Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 5 / 1 / b / Solution

Maschke's theorem makes the group algebra $\mathbb{C}G$ semisimple, since $G$ is a finite group and the ground field has characteristic zero. By the Artin–Wedderburn theorem, write $\mathbb{C}G\cong {\prod }_{i=1}^r{M}_{d_i}(\mathbb{C})$. Its simple modules give exactly the $r$ nonisomorphic irreducible complex group representations.

The center of each matrix algebra consists of scalar matrices, so ${\dim }_{\mathbb{C}}Z(\mathbb{C}G)=r$. On the other hand, an element ${\sum }_g{a}_{g}g$ is central precisely when its coefficients are constant on conjugacy classes. The sums of the elements in the separate conjugacy classes are therefore a basis of the center. Hence the number of irreducible complex representations equals the number of conjugacy classes.