Closed hyperbolic three-manifold groups contain no Z-squared subgroup

The lift of the complete metric to $\tilde{M}$ is complete: every geodesic upstairs projects by [F4] to a geodesic downstairs, which extends for all time; path lifting extends the original geodesic upstairs, and a local isometry preserves its geodesic equation.

By [F2], the complete simply connected universal cover $\tilde{M}$ is diffeomorphic to ${\mathbb{R}}^3$. Along each radial unit-speed geodesic from a fixed point, [F3] says the differential of the exponential map multiplies angular vectors by $\sinh r$, while its radial differential preserves length. Thus in polar coordinates the pulled-back metric is $d{r}^2+{\sinh }^2(r)g_{S^2}$, the standard hyperbolic metric; the exponential map identifies $\tilde{M}$ isometrically with ${\mathbb{H}}^3$. By [F4], $\mathrm{\Gamma }={\pi }_1(M)$ acts freely by deck isometries, preserving the lifted orientation.

Write $\pi :{\mathbb{H}}^3\to M$ for this covering. For points $x,y$ with different projections, choose neighborhoods with disjoint projections; no deck transformation carries a point of the first into the second. If their projections agree, choose a connected evenly covered neighborhood downstairs and the sheets containing $x,y$. A deck transformation maps a sheet onto a sheet over this neighborhood. At most one deck transformation carries the first specified sheet onto the second: two such maps send $x$ to the unique point of the second sheet over $\pi (x)$, and freeness makes them equal. For compact sets $C,D\subset {\mathbb{H}}^3$, these product neighborhoods cover $C\times D$; a finite subcover shows that only finitely many $\gamma$ satisfy $\gamma C\cap D\mathrm{\ne }\mathrm{\varnothing }$. This proves proper discontinuity on compact sets. Choose finitely many relatively compact open sets covering compact $M$, each with closure inside a connected evenly covered neighborhood, and lift each closure to one sheet. Their union $K$ is compact. The universal-cover deck group is transitive on each fiber by the identification in [F4], so $\mathrm{\Gamma }K={\mathbb{H}}^3$. This proves cocompactness.

Every nonidentity element of $\mathrm{\Gamma }$ is loxodromic. A Möbius map $z\mapsto (az+b)\mathrm{/}(cz+d)$ has fixed-point equation $c{z}^2+(d-a)z-b=0$ on the boundary sphere. If there are two distinct fixed points, conjugate them to $0,\mathrm{\infty }$; the map then has form $z\mapsto \lambda z$. If $\mathrm{\mid }\lambda \mathrm{\mid }=1$ and $\lambda \mathrm{\ne }1$, its extension fixes the axis joining $0$ to $\mathrm{\infty }$, contrary to the free deck action [F4]. If $\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\ne }1$, it preserves that axis and translates it nontrivially. If the fixed-point equation has a double root, conjugate it to $\mathrm{\infty }$; the map has form $z\mapsto az+b$, and its only fixed point is $\mathrm{\infty }$ precisely when $a=1$ and $b\mathrm{\ne }0$. This is parabolic: the upper-half-space distance formula gives $\cosh d((0,t),(b,t))=1+\mathrm{\mid }b{\mathrm{\mid }}^2\mathrm{/}(2t^2)$, so its displacement tends to zero as $t\to \mathrm{\infty }$. To exclude such a parabolic in a cocompact deck group, choose compact $K$ with $\mathrm{\Gamma }K={\mathbb{H}}^3$ and points $y_n$ whose displacement tends to zero. Choose $h_n\in \mathrm{\Gamma }$ so $x_n=h_n{y}_n\in K$ and put ${\gamma }_n=h_n\gamma h_n^{-1}$. The points $x_n$ and ${\gamma }_n{x}_n$ lie in a fixed compact neighborhood of $K$, so proper discontinuity permits only finitely many ${\gamma }_n$. Along a subsequence this element is constant and $x_n$ converges in $K$; the displacement limit makes it fix the limit point, contradicting freeness. Hence every nonidentity element has two fixed endpoints and preserves a unique axis with nonzero translation.

Fix $\gamma \mathrm{\ne }1$ and let $C_{\mathrm{\Gamma }}(\gamma )=\{\delta \in \mathrm{\Gamma }:\delta \gamma =\gamma \delta \}$. Commutation preserves the attracting and repelling endpoints of $\gamma$ individually, hence preserves its axis and acts there by a signed translation. The signed translation map $\tau :C_{\mathrm{\Gamma }}(\gamma )\to \mathbb{R}$ is a homomorphism. It is injective: an element in its kernel fixes every point on the axis, and [F4] makes it the identity. Its image is discrete: if infinitely many elements had bounded signed translations, their images of a fixed point on the axis would lie in a compact segment, contradicting the compact-set finiteness in step 3.1. Every discrete subgroup of $(\mathbb{R},+)$ is cyclic, so $C_{\mathrm{\Gamma }}(\gamma )$ is cyclic.

If $H\le \mathrm{\Gamma }$ were isomorphic to ${\mathbb{Z}}^2$, choose $\gamma \in H\setminus \{1\}$. Since $H$ is abelian, $H\subseteq C_{\mathrm{\Gamma }}(\gamma )$, contradicting step 5.1 because a cyclic group has no subgroup isomorphic to ${\mathbb{Z}}^2$. Therefore no such $H$ exists.