Group Theory: Proof of the cycle decomposition of the conjugate The following theorem is proven on this site:

Let $\pi$ and $\rho$ be permutations of $\{ 1,...n\}$. The cycle decomposition of $\rho \pi \rho^{-1}$ is obtained by replacing each integer $j$ in the cycle decomposition of $\pi$ with the integer $\rho (i)$.

I do not understand why $\rho(i)$ lying on the right of $\rho(\pi(i))$ implies that one can just replace every element $i$ in a cycle by $\rho(i)$ and then obtain a cycle decomposition of $\rho \pi \rho^{-1}$. 
Why is that so?

You only have to check it for cycles. Write $\pi = (\rho^{-1}(a_1),\rho^{-1}(a_2),\dots, \rho^{-1}(a_n))$. This is possible, because $\rho$ is a permutation. Now let $\rho\pi\rho^{-1}$ act on $a_i$ and you will see that indeed it is mapped to $a_{i+1}$. And that's exactly what you want.  
