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