Prove that $Z(S_n)=\{1\}$ for all $n\geq 3$.

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 1

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 $h$.

Answer 2

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 $\sigma\notin Z(S_n)$.