# 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!!

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.

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?

5. Originally Posted by mathmari
Ahh that means that my result is also correct, isn't it?
Yep.

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?

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

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.

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?

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

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.

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?