Conference rusure::math

>> Lets assume we have a set S = { a,b,1 } where 1 is the neutral element.
>> S has one operation defined it is called "o". So S with "o" is forming
>> a non commutative group G = (S,o).

     If you are talking about groups, I suppose you mean that 1
     is the "identity" element.  Your notation suggests that S
     is a three element set, but every three element group must
     be isomorphic to the integers modulo 3, which is commutative.

     The only example that readily comes to mind is for S to be
     an arbitrary group, and for R to be the integers, and for *
     to be "exponentiation".  For example, if a is in S and o is
     the operation on S

          a * 3 --> a o a o a

     What set are the results of a * c, a from S and c from R,
     supposed to be in?

    Ooops,
    
    you are right, but I am too... S and R have both an indefinite
    amount of elements. Sorry about that, so it should read:
    
    	S = { 1,a1,a2,.... } and R = { b1,b2,..... }
    
    The result of the "*" operation is an element of R
    
    	a1 * b1 = bi element R

    The more I look at this the more I figure there is no way to get
    this into a algbraic model without a change in the way the operations
    work, and the sets they are operating on.
        
    	Julian.
    
    
    

If you want to tie together operations between two groups, the concept of 
homomorphism is what you're looking for.

Consider two groups G and H based on the respective underlying sets S and R 
and the respective operations o and *.

A mapping f: S -> R is a group homomorphism from G to H if it preserves the
group structure, i.e.:

(1) maps G's neutral (identity) element into H's neutral element, and
(2) satisfies f(a o b) = f(a) * f(b)

The trick is to apply * to the homomorphic images in H of elements of G,
since * applies within H and not G.

There's _lots_ of material on homomorphisms in the standard texts on
algebra and group theory.  Probably more than enough to satisfy the most
jaded of palates.  Good luck!

David Ellis -- Secure Systems Group -- LTN2-2/C08 -- DTN 226-6784 

    Re: 3.
    
    I know about this, actually the problem I ran into is that I think
    (that's were I am not sure) you cannot use it for an operation which is
    not defined in S alone AND R alone. The other thing to remember
    is that R is not a group it is just a set, it has not even a operation
    defined with it.
    
    The "*" is only defined between both S and R, so if I have
    f: S -> R, f(a o b) = f(a) * f(b) then the "*" operation is not
    defined because it operates on S and R but not R or S alone.
    
    But can I use instead of f(a) * f(b) a similar construct like
    a * f(b). but that I think violates the homomorphism because it
    requires two groups which I don't have.
    
    	Julian.
    

    Off hand, I think this problem falls under the heading of category
    theory, where various morphisms and structures are studied in a very
    general way.

    - Jim

    Jim,
    
    do you know any literature about this. I don't have anything like
    this at home.
    
    	Julian.
    

    Firstly, I don't know of a general reference on categories since
    I haven't had the interest to persue it for its own sake.  But
    it seems to be a very general way of looking at morphisms of algebraic
    structures which preserve certain structural properties.  The idea
    is to have a set of algebraic structures with a related set of
    morphisms, and to use a so-called 'functor' to map this set to
    another set of structures, respecting some of the underlying structure.

    I think the mapping of topological properties of, say, a
    differentiable manifold, to an algebraic group (such as homology or
    homotopy groups) is an example of a functorial mapping.  The morphisms
    involved here would be, for example, the homomorphisms of homology
    groups under the action of a boundry operator.

    But in rereading your example, it really looks like a concrete
    example a group acting on a set; examples abound of this.
    (You did not say if your group had an inverse - if not is only a
    semigroup.)

    Consider the group of linear automorphisms of 3 dimensional
    Euclidean space.  The elements of this group have a concrete
    representation as non singular 3 by 3 matrices.  This is your S.
    These matrices also act on 3 dimensional vectors, and they are your R.

    If S1 is some subroup of S it will partition R into equivalence
    classes called the orbit of S1.  For instance, let S1 be those
    elements of S with determinant 1; the orbits will be concentric
    spheres in Euclidean space.

    Could you clarify what your are doing?

    - Jim

    A bit more information...
    
    I don't have any inverse elements, thus it is a semi-group.
    
    What I am trying to do is defining a message system, which allows
    4 different operations between objects. There are two different
    types of objects processes and process-rings. For processes you
    can have 2 different operations 'SEND' and 'RECV'. For process
    rings you can have the rest of the 4 operations 'CNX' and 'RMCNX'
    (which is connect and disconnect).
    
    Now a process can connect to a process-ring, as well as a process-
    ring can connected to another process-ring. 
    
    The operation CNX is defined between process rings or processes
    and a process ring. I would like to built the operations as
    semigroups (BTW thanks for the word I learned it as half-group).
    
    But the connect operation takes two different operands from two
    different sets (The process set and the process-ring set). What
    I basically would need is a function which allows to find a process
    for each process ring. I think I found one...in the last days
    I let you know.
    
    	Julian.

    A good book on abstract algebra is Algebra by Serge Lang in it he
    discusses the theory of catagories.  Good Luck!