## On subgroups of surface groups

It is known that a closed (ie. connected, compact without boundary) surface is homeomorphic to a sphere or to a connected sum of finitely-many tori or projective spaces. Let $S_g$ (resp. $\Sigma_g$) denote the connected sum of $g$ tori (resp. projective spaces). Because $\chi(A \sharp B)= \chi(A)+ \chi(B)-2$, we easily deduce the Euler characteristics $\chi(S_g)=2-2g$ and $\chi(\Sigma_g)=2-g$; the integer $g$ is called the genus.

A surface group is a group isomorphic to the fundamental group of one of these surfaces. If $G$ is a surface group, two cases happens: either $G$ is the fundamental group of an orientable surface $S_g$ of genus $g$ and

$G = \langle a_1,b_1, \dots, a_g,b_g \mid [a_1,b_1] \cdots [a_g,b_g]=1 \rangle$,

or $G$ is the fundamental group of a non-orientable surface $\Sigma_g$ of genus $g$ and

$G = \langle a_1, \dots, a_g \mid a_1^2 \cdots a_g^2=1 \rangle$.

(Just write $S_g$ or $\Sigma_g$ as a polygon with pairwise identified edges, and apply van Kampen’s theorem; for more information, see Massey’s book Algebraic Topology.)

Our main result is the following characterization of subgroups of a surface group:

Theorem 1: Let $G$ be the fundamental group of a closed surface $S$ and $H$ be a subgroup. Either $H$ has finite index $h$  in $G$ and $H$ is isomorphic to the fundamental group of a closed surface whose Euler characteristic is $h \cdot \chi (S)$, or $H$ is an infinite-index subgroup of $G$ and is free.

For this note, I was inspired by Jaco’s article, On certain subgroups of the fundamental group of a closed surface. The proof of property 2 below comes from Stillwell’s book, Classical topology and combinatorial group theory.

Essentially, the theorem above follows from Property 2 below:

Property 2: The fundamental group of a non-compact surface is free.

Combined with Property 3, let us first notice some corollaries of Theorem 1 on the structure of surface groups.

Property 3: The abelianization of $\pi_1(S_g)$ (resp. $\pi_1(\Sigma_g)$) is isomorphic to $\mathbb{Z}^{2g}$ (resp. $\mathbb{Z}^{g-1} \times \mathbb{Z}_2$). Therefore, two closed surfaces $S_1$ and $S_2$ are homeomorphic if and only if their fundamental groups are isomorphic.

Proof. In order to compute the abelianizations, it is sufficient to add all the possible commutators as relations in the presentations given above. Therefore,

$\pi_1(S_g)^{ab} \simeq \langle a_1,b_1 , \dots, a_g, b_g \mid [a_i,a_j] = [b_i,b_j ] = [a_i,b_i]=1 \rangle \simeq \mathbb{Z}^{2g}$

and, with $z: = a_1\cdots a_g$,

$\pi_1( \Sigma_g)^{ab} \simeq \langle a_1, \dots, a_{g-1},z \mid [a_i,a_j] = [a_i,z]=1, \ z^2=1 \rangle \simeq \mathbb{Z}^{g-1} \times \mathbb{Z}_2$.

Consequently, the $\pi_1(S_g)$ and $\pi_1( \Sigma_1)$ define a family of pairwise non-isomorphic groups, and the conclusion follows from the classication of closed surfaces. $\square$

Corollary 1: Let $G$ be the fundamental group of a closed surface different from the projective space. Then $G$ is torsion-free.

Proof. The abelianization of a surface group is cyclic if and only if it is the fundamental group of the projective plane; therefore, the same conclusion holds for the surface groups themself, and we deduce that $G$ has no cyclic subgroup of finite index. Consequently, the cyclic subgroup generated by a non trivial element of $G$ is free, that its order is infinite. $\square$

Corollary 2: Let $G$ be the fundamental group of a closed surface $S$ satisfying $\chi(S) \leq 0$ and $H$ be a subgroup generated by $k$ elements. If $k < 2- \chi (S)$ then $H$ is free.

