More precisely, what is the smallest exponent e such that, for every n, there exists a group of size at most Cn^e for some absolute constant C and with an n-dimensional irreducible complex representation? 


I know that e ≤ 3. For various special values of n we can attain the lower bound e ≥ 2. For example, if n+1 = q is a power of a prime then the group of affine linear transformations ax + b on Fq is doubly transitive, so the linearization of this group action decomposes into the trivial representation and an irreducible representation.


Edit: Oops. Reading Noah's comment, I probably should've justified the upper bound. I hope this example is correct: let zeta be an nth root of unity and consider the group generated by the diagonal matrix (zeta, zeta, ... zeta), the diagonal matrix (1, zeta, zeta^2, ... zeta^{n-1}), and a cyclic permutation matrix of order n. It can be verified that this is a group of order n^3 and that its elements span M\_n(C) , so the given representation is irreducible.



How small can a group with an n-dimensional irreducible complex representation be?
I [posted](https://mathoverflow.net/questions/834/arithmetic-progressions-without-small-primes) a question about primes in arithmetic progressions and was told my understanding is wrong. Although proving this would be harder than the generalized Riemmann hypothesis, the expectation of experts is that, for any n, there is a prime q which is 1 mod n and is O(n^{1+ \epsilon}). In that case, Z/n acting on Z/q would achieve e = 2+\epsilon.


It seems like the best result we are likely to prove is that the only way to beat e=3 is to find a small prime power which is 1 mod n.


