# Does $g+A\subseteq A+A$ imply $g\in A$?

Suppose that $$A$$ is a subset of a (large) finite cyclic group such that $$|A|=5$$ and $$|A+A|=12$$. Given that $$g$$ is a group element with $$g+A\subseteq A+A$$, can one conclude that $$g\in A$$?

• The condition "large" is weird: indeed, if there is a counterexample in a cyclic group $C_m$, then we get the same counterexample in $C_{m^n}$ for all $n$. Hence, either you want to omit it (but then you maybe know a counterexample), or you want to replace "large" with "with no small prime divisor". Or make some further assumption, e.g., that $A$ generates $C_m$. – YCor Apr 6 '20 at 10:35
• Why $5$, Seva? why $12$? What's the real question here? – Gerry Myerson Apr 6 '20 at 12:24
• Good question, Gerry; I actually expected someone to ask it!. If the answer were positive, I would be able to exclude one of the countless cases emerging in some large project. – Seva Apr 6 '20 at 12:27

Let $$G =Z_m, \ \ m>20$$ (say). Let $$A = \{0, 1, 3, 4, 6\}; g=2$$: Note that $$g+A=\{2, 3, 5, 6, 8\} \subseteq \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12\}=A+A$$ and that $$g \not\in A$$, $$|A|=5$$, and $$|A+A|=12$$. Note that this example works for all large finite cyclic groups, as the reduction modulo $$m$$ does not matter.