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Thread: Disjoint cycles...

  1. #1
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    Disjoint cycles...

    Q1) Write in the following as a single cycle or a product of disjoint cycles?

    A) ( 1 2 3 ) ^(-1) ( 2 4 ) ( 1 2 3 )

    B) ( 1 3 )^(-1) ( 2 4 ) ( 2 3 5 )^(-1)

    I am having some difficulties understanding the difference between the normal set and this inverse sets...I cant figure it out..
    Please help!

    Thanks!
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  2. #2
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    Quote Originally Posted by Vedicmaths View Post
    Q1) Write in the following as a single cycle or a product of disjoint cycles?

    A) ( 1 2 3 ) ^(-1) ( 2 4 ) ( 1 2 3 )
    I do this first one. First $\displaystyle (123)=(321)=(132)$ by the other post. Thus, we need to simplify $\displaystyle (132)(24)(123)$ remember this means $\displaystyle \sigma = (132)\circ (24) \circ (123)$. To do these problems we map $\displaystyle 1$ and then find $\displaystyle \sigma(1)$ then $\displaystyle \sigma(\sigma(1))$ until we complete the cycle. Note, $\displaystyle 1\to 2$ by applying $\displaystyle (123)$ first then $\displaystyle 2\to 4$ by applying $\displaystyle (24)$ second and $\displaystyle 4\to 4$ by applying $\displaystyle (132)$ third. Thus, $\displaystyle 1\to 4$ by $\displaystyle \sigma$. Now we find $\displaystyle \sigma(4)$. Note $\displaystyle 4\to 4$ by $\displaystyle (123)$ first then $\displaystyle 4\to 2$ under $\displaystyle (24)$ and finally $\displaystyle 2\to 1$ by $\displaystyle (132)$. Thus, $\displaystyle \sigma(\sigma(1)) = 1$ and it means we only write $\displaystyle (14)$ as we express $\displaystyle \sigma$ as a product of disjoint cycles. Next we do the same with $\displaystyle 2$. See that $\displaystyle 2\to 2$ under $\displaystyle \sigma$ and so $\displaystyle 3\to 3$. Which means the representation is $\displaystyle (14)(2)(3) = (14)$.
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