# Permutation representations of general linear groups over finite fields

Let $q=p^f$ be a prime power, $V$ is a $n$-dimensional vector space over $GF(q)$ and $G=GL(n,q)=GL(V)$. Is every transitive permutation representation $\rho$ of $G$ on $q^n-1$ points isomorphic to the natrual action $\tau$ of $G$ on $V\setminus\{0\}$?

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1. Yes, except for $(n,q)=(2,2)$ or $(2,3)$. 2. It depends how you define isomorphism for permutation representations. For $n \ge 3$, there are two such inequivalent representations, but their images are conjugate in the symmetric group. – Derek Holt Nov 8 '12 at 9:02
Thank you for your answer! It seems that the answer to question 1 will follow from the simplicity of $PSL(n,q)$ provided $(n,q)\neq(2,2),(2,3)$, but I don't get a proof yet. For $\rho$ is isomorphic to $\tau$ in question 2, I mean that there exist an automorphism $\alpha$ of $G$ and a bijection $\phi$ from the set $\Omega$ of $q^n-1$ points to $V\setminus\{0\}$ such that $g^\rho\phi=\phi(g^\alpha)^\tau$ for any $g\in G$. – Binzhou Xia Nov 8 '12 at 13:18
OK, so the answer to 2 is yes. See B.N. Cooperstein, Minimal degree for a permutation representation of a classical group, Israel J. Math. 30 (1978), 213-235. – Derek Holt Nov 8 '12 at 15:24
@DerekHolt The permutation representation $\tau$ has degree $q^n-1>\frac{q^n-1}{q-1}$, so it's not a minimal permutation representation. – Binzhou Xia Nov 9 '12 at 7:12
Yes sorry you are right! Ignore all my comments about 2. I believe the answer is no. I will try and send some examples later. – Derek Holt Nov 9 '12 at 8:33

Apologies for all of my incorrect comments about Question 2. The correct answer is yes for $q=2$, but no otherwise.
A transitive permutation representation of $G$ of degree $q^n-1$ corresponds to a subgroup of $G$ of index $q^n-1$. The stabilizer $H$ of a 1-dimensional subspace of $V$ is a maximal subgroup of $G$ of index $(q^n-1)/(q-1)$. It has a normal elementary abelian subgroup $N$ of order $q^{n-1}$, and $H/N$ is isomorphic to a direct product $C_{q-1} \times {\rm GL}_{n-1}(q)$. This is easy to see directly.
Thje complete inverse image in $H$ of the direct factor ${\rm GL}_{n-1}(q)$ is the stabilizer of a nonzero vector in $V$, and so corresponds to the natural action on vectors. But for $q>2$ there are several other subgroups of $H$ with the same index $q-1$, such as the inverse image in $H$ of $C_{q-1} \times {\rm SL}_{n-1}(q)$. Since these are not isomorphic to the stabilizer of a vector, their corresponding permutation representations are not isomorphic to the natural one.