How can I simplify this number theory problem? Let X = {1, 2, 3, 4, 5, 6} and σ=
\begin{bmatrix}
    1 & 2 & 3 & 4  & 5 & 6 \\
    6 & 4 & 3 & 5  & 2 & 1 \\
\end{bmatrix}
   Define a relation
∼ on X as follows: for x, y ∈ X, x ∼ y if there is a natural number k such that
$σ^k(x) = y$. Prove this is an equivalence relation and determine the equivalence
classes. Here $σ^k$ is the composition of σ with itself k times: $σ^2 = σ ◦ σ, σ^3 =σ ◦σ^2$, ...
For the transitive relation, I don't want to run through every pair catching transitivities, so I am just wondering if there is a nice a way in which I can show that the relation is transitive.
 A: Define $x \sim y$ if some power of $\sigma$ takes $x$ to $y$.  Let's show the relation is symmetric.  Suppose $x \sim y$. Then some power $\sigma^k$ of $\sigma$ takes $x$ to $y$, and so $\sigma^{-k} = \sigma^{-k+6}$ is a power of $\sigma$ takes $y$ to $x$. I've used the fact that $\sigma=(1,6)(2,4,5)(3)$ has order 6.  To show $\sigma$ is transitive, suppose $x \sim y$ and $y \sim z$. Then some power $\sigma^k$ takes $x$ to $y$ and some power $\sigma^\ell$ takes $y$ to $z$, and so $\sigma^{k+\ell}$ takes $x$ to $z$.  Thus, $x \sim z$.  The zeroth power of $\sigma$ takes each point $x$ to itself, and so the relation is also reflexive. This proves that the relation is an equivalence relation.  
Since every equivalence relation gives rise to a partition, the relation gives rise to a partition of $\{1,\ldots,6\}$.  The parts in this partition are called the orbits of the permutation group $\langle \sigma \rangle$. In this example, there are three orbits, namely $\{1,6\}, \{2,4,5\}$ and $\{3\}$.
A: So let's say $x \sim y$ and $y \sim z$.
Then, for some $k$ and $m$, $y=\sigma^k(x)$ and $z=\sigma^m(y)$.
Then, $$ z=\sigma^m(y)=\sigma^m(\sigma^k(x))=\sigma^{m+k}(x)$$. Thus $x \sim z$.
