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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.

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  • $\begingroup$ 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
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    $\begingroup$ @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
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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 $\prod_{i=1}^{r}( \mu(1) - c_{i})).$

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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ @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

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