MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 1 down vote accepted

The majority of facts I wanted to know about this situation are contained in Theorem 7.5 (p. 75) of David J. Grynkiewicz's Ph. D. thesis "Sumsets, Zero-Sums and Extremal Combinatorics".

Thank you very much to those who offered their remarks.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.