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Find a permutation representation

  1. Oct 26, 2008 #1
    1. The problem statement, all variables and given/known data
    Let G be the group S_3. Find the permutation representation of S_3. (Note: this gives an isomorphism of S_3 into S_6)

    3. The attempt at a solution
    Is there only ONE permutation representation, because the question asks for "the" p.r.
    I don't know where to start.
     
  2. jcsd
  3. Oct 26, 2008 #2
    For starters, what is a permutation representation?
     
  4. Oct 26, 2008 #3

    HallsofIvy

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    It can be shown that every finite group is isomorphic to some group of permutation.

    There is an easy way to do that: Write out the operation table for the group. Lable the elements "1", "2", ..., n according to how they are written along the top of the of the table. Now each row of the table is a permutation of 123...n Do you see how to assign each element of the group a corresponding permuation of 12...n?
     
  5. Oct 26, 2008 #4
    I get it know. This is just an exercise using Cayley's Theorem.
     
  6. Oct 27, 2008 #5
    I've written out the elements of S_3:
    {e, (123), (213), (12), (13), (23)}

    I'm sorry but I don't quite understand what you're describing. How do I write out the operation table?
     
  7. Oct 28, 2008 #6

    HallsofIvy

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    The "operation table" is a table with the elements of the group horizontally across the top and vertically along the left side. Where a column and row intersect is the product of the two elements. For example, "Klein's four group" with elements e, a, b, c being the identity, has operation group
    [tex]\begin{array}{ccccc} & e & a & b & c \\e & e & a & b & c \\ a & a & e & c & b \\ b & b & e & e & a \\ c & c & b & a & e\end{array}[/tex]
    If we replace "e", "a", "b", and "c" with "1", "2", "3", and "4" respectively that becomes
    [tex]\begin{array}{ccccc} & 1 & 2 & 3 & 4 \\ 1 & 1 & 2 & 3 & 4 \\ 2 & 2 & 1 & 4 & 3 \\ 3 & 3 & 4 & 1 & 2 \\ 4 & 4 & 3 & 2 & 1 \end{array}[/tex]
    The row starting with 1 is "1 2 3 4" so "1" takes 1234 into itself, the identity permutation. The row starting with 2 is "2143" so "2" take 1234 into that, the permutation (12)(34). The row starting with 3 is "3412" so "3" takes 1234 into that, the permutation (13)(24). The row starting with 4 is "4321" so "4" takes 1234 into that, the permutation (14)(23).

    The permutation representation of the Klein four group is {e, (12)(34), (13)(24), (14)(23)}.
     
  8. Oct 28, 2008 #7
    I've made the operation table, I don't see how the symmetry will work with 3 elements on top, and 3 elements on the side.

    I have that
    1 sends 123-->123
    this is the identity
    2 sends 123-->231
    this is (123)
    3 sends 123-->312
    this is (132)

    Did I set up my table correctly?
    Also, I don't see how I would get S_6 from this.

    Thank you for your help so far.
     
  9. Oct 29, 2008 #8

    HallsofIvy

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    S_3 does not contain 3 elements. It contains 3!= 6 elements- which is why it can be represented as a subgroup of S_6.
     
  10. Oct 29, 2008 #9
    I thought elements of S_3 were the permutations..no? I mean, in any case I did it wrong. So would I need to do it for entries 1 to 6 instead of what I did?

    Also, the operation table is ordered sort of like a symmetric matrix, right?
     
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