It is easy to see that for any finite Abelian group $G$ and any numbers $a,b$ with $|G|=ab$ there exist a subgroup $A\subset G$ and a subset $B\subset G$ such that $|A|=a$, $|B|=b$ and $G=A+B$, where $A+B=\{a+b:a\in A,\;b\in B\}$.

Problem. Is it true that for any finite abelian group $G$ and numbers $a,b$ with $ab\ge|G|$ there are two subsets $A,B\subset G$ of cardinlity $|A|\le a$ and $|B|\le b$ such that $A+B=G$?

Remark. The answer is affirmative if the group $G$ is cyclic.

It is easy to see that for any finite Abelian group $G$ and any numbers $a,b$ with $|G|=ab$ there exist a subgroup $A\subset G$ and a subset $B\subset G$ such that $|A|=a$, $|B|=b$ and $G=A+B$, where $A+B=\{a+b:a\in A,\;b\in B\}$.

Problem. Is it true that for any finite abelian group $G$ and numbers $a,b$ with $ab\ge|G|$ there are two subsets $A,B\subset G$ of cardinality $|A|\le a$ and $|B|\le b$ such that $A+B=G$?

Remark. The answer is affirmative if the group $G$ is cyclic.