[summary: In a symmetric group, if we are applying a collection of permutations which are each disjoint cycles, we get the same result no matter the order in which we perform the cycles.]
Consider two cycles $~$(a_1 a_2 \dots a_k)$~$ 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:
- $~$\sigma$~$, 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)$~$
- $~$\tau$~$, 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, $~$\sigma(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 $~$\tau(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.