- Admin
- #26

- Mar 5, 2012

- 8,736

- Thread starter mathmari
- Start date

- Admin
- #26

- Mar 5, 2012

- 8,736

- Thread starter
- #27

- Apr 14, 2013

- 4,036

What do you mean? I got stuck right now.

- Admin
- #28

- Mar 5, 2012

- 8,736

You wrote (6 12 1) here.I get the following from the cycle decomposition:

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

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

And you wrote (1 12 6) 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)$

Shouldn't they be the same?

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

- Thread starter
- #29

- Apr 14, 2013

- 4,036

Oh yes, you are right! I used theinverse accidentaly.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?

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