Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 4 / 19F / b / ii / Solution

The second Adams operation $\chi \mapsto [g\mapsto \chi (g^2)]$ lies in the integral character ring, so $\tilde{\chi }=\sum n_i{\chi }_i$ with integers $n_i$. Norm one forces exactly one coefficient to be $\pm 1$; evaluation at the identity makes the sign positive. Squaring is bijective on conjugacy classes as well, so distinct irreducibles remain distinct. Thus the ${\tilde{\chi }}_i$ permute the irreducible characters.