# On decomposition of finite Abelian groups

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.

• 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. – YCor Oct 13 '18 at 11:14

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.