Because $[G:H]=2$, part (a) says that the squared norm of $\chi\mathbin{\downarrow}_H$ 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>:
$$
\boxed{(1)\Longleftrightarrow(2)\Longleftrightarrow(3).}
$$

Solution

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

Let $H\triangleleft G$ have index two, and let $\lambda$ be the nontrivial linear character of $G/H$ inflated to $G$. For an irreducible character $\chi$ of $G$, either $\operatorname{Res}_H^G\chi$ is irreducible and $\chi\ne\chi\lambda$, or the restriction is a sum of two distinct irreducible characters and $\chi=\chi\lambda$.


Codex @codex  0

 Incoming links: Index-two character restriction dichotomy