# Permutations cycles

• May 4th 2008, 04:50 PM
alireza1989
Permutations cycles
Hi,
I know this is an easy problem, but I just can't understand how it is done. What happens when you take a cycle (permutation in disjoint cycle notation) to the power of something. I know that (1 2 3 4)^4 = (1 2 3 4)^28 = id. But in the case of something like (1 3 5 7)^2, or (3 4 5)^2, I don't know what to do.

Thanks you
• May 4th 2008, 04:59 PM
Jhevon
Quote:

Originally Posted by alireza1989
Hi,
I know this is an easy problem, but I just can't understand how it is done. What happens when you take a cycle (permutation in disjoint cycle notation) to the power of something. I know that (1 2 3 4)^4 = (1 2 3 4)^28 = id. But in the case of something like (1 3 5 7)^2, or (3 4 5)^2, I don't know what to do.

Thanks you

just go twice as opposed to once for the cycle

example:

for (1 3 5 7)^2

1 --> 3 --> 5, so 1 --> 5
5 --> 7 --> 1, so 5 --> 1

so one cycle is (1 5)

then, 3 --> 5 --> 7, so 3--> 7
and 7 --> 1 --> 3, so 7 --> 3

so another cycle is (3 7)

thus (1 3 5 7)^2 = (1 5)(3 7)