Splitting of Irreducible character when restricted to index $2$ subgroup over finite field In Exercise $15$ of chapter 2 of the book "The symmetric group" by B.E. Sagan, I can prove the following
Let $G$ be finite group and $H$ be an index $2$ subgroup. If $\chi$ is an irreducible character of $G$, then $\chi|_H$ is irreducible iff $\chi(g) \neq 0$ for some $g \not\in H$.
This is proved using character theory over $\mathbb{C}$ and the fact that a conjugacy class of  an element $h\in G$ in $H$ is as same as that of $G$ or splits in two conjugacy classes of same size.
Can this be proved for a finite field of characteristic $p$ as well, when $p\nmid |G|$?
 A: This is a very general result and holds at least when the characteristic is large enough ($p\nmid |G|$ makes no difference).
Let $k$ be an algebraically closed field of characteristic $\ne 2$. Let $\rho\colon G\to \mathrm{GL}_n(k)$ be an irreducible representation. Let $\chi\colon G\to k^\times$ be the representation $G\to G/H\cong\{\pm 1\}\subset k^\times$ (so we need $\mathrm{char}\ k\ne 2$). The following are equivalent:

*

*$\rho|_H$ is reducible

*$\rho\otimes \chi\cong\rho$
If the second condition is true, then $\mathrm{Tr}(\rho)(g) = \chi(g)\mathrm{Tr}(\rho)(g)=-\mathrm{Tr}(\rho)(g) = 0$ for all $g\in G\setminus H$.
Conversely, suppose that $\mathrm{char}\ k\ge n!$ so that $\rho$ is determined by its trace (you can probably get around this by playing with the entire characteristic polynomial). If $\mathrm{Tr}(\rho)(g) = 0$ for all $g\in G\setminus H$, then $\mathrm{Tr}(\rho)(g) = \chi(g)\mathrm{Tr}(\rho)(g)$ for all $g$ (if $g\in H$, then $\chi(g) = 1$, and if $g\notin H$, then both sides are $0$), so $\rho\cong\rho\otimes \chi$.
To prove the equivalence of $1$ and $2$:

*

*If $\rho|_H$ is reducible then let $\sigma$ be a subrepresentation of minimal dimension. By Frobenius reciprocity,
$$(\rho, \mathrm{Ind}_H^G\sigma) = (\rho|_H, \sigma) \ge 1.$$
Since $\rho$ is irreducible and by dimension counting, by Schur's lemma $\rho\cong \mathrm{Ind}_H^G\sigma$. But since $\rho|_H \cong (\rho\otimes\chi)|_H$, the same argument shows that $\rho\otimes\chi\cong \mathrm{Ind}_H^G\sigma$. So $\rho\cong\rho\otimes\chi$.


*If $\rho\cong\rho\otimes\chi$, suppose that $\rho|_H$ is irreducible. By Frobenius reciprocity and Schur's lemma,
$$(\rho, 
\mathrm{Ind}^G_H\rho|_H) = (\rho|_H, \rho|_H) = 1$$
so $\rho$ is a subrepresentation of $\mathrm{Ind}^G_H\rho|_H$. Write $\mathrm{Ind}^G_H\rho|_H = \rho \oplus \rho'$ with $\rho'\ne \rho$. So there exists $g\in G\setminus H$ such that $\mathrm{Tr}(\rho(g))\ne \mathrm{Tr}(\rho'(g))$. But since $H\lhd G$, we have $\mathrm{Tr}(\mathrm{Ind}^G_H\rho|_H(g))=0$. So
$$\mathrm{Tr}(\rho(g)) + \mathrm{Tr}(\rho'(g))=0.$$
In particular, $\mathrm{Tr}(\rho(g)) \ne 0$, contradicting $\rho\cong\rho\otimes\chi$.
