Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 133 / 2

Write an element of the Integer Heisenberg group as $(x,y,z)$. Matrix multiplication givesThe subgroupis normal and isomorphic to ${\mathbb{Z}}^2$. If $t=(1,0,0)$, thenThus, on the coordinate column $(y,z{)}^T$, conjugation by $t$ is the linear map with matrixEvery element has a unique expression $(0,y,z)t^x$, so

The commutators $[t,(0,y,z)]$ fill the central subgroup of matrices $(0,0,z)$, while the quotient by this subgroup is generated freely and abelianly by the images of $(1,0,0)$ and $(0,1,0)$. Equivalently, $(A-I){\mathbb{Z}}^2$ is the second coordinate axis. Therefore the abelianization is

Let $N={\mathbb{Z}}^2$ and let $t$ generate the other factor of ${\Gamma }_B=N{⋊}_B\mathbb{Z}$. For $n\in N$,After abelianization, the image of $N$ is thereforeIf $1$ is not an eigenvalue of $B$, then $\det (B-I)\ne 0$. Hence $(B-I)N$ is a full-rank sublattice of $N$ with finite index of a subgroup $|\det (B-I)|$. The displayed quotient, and therefore the image of the ${\mathbb{Z}}^2$ factor in ${\Gamma }_B^{\mathrm{ab}}$, is finite. In fact the abelianization of a semidirect product by the integers gives

TakeIts eigenvalues areNeither is a root of unity, so $1$ is not an eigenvalue of $C^k$ for any positive $k$.

Let $L$ be a finite-index subgroup of ${\Gamma }_C$, let $N={\mathbb{Z}}^2$, and project ${\Gamma }_C$ onto its $\mathbb{Z}$ factor. Then $N^{\prime }=L\cap N$ has finite index in $N$, while the image of $L$ is $k\mathbb{Z}$ for some $k>0$. Choose $s\in L$ projecting to $k$. Every element of $L$ has a unique expression $n{s}^j$ with $n\in N^{\prime }$, and conjugation by $s$ on $N^{\prime }$ is the restriction of $C^k$. ThusBecause $1$ is not an eigenvalue of $C^k$, part (b), applied to the finite-rank lattice $N^{\prime }$, shows that the image of $N^{\prime }$ in $L^{\mathrm{ab}}$ is finite. The quotient by that finite image is generated by the image of $s$, so $L^{\mathrm{ab}}$ is a finite extension of an infinite cyclic group. It is therefore a virtually cyclic group.