Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 104 / 2

The Integer Heisenberg groupis generated by the matrices with $(a,b,c)=(1,0,0)$ and $(0,1,0)$. Their group commutator is the nonidentity central matrix with $(a,b,c)=(0,0,1)$. Thus it is nonabelian and nilpotent of class two, while being finitely generated.

The countable direct sumis abelian and therefore nilpotent of class one. It is not finitely generated, whereas every polycyclic group is finitely generated. Hence it is nilpotent but not polycyclic.

The lamplighter groupis generated by one lamp switch and one translation. It is metabelian, hence solvable. Its base subgroup ${⨁}_{\mathbb{Z}}{C}_2$ is not finitely generated. Every subgroup of a polycyclic group is finitely generated, so the lamplighter group is not polycyclic.

Let $N$ be the subgroup whose upper-left block is $I_2$. It is a finitely generated nilpotent subgroup of the integer upper unitriangular group $U{T}_4(\mathbb{Z})$ and is normal in $L$. The block-diagonal matrixgenerates an infinite cyclic quotient, soFinitely generated nilpotent groups are polycyclic, and an extension of polycyclic groups is polycyclic. Hence $L$ is polycyclic.

Inside $L$, retain only $t$ and the entries in positions $(1,3)$ and $(2,3)$. They form a subgroupThe characteristic polynomial of $A$ is ${\lambda }^2-5\lambda +1$, so its eigenvalues areOne has modulus greater than one, and the resulting semidirect product has exponential growth. Every finitely generated virtually nilpotent group has polynomial growth, as does each of its finitely generated subgroups. Therefore $L$ cannot be virtually nilpotent.