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Abstract Algebra - Orbit of a permutation

  1. Oct 29, 2009 #1
    For this problem, I have to find all orbits of given permutation.

    [tex] \sigma: \mathbb{Z} \rightarrow \mathbb{Z}[/tex]

    Where,

    [tex]\sigma(n)=n-3 [/tex]

    Now, the problem is I do not know how to approach this permutation in the given format.

    All the permutations I dealt with were in the form:

    [tex]

    \mu = \left(
    \begin{array}{cc}
    1\ 2\ 3\ 4\ 5\ 6\\
    1\ 2\ 3\ 4\ 5\ 6
    \end{array}
    \right)

    [/tex]

    Which I understand. But I do not understand the sigma permutation first mentioned. I tried another example where I had an answer to σ(n)=n+2, but I did not understand how that answer was achieved.

    If someone can guide me with a start that'd be great.
     
  2. jcsd
  3. Oct 29, 2009 #2
    The sigma format makes it much harder to see what's going on with the permutation.

    All it really is though is sending the nth element to the n - 3 element. So 4 goes to 1, 3 goes to 6, 2 goes to 5...


    [tex]
    \mu = \left(
    \begin{array}{cc}
    1\ 2\ 3\ 4\ 5\ 6\\
    4\ 5\ 6\ 1\ 2\ 3
    \end{array}
    \right)
    [/tex]
     
  4. Oct 29, 2009 #3

    Office_Shredder

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    No that's not quite correct. Sigma is a map from the integers to the integers, so you can't express it like that. The map is what it says it is

    [tex]\sigma(3) = 3-3=0
    [/tex]
    [tex] \sigma(5) = 5-3= 2[/tex] and so forth. Basically sigma is just shifting the integers three places over
     
  5. Oct 29, 2009 #4
    Sorry i thought it was an element of the symmetric group of degree 6.
     
  6. Oct 29, 2009 #5
    I tried thinking of it that way. I originally thought n=1 -> -2, n=2 -> -1, n=3 -> 0, n=4 -> 1, then 1 -> -2. So that would be an orbit. So looking at the structure so to say, looked like each orbit consisted of 4 elements. But then the -2, -1, 0 kinda disappeared? I was not comfortable with it. Lets just say that.

    Ok, but what about [tex]\sigma(1) = 1-3=-2
    [/tex]

    1 is an integer, -2 is an integer, why doesn't that work?
     
  7. Oct 29, 2009 #6

    Office_Shredder

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    [tex] \sigma(1)=-2[/tex] is true.

    What is the definition of an orbit here? It helps to find out what the question is actually asking in this context
     
  8. Oct 29, 2009 #7
    Taken from First course in Abstract Algebra:

    Let [tex]\sigma[/tex] be a permutation of a set A. The equivalence classes in A determined by the equivalence relation (1) are the orbits of [tex]\sigma[/tex].

    Definition is a bit high on vocabulary, but I think I understand it.

    For example, in the following:

    [tex]

    \sigma = \left(
    \begin{array}{cc}
    1 \ 2 \ 3 \ 4 \ 5 \ 6 \ 7 \ 8\\
    4 \ 3 \ 2 \ 8 \ 5 \ 1 \ 7 \ 6
    \end{array}
    \right)

    [/tex]

    The orbits are:

    [tex]

    \{1, 4, 8, 6\} \ \ \{2, 3\} \ \ \{5\} \ \ \{7\}[/tex]

    But I am a bit confused how to determine an orbit with the form given in original post.
     
  9. Oct 29, 2009 #8

    Office_Shredder

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    So look at the orbit of 1. Let's see what some elements in it are

    [tex] \sigma(1)=-2, \sigma(-2)=-5, \sigma(-5)=-8,[/tex]

    and in the other direction

    [tex] \sigma(4)=1, \sigma(7)=4[/tex]

    noticing a pattern?
     
  10. Oct 29, 2009 #9
    Omg. It totally clicked. I am so contributing to this forum. Member by tomorrow night that's for sure. The ultimate source for tutoring. Best thing is, never yet have I gotten just an answer. Always an explanation to help me solve it myself. A lot of nice people with patience on here. Thank you very much! I am so excited, high hopes for the next test! :)

    Orbits will be:

    [tex]
    \{3n\ |\ n \ \epsilon \ \mathbb{Z}\} \ \ \{3n-1\ |\ n \ \epsilon \ \mathbb{Z}\} \ \ \{3n-2\ |\ n \ \epsilon \ \mathbb{Z}\}
    [/tex]
     
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