= Solution
The <Noncommutative Ruzsa triangle inequality> says that nonempty finite subsets $B,C,D$ of a <group> satisfy
$$
|B|\,|CD^{-1}|\leq |CB^{-1}|\,|BD^{-1}|.
$$
The <Ruzsa covering lemma> says that if $|CD|\leq K|D|$, then some $X\subseteq C$ with $|X|\leq K$ satisfies
$$
C\subseteq XDD^{-1}.
$$
Now let $A$ be a <symmetric subset of a group>, so $A^{-1}=A$ and $1\in A$. Apply the triangle inequality with the middle set $A$ to obtain, for $m\geq4$,
$$
|A|\,|A^m|\leq |A^{m-1}|\,|A^3|.
$$
The hypothesis therefore gives $|A^m|\leq K|A^{m-1}|$. Starting from $|A^3|\leq K|A|$ proves
$$
\boxed{|A^m|\leq K^{m-2}|A|\qquad(m\geq3).}
$$
In particular $|A^5|\leq K^3|A|$. Apply the covering lemma to $C=A^4$ and $D=A$. There is $X\subseteq A^4$ with $|X|\leq K^3$ such that
$$
A^4\subseteq XAA^{-1}=XA^2.
$$
The set $A^2$ is symmetric and contains the <identity element>, so this inclusion is exactly the covering condition showing that
$$
\boxed{A^2\text{ is a }K^3\text{-approximate group}.}
$$
Small doubling alone is insufficient in a <noncommutative group>. Let $H$ be a finite group, let $G=H*\langle x\rangle$ be its <free product> with an <infinite cyclic group>, and put
$$
A=H\cup\{x,x^{-1}\}.
$$
Then $A$ is symmetric and $|A^2|\leq5|H|+4=O(|A|)$, while $A^3$ contains the <double coset> $HxH$. Distinct pairs $(h_1,h_2)\in H^2$ give distinct reduced words $h_1xh_2$, so $|HxH|=|H|^2$. Letting $|H|\to\infty$ proves the <small doubling does not control tripling in a noncommutative group> phenomenon.
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