I have a finite-dimensional vector space $E$ over the finite prime field $\mathbf{F}\_p$ and a *solvable* subgroup $G\subset\mathbf{GL}(E)$ for which $E$ is an *irreducible* representation. Do the groups $H^1(G,E)$ and $H^2(G,E)$ both vanish ?


**Addendum** (2016/08/17) For those curious as to why I needed this result, see *Solvable primitive $p$-extensions* (<http://arxiv.org/abs/1608.04673>).



Solvable irreducible subgroups of the $\mathbf{GL}_n$ of $\mathbf{F}_p$ ($p$ prime)
It is true that if $M$ is any (solvable) group with $E \lhd M$ and $M/E \cong G$, (with the action of $G$ on $E$ specified by the given irreducible module action), then $E$ is complemented in $M$, and all complements to $E$ are conjugate. In fact, this does not require solvability of $G,$ only $p$-solvability. All this is well-known, but I outline the proof: Note that $O\_{p}(G) = 1$ since $G$ acts irreducibly on $E$, so that $O\_{p}(M) = E.$ Let $K = O\_{p,p^{\prime}}(M)$. Then $K$ has a unique conjugacy classes of Hall $p^{\prime}$-subgroups, say one of these is $L$, so by a Frattini-like argument we have $M = KN\_{M}(L)$.


Hence $M = ELN\_{M}(L) = EN\_{M}(L).$ Now $E$ is a minimal normal subgroup of $M$ since $G$ acts irreducibly on $E.$ However $E \cap N\_{M}(L) \lhd M$. We can't have $E \leq N\_{M}(L)$, otherwise $[E,L] = 1,$ whereas $G$ acts faithfully on $E$. Hence $E \cap N\_{M}(L) = 1$ and $G \cong N\_{M}(L)$.
 This argument actually shows that every complement to $E$ in $M$ has the form $N\_{M}(L)$ for some Hall $p^{\prime}$-subgroup $L$, of $O\_{p,p^{\prime}}(M)$ and $L$ is unique up to $M$-conjugacy, so all complements to $E$ are $M$-conjugate.


