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

For every $x\in B{C}^{-1}$, choose one representation $x=b_x{c}_x^{-1}$. DefineThe image determinesand the chosen representative then determines $a=(a{b}_x^{-1})b_x$. Thus $F$ is injective, and counting its domain and codomain proves the Noncommutative Ruzsa triangle inequality

Choose $a_0\in A$. Right multiplication by $a_0$ injects $A^2$ into $A^3$, soApply part i with anchor $A^{-1}$, $B=A$, and $C=A^2$. Since inversion preserves cardinality,Therefore $|A{A}^{-1}{A}^{-1}|\le K^2|A|$.

Apply part i again, now with anchor $A$, $B=A^2$, and $C=A^{-1}{A}^{-1}$. This givesThus $|A^4|\le K^3|A|$. Since $K\ge 1$, this proves the requested fourfold product bound from small tripling $|A^4|\le K^4|A|$.

Let $H$ be a finite group and form the free product $G=H\ast ⟨x⟩$. Put $A=H\cup \{x\}$. Thenso $|A^2|\le 3|H|+1<3|A|$.

On the other hand, $A^3$ contains the double coset $HxH$. Reduced-word uniqueness in the free product makes the mapinjective, so $|A^3|\ge |H{|}^2$. Taking finite groups $H$ of unbounded order keeps the doubling constant below three while $|A^3|/|A|$ tends to infinity. This realizes the small doubling does not control tripling in a noncommutative group phenomenon.