If $\alpha,\beta \in S_{n}$, and $\alpha\beta = \beta\alpha$, then $\beta$ permutes those elements left fixed by $\alpha$. Here is my solution. Let ${a_1,...,a_k}$ represent all the integers that are permuted by $\alpha$, and let ${a_{k+1},...,a_{k+j}}$, where $k+j \leq n$, be all the elements that are left fixed by $\alpha$. Since $\alpha$ and $\beta$ are bijections, using our assumption we can write $\beta = \alpha\beta\alpha^{-1}$. For each $a_{k+i}$, where $i=1,...,j$, we have $$\beta(a_{k+i})=(\alpha\beta\alpha^{-1})(a_{k+i})=(\alpha\beta)(a_{k+i})=\alpha(a_{k+m})=a_{k+m}$$ where $m=1,...,j$. Note, $\beta$ cannot send any $a_{k+i}$ to an element permuted by $\alpha$, otherwise we would have included it in the list of all integers permuted by $\alpha$. Thus, $\beta$ permutes elements left fixed by $\alpha$.
Does this convince anyone other than myself? How could I polish up this solution?

When you write $(\alpha\beta)(a_{k+i}) = \alpha(a_{k+m})$, you seem to be assuming what you are trying to prove.
I think it's better just to write $$\beta(a_{k+i})=(\alpha\beta\alpha^{-1})(a_{k+i})=(\alpha\beta)(a_{k+i})$$ so $\beta(a_{k+i})$ is left fixed by $\alpha$.
