Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
add comment

3 Answers 3

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.

share|improve this answer
    
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
add comment

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 $K$.

share|improve this answer
    
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
add comment

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

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.