Proof. If $\chi(S)=0$, $S$ is a torus or a Klein bottle, and the statement just says that its fundamental group is torsion-free, that follows from Corollary 1.

From now on, we suppose that $\chi(S)<0$. According to Property 3, the abelianization of $G$ has rank $2- \chi(S)$. We deduce that the smallest cardinality of a generating set of $G$ has cardinality $2- \chi(S)$; in particular, $H \subsetneq G$ because $k < 2- \chi(S)$. Suppose that $H$ has finite index $h$. Then $H$ is the fundamental group of a closed surface $\Sigma$ and $\chi(\Sigma)= h \cdot \chi(S)$. In the same way, necessarily $k \geq 2- \chi(\Sigma)$. Therefore,

$k \geq 2- \chi(\Sigma) = 2-h \cdot \chi(S) > 2 - \chi(S)$,

a contradiction. $\square$

Corollary 3: Let $G$ be the fundamental group of a closed surface $S$ satisfying $\chi(S)<0$. If $x,y \in G \backslash \{1\}$ commute, then there exist $z \in G$ and $m, n \in \mathbb{Z}$ such that $x=z^n$ and $y=z^m$.

Proof. Let $g \geq 2$. Then it is easy to find two epimorphisms

$\pi_1 (S_g ) \twoheadrightarrow \langle a_1 , b_1 , a_1 , b_2 \mid [ a_1, b_1 ] = [ a_2, b_2 ] ^{-1} \rangle \simeq \mathbb{F}_2 \underset{\mathbb{Z}}{\ast} \mathbb{F}_2$

and

$\pi_1 ( \Sigma_g ) \twoheadrightarrow \langle a_1, a_2 \mid a_1^2 = a_2^{-1} = 1 \rangle \simeq \mathbb{Z}_2 \ast \mathbb{Z}_2$.

Therefore, the groups $\pi_1(S_g)$ and $\pi_1(\Sigma_g)$ are not abelian for $g \geq 2$, and we deduce that the sphere, the projective plane and the torus are the only closed surfaces whose fundamental group is abelian.

Because $S$ satisfies $\chi(S)<0$, we conclude that $G$ has no finite-index abelian subgroup, and that the subgroup genereted by $\{ x,y \}$ is necessarily free; since $x$ and $y$ commute, the subgroup turns out to be cyclic and the conclusion follows. $\square$

Corollary 4: The commutator subgroup of a surface group is free.

Proof. Let $G$ be the fundamental group of a closed surface $S$. If $S$ is a projective space, then $G \simeq \mathbb{Z}_2$ and its commutator subgroup is trivial (in particular free). Otherwise, thanks to Property 3 we know that the abelianization of $G$ is infinite; therefore, the commutator subgroup is an infinite-index subgroup and so is free. $\square$

Proof of property 2. First, we notice that the fundamental group of a connected compact surface with boundary is free. If $S$ is such a surface, by gluing a disk along each boundary component, we get a closed surface $\overline{S}$; therefore, $S$ is homotopic to a punctured closed surface. Because a closed surface may be identified with a polygon whose edges are pairwise identified, it is not difficult to prove that a punctured closed surface is homotopic to a graph (by “enlarging the holes”); in particular, we deduce that $\pi_1(S)$ is free.

The figure below shows that the fundamental group a 2-punctured torus is a free group of rank three.

From now on, let $S$ be a non-compact surface. In order to prove that $\pi_1(S)$ is free, we want to find a sequence of compact surfaces with boundary

$S_1 \subset S_2 \subset \cdots \subset S$

such that $S= \bigcup\limits_{i \geq 1} S_n$ and  that the inclusions $S_i \hookrightarrow S_{i+1}$ are $\pi_1$-injective. It is sufficient to conclude since we proved above that the fundamental group of a compact surface with boundary is free.

