# Thread: Determine the cycle decomposition of the permutations Originally Posted by Klaas van Aarsen 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?  Originally Posted by Klaas van Aarsen Ah no, there is no difference.
The product of cycles is defined as the composition. They are the same. Ahh ok!!   Reply With Quote

2.

3. Originally Posted by mathmari 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?  Originally Posted by mathmari 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.   Reply With Quote Originally Posted by Klaas van Aarsen 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?   Reply With Quote

5. Originally Posted by mathmari Ahh that means that my result is also correct, isn't it?
Yep.   Reply With Quote Originally Posted by Klaas van Aarsen Yep. Ok!!

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

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

7. Originally Posted by mathmari 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.   Reply With Quote Originally Posted by Klaas van Aarsen 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?   Reply With Quote

9. Originally Posted by mathmari That means that since these are disjoint it is enough to reverse the elements inside the cycles?
Yep.   Reply With Quote Originally Posted by mathmari 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.   Reply With Quote

11. Originally Posted by mathmari 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.  Originally Posted by mathmari 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?  Reply With Quote

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