Sumsets with the property "$A+B=C$ implies $A=C-B$" 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 $A=C-B$?
 A: 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 $b_1,b_2\in B$.
A: 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 $|B|=1$.
