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I am interested in intransitive irreducible linear subgroups $G\subseteq\mathrm{GL}_n(\mathbb{F}_p)$ acting on $V-\{0\}=\mathbb{F}_p^n-\{0\}$ in the natural way, such that all of the orbits are very large, with constant density in $V$.

Actually I am wondering if there exists an infinite family of groups satisfying this property. Here $p$ is supposed to be a fixed prime.

The question can be formalized as follows:

Do there exist a prime $p>0$, a constant $c\in (0,1)$, and infinitely many integers $n\in\mathbb{N}^+$ and irreducibe linear groups $G_n\subseteq \mathrm{GL}_n(\mathbb{F}_p)$, such that each $G_n$ acts intransitively on $V_n-\{0\}:=\mathbb{F}_p^n-\{0\}$ in the natural way, and the size of every $G_n$-orbit is at least $cp^n$?

Transitive groups e.g. $G=\mathrm{GL}_n(\mathbb{F}_p)$ certainly have this property but I wonder if there exist infinitely many intransitive examples. I think one can easily reduce to the case that $G$ is a primitive linear group but then I don't know how to proceed.

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  • $\begingroup$ What about the natural copy of $\mathrm{AGL}_{n-1}(\mathbb{F}_p)$? Won't it have two orbits, of size $p^{n-1}-1$ and $p^n-p^{n-1}$? $\endgroup$
    – verret
    Commented Mar 29, 2016 at 8:18
  • $\begingroup$ @verret Oops. I think what I actually mean is an irreducible linear group so no proper subspace is preserved. Modified the question. And thanks. $\endgroup$
    – Zeyu
    Commented Mar 29, 2016 at 8:50
  • $\begingroup$ @verret btw, do you mean $\mathrm{AGL}_{n-1}(\mathbb{F}_p)\times \mathbb{F}_p^*$? $\endgroup$
    – Zeyu
    Commented Mar 29, 2016 at 8:59
  • $\begingroup$ Yes, that's what I meant. (I actually had the case $p=2$ in mind.) $\endgroup$
    – verret
    Commented Mar 29, 2016 at 13:21

1 Answer 1

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Take an odd prime $p$, let $q=p^n$ and consider $V=\mathbb{F}_q$ as a vector space over $\mathbb{F}_p$. For $G_n=(\mathbb{F}_q^*)^2$ you get $c=1/2$.

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  • $\begingroup$ Exercise: get $c=1/3$ for $p=2$ ($n$ even). Question: Did you want to assume $G_n$ algebraically closed by chance? $\endgroup$
    – Uri Bader
    Commented Mar 29, 2016 at 9:58

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