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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).

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

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  • $\begingroup$ 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. $\endgroup$
    – Martin Sleziak
    Sep 16, 2015 at 9:32
  • $\begingroup$ 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. $\endgroup$
    – Martin Sleziak
    Sep 16, 2015 at 9:35
  • $\begingroup$ 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? $\endgroup$
    – Sebastien Palcoux
    Sep 29, 2015 at 9:15

1 Answer 1

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Yes, it is the main result of the paper arXiv:1708.02565v1, Theorem 3.13.

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