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.

$\blacksquare$

Sources

• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 79$
• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 79 \gamma$
• John F. Humphreys: A Course in Group Theory (1996): $\S 9$: Proposition $9.8$