In a 1952 paper M. Hall proved that if $G=\{a_1,\ldots,a_n\}$ is an additive abelian group of order $n$ and $b_1,\ldots,b_n$ are elements of $G$ with $b_1+\ldots+b_n=0$ then we have $$\{a_{\sigma(i)}+b_i:\ i=1,\ldots,n\}=\{a_1,\ldots,a_n\}$$ for some $\sigma\in S_n$.

Motivated by this, I raise the following question.

Question. Let $G$ be an additive abelian group, and suppose that $A$ is a nonempty subset of $G$ with $A-A=\{a-a':\ a,a'\in A\}$ equal to $G$. If $b_1,\ldots,b_n\in G$ with $b_1+\ldots+b_n=0$, whether there are $a_1,\ldots,a_n\in A$ and a permutation $\sigma\in S_n$ such that $b_i=a_i-a_{\sigma(i)}$ for all $i=1,\ldots,n$?

I guess that this question has a positive answer. Your comments are welcome!


1 Answer 1


I guess, without further adjustments to the question, the answer is no. Consider the following:

Let $G = \mathbb{Z}$ and let $A = \{ 2z \; | \; z \in \mathbb{Z} \} \cup \{ 1 \}$. Now this set $A$ satisfies $A-A = G$.

However, by taking only odd integers for the $b_i$, say $b_1 = 1, b_2 = 3, b_3 = 5, b_4 = -9$, we obtain a contradiction since $a_i - a_{\sigma(i)}$ can only be an odd integer if $a_i$ or $-a_{\sigma(i)}$ equals $1$ or $-1$, respectively. So we cannot find the desired collection of $a_i$.

  • $\begingroup$ Can you find a counterexample with $G$ finite? $\endgroup$ Dec 20, 2020 at 10:10
  • 1
    $\begingroup$ What about taking $G = \mathbb{Z} / 2n\mathbb{Z}$ for some sufficiently large $n$ and the rest as mentioned above, i.e. $A = \{ [2z] \; | \; z \in \mathbb{Z} \} \cup \{ [1] \}$ and $b_1 = [1]$, etc. As before, $A - A = G$ and the same parity argument regarding $a_i - a_{\sigma(i)}$ works as well. $\endgroup$
    – karlheuer
    Dec 20, 2020 at 10:43

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