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Let $p\geq 5$ be a prime and $\mathbf{F}_p$ a finite field of characteristic $p$. A subgroup of ${\rm GL}_n(\mathbf{F}_p)$ is full if it contains ${\rm{SL}}_n(\mathbf{F}_p)$. When $n=2$, we have the following nice characterisation of full subgroups, cf. P.182,Theorem 2.1 in [Lang, S., 2001. Introduction to modular forms (Vol. 222).]

Let $G$ be a subgroup of ${\rm GL}_2(\mathbf{F}_p)$. Then $G$ is a full subgroup if and only if the order of $G$ is divisible by $p$ and $G$ is non-solvable.

My question is following:

Let $G$ be a subgroup of ${\rm GL}_n(\mathbf{F}_p)$ for some $n\geq3$. Assume that the order of $G$ is divisible by $p$ and $G$ is non-solvable. Is it possible to add some slight conditions to ensure that $G$ is a full subgroup?

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    $\begingroup$ "$\mathbf{F}_p$ is a finite field of characteristic $p$": no, you mean, of cardinal $p$. Otherwise first it would be very non-standard notation, and also the quoted theorem from Lang would be false for a proper extension of the field of order $p$. $\endgroup$
    – YCor
    Commented Mar 24, 2023 at 1:04
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    $\begingroup$ Vague question, but I think the answer is no. Just assuming $G$ divisible by $p$ and $G$ non-solvable, you have examples like ${\rm GL}_{n_1}(\mathbf{F}_p) \times \cdots \times {\rm GL}_{n_t}(\mathbf{F}_p) < {\rm GL}_n(\mathbf{F}_p)$ for $n = n_1 + \cdots + n_t$. So I suppose you should assume that $G$ is irreducible. Then you have imprimitive subgroups like ${\rm GL}_{a}(\mathbf{F}_p) \wr S_b$ with $n = ab$. So then maybe you want to assume that $G$ is primitive, etc.. It may be helpful for you to look up Aschbacher's theorem on maximal subgroups of classical groups. $\endgroup$
    – spin
    Commented Mar 24, 2023 at 1:35
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    $\begingroup$ The question is peculiar because "full" subgroups are not mysterious. Since they contain the derived subgroup $\mathrm{SL}_n(p)$ they are in 1:1 correspondence with subgroups of $\det(\mathrm{GL}_n(p)) \cong C_{p-1}$. Really it seems what you are trying to understand is nonsolvable linear groups of $p$-divisible order, of which there are an ocean of possibilities. $\endgroup$
    – Sean Eberhard
    Commented Mar 24, 2023 at 13:03
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    $\begingroup$ The case $n =2$ is completely exceptional because a subgroup of ${\rm GL}(2,p)$ with more that one Sylow $p$-subgroup contains ${\rm SL}(2,p)$. You need much more stringent conditions for larger $n$. Of course, a concise statement would be "a subgroup of ${\rm GL}(n,p)$ is full (for $n >1$) if and only if it contains all commutators", but that is not helpful. $\endgroup$
    – Geoff Robinson
    Commented Mar 24, 2023 at 16:05

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