# Determine the cycle decomposition of the permutations

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• February 10th, 2020, 01:41
mathmari
Re: Determine the cycle decomposition of the permutations
Quote:

Originally Posted by Klaas van Aarsen
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. (Worried)
Why is it in the list twice?

We start from the right-most cycle, right? Then we have the element $2$ and see that this goes to $4$ and $4$ goes to $9$, so $2$ goes to $9$.
At the second cycle we have again the element $2$. When we reach this cycle do we check again the element $2$ ? (Wondering)
• February 10th, 2020, 04:58
Klaas van Aarsen
Re: Determine the cycle decomposition of the permutations
Quote:

Originally Posted by mathmari
We start from the right-most cycle, right? Then we have the element $2$ and see that this goes to $4$ and $4$ goes to $9$, so $2$ goes to $9$.
At the second cycle we have again the element $2$. When we reach this cycle do we check again the element $2$ ?

No. (Shake)
When we consider the image of an element such as $2$, we must evaluate all cycles from right to left. We are not supposed to start in the middle. (Worried)
• February 10th, 2020, 10:45
mathmari
Re: Determine the cycle decomposition of the permutations
Ok !!

Now I think I got the right results :

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

$\pi_5=\pi_2\circ\pi_3=(1\ 4) \ (2\ 8\ 11\ 12\ 6\ 7)\ (13\ 14)$

$\pi_6=\pi_2\circ\pi_1=(3 \ 4 \ 9 \ 12\ 1) \ (2\ 5\ 8\ 6\ 7)\ (13 \ 14)$

For the signum we have to write each permutation as a product of cycles of two elements, right? How can we do that? (Wondering)
• February 10th, 2020, 13:02
Klaas van Aarsen
Re: Determine the cycle decomposition of the permutations
Quote:

Originally Posted by mathmari
For the signum we have to write each permutation as a product of cycles of two elements, right? How can we do that?

We can write $(a\,b\,c\,d)$ for instance as either $(a\,d)(a\,c)(a\,b)$ or $(a\,b)(b\,c)(c\,d)$. (Thinking)

Note that more generally a cycle with an even number of elements can be written as an odd number of 2-cycles (signum -1).
And a cycle with an odd number of elements can be written as an even number of 2-cycles (signum +1). (Nerd)
• February 20th, 2020, 02:13
mathmari
Re: Determine the cycle decomposition of the permutations
I am looking aagain at the previous question, at the calculation of $\pi_4$. Do we get the same results if we use the initially given $\pi_1$ and $\pi_2$ and the same if we use the cycle decomposition of them? (Wondering)

I get the following from the cycle decomposition:

$\pi_1=(8 \ 6 \ 12)\ (1 \ 9 \ 4 \ 2 \ 3 )$

$\pi_2=(7 \ 8 \ 5 \ 4 \ 2)\ (6 \ 12 \ 1 ) \ (14 \ 13)$

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

$14\rightarrow 13$

$13\rightarrow 14$

$6\rightarrow 12\rightarrow 8$

$12\rightarrow 1\rightarrow 9$

$1\rightarrow 6\rightarrow 12$

$7\rightarrow 8\rightarrow 6$

$8\rightarrow 5$

$5\rightarrow 4\rightarrow 2$

$4\rightarrow 2\rightarrow 3$

$2\rightarrow 7$

$9\rightarrow 4$

$3\rightarrow 1$

So $\pi_4=(14 \ 13) \ (6 \ 8 \ 5 \ 2 \ 7) \ (12 \ 9 \ 4 \ 3 \ 1)$.

And from the initially given forms:

$\pi_1=(1 \ 2\ 4 \ 9)\ (1\ 3)\ (6 \ 8 \ 12)$

$\pi_2=(2 \ 4\ 5 \ 8 \ 7)\ ( 1 \ 12 \ 6) \ (13 \ 14)$

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

$13 \rightarrow 14$

$14\rightarrow 13$

$1\rightarrow 12 \rightarrow 6$

$12\rightarrow 6\rightarrow 8$

$6\rightarrow 1 \rightarrow 3$

$2\rightarrow 4\rightarrow 9$

$4\rightarrow 5$

$5\rightarrow 8\rightarrow 12$

$8\rightarrow 7$

$7\rightarrow 2\rightarrow 4$

$3\rightarrow 1\rightarrow 2$

$9\rightarrow 1$

So $\pi_4=(13 \ 14) \ (1 \ 6 \ 3 \ 2 \ 9 ) \ (12 \ 8 \ 7 \ 4 \ 5)$.

Have I done something wrong or is it correct that we get different results? (Wondering)
• February 20th, 2020, 05:37
Klaas van Aarsen
Re: Determine the cycle decomposition of the permutations
The original $\pi_2$ was already disjoint and contained (1 12 6). But it's not like that in both your cases, is it? (Worried)
• February 20th, 2020, 19:24
mathmari
Re: Determine the cycle decomposition of the permutations
Quote:

Originally Posted by Klaas van Aarsen
The original $\pi_2$ was already disjoint and contained (1 12 6). But it's not like that in both your cases, is it? (Worried)

What do you mean? I got stuck right now. (Wondering)
• February 20th, 2020, 20:18
Klaas van Aarsen
Re: Determine the cycle decomposition of the permutations
Quote:

Originally Posted by mathmari
I get the following from the cycle decomposition:

$\pi_1=(8 \ 6 \ 12)\ (1 \ 9 \ 4 \ 2 \ 3 )$

$\pi_2=(7 \ 8 \ 5 \ 4 \ 2)\ (6 \ 12 \ 1 ) \ (14 \ 13)$

You wrote (6 12 1) here.

Quote:

Originally Posted by mathmari
And from the initially given forms:

$\pi_1=(1 \ 2\ 4 \ 9)\ (1\ 3)\ (6 \ 8 \ 12)$

$\pi_2=(2 \ 4\ 5 \ 8 \ 7)\ ( 1 \ 12 \ 6) \ (13 \ 1 4)$

And you wrote (1 12 6) here.
Shouldn't they be the same? (Wondering)

Did you perhaps accidentally write $\pi_2^{-1}$ instead of the decomposition? (Wondering)
• February 21st, 2020, 01:43
mathmari
Re: Determine the cycle decomposition of the permutations
Quote:

Originally Posted by Klaas van Aarsen
You wrote (6 12 1) here.

And you wrote (1 12 6) here.
Shouldn't they be the same? (Wondering)

Did you perhaps accidentally write $\pi_2^{-1}$ instead of the decomposition? (Wondering)

Oh yes, you are right! I used theinverse accidentaly. (Tmi)

So we get the same result using the initial permutation and the cycle decomposition of it.
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