Existence of universal bound related to characters Let $G$ be a finite group and $\lambda \in \text{Irr}(G)$ an irreducible complex character of $G$.
Let $m(\lambda) := \min \{ \vert G : H \vert \mid H \leq G, \lambda\vert_H \text{ has a linear component}\}.$
Is there a universal constant $c$, independent of $G$ and $\lambda$, such that
$$ \frac{m(\lambda)}{\lambda (1_G)} \leq c? $$
When we consider only monomial groups such a constant is given by 1.
Is there such a constant when we consider only solvable or simple groups?
 A: There is no such universal constant for solvable groups, at least. Essentially this is because $m(\chi)$ as defined in the question, is multiplicative. That is, if $G_1$ and $G_2$ have coprime orders, and if $\chi_i\in\mathrm{Irr}(G_i)$ for $i=1,2$ then $\chi_1\times\chi_2\in\mathrm{Irr}(G_1\times G_2)$ satisfies $m(\chi_1\times\chi_2)=m(\chi_1)m(\chi_2).$
This follows because the subgroups of $G_1\times G_2$ are just the $H_1\times H_2$ where $H_i\subseteq G_i,$ and $(\chi_1\times\chi_2)_{H_1\times H_2}=(\chi_1)_{H_1}\times (\chi_2)_{H_2}$ contains a linear constituent if and only if $(\chi_i)_{H_i}$ contains a linear constituent for $i=1,2.$ Hence if $c(\chi)=m(\chi)/\chi(1)$ then in the situation above, also $c(\chi_1\times\chi_2)=c(\chi_1)c(\chi_2).$
For odd primes $p>q$ with $q$ dividing $p+1,$ it is known that there exists a solvable group $G$ of order $|G|=p^3q$ and a primitive $\chi\in\mathrm{Irr}(G)$ of degree $p.$ If $H\subseteq G$ and $\chi_H$ contains a linear constituent then $|G:H|>p$ since $\chi$ is not monomial, so $|G:H|\ge pq$ and $c(\chi)\ge q$ (in fact equality holds). For any $n\ge 1$ we can find $n$ such groups with pairwise coprime orders, for example as follows: Start with $|G_1|=5^3\cdot 3,$ and for $n\ge 2,$ choose $q_n$ odd and prime to $\prod_{i=1}^{n-1}|G_i|$ and $p_n$ with $p_n=-1$ mod $q_n$ and $p_n=1$ mod $\prod_{i=1}^{n-1} |G_i|.$ Setting $G=\prod_1^n G_i$ then $\chi=\prod_1^n\chi_i$ (where $\chi_i\in\mathrm{Irr}(G_i)$ is any primitive character of $G_i$ as mentioned above) has $c(\chi)\ge \prod_1^n q_i\ge 3^n.$
Since simple groups can't have pairwise coprime orders this doesn't work there, though I guess that there is no universal bound in that case.
