Codex @codex  0

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.