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Abstract algebra module

  1. Feb 27, 2012 #1
    http://i111.photobucket.com/albums/n149/camarolt4z28/untitled.jpg

    G/N is the set of all left cosets of N in G.

    I don't understand the notation.

    a) The permutations are (1,2), (2,3), (3,1). What are the left cosets - <1>, <2>, <3>? That doesn't make sense with permutations.

    b) I have no idea. N = mZ is simply all integer multiples of m. How do I find the left cosets?
     
  2. jcsd
  3. Feb 27, 2012 #2

    Dick

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    You find left cosets by operating on the subgroup on the left by members of your group. For the first one beta isn't three permutations, it's only one. In cycle notation it's (123). And the group N is the subgroup generated by beta. What's that?
     
  4. Feb 27, 2012 #3
    It's the same thing. (123) is simply more concise. The left cosets are generated by gN, i.e. each element of g is composed with each element of N. Are you saying it's <123>?
     
  5. Feb 27, 2012 #4

    Dick

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    No, I'm asking what N is. It's should be a subgroup with three permutations in it.
     
  6. Feb 27, 2012 #5
    (1), (2), (3)?
     
  7. Feb 27, 2012 #6

    Dick

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    Are you using cycle notation? Aren't those all the identity permutation?
     
  8. Feb 27, 2012 #7
    Yes. For these two problems, I don't see how you can take the composition gN.
     
  9. Feb 27, 2012 #8

    Dick

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    For the first problem, you haven't figured out what N is yet. It's the subgroup generated by the permutation (123). That's a permutation of order three and it's one of the permutations in N. What are the other two?
     
  10. Feb 27, 2012 #9
    What about (1,3) and (1,2)?
     
  11. Feb 27, 2012 #10

    Dick

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    Are you just guessing? The product of (123) and (123) must be in the subgroup, mustn't it?
     
  12. Feb 27, 2012 #11
    Oh, shoot. We're taking the generation <(123)>, right?
     
  13. Feb 27, 2012 #12

    Dick

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    Sure are.
     
  14. Feb 27, 2012 #13
    Wait, so it's really the composition N of G. We're looking at the generator of the entire permutation.
     
  15. Feb 27, 2012 #14

    Dick

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    I'm not sure what you are saying. It's the subgroup generated by (123). For any element a, <a> is just all of the powers of a and its inverse.
     
  16. Feb 27, 2012 #15
    The permutation beta is fully defined by (123). You're sticking that in the generator. I need to to look at the (123)n until I find all the distinct left cosets.

    This abstract algebra modulus is part of Survey for Undergrad Math. It's a preparatory class for the math field test. I took Abstract last summer and did well, but some of this stuff we didn't cover.

    I just did the problem where the field is R, but I haven't dealt with a polynomial defined over a congruence class.
     
  17. Feb 27, 2012 #16

    Dick

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    You are still thinking about this wrong. Finding (123)^n is just going to give you one coset. N itself. There's another one.

    I would just do the division algorithm and substitute arithmetic mod 3 for real arithmetic.
     
  18. Feb 27, 2012 #17
    How about the inverse - (321)?

    Okay. Let me try that for the division algorithm problem.
     
    Last edited: Feb 27, 2012
  19. Feb 27, 2012 #18

    Dick

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    Ok! Yes, N={e,(123),(321)} where e is the identity permutation. That's the coset eN. Now there's another coset, pick another element g in S3 that's not in N and find gN.
     
  20. Feb 27, 2012 #19
    How did you know the subgroup N has only three elements - because of the order of S3? Each subgroup must have an order that's a multiple of the order of S3. So of course each subgroup has to have the identity and an inverse of a given element.

    If S3 is the set of all permutations, then I can pick any permutation that's not in N. (231)?
     
  21. Feb 27, 2012 #20

    Dick

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    I knew it because (123) is a 3-cycle, (123)^3=e. (123)^2=(321). And (231) IS in N. It's the same as (123).
     
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