Disjoint cycles commute in symmetric groups

[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.