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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, andThen there are subgroupswith $H\subseteq S^{\lfloor n/2\rfloor }$, 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 $C=G$.^[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.

Put $t=\lceil \sqrt{n}\rceil$ and consider the powerswhere $L$ is maximal subject to $5^{L+1}t\le n/20$. For $n$ sufficiently large in terms of $d$, one has $L+1\ge \frac{1}{3}{\log }_{5}n$. Since the successive growth ratios telescope,Thus one ratio is at most $n^{d/(L+1)}\le 5^{3d}$. For the corresponding $m=5^{i}t$,

Set $B=S^m$. Then $|B^3|\le O_d(1)|B|$, so the small-tripling argument from Question 1(b) makesan $O_d(1)$-approximate group. Apply the Breuillard-Green-Tao structure theorem for approximate groups. It gives subgroupssuch that$C/H$ is a nilpotent group of class $O_d(1)$, and $A$ is covered by $O_d(1)$ left cosets of $C$.

It remains to pass from a covering to an index bound. The ball $S^m\subseteq A$ meets only $O_d(1)$ vertices of the Schreier graph of $G/C$. If $G/C$ had more vertices, a simple path from $C$ would give more than that many distinct cosets within distance $O_d(1)$. Since $m\ge \sqrt{n}$ and $n$ is sufficiently large, this is impossible. Therefore