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?
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