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Homework Help: Cayley Tables and algebraic structures

  1. Nov 28, 2008 #1
    I have a question regarding Cayley tables, specifically using the composition operator for this particular problem.

    I had to miss class the other week and was just now sitting down to catch up on my homework for this class when I was hit with the algebraic structures section. They have a Cayley table with the composition operator. It goes like this

    o ] a | b | c | d | < row of operation
    a ] a | b | c | d |
    b ] b | a | d | c |
    c ] c | d | a | b |
    d ] d | c | b | a |

    Hopefully you can decipher that table :P They ask a series of questions for the problem such as, is it an algebraic structure, name the identity element, associative, commutative etc. I can answer all those questions by looking at the table, I just can't figure out how they have c o b = d.

    I have studied the definition of composite, looked back at identity functions, even checked out some other cayley tables for other operators. I see the pattern, but I just cant see why they get that pattern. Is there another way to do it? Because I see the ring pattern for the identity function, but why is c o c = a!!!
    I'm sure as soon as someone explains it to me I will bang my head, but I can't seem to get it right now :/. I'm a geology major (if that explains my lack of understanding :P - math minor) so go easy on me :)
    Last edited: Nov 28, 2008
  2. jcsd
  3. Nov 28, 2008 #2


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    Content in your post is usually a good start
  4. Nov 28, 2008 #3
    Yea I thought I had tabbed into this window and went to hit shift and hit enter instead :) Im in an awkward position with my laptop :eek:
  5. Nov 29, 2008 #4


    Staff: Mentor

    Tables like these are typically used for whatever the operation on a group. Apparently the group you're working with consists of four functions, and the operation is function composition.

    The table defines how the members of the group interact via the operation, so you don't need to figure out how they got, for example, c o b = d. Just take it as being defined this way.

    You can look at the table and see whether the operations are associative or commutative by noting whether x o (y o z) is the same as (x o y) o z for any choices of x, y, and z (associative). You can also see whether x o y is the same as y o x for any choices of x and y (commutative).

    It looks like a is the identity, since a o a = a, a o b = b, a o c = c, and a o d = d, and also a o a = a, b o a = b, c o a = c, and d o a = a.
  6. Nov 29, 2008 #5
    Mark is absolute right. I only wish to add a piece.

    If you want to consider a concrete example, Succubus, as you asked "why is c o c = a!!!", consider the abelian 4-element group [tex]G = \mathbf{Z}_2\times \mathbf{Z}_2[/tex], with the group operation being component-wise addition modulo 2. That is, G = {00, 01, 10, 11}, where 00 is the identity (00 + 01 = 01, etc), and x+x=00 for each x in G (e.g., 10 + 10 = 00).

    Consider the functions [tex]f_t:G\to G[/tex] defined by [tex]f_t(x) = x + t[/tex] (with addition as defined above). Let a, b, c and d be f_t for t=00, t=01, t=10 and t=11, respectively.

    Thus a(x) = x + 00 for each x in G; b(x) = x + 01; c(x) = x + 10; d(x) = x + 11.

    Then {a,b,c,d} forms a group under function composition. In particular, for each x in G, we have (c o c)(x) = c(x + 10) = (x + 10) + 10 = x = a(x). Hence c o c = a.
    Last edited: Nov 29, 2008
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