[math-fun] AB = B.AA seems to imply commutativity?
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 . There are 85196 non-isomorphic tables of 5 elements that satisfy the rule. (Roughly 5^10/5!). It's clear enough that an idempotent table (satisfying the rule AA=A) will mean that AB = B.AA can be morphed to AB = BA . And it's also clear that any table which is both idempotent and commutative will also satisfy AB = B.AA . However, there are lots of tables that are not fully idempotent. Empirically, and surprisingly (to me at least), all 85196 tables are commutative. Ditto for smaller tables. And also, apparently there's always at least one idempotent element. Anyway, if it can be proven that all *-tables are commutative, and that the algebraic relation AB=B.AA does not algebraically imply AB=BA, this would provide a fine example of a truth about the system that can't be proven within the system. (Maybe because it's only true in the finite case?) Rich
On Tue, 14 Feb 2006, Schroeppel, Richard wrote:
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 .
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Empirically, and surprisingly (to me at least), all 85196 tables are commutative. Ditto for smaller tables. And also, apparently there's always at least one idempotent element.
In fact, this axiom implies commutativity, since we have ab = b(aa) = (aa)(bb) = (bb)[(aa)(aa)] = (bb)[a(aa)] = (bb)(aa) = a(bb) = ba. Therefore, we also have ab = (aa)b, so, while not everything has to be idempotent, the multiplication action of every element is the same as that of its square. Of course, without associativity, this does not mean that the multiplication action is idempotent! David
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David P. Moulton -
Schroeppel, Richard