Take a triangulation of $S$ so that $S$ may be identified with a simplicial complex; let $\Delta_1,\Delta_2, \dots$ denote the 2-simplexes. Without loss of generality, we may suppose that $\Delta_{i+1}$ is adjacent to one of the simplexes $\Delta_1, \dots , \Delta_i$; otherwise, number the simplexes by taking first the simplexes adjacent to a vertex $P$ cyclically, then the simplexes within $1$ from $P$, then the simplexes within $2$ from $P$, etc.

Notice that a connected union of simplexes may not be a surface; however, if $\Delta_i'$ denotes the union of $\Delta_i$ with a small closed ball around each of its vertices, a connected union of $\Delta_i'$ is automatically a compact surface with boundary. Now, we construct the surfaces $S_n$ by induction:

Let $S_1=\Delta_1'$. Now suppose that $S_n$ is given. If $\Delta_{n+1}' \subset S_n$, then set $S_{n+1}=S_n$; otherwise, define $S_{n+1}$ as the union of $S_n$ with $\Delta_{n+1}'$ and with the disks bounding a boundary component of $S_n \cup \Delta_{n+1}'$.

Because $S_n$ is a union of $\Delta_i'$, we know that it is a compact surface with boundary. To conclude, it is sufficent to prove that the inclusion $S_i \hookrightarrow S_{i+1}$ is $\pi_1$-injective. To do that, just notice that $\Delta_{n+1}'$ is attatched on the boundary of $S_n$ in six possible ways:

In the first three cases, $S_{n+1}= S_n \cup \Delta_{n+1}'$ because the boundary component bounds a disk in $S$ if and only if it bounds a disk in $S_n$. So $S_{n+1}$ clearly retracts on $S_n$ so that the inclusion $S_n \hookrightarrow S_{n+1}$ induces an isomorphism $\pi_1(S_n) \simeq \pi_1(S_{n+1})$.

In the three last cases, up to homotopy, we just glue one or two edges on a boundary component, and using van Kampen’s theorem, we notice that a free basis of $\pi_1(S_{n+1})$ is obtained by adding one or two elements to a free basis of $\pi_1(S_n)$. $\square$

Proof of theorem 1. Let $\overline{S} \to S$ be the covering associated to the subgroup $H$; in particular, $\overline{S}$ is a surface of fundamental group $H$. If $H$ is a finite-index subgroup, $\overline{S}$ is compact and the conclusion follows. Otherwise, $\overline{S}$ is not compact, and the conclusion follows from Property 2. $\square$

## Freudenthal Compactification

This note is dedicated to Freudenthal compactification, a kind of compactification for some topological spaces. In particular, it will be noticed how such a compactification leads to the number of ends, a nice topological invariant, and how it can be used to classify compactifications with finitely many points at infinity. The proof of theorem 1 is mainly based on Baues and Quintero’s book, Infinite homotopy theory.

First of all, we introduce the spaces we will work with.

Definition: A generalized continuum is a locally compact, connected, locally connected, $\sigma$-compact, Hausdorff topological space.

Typically, a generalized continuum may be think of as a locally finite CW complex or as a topological manifold.

Let $X$ be a generalized continuum. Because $X$ is $\sigma$-compact, it is an increasing union of compact subspaces $(K_n)$; moreover, by locally compactness, we may suppose that $K_n \subset \mathrm{int} ~ K_{n+1}$ for all $n \geq 1$. Such a family of compact subspaces is called an exhausting sequence.

Definition: Let $X$ be a generalized continuum and $(K_n)$ be an exhausting sequence. An end $\epsilon$ is a decreasing family $(C_n)$ of subspaces such that $C_n$ is an unbounded (ie. not relatively compact) connected component of $X \backslash K_n$.

Let $E(X)$ denote the set of ends of $X$ and $\overline{X} = X \cup E(X)$. If $\epsilon=(C_n)$ is an end and $U \subset X$ is an open subspace, $\epsilon < U$ will mean that $C_n \subset U$ for some $n\geq 1$.

