Show that \[ Z(G)=\bigcap_{a\in G}C_G(a). \]
Solution
An element \(z\in G\) lies in the center exactly when it commutes with every element of \(G\): \[ z\in Z(G) \iff za=az\text{ for all }a\in G \iff z\in C_G(a)\text{ for all }a\in G. \] The last condition is precisely \[ z\in\bigcap_{a\in G}C_G(a). \] Hence the two subgroups are equal.