The <Solymosi sum-product theorem over the complex numbers> states that if finite $U,V,W\subseteq\mathbb C$ satisfy $U\ne\{0\}$ and $W\ne\{0\}$, then
$$
\boxed{|U+V|\,|UW|
\geq\frac1{56}|U|^{3/2}|V|^{1/2}|W|^{1/2}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 149 / 3 / b / Solution

Write $N=[G,G]$ and identify it with $(\mathbb C,+)$ as in part (a). If every two members of $A$ commuted, then $\langle A\rangle$ would be <Abelian>, contrary to hypothesis. Thus some commutator of two members of $A$ is a nonidentity translation in $A^4\cap N$. Consequently the translation-coordinate set
$$
T=\{c\in\mathbb C:f_{1,c}\in A^4\cap N\}
$$
contains both zero and a nonzero element.

The identity
$$
f_{a,b}\circ f_{1,c}\circ f_{a,b}^{-1}=f_{1,ac}
$$
shows that $T\pi(A)$ is contained in the translation coordinates of $A^6\cap N$, while $T+T$ is contained in those of $A^8\cap N$. The <intersection of an approximate group power with a subgroup> therefore gives
$$
|T+T|\leq K^{O(1)}|T|,
\qquad
|T\pi(A)|\leq K^{O(1)}|T|.
$$
Apply the <Solymosi sum-product theorem over the complex numbers> with $U=V=T$ and $W=\pi(A)$. Since $1\in\pi(A)$, its hypotheses hold, and
$$
K^{O(1)}|T|^2
\geq |T+T|\,|T\pi(A)|
\geq\frac1{56}|T|^2|\pi(A)|^{1/2}.
$$
Cancelling $|T|^2$ and absorbing the absolute constant proves
$$
\boxed{|\pi(A)|\leq K^{O(1)}.}
$$

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 149 / 3 / c / Solution

{c}

For finite $U,V,W\subseteq\mathbb C$ with $U\ne\{0\}$ and $W\ne\{0\}$,
$$
|U+V|\,|UW|
\geq\frac1{56}|U|^{3/2}|V|^{1/2}|W|^{1/2}.
$$
In particular, addition and multiplication cannot both expand a nontrivial finite set of <complex numbers> only slightly.

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