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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?

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up vote 7 down vote accepted

${\rm GL}(k,2)$ 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.

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Thanks for your answer! Yes, the only case that works is $k=2$, and it is because of exactly what you said: solvable groups of $GL(k,2)$ acting transitively on $V-\{0\}$ have order at most $k(2^k-1)$. This is proved in Passman's Permutation Groups. But it seems to require quite a lot of machinery to set up. Do you know of a shorter proof? – user641 Oct 3 '11 at 12:27
I don't think it's too hard to set up in this situation. You can choose an arbitrary nonzero vector as the 1 element of the field. Then a generator of $N$ acts as a $p$-cycle on nonzero vectors, which you can denote by $(1,t,t^2,\ldots,t^{p-1})$, which gives you the multiplicative structure of the field. The fact that $N$ is acting linearly on $V$ implies the distributive law in the field. The stabiliser of $0$ and $1$ in the group induces automorphisms of both $V$ and $N$, so it respects both the additive and multiplicative structure of the field, and hence induces field automorphisms. – Derek Holt Oct 3 '11 at 15:39
Wow, that is incredibly elegant! Passman's proof is much longer, but probably because he (i) includes all details and (ii) treats a much more general case. And yes you are right; the only prime power $q^m$ such that $q^m+1$ is a power of $2$ is $m=1$, $q$ a Mersenne prime: Passman also proves this. – user641 Oct 3 '11 at 16:40

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