Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 2 / 19I (Representation Theory)

Let the distinct values of the faithful character $\chi$ be $a_1,\dots ,a_r$. If an irreducible character $\psi$ occurred in none of $1,\chi ,\dots ,{\chi }^{r-1}$, thenfor $0\le n<r$. The Vandermonde matrix is invertible, so every $S_j=0$. But faithfulness implies that $\chi (g)=\chi (1)$ only for $g=1$: in a unitary realization, equality of the trace with the dimension forces every eigenvalue to be one. For that fiber $S_j=\psi (1)>0$, a contradiction.

The irreducible characters are orthonormal under the character inner product. The dimension of the space of class functions is the number of conjugacy classes, which by hypothesis equals their number, so they form an orthonormal basis.

Conjugate elements have equal values under every character. Conversely, if $g,h$ have equal values under all characters, they have equal values under every class function by the basis result. Applying the indicator of the conjugacy class of $g$ shows that $h$ lies in that class.

Let $K=\ker \chi$, which is nontrivial. Every tensor power affording ${\chi }^n$ is trivial on $K$, so each of its irreducible constituents has $K$ in its kernel. Some irreducible representation is nontrivial on $K$; otherwise the regular representation, which contains every irreducible, would also be trivial on $K$. Choose its character $\psi$. Then $\psi$ is a constituent of no ${\chi }^n$, and