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Math Help - cycles

  1. #1
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    Question cycles

    Write as a single cycle or product of disjointed cycles and as a product of 2-cycles.
    A) alph*beta where
    alph=
    1 2 3 4 5 6 7 8
    2 3 4 5 1 7 8 6

    beta=
    1 2 3 4 5 6 7 8
    1 3 8 7 6 5 2 4

    I found that alph*beta=
    1 2 3 4 5 6 7 8
    2 4 6 8 7 1 3 5
    but I dont know how to get the rest?

    B)
    (1 2 3)^-1 (2 3) (1 2 3)
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  2. #2
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    Quote Originally Posted by mandy123 View Post
    Write as a single cycle or product of disjointed cycles and as a product of 2-cycles.
    A) alph*beta where
    alph=
    1 2 3 4 5 6 7 8
    2 3 4 5 1 7 8 6

    beta=
    1 2 3 4 5 6 7 8
    1 3 8 7 6 5 2 4
    Look at how the elements go:
    \alpha(\beta (1)) = \alpha(1) = 2
    \alpha(\beta (2)) = \alpha(3) = 4
    \alpha(\beta(4)) = \alpha (7) = 8
    \alpha(\beta(8)) = \alpha (4) = 5
    \alpha( \beta(5)) = \alpha (6) = 7
    \alpha(\beta(7)) = \alpha (2) = 3
    \alpha(\beta(3)) = \alpha(8) = 6
    \alpha(\beta(6)) = \alpha(5) = 1
    Therefore (12485736) is \alpha \beta
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mandy123 View Post
    Write as a single cycle or product of disjointed cycles and as a product of 2-cycles.
    A) alph*beta where
    alph=
    1 2 3 4 5 6 7 8
    2 3 4 5 1 7 8 6

    beta=
    1 2 3 4 5 6 7 8
    1 3 8 7 6 5 2 4

    I found that alph*beta=
    1 2 3 4 5 6 7 8
    2 4 6 8 7 1 3 5
    but I dont know how to get the rest?
    right.

    now, start the cylcle by starting with 1, so you write (1

    now, follow where it leads until you get back to 1. that, see that 1-->2, 2-->4, 4-->8, 8 -->5, 5-->7, 7-->3, 3--> 6, 6 --> 1

    so you have

    (1 2 4 8 5 7 3 6)

    and that's everybody. now, break it up into transpositions (2-cycles) and you can take it from there

    note that (a_1~a_2~a_3~ \cdots a_n) = (a_1~a_2)(a_2~a_3) \cdots (a_{n - 1}~a_n)
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