Answer to Question #141709 in Abstract Algebra for Aaru

Question

Any two disjoint permutation commute

Answer 1

Any permutation is regarded as an element of the symmetric group $S(X)$ where $X$ is a set of elements on which we permute.Hence let $\sigma,\tau$ be two such permutations. They are disjoint means, the elements not fixed by $\sigma$ are fixed by $\tau$ and vise-versa. We need to show $\sigma\tau=\tau\sigma.$ Let $a$ be an element of the set $X$. Now if $\sigma,\tau$ both fixes $a,$ then $\sigma\tau(a)=\sigma(a)=a=\tau\sigma(a)$ . Now let $\tau$ doesn't fix it . Then by above discussion $\sigma$ has to fix it. Let $\tau(a)=b$ . Then $\tau\sigma(a)=\tau(a)=b.$ Again, $\sigma\tau(a)=\sigma(b).$ Now $\tau$ is a permutation i.e. a bijection. Hence $\tau(b)\neq\tau(a)=b.$ Hence $\tau$ doesn't fix $b.$ Hence $\sigma$ must fix $b$ or $\sigma(b)=b.$ Hence $\sigma\tau(a)=\tau\sigma(a).$ Similarly the reult holds for the case when $\sigma$ doesn't fix $a.$ Hence $\tau\sigma=\sigma\tau.$ Hence done