If \(F\) is a finitely generated free group and \(N\) is a nontrivial normal subgroup of infinite index, show, using covering spaces, that \(N\) is not finitely generated.
Solution
1. Let \(F = \pi_1(X)\) where \(X\) is a finite wedge of circles (a finite graph with one vertex).
a finitely generated free group is the fundamental group of a finite wedge of circles.
2. \(N\) corresponds to a connected covering space \(p: \tilde X \to X\) with \(p_*(\pi_1(\tilde X)) = N\).
the fundamental theorem of covering spaces.
3. Since \(N\) is normal, \(\tilde X\) is a normal (regular) covering, and its deck group is \(F/N\), which is infinite (since \(N\) has infinite index).
a normal subgroup corresponds to a regular covering, and the deck group is \(F/N\).
4. \(\tilde X\) is a graph (a covering of a graph is a graph), and it is infinite.
the deck group \(F/N\) is infinite, so \(\tilde X\) has infinitely many sheets, hence infinitely many vertices.
5. \(\pi_1(\tilde X) = N\) is the fundamental group of an infinite graph.
2.
6. The fundamental group of an infinite graph is not finitely generated.
6.1. An infinite graph has infinitely many edges.
a finite graph has finitely many edges; \(\tilde X\) is infinite, so it has infinitely many edges.
6.2. The fundamental group of a graph is free on the edges not in a maximal tree.
standard fact.
6.3. An infinite graph has a maximal tree whose complement has infinitely many edges, so its fundamental group is free on infinitely many generators.
1 and 2.
6.4. Hence \(\pi_1(\tilde X)\) is not finitely generated.
a free group on infinitely many generators is not finitely generated.
7. Therefore \(N\) is not finitely generated.
5 and 6.
8. Q.E.D.
7.