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    #1
    Hey!!

    We have the following permutations in $\text{Sym}(14)$ :

    - $\pi_1=(1 \ 2\ 4 \ 9)\circ(1 \ 3)\circ (6 \ 8\ 12)$
    - $\pi_2=(2 \ 4\ 5 \ 8\ 7)\circ (1 \ 12 \ 6)\circ \ (13 \ 14)$
    - $\pi_3=(1 \ 4 \ 5\ 8 \ 11)\circ (2 \ 4\ 6 \ 5 \ 1)$



    1. Determine the cycle decomposition of $\pi_1, \pi_2, \pi_3$.
    2. Determine $\pi_1^{-1}, \pi_2^{-1}, \pi_3^{-1}$.
    3. Determine $\pi_4=\pi_1\circ \pi_2$, $\pi_5=\pi_2\circ\pi_3$, $\pi_6=\pi_2\circ\pi_1$.
    4. Determine the signum of $\pi_1, \pi_2, \pi_3, \pi_4, \pi_5, \pi_6$.



    1. We consider the composition fromright to left, or not? Thenfrom the last we consider the element $6$ that goes $8$ and since there is no other $8$ previously we have that $6\rightarrow 8$.

    Is this the idea to get the cycles?

    2. We get the inverse permutation by the cycle decompsition, right?

    Last edited by mathmari; February 9th, 2020 at 11:21.

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    #2
    Quote Originally Posted by mathmari View Post
    1. We consider the composition fromright to left, or not? Thenfrom the last we consider the element $6$ that goes $8$ and since there is no other $8$ previously we have that $6\rightarrow 8$.

    Is this the idea to get the cycles?
    Hey mathmari!!

    Yes, this is the way to decompose the permutation into disjoint cycles.

    Quote Originally Posted by mathmari View Post
    2. We get the inverse permutation by the cycle decompsition, right?
    We can choose.
    We get the inverse by reversing each cycle in combination with applying the constituent cycles in reverse.
    We can do this with the permutation as given, or with the decomposition.

    We can for instance already write:
    $$\pi_1^{-1}=(12 \ 8\ 6)\circ(3 \ 1)\circ(9 \ 4\ 2 \ 1) $$
    It may be intended to do this with the decomposition of $\pi_1$ though.

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    #3 Thread Author
    Quote Originally Posted by Klaas van Aarsen View Post
    Yes, this is the way to decompose the permutation into disjoint cycles.
    Aren't the first two permutations already the decomposition into disjoint cycles?

    As for $\pi_3$ we have the following:
    $1$ goes to $2$
    $5$ goes to $1$ and $1$ goes to $4$, so $5$ goes to $4$
    $6$ goes to $5$ and $5$ goes to $8$, so $6$ goes to $8$
    $4$ goes to $6$
    $2$ goes to $4$ and $4$ goes to $5$, so $2$ goes to $5$
    $11$ goes to $1$
    $8$ goes to $11$

    Is this correct?

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    #4
    Quote Originally Posted by mathmari View Post
    Aren't the first two permutations already the decomposition into disjoint cycles?
    Yes.
    But the third permutation could still conceivably overlap both of them, so that they might not actually end up as disjoint cycles in the decomposition.

    Quote Originally Posted by mathmari View Post
    As for $\pi_3$ we have the following:
    $1$ goes to $2$
    $5$ goes to $1$ and $1$ goes to $4$, so $5$ goes to $4$
    $6$ goes to $5$ and $5$ goes to $8$, so $6$ goes to $8$
    $4$ goes to $6$
    $2$ goes to $4$ and $4$ goes to $5$, so $2$ goes to $5$
    $11$ goes to $1$
    $8$ goes to $11$

    Is this correct?
    Yep.
    Can we write it as disjoint cycles?

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    #5 Thread Author
    Quote Originally Posted by Klaas van Aarsen View Post
    Yep.
    Can we write it as disjoint cycles?
    So we get $\pi_3=(1 \ 2 \ 5 \ 4 \ 6 \ 8 \ 11)$, or not?

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    #6
    Quote Originally Posted by mathmari View Post
    So we get $\pi_3=(1 \ 2 \ 5 \ 4 \ 6 \ 8 \ 11)$, or not?
    Yep.

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    #7 Thread Author
    Quote Originally Posted by Klaas van Aarsen View Post
    Yep.
    Therefore, the cycle decompositions of the permutations are:
    - $\pi_1=(1 \ 2\ 4 \ 9) \ (1 \ 3)\ (6 \ 8\ 12)$
    - $\pi_2=(2 \ 4\ 5 \ 8\ 7)\ (1 \ 12 \ 6) \ (13 \ 14)$
    - $\pi_3=(1 \ 2 \ 5 \ 4 \ 6 \ 8 \ 11)$

    Right?

    At the first two do we consider the product of the cycles or the composition as we had initially?

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    #8
    Quote Originally Posted by mathmari View Post
    Therefore, the cycle decompositions of the permutations are:
    - $\pi_1=(1 \ 2\ 4 \ 9) \ (1 \ 3)\ (6 \ 8\ 12)$
    - $\pi_2=(2 \ 4\ 5 \ 8\ 7)\ (1 \ 12 \ 6) \ (13 \ 14)$
    - $\pi_3=(1 \ 2 \ 5 \ 4 \ 6 \ 8 \ 11)$

    Right?
    The cycles in $\pi_1$ are not disjoint are they?
    Isn't $1$ in multiple cycles?

    Quote Originally Posted by mathmari View Post
    At the first two do we consider the product of the cycles or the composition as we had initially?
    Huh?
    Which 'first two'?
    Which question are we talking about?

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    #9 Thread Author
    Quote Originally Posted by Klaas van Aarsen View Post
    The cycles in $\pi_1$ are not disjoint are they?
    Isn't $1$ in multiple cycles?
    Oh yes, you are right!

    So, the cycle decompositions of the permutations are:
    - $\pi_1=(12 \ 6 \ 8)\ (3 \ 2)\ (9 \ 1 \ 2 \ 4)$
    - $\pi_2=(2 \ 4\ 5 \ 8\ 7)\ (1 \ 12 \ 6) \ (13 \ 14)$
    - $\pi_3=(1 \ 2 \ 5 \ 4 \ 6 \ 8 \ 11)$

    Right?


    Quote Originally Posted by Klaas van Aarsen View Post
    Huh?
    Which 'first two'?
    Which question are we talking about?
    My question is: Is there a difference between $(12 \ 6 \ 8)\ (3 \ 2)$ and $(12 \ 6 \ 8)\circ (3 \ 2)$ ? Is the product and the composition symbol equivalent?

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    #10
    Quote Originally Posted by mathmari View Post
    Oh yes, you are right!

    So, the cycle decompositions of the permutations are:
    - $\pi_1=(12 \ 6 \ 8)\ (3 \ 2)\ (9 \ 1 \ 2 \ 4)$
    Doesn't $\pi_1$ have two cycles with $2$ in them now?

    Quote Originally Posted by mathmari View Post
    My question is: Is there a difference between $(12 \ 6 \ 8)\ (3 \ 2)$ and $(12 \ 6 \ 8)\circ (3 \ 2)$ ? Is the product and the composition symbol equivalent?
    Ah no, there is no difference.
    The product of cycles is defined as the composition. They are the same.

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