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Permutation Representation

  1. Feb 28, 2008 #1
    1. The problem statement, all variables and given/known data

    a. Find the permutation representation of a cyclic group of order n.
    b. Let G be the group S3. Find the permutation representation of S3.

    2. Relevant equations

    n/a

    3. The attempt at a solution

    I unfortunately have not been able to come up with a solution. I really don't understand what it means by "permutation representation". If someone can give me a definition or explanation, I think I can go from there.

    Thank you.
     
  2. jcsd
  3. Feb 28, 2008 #2

    Dick

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    I would like it better if they said 'find A permutation representation' rather than 'THE permutation representation'. Maybe it will help to just do an example. Take the cyclic group Z(3). It has three elements {0,1,2}. An element of the group, g, acts on an element x by taking x -> gx. So a representation of 0 is the permutation (0)(1)(2) (the identity permutation). 1 becomes (012) and 2 becomes (021). The easiest way to represent S(3) is by ALL possible permutations on {0,1,2}. The previous representation of Z(3) represents it as a subgroup of S(3). Now, you could represent S(3) by a group of permutations on the six elements that compose S(3) just as we did for Z(3). But that's much harder to write down. That's why I wish they had said 'A permutation representation'. A group can have many permutation representations. Some are more economical than others. Does that make any sense at all?
     
  4. Feb 28, 2008 #3
    Hmm... I think I kind of understand. Then again, I'm also really tired, so who knows?

    So, for S3, we'd have (1)(2)(3),(12)(3),(13)(2),(1)(23),(123) ?

    Or am I totally off?
     
  5. Feb 28, 2008 #4

    Dick

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    I'm tired too. But there are six elements in S3. What about (132)? That doesn't make you 'totally off'.
     
  6. Feb 28, 2008 #5

    Dick

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    Maybe I'm skipping something. A 'representation' of a group is a 'model' of it. I can compute the sum of two elements in Z(3) by computing the product of the permutations that 'represent' them. Same for S(3).
     
  7. Feb 28, 2008 #6
    Okay, so I only missed one? And we get our 3-cycles through multiplying the 2-cycles?

    And as far as the cyclic group of order n... How is that specifically different from Sn?

    Thank you so much for all of your help!!!
     
  8. Feb 28, 2008 #7
    Sorry, I think I'm really confused, after reading the post directly before my previous post. I swear I'm trying to understand.

    Thank you.
     
  9. Feb 28, 2008 #8

    Dick

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    It's not really about cycles. Here's Z(4). (0)(1)(2)(3),(0123),(02)(13),(0321). That's making it look hard. But there are ways to represent permutations without using cycles. Just name the base set and give a rule for what maps into what to describe a permutation.
     
  10. Feb 29, 2008 #9

    HallsofIvy

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    Write down the "operation table" for the group. Let's say that "e" is the identity and you have elements a, b, etc. Notice that multiplying a by each member of the group gives the row labled "a" on your operation table. Each of the members of the group appears once and only once on that row. That is, the row is a "permutation" of the top row just listing the elements of the group. If you replace e, a, b, etc. with 0, 1, 2, etc. each row is a permutation of {0,1,2, ...}. That is the permutation representation.

    Take, for example, the Klein 4-group. Its operation table is
    __e a b c
    e e a b c
    a a e c b
    b b c e a
    c c b a e

    Now replace those with 0, 1, 2, 3 and the table is
    __0 1 2 3
    0 0 1 2 3
    1 1 0 3 2
    2 2 3 0 1
    3 3 2 1 0
    That is, 0, the identity, takes {0, 1, 2, 3} to {0, 1, 2, 3}, the identity permutation.
    1 takes {0, 1, 2, 3} to {1, 0, 3, 2}, the permutation (0 1)(2 3)
    2 takes {0, 1, 2, 3} to {2, 3, 0, 1}, the permutation (0 2)(1 3)
    3 takes {0, 1, 2, 3} to {3, 2, 1, 0}, the permutation (0 3)(1 2).

    Those four permutations (out of 4!= 24 possible permutations of {0, 1, 2, 3}) are the permutation representations of the elements of the Klein 4-group.
     
  11. Feb 29, 2008 #10
    Okay, now that it's morning, this makes much more sense. Thanks for the clarification HallsofIvy.

    I guess now, my only question is, does a 4 element permutation have a 4 element permutation representation or a 4!=24 element permutation?

    Thank you so much!
     
  12. Feb 29, 2008 #11

    Dick

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    Both. You can represent S(4) as the permutations on four arbitrary elements or you can represent it as permutations of the 24 group elements. I think the former is easier.
     
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