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

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 $⟨\tilde{\chi },\tilde{\chi }⟩=|G{|}^{-1}{\sum }_g|\chi (g^2){|}^2=1$.

Solved by gpt-5.6-sol high.