How to prove that $\alpha\beta =\beta\alpha$ if $\alpha$ and $\beta $ are disjoint permutations of $S_n$ symmetric group of $n$ elements. Two permutations are said to be disjoint if every elment moved by one is fixed by another permutation.
My attempt : Let any $x \in [n]$,
- if $\beta$ fixes element $x$, $\alpha\beta(x)$ it will be $\alpha(x)$ let's say $\alpha(x) = y \in [n]$, now we need to show that $\beta\alpha(x)$ is equal to $y$, $\beta(y)$ will be $y$ as $y$ will be moved by $\alpha$.
Is there better way to prove this? I will go by case by case.
If the elements that are not fixed by $\beta$ are fixed by $\alpha$ and viceversa, it is intuitively clear that it does not make any difference which we use first. To get a rigorous proof, the only tricky point is that, since we are dealing with permutations, an element is fixed if and only if its image is fixed (the "only if" direction is trivially true always, but the "if" direction requires injectivity).
We can subdivide $[n]$ in a disjoint union of the elements $A$ moved by $\alpha$, the elements $B$ moved by $\beta$ and the elements $C$ fixed by both. On $C$ they are both the identity, and so is their composition. For $x \in A$ (hence also $\alpha(x) \in A$, by the above) we have that $\beta(\alpha(x)) = \alpha(x) = \alpha(\beta(x))$. Conversely, for $y \in B$ we have $\alpha(\beta(y)) = \beta(y) = \beta(\alpha(y))$. That's more or less your proof, just with some more detail.