Question: I'm going to type down here the answer I found in a manual solutions. However I have a question about it. And I hope I can get help here on this site. Well, I saw that there are several other issues that were closed because they were classified as duplication. Although there may be different doubts on the same question.

**P.S. My doubts are in ${\color{red}{\text{Color red}}}.$**

**RIGHT ANSWER - MANUAL SOLUTION**

Let $g$ be a element in the center of $Z(S\_n)$. Suppose by contradiction that $g\neq 1$ and let be $i\in \{1, \cdots, n\}$ such that $g(i)=j\neq i$. Because of $n\geq 3$ there is $k\in \{1, \cdots, n\}$ where $k\neq i$ and $k \neq j$. By hypothesis, the 3-cycle $x=(ijk)$ comutes with $g$. Then $g(ijk)=(ijk)g$. Applying to $i$ we obtain that $g(j)=k$. On the other hand, $g(ij)=(ij)g$ too, and applying to $i$ we find $g(j)=i$. Contradiction.

${\color{red}{\text{I can't understand why to prove that $g(ijk)$ is not always equal to $ (ijk) g $ it was necessary to use}}}$

${\color{red}{\text{data obtained from $g(ij) = (ij)g$? Since I have $ x = (ijk) \neq (ij) $.}}}$

${\color{red}{\text{ This answer does not make sense to me.}}}$

Answer: 
This comes down to understanding the operation of conjugation in $S\_n$. If $g$ is a permutation that maps $i$ to $j$ then $h g h^{-1}$ maps $h(i)$ to $h(j)$:

$$ (hgh^{-1})(h(i)) = (hg)(i) = h(j).$$

So what you want is a permutation such that $h(i) = i$ and $h(j) \ne j$. Then $$(hgh^{-1})(i) = h(j) \ne g(i) = j$$ so $hgh^{-1} \ne g$. So the simple solution is to take $h = (jk)$.

That's the simple solution. I don't know what the person writing your solution was trying to do. Maybe they thought that "$i$" should appear in $h$.
For any thing other than the identity, there's something it doesn't commute with. Take $\sigma\in S\_n$. If $\sigma(i)\ne i$, take $(ik)$, where $k\ne \sigma(i),i$. (You can do this because $n\ge3$.) Then $\sigma(ik)\ne(ik)\sigma$. Thus $\sigma\notin Z(S\_n)$.