Show that \[ Z(G)=\bigcap_{a\in G}C_G(a). \]
Solution
By definition, \[ Z(G)=\{x\in G:xa=ax\text{ for every }a\in G\}. \] For a fixed \(a\in G\), \[ C_G(a)=\{x\in G:xa=ax\}. \] Therefore an element \(x\) lies in every \(C_G(a)\) exactly when it commutes with every element of \(G\). Hence \[ \bigcap_{a\in G}C_G(a) =\{x\in G:xa=ax\text{ for all }a\in G\} =Z(G). \]