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Determine the cycle decomposition of the permutations

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,711
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)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,028
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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,711
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? (Wondering)

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

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,028
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.