Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 2 / 19I / b / iii / Solution

Here the restriction has squared norm two, so part (b)(i) gives $\chi \lambda =\chi$. ConsequentlyBy Frobenius reciprocity, each induced character on the left contains $\chi$, because each ${\psi }_i$ is a character constituent of $\chi {\downarrow }_H$. Positivity of irreducible multiplicities then forcesIf $\varphi {\downarrow }_H$ contains either ${\psi }_i$, Frobenius reciprocity makes $\varphi$ a constituent of ${\mathrm{Ind}}_H^G{\psi }_i=\chi$. Since both are irreducible,