The Freudenthal compactification of $X$ is defined as $\overline{X}$ endowed with the topology generated by

$\{ U \cup \{ \epsilon \in E(X) \mid \epsilon < U \} \mid U \subset X \ \mathrm{open} \}$.

First, notice that $\overline{X}$ does not depend on the sequence of compact subspaces we chose. Indeed, it is possible to consider the increasing functions $f$ associating to any compact $K \subset X$ an unbounded connected component $f(K)$ of $X \backslash K$, and then to define a space $\tilde{X}= X \cup F(X)$, where $F(X)$ is the set of functions we just mentionned, endowed with the topology generated by

$\{U \cup \{ f \in F(X) \mid f(K) \subset U \ \mathrm{for \ some \ compact} \ K \subset X \} \mid U \subset X \ \mathrm{open} \}$.

Clearly, the restriction of a function $f \in F(X)$ to the family $(K_n)$ we fixed to construct $\overline{X}$ defines an end of $X$, so there exists a natural continuous map $\tilde{X} \to \overline{X}$. Conversely, an end $\epsilon = (C_n)$ naturally corresponds to a function of $F(X)$: if $K \subset X$ is compact, there exists some $n \geq 1$ such that $K \subset K_n$, and $\epsilon(K)$ may be defined as the unbounded connected component of $X \backslash X$ containing $C_n$. Therefore, the map $\tilde{X} \to \overline{X}$ turns out to be a homeomorphism. In particular, $\overline{X}$ does not depend on the sequence $(K_n)$ since neither does $\tilde{X}$.

Theorem 1: Let $X$ be a generalized continuum. Then $\overline{X}$ is a compactification of $X$, that is $X$ is a(n) (open) dense subset of $\overline{X}$ and $\overline{X}$ is compact.

We will prove the theorem only at the end of this note. Before that, we will mention two nice consequences of our construction. The first one is that Freudenthal compactification gives a topological invariant: the number of ends, $e(X)= |E(X)|$.

Theorem 2: Let $X,Y$ be two generalized continua. If $\varphi : X \to Y$ is a homeomorphism, then $\varphi$ extends to a homeomorphism $\overline{\varphi} : \overline{X} \to \overline{Y}$; in particular, $\overline{\varphi}$ induces a homeomorphism $E(X) \to E(Y)$.

Proof. Let $(K_n)$ be an exhausting sequence for $X$. Then $(\varphi(K_n))$ is an exhausting sequence for $Y$. Moreover, if $(C_n)$ is an end of $X$ then $(\varphi(C_n))$ is naturally an end of $Y$: it defines our extension $\overline{\varphi} : \overline{X} \to \overline{Y}$. We easily check that $\overline{\varphi}$ is a homeomorphism. $\square$

Corollary: The number of ends is a topological invariant.

Of course, two non-homeomorphic spaces may have the same number of ends (eg. $e(\mathbb{R}^n)=1$ for all $n \geq 2$). However, the number of ends appears to be easy to compute in several situations, so it may be a simple way to prove that two spaces are not homeomorphic. For example:

Property: Let $n,m \geq 2$ and $S \subset \mathbb{R}^n$, $R \subset \mathbb{R}^m$ be two finite subsets. If $|S| \neq |R|$ then $\mathbb{R}^n \backslash S$ and $\mathbb{R}^m \backslash R$ are not homeomorphic.

Proof. Just notice that $e(\mathbb{R}^n \backslash S)= |S|$ and similarly $e(\mathbb{R}^m \backslash R)= |R|$. $\square$

Our second consequence of Freudenthal compactification is the classification of all possible compactifications whose remainders are finite (the remainder of a compactification $Y$ of a space $X$ is $Y \backslash X$; sometimes, the points of the remainder are also called points at infinity).

Theorem 3: Let $X$ be a generalized continuum and $Y$ be a compactification of $X$ whose remainder is finite. Then $Y$ is obtained from $\overline{X}$ by identifying some ends.

Therefore, Freudenthal compactification can be viewed as a maximal compactification among compactifications with finitely many points at infinity.

