TITLE: Minimal size of subsets $A,B$ in a finite group $G$ such that $AB=G$
QUESTION [12 upvotes]: Given a finite group $G$ with $N$ elements what are the "smallest" possible subsets
$A,B$ of $G$ such that $G=AB$ (ie. every element of $G$ is a product of an element in $A$
and an element in $B$)? 
We have of course $\sharp(A)\sharp(B)\geq N$
but can one always find two subsets $A,B$ of size roughly $\sqrt N$ with the property $G=AB$?
Since I do not know the optimal answer a good definition of "roughly" 
should be part of an answer (one can surely not do better than
$A,B$ of size $O(1)+\sqrt N$ but I guess this is much to optimistic.
It is perhaps more reasonable to hope for $\sqrt N$ times something bounded).
The optimal answer $A,B$ of size (at most) $1+\sqrt N$ works for cyclic groups: take 
$A=\mathbb N\left\lceil \sqrt N\right\rceil\cap \{0,\dots,N-1 \}$
and $ B= \{ 0,1,\dots,\left\lfloor \sqrt{N} \right \rfloor \} $ in $\mathbb Z/N\mathbb Z=
\{0,\dots,N-1\}$.
On the other hand, if $G$ has a (not-necessarily normal) subgroup $H$ of size roughly $\sqrt{N}$ one can 
consider $A=H$ and take for $B$ coset-representantives.
Combining the two constructions we get a nearly optimal result for abelian groups.
Is there for example a sequence of "bad" groups $G_i$ such that, say
$$\left((1+\sharp(A_i))(1+\sharp(B_i))-\sharp(G_i)\right)\Big/\sqrt{\sharp(G_i)}$$
is unbounded for every (sequence of) subsets $A_i,B_i\subset G_i$ 
such that $G_i=A_iB_i$? (The existence of such a sequence would of course
the hope for the existence of $A,B$ of size $O(1)\sqrt{\sharp(G)}$ (with $AB=G$)
in every finite group $G$.

REPLY [10 votes]: For square root of N times something bounded, this can be done with A=B (and the bound
being 4 over the square root of 3) according to
Kozma and Lev, Bases and decomposition numbers of finite groups
Arch. Math. (Basel) 58 (1992), 417-424
which seems to do as you say, namely find a subgroup of order about the square root of
N (via an appeal to CFSG).
It wouldn't surprise me if the O(1)+square root of N case is open.