Determine the covering space of \(S^1 \vee S^1\) corresponding to the subgroup of \(\pi_1(S^1 \vee S^1)\) generated by the cubes of all elements. The covering space is 27-sheeted and can be drawn on a torus so that the complementary regions are nine triangles with edges labeled \(aaa\), nine triangles with edges labeled \(bbb\), and nine hexagons with edges labeled \(ababab\).
Solution
Let \[ F=F(a,b)=\pi_1(S^1\vee S^1) \] and let \(N\) be the subgroup generated by all cubes \(w^3\), \(w\in F\). Since conjugates of cubes are cubes, \[ g w^3 g^{-1}=(gwg^{-1})^3, \] the subgroup \(N\) is normal.
1. The quotient \(G=F/N\) is the free two-generator group of exponent \(3\).
By construction every element of \(G\) satisfies \[ x^3=1. \] Conversely, \(N\) is the smallest normal subgroup with this property, since any quotient of \(F\) in which every element has cube \(1\) must kill every \(w^3\).
2. In \(G\), the commutator\[ c=[a,b]=aba^{-1}b^{-1} \]is central.
We first prove a general calculation valid in every group of exponent \(3\). Write \[ X=x^{-1},\qquad Y=y^{-1}. \] Then, using \(x^2=X\), \(y^2=Y\), and the fact that the cube of every word is \(1\), \[ \begin{aligned} [[x,y],y] &=xyXY\,y\,(xyXY)^{-1}Y\\ &=xyXY\,yyxYXY\\ &=xyXyxx yx(XY)^3\\ &=xyXyXyx\\ &=xx(Xy)^3x\\ &=x^3\\ &=1. \end{aligned} \] Here the third and fifth lines only insert cube relations and use \(x^2=X\), while the other steps are free cancellation. Thus \([x,y]\) commutes with \(y\). Interchanging \(x\) and \(y\) shows that \([y,x]=[x,y]^{-1}\) commutes with \(x\), hence \([x,y]\) commutes with \(x\) as well. Applying this with \(x=a\), \(y=b\) proves that \(c\) commutes with both generators and is therefore central in \(G\).
3. Every element of \(G\) has the form\[ a^i b^j c^k, \qquad i,j,k\in\{0,1,2\}. \]
Since \(c\) is central and \[ ab=cba, \] one may move all \(a\)’s to the left of all \(b\)’s at the cost of powers of \(c\). The exponent-three relations reduce the exponents modulo \(3\). Thus \(|G|\le27\).
4. The group \(G\) has at least \(27\) elements.
Let \[ U=UT_3(\mathbf F_3) =\left\{ \begin{pmatrix} 1&x&z\\ 0&1&y\\ 0&0&1 \end{pmatrix}:x,y,z\in\mathbf F_3 \right\}. \] This group has \(27\) elements. It is generated by \[ A=\begin{pmatrix}1&1&0\\0&1&0\\0&0&1\end{pmatrix}, \qquad B=\begin{pmatrix}1&0&0\\0&1&1\\0&0&1\end{pmatrix}. \] Every element of \(U\) has cube \(1\), so the assignment \[ a\mapsto A, \qquad b\mapsto B \] induces a surjection \[ G\twoheadrightarrow U. \] Hence \(|G|\ge27\). Together with 3, \[ |G|=27 \] and \(G\cong U\).
5. The required covering is the labelled Cayley cover of \(G\) with generators \(a,b\).
Take one vertex \(v_g\) for each \(g\in G\). For every \(g\), attach an oriented \(a\)-edge \[ v_g\to v_{ga} \] and an oriented \(b\)-edge \[ v_g\to v_{gb}. \] Map all \(a\)-edges around the \(a\)-circle and all \(b\)-edges around the \(b\)-circle. As in the standard Cayley-cover construction, each vertex has exactly one incoming and outgoing lift of each oriented base edge, so this is a connected normal \(27\)-sheeted cover. Its based subgroup is the kernel of \[ F\to G=F/N, \] hence exactly \(N\).
6. The \(a\)-edges decompose into nine directed \(3\)-cycles, and the \(b\)-edges do likewise.
Right multiplication by \(a\) and by \(b\) both have order \(3\) and act freely on the \(27\) vertices. Hence each action has \[ 27/3=9 \] orbits, each of length \(3\). These give nine triangular \(aaa\) cycles and nine triangular \(bbb\) cycles.
7. The alternating word \(ab\) also has order \(3\), so the alternating \(a,b\) edges form nine hexagonal boundary cycles labelled\[ ababab. \]
The element \(ab\) has cube \(1\) in the exponent-three quotient. Starting at a vertex and alternately traversing an \(a\)-edge and a \(b\)-edge returns after three such pairs, producing a six-edge cycle. Again there are \(27/3=9\) orbits under right multiplication by \(ab\).
8. These cycles give Hatcher’s torus drawing of the covering graph.
Attach a \(2\)-cell to each of the nine \(aaa\) cycles, each of the nine \(bbb\) cycles, and each of the nine \(ababab\) cycles. The resulting cell complex has \[ V=27, \qquad E=54, \qquad F=27, \] so \[ \chi=27-54+27=0. \] The labelled incidence pattern is connected and each edge is incident with two of these faces, giving a closed orientable surface of Euler characteristic \(0\), hence a torus. Removing the interiors of these faces leaves exactly the covering graph from 5, with the stated nine \(aaa\) triangles, nine \(bbb\) triangles, and nine \(ababab\) hexagons as complementary regions.
9. Therefore the subgroup generated by all cubes corresponds to the \(27\)-sheeted normal Cayley covering described above.
This follows from <1>4–<1>8.