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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 $G=AB$.

Considering https://math.stackexchange.com/questions/882859/groups-of-order-n2-that-have-no-subgroup-of-order-n?lq=1,

$n$ 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.

Thanks in advance for your feedback.

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  • $\begingroup$ 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 $1$.) $\endgroup$
    – Aaron Meyerowitz
    Commented Aug 9, 2014 at 0:06

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