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With group commutator convention $[a,b]=a^{-1}{b}^{-1}ab$, the displayed group presentation describes the fundamental group of the Borromean rings complement. Its abelianization is ${\mathbb{Z}}^3$, since each relator is a group commutator and hence imposes no abelian relation. It embeds discretely in the Gaussian Bianchi group by sending the meridians to $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$, $\left(\begin{array}{cc}1& 0\\ 2i& 1\end{array}\right)$ and $\left(\begin{array}{cc}i& 1\\ 2i& 2-i\end{array}\right)$.

We use $[a,b]=a^{-1}{b}^{-1}ab$. To obtain a Wirtinger presentation, choose an oriented regular knot diagram, cut its strands at undercrossings, and give each resulting arc a positively oriented meridian of a knot. The complement of the arcs above the projection plane has a free fundamental group on these meridians. At a crossing, pushing an understrand meridian past the overstrand conjugates it by the overstrand meridian. Thus, if $a$ is the overstrand generator and $b,c$ are the incoming and outgoing understrand generators, the relation is $c=a^{ϵ}b{a}^{-ϵ}$, where $ϵ$ is determined by the crossing orientation. Applying the Seifert-van Kampen theorem while adjoining the crossing neighborhoods gives one such relation per crossing. The remaining three-dimensional region adds no fundamental-group relation. For a connected diagram one crossing relation is redundant. This derives a group presentation from any tame knot diagram, rather than just specifying a recipe without its topological justification.

For the Borromean rings, choose three meridians $x,y,z$ and call the other three arc meridians $x^{\prime },y^{\prime },z^{\prime }$. Base paths and orientations can be chosen so the six crossing relations readEliminate the primed generators with the first row. The second row says respectively that $x$ commutes with $(z^{\prime }{)}^{-1}z=[y^{-1},z]$, that $y$ commutes with $(x^{\prime }{)}^{-1}x=[z^{-1},x]$, and that $z$ commutes with $(y^{\prime }{)}^{-1}y=[x^{-1},y]$. The last relation is redundant, just as one crossing relation is redundant in the Wirtinger presentation. Alternatively the Hall-Witt identity, after conjugating its three factors, shows that any two of these cyclic relations imply the third. Hence the Borromean link group isBoth relators have zero exponent sum in each generator. Passing from the free group to its abelianization therefore imposes no relations beyond commutation, andEquivalently the three linking-number maps to $\mathbb{Z}$ independently detect the three meridians.

Here is an explicit representation, followed by a faithfulness argument:Each determinant is one. Direct multiplication verifies the two relators. For example $[Y^{-1},Z]=\left(\begin{array}{cc}-1& 2i\\ 0& -1\end{array}\right)$, which commutes with $X$. Merely checking those relators would not prove faithfulness, so we identify the image's complete group presentation.

Write $B=PS{L}_2(\mathbb{Z}[i])$, the Gaussian Bianchi group, and use its four generators $T,U,L,A$ from the allowed presentation. In $B$ the three proposed generators areThe last equality is in $PS{L}_2$, so an overall minus sign in matrix multiplication is immaterial. Reduce matrices modulo $4$ in $\mathbb{Z}[i]$, and then quotient by $\{\pm I\}$. The finite image $Q$ has order $1536$: the residue field is ${\mathbb{F}}_2$, $S{L}_2({\mathbb{F}}_2)$ has order six, and each of the three subsequent local-ring layers has a kernel of order $2^3$, before the final quotient by $\{\pm I\}$. Elementary elimination over this local ring shows that the images of $T,U,A$ generate it. The subgroup $\bar{H}=⟨\bar{X},\bar{Y},\bar{Z}⟩$ has order $64$, as multiplication in this finite ring verifies. Let $K$ be its inverse image in $B$; then $[B:K]=24$.

For completeness the following gives the coset computation needed to check the subgroup presentation explicitly. Cosets are right cosets $Kr$; columns record right multiplication by $T,U,L,A$. Representatives are words in the allowed generators.Apply the Reidemeister–Schreier theorem: for a transition $j{⟶}^{g}k$, use generator $r_{j}g{r}_k^{-1}$, omitting the 23 tree transitions. Rewrite each of the eight allowed ambient relators starting at each of the 24 rows. There are 73 remaining Schreier generators. Adjoin the names $x=T$, $y=A{U}^{-2}{A}^{-1}$, $z=A{T}^{-2}{A}^{-1}L{U}^{-1}$ and eliminate the Schreier generators using relations in which they occur once. The remaining relations, with an uppercase letter denoting the inverse of the corresponding lowercase one, are the wordsThis table and the rewriting rule determine every relation without an appeal to a guessed presentation. In this notation cyclically reduced versions of the two desired relators are $r=ZYzXZyzYxy$ and $q$. Elementary substitutions of relator segments, with cyclic reduction at each step, giveusing $r,q$; conversely $r⟶p⟶1$ using $p,q$. These substitutions show equality of the two normal closures. For example the first substitution replaces $xyX$ in the cyclic rotation of $p$ by $zxZyzXZ$, as allowed by $q$, and then the initial $ZYz$ is replaced by $YXyZYzx$ using $r$. Thus $K$ has exactly the presentation of $G$, with $x,y,z$ represented by $X,Y,Z$. In particular these three matrices generate $K$, and

To identify the actual link complement as a hyperbolic three-manifold, rather than only on a manifold with an isomorphic group, cut along the projection-plane regions, making the usual small crossing detours above and below. The octahedral projection graph has eight triangular regions. The upper and lower pieces are ideal octahedra; corresponding faces are matched with one-third-turn corner permutations, with opposite directions on adjacent checkerboard faces. Reversing these cuts reconstructs the link complement. Give both pieces the regular ideal hyperbolic space octahedron metric. All triangular face maps are isometries and each edge cycle has four dihedral angles $\pi /2$, so the metric has no edge singularity. Equal horospherical truncations match in Euclidean squares and assemble into three flat tori, giving complete torus cusps. The compact truncated remainder and these cusps show completeness and finite volume. Thus the Borromean complement is a complete finite-volume hyperbolic three-manifold. This also supplies a geometric discrete faithful holonomy, independently of the subgroup calculation.