Question

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) ?

Answer 1

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.

Answer 2

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 $5^3 \cdot SL_3(5)$.

Answer 3

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 $i>0$.