Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 149 / 4

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

Apply part (a). The subgroup $H$ is finite because it lies in the finite set $S^{\lfloor n/2\rfloor }$. Conjugation gives a homomorphismIts kernel $C_0$ has finite index in $C$ and centralizes $H$. Since $C_{0}H/H$ is a subgroup of the $O_d(1)$-step nilpotent group $C/H$, it is itself nilpotent of class $O_d(1)$. Hencefor some $s=O_d(1)$. As $C_0$ centralizes $H$, one more group commutator vanishes, so ${\gamma }_{s+2}(C_0)=\{1\}$. Thus $C_0$ is nilpotent of class at most $s+1=O_d(1)$.

Both $[G:C]$ and $[C:C_0]$ are finite, soThis is the Gromov theorem on groups of polynomial growth in the form needed here.

The Cayley graph ${\mathrm{Cay}}_S(G)$ is a connected graph with $|G|$ vertices. A shortest path never repeats a vertex, so it has at most $|G|-1$ edges. Therefore

Fix $\epsilon >0$ and set $d=2/\epsilon$. Suppose, towards a contradiction, thatwhere $\lambda$ will absorb constants depending only on $\epsilon$. Put $n=\lfloor |G{|}^{\epsilon }\rfloor$. Once $\lambda$ is large enough, $n\ge N(d)$, $n<D$, andPart (a) yields $H⊴C<G$ with $[G:C]=O_{\epsilon }(1)$, $H\subseteq S^{\lfloor n/2\rfloor }$, and $C/H$ nilpotent of class $O_{\epsilon }(1)$.

The subgroup core ${\bigcap }_{g\in G}gC{g}^{-1}$ is normal in $G$ and has index at most $[G:C]!$. Since $G$ is a simple group, the core is either $\{1\}$ or $G$. In the first case $|G|\le [G:C]!=O_{\epsilon }(1)$, which is excluded by increasing $\lambda$. Hence the core is $G$, so $C=G$.

Now $H⊴G$. Since $n/2<D$, the ball $S^{\lfloor n/2\rfloor }$ is not all of $G$, so $H\ne G$. Simplicity gives $H=\{1\}$, and therefore $G=C/H$ is a nilpotent group. A nontrivial finite nilpotent group has nontrivial center of a group; simplicity would force that center to be all of $G$, making $G$ Abelian. This contradicts the assumption that $G$ is non-abelian. Consequently