is it true that a certain induction of an H-irrep $\psi$ is irreducible iff there's a G-conjugacy class on which $\chi_\psi$ is nonconstant

Question

Assume $H\subset G$ is a subgroup of index $2$. Let a G-conjugacy class mean the elements conjugate to a fixed h in G. Prove or disprove whether the induced representation $Ind_H^G (\psi)$ of an H-irrep $\psi$ is irreducible iff there's a G-conjugacy class on which $\chi_\psi$ is nonconstant.

By definition, the induced representation acts on a direct sum of two vector spaces that are isomorphic to the two cosets of H, say $g_i H$ for $i=1,2$. Normally, characters are constant on conjugacy classes. If $\chi_\psi$ is not constant on a conjugacy class, I'm not sure how this relates to the irreducibility of $\psi$. I know that for each $g\in G$ and each $i, \exists h_i\in H, j(i)\in \{1,2\}$ so that $g g_i = g_{j(i)} h_i$ and $g\cdot \sum_{i=1}^2 g_i v_i = \sum_{i=1}^2 g_{j(i)} \psi(h_i) v_i$, but again I'm not sure how to proceed from here.

Answer 1

Let's fix a representative $g$ of the coset $G\setminus H$ and write $\chi=Ind_H^G(\psi)$. As $H\unlhd G$ we have $\chi(x)=0$ whenever $x\notin H$. But for all $x\in H$, we have $\chi(x)=\psi(x)+\psi^g(x)$, where $\psi^g(x)=\psi(g^{-1}xg)$. As $\psi$ was known to be irreducible, so is $\psi^g$.

Consequently $$ \begin{aligned} \langle\chi,\chi\rangle_G&=\frac1{|G|}\sum_{x\in G}\chi(x)\overline{\chi(x)}\\ &=\frac1{|G|}\sum_{x\in H}\chi(x)\overline{\chi(x)}\\ &=\frac1{2|H|}\sum_{x\in H}(\psi(x)+\psi^g(x))\overline{(\psi(x)+\psi^g(x)})\\ &=\frac12\left(\langle\psi,\psi\rangle_H+\langle\psi,\psi^g\rangle_H+\langle\psi^g,\psi\rangle_H+\langle\psi^g,\psi^g\rangle_H\right). \end{aligned} $$ Because $\psi$ and $\psi^g$ are irreducible we know that $$ \langle\psi,\psi\rangle_H=1=\langle\psi^g,\psi^g\rangle_H. $$ Furthermore, $\langle\psi,\psi^g\rangle_H$ is either $1$ or $0$ according to whether $\psi=\psi^g$ or not.

So we have shown that $\langle\chi,\chi\rangle_G$ is either $2$ or $1$ according to whether $\psi=\psi^g$ or not. This answers the question in the affirmative.

Answer 2

We proceed by contrapositive. Suppose that $\psi^G:=Ind_H^G(\psi)$ is constant over $\text{cl}_G(h)$ for every $h\in H$. Let $g\in G\setminus H$. By Frobenius reciprocity it holds that $[\psi^G,\psi^G]=[(\psi^G)_H,\psi]$. Let $h\in H$. Then $$(\psi^G)_H(h)=\psi(h)+\psi(ghg^{-1})=\psi(h)+\psi(h)=2\psi(h).$$ Thus $(\psi^G)_H=2\psi$ and $[\psi^G,\psi^G]=2$, so $\psi^G$ is not irreducible.

On the other hand, if $\psi^G$ is not irreducible then $1\not=[\psi^G,\psi^G]=[(\psi^G)_H,\psi]\leq2$, since $\psi^G(1)=2$. Then $(\psi^G)_H=2\psi$. Let $h\in H$ and let $g\in G$. Then, $$2\psi(h^g)=(\psi^G)_H(h^g)=\psi^G(h^g)=\psi^G(h)=(\psi^G)_H(h)=2\psi(h),$$ as $\psi^G$ is a class function. Then $\psi(h^g)=\psi(h)$ and we are done.