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 $g\in A$?

  • 2
    $\begingroup$ 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 $C_m$. $\endgroup$
    – YCor
    Apr 6 '20 at 10:35
  • 6
    $\begingroup$ Why $5$, Seva? why $12$? What's the real question here? $\endgroup$ Apr 6 '20 at 12:24
  • 1
    $\begingroup$ 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. $\endgroup$
    – Seva
    Apr 6 '20 at 12:27

This is not the case. Here is a counter example.

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.


This site is temporarily in read only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .