Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $V\setminus\{0\}$?

Thanks in advance!

share|cite|improve this question
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
up vote 3 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.