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 $i \in \Fix \rho$.

Then:
 * $\map {\sigma \rho} i = \map \sigma i$

whereas:
 * $\map {\rho \sigma} i = \map \rho {\map \sigma i}$

$\map \sigma i \notin \Fix \rho$.

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

But it was previously established that $i \in \Fix \rho$.

This is a contradiction.

Therefore:
 * $\map \sigma i \in \Fix \rho$

and so:
 * $\map {\rho \sigma} i = \map \sigma i = \map {\sigma \rho} i$

Let $i \notin \Fix \rho$.

Then:
 * $i \in \Fix \sigma$

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