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    #11 Thread Author
    Quote Originally Posted by Klaas van Aarsen View Post
    Doesn't $\pi_1$ have two cycles with $2$ in them now?
    Ah yes!

    I tried that again and now I get $(12 \ 6 \ 8)\ (3 \ 2 \ 4 \ 9 \ 1)$ but there must be a mistake (but I don't know wheat I have done wrong) because at the initlal permutation $1$ goes to $2$ but not here.

    From the initial permutation $\pi_1$ we get:

    $12 \rightarrow 6$
    $6 -\rightarrow 8$
    $8 \rightarrow 12$
    $3 \rightarrow 1 \rightarrow 2$
    $9 \rightarrow 1$
    $1 \rightarrow 2$
    $2 \rightarrow 4$
    $4 \rightarrow 9$

    Or not?


    Quote Originally Posted by Klaas van Aarsen View Post
    Ah no, there is no difference.
    The product of cycles is defined as the composition. They are the same.
    Ahh ok!!

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    #12
    Quote Originally Posted by mathmari View Post
    Ah yes!

    I tried that again and now I get $(12 \ 6 \ 8)\ (3 \ 2 \ 4 \ 9 \ 1)$ but there must be a mistake (but I don't know wheat I have done wrong) because at the initlal permutation $1$ goes to $2$ but not here.

    From the initial permutation $\pi_1$ we get:

    $12 \rightarrow 6$
    $6 -\rightarrow 8$
    $8 \rightarrow 12$
    $3 \rightarrow 1 \rightarrow 2$
    $9 \rightarrow 1$
    $1 \rightarrow 2$
    $2 \rightarrow 4$
    $4 \rightarrow 9$

    Or not?
    Quote Originally Posted by mathmari View Post
    We have the following permutations in $\text{Sym}(14)$ :

    - $\pi_1=(1 \ 2\ 4 \ 9)\circ(1 \ 3)\circ (6 \ 8\ 12)$
    Don't we have $1\to 3$?

    Anyway, here's a slightly different way to find it.
    Let's start with $1$.
    $1 \to 3$ for $(1\,3\,\ldots)$.
    $3\to 1 \to 2$ for $(1\,3\,2\,\ldots)$.
    $2\to 4$ for $(1\,3\,2\,4\,\ldots)$.
    $4\to 9$ for $(1\,3\,2\,4\,9\,\ldots)$.
    $9\to 1$, which completes the cycle $(1\,3\,2\,4\,9)$.
    Now we start a new cycle with the first missing number, which is $6$.
    $6\to 8$ for $(6\,8\,\ldots)$.
    $8\to 12$ for $(6\,8\,12\,\ldots)$.
    $12\to 6$, which completes the cycle $(6\,8\,12)$.
    So $\pi_1=(1\,3\,2\,4\,9)(6\,8\,12)$.

    And yes, that is the same decomposition that you have, although the numbers are in a different order.

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    #13 Thread Author
    Quote Originally Posted by Klaas van Aarsen View Post
    Don't we have $1\to 3$?

    Anyway, here's a slightly different way to find it.
    Let's start with $1$.
    $1 \to 3$ for $(1\,3\,\ldots)$.
    $3\to 1 \to 2$ for $(1\,3\,2\,\ldots)$.
    $2\to 4$ for $(1\,3\,2\,4\,\ldots)$.
    $4\to 9$ for $(1\,3\,2\,4\,9\,\ldots)$.
    $9\to 1$, which completes the cycle $(1\,3\,2\,4\,9)$.
    Now we start a new cycle with the first missing number, which is $6$.
    $6\to 8$ for $(6\,8\,\ldots)$.
    $8\to 12$ for $(6\,8\,12\,\ldots)$.
    $12\to 6$, which completes the cycle $(6\,8\,12)$.
    So $\pi_1=(1\,3\,2\,4\,9)(6\,8\,12)$.

    And yes, that is the same decomposition that you have, although the numbers are in a different order.
    Ahh that means that my result is also correct, isn't it?

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    #14
    Quote Originally Posted by mathmari View Post
    Ahh that means that my result is also correct, isn't it?
    Yep.

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    #15 Thread Author
    Quote Originally Posted by Klaas van Aarsen View Post
    Yep.
    Ok!!

    As for the inverse permutations, do we reverse the cycles and also the elements of each cycle?

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

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


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    #16
    Quote Originally Posted by mathmari View Post
    Ok!!

    As for the inverse permutations, do we reverse the cycles and also the elements of each cycle?

    So do we get the following?

    - $\pi_1^{-1}=(1 \ 9 \ 4 \ 2 \ 3) \ (8 \ 6 \ 12)$
    - $\pi_2^{-1}=(14 \ 13) \ (6 \ 12 \ 1) \ (7 \ 8 \ 5 \ 4 \ 2)$
    - $\pi_3^{-1}= (11 \ 8 \ 6 \ 4 \ 5 \ 2 \ 1)$
    Yep.

    Btw, since the cycles are disjoint, their order does not matter.

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    #17 Thread Author
    Quote Originally Posted by Klaas van Aarsen View Post
    Yep.

    Btw, since the cycles are disjoint, their order does not matter.
    That means that since these are disjoint it is enough to reverse the elements inside the cycles?

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    #18
    Quote Originally Posted by mathmari View Post
    That means that since these are disjoint it is enough to reverse the elements inside the cycles?
    Yep.

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    #19 Thread Author
    Quote Originally Posted by mathmari View Post
    3. Determine $\pi_4=\pi_1\circ \pi_2$, $\pi_5=\pi_2\circ\pi_3$, $\pi_6=\pi_2\circ\pi_1$.
    Do we just write each permutation or is it maybe meant to find the cycle decomposition also of these ones?

    We have the following:

    $\pi_4=\pi_1\circ \pi_2=(12 \ 6 \ 8)\ (3 \ 2 \ 4 \ 9 \ 1)(2 \ 4\ 5 \ 8\ 7)\ (1 \ 12 \ 6) \ (13 \ 14)$

    From that we get:

    14 -> 13

    13-> 14

    6 -> 1 -> 3

    1 -> 12 -> 6

    12 -> 6 -> 8

    7 -> 2 -> 4

    2 -> 4 -> 9

    4 -> 5

    5 -> 8 -> 12

    8 -> 7

    1 -> 3

    3 -> 2

    2 -> 4

    4 -> 9

    9 -> 1

    8 -> 12

    12 -> 6

    6 -> 8

    Have I done that correctly? Because now for example the element $2$ goes to two different elements.

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    #20
    Quote Originally Posted by mathmari View Post
    Do we just write each permutation or is it maybe meant to find the cycle decomposition also of these ones?
    I suspect that when they write 'determine' that they mean to find a cycle decomposition.
    Otherwise the answer would be trivial - that is, we could simply write the given transformations after each other.

    Quote Originally Posted by mathmari View Post
    We have the following:

    $\pi_4=\pi_1\circ \pi_2=(12 \ 6 \ 8)\ (3 \ 2 \ 4 \ 9 \ 1)(2 \ 4\ 5 \ 8\ 7)\ (1 \ 12 \ 6) \ (13 \ 14)$

    From that we get:

    14 -> 13
    ...
    6 -> 8

    Have I done that correctly? Because now for example the element $2$ goes to two different elements.
    I didn't check them all, but indeed it can't be right that $2$ is in this list twice, nor that it maps to two different elements.
    Why is it in the list twice?

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