# 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.