Disjoint cycles commute

Statement

Let $\sigma$ and $\tau$ be cycles in the symmetric group $\operatorname{Sym}(X)$ whose sets of moved points are disjoint. Then $\sigma \tau = \tau \sigma$.

Remarks

A point moved by $\sigma$ is fixed by $\tau$, and $\sigma$ maps it to another point that $\tau$ fixes. So both products send it to the same point. The same argument works for the points moved by $\tau$, and the points fixed by both are fixed by both products. Cycles that are not disjoint need not commute, for example $(1\ 2)(1\ 3) \neq (1\ 3)(1\ 2)$.

Explanations

L1 human

Two disjoint cycles move different points. Each cycle does its work without touching the points of the other cycle. So the order in which you do them does not matter.