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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 $G=EH$.

Q1 Where can I read about such non-split extensions? What are the main results about such groups?

Given that this is rather general, let me propose a more specific version:

Q2 What can one say if one assumes that $|E|=q^d$?

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 - 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.

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?

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.

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  • $\begingroup$ Full answers to Q1, Q2 may not be known, but the issue here might be studied in the broader framework of existence of Levi factors in connected algebraic groups in prime characteristic which are neither solvable nor semisimple. See in particular the work of George McNinch indicated in his answer to an older question of mine: mathoverflow.net/questions/22118. $\endgroup$ Nov 26, 2013 at 16:48
  • $\begingroup$ @JimHumphreys, Thanks for your comment, I'll look up George's work. Could you just clarify what you mean when you say `Levi factor'... Are you talking about a subgroup $L$ of a parabolic $P$ for which $P=Rad(P).L$ and which is minimal with respect to this property? $\endgroup$
    – Nick Gill
    Nov 26, 2013 at 17:16
  • $\begingroup$ The conjugation action of $G$ on its abelian normal subgroup $E$ factors through a $G/E$-action on $E$, so the given extension structure makes $E$ into an ${\rm{SL}}_d(q)$-module. If you specify this module structure then the set of isomorphism classes of such extensions is in bijection with ${\rm{H}}^2({\rm{SL}}_d(q),E)$, so you're asking to compute degree-2 cohomology for ${\rm{SL}}_d(q)$ acting on $\mathbf{F}_p$-vector spaces; sounds hard! The case $|E| = q^d$ isn't special, but natural cases of interest are $\mathbf{F}_q^d$ with usual action and ${\mathfrak{sl}}_d(q)$ with adjoint action. $\endgroup$
    – user76758
    Nov 26, 2013 at 17:27
  • $\begingroup$ @Nick: As the preceding comment points out, cohomology questions are involved here; so it's useful to consult the paper by Cline-Parshall-Scott, van der Kallen and their references: gdz.sub.uni-goettingen.de/dms/load/img/?PPN=GDZPPN002093200 $\endgroup$ Nov 26, 2013 at 19:01
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    $\begingroup$ @Jim Humphreys: This is addressed more to you as the expert on representations of groups of L:ie type in natural characteristic. is it not true that, if $|E|=q^d$ as in Qn 2, the induced module action would have to be either trivial or the natural module or its dual (possibly twisted by a field automorphism)? The basic modules for ${\rm SL}_d(q)$ are all defined over ${\mathbb F}_q$, and tensor products that could be written over smaller fields would have order larger than $q^d$. For the natural module, the only nonsplit extension is for ${\rm SL}_5(2)$. $\endgroup$
    – Derek Holt
    Nov 26, 2013 at 20:35

2 Answers 2

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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.

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  • $\begingroup$ Derek, this is a great answer, thank you. I will need some time to digest what you've written, and I'll also leave time for others to answer before I accept... $\endgroup$
    – Nick Gill
    Nov 27, 2013 at 9:28
  • $\begingroup$ I hope it's helpful and correct! You seem to have accepted already though! $\endgroup$
    – Derek Holt
    Nov 27, 2013 at 9:39
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Here are some further comments, to supplement Derek's focused answer to Q2 as well as my own somewhat cryptic comments.

  1. Q2 is fairly narrow, so the kind of ad hoc treatment Derek suggests is quite appropriate. Note here that you do need to have some information about the $p$-modular representations, for which the dimension bound is crucial. In general, even for finite special linear groups not much is known about the details of their modular representations.

  2. I share Derek's view that Q1 is too general to admit a sensible answer. However, there has been a lot of organized study of extensions and cohomology for finite groups of Lie type, including subtle connections with the corresponding ideas for ambient algebraic groups. Here the 1977 Invent. Math. paper on rational and generic cohomology for which I gave a link is basic. (This paper combines independent work by CPS and by van der Kallen, whom I had the privilege to introduce to each other soon afterward at an Oberwolfach meeting.) For a much more recent treatment of issues related to $H^2$ along with a useful reference list, see the paper (later published in J. Algebra) here. The earliest work tends to rely just on finite group techniques, but by now it's natural to think also about algebraic groups and to deal with general groups of Lie type. There are of course some special features of your groups which might make things easier, but it's a good idea to have the general picture in mind. (For the algebraic groups, the possible non-existence of Levi factors will be one related theme.)

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  • $\begingroup$ Thanks for this Jim - I'll follow up the leads you suggest. (And, ignore my comment above asking which CPS paper - for some reason I didn't see the link first time around.) $\endgroup$
    – Nick Gill
    Nov 28, 2013 at 9:38

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