# Normal subgroups of free groups: finitely generated $\implies$ finite index.

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.

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.

• 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! Jan 20, 2014 at 17:14
• @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. Jan 18, 2020 at 19:49

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?

• 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! Jul 24, 2012 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, 2012 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. Jul 24, 2012 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, 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.

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.