Given a finite group $G$, I am interested in finding a non-trivial proper subgroup $H$ of $G$ such that $\mathrm{Ind}_H^G\mathbf{1}$ contains all the irreducible representations of $G$, that is, permutation representation of $G$ on $G/H$ contains all the representations of $G$. I can see that if $G$ is abelian then this is not possible. Because, if $G$ is abelian and $H$ is such a subgroup then $\mathrm{Ind}_H^G\mathbf{1}$ will contain all $|G|$ many representations but $\mathrm{dim}(\mathrm{Ind}_H^G\mathbf{1})=[G: H]$ which is a contradiction.

Now, I am thinking about nilpotent groups. Given a finite nilpotent group $G$, is it possible to have such a subgroup? What about solvable groups?

I was also looking at this problem from another point of view. If $\mathrm{Ind}_H^G\mathbf{1}$ contains all the irreducible representations of $G$, then by Frobenius reciprocity we can say that for every irreducible representation $(\rho, V)$ of $G$, $V^{H}\neq {0}$. Hence, there is a one-one correspondence between the simple modules for the group algebra $\mathbb{C}[G]$ and the simple modules for the Hecke algebra $e_H\mathbb{C}[G]e_H$ (where $e_H=\frac{1}{|H|}\underset{h\in H}{\sum}h$). And I want to ask if there is some sort of classification of groups having subgroup with the property discussed above.

Also, given a compact connected Lie group $G$, when does it possess a closed subgroup $H$ such that $L^2(G/H)$ contains all the representations of $G$?

I posted this in MSE. Got no answer. That is why I am posting it here.