# About a sumset in $\mathbb{Z}_{2k}$

Suppose $U\subset\mathbb{Z}_{2k}$ with $|U|=k$. Let $U^c$ denote the complement of $U$.

Let $v\in \mathbb{Z}_{2k}^{\times}$. How much is it known about $U+vU^c$?

For example: When $U+v\complement U = \mathbb{Z}_{2k}$?

Yet another example: if $U$ is aperiodic (i. e., its isotropy group is trivial), Kemperman's theorem implies $U-\complement U = \mathbb{Z}_{2k}\setminus \{0\}$. Anything beyond that?

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Can you make your question more specific? What information do you want to know about $U+vU^c$? It is certainly possible for $U+vU^c$ to equal all of $\mathbb Z_{2k}$, and also possible for it to equal just $U$ again: take $U=2\mathbb Z_{2k}$ and $v$ odd or even, respectively. – Greg Martin Feb 9 '13 at 6:17
The element $v$ cannot be even, for it belongs to $\mathbb{Z}_{2k}^{\times}$. – Lurgul Feb 11 '13 at 23:39
Your yet another example seems incorrect. Greg Martin's comment might contain a minor misformulation (with the odd and even) but it is very much to the point. Take U as he suggested. Then your group is the union of U and the other coset modulo U, write it e+U. So U - C U = U - (e+U) = - e + U = e + U. And this is the same for this U for any v. What you say should/could be true if you assume U to be aperiodic. Likely if U is periodic (ie there is a nontriv subgroup such that U = U + H) you could mod out by the stabilizer H (which is also stab of the complement) and so reduce to aperiodic. – user9072 Feb 16 '13 at 9:41
Ok! Thank you very much, quid. The observation has been edited accordingly. – Lurgul Feb 16 '13 at 19:39