Why are the orbits disjoint? I'm reading Beardon's, Algebra and Geometry:



I'm a bit confused: How can the orbits be pairwise disjoint sets? I understand that the orbits might be disjoint if we're (for example) talking about the trivial permutation:
$$\begin{pmatrix} 1 & 2 & 3 & \dots & 5 \\
1 & 2 & 3 & \dots & 5\end{pmatrix}$$
But in general, I understand that some of the orbits might be disjoint while others might not be. The problem for me is that it seems that he worded it like "they must be" and I guess that's not generally true.
 A: Suppose two orbits $O(k)$ and $O(\ell)$ intersect.  That means there are integers $m$ and $n$ such that $\rho^m(k)=\rho^n(\ell)$.  But then notice that $$\rho^s(k)=\rho^{s-m}(\rho^m(k))=\rho^{s-m}(\rho^n(\ell))=\rho^{s-m+n}(\ell)$$ for any integer $s$.  Since $O(k)=\{\rho^s(k):s\in\mathbb{Z}\}$, this shows that $O(k)\subseteq O(\ell)$.  Reversing the roles of $k$ and $\ell$, we get $O(\ell)\subseteq O(k)$ as well, so $O(k)=O(\ell)$.
A: think of actually enacting the permutation on a set of $n$ colored balls arranged in a set of cups numbered $1$ to $n$.
any balls in positions unaffected by the permutation obviously form disjoint orbits of size 1. so we need only consider derangements - permutations in which each ball changes position. 
take the first ball ($B_1$) in hand, leaving an empty cup. move the ball to its new position, taking in hand the displaced ball ($B_2$). if the permutation calls for this ball into be placed in the empty cup then $B_1$ and $B_2$ form an orbit of size 2. Otherwise place $B_2$ in its new position displacing $B_3$. continue until a ball ($B_k$) is placed in the empty cup. then $B_1, B_2,\dots,B_k$ form an orbit of size $k$. 
if any balls remain, repeat the process beginning with ball $B_{k+1}$. it should be clear that all the orbits formed in this way are disjoint.
