I am looking at what should be a simple exercise in geometric group theory. I have reduced the problem to just completing an exercise from Hatcher, Section 1.B page 87:
7. 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.
A finitely generated free group can be realised as the fundamental group of a wedge of circles, so it seems I should be looking at the covering space of this bouquet induced by the infinite-index normal subgroup $N$. Since it is a normal subgroup, I know the group of deck transformations of my covering space is naturally isomorphic to the subgroup itself. Supposing that $N$ is finitely generated, I would like to lift its generating loops to the covering space, I will get, because of the infinite-index, loops starting at all the fibers of my base point. I would like from this to get that the group of deck transformations is finitely generated, but I can't see it.
Best Answer
The argument I would like to propose is as follows:
Fix a wedge of circles representing the free group F. Consider the cover space X representing the normal subgroup N. This is a regular cover space, which implies that the quotient group F/N acts transitively on X.
As N is finitely generated, then the cover space X which is an infinite graph has the following structure: after droping finitely many infinite trees, we get a compact subgraph C whose fundamental group is N.
Since F/N is infinite and acts transtively action on X, it follows that C has to be a tree. So X will be also a tree, which has the trivial fundamental group. This is the contradiction.
N.B. A covering with finitely generated fundamental group does not have to be compact.