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This post is a dual version for the Generalization of a theorem of Øystein Ore in which it's proved:
Theorem: Let $[H, G]$ be a distributive interval of finite groups. Then $\exists g \in G$ such that $\langle H,g \rangle = G$.

Definition: Let $W$ be a representation of $G$, $K$ a subgroup of $G$, and $X$ a subspace of $W$.
Let the fixed-point subspace $W^{K}:=\{w \in W \ \vert \ kw=w \ , \forall k \in K \}$.
Let the pointwise stabilizer subgroup $G_{(X)}:=\{ g \in G \ \vert \ gx=x \ , \forall x \in X \}$.

Definition: $[H,G]$ is called linearly primitive if $\exists V$ irred. complex repr. of $G$ with $G_{(V^H)} = H$. Remark: $[1,G]$ is linearly primitive iff $G$ is linearly primitive.

Question: Let $[H, G]$ be a distributive interval of finite groups. Is $[H,G]$ linearly primitive?

Remark: The case $H = 1$ is true because $[1,G]=\mathcal{L}(G)$ is distributive iff $G$ is cyclic, but an abelian group is linearly primitive iff it is cyclic. It follows that it's also true if $H \triangleleft G$.

It's true by GAP for $|G:H|<32$ or $\vert G \vert \le 1000$ or $G$ perfect with $\vert G \vert < 10080$ (except $7680$).

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Yes.

This was proved in the planar algebra framework, see arXiv:1704.00745, Corollary 6.10.

For a self-contained group-theoretic proof, see arXiv:1708.02565.

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