A very naive question : I just learned that there is a nonsplit extension of $GL_3(F_2)$ by $F_2^3$ (with standard action). It can be realized as the subgroup of the automorphism group $G_2$ of CayleyGraves octaves (edit: octonions) that preserve up to sign the basis $e_i$, $i=1..7$of imaginary octaves. Does this happen for other values of $(n,q)$ (as in the title) ?

2$\begingroup$ Shouldn't this essentially always happen? Look at $V$ an $n$ dimensional vector space over $F_q$. Take the collection of affine transformations. This is a group, and it's an extension of this sort (or do I have the order backwards?) and it won't be split. $\endgroup$– Charles SiegelJun 24 '10 at 19:35

3$\begingroup$ The affine transformations are a semidirect product, so the exact sequence does split. $\endgroup$– Andy PutmanJun 24 '10 at 19:39

2$\begingroup$ I assume the OP means the octonions. $\endgroup$– Qiaochu YuanJun 24 '10 at 20:06

2$\begingroup$ Sorry, I said CayleyGraves to avoid ambiguity, and spread it instead. I meant octonions, the only eightdimensional normed division algebra over the reals. $\endgroup$– BS.Jun 24 '10 at 22:39

2$\begingroup$ Charles, for groups and modules, the meaning is the same: an extension $A\to B\to C$ splits if the second map admits a right inverse $C\to B.$ $\endgroup$– Victor ProtsakJun 25 '10 at 7:25
This never happens for finite fields $F \neq \mathbb{F}_2$. If a group $G$ acts on an abelian group $M$, then short exact sequences
$1 \rightarrow M \rightarrow \Gamma \rightarrow G \rightarrow 1$
are classified by elements of $H^2(G;M)$. It is thus enough to show that if $F \neq \mathbb{F}_2$ is a finite field and $V = F^n$, then $H^2(GL_n(F);V)=0$. In fact, we will show that $H^k(GL_n(F);V)=0$ for all $k$.
We have a short exact sequence
$1 \rightarrow F^{\times} \rightarrow GL_n(F) \rightarrow PGL_n(F) \rightarrow 1.$
Associated to this is the HochschildSerre spectral sequence in cohomology with coefficients in $V$. The $E_2$term is $H^p(PGL_n(F);H^q(F^{\times};V))$. The key fact here is that $H^q(F^{\times};V)=0$ for all $q$.
On page 58 of Brown's book on group cohomology, there is a calculation of the cohomology of finite cyclic groups with nontrivial coefficients. In the case we're considering, it goes as follows. Define $N = \sum_{x \in F^{\times}} x \in \mathbb{Z}[F^{\times}]$ (of course, $N$ acts as $0$ on $V$, but forget that for the moment). We then get a map $N : V \rightarrow V$ whose image lies in the ring of invariants $V^{F^{\times}}$ and which satisfies $N(gv)=N(v)$ for all $g \in F^{\times}$ and all $v \in V$. Let $V_{F^{\times}}$ be the ring of coinvariants, ie the quotient of $V$ by the subspace spanned by $\langle g vv\ \ g \in F^{\times},\ v \in V\rangle$. We get an induced map $\overline{N}:V_{F^{\times}} \rightarrow V^{F^{\times}}$. The result then is that $H^0(F^{\times};V) = V^{F^{\times}}$, that $H^i(F^{\times};V) = ker\ \overline{N}$ for $i \geq 1$ odd, and that $H^i(F^{\times};V) = coker\ \overline{N}$ for $i \geq 1$ even. But clearly $V^{F^{\times}} = 0$, and since $F$ is not the field with $2$ elements we also have $V_{F^{\times}} = 0$. The result follows.

2$\begingroup$ Computing H^2 of even modestly sized finite groups with twisted coefficients is pretty computationally intensive. The real problem is that a 2cocycle is not determined by its values on a generating set for the group, so you have to actually keep track of EVERY pair of elements of the group. An amusing size effect of this is a theorem of Gordon that says that there is no algorithm for computing even untwisted H^2 of a group given by generators and relations (of course, it is easy to compute twisted H^1 of such a group...). $\endgroup$ Jun 25 '10 at 0:24

2$\begingroup$ You could use the "cohomolo" package for n ≤ 6. H^2(GL(n,2),2^n) has dimension 1 when n=3,4,5, and 0 for n=1,2,6. Perhaps those calculations suffice. In case you are curious, H^1(GL(n,2),2^n) has dimension 1 when n=3, and 0 for n=1,2,4,5,6,7,8. You definitely don't need to keep track of every pair for a cocycle. For GL(n,2), you only need polynomially many values in n, even if you want to efficiently to compute any other value (and not just know it is determined). I think C*n^4 should suffice, and the calculations to get the cohomology are somewhere in the n^8 to n^12 range. $\endgroup$ Jun 25 '10 at 1:03

1$\begingroup$ As a side note, this might be the longest ontopic comment thread I've seen in a long time! Though this comment probably breaks that... $\endgroup$ Jun 25 '10 at 2:52

