4
$\begingroup$

This question was previously asked at Math.SE, but didn't receive much attention.

Let $G$ be an infinite finitely generated group and suppose that $G$ is linear, say $G \leq \operatorname{GL}_n(K)$ for some field $K$.

Question: Are there infinitely many primes $p$ such that $p$ divides $[G:N]$ for some normal subgroup $N < G$ of finite index?

This seems to be true when $\operatorname{char} K = 0$, by 10.4 in "Infinite Linear Groups" by Wehrfritz.

In case the formulation of the question is unclear, here is a rephrasing. Let $\mathscr{N}$ be the set of all normal subgroups of $G$ with finite index. Is the following set of primes infinite? $$\{p \in \mathbb{Z}: p \text{ is a prime, and } p \mid [G:N] \text{ for some } N \in \mathscr{N} \}$$

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ [This is almost already said in your post] If $K$ has characteristic zero, for all but finitely many $p$, $G$ is virtually residually $p$. Hence for every large enough prime $p$, there exists $m$ such that $G$ has subgroups of index $mp^k$ for infinitely many values of $k$. — If $K$ has characteristic $p$, $G$ might be only virtually residually-$\ell$ for $\ell=p$ so this approach fails. $\endgroup$
    – YCor
    Commented Jun 14, 2021 at 8:09
  • 1
    $\begingroup$ This might be true anyway using congruences subgroups, some Weisfeiler-like phenomenon and some ring theory (every infinite f.g. $F_p$-algebra has residual fields of arbitrary large cardinal). I'll think about it. $\endgroup$
    – YCor
    Commented Jun 14, 2021 at 8:13

1 Answer 1

Reset to default
3
$\begingroup$

For $G$ a group, let $X(G)$ be the set of $n\ge 1$ such that $n$ divides the index of some finite-index normal subgroup of $G$. Then "normal" can be skipped in the definition (since every finite-index subgroup contains a normal one). In particular, if $H$ has finite index in $G$ then $X(H)\subseteq X(G)$.

The question is

If an infinite finitely generated group $G$ is linear over a field, does it follow that $X(G)$ contains infinitely many primes?

The answer is already to be yes when $K$ has characteristic zero, so let's assume $K$ has characteristic $p>0$. We can suppose for the moment that $K$ is algebraically closed. Because of the "finite index" remark, it is enough to assume that $G$ has a connected Zariski closure. and, hence, passing to a quotient, we can suppose that the Zariski closure of $G$ is either simple or abelian 1-dimensional.

If the Zariski closure is abelian 1-dimensional, then the abelianization of $G$ is infinite, and hence $X(G)=\mathbf{N}_{\ge 1}$. Hence we can suppose that the Zariski closure of $G$ is simple.

We have the following remark:

Lemma: if $W$ is normal finite in $H$ then for every prime $p>|W|$, $p\in X(H)$ implies $p\in X(H/W)$.$\Box$

Hence we can assume that the Zariski closure $S$ of $G$ is simple and simply connected.

Next we use a strong approximation result, essentially due to Pink in positive characteristic (see Theorem 6.1 here):

Let $L$ be a global field of char. $p\ge 0$, $S\subset\mathrm{GL}_d$ be a simply connected simple algebraic $L$-subgroup, and $\Gamma\subset S(L)$ be a Zariski-dense subgroup. Then for a set of natural density 1 (w.r.t. the norm) of primes $P$ of $O_L$, the projection "modulo $P$" of $\Gamma$, which is well-defined for large enough $P$, is onto $S(O_K/P)$.

So the answer should be positive: otherwise, for some finite set $\Xi$ of primes, for overwhelming all primes $P$, denoting by $q_P$ the cardinal of $O_K/P$, we would have that the cardinal of $S(\mathbf{F}_{q_P})$ is divisible only by primes in $\Xi$. This sounds highly unlikely, but I'm not well-enough acquainted in number theory to readily discard this.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thank you for the answer. Some details, just to make sure I understand the beginning correctly. Let $S$ be the Zariski closure of $G$. Then $S/S^\circ$ is finite. Thus $G/G \cap S^\circ$ is finite, so $G \cap S^\circ$ is finitely generated. Furthermore, The Zariski closure of $G \cap S^\circ$ has the same dimension as $S^\circ$, so it must be equal to $S^\circ$. So we can assume that $G$ has connected Zariski closure $S$. $\endgroup$
    – spin
    Commented Jun 17, 2021 at 9:42
  • $\begingroup$ I should have mentioned that my question was motivated from looking at a proof in MR0387428. There they claim the positive answer follows from the result of Platonov (10.4 in Wehrfritz), but in positive characteristic this does not seem to be clear. $\endgroup$
    – spin
    Commented Jun 17, 2021 at 9:47
  • $\begingroup$ For the first argument: in an arbitrary topological space, if a subset $X$ has closure $Y$ and $Y'$ is clopen in $Y$, then $X\cap Y'$ has closure $Y'$ (so no argument using irreducibility or dimension is needed). So, yes, this allows to reduce to connected Zariski closure. $\endgroup$
    – YCor
    Commented Jun 17, 2021 at 19:15
  • $\begingroup$ I see, thanks, that argument seems better. $\endgroup$
    – spin
    Commented Jun 18, 2021 at 1:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .