Permutation Induces Equivalence Relation

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

Let $$\pi \in S_n$$.

Let $$\mathcal{R}_\pi$$ be the relation defined by:

$$i \mathcal{R}_\pi j \iff \exists k \in \mathbb{Z}: \pi^k \left({i}\right) = j$$

Then $$\mathcal{R}_\pi$$ is an equivalence.

Corollary
It follows that $$i \mathcal{R}_\pi j$$ if $$i$$ and $$j$$ are in the same cycle of $$\pi$$.

Proof
Let $$\pi \in S_n$$.

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

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

Checking in turn each of the critera for equivalence:

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

Therefore $$\exists k \in \mathbb{Z}: \pi^k \left({i}\right) = i$$.

Symmetric
Let $$\pi^k \left({i}\right) = j$$.

Because $$\pi^r = e$$, we have $$\pi^r \left({i}\right) = i$$ (from above).

Thus $$\pi^{k-r} \left({j}\right) = i$$.

Transitive
Let $$\pi^{s_1} \left({i}\right) = j, \pi^{s_2} \left({j}\right) = k$$.

Then $$\pi^{s_1 + s_2} \left({i}\right) = k$$.