1$\begingroup$ I think most of it is folklore. H^2 is harder than H^1. H^1 depends on the number of generators, H^2 more on the number of relations (rewrite rules). In the soluble world, the number of rewrite rules is the square of the number of generators (composition length). In the finite world, if the field size of composition factors is bounded, then the number of rewrite rules stays polynomial in the max rank and composition length. Squier, Groves, and Anick described using rewriting systems for cohomology, but I guess people didn't know it actually worked until recently. $\endgroup$ Jun 25 '10 at 5:17

1$\begingroup$ Many thanks to Andy and Jack for your very interesting answer and comments. This is definitely an answer ! $\endgroup$– BS.Jun 25 '10 at 11:39
There is a famous nonsplit extension called the "Dempwolff group", $2^5 \cdot GL_5(2) = 2^5 \cdot SL_5(2)$. And apparently this is the largest case for which it happens, as you can see from the Wikipedia page http://en.wikipedia.org/wiki/Dempwolff_group.
If you consider $SL_n$ rather than $GL_n$, there are more nonsplit extensions, for example $5^3 \cdot SL_3(5)$.

$\begingroup$ If I could, I would credit you the answer, since it completely answers my question. $\endgroup$– BS.Jan 13 '11 at 22:29
There's an elementary (noncohomological) grouptheoretic explanation for why $GL(n,q)$ must split over its natural module in odd characteristic. Suppose $N \triangleleft G$, where $G/N \cong GL(n,q)$ with $q$ odd and $N$ is isomorphic to the natural module for $G/N$. The negative of the identity matrix in $GL(n,q)$ is a central element of order $2$ that acts on $N$ with no nonidentity fixed points. Thus there exists $T \triangleleft G$ with $N \subseteq T$ such that $T/N$ has order $2$, and if $S$ is a Sylow $2$subgroup of $T$ then ${\bf C}_N(S) = 1$. Now let $K = {\bf N}_G(S)$. By the Frattini argument, $G = TK = NSK = NK$. Also, $N \cap K = 1$ since $T$ must centralize $N \cap K$. Thus $K$ is the desired complement for $N$ in $G$.

2$\begingroup$ This is a very nice argument, and it seems to me that it works for all $q>2$: just replace $I$ by $aI$ where $a\in \mathbf{F}_q^*$ has prime order $p$ (multiplicatively, so $p$ divides $q1$). $\endgroup$ Jul 10 '15 at 22:34

$\begingroup$ @LadischRight. I should have thought of that. $\endgroup$ Jul 11 '15 at 1:15
I'm writing this as an "answer" because (a) there are a number of comments, and (b) I don't know if it would fit in a comment.
Let $F$ a finite field, and let $V$ a finite dimensional $F$vector space, and view $V$ as an $F^\times$module via multiplication. Then as pointed out in Andy Putman's answer, $H^i(F^\times,V) = 0$ for all $i \ge 0$ provided $F > 2$.
Well, it is clear enough under the assumption $F>2$ that $H^0(F^\times,V) = V^{F^\times} = 0$. For the higher cohomology vanishing, there is no need to use the description of "cohomology of cyclic groups" to obtain this vanishing; the point is just that $F^\times$ is invertible in $F$. Use the following generality:
Let $H$ be a subgroup of finite index $n$ in a group $G$. If $M$ is a $\mathbf{Z}G$module, then $\operatorname{Cor} \circ \operatorname{Res}$ is multiplication by $n$ on $H^\bullet(G,M)$, where $\operatorname{Cor}:H^\bullet(H,M) \to H^\bullet(G,M)$ denotes the corestriction and $\operatorname{Res}:H^\bullet(G,M) \to H^\bullet(H,M)$ the restriction; see e.g. Serre's Local Fields VII.7, VIII.2.
Let now $k$ be a commutative ring (with 1), suppose that $H=1$ and that $n = [G:1]= G$ is invertible in $k$. If $M$ is a $kG$module (i.e. a $k$module with $k$linear $G$ action), then all $H^i(G,M)$ are $k$modules and $H^i(H,M) = H^i(1,M) = 0$ for $i>0$. For $i>0$, the preceding result shows these $k$modules to be annihilated by the unit $n$ of $k$; thus $H^i(G,M) = 0$ for $i>0$.
To apply this result in the original setting, take $k=F$, $M=V$ and $G=F^\times$; we find that $H^i(F^\times,V) = 0$ for $i>0$.

1$\begingroup$ For some strange reason, mathematicians have trouble spelling my last name... $\endgroup$ Jun 25 '10 at 14:01

$\begingroup$ oops  very sorry about that! (Thanks for the edit...) $\endgroup$ Jun 25 '10 at 14:19

4$\begingroup$ And a completely straightforward generalization of Andy's argument: Let $R$ be a ring and $V$ an $R$module. When can we say $H^i(Aut(V),V)=0$ for all $i$? It's enough if $Aut(V)$ has a normal subgroup $G$ such that $H^i(G,V)=0$ for all $i$. As long as no nonzero element of $R$ annihilates $V$, the units in the center of $R$ will inject into the center of $Aut(V)$. So then it suffices if $R$ has a central element $u$ such that both $u$ and $1u$ are units. Here $u$ may have finite or infinite order. $\endgroup$ Jun 25 '10 at 15:14
