Disjoint Permutations Commute

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

Let $\rho, \sigma \in S_n$ such that $\rho$ and $\sigma$ are disjoint.

Then $\rho \sigma = \sigma \rho$.

Proof
Let $\rho$ and $\sigma$ be disjoint permutations.

Let $i \in \operatorname{Fix} \left({\rho}\right)$.

Then:
 * $\sigma \rho \left({i}\right) = \sigma \left({i}\right)$

whereas:
 * $\rho \sigma \left({i}\right) = \rho \left({\sigma \left({i}\right)}\right)$

$\sigma \left({i}\right) \notin \operatorname{Fix} \left({\rho}\right)$.

Then because $\sigma$ and $\rho$ are disjoint it follows that:

But it was previously established that $i \in \operatorname {Fix} \left({\rho}\right)$.

This is a contradiction.

Therefore:
 * $\sigma \left({i}\right) \in \operatorname{Fix} \left({\rho}\right)$

and so:
 * $\rho \sigma \left({i}\right) = \sigma \left({i}\right) = \sigma \rho \left({i}\right)$

Let $i \notin \operatorname{Fix} \left({\rho}\right)$.

Then:
 * $i \in \operatorname{Fix} \left({\sigma}\right)$

and the same proof can be performed with $\rho$ and $\sigma$ exchanged.