Here the restriction has squared norm two, so part (b)(i) gives $\chi\lambda=\chi$. Consequently
$$
\operatorname{Ind}_H^G\psi_1+
\operatorname{Ind}_H^G\psi_2
=\operatorname{Ind}_H^G(\chi\mathbin{\downarrow}_H)
=\chi+\chi\lambda=2\chi.
$$
By <Frobenius reciprocity>, each induced character on the left contains $\chi$, because each $\psi_i$ is a <character constituent> of $\chi\mathbin{\downarrow}_H$. Positivity of irreducible multiplicities then forces
$$
\operatorname{Ind}_H^G\psi_1
=\operatorname{Ind}_H^G\psi_2=\chi.
$$
If $\phi\mathbin{\downarrow}_H$ contains either $\psi_i$, Frobenius reciprocity makes $\phi$ a constituent of $\operatorname{Ind}_H^G\psi_i=\chi$. Since both are irreducible,
$$
\boxed{\phi=\chi.}
$$

Solution

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

An irreducible character $\varphi$ is a constituent of a character $\chi$ when its multiplicity $\langle\chi,\varphi\rangle$ is positive.


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 Incoming links: Character constituent