2. Let $\displaystyle \alpha, \beta \in S_n$ be two disjoint cycles. Clearly the $\displaystyle j \in \{1,2,...,n\}$ which are fixed by both are fixed by the product, in which ever order it is taken. Now consider those $\displaystyle j \in \{1,2,...,n\}$ which are not fixed by both. Since $\displaystyle \beta$ and $\displaystyle \alpha$ are disjoint, for a fixed $\displaystyle j \in \{1,2,...,n\}$ which is not fixed by both, exactly one of $\displaystyle \beta$ or $\displaystyle \alpha$ moves $\displaystyle j$. So, without loss of generality, let $\displaystyle \alpha(j)=k, \beta(j)=j$ with $\displaystyle j\neq k$. Since $\displaystyle k$ is part of the cycle $\displaystyle \alpha$, it must not be part of the cycle $\displaystyle \beta$ because $\displaystyle \beta$ and $\displaystyle \alpha$ are disjoint. Then $\displaystyle \alpha\beta(j)=\alpha(j)=k$ and $\displaystyle \beta\alpha(j)=\beta(k)=k$, so both $\displaystyle \beta\alpha$ and $\displaystyle \alpha\beta$ are the same function on $\displaystyle \{1,2,...n\}$.