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Homework Help: Cycle notation for permutations?

  1. May 20, 2010 #1
    1. The problem statement, all variables and given/known data

    Why are the permutations (1 4 2 6)(2 3 4 5) and (1 4 5 6)(2 3) equal? It seems to me as if the first pair of 4-cycles want to permute 4->2 and 4->5, yielding a contradiction, but I suspect I've misunderstood something about composite cycles.

    I suspect it has something to do with dividing the cycles into [tex]\alpha[/tex] = (1 4 2 6) and [tex]\beta[/tex] = (2 3 4 5) and investigating the composite function [tex]\alpha \beta[/tex] (or [tex]\beta \alpha[/tex]?), but I can't seem to get the result right.
     
  2. jcsd
  3. May 20, 2010 #2
    I think I figured it out. Go from the right cycle and move left. In the case of (1 4 2 6)(2 3 4 5), starting off with 2 in the right cycle, we get 2 -> 3, and 3 is unchanged by the left cycle. Then, 3 -> 4 and 4 -> 2 in the left cycle, so 4 -> 2. Then, 4 -> 5, and 5 is unchanged by the left cycle, and so on. We end up with (2 3)(4 5 6 1), which is equivalent to what I stated in my previous post.

    So in other words, the algorithm seems to be to start with the cycle to the right, permute a "member" X in it to whatever member is one place to the right of X, and if possible, permute the result again in the cycle one step to the left, and repeat, until you can't permute it through any more cycles. Repeat for additional members in the cycle to the right. Once you're out of members in the cycle to the right, check for members in the cycle one step left of it that have not yet been permuted into anything, permute them, and so on.

    I guess this "makes sense" in relation to how composite functions work; I just wish my book (Discrete Mathematics by Biggs) had been more explicit about it. Let me know if I'm wrong!
     
    Last edited: May 20, 2010
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