
MHB Master
#21
February 10th, 2020,
01:41
Thread Author
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.
Why is it in the list twice?
We start from the rightmost 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 $2$ ?

February 10th, 2020 01:41
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MHB Seeker
#22
February 10th, 2020,
04:58
Originally Posted by
mathmari
We start from the rightmost 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.
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.

MHB Master
#23
February 10th, 2020,
10:45
Thread Author
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?

MHB Seeker
#24
February 10th, 2020,
13:02
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 $(a\,b)(b\,c)(c\,d)$.
Note that more generally a cycle with an even number of elements can be written as an odd number of 2cycles (signum 1).
And a cycle with an odd number of elements can be written as an even number of 2cycles (signum +1).