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

Apply part i again, now with anchor $A$, $B=A^2$, and $C=A^{-1}A^{-1}$. This gives
$$
|A|\,|A^4|
\leq|AA^{-1}A^{-1}|\,|A^3|
\leq K^3|A|^2.
$$
Thus $|A^4|\leq K^3|A|$. Since $K\geq1$, this proves the requested <fourfold product bound from small tripling> $|A^4|\leq K^4|A|$.

Solution

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

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

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