# Is a boolean interval of finite groups linearly primitive?

Let $[H,G]$ be an interval of finite groups.

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

The interval $[H,G]$ is called linearly primitive if $\exists V$ irred. complex repr. of $G$ with $G_{(V^H)} = H$.
Remark: We recover the usual definition of "linearly primitive" for groups by taking $H =\{ e \}$.

Definition: The subset lattice of $\{1, \dots , n \}$ is called the boolean lattice $B_n$ (see $B_3$ below).

Question: Is a boolean interval of finite groups linearly primitive?

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• Is perhaps this result of Tůma useful for answering this question? Every algebraic lattice is isomorphic to an interval in the subgroup lattice of some group. See DOI: 10.1016/0021-8693(89)90171-3, Google, Google Books, Google Scholar. – Martin Sleziak Sep 16 '15 at 9:32
• Sorry, now I noticed that your question is about finite groups. But I will keep comment, just in case it is useful for other users reading this question. – Martin Sleziak Sep 16 '15 at 9:35
• Thanks. Anyway, the lattice $B_n$ is also realized by an inclusion of finte group (for example $G=\mathbb{Z}/p_1 \cdots p_n$ and $H=\{ e \}$). We can reformulate the question by: Can a $B_n$ lattice be realizable by a non linearly primitive inclusion? – Sebastien Palcoux Sep 29 '15 at 9:15