For any $n$, the group ${\rm GL}(n,\Bbb Z)$ has a natural action on $\Bbb Z^n$. Modding out a prime $p$ yields an action on the vector space $F_p^n$, where $F_p$ is the finite field with $p$ elements. Projectivising gives an action on the finite projective geometry $PG_{n-1}(p)$. Therefore, every subgroup of ${\rm GL}(n,\Bbb Z)$ has a natural action on $PG_{n-1}(p)$, for every prime $p$.

Question: Do there exist arbitrarily large primes $p$ such that, for some $n>1$, there is a finite subgroup of ${\rm GL}(n,\Bbb Z)$ whose natural action on the points of $PG_{n-1}(p)$ is transitive?

Motivation: in https://arxiv.org/abs/1502.06114, Corollary 6.6, I defined a function $k_n$; in my proof of its existence, this function grows with $n$. The referee has asked the reasonable question of whether or not the actual function is unbounded. I believe this is equivalent to the question I'm asking here, but I don't know the answer.

  • $\begingroup$ Since scalar multiplication by any $a\in (\mathbb Z/(p\mathbb Z))^*$ is invertible, your question is the same as asking whether $GL(n,\mathbb Z/p\mathbb Z)$ acts transitively on $F_p^n\setminus\{0\}$, right? $\endgroup$ – Anthony Quas Oct 18 '16 at 16:13
  • 3
    $\begingroup$ @AnthonyQuas No, because GL(n, Z/pZ) is not a subgroup of GL(n,Z). $\endgroup$ – Joy Morris Oct 18 '16 at 16:14
  • $\begingroup$ Oh yes - I see that now. $\endgroup$ – Anthony Quas Oct 18 '16 at 16:16
  • $\begingroup$ @Joy Morris: Just to clarify, are you asking if there are finite subgroups of $GL(n,\mathbb{Z})$ which act transitively on $\mathbb{F}_p^n\setminus \{0\}$ (modulo scalars)? $\endgroup$ – Venkataramana Oct 18 '16 at 16:45
  • $\begingroup$ @Venkataramana: Yes, that is the question. $\endgroup$ – Dave Witte Morris Oct 18 '16 at 16:53

Probably final revision: I am indebted to Dave Witte-Morris, who added a reference to a refinement of Zsigmondy's Theorem by W. Feit, of which I was unaware, and pointed out that consequently, a complete answer to the question implicitly followed from what was previously written.

In fact, going beyond Dave Witte-Morris's original suggestion (but still making use of Feit's result in its more precise statement as his Theorem A), we may obtain the sharper conclusion that if there is a transitive action of the form requested by the OP, then $p \leq 3$, so I now incorporate this observation.

The order of a finite subgroup $X$ of ${\rm PGL}(n,\mathbb{Z})$ is always a divisor of $\frac{(2n)!}{2}$( by the result of Blichfeldt or Minkowski mentioned below, applied to the preimage $G$ of $X$ in ${\rm GL})$. In particular, the order of a finite subgroup of ${\rm PGL}(2,\mathbb{Z})$ divides $12.$ However, we note that ${\rm GL}(2,\mathbb{Z})$ has no quaternion subgroup of order $8,$ from which it easily follows that a finite subgroup of ${\rm GL}(2,\mathbb{Z})$ has a normal $2$-complement (necessarily of order dividing $3$), and also that ${\rm PGL}(2,\mathbb{Z})$ has no subgroup of order $12.$

Now ${\rm GL}(6,\mathbb{Z})$ contains no element of order $31,$ so we need not concern ourselves with the case $p =5, n = 6.$

A finite group $H$ which acts transitively on a set of $\frac{p^{n}-1}{p-1}$ points with $n >1$ has order at least $p+1$ (and if $n =2$, then such a group has order divisible by $p+1$). Hence if there is a prime $p$ as asked for in the question, we certainly have $2p < (2n)!$. If $n = 2,$ we can exclude $p = 7,$ since $p+1 =8$ does not divide the order of ${\rm PGL}(2,\mathbb{Z}).$ If $X$ is a finite subgroup of ${\rm PGL}(2,\mathbb{Z})$ of order $6,$ then $X$ is the image of a subgroup $G \cong ( \mathbb{Z}/2 \mathbb{Z}) \times S_{3}$ in which every non-central involution has the eigenvalue $1$. Thus $X$ does not act regularly on ${\rm PG}_{1}(5),$ and hence does not act transitively as $|X| = 6 =|{\rm PG}_{1}(5)|.$

We can also exclude the possibility $p =11$ when $n =2,$ since no subgroup of ${\rm PGL}(2,\mathbb{Z})$ has order $12.$

Hence we may ( and do, from now) assume that $n >2$ and $p \geq 5.$ Furthermore, if $ p = 5,$ we may suppose that $n \neq 6,$ so we do.

