# Thread: Permutations

1. ## Permutations

Im not sure how to answer the follwing question on permutations. Help would be very much appreciated. Thank you

2. How about f = (1 2 3)(4 5)?

3. how would that be written as a matrix?

4. Permutations are NOT normally written as matrices. (1 2 3)(4 5) is the matrix that maps 1 to 2, 2 to 3, then 3 back to 1, 4 to 5, and 5 back to 4. It can also be written as $\displaystyle \begin{pmatrix}1 & 2 & 3 & 4 & 5 \\3 & 1 & 2 & 5 & 4\end{pmatrix}$.

If you insist upon writing it as a matrix, it would be
$\displaystyle \begin{bmatrix}0 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0\end{bmatrix}$

Do you see why?

5. lool, I meant the first thing you drew which looks similar to a matrix. But back to the question, how would you know that is the correct permutation?

6. Originally Posted by iPod
how would you know that is the correct permutation?
Do the calculations.

7. Ive tried , however I don't know how to calculate an answer...I hoped somone could show me a method of working this out
(unless you mean trial and error of a number of permutations, since that would take a very long time since there a 5!=120 different possibilites.)

8. I know that ackbeets answer is correct since Ive used it, but i want to know how he got that answer..

9. Well, you needed something with periodicity 6 and not less than that. The simple permutation (1 2 3 4 5) wouldn't work, because it would have periodicity 5. So I thought to myself 3 x 2 = 6. If I can dream up a permutation that is a combination of one that is period 2 and one that is period 3, then the first time the 3-period permutation came around, the other permutation would not have arrived back at the identity, because 2 and 3 are relatively prime. So I'd have to go back and do the 3-period permutation again. Then, after the second time through, the 2-period permutation would be back to the starting-place. And that's how I arrived at (1 2 3)(1 2).

Make sense?

10. so you split a the set {1 2 3 4 5} into 2 seperate parts (1 2 3) & (4 5) and each would have its own periodicity, 3 and 2 - thus they have a periodicity of 6 when concidered together. The 2-cycle permutation would already be correct since we are dealing with 'functioning it 6 times' which is an even number, and thus would end up to the initial number (ie; 4 -> 5 -> 4), So we just concentrate on the 3-cycle permutation. So from then on it kinda seems like trial and error until you find the correct permutation, concidering the fact that each number (1 2 3) leads to a distinct number. So we have narrowed our choices down to (1 2 3)(4 5), (1 3 2)(4 5), (2 1 3)(4 5), (3 2 1)(4 5), (3 1 2)(4 5) etc etc and keep going if you find the right one?

Please do correct me if I am incorrect.

11. (1 2 3) = (3 1 2) = (2 3 1), and

(1 3 2) = (2 1 3) = (3 2 1).

And, in fact, ANY of these would work, because they're all periodicity 3, which is really the only feature that you require here. Also, (4 5) = (5 4).

12. Oh yes of course!
It seems to be much clearer to me now, I may be able to handle questions like this in the future on my own now concidering your approach.
Thank you v much

13. You're welcome. Have a good one!