Let be $G$ a finite solvable group and $M$ a minimal normal subgroup of $G$. The group $M$ is an elementary abelian $q$-group, for a prime $q$ that we assume to be odd. Suppose that $1_M \neq\lambda \in \hat M$ is $G$-invariant. Let be $p$ an odd prime, different from $q$, call $\pi=\{2,p\}$ and take $H \in Hall_\pi(G)$. Then there is a canonical extension $\mu$ of $\lambda$ to $HM$ (see 8.16 of Isaacs' book "Character theory of finite groups"). Let be $\varphi$ an irreducible contituent of the induced character $\mu^G$. Is it true that no prime in $\pi$ divides $\varphi(1)$ or, at least, $p$ does not divide $\varphi(1)$?
Motivation: the context is degrees of real characters. With the previous notation, suppose that $G$ has an automorphism $\tau$, of order a power of $2$, that fixes $M$ and such that $\lambda^\tau=\overline \lambda$. I want to find an irreducible character $\varphi$ of $G$ over $\lambda$ such that $\varphi(1)$ is a $p'$-number and to $\varphi^\tau=\overline \varphi$. Suppose that we find out that $\sigma$ fixes $HM$, so by canonicity $\mu^\tau=\overline \mu$. Since the the operation $\sigma \colon\varphi \to \overline\varphi^x$ is a permutation of the set $X$ of irreducible contituents of $\mu^G$, $\sigma$ has order a $2$-power, and $2$ does not divide $\mu^G(1)$, then there is $\varphi \in X$ that is $\sigma$-fixed, namely $\varphi^x=\overline \varphi$. I am wondering if such a $\varphi$ is a good candidate for having a $p'$-degree.