= One-scale virtual nilpotence theorem
For every $d>0$ there is $N(d)$ such that the following holds. Suppose a <group> $G$ has a finite symmetric <generating set> $S$ containing the identity, and
$$
|S^n|\leq n^d|S|\qquad\text{for some }n\geq N(d).
$$
Then there are <subgroups>
$$
H\trianglelefteq C\leq G,
$$
with $H\subseteq S^{\lfloor n/2\rfloor}$, bounded <index of a subgroup> $[G:C]=O_d(1)$, and <nilpotent> <quotient group> $C/H$ of class $O_d(1)$. In particular $H$ is finite. The containment allows \b[$C=G$].<The structure of approximate groups by Emmanuel Breuillard and others>[Proof of Corollary 11.9, pages 68–69; adjusting its fixed intermediate-scale constants gives the displayed half-radius containment]{ref}
A geometric-scale pigeonhole argument finds an intermediate <word metric ball> with bounded growth. The <Breuillard-Green-Tao structure theorem for approximate groups> then supplies the finite subgroup and the bounded-index nilpotent quotient.
For example, take a <cyclic group> of arbitrarily large prime order and $S=G$. The growth hypothesis holds, but every proper <subgroup> has index equal to that prime. Requiring a proper $C$ would therefore contradict an index bound depending only on $d$; choosing $C=G$ gives the valid nilpotent conclusion.
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