Proof. As above, let $\overline{X}$ denote the Freudenthal compactification of $X$. Let

$\overline{X} = X \cup \{ p_1, \dots, p_r \}$ and $Y= X \cup \{ q_1,\dots, q_s \}$.

For all $1 \leq i \leq s$, let $U_i \subset Y$ be an open neighborhood of $q_i$; we may suppose that $U_i \cap U_j = \emptyset$ if $i \neq j$. Let

$C= Y \backslash \bigcup\limits_{i=1}^s U_i \subset X$;

notice that $C$ is compact. For all $1 \leq i \leq r$, let $V_i \subset \overline{X}$ be a connected neighborhood of $p_i$ disjoint from $C$; again, we may suppose that $V_i \cap V_j = \emptyset$ if $i\neq j$.

Because $V_i$ is connected, there exists $j$ such that $V_i \subset U_j$; moreover, because $U_i \cap U_j = \emptyset$ if $i \neq j$, such a $j$ is unique. Therefore, a map $f : \overline{X} \to Y$ may be defined by $f(p_i)=q_j$ and $f_{|X}= \mathrm{Id}_{X}$.

Let $I_j = \{ 1 \leq i \leq r \mid f(p_i)=q_j \}$. Noticing that

$f^{-1}(U_j)= \bigcup\limits_{i \in I_j} V_i \cup \{ p_i \}$

is open in $\overline{X}$ since $i \in I_j$ if and only if $p_i < U_j$, we deduce that $f$ is continuous. Moreover, $f$ is onto.

Let $\tilde{X}$ denote the space obtained from $\overline{X}$ by identifying two ends $p_i$ and $p_j$ whenever $f(p_i)=f(p_j)$.

By construction, $\tilde{f} : \tilde{X} \to Y$ is well-defined and into. Then, $\tilde{f}$ is continuous since $\pi$ is open, and onto since $f$ is onto. Finally, $\tilde{f}$ is a continuous bijection between two compact spaces: it is necessarily a homeomorphism. $\square$

We illustrate the classification given above with the following example:

Corollary 1: Let $X$ be a generalized continuum. Then $X$ has only one compactification with one point at infinity: its Alexandroff compactification.

Corollary 2: Let $X$ be a generalized continuum and $Y$ be a compactification of $X$ whose remainder is finite. Then $|Y \backslash X | \leq e(X)$.

In particular, we get that the circle $\mathbb{S}^1$ and the segment $[0,1]$ are the only compactifications of $\mathbb{R}$ with finitely many points at infinity. However, there is a lot of possible compactifications of $\mathbb{R}$ with infinitely many points; more precisely, the following result can be found in K. Magill’s article, A note on compactification:

Theorem: Let $K$ be Peano space. Then there exists a compactification $X$ of $\mathbb{R}$ whose remainder is homeomorphic to $K$.

For example, if $\Gamma \subset [0,+ \infty) \times \mathbb{R}$ denotes the graph of the function $x \mapsto \sin(1/x)$, then $\Gamma$ is homeomorphic to $\mathbb{R}$, and $X = \Gamma \cup \{ 0 \} \cup [ -1 , 1 ]$ is a compactification of $\mathbb{R}$ whose remainder is homeomorphic to $[0,1]$.

See also B. Simon’s article Some pictoral compactification of the real line to find a compactification of $\mathbb{R}$ whose remainder is homeomorphic to the torus $\mathbb{T}^2$.

From now on, we turn to the proof of theorem 1.

Lemma: Let $X$ be a generalized continuum. Then $E(X)$ is a compact subspace of $\overline{X}$.

Proof of Theorem 1: Fistly, the injection $X \hookrightarrow \overline{X}= X \cup E(X)$ is clearly a homeomorphism onto its image; therefore, from now on, $X$ will be confunded with its image in $\overline{X}$. Then $X$ is clearly dense in $\overline{X}$, so to conclude the proof, we only need to prove that $\overline{X}$ is compact.

