Use covering space theory to prove that any finitely presented group \(G\) has only finitely many subgroups of index \(3\).
Solution
Choose a finite presentation \[ G\cong\langle g_1,\ldots,g_n\mid r_1,\ldots,r_m\rangle. \]
1. There is a connected finite CW complex \(X\) with basepoint \(x_0\) such that\[ \pi_1(X,x_0)\cong G. \]
Take one \(0\)-cell, attach one oriented \(1\)-cell for each generator \(g_i\), and attach one \(2\)-cell along the loop spelling each relator \(r_j\). The cellular form of the Seifert–van Kampen theorem gives \[ \pi_1(X,x_0) \cong \langle g_1,\ldots,g_n\mid r_1,\ldots,r_m\rangle \cong G. \] This is the presentation complex of the chosen finite presentation.
2. Every subgroup \(H\le G\) of index \(3\) determines a connected \(3\)-sheeted covering\[ p_H:(\widetilde X_H,\widetilde x_H)\longrightarrow(X,x_0) \]whose associated subgroup is \(H\).
A CW complex is path-connected, locally path-connected, and semilocally simply connected. Hence the covering-space classification theorem applies to \(X\). Under the identification in 1, the subgroup \(H\le\pi_1(X,x_0)\) corresponds to a connected based covering for which \[ (p_H)_*\pi_1(\widetilde X_H,\widetilde x_H)=H. \] The number of sheets is the subgroup index, so \[ |p_H^{-1}(x_0)|=[G:H]=3. \]
3. The covering in 2 gives a homomorphism\[ \rho_H:G\longrightarrow S_3 \]such that, after labeling the chosen point \(\widetilde x_H\) of the fiber by \(1\),\[ H=\{g\in G:\rho_H(g)(1)=1\}. \]
The fundamental group acts by monodromy on the fiber \(p_H^{-1}(x_0)\): a loop at \(x_0\) sends a point of the fiber to the endpoint of its lift beginning at that point. After labeling the three fiber points by \(\{1,2,3\}\) with \(\widetilde x_H\) labeled \(1\), this action is a homomorphism \[ \rho_H:\pi_1(X,x_0)\longrightarrow\operatorname{Sym}(\{1,2,3\})=S_3. \] Using 1, regard it as a homomorphism from \(G\).
An element \(g\in G\) fixes \(1\) under this action exactly when a representing loop lifts from \(\widetilde x_H\) to a loop ending again at \(\widetilde x_H\). By the lifting criterion for the associated subgroup, this happens exactly when \[ g\in(p_H)_*\pi_1(\widetilde X_H,\widetilde x_H)=H. \] This proves the displayed formula for \(H\).
4. There are only finitely many homomorphisms\[ G\longrightarrow S_3. \]
A homomorphism out of \(G\) is determined by the images of the finite generating set \[ g_1,\ldots,g_n. \] Each generator has only \[ |S_3|=6 \] possible images. Thus there are at most \[ 6^n \] possible assignments to the generators, and only some of these satisfy the relators and extend to homomorphisms. Hence the set \(\operatorname{Hom}(G,S_3)\) is finite.
5. The group \(G\) has only finitely many subgroups of index \(3\).
By 3, every index-\(3\) subgroup is the inverse image of a point stabilizer in \(S_3\) under a homomorphism \[ \rho:G\to S_3. \] There are only finitely many such homomorphisms by 4, and \(S_3\) has only three point stabilizers in its natural action on \(\{1,2,3\}\). Therefore only finitely many subgroups can arise. In fact the argument gives the crude bound \[ \#\{H\le G:[G:H]=3\}\le 3\cdot6^n. \]