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Products of permutations

  1. Mar 3, 2008 #1
    I dont have any example of this in my notes:

    I have the permutation p1 = (1 5)(2 4 6 3)

    and the permutation p2 = (1 3 7 4)(2)(5 8 6)

    and I have to find p2 * p1

    i THINK you take it in the form p1 then p2, like:


    (1 5)(2 4 6 3)(1 3 7 4)(2)(5 8 6) = ...

    But them I'm stuck.

    I tried googling some info, but it was all too vague.

    I had an attempt which ended up in:

    1 -> 5 in p1, then 5 -> 8 in p2 so 1 -> 8

    etc

    so p2*p1 = (1 8)(2 1 8)(3 2 1 8)(4 5 3 2 1 8)(6 7)

    = (4 5 3 2 1 8)(6 7)

    Looks right? :P
     
  2. jcsd
  3. Mar 3, 2008 #2

    HallsofIvy

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    Take it "one step at a time". p2 is (1 3 7 4)(2)(5 8 6) and p1 is (1 5)(2 4 6 3).
    p1 takes 1 to 5 and then p2 takes 5 to 8: together p1p2 takes 1 to 8.
    p1 takes 2 to 4 and then p2 takes 4 to 1: p1p2 takes 2 to 1.
    p1 takes 3 to 2 and then p2 takes does not change 2: p1p2 takes 3 to 2.
    p1 takes 4 to 6 and then p2 takes 6 to 5: p1p2 takes 4 to 5.
    p1 takes 5 to 1 and then p1 takes 1 to 3: p1p2 takes 5 to 3.
    p1 takes 6 to 3 and then p2 takes 2 to 7: p1p2 take 6 to 7
    p1 does not change 7 but p2 takes 7 to 4: p1p2 takes 7 to 4
    p1 does not change 8 but p2 takes 8 to 6: p1p2 takes 8 to 6
    (That's how I am interpreting the fact that neither 7 nor 8 appear in the definition of p1.)

    Now look for cycles: 1 changes to 8 which changes to 6 which changes to 7 which changes to 4 which changes to 5 which changes to 3 which changes to 2 which changes to 1: a cycle is (18674532). Since that includes every number from 1 to 8, p1p2 is that cycle: p1p2= (18674532).
     
  4. Mar 3, 2008 #3
    I just tried to comnpute p2*p2

    (1 3 7 4)(2)(5 8 6)(1 3 7 4)(2)(5 8 6)

    To show where i went wrong ill miss out some steps and jum straight to it.

    I got 1 -> 3 then 3 -> 7 : so 1->7

    I also got 7 -> 4 then 4 -> 1 : so 7->1

    when starting off my cycle: 1 goes to 7, then 7 goes to 1...

    so my cycle would look like (17...)

    but it goes back to 1, does the cycle have to be in one set of brackets?

    otherwise I can do it as (17)(...)

    thanks!
     
  5. Mar 3, 2008 #4

    CompuChip

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    You are right, so p2 * p2 starts with (17)
    Then you open a new pair of brackets and start with a number you haven't looked at yet. So take for example 2, then p2 * p2 will become
    (17)(2...)
    If you complete the cycle while still not having had all the numbers 1 up to 8, open a new bracket again, etc.
     
  6. Mar 3, 2008 #5
    I now how to write p2 as a product of tranbspositions :P

    so (1 3 7 4)(2)(5 8 6)

    = (1 3) (3 7) (7 4) ( ? ) (5 8) (8 6)

    what goes in the bracket of ( ? ), i believe it needs to have 2 values, so (2 2)?
     
  7. Mar 3, 2008 #6
    What happens to the 4? What happens to the 6?
    2 goes to 2, so it's just (2)
     
  8. Mar 3, 2008 #7
    Well i have an example in my book, and it says:

    ( a1, a2, a3, ... , a(k-1), a(k)) = (a1 a2) (a2 a3) (a3 a4) ... (a(k-1) a(k))

    Which suggests I would leave the final values inside the bracket alone, as there is no (a(k) a1) term.

    And the example I have is:

    (2 4 6 7 3)(2 5 1 3)(7 4 8)

    = (1 2 5)(3 4 8)(6 7)

    = (1 2)(2 5)(3 4)(4 8)(6 7)

    Which also leaves out the 5 and 8 (final values of the brackets), is this wrong or?
     
  9. Mar 3, 2008 #8
    Our Professor had us putting in the last thing as well. I think it's just a difference in the method...My professor wanted it written out like (1 3) (3 7) (7 4) ( 4 1)(2) (5 8) (8 6)(6 5)
    I looked in my book and your form of the transpositions isn't in there.

    I think you're good.
    CC
     
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