Exercise 4.3.8 (Center of $S_n$)

Proof. Since $S_2\simeq Z_2$ is abelian, $Z(S_2)=S_2$. We suppose now $n\ge 3$.

Suppose that $\sigma \in Z(S_n)$. Then $\mathit{𝜎𝜏}{\sigma }^{-1}=\tau$ for every $\tau \in S_n$.

Let $i$ be any element of $[[1,n]]$. Since $n\ge 3$, we can find $j,k\in [[1,n]]$ such that $i,j,k$ are distinct.

If $\tau =(ij)$, we obtain

thus

Similarly, for $\tau =(ik)$, we obtain

where $\sigma (i),\sigma (j),\sigma (k)$ are distinct.

Therefore

This shows that $\sigma (i)=i$. This is true for every $i\in [[1,n]]$, so $\sigma ={\mathrm{id}}_{[[1,n]]}=()$. Of course $()\in Z(S_n)$, thus

is the trivial subgroup. □