Prove that the center of the alternating group \(A_n\) is trivial for all \(n \ge 4\): \[Z(A_n) = \{e\} \quad \text{for all } n \ge 4.\]
Solution
Let \(\sigma\in Z(A_n)\) with \(n\ge4\). For every \(3\)-cycle \((i\,j\,k)\in A_n\), centrality gives \[ \sigma(i\,j\,k)\sigma^{-1}=(i\,j\,k). \] But conjugation acts on a cycle by applying \(\sigma\) to its entries, so \[ (\sigma(i)\,\sigma(j)\,\sigma(k))=(i\,j\,k). \] In particular, \(\sigma\) preserves the underlying set \(\{i,j,k\}\) of every \(3\)-subset of \(\{1,\dots,n\}\).
Fix \(i\). The intersection of all \(3\)-subsets containing \(i\) is exactly \(\{i\}\) when \(n\ge4\). Since \(\sigma\) preserves each such \(3\)-subset, it preserves their intersection, so \(\sigma(i)=i\). This holds for every \(i\), hence \(\sigma=e\).
Therefore \[ Z(A_n)=\{e\}\qquad(n\ge4). \]