# Combination of two recent problems about finite groups of square orders

Is there any finite group $G$ of order $n=m^2$ satisfying the following conditions?

(a) There is no subgroup of order $m$;

(b) There exist subsets $A$, $B$ such that $|A|=|B|=m$ and $G=AB$.

$n$ must be $\geq 24^2=576$ ($m=24, 28, 30, \cdots$). Also, if answer of this problem is positive, then we obtain a counterexample for the problem Factorization of a finite group by two subsets.

• Since there is only one example for $m=24$ (according to the mentioned problems) maybe you should start by looking at it. You need to find subsets $A,B$ with $AA^{-1} \cap BB^{-1}=\{{1\}}$. It would help to have the two sets intersected be small. Hence one might try things like finding a subgroup of order $12$ (if there are such) and taking $A$ as a union of two (right) cosets. Then $AA^{-1}$ would have order at most $12+24\cdot 12=300.$ To do the same thing for $B$ you would need a different subgroup disjoint from the first (except $1$.) – Aaron Meyerowitz Aug 9 '14 at 0:06