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.


4 Answers 4


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?

  • $\begingroup$ 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! $\endgroup$ Jul 24, 2012 at 22:11
  • 1
    $\begingroup$ 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). $\endgroup$
    – user641
    Jul 24, 2012 at 22:36
  • $\begingroup$ 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. $\endgroup$ Jul 24, 2012 at 22:43
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    $\begingroup$ 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... $\endgroup$
    – user641
    Jul 24, 2012 at 22:51

Let $F$ be the finitely generated free group, say with $n$ generators, so that if $X$ is the wedge of $n$ circles with wedge point $x_0$, $\Pi_1(X,x_0)\cong F$. We may then construct the covering space $\tilde{X}$ corresponding to $N$, which we know to be a connected graph, where each vertex projects to the wedge point and each edge projects to one of the circles in $X$. As usual, we may take $T$ a maximal tree inside of $\tilde{X}$ and use the fact that the quotient $\tilde{X}/T$ is homotopoy equivalent to $\tilde{X}$, and so has the same fundamental group, namely $N$.

Everything we have done so far may be done in complete generality fo any $N\leq \Pi_1(X)$. We now prove that if $N$ is as given, it is not finitely generated.

The first step is to recall that the fundamental group of $\tilde{X}/T$ is always free with cardinality equal to the cardinality of edges in $\tilde{X}\setminus T$ (since the quotient collapses all of the verticies and edges inside $T$ down to a single point, so all that is left is a wedge of circles, with each circle corresponding to an edge not in $T$). So if $N$ were finitely generated, there would be only finitely many edges in $X\setminus T$. We will now show that if $N$ is also normal and non-trivial and of infinite index, then $T$ contains a loop, contradicting the fact that $T$ is a tree, so we win.

Since $N$ is normal, for any vertex $v \in\tilde{X}$, $p_*\Pi_1(\tilde{X},v)=N$, and since $N$ is non-trivial, there is some non-trivial loop $\gamma$ in $X$ based at $x_0$ corresponding to an element of $N$. This may be expressed in terms of a reduced word, say of length $k\geq 1$. (We count e.g. $a^2$ as having length $2$.) We will find a vertex $v$ such that the lift $\tilde{\gamma}$ of $\gamma$ at $v$ lies inside of $T$. But then since $[\gamma] \in p_*\Pi_1(\tilde{X},v) = N$ is non-trivial, $\tilde{\gamma}$ is a non-trivial loop in $T$. (It is a loop since $\gamma$ lies in $p_*\Pi_1(\tilde{X},v)$).

To find $v$, let $V_0$ be the set of verticies such that one of the edges of that vertex lies outside of $T$. Let $V_i$ be the set of verticies of distance at most $i$ from some element of $V_0$ (i.e. the set of verticies that can be reached from following a path starting at $V_0$ that travels no more than $i$ edges). Since $N$ is of infinite index, $\tilde{X}$ has infinitely many verticies, but since each vertex has finitely many edges (each vertex must have one arrow going in and one going out for each generator of $F$, so has at most $2n$ edges) $V_i$ is always a finite set. Thus we may take any $v \in \tilde{X} \setminus V_{k}$. Since $\gamma$ is a word of length $k$, the lift of $\gamma$ at $v$ is a path of length $k$ and so never hits a vertex in $V_0$. But this means that all of the edges travelled by $\tilde{\gamma}$ lie inside of $T$, as claimed.


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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    $\begingroup$ 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! $\endgroup$
    – Bogdan
    Jan 20, 2014 at 17:14
  • $\begingroup$ @Bogdan because $N=\pi_1(C)=\pi_1(X)$, if $C$ were a tree, then so is $X$, as every space in sight is a graph. $\endgroup$
    – Mr. Brown
    Jan 18, 2020 at 19:49

Let$$X = \bigvee_{i=1}^n S_i^1,$$the wedge sum of $n$ circles. Then $\pi_1(X) = F$. Let $X_N$ be the covering space corresponding to $N \unlhd F$. Note that $p: X_N \to X$ is normal. Assume, for the sake of contradiction, that $\pi_1(X_N)$ is finitely generated by $m$ elements.

Since the index of $N$ is infinite, $X_N$ must cover $X$ by infinitely many sheets. So $X_N$ consists of an infinite tree together with $m$ loops $e_1, e_2, \dots, e_m$. (The infinite tree gives the necessary sheets and the $m$ loops generate the fundamental group.)

Let $\widetilde{\gamma}$ be a path in $X_N$ based at $x_0^N \in X_N$ with $p(\widetilde{\gamma}) = \gamma$ a nontrivial path in $X$. Since the cover is normal, there exists a deck transformation mapping $x_0^N$ to any other preimage under $p$ of $x_0 \in X$. (Why?) Let $X_1^N$ be a vertex in $p^{-1}(x_0)$ farther away (using the notion of "height" from page 85 of Hatcher) from the $e_i$ than is $x_0^N$. Then there exists a deck transformation $f: X_N \to X_N$ with $f(x_0^N) = x_1^N$. But $f(\widetilde{\gamma})$ contains more edges which are loops in $X$ of $X_N$ than $\widetilde{\gamma}$, and so $f$ can not be a covering isomorphism. This contradicts the definition of a deck transformation.

Therefore, $N$ is not finitely generated.


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