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.

Parents:

  • Symmetric group

    The sym­met­ric groups form the fun­da­men­tal link be­tween group the­ory and the no­tion of sym­me­try.