Let $G$ be a finite group, $N$ a subgroup of index $2$, and $V$ an irreducible representation of $N$.
Question: How $\mathrm{Ind}_{N}^{G}(V)$ decomposes? Can we write explicitly the irreducible components?
Context: this question is motivated by a dual version of Ore's theorem. Let $[H,G]$ be a boolean interval of finite groups and $N \in [H,G]$ with $|G:N| = 2$ and $[H,N]$ linearly primitive (as interval). Then the answer of Derek Holt helped me to prove that $[H,G]$ is also linearly primitive.
Corollary (6.19) of Isaacs, Character Theory of Finite Groups says that, if $N$ is normal in $G$ with $|G:N|=p$ prime and $\chi \in {\rm Irr}(G)$, then $\chi_N$ is either irreducible or a sum of $p$ distinct conjugate irreducible characters.
Let $\psi \in {\rm Irr}(N)$.
If $\psi = \chi_N$ for some $\chi \in {\rm Irr}(G)$, then $\psi^G$ is the sum of $p$ distinct irreducible characters of $G$, all having restriction $\psi$. Conversely, if $\psi^G$ is the sum of $p$ irreducible characters of $G$, then by frobenius reciprocity they must all reduce to $\psi$. In this case $\psi$ is fixed by conjugation by all elements of $G$.
Otherwise $\psi^G=\chi$ is irreducible. Then the second case of the result in Isaacs must apply to $\chi$, so $\chi$ has $p$ distinct conjugates under $G$.
This is all assuming that we are talking about complex representations. In the more general case it could be possible that $\psi^G$ was irreducible and reduced to to the sum of two copuies of $\psi$, but I have not thought about that.