Disjoint cycles commute in symmetric groups

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, $\sigma \left(a_i\right)=(b_1{b}_2\dots b_m)\left[\right(a_1{a}_2\dots a_k\left)\right(a_i\left)\right]=(b_1{b}_2\dots b_m)\left(a_{i+1}\right)=a_{i+1}$ (taking $a_{k+1}$ to be $a_1$), while $\tau \left(a_i\right)=(a_1{a}_2\dots a_k)\left[\right(b_1{b}_2\dots b_m\left)\right(a_i\left)\right]=(a_1{a}_2\dots a_k)\left(a_i\right)=a_{i+1}$, 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.