If $\alpha$ and $\beta$ are disjoint permutations, then $\alpha \beta = \beta \alpha$. Consider the permutations on the set $I = \{1,2,3....,n\}$. If $\alpha$ and $\beta$ are disjoint permutations on $I$, then they commute with respect to function composition. 

My Proof Attempt:
Let $\alpha$ and $\beta$ be disjoint permutations. Then, $\forall k \in I: \alpha(k) = k \lor \beta(k) = k$. 
Suppose that $\alpha(k) = k$. Then:
$\forall k \in I: \alpha(\beta(k)) = \beta(k)$
$\forall k \in I: \beta(\alpha(k)) = \beta(k)$
Hence:
$\alpha \beta = \beta \alpha$
Now, suppose that $\beta(k) = k$. Then:
$\forall k \in I: \alpha(\beta(k)) = \alpha(k)$
$\forall k \in I: \beta(\alpha(k)) = \alpha(k)$
Hence:
$\alpha \beta = \beta \alpha$
This proves that they commute under function composition. 
Could someone check my proof above and see if it works or not? Otherwise, how could I improve it? How could I improve my style of presentation? Is it too untidy or unreadable?
 A: The idea of the proof below is taken from your proof. 
This presentation might not the be the best one, and you might find different style of proofs that you might appreciate more, but still hope that this example may help you :



*

*It suffices to show that :
$$\text{For each }k\in I\text{, we have }\alpha\beta(k)=\beta\alpha(k)$$
(If you want to be more clear, you may also add "Since $S_n$ acts faithfully on the set $I$")

*Since $\alpha,\beta$ are disjoint, we have the following :
$$\text{Each element in }I\text{ is fixed either by }\alpha\text{ or by }\beta.$$

*Pick $k\in I$. Interchange $\alpha,\beta$ if necessary, we may assume $k$ is fixed by $\alpha$. We have
$$\beta\alpha(k)=\beta(k)$$
If we can show that either $\alpha$ fixes $\beta(k)$ or $\beta$ fixes $k$, then we are done, since either cases implies
$$\beta(k)=\alpha\beta(k)$$

*Suppose $\alpha$ doesn't fix $\beta(k)$, then we have $\beta(\beta(k))=\beta(k)$ (since $\alpha,\beta$ are disjoint)
but since $\beta$ is a permutation, we must have from the identity $\beta(\beta(k))=\beta(k)$ that $\beta(k)=k$.
Therefore, we are done.
