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.


 * If $$i \in \operatorname{Fix} \left({\rho}\right)$$, then $$\sigma \rho \left({i}\right) = \sigma \left({i}\right)$$, whereas $$\rho \sigma \left({i}\right) = \rho \left({\sigma \left({i}\right)}\right)$$.

Since $$\rho$$ and $$\sigma$$ are disjoint, then $$\sigma \left({i}\right) \in \operatorname{Fix} \left({\rho}\right) \implies \rho \sigma \left({i}\right) = \sigma \left({i}\right) = \sigma \rho \left({i}\right)$$.


 * If $$i \notin \operatorname{Fix} \left({\rho}\right)$$, then that means $$i \in \operatorname{Fix} \left({\sigma}\right)$$, and the same proof can be performed with $$\rho$$ and $$\sigma$$ exchanged.