I wanted to prove that the center of the symmetric group $S_n$, $n\geq 3$ is trivial. Is my argument correct?
Suppose $\alpha\in Z(S_n)$, that is $\alpha\beta=\beta\alpha$ for all $\beta\in S_n$. We can consider, for instance, $\beta_j=(1\,2\dots j-1\,\, j+1\dots n)$ a cycle of length $n-1$ that only fixes $j$ (we know such cycle exists because $n\geq 3$).
$$\beta_j\alpha=\alpha\beta_j\Rightarrow \beta_j\alpha(j)=\alpha\beta_j(j)\Rightarrow \beta_j(\alpha(j))=\alpha(j)$$
so $\beta_j$ fixes $\alpha(j)$, but $j$ is the only letter fixed by $\beta_j$, therefore $\alpha(j)=j$.
Analogously, if we consider $\beta_i$ for any $i=1,2,3\dots,n$ we get $\alpha=id$. Hence $Z(S_n)=\{id\}$.
Thanks in advance!
