Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 2 / 19I (Representation Theory)

By character orthogonality, the squared multiplicities are the character inner product norm of the restriction of a character:Since $\chi$ is an irreducible character, its norm on $G$ is one, and henceThe omitted summands are nonnegative, so the character restriction norm bound is an equality precisely when

Because $[G:H]=2$, part (a) says that the squared norm of $\chi {\downarrow }_H$ is either one or two. It is one exactly when the restriction is irreducible, whereas it is two exactly when $\chi$ vanishes throughout $G\setminus H$. Thus condition (1) is equivalent to the existence in condition (2).

On $H$ we always have $\chi \lambda =\chi$, while outside $H$ we have $\chi \lambda =-\chi$. Therefore $\chi \lambda =\chi$ exactly when $\chi$ vanishes outside $H$. This proves all three equivalences in the index-two character restriction dichotomy:

The permutation character of a coset action on $G/H$ is $1_G+\lambda$. The tensor identity for an induced character therefore givesIf $\psi {\downarrow }_H=\chi {\downarrow }_H$ is irreducible, then Frobenius reciprocity givesBoth $\chi$ and $\chi \lambda$ are irreducible, so

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,

The normis a positive integer, because it is the sum of the squares of the multiplicities of the irreducible constituents. The character restriction norm bound and $[G:K]=3$ make it at most three. It is therefore $1$, $2$, or $3$.

All three values occur: