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.

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    $\begingroup$ $G$ must be assumed to be finite. Or at least $A$ must be. $\endgroup$
    – Arthur
    Aug 21, 2019 at 11:58

1 Answer 1


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 $G=\mathbb{Z},A=\mathbb{N}$.


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