Let $\{ O_i \mid i \in I \}$ be an open covering of $\overline{X}$. Because $E(X)$ is compact, there exists $J_1 \subset I$ finite such that for all $\epsilon \in E(X)$, there exist $j \in J_1$ and an unbounded connected component $C_j \subset O_j$ of $X \backslash K_{n_j}$ such that $\epsilon < C_j$. If $p= \max\limits_{j \in J_1} n_j$, then $\{ O_j \mid j\in J_1 \}$ is a finite subcovering of the unbounded connected components of $X \backslash K_p$.

Notice that $K_p \subset \mathrm{int} ~ K_{p+1}$ implies that $K_p \cap \partial K_{p+1} = \emptyset$, so a connected component of $X \backslash K_p$ intersects $\partial K_{p+1}$ or is included in $K_{p+1}$. But $\partial K_{p+1}$ is compact, so there are only finitely many components of $X \backslash K_p$ intersecting $\partial K_{p+1}$. Therefore, if $K$ denotes the union of $K_{p+1}$ with the closure of the bounded components of $X \backslash K_p$ intersecting $\partial K_{p+1}$, then $K$ is compact. In particular, there exists $J_2 \subset I$ finite such that $\{ O_j \mid j \in J_2 \}$ is a finite subcovering of $K$.

Finally, we deduce that $\{ O_j \mid j \in J_1 \cup J_2 \}$ is a finite subcovering of $\overline{X}$. Hence $\overline{X}$ is compact. $\square$

Proof of the lemma. Let $(K_n)$ be an exhausting sequence and let $\pi(K_n)$ denote the set of unbounded connected components of $X \backslash K_n$. First notice that $\pi(K_n)$ is finite. Indeed, if $C \in \pi(K_n)$, it cannot be included in $K_{n+1}$ so it intersects $\partial K_{n+1}$; the conclusion follows by compactness of $\partial K_{n+1}$. Therefore, if each $\pi(K_n)$ is endowed with the discrete topology, from Tykhonov theorem the cartesian product $\prod\limits_{n \geq 1} \pi(K_n)$ is compact. Because there is a natural map

$\phi : E(X) \to \prod\limits_{n \geq 1} \pi(K_n)$,

it is sufficient to prove that $\phi$ is a homeomorphism onto its image and that its image is closed. In order to show that $\phi$ is continuous, we take an open set $O \subset \prod\limits_{n \geq 1} \pi(K_n)$ of the form

$\{ (C_n) \mid C_{n_1}= C_1 \subset \cdots \subset C_{n_r} = C_r \}$

for some $r \geq 1$ and $C_1 \in \pi(K_1), \dots, C_r \in \pi(K_r)$ fixed. Now, if

$\Omega = \{ \epsilon \in E(X) \mid \mathrm{for \ all} \ 1 \leq i \leq r, \ \epsilon < C_{n_i} \} = \bigcap\limits_{i=1}^r \{ \epsilon \in E(X) \mid \epsilon < C_{n_i} \}$,

then $\Omega$ is open since

$\Omega = \bigcap\limits_{i=1}^r E(X) \cap \{ C_{n_i} \cap \{ \epsilon \} \mid \epsilon < C_{n_i} \}$,

and $\phi(\Omega) \subset O$. We deduce that $\phi$ is continuous. In order to show that $\phi$ is an open map, we take an open subspace $V \subset E(X)$ of the form $\{ \epsilon \mid \epsilon < U \}$ for some open subspace $U \subset X$, and we notice that

$\phi (U) = \{ (C_n) \mid \exists i \in \mathbb{N}, \ C_i \subset U \} = \bigcup\limits_{ i \in \mathbb{N} } \bigcup\limits_{ C \in \omega_i } \{ (C_n) \mid C_i = C \}$

where $\omega_i$ is the set of elements of $\pi(K_i)$ included in $U$. Therefore, $\phi(U)$ is open, and we deduce that $\phi$ is an open map. We just proved that $\phi$ is a homeomorphism onto its image. Now we want to prove that $\phi(E(X))$ is closed. We have

$\phi (E(X)) = \left\{ (C_n) \in \prod\limits_{n \geq 1} \pi(K_n) \mid \forall i \leq j, C_j \subset C_i \right\}$.

