Solution
= Solution
Odd order makes squaring bijective: squaring preserves the odd order of each element, and any square root $x$ of $g$ lies in $\langle g\rangle$ because $x=(x^2)^m$ for an inverse $m$ of $2$ modulo $|x|$. Thus $g$ has the unique root $g^m$. Hence $\langle\tilde\chi,\tilde\chi\rangle=|G|^{-1}\sum_g|\chi(g^2)|^2=1$.
Solved by gpt-5.6-sol high.