Hessenberg: $\kappa \otimes \kappa =\kappa$ for every infinite cardinal $\kappa$, proved in ZF from the canonical well-order of $\kappa \times \kappa$

For an ordinal $\alpha$ define $(\xi ,\eta )⊲({\xi }^{\mathrm{\prime }},{\eta }^{\mathrm{\prime }})$ on $\alpha \times \alpha$ to hold when $\max (\xi ,\eta )\in \max ({\xi }^{\mathrm{\prime }},{\eta }^{\mathrm{\prime }})$, or the two maxima are equal and $\xi \in {\xi }^{\mathrm{\prime }}$, or the two maxima are equal, $\xi ={\xi }^{\mathrm{\prime }}$ and $\eta \in {\eta }^{\mathrm{\prime }}$; this is the lexicographic order on the triple $(\max (\xi ,\eta ),\xi ,\eta )$ of ordinals, hence irreflexive, transitive and trichotomous by [L8], and a nonempty $S\subseteq \alpha \times \alpha$ has a $⊲$-least element obtained by taking in turn the $\in$-least maximum occurring in $S$, then the $\in$-least admissible $\xi$, then the $\in$-least admissible $\eta$, each of which is the least element of a nonempty set of ordinals and so is determined rather than chosen; therefore $⊲$ well-orders $\alpha \times \alpha$.

For $(\xi ,\eta )\in \alpha \times \alpha$ put $\gamma =\max (\xi ,\eta )\cup \{\max (\xi ,\eta )\}$, the successor of the maximum; then every $({\xi }^{\mathrm{\prime }},{\eta }^{\mathrm{\prime }})⊲(\xi ,\eta )$ has $\max ({\xi }^{\mathrm{\prime }},{\eta }^{\mathrm{\prime }})\subseteq \max (\xi ,\eta )\in \gamma$, so ${\xi }^{\mathrm{\prime }},{\eta }^{\mathrm{\prime }}\in \gamma$, and the $⊲$-initial segment of $\alpha \times \alpha$ below $(\xi ,\eta )$ is contained in $\gamma \times \gamma$.

Base value: $\omega \times \omega \approx \omega$ by [L5], so $\mid \omega \times \omega \mid =\mid \omega \mid =\omega$ by [L1] and [L4].

Lower bound: for any ordinal $\mu$ with $0\in \mu$ the map $\xi \mapsto (\xi ,0)$ is an injection $\mu \to \mu \times \mu$, so $\mid \mu \mid \le \mid \mu \times \mu \mid$ by [L3], and $\mid \mu \mid =\mu$ when $\mu$ is a cardinal.

Now let $\mu$ be an infinite cardinal with $\omega \in \mu$, and assume the induction hypothesis that $\mid \nu \times \nu \mid =\nu$ for every infinite cardinal $\nu \in \mu$; for $(\xi ,\eta )\in \mu \times \mu$ and $\gamma$ as in step 1.2 we have $\gamma \in \mu$, because $\mu$ is a limit ordinal by [L4] and $\max (\xi ,\eta )\in \mu$, and moreover $\mid \gamma \times \gamma \mid \in \mu$: if $\gamma \in \omega$ then $\mid \gamma \times \gamma \mid =\gamma \otimes \gamma \in \omega \subseteq \mu$ by [L4], while if $\omega \subseteq \gamma$ then $\nu =\mid \gamma \mid$ satisfies $\omega \le \nu \le \gamma \in \mu$ by [L1] and [L3], so $\nu$ is an infinite cardinal in $\mu$ and $\gamma \times \gamma \approx \nu \times \nu$ by [L1] and [L2], whence $\mid \gamma \times \gamma \mid =\mid \nu \times \nu \mid =\nu \in \mu$.

Under the same hypothesis, the $⊲$-initial segment $I$ of $\mu \times \mu$ below any $(\xi ,\eta )$ has order type in $\mu$: $I\subseteq \gamma \times \gamma$ by step 1.2 and $I$ is well ordered by the restriction of $⊲$ by step 1.1, so $\mid I\mid \le \mid \gamma \times \gamma \mid \in \mu$ by [L3] and step 2.1; and if the order type $\theta$ of $I$ satisfied $\mu \subseteq \theta$ then $\mu ⪯\theta \approx I$ by [L6] and [L9], giving $\mu \le \mid I\mid$ by [L3], which contradicts $\mid I\mid \in \mu$.

Under the same hypothesis, $\mid \mu \times \mu \mid =\mu$: let $\delta$ be the order type of $(\mu \times \mu ,⊲)$ and $g:\delta \to \mu \times \mu$ the inverse of the collapsing isomorphism ([L6]); if $\mu \in \delta$ then $g$ carries the initial segment of $\delta$ below $\mu$, which is $\mu$ itself, onto the $⊲$-initial segment below $g(\mu )$, whose order type would then be $\mu$, contradicting step 3.1; so $\delta \subseteq \mu$ and $\mid \mu \times \mu \mid =\mid \delta \mid \le \delta \le \mu$ by [L1], while step 1.4 gives $\mu \le \mid \mu \times \mu \mid$.

Apply [L7] to the well-order $(\kappa \cup \{\kappa \},\in )$ of [L8] and to $S=\{\mu \in \kappa \cup \{\kappa \}:\mu$ is not an infinite cardinal, or $\mid \mu \times \mu \mid =\mu \}$: a $\mu$ below which everything lies in $S$ is in $S$, trivially if $\mu$ is not an infinite cardinal, by step 1.3 if $\mu =\omega$, and by step 4.1 otherwise; hence $S=\kappa \cup \{\kappa \}$, so $\kappa \in S$ and $\kappa \otimes \kappa =\mid \kappa \times \kappa \mid =\kappa$.