# coset of a group

Assume that $$G$$ is an abelian group and $$A$$ is a subset of $$G$$ such that $$|A+A|=|A|$$ then $$A$$ is a coset of a subgroup. I tried that if $$0 \in A$$ then $$A+A =A$$ but I can't go any further. Could someone help me? sorry if my English isn't well.

• $G$ must be assumed to be finite. Or at least $A$ must be. Aug 21, 2019 at 11:58

Choose $$a_0\in A$$, clearly $$A=a_0+B$$ where $$B=A-a_0$$. We show that $$B$$ is a subgroup (i.e. contains $$0$$, closed under addition and under taking inverse).
It is given that $$|A+A|=|A|$$ and therefore $$|B+B|=|B|$$. Moreover since $$a_0\in A$$ we have that $$0\in B$$.
Let $$b,b'\in B$$ we claim that $$b+b'\in B$$. If not, then $$\{b+0:b\in B\}$$ and $$b+b'$$ are in $$B+B$$ but then $$|B+B|>|B|$$. Therefore $$B$$ is closed under addition.
Finally, let $$b\in B$$ and consider $$\{b+b':b'\in B\}$$. This set is of size $$|B|$$ and since $$B+B\subseteq B$$ we conclude that this set must be equal to $$B$$. In particular we conclude that $$\{b+b':b'\in B\}$$ contains zero. We thus have that $$-b\in B$$.
This proves that $$B$$ is a subgroup.
*As mentioned by @Arthur's comment. If $$A$$ is infinite the argument doesn't work, as a counter-example one can take $$G=\mathbb{Z},A=\mathbb{N}$$.