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

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- Oct 23rd 2010, 02:46 PMiPodPermutations
Im not sure how to answer the follwing question on permutations. Help would be very much appreciated. Thank you

- Oct 23rd 2010, 05:53 PMAckbeet
How about f = (1 2 3)(4 5)?

- Oct 24th 2010, 02:09 PMiPod
how would that be written as a matrix?

- Oct 24th 2010, 03:04 PMHallsofIvy
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? - Oct 24th 2010, 03:15 PMiPod
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?

- Oct 24th 2010, 03:20 PMPlato
- Oct 24th 2010, 03:23 PMiPod
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.) - Oct 24th 2010, 03:45 PMiPod
I know that ackbeets answer is correct since Ive used it, but i want to know how he got that answer..

- Oct 25th 2010, 01:44 AMAckbeet
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? - Oct 25th 2010, 02:09 AMiPod
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. - Oct 25th 2010, 02:15 AMAckbeet
(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). - Oct 25th 2010, 02:18 AMiPod
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 - Oct 25th 2010, 02:22 AMAckbeet
You're welcome. Have a good one!