Let $A = \{ \{0\}, \{0,1\} \}$. Let $\bar{A}$ be the family of sets generated by the Cartesian product on $A$.
This is a magma $(\bar{A}, \times)$ that has what I am calling a "double cancellative" operation, because if $A \times B = C \times D$, then $A = C$ and $B = D$.
Is there a good, deep reason why an operation that is "double cancellative" seems to break every single property we are usually interested in?
If we ask that a magma $(M,*)$ be associative with double cancellation, it's the trivial group. If we ask for commutativity, it's the trivial group. If we ask that it have inverses, it's the trivial group. If we even ask for an identity, it's the trivial group. If we ask for a quasigroup, it's the trivial group, if we ask the operation to be medial, it's the trivial group, left or right semimedial, trivial group, distributive, it's the trivial group. I can't seem to ask anything of a double-cancellative operation without destroying everything!
But, it's not like this is such a strange idea, since Cartesian product is a common operation. Why is this property so degenerative on an algebraic structure?
My intuition hints that double-cancellation is a very "restrictive" property (we're exactly saying that for any $A$, $B$, $A * B$ can't be represented as any other product of two elements), and so any other property being placed innately on the operation is either already implied by double-cancellation or contrary to it. Is there a way to show this?