TITLE: How small can a group with an n-dimensional irreducible complex representation be?
QUESTION [18 upvotes]: 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.

REPLY [2 votes]: Note that for every odd prime $p$, the finite group ${\rm PSL}(2,p)$ has an irreducible complex character of degree $p$, and has order $\frac{p(p^{2}-1)}{2}$.  Hence if $n = p_{1}^{a_{2}}\ldots p_{r}^{a_{r}}$ where each $a_{i} \in \mathbb{N}$ and the $p_{i}$ are distinct odd primes, then there is a finite group $G$ of order less than 
$\frac{n^{3}}{2^{t}}$, where $t$ is the number of odd prime divisors of $n$ (counting multiplicities), which has a faithful complex irreducible character of degree $n$. The group $G$ is a direct product of groups of the form ${\rm PSL}(2,p_{i})$
($a_{i}$ factors ${\rm PSL}(2,p_{i})$ for each $i$). If $n$ is as above, and $h = 2^{a}n$ for some positive integer $a$, then we may take the direct product of $a$ copies of ${\rm SL}(2,2)$ with the previous group to obtain a group of order less than $\left(\frac{3}{4}\right)^{a} \frac{h^{3}}{2^{t}}$ with an irreducible character of degree $h$ (and if $h = 2^{a}$, a case which, strictly speaking, we have not yet covered, we obtain a group of order $\frac{2^{3a}3^{a}}{4^{a}}$ with a faithful complex irreducible character of degree $2^{a}$). Actually, since $S_{4}$ has an irreducible character of degree $4$, 
we can find a group of order $(24)^{\frac{a}{2}}$ with an irreducible character of degree $2^{a}$ when $a \geq 2$ is even, which improves $6^{a}.$ Likewise, we can find a group of order $ 6 \times (24)^{\frac{a-1}{2}}$ with an irreducible character of degree $2^{a}$ when $a \geq 1$ is odd, which improves $6^{a}$ for $a \geq 3$ So this improvement could also be used to improve cases when $h$ above is even).
This is only a marginal improvement of the bound obtained by the construction in the question, and does not impact on the optimal choice of $e$ asked for.