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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 :)

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    See [this question](http://math.stackexchange.com/questions/719915/find-the-center-of-the-symmetry-group-s-n). – 2017-02-26
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    Possible duplicate of [Find the center of the symmetry group $S_n$.](https://math.stackexchange.com/questions/719915/find-the-center-of-the-symmetry-group-s-n) – 2017-10-19
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    Since this was a proof verification Question (in spirit, though not originally so tagged) and since the posted Answer points out a legitimate gap in the proposed proof, I'm voting to leave open. – 2017-10-19

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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 $Z(S_2)=S_2$.

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    Thanks, I just realized that the order of the elements in the original cycle doesn't have to be exactly the same, then there could be many options for x's. :) – 2017-02-26