Permutation Induces Equivalence Relation

Theorem
Let $S_n$ denote the symmetric group on $n$ letters.

Let $\pi \in S_n$.

Let $\RR_\pi$ be the relation defined by:


 * $i \mathrel {\RR_\pi} j \iff \exists k \in \Z: \map {\pi^k} i = j$

Then $\RR_\pi$ is an equivalence relation.

Proof
Let $\pi \in S_n$.

First we note that, from Element of Finite Group is of Finite Order, every element of a finite group has finite order.

Thus $\pi$ has finite order, so:
 * $\exists r \in \Z: \pi^r = e$

Checking in turn each of the criteria for equivalence:

Reflexive
From above, $\exists r \in \Z: \pi^r = e$.

Therefore $\exists k \in \Z: \map {\pi^k} i = i$.

So $\RR_\pi$ is reflexive.

Symmetric
Let $\map {\pi^k} i = j$.

Because $\pi^r = e$, we have $\map {\pi^r} j = j$ (from above).

Thus $\map {\pi^{r - k} } j = i$.

So $\RR_\pi$ is symmetric.

Transitive
Let $\map {\pi^{s_1} } i = j, \map {\pi^{s_2} } j = k$.

Then $\map {\pi^{s_1 + s_2} } i = k$.

So $\RR_\pi$ is transitive.

All criteria are met, and so $\RR_\pi$ is an equivalence relation.