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Let $K=PSL_2(q)$ where $q=p^a$ for some odd prime $p$, and let $G$ be a group such that $G/O(G)\cong K$. (Here $O(G)$ is the largest odd-order normal subgroup of $G$.)

I have a homomorphism $\phi: G\to PGL_n(\overline{\mathbb{F}_r})$ whose image is non-solvable. (Here $r$ is a prime distinct from $p$.) I am interested in giving a lower bound for $n$.

In the case where $O(G)$ is trivial, a classical result originating with work of Frobenius gives a sharp lower bound for $n$, namely $\frac12(q-1)$. (Provided $q\neq9$, but let's ignore this exception.)

I'd like to prove that this bound is best possible, i.e.

Q1. For arbitrary $G$ of the given form, prove that $n\geq \frac12(q-1)$.

I reckon I can do this by brute force using Aschbacher's classification of subgroups of $GL_n(q)$ but I'd prefer a more elegant solution. A couple of easy reductions allow me to assume that $G$ acts absolutely irreducibly, that $G$ is center-less, and that $G$ is a minimal non-split extension of $K$, i.e. there does not exist a proper subgroup $H$ such that $H$ is an extension of $K$.

These reductions led me to wonder about a related (but probably much harder) question:

Q2. What are the center-less minimal non-split extensions of $K$?

I know that examples of these things exist where $G$ is not isomorphic to $K$ (see, for instance this MO question, in particular an example mentioned in the answer of @nkrempel). However I cannot find a systematic treatment of these things in the literature.

Final note: the word `extension' has two meanings in group theory. In the definition I'm using the group $G$ is an example of an extension of $K$.

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I would guess that the most efficient approach to this is to use the arguments used in the proof of the Aschbacher Theorem, which you can do without just citing the theorem and ploughing through the cases. There is a nice proof of a basic version of Aschbacher's Theorem in Theorem 3.5 (page 85) of "The Finite Simple Groups" by Robert A. Wilson.

It goes roughly as follows. Let $N$ be the socle of the image of $G$ in ${\rm PGL}_n(r^k)$ (for suitable $k$). So $N$ will have odd order in your case. If $N$ is not homogeneous, then $G$ acts imprimitively, so you get a reduction in degree or to a permutation group. If $N$ is homogeneous but reducible, then you get a tensor product decomposition for $G$, which again gives you a degree reduction. The same applies if $N$ is not the unique minimal normal subgroup, and if it is, then you are in the symplectic normalizer case, which again gives you an easy degree reduction.

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Derek, this is really interesting, thank you. I don't have a copy of Wilson's book but will try and lay my hands on one and follow this up. I presume "homogeneous" here means "product of a bunch of isomorphic simple groups"? (I'm going to delay accepting your answer while I follow this up, and so that others can contribute.) –  Nick Gill Nov 21 '12 at 15:53
    
Google books is helping me... and has alerted me to the fact that my definition of `homogeneous' was wrong, one needs the representation to be a direct sum of isomorphic irreducibles –  Nick Gill Nov 21 '12 at 16:05
    
Yes, homogeneous was referring to the module defined by the representation, not the group. –  Derek Holt Nov 21 '12 at 17:55

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