If $\alpha$ and $\beta$ are cardinal numbers such that $0\neq \beta \leq \alpha$ and $\alpha$ is infinite, then $\alpha\beta=\alpha.$
Sketch: Let $A$ be an infinite set with $|A|=\alpha$ and let $\mathcal{F}$ be the set of all functions $f:X\times X\rightarrow X,$ where $X$ is an infinite subset of $A$. We can show that $\mathcal{F}\neq \emptyset.$ Partially order $\mathcal{F}$ by extension and use Zorn's Lemma to obtain a maximal element $g:B\times B\rightarrow B.$ To complete the proof we shall show that $|A|=|B|=\alpha.$
Suppose $|A-B|>|B|.$ Then there is a subset $C$ of $A-B$ such that $|C|=|B|.$ Verify that $|C|=|B|=|B\times B|=|B\times C|=|C\times B|=|C\times C|$ and that these sets are mutually disjoint. Consequently we have
$|(B\cup C)\times (B\cup C)|=|B\times B|+ |B\times C|+ |C\times B|+ |C\times C|=(|B|+|B|)+(|C|+|C|)=|B|+|C|=|B\cup C|.$
Then there is a bijection $h: (B\cup C)\times (B\cup C)\rightarrow B\cup C.$ He says that contradicts the maximality of $g$ in $\mathcal{F}.$
Why? What is the relation between $h$ and $g$?
I know that $B$ is a subset of $B\cup C$, but I should show that $h|B=g.$ Right?