Does the Product of Cycles Hold for Permutations?

[ForMyThunder]

Homework Statement

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

Homework Equations

N/A

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 = (i_m1i_m2...i_mk) be a cycle and let D = (i₁i₂...i_r). Then, for m_k $$\neq$$ r,

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

Let D` = (C(i<sub>1</sub>)C(i<sub>2</sub>)...C(i<sub>r</sub>)), then i<sub>mk</sub>$$\stackrel{}{D`\rightarrow}$$i_mk+1

 

[ForMyThunder]

I accidentally made two threads. Just ignore this one.

 

[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).