> [!NOTE] Theorem (Disjoint Permuations commute) > Let $S_{n}$ be the [[Symmetric Groups of Finite Degree|nth symmetric group]]. Let $\alpha,\beta\in S_{n}$ be disjoint [[Permutation of Finite Degree|permutations]]. Then $\alpha \beta=\beta\alpha.$ > Proof. First show that disjoint cycles commute. Let $\alpha = (a_{1},\dots,a_{k}), \quad \beta=(b_{1},b_{2},\dots,b_{l})$be [[Cyclic Permutation of n Letters|cycles]] such that $a_{i}\neq b_{j}$ for $i=1,2,\dots,k$ and $j=1,2,\dots,l.$ Consider the following cases: Suppose $x$ does not equal any of the $a_{i}$ or $b_{j}$. So $\beta (x) = x$ and $\sigma( x) = x$. Hence $(\alpha\beta)(x)=\alpha(\beta(x))=\alpha(x) = x = \beta(x)=\beta (\alpha(x)) = (\beta\alpha)(x)$ Now WLOG suppose $x = a_{i}$ for some $1 \leq i \leq k$. Thus since $\beta$ fixes all $a_{i},$ $x$ does not equal any of the $b_{j}$ so $\beta (a_{i}) =a_{i}.$ Hence $(\alpha\beta)(a_{i})=\alpha(\beta(a_{i}))=\alpha(a_{i})=a_{i+1}$Similarly $(\beta\alpha)(a_{i})=\beta(\alpha(a_{i}))=\beta(a_{i+1})=a_{i+1}.$Hence $(\alpha\beta)(x)=(\beta \alpha)(x).$ Then by [[Existence of Disjoint Cycle Decomposition for Permutations of n Letters]], disjoint permutations commute.