Assume $|A+A|\le K|A|$ and put
$$
H=2A-2A.
$$
This set is symmetric, contains zero, and the <Plünnecke-Ruzsa inequality> gives
$$
|H|\le K^4|A|.
$$
It also contains $A-A$, so for every $x\in A$,
$$
A\subseteq H+x.
$$

To control $H+H=4A-4A$, observe again by Plünnecke-Ruzsa that
$$
|(H+H)+A|=|5A-4A|\le K^9|A|.
$$
Apply the <Ruzsa covering lemma> with $S=H+H$ and $T=A$. There is a set $Y$ with $|Y|\le K^9$ such that
$$
H+H\subseteq Y+A-A\subseteq Y+H.
$$
Consequently $H$ is a $K^9$-<approximate group>. Taking a sufficiently large absolute constant $C$ gives
$$
\boxed{|H|\le CK^C|A|,qquad H\text{ is a }CK^C\text{-approximate group},qquad A\subseteq H+x\ (x\in A).}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 129 / 1 / iii / Solution

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{wiki}

If finite sets satisfy $|S+T|\leq K|T|$, then a maximal family of disjoint translates $x+T$, with $x\in S$, has at most $K$ members and gives
$$
S\subseteq X+T-T,
\qquad |X|\leq K.
$$


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 Incoming links: Ruzsa covering lemma