Consider two cycles and $(b_{1} b_{2} \dots b_{m})$ in the symmetric group $S_{n}$ , where all the $a_{i}, b_{j}$ are distinct.

Then it is the case that the following two elements of $S_{n}$ are equal:

$σ$ , which is obtained by first performing the permutation notated by $(a_{1} a_{2} \dots a_{k})$ and then by performing the permutation notated by $(b_{1} b_{2} \dots b_{m})$

$τ$ , which is obtained by first performing the permutation notated by $(b_{1} b_{2} \dots b_{m})$ and then by performing the permutation notated by $(a_{1} a_{2} \dots a_{k})$

Indeed, $σ (a_{i}) = (b_{1} b_{2} \dots b_{m}) [(a_{1} a_{2} \dots a_{k}) (a_{i})] = (b_{1} b_{2} \dots b_{m}) (a_{i + 1}) = a_{i + 1}$ (taking $a_{k + 1}$ to be $a_{1}$ ), while $τ (a_{i}) = (a_{1} a_{2} \dots a_{k}) [(b_{1} b_{2} \dots b_{m}) (a_{i})] = (a_{1} a_{2} \dots a_{k}) (a_{i}) = a_{i + 1}$ , so they agree on elements of $(a_{1} a_{2} \dots a_{k})$ . Similarly they agree on elements of $(b_{1} b_{2} \dots b_{m})$ ; and they both do not move anything which is not an $a_{i}$ or a $b_{j}$ . Hence they are the same permutation: they act in the same way on all elements of ${1, 2, \dots, n}$ .

This reasoning generalises to more than two disjoint cycles, to show that disjoint cycles commute.