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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.

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  • $\begingroup$ If $G$ has exponent 3 and $A=\{1,a\}$ has cardinal 2, then the injectivity of the sum map $A\times B\to G$ implies its injectivity on $\{1,a,a^2\}\times B$. This yields many further counterexamples. $\endgroup$ – YCor Oct 13 '18 at 11:14
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I don't think it's true for $G=\mathbb{F}_2^3$ and $a=b=3$.

If there were such sets $A$ and $B$, they must have exactly three elements each.

By applying a translation and a group automorphism, we may as well take $$A=\{(0,0,0),(1,0,0),(0,1,0)\}.$$

Then $B$ either has at most one element with third coordinate zero, in which case $A+B$ has at most three of the four elements with third coordinate zero, or it has at most one element with third coordinate one, in which case $A+B$ has at most three of the four elements with third coordinate one.

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  • $\begingroup$ Yes indeed, you're right. $\endgroup$ – YCor Oct 13 '18 at 11:11

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