# Permutation help.

1. Feb 10, 2010

### Seb97

1. The problem statement, all variables and given/known data
need to compute this permutation in S10 (4 2 1)(5 4 9 10)(2 3 4)(7 1)(3 6)

2. Relevant equations

3. The attempt at a solution
I can compute it when i put it into 2 rows eg
1 2 3 4 5 6 7 8 9 10
4 1 3 2 5 6 7 8 9 10 thats equal to ( 4 2 1)
but doing this out with the above permutation isnt pratical. I know you have to start by the right but I not sure what to do with it. Any help would be much appreciated.

2. Feb 10, 2010

### ystael

Why isn't it practical? Do the two-row thing for each permutation you have, starting from the right one at the top, and taking the second row of each cycle as the first row of the next cycle; the last row is the second row of the two-row notation for the product. Then compute the cycle decomposition of the result. It's a perfectly sensible way to multiply permutations.

3. Feb 10, 2010

### Seb97

Hi thanks for the reply. Ah im not sure if i compute it the same way as you do.
The way I do it is take the 2 cycles at the end to begin with

1 2 3 4 5 6 7 8 9 10) (1 2 3 4 5 6 7 8 9 10
7 2 3 4 5 6 1 8 9 10) (1 2 6 4 5 3 7 8 9 10
thats what you need to write before you ever need to start computing the first cycle. Well thats what im used to but its impractical. We have another question where there are more cycles. So im guessing there is a better method. Is that the way you would do it?

4. Feb 12, 2010

### ystael

No, your computation looks something like this -- for example, let's compute (6 3 5)(5 2 1)(1 3) in $$\mathfrak{S}_6$$:

1 2 3 4 5 6
3 2 1 4 5 6 # (1 3)
3 1 5 4 2 6 # (5 2 1)
5 1 6 4 2 3 # (6 3 5)

Apply each cycle in turn, from right to left. The bottom row is your result, so the two-line form of the product (6 3 5)(5 2 1)(1 3) is:

1 2 3 4 5 6
5 1 6 4 2 3

From this we can find the cycle decomposition, (1 5 2)(3 6).

5. Feb 12, 2010

### Seb97

It was due for today but I got the right answer the way I did it I checked with a few people and they all have done it your way. Altough my way is correct its a bit impractical.
In the first part you have below. I understand all the stuff you have written the only bit I cant get is how you turned the four rows into the cycle.
1 2 3 4 5 6
5 1 6 4 2 3