I wonder if this questions can be done by induction.
$S_3 = \{(1),(12),(13),(23),(123),(132)\}$
$Z(S_3)$ contains all the elements in $S_3$ that commutes with all the element in $S_3$
We can easily proof the base case for $n=3$
However, I have a hard time proving the induction step.
Assume $Z(S_k) = \{1\}$ then now we need to show it works for $Z(S_{k+1})$
The thing is $S_{k+1}$ contains $(k+1)|S_{k}|$ elements which is $(k+1)k!$
But can I go anywhere from here ?
Best Answer
Here's an induction "proof". I'll explain the scare quotes at the end.
Base Case: verify $S_3$.
Induction: Assume for some $n$ than $Z(S_n) = \{e\}$. We want to show that $Z(S_{n+1}) = \{ e \} $. Let $\sigma \in S_{n+1}$ be some non-identity permutation. If $\sigma \in S_n$ then $\sigma$ is not in the center by the induction hypothesis. Therefore assume $\sigma \not\in S_n$, so $\sigma$ does not leave $n+1$ fixed (here I am identifying $S_{n+1}$ with the symmetry group of the set $\{ 1, 2, \ldots , n+1\}$). Now let $a = \sigma(n+1)$ and let $b$ be some element of $\{ 1, 2, \ldots , n+1\}$ not equal to $a$, which is possible (why can we choose $b$ like this?). Let $\tau = (ab)$. Then:
$$ \tau\sigma(n+1) = \tau(a) = b $$ $$ \sigma\tau(n+1) = \sigma(n+1) = a $$ $$ a \neq b $$
This implies $\sigma\tau \neq \tau\sigma$ (why?). Hence for any $\sigma$ we can choose some $\tau$ which does not commute with it.
This is essentially a proof which does not rely on induction, just replace $n+1$ by some element not fixed by $\sigma$, which is possible if $\sigma \neq e$.