If $n\geq3$, show that if $\alpha \in S_n$ commutes with every $\beta \in S_n$, then $\alpha = (1)$

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I've been stuck in this problem for a while and I don't really know how to start. I was thinking in something like this:

Suppose that $\alpha \beta=\beta \alpha$ and $\alpha\neq{(1)}$, then $\alpha, \beta$ are disjoint permutations. Suppose that exists $i,j$ such that $\alpha(i)=j$ then $\beta(i)=i$, and $\alpha(\beta(i))=j$ but $\beta(\alpha(i))=\beta(j)=j$.

I don't know what I'm missing or if I'm doing something that doesn't make sense. Any hint on how should I start would be really appreciated.

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Suppose the permutation $\alpha\ne(1)$. So for some $i$ we have $\alpha(i)=j$ with $j\ne i$.

We now consider two cases.

Case 1. $\alpha(j)=i$. Take $k\ne i$ and $k\ne j$. If we apply $(ik)$ followed by $\alpha$, then $j$ goes to $i$. But if we apply $\alpha$ followed by $(ik)$, then $j$ goes to $k$. So the transposition $(ik)$ does not commute with $\alpha$.

Case 2. $\alpha(j)=k$ where $k\ne i$ and $k\ne j$. If we apply $(jk)$ followed by $\alpha$, then $i$ goes to $j$. But if we apply $\alpha$ followed by $(jk)$, then $i$ goes to $k$. So $(jk)$ does not commute with $\alpha$.