2-transitive action on vector space of char. 2 Suppose $V$ is an elementary abelian 2-group of order $2^k$, with $k>2$.  Let $H\le GL(k,2)$ be a solvable group of automorphisms of $V$.  How does one prove that $H$ cannot act 2-transitively on the non-zero vectors of $V$?
It seems counterintuitive for such an action to exist, since a point stabilizer in $H$ would have an orbit of size $|V|-2$. And of course $GL(2,k)$ is simple in this case.  But I don't see how to proceed. I actually wonder if the solvability of $H$ is necessary; in particular:
What would happen if we dropped the solvability hypothesis?
 A: ${\rm GL}(k,2)$ itself acts 2-transitively on $V \setminus \{0\}$, so solvability of $H$ is certainly necessary. In fact, taking $k=2$ is a counterexample to what you want to prove, because ${\rm GL}(2,2)$ is solvable.
However, I think that is the only solvable counterexample. As a 2-transitive group, $H$ is primitive, and so any nontrivial normal subgroup of $H$ acts transitively. Let $N$ be a minimal normal subgroup of $H$. Then $H$ solvable implies $N$ is elementary abelian of order $p^m$ for some prime $p$. Then $N$ must act regularly on $V \setminus \{0\}$, so we have $2^k - 1 = p^m$.
I am not so strong on number theory, but I believe that is only possible for $m=1$, with $p$ a Mersenne prime. If $m$ is even, then $p^m+1 \equiv 2 \bmod 4$, so that's not possible. But for $m>1$ odd, Zsigmondy's Theorem says that there is a prime dividing $p^m+1$ but not $p+1$, so $p^m+1$ cannot be a power of 2.
So $N$ has prime order. It can be shown then that $V$ can be identified with the field of order $2^n$, where the action of $N$ on $V$ corresponds to multiplication in the field, and the normalizer of $N$ in ${\rm GL}(k,2)$ modulo $N$ is isomorphic to the Galois group of the field - so it is cyclic of order $k$. Hence it acts transitively by conjugation  on $N \setminus \{1\}$ only when $k=2$ and $|N|=3$. I can provide more details or perhaps a reference for that argument if necessary. 
