Oh, sorry - I missread your first question. To solve your question, you need to think about how long cycles square. Basically, they will either square to give another long cycle (if they have odd length, say 5) or split in two and give two m-cycles, of they have length 2m (e.g.

). Can you generalise? What happens when you take a long cycle of length

to a power of

where

and when

? Basically, you should see that the power of a long cycle will always consist of a number of short cycles

*all of the same length*. The permutation you are looking at is a product of a two-cycle and a three-cycle, so this can't happen.

. It means conjugating the cycle by sigma. Sorry - it is standard, I just forget that not everyone knows such things...