# Permutation and cycles

1. Dec 29, 2009

### ForMyThunder

1. The problem statement, all variables and given/known data

Let P be a permutation of a set. Show that P(i1i2...ir)B-1 = (P(i1)P(i2)...P(ir))

2. Relevant equations

N/A

3. The attempt at a solution

Since P is a permutation, it can be written as the product of cycles. So I figured that showing that the above equation holds for cycles will be sufficient to show that it holds for all permutations.

Let C = (im1im2...imk) be a cycle and let D = (i1i2...ir). Then, for mk $$\neq$$ r,

imk$$\stackrel{C^{-1}}{\rightarrow}$$imk-1$$\stackrel{D}{\rightarrow}$$imk-1+1$$\stackrel{C}{\rightarrow}$$imk+1

Let D = (C(i1)C(i2)...C(ir)), then imk$$\stackrel{}{D\rightarrow}$$imk+1
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 29, 2009

### ForMyThunder

I accidentally made two threads. Just ignore this one.

3. Dec 29, 2009

### Dick

Well, for example (P(i1)P(i2)...P(ir)) maps P(i1) to P(i2). You want to show P(i1,i2,...ir)P^(-1) does the same thing. Try it. Evaluate P(i1,i2,...ir)P^(-1)P(i1).