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.

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