Q on associative binary operation

[John Washburn]

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 at hotmail.com] 
Sent: Thursday, August 28, 2003 2:36 PM
To: timcmay at got.net; cypherpunks at 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 at got.net>
>To: cypherpunks at 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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