Prove that $Z(S_n) =$ {$e$} I've read some proofs of that statement but I would like to receive a feedback on my own proof:
Let $x \in Z(S_n)$, then it particularly exchanges with a cycle of order n.
Now let $\phi$ be a cycle of order n, then:
$x\phi = \phi x$
Multiplying each side by $x^{-1}$ gives :
$x\phi x^{-1} = \phi$
Then, if $\phi = (a_1,a_2,a_3,....,a_n)$, then $x\phi x^{-1}=(x(a_1),x(a_2),....,x(a_n))$
That gives us that $\forall i \in [1,n]: x(a_i) = a_i$, hence $x = id$.
Is it complete? Am I missing something?
Btw, sorry for my english :)

You are missing something. Consider for example $x=\phi$ - you can't conclude that $x=id$.
What you can actually conclude is that $\phi^m$ is a cycle in $x$ for some $m\in\mathbb{Z}$.
You can use this on any two transpositions $(a,b)\ne(b,c)$ to show that $x$ fixes $a,b,c$. Alternatively you can use those transpositions to show that $x$ fixes setwise $\{a,b\}$ and $\{b,c\}$, so fixes $b$. Notice that this assumes $n\ge 3$ - this is necessary as $Z(S_2)=S_2$.
