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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?

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up vote 12 down vote accepted

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.

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OK, I guess I just have to accept this one, because it's extremely beautiful in its simplicity, and makes me most embarrassed for not thinking of it! – Vladimir Dotsenko Mar 20 '12 at 17:31

For a (finite) group $G$, let $R(G)$ be the smallest dimension of any faithful (possibly projective) representation of $G$ over any field. There are many results in the literature giving lower bounds for $R(G)$ for the various classes of finite simple groups.

For example, for $n \ge 9$, we have $R(A_n) = n-2$. So, for $n \ge 9$ and $n > N+2$, $A_n$ is not a subgroup of ${\rm GL}(n,K)$ for any field $K$. I can look up references if you like.

Similarly, for all sufficiently large $d$, if $G$ is an irreducible quasi-simple classical group of dimension $d$ then, $R(G) = d$. So, for example, for large enough $d$, ${\rm GL}(d,2)$ is not a subgroup of ${\rm GL}(d-1,K)$ for any field $K$.

I believe there are also bounds as functions of $d$ on things the derived length of solvable groups with faithful representations of degree $d$ so, for any fixed $d$, there exist solvable (and probably also nilpotent) groups that do not embed into ${\rm GL}(d,K)$ for any $K$.

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That's very nice! For some reason I never fully realised there is a uniform bound in terms of a field, though I must have experienced some evidence of that. – Vladimir Dotsenko Mar 20 '12 at 17:30

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 ${\rm GL}(N,q)$.

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Oh yes, that's very close to what I was thinking about. Now I will wonder forever why I thought there was a problem. Thanks! – Vladimir Dotsenko Mar 20 '12 at 17:28

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