Let $(G,+)$ be an abelian group and $A$, $B$ and $C$ be finite subsets of $G$ with $A+B=C$. One may conclude that $A\subset C-B$. However, $A$ need not be equal to $C-B$. What is a necessary and sufficient condition to have $A=C-B$?
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3$\begingroup$ You may want to take the following into account while looking for a characterization of the pairs $(A,B)$ with $A=(A+B)−B$: Assume $G$ is finite (and possibly non-abelian). It's folklore (see, e.g., Lemma 3.1.2 in Chap. 3 of Geroldinger & Ruzsa's Combinatorial Number Theory and Additive Group Theory) that, if $A$ and $B$ are subsets of $G$ with $|G|<|A|+|B|$, then $A+B=G$, with the result that $A=(A+B)−B$ iff $A=G$. $\endgroup$– Salvo TringaliJul 15, 2022 at 23:12
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$\begingroup$ Thanks! Yes such pairs are characterized in the case at least one of $A$ or $B$ is large enough. I was thinking that for small sets, possibly $A\cup B$ being a Sidon set might give us a hint. $\endgroup$– ShahabJul 15, 2022 at 23:49
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2 Answers
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Regardless of whether $A,\ B$, and $G$ are finite or infinite, the necessary and sufficient condition is that $B$ lies in a coset of the period (or stabilizer) of $A$, defined to be the subgroup of all those group elements $g$ with $A+g=A$.
To see this, notice that $A\subseteq (A+B)-B$ holds true in a trivial way, so that your condition reduces to $(A+B)-B\subseteq A$, which is thus equivalent to $A+(b_1-b_2)=A$ for any $b_1,b_2\in B$.
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$\begingroup$ Thank you. May I include this as a remark in my paper?(of course with addressing it in the acknowledgment) $\endgroup$– ShahabJul 16, 2022 at 22:02
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$\begingroup$ @Shahab: This is a very basic observation, you certainly can use it without any acknowledgement. $\endgroup$– SevaJul 17, 2022 at 5:26
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The condition you want is very restrictive, at least when $G$ is finite. Notice that for any two non-empty subsets $X$ and $Y$ of $G$, one has $|X\pm Y|\geq |X|$. Indeed, if $G$ is a cyclic group of prime order, by the Cauchy-Davenport Theorem, the equality is achieved if and only if $|Y|=1$ or $X=G$. Now notice that $|C=A+B|\geq |A|$ and $|C-B|\geq |C|$. So if $C-B=A$, then we have $|C|=|A|$ and equality should be achieved in both $|A+B|\geq |A|$ and $|C-B|\geq |C|$. In the case of cyclic groups of prime order, the only possibilities are $A=C=G$ or $|B|=1$.
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1$\begingroup$ Thanks! Cauchy-Davenport theorem has a generalization for arbitrary abelian groups as follows: $|A+B|\geq \min\{p(G), |A|+|B|-1\}$, where $p(G)$ stands for the size of the smallest nontrivial subgroup of $G$. So we may repeat your argument for small subsets of abelian groups. However, when $|A|+|B|-1>p(G)$ we may need another argument. $\endgroup$– ShahabJul 16, 2022 at 1:36
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