Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Multiplying non-disjoint permutation cycles

  1. Apr 10, 2007 #1
    Maybe I'm just being dense, but I've been having issues with the multiplying non-disjoint permutation cycles (as you may have guessed from the topic title). Simple products like (1, 4, 5, 6)(2, 1, 5) [an example from my textbook], as well as in the opposite order. Mayhap that I'm tired.

    Disjoint cycles are quite easy, so I'm not sure what's going on in my head that fails me so.

  2. jcsd
  3. Apr 10, 2007 #2


    User Avatar
    Homework Helper

    Just keep track of each element, keeping in mind that the order is reversed, eg, ab is the permutation b followed by the permutation a. So, in your example, 1 is sent by the second cycle (in the order they're written) to 5, then 5 is sent by the first to 6, so 1 ends up at 6. 6 is unmoved by the second cycle, then moved by the first to 1. Thus one of the cycles in the product is the transposition (1 6) (usually these are written without commas).
  4. Apr 10, 2007 #3
    Here are some links with nice tutorials:

    1) Permutation groups in the blog "The unapologetic mathematician" here
    This is where I first learned about the cycle notation.

    2) Tom Davis Rubik's Cube notes contains a chapter on the
    cycle notation. Read the next pages too.

    3) Short article at Wolfram Mathworld

    4) Sarah Kitchen's notes on Permutation groups and polynomials. Read the chapter on "Cycle Decomposition" on page 2 and 3.

    5) Jaap's puzzle explains permutations with diagrams
    Last edited: Apr 10, 2007
  5. Apr 10, 2007 #4
    I just found an awesome interactive flash video on permutations.
    First of all watch the graphics here.

    Then watch the tutorials on how to use the interactive flash video:
    Short tutorial. Watch the first tutorial "Making Shuffles"
    and then try out producing some shuffles with the
    Shuffle factory program. Then watch the second tutorial "Manipulating Shuffles".
    You should then be able to calculate
    your example (1 4 5 6)(2 1 5) with the Shuffle Factory Program.

    (i) Just start by creating the (1 5 2) shuffle and the (1 4 5 6) shuffle.
    Note that (2 1 5) is the same as (1 5 2).
    Of course, both shuffles have to consist of "6 pearls".
    Make a copy of both and put them on the right side.

    (ii) Now work with the copies:
    Put the (1 4 5 6) below the (2 1 5) shuffle like this:

    (1 4 5 6)
    (2 1 5)

    Then merge them together and see the result by first clicking on the PLAY button
    in the right bottom corner of the merged shuffle. You can follow the path of the pearls.
    Then press the button in the left bottom corner of the shuffle to finally merge the two shuffles together.
    See the result.

    This is really fun!
    Last edited: Apr 10, 2007
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Multiplying non-disjoint permutation cycles
  1. Multiplying permutations (Replies: 10)

  2. Disjoint Cycles (Replies: 5)