[This question](https://mathoverflow.net/questions/91649/subgroups-of-gl2-q) on subgroups of $GL(2,q)$ asked by [Jan](https://mathoverflow.net/users/18394/jan), and especially wonderful answers to it given by [Geoff Robinson](https://mathoverflow.net/users/14450/geoff-robinson), [Ralph](https://mathoverflow.net/users/10194/ralph), and [Will Sawin](https://mathoverflow.net/users/18060/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?



Subgroups of $GL(k,q)$ for bounded $k$
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.


