Rich writes: << I've been exploring the binary operator *, which satisifies the rule A*B = B*(A*A). I like to write it in dot notation as AB = B.AA .
(Groups won't help since any group satisfying AB = B.AA is clearly trivial.) << There are 85196 non-isomorphic tables of 5 elements that satisfy the rule. (Roughly 5^10/5!). Empirically, and surprisingly (to me at least), all 85196 tables are commutative. Ditto for smaller tables. . . . . . . (Maybe because it's only true in the finite case?)
I just learned the word "magma" for a set S together with a binary operation *. Call this magma (S,*). QUESTION for Rich: Or are you actually assuming |S| <= 5 Or, as seems more likely, are you considering all magmas (S,*) that are *generated* by <= 5 elements of S ? --> If the latter, any idea how to tell which tables yield a finite S vs. an infinite S ? ---------------------------------------------------------------------------------- In particular there is a "free magma" FM(S) on any set of generators S. (Cf. Wikipedia, "Magma".) This is of course all finite strings of elements of S with all possible parenthesizations (or in Rich's notation, all periodizations), and A*B = A.B is just concatenation. Then there should exist a "quotient magma" which is just FM(S) with specified identifications, such as M = FM({x,y}) / AB ~ B.AA If this M = FM({x,y}) / AB ~ B.AA turns out to be commutative, then it should be true that any magma with the axiom AB = B.AA is commutative. QUESTION: Is this M commutative? --Dan