TITLE: Decomposing a finite group as a product of subsets
QUESTION [18 upvotes]: My friend Wim van Dam asked me the following question:

For every finite group $G$, does there exist a subset $S\subset G$ such that $\left|S\right| = O(\sqrt{\left|G\right|})$ and $S\times S = G$?  Also, can we describe such an $S$ explicitly?

I believe it's not hard to show that if we choose $S$ uniformly at random, then $\left|S\right| = O(\sqrt{\left|G\right| \log \left|G\right| })$ suffices.  But this still leaves the problems of giving an explicit construction and of removing the $\log \left|G\right|$ term.
Of course, if $G$ happens to have a subgroup $H$ of size $\approx \sqrt{\left|G\right|}$, then we just take the union of $H$ with a collection of coset representatives and are done.  So, this suggests the possibility of a win-win analysis, where we identify the relatively rare finite groups that don't have a subgroup of the right size and handle them in a different way.  Work on approximate subgroups might also be relevant.
In the likely event that this has already been solved, just a link to the paper would be great (a half hour of googling didn't succeed).

REPLY [12 votes]: It was brought to my attention by Noga Alon that my previous answer (which I keep to avoid any confusion) was in fact incorrect: the Rohrbach conjecture got solved completely by Finkelstein, Kleitman, and Leighton in 1988 ("Applying the classification theorem for finite simple groups to minimize pin count in uniform permutation architectures"), and independently by Kozma and Lev in 1992 ("Bases and decomposition numbers of finite groups"). 
Explicitly, there exists a subset $S\subseteq G$ of size  $|S|\le(4/\sqrt 3)|G|^{1/2}$ such that every element of $G$ is representable as a product of two elements from $S$.