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, \(\sigma(a_i) = (b_1 b_2 \dots b_m)[(a_1 a_2 \dots a_k)(a_i)] = (b_1 b_2 \dots b_m)(a_{i+1}) = a_{i+1}\) (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.