A question on groups having a subgroup which fixes a vector in every irreducible representations

Question

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.

Answer 1

Let us write ($*$) for the property that a subgroup $H$ of a finite group $G$ has nonzero fixed vectors in every representation of $G$.

Proposition 1: if $G$ is nilpotent, then no nontrivial subgroup $H$ of $G$ satisfies ($*$).

Proof: Assume such an $H$ exists. We may assume that $H$ is cyclic, generated by some nontrivial $h\in G$. Let $1=G_0 \triangleleft G_1\triangleleft \dots \triangleleft G_n = G$ be a central series. For some $i$, the element $h$ belongs to $G_{i+1}$ but not to $G_i$. Therefore, the image of $h$ in $G/G_i$ is nontrivial, but central. For every irreducible representation of $G/G_i$, the element $h$ has fixed points, but since it is central it acts trivially (Schur's lemma); therefore $h$ is trivial in $G/G_i$, a contradiction.

More generally, let $G$ be a finite group. We have the following simple obstructions:

• If $N\triangleleft G$ is a normal subgroup and $H$ satisfies ($*$), then so does the image of $H$ in $G/N$.
• A nontrivial central subgroup $H$ of $G$ never satisfies ($*$).
• Combining these two, a subgroup $H$ whose image is nontrivial and central in some quotient $G/N$ never satisfies ($*$).

There is a converse for subgroups of order $2$.

Proposition 2: Let $H$ be generated by an element $h$ of order $2$. Assume that the image of $h$ in every quotient $G/N$ is either trivial or noncentral. Then $H$ satisfies ($*$).

Proof: Let $(V,\rho)$ be an irreducible representation of $G$ and let $N=\ker\rho$. If $h$ is trivial in $G/N$, then $h$ acts trivially on $V$ and therefore $H$ has nonzero fixed points in $V$. If $h$ is noncentral in $G/N$, then $\rho(h)$ has at least two distinct eigenvalues, so it has $1$ as an eigenvalue, i.e. $H$ has fixed points in $V$.

Corollary 3: Let $G$ be a nonabelian finite simple group, and let $H$ be a subgroup of order $2$. Then $H$ satisfies ($*$).

Proof: there is no nontrivial quotient, and $Z(G)$ is trivial. So Proposition 2 applies.

Corollary 4: Let $G = S_n$ for $n\ge 4$, and let $H$ be generated by a double transposition. Then $H$ satisfies ($*$).

Proof: $H$ is not central, and belongs to all the proper normal subgroups. So Proposition 2 applies.