# Prove that if $|A+A| \leq K|A|$ then $2A - 2A$ is a $K^{16}$-approximate group.

Let $$A$$ be a finite subset of an abelian group, $$G$$ (call the operation addition). We say $$A$$ is a $$K$$-approximate group if:

1) $$e_G \in A$$

2) $$A^{-1} = \{ a^{-1} \mid a \in A \} = A$$

3) $$\exists X \subset G, \; |X| \leq K$$ such that: $$2A \subset X+A$$

Where: $$A+A = \{a+b \mid a,b \in A\}$$

I am asked to show that if $$|A+A| \leq K|A|$$, then $$2A - 2A$$ is a $$K^{16}$$-approximate group.

To this end, I am not entirely sure where to start. The first two properties fall out reasonably easily.

I believe it is well known that $$|2A - 2A| \leq K^4|A|$$

I am aware of a result that allows me to find an $$X \subset G, \; |X| \leq K^4$$ such that $$nA - A \subset (n-1)X + A - A$$, which I believe implies:

$$2A \subset X+A$$

What can I do now though? I don't see how I can get the required subset of $$G$$, and the corresponding bound on the size?

• What is $K{}{}{}?$ – Thomas Andrews Feb 11 at 19:39
• @ThomasAndrews $K$ is a constant such that $|A+A| \leq K|A|$ – user366818 Feb 11 at 19:43

In $$nA-A\subset(n-1)X+A-A$$, take $$n=2$$ to get $$2A-A\subset X+A-A.$$ It follows that $$2A-2A \subset X+A-2A = X - (2A-A) \subset X-X+A-A.$$ This yields $$4A-4A \subset 2(X-X) + (2A-2A).$$ In view of $$|2(X-X)|\le|X|^4\le K^{16}$$, this shows that $$2A-2A$$ is a $$K^{16}$$-approximate group