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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.

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2 Answers 2

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If $N$ is normal, the associated covering space is regular. That means the degree of each vertex is the same, etc. If $N$ was finitely generated, the covering space would be compact (can you see why?); what do you know about the number of sheets in such a situation?

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I was just thinking about this problem myself and came across this hint. I can't see why this implies that the covering space is compact... If we realize the free group as the fundamental group of a (finite) bouquet of circles, then any covering graph has the property that the degree of each vertex is the same, right? So what else does regularity tell us? Thanks! –  Justin Campbell Jul 24 '12 at 22:11
    
Regularity is much stronger: it implies there is "maximal symmetry" in your graph: an automorphism taking any vertex to any other vertex (said in another way, the fundamental group acts transitively on the fibers). –  user641 Jul 24 '12 at 22:36
    
I see. This is true for arbitrary covering spaces, but what does it say combinatorially for graphs? I just don't see how this implies finiteness of the covering graph here. –  Justin Campbell Jul 24 '12 at 22:43
    
Let me try and say it a different way. Suppose your covering graph was infinite-sheeted. Then saying that $N$ was finitely generated would mean a maximal tree misses only finitely many edges in this covering graph. But regularity implies any edge missed at one vertex is missed at every vertex; thus there are only finitely many vertices... –  user641 Jul 24 '12 at 22:51

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.

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Hi, thanks for your ideas, your answer was helpful! However, I don't think it is quite right as you wrote it. If $C$ is a tree, why does $X$ have to be a tree ? For example, $X$ could be a "ladder", you cut it in half vertically and remove the right side which is a tree, the remaining left part $C$ is also a tree, but $X$ is not. And when you mean the action is transitive, I believe you mean on each fiber and not on $X$. Anyways, thank you for the answer! –  Bogdan Jan 20 at 17:14

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