Now, if $i \leq j$, let $\phi_{ij}$ denotes the map that sends a connected component of $X \backslash K_j$ to the connected component of $X \backslash K_i$ containing it. Because $C_j \subset C_i$ is equivalent to $\phi_{ij} (C_j)=C_i$, we deduce that

$\phi(E(X))= \bigcap\limits_{i \leq j} M_{ij}$,

where

$M_{ij} = \left\{ C \in \prod\limits_{n \geq 1} \pi(K_n) \mid \phi_{ij} \circ \mathrm{pr}_j (C)= \mathrm{pr}_i (C) \right\}$.

But $\phi_{ij} : \pi(K_j) \to \pi(K_i)$ is obviously continuous, $\mathrm{pr}_i$ is continuous by definition and $\pi(K_i)$ is Hausdorff, so $M_{ij}$ is closed. We conclude that $\phi(E(X))$ is closed, and finally compact. $\square$

## Graphs and Free Groups

Algebraic topology gives a strong link between graphs and free groups. Namely, the fundamental group of a graph is a free group, and conversely, any free group is the fundamental group of a graph. Surprisingly, using the framework of graphs and coverings gives a lot of information about free groups. Our aim here is to exhib some purely algebraic results about free groups and to prove them using graphs.

Our main references are Massey’s book, Algebraic Topology. An Introduction, Hatcher’s book, Algebraic Topology, and Stallings’ article, Topology of finite graphs. We refer to these references for the results of algebraic topology we use below.

Theorem 0: The fundamental group of a graph is free. Conversely, a free group is the fundamental group of a graph.

Proof. Let $\Gamma$ be a graph. If $T \subset \Gamma$ is a maximal subtree, then $\Gamma/T$ is a bouquet of $\kappa$ circles and $\pi_1(\Gamma) \simeq \pi_1( \Gamma /T)$. The universal covering of $\Gamma /T$ is a regular tree of degree $\kappa$, so it is also the usual Cayley graph of the free group $F$ of rank $\kappa$. Because a group acts freely on its Cayley graph, we deduce that $\pi_1(\Gamma) \simeq F$.

Conversely, let $F$ be a free group of rank $\kappa$. With the same argument, $F$ is the fundamental group of a bouquet of $\kappa$ circles. $\square$

Notice that in the first part of the proof, $\kappa$ is the cardinal of edges of $\Gamma$ not in $T$. In particular, the fundamental group of a finite graph is a free group of finite rank.

Theorem 1: A subgroup $H$ of a free group $F$ is free. Moreover, if $[F:H], \mathrm{rk}(F) < + \infty$ then

$\mathrm{rk}(H)=1+[F:H] \cdot (\mathrm{rk}(F)-1)$.

Proof. We view $F$ as the fundamental group of a graph $X$. Let $Y \twoheadrightarrow X$ be a covering satisfying $\pi_1(Y) \simeq H$. In particular, the covering induces a structure of graph on $Y$ making the covering cellular; in particular, $H$ is a free group.

If $n=[F:H] < + \infty$, the covering $Y \twoheadrightarrow X$ is $n$-sheeted, hence $\chi(Y)=n \chi(X)$. If $T \subset X$ is a maximal subtree, then $X$ is the disjoint union of $T$ and $\mathrm{rk}(F)$ edges, hence

$\chi(X)= \chi(T)- \mathrm{rk}(F)=1- \mathrm{rk}(F)$;

in the same way, $\chi(Y)= 1- \mathrm{rk}(H)$. Therefore, $\mathrm{rk}(H)=1+n (\mathrm{rk}(F)-1)$. $\square$

Theorem 2: A non-abelian free group contains a free group of infinite rank.

We already proved theorem 2 using covering spaces in the previous note A free group contains a free group of any rank.

