Use the <character formula for an induced representation>. For $g\in G$,
$$
\begin{aligned}
\chi(g)(\phi\!\uparrow_H^G)(g)
&=\frac1{|H|}\sum_{\substack{x\in G\\x^{-1}gx\in H}}
\chi(g)\phi(x^{-1}gx)\\
&=\frac1{|H|}\sum_{\substack{x\in G\\x^{-1}gx\in H}}
\chi(x^{-1}gx)\phi(x^{-1}gx)\\
&=\bigl((\chi\!\downarrow_H)\phi\bigr)\!\uparrow_H^G(g).
\end{aligned}
$$
The middle equality uses that a <character of a representation> is constant on conjugacy classes. This proves the <tensor identity for an induced character>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 160 / 2 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-160/2/c/solution