Let \(F(a,b)\) be the free group of two generators. What are the presentation complex and the Cayley complex for \(F(a,b)\)? Find an index \(4\) subgroup and the covering space that corresponds to it.
Solution
The presentation complex for \[ F(a,b)=\langle a,b\mid\ \rangle \] is the wedge of two circles \(S^1_a\vee S^1_b\): one vertex and two oriented \(1\)-cells, with no \(2\)-cells. Its Cayley complex is therefore just its Cayley graph, the \(4\)-regular tree with vertices indexed by \(F(a,b)\) and directed \(a\)- and \(b\)-edges \[ g\longrightarrow ga,\qquad g\longrightarrow gb. \]
For an index-four subgroup, define \[ \phi:F(a,b)\twoheadrightarrow\mathbb Z/4, \qquad \phi(a)=1,\quad \phi(b)=0, \] and let \(H=\ker\phi\). Then \([F(a,b):H]=4\).
The corresponding based covering graph has vertices \(v_0,v_1,v_2,v_3\). For each \(i\pmod4\), there is an \(a\)-edge from \(v_i\) to \(v_{i+1}\), and there is a \(b\)-loop at every \(v_i\). The covering map sends every \(a\)-edge to the \(a\)-circle and every \(b\)-loop to the \(b\)-circle. It is visibly a connected four-sheeted cover of the two-petal rose.
With basepoint \(v_0\), a free basis for \(H\) is \[ a^4,\quad b,\quad aba^{-1},\quad a^2ba^{-2},\quad a^3ba^{-3}. \] In particular \(H\) has rank \(5=1+4(2-1)\), as the covering graph also shows.