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Abstract Algebra

  1. Dec 21, 2008 #1
    1. This is not a question it's an example.



    2.The permutation (123)= (13)(12)= (13)(23)(12(13)= (23)(13)(12)(13)(12)(23) is even.




    3. I got the frist one because it is the product of tranposition...I just don't get the rest. I know that it is even depending on the number of transposition it contains, getting them is my problem.
     
  2. jcsd
  3. Dec 21, 2008 #2
    what is your question?
    There is a theorem that says, that every permutation can be writen as a product of transpositions.

    [tex](d_1d_2.....d_n)=(d_nd_1)(d_nd_2).....(d_nd_{n-1})[/tex]

    But you have to understand that there are many ways of rewriting a permutation as a product of transpositions, i.e

    [tex]\theta=\alpha_1\alpha_2.....\alpha_r=\beta_1\beta_2....\beta_s=\delta_1\delta_2...\delta_m[/tex]

    but all r,s and m have the same parity, that is they are either odd or even.(there is a theorem that establishesh this fact as well).
     
    Last edited: Dec 22, 2008
  4. Dec 22, 2008 #3

    HallsofIvy

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    The "rest" of what? There exist an infinite number of ways any permutation can be written as a product of transpositions.
     
  5. Dec 22, 2008 #4
    want to figure out whether a permutation is odd or even...my textbook gave the example (123) = (12)(13), (123) = (13)(23)(12)(13), (123)= (23)(13)(12)(13)(12)(23) they called it the factorization as a product of transformations.
    Now I see that (123) is even because for each factorization there is an even number of transposition, what I don't get is how to derive the factorization of transformations.

    Here is another example:(1235) = (15)(24)(24)(13)(23)(23)(12), (1235) = (13)(24)(35)(14)(24), (1235) = (15)(13)(12).....
    ....I'm trying to determine whether (12) (134) (152) this is even or odd, In the examples above I only hadone permutation , now I have cycles....
     
  6. Dec 22, 2008 #5

    HallsofIvy

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    (1235) means "1 becomes 2, 2 becomes 3, 3 becomes 5, 5 becomes 1 and all others remain the same": (12345) is changed to (23541).

    You can write that as transpositions by: (12)(13)(15)- I got that by thinking "first 1 and 2 swap. Now that 1 is in the old "2" place, it swaps with 3, once it is the "3" place it swaps with 5: first (12345) becomes (21345) then (23145) then (23541). That is your third option. Because that is 3 (an odd number of) transpositions this is an odd permutation. But notice that all of your possible factorizations has an odd number of transpositions: 7, 5, and 3.

    To factor (12)(134)(152), note that the first (12) is already a transposition. (134) can be factored as (13)(14) and (152) as (15)(12). (12)(134)(152)= (12)(13)(14)(15)(12) which has 5 transpositions- this is an odd permutation.
     
  7. Dec 22, 2008 #6
    In case you have a permutation written as a product of say t disjoint cycles, then you can tell whether the permutation is odd or even by just counting the lengths of each cycle and subtracting from it the number of cycles. That is say

    [tex]\theta=\alpha_1\alpha_2\alpha_3....\alpha_t[/tex] are t disjoint cycles each with length [tex]s_1,s_2,....,s_t[/tex] then this permutation is odd or even depending on whether the following is odd or even

    [tex]s_1+s_2+...+s_t-t[/tex]
     
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