A strengthening of Zsigmondy's theorem proved by Feit as his Theorem A shows (as $p \geq 5$ and $n >2,$ with $n \neq 6$ if $p=5$), that there is a "large" Zsigmondy prime divisor $q$ of $\frac{p^{n}-1}{p-1}$, which means we can assume either that $q > n + 1$ or else that $p^n - 1$ is divisible by $q^2$. Then $q$ divides $\frac{p^{n}-1}{p-1}$ and $p$ has multiplicative order $n$ in $\mathbb{Z}/q\mathbb{Z},$ so that $q \equiv 1$ (mod $n$).

However, by Cauchy's theorem, $G$ contains an element, say $x,$ of order $q.$ But $\langle x \rangle$ has no non-trivial irreducible representation of degree less than $q-1$ over $\mathbb{Q},$ so we must have $n = q-1$ (this also implicitly shows that $G$ contains no element of order $q^{2}).$ Thus $x$ has trace $-1$ in the given representation, and a Lemma of Blichfeldt (or maybe Minkowski, as J-P. Serre attributes it) shows that $\langle x \rangle$ must be a Sylow $q$-subgroup of $G,$ so that, in particular, $\frac{p^{n}-1}{p-1}$ is not divisible by $q^{2}.$ Since $q$ is a Zsigmondy prime, we know that $p - 1$ is not divisible by $q$, so this implies that $p^{n}-1$ is not divisible by $q^{2}.$ This contradicts the fact that $q$ is a "large" Zsigmondy prime.

( For the sake of completeness, I outline the argument of Blichfeldt used, or an alternative using basic character theory: Let $Q$ be a Sylow $q$-subgroup of $G,$ and let $\chi$ denote the character of $G$ afforded by the given representation. Then $Q$ has exponent $q,$ and we have see that $\chi(x) = -1$ for each non-identity element of $Q$. Now the (necessarily integral) multiplicity of the trivial character in ${\rm Res}^{G}_{Q}(\chi)$ is given by $\frac{1}{|Q|} ( (q-1) - (|Q|-1)).$ Hence $|Q|$ divides $|Q|-q$ which forces $|Q| = q.$ The result of Blichfeldt alluded to is the observation that if $\mu$ is a faithful complex character of a finite group $X$ and $c_{1},c_{2}, \ldots, c_{r}$ are all the distinct values assumed by $\mu$ on non-identity elements of $X$ then $|X|$ divides $\prod_{i=1}^{r}( \mu(1) - c_{i})).$

  • $\begingroup$ en.wikipedia.org/wiki/Zsigmondy%27s_theorem $\endgroup$ – YCor Oct 19 '16 at 16:13
  • $\begingroup$ We can thus wonder in general, for prime $q$, what is the maximal order of a finite subgroup of $GL_{q-1}(\mathbf{Z})$ of order divisible by $q$. (This in particular discards subgroups of maximal order, namely $2^{q-1}(q-1)!$ if $q\ge 13$, see mathoverflow.net/questions/168292) $\endgroup$ – YCor Oct 19 '16 at 16:23
  • 1
    $\begingroup$ It seems to me that once you have $n+1 = q$ is prime (and not two) you can run virtually the same argument on a Zsygmondy prime dividing $p^m-1$ where $m = n/2$ to get that $m+1$ is also prime. $\endgroup$ – Nate Oct 19 '16 at 19:52
  • 1
    $\begingroup$ Thinking twice, we can get a twisted form by considering the action of $\mathrm{Alt}_q$ on the subgroup of $\mathbf{Z}^q$ of those $(n_1,\dots,n_q)$ with $\sum n_i=0$ and $q| n_i-n_j$ for all $i$; this is an action on a certain free abelian group of rank $q$. It is not integrally conjugate to the standard action, because the action preserves an additive subgroup of index $q$ (namely those $q$-tuples of $q\mathbf{Z}$ with sum 0). Anyway I doubt it's relevant to the original question (my mistake! but I like digressions). $\endgroup$ – YCor Oct 20 '16 at 7:04
  • 1
    $\begingroup$ @GeoffRobinson I edited your answer, because the first half of your argument provides a complete proof if you replace Zsigmondy's theorem with an improved version due to Feit. Well done! $\endgroup$ – Dave Witte Morris Oct 21 '16 at 7:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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