In Example 1.44 let \(d\) be the greatest common divisor of \(m\) and \(n\), and let \(m' = m/d\) and \(n' = n/d\). Show that the graph \(T_{m,n}/K\) consists of \(m'\) vertices labeled \(a\), \(n'\) vertices labeled \(b\), together with \(d\) edges joining each \(a\) vertex to each \(b\) vertex. Deduce that the subgroup \(K \subset G_{m,n}\) is free on \(dm'n' - m' - n' + 1\) generators.
Solution
Recall from Example 1.44 that \[ G_{m,n}=\langle a,b\mid a^m=b^n\rangle \] acts on the tree \(T_{m,n}\), and that \[ K=\ker\chi, \] where after scaling the translation action to have image \(\mathbb Z\) one may take \[ \chi(a)=n', \qquad \chi(b)=m', \] with \[ m=dm',\qquad n=dn',\qquad \gcd(m',n')=1. \]
1. The quotient graph \(T_{m,n}/K\) has \(m'\) vertices of the type labeled \(a\) and \(n'\) vertices of the type labeled \(b\).
In the Bass–Serre tree, the vertices labeled \(a\) are the cosets of the subgroup \(\langle b\rangle\), while the vertices labeled \(b\) are the cosets of \(\langle a\rangle\). Passing to \(K\)-orbits and using \[ G_{m,n}/K\cong\mathbb Z \] gives \[ K\backslash G_{m,n}/\langle b\rangle \cong \mathbb Z/\chi(\langle b\rangle) \cong \mathbb Z/m'\mathbb Z, \] so there are \(m'\) \(a\)-vertices. Similarly, \[ K\backslash G_{m,n}/\langle a\rangle \cong \mathbb Z/n'\mathbb Z, \] so there are \(n'\) \(b\)-vertices.
2. The quotient graph has\[ dm'n' \]edges.
The stabilizer of an edge in \(T_{m,n}\) is the central subgroup \[ Z=\langle a^m\rangle=\langle b^n\rangle. \] Since \[ \chi(a^m)=mn'=dm'n', \] we have \[ \chi(Z)=dm'n'\mathbb Z. \] Therefore \[ K\backslash G_{m,n}/Z \cong \mathbb Z/(dm'n')\mathbb Z, \] so there are \(dm'n'\) edge orbits.
3. Each \(a\)-vertex is joined to each \(b\)-vertex by exactly \(d\) edges.
The quotient action of \(G_{m,n}/K\cong\mathbb Z\) is transitive on the vertices of each fixed type modulo the respective stabilizers, so the number of edges joining a given pair of vertex orbits is constant. There are \[ m'n' \] pairs consisting of one \(a\)-vertex and one \(b\)-vertex, while 2 gives \(dm'n'\) edges total. Hence each pair is joined by exactly \[ \frac{dm'n'}{m'n'}=d \] edges. Thus \(T_{m,n}/K\) has exactly the asserted form.
4. The action of \(K\) on \(T_{m,n}\) is free, so\[ K\cong\pi_1(T_{m,n}/K). \]
The kernel of the \(G_{m,n}\)-action on the tree is the central subgroup \[ Z=\langle a^m\rangle. \] But \[ K\cap Z=1 \] because \(\chi(a^m)=dm'n'\ne0\). Vertex stabilizers are conjugates of the cyclic vertex groups, and their intersection with \(K\) is likewise trivial since \(\chi(a)\) and \(\chi(b)\) are nonzero. Hence \(K\) acts freely on the tree. A group acting freely on a tree is naturally the fundamental group of its quotient graph.
5. Therefore \(K\) is free of rank\[ \boxed{dm'n'-m'-n'+1.} \]
The quotient graph is connected and finite, with \[ E=dm'n', \qquad V=m'+n'. \] The fundamental group of a connected finite graph is free of rank \[ E-V+1. \] Using 4 gives \[ \operatorname{rank}K =dm'n'-m'-n'+1. \]