# A combinatorial problem on abelian groups

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 $$i=1,\ldots,n$$?

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