If a finite subset $A$ of a group satisfies $|A^3|\leq K|A|$, then repeated use of the <Noncommutative Ruzsa triangle inequality> gives
$$
|A^4|\leq K^3|A|\leq K^4|A|.
$$

Fourfold product bound from small tripling

For every $x\in BC^{-1}$, choose one representation $x=b_xc_x^{-1}$. Define
$$
F:A\times BC^{-1}\longrightarrow AB^{-1}\times AC^{-1},
\qquad
F(a,x)=(ab_x^{-1},ac_x^{-1}).
$$
The image determines
$$
(ab_x^{-1})^{-1}(ac_x^{-1})=b_xc_x^{-1}=x,
$$
and the chosen representative then determines $a=(ab_x^{-1})b_x$. Thus $F$ is injective, and counting its domain and codomain proves the <Noncommutative Ruzsa triangle inequality>
$$
|A|\,|BC^{-1}|\leq|AB^{-1}|\,|AC^{-1}|.
$$

Solution

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

{c}

For nonempty finite subsets $A,B,C$ of an arbitrary group,
$$
|A|\,|BC^{-1}|\leq|AB^{-1}|\,|AC^{-1}|.
$$
Choose one representation $x=b_xc_x^{-1}$ for each $x\in BC^{-1}$. The map $(a,x)\mapsto(ab_x^{-1},ac_x^{-1})$ is injective.


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 Incoming links: Noncommutative Ruzsa triangle inequality