# 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.