Throughout this question, the following notation holds: Let $q$ be a power of a prime $p$, and let $d>4$ be a positive integer. Let $G$ be a finite group with a normal subgroup $E$ which is an elementary-abelian $p$-group, and such that $G/E\cong\mathrm{SL}\_d(q)$. 


I am interested in the situation where this extension is **non-split**, i.e. when there is no subgroup $H<G$ such that $H\cong \mathrm{SL}\_d(q)$ and $G=EH$.



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> Q1 Where can I read about such non-split extensions? What are the main results about such groups?
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Given that this is rather general, let me propose a more specific version:



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> Q2 What can one say if one assumes that $|E|=q^d$?
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**Note**: There are a number of results in the literature for a situation similar to this more specific situation. The difference is that people usually consider what happens when the quotient $G/E$ is isomorphic to $\mathrm{GL}\_d(q)$, rather than $\mathrm{SL}\_d(q)$. Some relevant results are listed in [this MO question](https://mathoverflow.net/questions/29417/non-split-extensions-of-gl-nf-q-by-f-qn/51970#51970) - note especially that the answer of @ndkrempel suggests that the situation I am considering allows for significantly different behaviour.


Finally, a question about how such groups might embed into larger linear groups. 



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> Q3. Suppose that $G$ is isomorphic to a subgroup of $\mathrm{SL}\_n(q)$. Is it true that if $d>\frac34n$, then the extension is split?
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**Note**: I chose the number $\frac34$ out of thin air. I'd be happy to hear of a `yes' answer, with that $\frac34$ replaced by your favourite positive real number.



On non-split extensions of $\mathrm{SL}_d(q)$
I will try and answer Question 2, although I only have a superficial knowledge of the representation theory involved. I claim that, when $|E|=q^d$, the induced module action of ${\rm SL}\_d(q)$ on $E$ must either be trivial, or it must be quasi-equivalent to the action on the natural module. (Quasi-equivalent means equivalent after applying a group automorphism.)


By results of Steinberg, the irreducible modules of groups $X(q)$ of Lie type (such as ${\rm SL}\_d(q)$) with $q=p^e$, over fields of characteristic $p$, are tensor products of certain basic modules, where each tensor factor is twisted by a different field automorphism. The smallest field over which the basic modules can be written is ${\mathbb F}\_q$. If $q=r^f$, $\phi$ is a field automorphism of order $f$, and $M$ is a basic module, then the module $M \otimes M^\phi \otimes M^{\phi^2} \otimes \cdots \otimes M^{\phi^{f-1}}$ can be written over ${\mathbb F}\_r$, and this is the only way in which the irreducible modules can be written over smaller fields. (This is for the untwisted groups. The theory is a little more complicated for the twisted groups such as ${\rm SU}\_d(q)$.)


For ${\rm SL}\_d(q)$, the smallest nontrivial basic modules are the natural module and its dual, with dimension $d$ over ${\mathbb F}\_q$. (The next smallest if the exterior square of the natural module, with dimension $d(d-1)/2$.) It seems clear that there is no way of taking tensor products of modules twisted by field automorphisms as described above that will result in an irreducible module with dimension at most $ed$ over ${\mathbb F}\_p$. For example, if $q=p^2$ and $M$ is the natural module, then $M \otimes M^\phi$ can be written over ${\mathbb F}\_p$, but it has dimension $d^2$, which is greater than $2d$ for $d>2$.


For $d>4$, the Schur multiplier of ${\rm SL}\_d(q)$ is not divisible by $p$, so no non-split extensions can arise when the action on $E$ is trivial. (But there are some exceptions when $d \le 4$.)


The second cohomology groups of ${\rm SL}\_d(q)$ on the natural module (and also on its exterior powers) were computed in the two papers:


G.W. Bell, *On the cohomology of the finite special linear groups I and II*, J. Algebra 54, 216-238, 239-259.


For $d>4$, the only nonsplit extension (the Dempwolff group) arises for ${\rm SL}\_5(2)$ with $|E|=2^5$.


I think Qn 1 is too general to answer sensibly, but note that, for any ${\rm SL}\_d(q)$, there must be some irreducible module over ${\mathbb F}\_p$ with nontrivial second cohomology, so there must exist an $E$ with a nonsplit extension.


As for Qn 3, I would certainly bet on the answer being yes, because it is hard to see how you could fit even a split extension in ${\rm SL}\_n(q)$ with $d > 3n/4$ if the module $E$ was not trivial or natural, but it might need a bit of hard work to prove it.


