# Determine the cycle decomposition of the permutations

#### Klaas van Aarsen

##### MHB Seeker
Staff member
The original $\pi_2$ was already disjoint and contained (1 12 6). But it's not like that in both your cases, is it?

#### mathmari

##### Well-known member
MHB Site Helper
The original $\pi_2$ was already disjoint and contained (1 12 6). But it's not like that in both your cases, is it?
What do you mean? I got stuck right now.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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.

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?

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

#### mathmari

##### Well-known member
MHB Site Helper
You wrote (6 12 1) here.

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

Did you perhaps accidentally write $\pi_2^{-1}$ instead of the decomposition?
Oh yes, you are right! I used theinverse accidentaly.

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