After David's nice little proof below, he writes

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!

Maybe a bit more can be said, or at least said differently. aa=a(aa)=(aa)(aa) So squares of elements are idempotent. Mark -----Original Message----- From: math-fun-bounces+mdtorge=sandia.gov@mailman.xmission.com [mailto:math-fun-bounces+mdtorge=sandia.gov@mailman.xmission.com] On Behalf Of David P. Moulton Sent: Tuesday, February 14, 2006 7:38 AM To: math-fun Subject: Re: [math-fun] AB = B.AA seems to imply commutativity? 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 _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun