A proof about $Z(A_n) = \{ e \}, \forall n \geq 4$.. I have to demonstrate that:

Show $Z(A_n) = \{ e \}, \forall n \geq 4$.

I think I solved this problem, but I am not sure. This is my solution:

Let's say $Z(A_n) \neq \{ e \}, \forall n \geq 4$. Then I have two cases:
1) $n = 2k, k \in \Bbb Z$
2) $n = 2k+1, k \in \Bbb Z$. 
I tried only case 1), I think case 2) work the same as well.
    I choose:
  $$a =   \begin{pmatrix}
    1 & 2 & 3 & \cdots & n-2 & n-1 & n\\
    2 & 3 & 4 & \cdots &  n-1  & 1 & n
  \end{pmatrix},$$ where $a \in Z(A_n)$
  and
  $$\sigma =   \begin{pmatrix}
    1 & 2 & 3 & \cdots & i & \cdots & n-1 & n\\
    2 & 3 & 4 & \cdots & i+1 & \cdots & n & 1
  \end{pmatrix},$$ where $\sigma \in A_n.$
Then we have $a(\sigma(n-1)) = a(n) = n$ and $\sigma(a(n-1)) = \sigma(1)=2$ so $a(\sigma(1)) \neq \sigma(a(2))$, contradiction.

 A: Not sure why the other answer had a -1 as it looks correct to me, but I'll give a hint of slightly faster proof:
Suppose $1\ne\sigma\in Z(A_n)$ so there is some $i\in\{1,\ldots,n\}$ with $\sigma(i)\ne i$.
Choose some $\rho\in A_n$ with $\rho(i)=i$ and $\rho(\sigma(i))\ne\sigma(i)$. Use $(\rho\sigma\rho^{-1})(i)$ to show $\rho\sigma\rho^{-1}\ne\sigma$ completing the proof by contradiction. 
A: My solution its the next: Remember that if $\alpha, \beta \in S_{n}$ and $\beta$ has the descompocition in disjoin cyles $\beta = (x_{1}, \ldots, x_{k})(y_{1}, \ldots, y_{l}) \cdots (z_{1}, \ldots, z_{m})$ then $\alpha \beta \alpha^{-1} = (\ \alpha(x_{1}), \ldots, \alpha(x_{k})\ )(\ \alpha(y_{1}), \ldots, \alpha(y_{l})\ ) \cdots (\ \alpha((z_{1}), \ldots, \alpha(z_{m})\ )$. 
Let $\sigma \in A_{n}$, $\sigma \neq id$ with descompocition in disjoin cycles $(x_{1}, \ldots, x_{k})(y_{1}, \ldots, y_{l}) \cdots (z_{1}, \ldots, z_{m})$.
Case 1) There are two non-trivial cycles $ (x_{1}, \ldots, x_{k})$ and $ (y_{1}, \ldots, y_{l}) $ then $ x_{1}, x_{ 2}, y_{1}, y_{2} $ are different from each other. If $\tau = (x_{1},y_{1},y_{2})$ you get that:
\begin{equation*}
\tau \sigma \tau^{-1} = (y_{1}, x_{2}, x_{3}, \ldots)(y_{2}, x_{1}, y_{3}) \neq \sigma
\end{equation*}
The other case is similar and the same idea is used.
A: Here's another try.   Since $Z(A_n)\triangleleft A_n$,  and for $n\ge5$ we have $A_n$ simple, we are done in this case. 
If $n=4$, we need to do something else.  If you study $A_4$,  its only normal subgroup turns out to be $\{e, (12)(34), (13)(24), (14)(23)\}$..  But, for instance,  $(13)(14)(12)(34)=(124)\ne(132)=(12)(34)(13)(14)$.  So the center must be trivial. 
A: Select $\sigma\in Z(A_n)$, since $n\geqslant 4$ take the even 3-cycles $(1,2,3)$. Now, take $\sigma(1,2,3)\sigma^{-1}=(1,2,3)$, then for $\sigma(1)$, $\sigma(1,2,3)\sigma^{-1}$ returns $\sigma(2)$ and so $(1,2,3)$ must send $\sigma(1)$ to $\sigma(2)$ and since $(1,2,3)$ move only $1,2,3$ we conclude $\sigma(1)\in \{1,2,3\}$. Using $(1,2,4)$ instead of $(1,2,3)$ we conclude in the same way as before that $\sigma(1)\in\{1,2,4\}$, and finally using $(1,4,3)$, $\sigma(1)\in\{1,3,4\}$, so $$\sigma(1)\in\{1,2,3\}\cap\{1,2,4\}\cap\{1,3,4\}=\{1\}\rightarrow \sigma(1)=1$$ in the same way you can prove that $\sigma(i)=i$, for $i=2,\cdots,n$, thus $\sigma=e$
