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

Question

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.

Answer 1

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 homotopy 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 for 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 vertices 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 $\tilde{X}\setminus T$. 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 vertices such that one of the edges of that vertex lies outside of $T$. Let $V_i$ be the set of vertices of distance at most $i$ from some element of $V_0$ (i.e. the set of vertices 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 vertices, 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.

Answer 2

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?

Answer 3

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.

Answer 4

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.

Answer 5

I know this is an old question, but it popped up on my feed today. I gave an answer many years ago, and from the comments it seems it was more confusing than helpful. I wanted to post a new answer, because I think the graph-theoretic approach here is too beautiful to not capture. All of the below is from Stallings's Topology of Finite Graphs paper.

Let $F$ be a finitely generated free group of rank $n$. Then $F$ can be represented by a finite wedge of $n$ circles, where we color each circle with a generator. For example, if $F=\langle\alpha,\beta,\gamma\rangle$, then we would have:

Any finitely generated subgroup of $F$ can be represented by a graph immersion, as follows. Let $1<H\le F$ be generated by $\{w_1,\ldots,w_m\}$, where each $w_i$ is a word in the $n$ generators of $F$. Then draw a graph that is the wedge of $m$ circles, where the $i$th circle is segmented to correspond to the word $w_i$. For example, if $H=\langle\alpha^2\beta, \gamma^2,\alpha\beta\gamma\rangle$, then we get the following graph for $H$:

A folding is identifying two edges in a graph that have the same starting/ending point, and the same color. Since this is a finite graph, after a finite number of foldings, we will end up with an immersion:

An immersion is one where every vertex has at most one incoming/outgoing edge of a given color. This is weaker than a covering, where every vertex has exactly one incoming/outgoing edge of a given color. But that points to how to turn every immersion into a covering: simply add the missing edges. Example for $H$:

This new (covering) graph corresponds to a subgroup $K\le F$. Since the graph is finite, $K$ is finitely generated and the covering is finite-sheeted, so $[F:K]$ is finite. Because we didn't add any new vertices to this covering graph, a spanning tree for the original (immersion) graph is still a spanning tree for the new (covering) graph. So the edges excluded from this tree gives a free basis for $K$, which contains a free basis for $H$ as a subset. This means that we have $K$ freely generated by a set $\{a_1,\ldots,a_r,b_1,\ldots,b_s\}$ where $H$ is generated by the $\{a_i\}$. In our example, we have \begin{equation} K = \langle \alpha^2\beta,\alpha\beta\gamma,\gamma^2,\beta^2, \alpha\gamma\beta, \alpha\gamma^{-1}\beta, \gamma\beta\alpha^{-1},\beta\alpha\gamma, \gamma\alpha \rangle \end{equation} More generally we have proved

Theorem (Hall): If $H\le F$ is a finitely generated subgroup, then there exists a finite index subgroup $K\le F$ containing $H$ where $H$ is a free factor of $K$.

Now if $s>0$ above, then note that $b_1^{-1}a_1b_1\not\in H$, so $H$ is not normal in $K$. This means that if $H\lhd G$, then we must have $s=0$, so that $H=K$, and the original immersion representing $H$ was already a covering. In other words, we have

Theorem (Greenberg): If $1<H\lhd F$ is finitely generated, then $[F:H]$ is finite.