TITLE: Find the center of the symmetry group $S_n$.
QUESTION [30 upvotes]: Find the center of the symmetry group $S_n$.

Attempt:
By definition, the center is $Z(S_n) = \{ a \in S_n : ag = ga \forall\  g \in S_n\}$. Then we know the identity $e$ is in $S_n$ since there is always the trivial permutation. Suppose $a$ is in $S_n$, but not equal to identity. Now we can imagine the permutation as bijective function that maps from $\{1,2,\dotsc,n\}$ to $\{1,2,\dotsc,n\}$.
So suppose $p$ is a permutation map. Then $p$ maps from a location $i$ to a location $j$. Take $p(i) = j$ where $i\neq j$. Let $k$ be in $\{1,2,\dotsc,n\}$, where $k$, $i$ and $j$ are all different elements. The cycle $r = (jk)$, then we will see if this commutes. 
$rp(i) = rj$  
Can someone please help me, I am stuck?  Thank you.

REPLY [2 votes]: Let $h $ be in the center. Then $gh = hg $ for all $g ∈ S_n$. Let $g \in G_x,$ the
stabilizer of $x \in X$ (we realize $S_n$ as the group of permutations on a finite set $X$ with n elements). Then $g · hx = hx$ for all $g \in G_x$. So $g \in G_{hx}$, as well. This means that $G_x \le G_{hx}$ for all $x \in  X$. Suppose $hx \neq x$ for some $x \in X$. Then every permutation of $X$ that fixes $x$ must also fix $hx$. Since $n > 3$, let $x, hx, y \in X,$ all
distinct. Define $\sigma : X \to X$ by the following:
$x \to x$
$hx \to y$
$y \to hx,$
and $\sigma$ fixes everything else. Then $\sigma$ is a permutation of $X$ that fixes $x$ but does not fix $hx$. This is a contradiction. Hence it must not be the case that $hx \ne x$ for some
$x \in X$, that is, $hx = x$ for all $x \in X$. So the center is trivial