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).

                                                                     enter image description here

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

  • $\begingroup$ 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. $\endgroup$ 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. $\endgroup$ 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? $\endgroup$ Sep 29, 2015 at 9:15

1 Answer 1


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.