TITLE: Subgroups of $GL(k,q)$ for bounded $k$
QUESTION [7 upvotes]: This question on subgroups of $GL(2,q)$ asked by Jan, and especially wonderful answers to it given by Geoff Robinson, Ralph, and Will Sawin showing that "almost no finite groups" inject in $GL(2,q)$ made me wonder (completely recreationally, I have to admit) whether there exists $N$ such that every finite group, or "most finite groups" inject in $GL(N,q)$. 
Probably no such $N$ exists, but the ideas I had when thinking about the $N=2$ case use the specifics of the $2\times2$-situation way too much. Is it true, for instance that, along the lines of Ralph's and Will's answer, an abelian $p$-subgroup of $GL(N,q)$ may only have a bounder number of cyclic factors?

REPLY [12 votes]: If $p$ is prime, the least dimension for a faithful representation of $(\mathbf{Z}/p)^d$ over any field of characteristic $\neq p$ is $d$. The argument is very easy, as you can assume the field algebraically closed and diagonalize. 
It follows that if $G$ is a group containing isomorphic copies of $(\mathbf{Z}/p)^d$ for two distinct primes $p$ (e.g. $(\mathbf{Z}/6)^d$) then its smaller faithful representation over some field is of dimension $d$.
Note: for $(\mathbf{Z}/6)^d$, on the other hand there is a faithful representation in dimension two over a commutative ring (the product of finite fields $F_{2^d}\times F_{3^d}$), and looking at representations in arbitrary commutative rings seems more natural for obvious stability reasons.
 On the other hand, if $S$ is finite, simple non-abelian and has a faithful rep. in dimension $d$ over a commutative ring, then it has one over a finite field.

REPLY [9 votes]: Yes, the idea that works for ${\rm GL}(2,q)$ with Abelian subgroups works with higher derived lengths for other ${\rm GL}(N, q).$ If $p$ is a prime which does not divide $q$ then the Sylow $p$-subgroups of ${\rm GL}(n,q)$ are monomial (up to equivalence) over some extension field. An easy indction argument (no pun intended) show that a monomial $p$ goup with a faithful representation of degree $p^{k}$ or less has derived length at most $k+1.$ Hence the Sylow $p$-subgroups of ${\rm GL}(N,q)$ have derived length at most $1 +\log_{p}(N)$ for all such $p.$ If $q$ is a power of $p,$ the bound on the derived length of a Sylow $p$-subgroup of ${\rm GL}(N,q)$ is similar. The nilpotence class of the upper unitriangular group is at most $N-1$, so the derived length is at most $\log_{2}(N),$ because for any nilpotent group $U,$ we have $U^{(k)} \leq L_{2^{k}}(U),$ where $U^{(k)}$ is the $k$-the term of the derived series and $L_{m}(U)$ is the $m$-th term of the lower central series. Hence  in particular, no $p$-group of derived length $2 + \log_2(N)$ or more is a subgroup of any ${\rm GL}(N,q)$.