Show that disjoint cycles commute.
Hint
Each cycle maps its own support to itself and fixes every point outside it.
Solution
1. Let \(\sigma\) and \(\tau\) be cycles with disjoint supports \(A = \supp(\sigma)\) and \(B = \supp(\tau)\), so \(A \cap B = \emptyset\), and let \(x\) be any point.
2. \(\sigma\) maps \(A\) to itself, and \(\tau\) maps \(B\) to itself.
If \(a \in A\) then \(\sigma(a) \neq a\), so \(\sigma^{-1}(\sigma(a)) = a \neq \sigma(a)\), which means \(\sigma(a)\) is not fixed by \(\sigma^{-1}\). But \(\sigma\) and \(\sigma^{-1}\) have the same support (a point is moved by \(\sigma\) iff it is moved by \(\sigma^{-1}\)), so \(\sigma(a) \in A\).
3. If \(x \in A\) then \((\sigma\tau)(x) = (\tau\sigma)(x)\).
3.1. \(x \notin B\), so \(\tau(x) = x\) and \((\sigma\tau)(x) = \sigma(x)\).
3.2. \(\sigma(x) \in A\) by step 2, so \(\sigma(x) \notin B\) and \(\tau(\sigma(x)) = \sigma(x)\).
3.3. Both composites give \(\sigma(x)\).
4. If \(x \in B\) then \((\sigma\tau)(x) = (\tau\sigma)(x)\).
Exchange the roles of \(\sigma\) and \(\tau\) in step 3.
5. If \(x \notin A \cup B\) then both composites fix \(x\).
\(\sigma\) and \(\tau\) each fix every point outside their own support.
6. Q.E.D.
Steps 3 through 5 cover every point, so \(\sigma\tau = \tau\sigma\).