Solvable irreducible subgroups of the $\mathbf{GL}_n$ of $\mathbf{F}_p$ ($p$ prime) 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).
 A: Here's a slightly different answer, less group-theoretic and more representation-theoretic than Geoff's.
Rephrasing your question in terms of $\mathbb{F}_pG$-modules, you are asking about a faithful irreducible $\mathbb{F}_pG$-module $E$ for a finite solvable group $G$. 
Let $N$ be a minimal normal subgroup of $G$. Since $G$ is solvable, $N$ is an elementary abelian $q$-group for some prime $q$. By Clifford's Theorem, the restriction of $E$ to $N$ is semisimple.
If $q=p$, then the restriction of $E$ to $N$ is trivial, since the only simple $\mathbb{F}N$-module for a $p$-group $N$ is the trivial module. But this contradicts the faithfulness of $E$.
If $q\neq p$ then $e_N=\frac{1}{\vert N\vert}\sum_{n\in N}n$ is a central idempotent of $\mathbb{F}_pG$. Let $f_N=1-e_N$.
$E=Ee_N\oplus Ef_N$, and since $E$ is irreducible, either $E=Ee_N$ or $E=Ef_n$.
If $E=Ee_N$ then, since the trivial $\mathbb{F}_pN$-module is the only irreducible $\mathbb{F}_pN$-module which is not annihilated by $e_N$, $N$ acts trivially on $E$, again contradicting the faithfulness of $E$.
So $E=Ef_N$, which means that $E$ and the trivial module $\mathbb{F}_p$ are in different blocks of $\mathbb{F}_pG$, and so $H^i(G,E)=\operatorname{Ext}^i_{\mathbb{F}_pG}(\mathbb{F},E)=0$ not only for $i=1,2$ but for all $i$.
A: 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.
A: The question is whether every extension $\Gamma$ of $G$ by the
$G$-module $E$ splits, and whether any two sections of the split extension
are conjugate.  So, as Geoff says, we have to show that $E$ is
complemented in $\Gamma$, and any two complements are conjugate.
Here is an elementary argument sent by a correspondent :
Since the quotient $G=\Gamma/E$ is soluble, a minimal normal subgroup
of $G$ is an elementary abelian $q$-group $N/E$ for some prime $q$
different from $p$.  By Sylow's Theorems there is a complement $Q$ to
$E$ in $N$.  Consider the normaliser $H$ of $Q$ in $\Gamma$.  By the
Frattini argument (simply the fact that $Q$ is a Sylow $q$-subgroup of
$N$, so $N$ acts transitively, hence $\Gamma$ also acts transitively,
by conjugation on the set of conjugates of $Q$ in $N$, which is the
set of conjugates of $Q$ in $\Gamma$ since $N$ is normal in $\Gamma$),
$\Gamma = H.E$.  But the intersection of $H$ and $E$ is trivial (any
element of $E$ that normalises $Q$ centralises it since $E$ is normal
in $QE$, whereas $E$ is its own centraliser in $\Gamma$).  Thus $H$ is
a complement of $E$ in $\Gamma$ and moreover, any complement of $H$ in
$\Gamma$ has a conjugate of $Q$ as a normal subgroup and therefore is
conjugate to $H$.  That is to say, there is a unique conjugacy class
of complements of $E$ in $\Gamma$ (all of them maximal proper
subgroups of $\Gamma$).
Addendum.  The correspondent has further provided the following references :
on p.55 of `Finite Soluble Groups' by Doerk & Hawkes (De Gruyter 1992),
on p.102 of `Group Theory II' by Suzuki (Springer 1986),  and
on p.159 of `Endliche Gruppen I' by Huppert (Springer 1967). 
