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) ?



Non-split extensions of $GL_n(F_q)$ by $F_q^n$ ?
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.


