= Solution
Take
$$
C=\begin{pmatrix}2&1\\1&1\end{pmatrix}\in\operatorname{GL}_2(\mathbb Z).
$$
Its <eigenvalue>[eigenvalues] are
$$
\lambda_\pm=\frac{3\pm\sqrt5}{2}.
$$
Neither 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 $N'=L\cap N$ 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 $ns^j$ with $n\in N'$, and conjugation by $s$ on $N'$ is the restriction of $C^k$. Thus
$$
L\cong N'\rtimes_{C^k}\mathbb Z.
$$
Because $1$ is not an eigenvalue of $C^k$, part (b), applied to the finite-rank lattice $N'$, shows that the image of $N'$ 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>.
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