Show that a finitely generated group has only a finite number of subgroups of a given finite index. [First do the case of free groups, using covering spaces of graphs. The general case then follows since every group is a quotient group of a free group.]
Solution
1. Let \(F\) be a free group of rank \(r\), realized as \(\pi_1\) of a wedge of \(r\) circles (a finite graph \(\Gamma\)).
a free group is the fundamental group of a finite graph.
2. A subgroup \(H \le F\) of index \(n\) corresponds to a connected \(n\)-fold covering space \(\tilde\Gamma \to \Gamma\).
covering space theory (subgroups of \(\pi_1\) correspond to connected covers).
3. An \(n\)-fold cover of a finite graph \(\Gamma\) is itself a finite graph with \(n \cdot |V(\Gamma)|\) vertices and \(n \cdot |E(\Gamma)|\) edges.
each vertex/edge of \(\Gamma\) has \(n\) preimages.
4. There are only finitely many such covering graphs (up to isomorphism), since there are finitely many graphs with a fixed finite number of vertices and edges.
3 (the number of vertices and edges is fixed, so only finitely many combinatorial types exist).
5. Hence \(F\) has only finitely many subgroups of index \(n\).
2 and 4.
6. Now let \(G\) be a finitely generated group, written as \(G = F/N\) for a free group \(F\) (of finite rank) and normal subgroup \(N\).
every finitely generated group is a quotient of a finitely generated free group.
7. Subgroups of \(G\) of index \(n\) correspond bijectively to subgroups \(H \le F\) with \(N \le H\) and \([F : H] = n\).
the correspondence theorem for the quotient \(F \to F/N = G\).
8. These \(H\) are among the (finitely many) index-\(n\) subgroups of \(F\).
5 and 7.
9. Hence \(G\) has only finitely many subgroups of index \(n\).
7 and 8.
10. Q.E.D.
5 and 9.