Non-split extensions of $GL_n(F_q)$ by $F_q^n$ ? A very naive question :
I just learned that there is a non-split 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 Cayley-Graves 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) ?
 A: 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 $i>0$.
A: 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 Hochschild-Serre 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 $\langle g v-v\ |\ g \in F^{\times},\ v \in V\rangle$.  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.
A: There is a famous non-split 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 non-split extensions, for example $5^3 \cdot SL_3(5)$.
A: There's an elementary (non-cohomological) group-theoretic 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 $G$.
