Conjugation of Cycles in Permutation Groups: Proving the Property with Examples

[gtfitzpatrick]

Homework Statement

show that if $(x_1 x_2 ... x_k)$ is a cycle in $S_n$ ( $k \leq n$ ) and $\pi$ is any permutation in $S_n$ then $\pi (x_1 x_2 ... x_k) \pi ^{-1} = ( \pi(x_1) \pi(x_2) ... \pi(x_k) )$

Homework Equations

The Attempt at a Solution

firstly is this question right?

i multiplied both sides by $\pi^{-1}$ and get

$$(x_1 x_2 ...x_k)\pi^{-1} = \pi^{-1} (\pi(x_1) \pi(x_2) ...\pi(x_k))$$

$$= \pi^{-1}\pi(x_1) \pi^{-1}\pi(x_2) ...\pi^{-1}\pi*x_k))$$
$$=(x_1 x_2 ... x_k)$$
which obviously isn't right?

 

[Billy Bob]

The question is right.

You are making an error when you say $$\pi^{-1}(\pi(x_1),\pi(x_2),\dots,\pi(x_k))=(\pi^{-1}\pi(x_1),\pi^{-1}\pi(x_2),\dots,\pi^{-1}\pi(x_k))$$.

Try an example. $$\pi=(1 2 3),\sigma=(x_1,x_2,x_3,x_4,x_5)=(8 2 4 3 5), \pi^{-1}=(3 2 1)$$.