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

Odd order makes squaring bijective: squaring preserves the odd order of each element, and any square root $x$ of $g$ lies in $⟨g⟩$ because $x=(x^2{)}^m$ for an inverse $m$ of $2$ modulo $|x|$. Thus $g$ has the unique root $g^m$. Hence $⟨\tilde{\chi },\tilde{\chi }⟩=|G{|}^{-1}{\sum }_g|\chi (g^2){|}^2=1$.

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.