Infinite, finitely generated linear group has indices of subgroups divisible by infinitely many primes 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? $$\{p \in \mathbb{Z}: p \text{ is a prime, and } p \mid [G:N] \text{ for some } N \in \mathscr{N} \}$$
 A: 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.
