Let \(F\) be the free group on two generators and let \(F'\) be its commutator subgroup. Find a set of free generators for \(F'\) by considering the covering space of the graph \(S^1 \lor S^1\) corresponding to \(F'\).
Solution
Let \[ F=\langle a,b\rangle \] be the free group on two generators. Its commutator subgroup is the kernel of the abelianization map \[ F\longrightarrow\mathbb Z^2, \qquad a\longmapsto(1,0), \quad b\longmapsto(0,1). \]
1. The covering graph corresponding to \(F'\) is the square lattice with vertex set\[ \mathbb Z^2, \]horizontal \(a\)-edges\[ (m,n)\longrightarrow(m+1,n), \]and vertical \(b\)-edges\[ (m,n)\longrightarrow(m,n+1). \]
The cover corresponding to a normal subgroup \(H\triangleleft F\) is the Cayley graph of the quotient \(F/H\) with respect to the images of the free generators. Here \[ F/F'\cong\mathbb Z^2, \] so the Cayley graph is exactly the square lattice.
2. Let \(T\) be the subgraph consisting of all horizontal edges together with all vertical edges on the line \(m=0\).Then \(T\) is a spanning tree of the lattice graph.
Every vertex \((m,n)\) is joined to \((0,0)\) by first moving vertically along the line \(m=0\) to \((0,n)\) and then horizontally to \((m,n)\), so \(T\) is connected and spanning. Any cycle would have to contain a vertical edge away from the line \(m=0\), but \(T\) contains no such edge. Hence \(T\) is acyclic.
3. The edges outside \(T\) are precisely the vertical \(b\)-edges\[ e_{m,n}:(m,n)\to(m,n+1) \]with\[ m\ne0, \qquad n\in\mathbb Z. \]
All horizontal edges belong to \(T\), and among vertical edges exactly those on the axis \(m=0\) belong to \(T\).
4. The basis loop associated to \(e_{m,n}\) reads the word\[ \gamma_{m,n} =b^n a^m b a^{-m}b^{-n-1}. \]
Starting at \((0,0)\), the unique path in \(T\) to \((m,n)\) first traverses \(b^n\) along the vertical axis and then \(a^m\) along the horizontal row. Next traverse the non-tree edge \(b\) from \((m,n)\) to \((m,n+1)\). The unique return path in \(T\) goes horizontally by \(a^{-m}\) to \((0,n+1)\) and then vertically by \(b^{-n-1}\) to the origin. Concatenating gives the displayed word.
5. Therefore\[ \boxed{ F'=F\bigl(\gamma_{m,n}:m\in\mathbb Z\setminus\{0\},\ n\in\mathbb Z\bigr) } \]freely, where\[ \gamma_{m,n}=b^n a^m b a^{-m}b^{-n-1}. \]
For any connected graph, choosing a maximal tree gives a free basis for the fundamental group indexed by the edges outside the tree. By 1 the fundamental group of this covering graph is exactly the subgroup \(F'\), and by <1>2–<1>4 the corresponding basis is the displayed family. Each word has trivial abelianization, as expected.