beta\cdot 2+\alpha\cdot 2$ implies that $m=n$, $\alpha_k=\beta_k$ for all $k$, and $a_k=b_k$ for $k<n$. But then using that criterion in the other direction (or just directly adding the Cantor normal forms), we conclude that $\alpha+\beta=\beta+\alpha$.

Now we prove statement (2).

Theorem: Let $\alpha$ and $\beta$ be ordinals such that $\alpha^2\beta^2=\beta^2\alpha^2$. Then $\alpha\beta=\beta\alpha$.

Proof: If $\alpha$ is finite, it is easy to see that $\alpha^2\beta^2=\beta^2\alpha^2$ iff either $\beta$ is finite or $\alpha$ is $0$ or $1$, in which case we have $\alpha\beta=\beta\alpha$. So we assume $\alpha$ is infinite, and in the same way we may assume $\beta$ is infinite.

Write $\alpha$ and $\beta$ in Cantor normal form as $$\alpha=\omega^{\alpha_n}\cdot a_n+\cdots+\omega^{\alpha_1}\cdot a_1+a_0$$ and $$\beta=\omega^{\beta_m}\cdot b_m+\cdots+\omega^{\beta_1}\cdot b_1+b_0.$$ Then we have the Cantor normal form $$\alpha^2=\omega^{\alpha_n+\alpha_n}\cdot a_n+\dots+\omega^{\alpha_n+\alpha_1}\cdot a_1 + \omega^{\alpha_n}\cdot a_na_0+\omega^{\alpha_{n-1}}\cdot a_{n-1}+\dots+\omega^{\alpha_1}\cdot a_1+a_0$$ if $\alpha$ is a successor (i.e., $a_0>0$) and $$\alpha^2=\omega^{\alpha_n+\alpha_n}\cdot a_n+\dots+\omega^{\alpha_n+\alpha_1}\cdot a_1$$ if $\alpha$ is a limit (i.e., $a_0=0$), and similarly for $\beta^2$.

If $\alpha$ and $\beta$ are both successor ordinals, then by my answer here there exists an ordinal $\gamma$ and natural numbers $i$ and $j$ such that $\alpha^2=\gamma^i$ and $\beta^2=\gamma^j$. If $i$ and $j$ are even we get $\alpha=\gamma^{i/2}$ and $\beta=\gamma^{j/2}$ (we can recover $\alpha$ from the formula for $\alpha^2$ above so $\alpha$ is the unique square root of $\alpha^2$) and so $\alpha\beta=\beta\alpha$. To handle the odd case, note that a successor ordinal whose Cantor normal form has $N+1$ terms can be written (uniquely) as an $i$th power iff its Cantor normal form is "$N/i$-periodic" in the sense that $N=di$ for some $d$, with term $dx+y$ from the right (starting from $0$) having the form $\omega^{\gamma_d\cdot x+\gamma_y}\cdot c_y$ (except that when $y=0$, the coefficient is instead $c_{N}c_0$ for $0<x<i$ and $c_{N}$ for $x=i$). Indeed, this is exactly what you get from raising $\omega^{\gamma_d}\cdot c_N+\omega^{\gamma_{d-1}}\cdot c_{d-1}+\dots+c_0$ to the $i$th power. It follows from this description that if a successor ordinal is both an $i$th power and an $i'$th power, it must also be an $\operatorname{lcm}(i,i')$th power (since the Cantor normal form will be "$\gcd(N/i,N/i')$-periodic", by an argument similar to the proof in the answer linked above). So if $i$ is odd, $\alpha^2=\gamma^i$ implies that $\gamma^i$ is actually a $2i$th power and so $\gamma$ has a square root $\delta$. We then see that $\alpha=\delta^i$ and $\beta=\delta^j$ and so again $\alpha\beta=\beta\alpha$.

If one if $\alpha$ and $\beta$ is a successor and the other is a limit, let us say $\alpha$ is a successor. Then the Cantor normal form for $\alpha^2$ has $2n+1$ terms including a final $a_0$ term and the Cantor normal form for $\beta^2$ has $m$ terms. When we multiply $\alpha^2\beta^2$, we get one term for each term of $\beta^2$, for just $m$ terms. When we multiply $\beta^2\alpha^2$, we get one term for each term of $\alpha^2$ before the last one, and then also $m$ more terms from $\beta^2\cdot a_0$, for a total of $2n+m$ terms. Since we are assuming $\alpha$ is infinite, $n>0$, so $2n+m\neq m$ and we cannot have $\alpha^2\beta^2=\beta^2\alpha^2$.

Finally, if $\alpha$ and $\beta$ are both limit ordinals, let us see what we can conclude from $\alpha^2\beta^2=\beta^2\alpha^2$ using the criterion for commutativity of multiplication in this answer (note that the criterion there is necessary only for limit ordinals; see my comments there). We find that $m=n$, and for each $k$ from $1$ to $n$, $a_k=b_k$ and $$\alpha_n+\alpha_n+\beta_n+\beta_k=\beta_n+\beta_n+\alpha_n+\alpha_k.$$ In the case $k=n$, this tells us $\alpha_n+\beta_n=\beta_n+\alpha_n$ by the Lemma. So we can rearrange the terms above to get $$\alpha_n+\beta_n+\alpha_n+\beta_k=\alpha_n+\beta_n+\beta_n+\alpha_k.$$ Since addition is left-cancellative, this implies $\alpha_n+\beta_k=\beta_n+\alpha_k$. Using the commutativity criterion in the other direction, we conclude that $\alpha\beta=\beta\alpha$.

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 $\alpha\beta=\beta\alpha$.