Let $H$ be a finite group and form the <free product> $G=H*\langle x\rangle$. Put $A=H\cup\{x\}$. Then
$$
A^2\subseteq H\cup Hx\cup xH\cup\{x^2\},
$$
so $|A^2|\leq3|H|+1<3|A|$.

On the other hand, $A^3$ contains the double coset $HxH$. Reduced-word uniqueness in the free product makes the map
$$
H\times H\longrightarrow HxH,
\qquad(h_1,h_2)\longmapsto h_1xh_2
$$
injective, so $|A^3|\geq|H|^2$. Taking finite groups $H$ of unbounded order keeps the doubling constant below three while $|A^3|/|A|$ tends to infinity. This realizes the phenomenon.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 129 / 1 / iii / Solution

Let $H$ be a finite subgroup and choose $x$ with $H\cap xHx^{-1}=\{1\}$. For $A=H\cup\{x\}$, the set $A^2$ has size at most $3|H|+1$, while $A^3$ contains the double coset $HxH$ of size $|H|^2$. Thus bounded doubling alone gives no tripling bound in arbitrary groups.


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 Incoming links: Small doubling does not control tripling in a noncommutative group