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Group action on a set?

  1. Feb 16, 2009 #1

    tgt

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    1. The problem statement, all variables and given/known data
    Up to isomorphism, how many transitive Z6={0,1,2,3,4,5} sets X are there? In other words Z6 acting on X. How many X, up to isomorphism are there?


    2. Relevant equations
    A key theorem is Let X be a G-set and let x in X. Then |Gx|=(G:G_{x}).


    3. The attempt at a solution
    I found 1 set with 1 element. 1 set with 2 elements. 2 sets with 3 elements. 5*5! sets with 6 elements.

    However the answer only had 1 set with 1 element. 1 set with 2 elements. 1 set with 3 elements.

    Surely there should be at least a set with 6 elements?
     
  2. jcsd
  3. Feb 16, 2009 #2

    Hurkyl

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    There indeed should be a transitive G-set with 6 elements. Unless, of course, you missed a condition of the problem....


    As for your counts, remember that you were asked to count isomorphism classes....
     
  4. Feb 16, 2009 #3

    tgt

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    Didn't leave out any of the problem. It is on p196 q17 of Fraleigh's book.

    Yes, I see up to isomorphism. Take the set with 3 elements. I am able to produce to different multiplication tables with the same three elements in the set. That must mean two non isomorphic sets.
     
  5. Feb 16, 2009 #4

    Hurkyl

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    I'm not convinced. What are your two examples?
     
  6. Feb 17, 2009 #5

    tgt

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    Attached is the two tables. Once the fixed elements and the multiples of 1 line is determined, the rest of the table is self explanatory by the axiom that a(bx)=(ab)x.
     

    Attached Files:

  7. Feb 19, 2009 #6

    tgt

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    What do people think?
     
  8. Feb 23, 2009 #7

    tgt

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    Hurkyl, have you disappeared?
     
  9. Feb 23, 2009 #8

    Hurkyl

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    I didn't want to download a .doc file, so I was hoping someone else who wasn't bothered would chime in.
     
  10. Feb 28, 2009 #9

    tgt

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    Strange reason? Are you worried about your memory?

    Anyway, here it is without the .doc file

    * a b c
    0 a b c
    1 b c a
    2
    3 a b c
    4
    5


    * a b c
    0 a b c
    1 c a b
    2
    3 a b c
    4
    5

    The above are the two different tables hence two different isomorphisms.
     
  11. Mar 1, 2009 #10

    Hurkyl

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    But those are isomorphic Z6-sets. One isomorphism is

    a |--> a
    b |--> c
    c |--> b
     
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