## Sym(n) is centerless if n is at least 3

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

First we prove a lemma.

Lemma: Let $n \geq 3$. If $\sigma \in S_n$ is not the identity, then there exists $\tau \in S_n$ such that $\sigma\tau \neq \tau\sigma$. Proof: Since $\sigma \neq 1$, there exist $\alpha,\beta \in \{1,2,\ldots,n \}$ such that $\sigma(\alpha) = \beta$. Since $n \geq 3$, there exists an element $\gamma \in \{1,2,\ldots,n\}$ distinct from $\alpha$ and $\beta$. Let $\tau = (\beta\ \gamma)$. Note that $(\sigma \circ \tau)(\alpha) = \sigma(\tau(\alpha)) = \sigma(\alpha) = \beta$ while $(\tau \circ \sigma)(\alpha) = \tau(\sigma(\alpha)) = \tau(\beta) = \gamma$. Hence $\sigma\tau \neq \tau\sigma$. $\square$

Thus, if $\sigma \in S_n$ is not the identity, $\sigma \notin Z(G)$. Hence $Z(G) = 1$.