# Find the center of the symmetry group $S_n$.

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. $rp(i) = rj$

• (Note that n=2 will be a special case.) $\;$ – user57159 Mar 20 '14 at 15:09
• @RickyDemer, also $n = 0$ and $n = 1$. :-) – LSpice Aug 29 '15 at 18:56
• @LSpice : $\;\;\;$ How so? $\:$ It seems like in those cases, the center will still be the trivial subgroup. $\hspace{.65 in}$ – user57159 Aug 29 '15 at 19:11
• @RickyDemer, yes, you are right! I seem to have been thinking of the much weaker statement "$S_n$ is not Abelian unless $n \le 2$." – LSpice Aug 30 '15 at 15:54

If $n=2$; $S_2$ is cyclic of order $2$, so it is abelian and $Z(S_2)=S_2$.

Suppose $n>2$. If $\sigma \in S_n$ is not the identity, then it moves at least one letter $i$, say $\sigma(i)=k$ and since $i\neq k$, it also moves $k$, say $\sigma(k)=j$. Can you produce a permutation (a simple one, don't think too hard) that doesn't commute with $\sigma$?

Spoiler For example, say $\sigma(i)=k$ and $\sigma(k)=i$; (so $i=j$), and $\sigma$ is of the form $\sigma=(ij)\tau$, with $\tau(ij)=\tau(ji)$. Note that $\tau$ fixes $i,j$, and cannot map something to $i$ or $j$. Then consider $(i\ell)$, a transposition. Then $\sigma(i\ell)$ doesn't map $i$ to $j$: if $\tau$ moves $\ell$, it moves $i$ to something different from $j$; and if $\tau$ doesn't move $\ell$, $i\to\ell$ -- but $(i\ell)\sigma$ maps $i\to j$, so $(i\ell)\sigma\neq \sigma(i\ell)$.

It remains you see what happens when $i\neq j$; but it shouldn't be too hard either.

Here is a different proof based on the fact that the center of any group is precisely the set of elements whose conjugacy classes are singletons.

For $S_n$, the conjugacy classes are in bijection with partitions of $n$ (since conj. classes are determined by cycle type.) We'll focus on $n\geq 3$ since $S_2$ is abelian. So fix an arbitrary non-increasing partition $\lambda_1,\cdots,\lambda_k$ of $n$ whose largest part is $>1$ (otherwise it's the conj. class of the identity).

The conjugacy class of $\sigma = (1\cdots\lambda_1)(\lambda_1{+}1\ \cdots\ \lambda_1{+}\lambda_2)\cdots((\sum_{i=1}^{k-1}\lambda_i){+}1\ \cdots\ n)$ has an element distinct from $\sigma$ since we can switch 1 & 3 (remember $\lambda_1>1$ and $n\geq 3$). Thus there are no nontrivial conjugacy classes with only 1 element for $n\geq 3$, so the center of $S_n$, $n\geq 3$, is trivial.