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.



Transitive actions of finite subgroups of ${\rm GL}(n,\Bbb Z)$ on projective geometries
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](https://doi.org/10.1090/S0002-9939-1988-0915710-1) 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 $\prod\_{i=1}^{r}( \mu(1) - c\_{i})).$


