
MHB Master
#1
February 9th, 2020,
08:16
Hey!!
We have the following permutations in $\text{Sym}(14)$ :
 $\pi_1=(1 \ 2\ 4 \ 9)\circ(1 \ 3)\circ (6 \ 8\ 12)$
 $\pi_2=(2 \ 4\ 5 \ 8\ 7)\circ (1 \ 12 \ 6)\circ \ (13 \ 14)$
 $\pi_3=(1 \ 4 \ 5\ 8 \ 11)\circ (2 \ 4\ 6 \ 5 \ 1)$
1. Determine the cycle decomposition of $\pi_1, \pi_2, \pi_3$.
2. Determine $\pi_1^{1}, \pi_2^{1}, \pi_3^{1}$.
3. Determine $\pi_4=\pi_1\circ \pi_2$, $\pi_5=\pi_2\circ\pi_3$, $\pi_6=\pi_2\circ\pi_1$.
4. Determine the signum of $\pi_1, \pi_2, \pi_3, \pi_4, \pi_5, \pi_6$.
1. We consider the composition fromright to left, or not? Thenfrom the last we consider the element $6$ that goes $8$ and since there is no other $8$ previously we have that $6\rightarrow 8$.
Is this the idea to get the cycles?
2. We get the inverse permutation by the cycle decompsition, right?
Last edited by mathmari; February 9th, 2020 at 11:21.

February 9th, 2020 08:16
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MHB Seeker
#2
February 9th, 2020,
08:38
Originally Posted by
mathmari
1. We consider the composition fromright to left, or not? Thenfrom the last we consider the element $6$ that goes $8$ and since there is no other $8$ previously we have that $6\rightarrow 8$.
Is this the idea to get the cycles?
Hey mathmari!!
Yes, this is the way to decompose the permutation into disjoint cycles.
Originally Posted by
mathmari
2. We get the inverse permutation by the cycle decompsition, right?
We can choose.
We get the inverse by reversing each cycle in combination with applying the constituent cycles in reverse.
We can do this with the permutation as given, or with the decomposition.
We can for instance already write:
$$\pi_1^{1}=(12 \ 8\ 6)\circ(3 \ 1)\circ(9 \ 4\ 2 \ 1) $$
It may be intended to do this with the decomposition of $\pi_1$ though.

MHB Master
#3
February 9th, 2020,
11:27
Thread Author

MHB Seeker
#4
February 9th, 2020,
11:36
Originally Posted by
mathmari
Aren't the first two permutations already the decomposition into disjoint cycles?
Yes.
But the third permutation could still conceivably overlap both of them, so that they might not actually end up as disjoint cycles in the decomposition.
Originally Posted by
mathmari
As for $\pi_3$ we have the following:
$1$ goes to $2$
$5$ goes to $1$ and $1$ goes to $4$, so $5$ goes to $4$
$6$ goes to $5$ and $5$ goes to $8$, so $6$ goes to $8$
$4$ goes to $6$
$2$ goes to $4$ and $4$ goes to $5$, so $2$ goes to $5$
$11$ goes to $1$
$8$ goes to $11$
Is this correct?
Yep.
Can we write it as disjoint cycles?

MHB Master
#5
February 9th, 2020,
11:41
Thread Author
Originally Posted by
Klaas van Aarsen
Yep.
Can we write it as disjoint cycles?
So we get $\pi_3=(1 \ 2 \ 5 \ 4 \ 6 \ 8 \ 11)$, or not?

MHB Seeker
#6
February 9th, 2020,
11:49
Originally Posted by
mathmari
So we get $\pi_3=(1 \ 2 \ 5 \ 4 \ 6 \ 8 \ 11)$, or not?
Yep.

MHB Master
#7
February 9th, 2020,
14:05
Thread Author

MHB Seeker
#8
February 9th, 2020,
14:12
Originally Posted by
mathmari
Therefore, the cycle decompositions of the permutations are:
 $\pi_1=(1 \ 2\ 4 \ 9) \ (1 \ 3)\ (6 \ 8\ 12)$
 $\pi_2=(2 \ 4\ 5 \ 8\ 7)\ (1 \ 12 \ 6) \ (13 \ 14)$
 $\pi_3=(1 \ 2 \ 5 \ 4 \ 6 \ 8 \ 11)$
Right?
The cycles in $\pi_1$ are not disjoint are they?
Isn't $1$ in multiple cycles?
Originally Posted by
mathmari
At the first two do we consider the product of the cycles or the composition as we had initially?
Huh?
Which 'first two'?
Which question are we talking about?

MHB Master
#9
February 9th, 2020,
15:21
Thread Author
Originally Posted by
Klaas van Aarsen
The cycles in $\pi_1$ are not disjoint are they?
Isn't $1$ in multiple cycles?
Oh yes, you are right!
So, the cycle decompositions of the permutations are:
 $\pi_1=(12 \ 6 \ 8)\ (3 \ 2)\ (9 \ 1 \ 2 \ 4)$
 $\pi_2=(2 \ 4\ 5 \ 8\ 7)\ (1 \ 12 \ 6) \ (13 \ 14)$
 $\pi_3=(1 \ 2 \ 5 \ 4 \ 6 \ 8 \ 11)$
Right?
Originally Posted by
Klaas van Aarsen
Huh?
Which 'first two'?
Which question are we talking about?
My question is: Is there a difference between $(12 \ 6 \ 8)\ (3 \ 2)$ and $(12 \ 6 \ 8)\circ (3 \ 2)$ ? Is the product and the composition symbol equivalent?

MHB Seeker
#10
February 9th, 2020,
15:30
Originally Posted by
mathmari
Oh yes, you are right!
So, the cycle decompositions of the permutations are:
 $\pi_1=(12 \ 6 \ 8)\ (3 \ 2)\ (9 \ 1 \ 2 \ 4)$
Doesn't $\pi_1$ have two cycles with $2$ in them now?
Originally Posted by
mathmari
My question is: Is there a difference between $(12 \ 6 \ 8)\ (3 \ 2)$ and $(12 \ 6 \ 8)\circ (3 \ 2)$ ? Is the product and the composition symbol equivalent?
Ah no, there is no difference.
The product of cycles is defined as the composition. They are the same.