The term is non-Abelian. You are deeply confusing the concept of operation and relationship. Normally this is not a problem. But, if you are going to use the terms associative and transitive, then you must keep the 2 concepts distinct. Group theory 101. First, relationships. A relationship ~ on the members {a,b,c} of a (possibly infinite) set S is: Reflexive: if a~a for all a in S Symmetric: if a~b, is true then b~a is true for all a and b in S Transitive: if a~b and b~c are true then a~c is true for all a,b,c in S An equivalence relationship is a relationship which is Reflexive, Symmetric and Transitive. You are denoting an equivalence relationship with the = symbol. Second, operations/operators. An operation (maps/combines/mixes) 1 or more elements of a (possibly infinite) set S to a single element of S. The most common types of operations are: Unitary: one element mapped to 1 element. E.g. the negation, conjugation or exponentiation operations. Binary: two elements mapped to 1 element. E.g. addition, multiplication, subtraction, division, convolution operators. More than 2: At this point the operator is generally called a mapping. A binary, operation, *, can have none, some or all of the following properties: Closure: For all a and b in S the a*b is also in S Associative: For all a, b, and c in S the (a*b)*c is equivalent to a*(b*c). Note the dependence on a relationship to define this property of an operator. Commutative: For all a and b in S, (a*b) is equivalent to (b*a). Note again, the dependence on a relationship to define this property of an operator. Groups require only 5 things: A set of at least one element. (The 1-element group is the trivial group) A single, binary operation which operates on the elements of the set. The binary operation is closed. The binary operation is associative. There exists at least one element, e, of the set (an identity element) for the operation such that for all a in S e*a=a*e=a. Where = is some equivalence relation); If the group, by some chance, has an operator which is also commutative, the group is Abelian. If the operation is not commutative, then the group is non-Abelian. Add an additional commutative, operation (and identity element) and the group becomes a ring. If the second operation happens to be invertible the ring, becomes a field. See: http://mathworld.wolfram.com/Group.html http://mathworld.wolfram.com/Ring.html http://mathworld.wolfram.com/FieldAxioms.html http://mathworld.wolfram.com/AbelianGroup.html for more details. -----Original Message----- From: Tyler Durden [mailto:camera_lumina@hotmail.com] Sent: Thursday, August 28, 2003 2:36 PM To: timcmay@got.net; cypherpunks@minder.net Subject: Re: Q on associative binary operation Yeah, kinda bizarre. There's also an ambiguity that prevents one from saying Q is associative. Is the table defined for both directions of *? In other words, is the table meant to imply values for both x*y (ie, left*top) as well as y*x (top*left)? For most objects x*y will not equal y*x (indeed, one may be undefined as in the case of matrix multiplication). Anyone remember the group theoretic term for these kinds of groups? -TD

From: Tim May <timcmay@got.net> To: cypherpunks@lne.com Subject: Re: Q on associative binary operation Date: Thu, 28 Aug 2003 10:41:51 -0700

On Thursday, August 28, 2003, at 12:14 AM, Sarad AV wrote:

...

hi,

Table shown is completed to define 'associative' binary operation * on S={a,b,c,d}.

*|a|b|c|d --------- a|a|b|c|d --------- b|b|a|c|d --------- c|c|d|c|d --------- d|d|c|c|d

The operation * is associative iff (a*b)*c=a*(b*c) for all a,b,c element of set S.

So can (a*d)*d=a*(d*d)=d considered as associative over * for this case as per definition?

This looks like a homework assignment for a class.

If a, b, c, and d are all arbitrary symbols, most substitutions (such as a = 2, b = 5, etc.) certainly will _not_ give (a*d)*d=a*(d*d)=d as a true

statement. Only special values of a and d will make that true. (For example, a = 1 and d = 1 makes (1*1)*1=1*(1*1)=1. Other values may as well. Finding them is up to you.

Lastly, your English is again unclear. "So can (a*d)*d=a*(d*d)=d considered as associative over * for this case as per definition?" isn't a proper English sentence.

Why do you keep posing these problems to the list? Are they homework problems? Do you think the list is just too quiet and needs ill=phrased

puzzlers to keep it occupied?

Did you ever do the coin flip experiment we suggested you do?

Are you an AI which has failed the Turing Test and escaped?

--Tim May

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