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I am reading the Kleidman-Liebeck book ("The subgroup structure of the finite classical groups") which is about the Aschbacher classification of maximal subgroups of the classical almost simple groups. But thinking just about natural actions of classical groups I got stuck in the following problem.

Consider the group $GL(n,q)$ of invertible $\mathbb{F}_q$-linear maps $\mathbb{F}_q^n \to \mathbb{F}_q^n$, where $n$ is a positive integer and $q=p^k$ is a prime power. This group acts on $V := \mathbb{F}_q^n$ in the natural way, and this action is $\mathbb{F}_q$-linear, in particular it is $\mathbb{F}_p$-linear. $\mathbb{F}_p$-linearity yields an injective homomorphism $GL(n,q) \to GL(nk,p)$, and I will identify $GL(n,q)$ with a subgroup of $GL(nk,p)$ by means of this homomorphism. Let $W$ be a proper $\mathbb{F}_p$-subspace of $V$, and let $H$ be the stabilizer of $W$ in $GL(nk,p)$, i.e. the set of invertible $\mathbb{F}_p$-linear transformations $f$ of $V$ such that $f(W)=W$.

Since $GL(n,q)$ acts transitively on non-zero vectors, $L := H \cap GL(n,q)$ is a proper subgroup of $GL(n,q)$, so it is contained in a maximal subgroup of $GL(n,q)$ which belongs to one of the Aschbacher classes. Which one?

If $W$ spans a proper $\mathbb{F}_q$-subspace of $V$ then $L$ is of parabolic type (class $\mathcal{C}_1$).

What happens if $\langle W \rangle_{\mathbb{F}_q} = V$? My guess is that in this case $L$ falls in class $\mathcal{C}_5$ (subfield stabilizers), but unfortunately I do not know enough about the linear groups to know how to think about this or where to look at.

I have been thinking about this and I asked people, but the question does not seem easy. This is just to say that I hope it is not trivial for experts, in which case I apologize.

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It seems to me that this is almost the definition of the maximal subgroups in class $\mathcal{C}_5$, but perhaps I am overlooking something. What definition for $\mathcal{C}_5$ are you using? –  Tom De Medts Jan 23 '12 at 11:04
    
$\mathcal{C}_5$ consists of subgroups of the form $GL(n,q^{1/r})$ where $q=p^k$ and $r$ is a prime divisor of $k$, and the inclusion is realized as follows: taking a $\mathbb{F}_q$-basis $B$ of $V$, and called $V_0$ the $\mathbb{F}_{q^{1/r}}$-span of $B$, every invertible $\mathbb{F}_{q^{1/r}}$-linear map $V_0 \to V_0$ extends uniquely to an element of $GL(n,q)$. The problem is that an invariant $\mathbb{F}_p$-subspace spanning $V$ over $\mathbb{F}_q$ may not contain an invariant $\mathbb{F}_p$-subspace spanned by an $\mathbb{F}_q$-basis. Right? –  Martino Garonzi Jan 23 '12 at 12:04
    
But $\operatorname{dim}_{\mathbb{F}_p}(V_0) = n$, right? So then you can take any $\mathbb{F}_q$-basis of $V$ contained in $V_0$ (which exists because $V_0 \subset V$ spans $V$), and this basis will then also be an $\mathbb{F}_p$-basis for $V_0$ because it consists of $n$ linearly independent vectors. Or? –  Tom De Medts Jan 23 '12 at 13:51
    
Yes of course, but what I am saying is that for a given $g \in GL(n,q)$, a generic $g$-invariant $\mathbb{F}_p$-subspace which spans $V$ over $\mathbb{F}_q$ (which in particular has $\mathbb{F}_p$-dimension at least $n$ and at most $kn$) is not forced to contain an $n$-dimensional $g$-invariant $\mathbb{F}_p$-subspace. –  Martino Garonzi Jan 23 '12 at 14:41
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I believe that the stabilizer $H$ of a subspace must be either reducible as a subgroup of ${\rm GL}(n,q)$ or it can be written over a proper subfield of ${\mathbb F}_q$.

Suppose that it is irreducible, so we have an irreducible representation $\rho$ of $H$ over ${\mathbb F}_q$. If I am remembering this correctly then, if you regard this $\rho$ first as a representation of degree $kn$ over ${\mathbb F}_p$ and then think of that as a representation of degree $kn$ over ${\mathbb F}_q$, then the resulting representation is $\rho + \rho^\phi + \cdots + \rho^{\phi^{k-1}}$, where $\phi$ is the Galois automorphism of order $k$ of ${\mathbb F}_q$.

If $H$ is not writeable over a proper subfield of ${\mathbb F}_q$, then the representations $\rho, \rho^\phi, \ldots, \rho^{\phi^{k-1}}$ are all inequivalent. But since $H$ stabilizes a proper subspace over ${\mathbb F}_p$, the restriction to that subspace as a ${\mathbb F}_q$ representation would be equivalent to a sum of a proper subset of $\{\rho, \rho^\phi, \ldots, \rho^{\phi^{k-1}}\}$. But then, since that sum is writeable over ${\mathbb F}_p$, it must be stabilized by $\phi$, which is not possible is $\rho, \rho^\phi, \ldots, \rho^{\phi^{k-1}}$ are all inequivalent.

I expect an expert in representation theory would be able to explain it better than I have, but I think this is essentially correct!

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