 Solution

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.