We know from theorem 1 that any finite-index subgroup of a finitely generated free group is finitely generated. Of course, the converse is false in general: $\langle a \rangle$ is a finitely-generated subgroup of infinite index in $\mathbb{F}_2 = \langle a,b \mid \ \rangle$. However, the converse turns out to be true for normal subgroups.

Theorem 3: Let $F$ be a free group of finite rank and $H \leq F$ be a non-trivial normal subgroup. Then $H$ is finitely generated if and only if it is a finite-index subgroup.

Proof. Suppose that $H$ is an infinite-index subgroup. Let $X$ be a bouquet of $\mathrm{rk}(F)$ circles so that $F$ be the fundamental group of $X$ and let $Y \twoheadrightarrow X$ be a covering such that $H$ is the fundamental group of the graph $Y$. Noticing that a deck transformation associated to the covering is a graph automorphism of $Y$, we deduce that the group $\mathrm{Aut}(Y)$ of graph automorphisms is vertex-transitive since the covering is normal.

Let $T \subset Y$ be a maximal subtree. Because $H$ is non-trivial, there exists a cycle $C$ in $Y$. Moreover, because $H$ is an infinite-index subgroup the graph $Y$ is infinite and there exist vertices $v_0,v_1, \dots$ of $Y$ satisfying $v_0 \in C$ and $d(v_n,v_m) > \ell g (C)$ for all $n \neq m$; let $\varphi_n \in \mathrm{Aut}(Y)$ be a graph automorphism sending $v_0$ to $v_n$. Then $(\varphi_n(C))$ is a family of disjoint cycles, so there exist infinitely-many edges of $Y$ not in $T$. Therefore, $H$ is not finitely generated. $\square$

Theorem 4: A free group of finite rank $F$ is subgroup separable. In particular, it is residually finite and Hopfian (ie. every epimorphism $F \twoheadrightarrow F$ is an isomorphism).

It is precisely the content of the note How Stallings proved finitely-generated free groups are subgroup separable. Using the same ideas, we are able to prove the following result:

Theorem 5: (Hall) Let $F$ be a free group of finite rank and $H \leq F$ be a finitely generated subgroup. Then $H$ is a free factor in a finite-index subgroup of $F$.

Proof. Let $X$ be a bouquet of $n:= \mathrm{rk}(F)$ circles so that $\pi_1(X) \simeq F$ and let $Y \to X$ be a covering so that $\pi_1(Y) \simeq H$. If $T \subset Y$ is a maximal subtree, because $H$ is finitely generated by assumption, there exist only finitely-many edges of $Y$ not in $T$; let $D \subset Y$ be a finite connected graph containing these edges and such that $T \cap D$ is a maximal subtree of $D$. Let $Z$ denote the canonical completion $C(D,X)$.

Now $\pi_1(Z)$ a finite-index subgroup of $\pi_1(X) \simeq F$ and $\{ \text{edges of} \ Z \ \text{not in} \ T \cap D \}$ is a free basis. By construction, this basis contain a free basis of $H$, so $H$ is a free factor in $\pi_1(Z)$. $\square$

Theorem 6: (Howson) Let $S_1$ and $S_2$ be two finitely-genereated subgroups of a free group $F$. Then $S_1 \cap S_2$ is finitely-generated.

It is precisely the content of the note Fiber product of graphs and Howson’s theorem for free groups. We conclude with an application holding not only for free groups of finite rank but for any finitely generated group! The clue is that any group is the quotient of a free group.

Theorem 7: A finitely generated group has finitely many subgroups of a given finite index.

Proof. Let $F$ be a free group of finite rank and $n \geq 1$. Clearly, a finite graph has finitely many $n$-sheeted covering (because such a covering is a graph with a fixed number of vertices and edges), so $F$ has finitely many conjugacy classes of groups of index $n$; moreover, a finite-index subgroup’s conjugacy class is necessarily finite (if a subgroup $H$ has index $n$, then $xH=yH$ implies $xHx^{-1}=yHy^{-1}$). Therefore, theorem 7 holds for finitely generated free groups.

Let $G$ be a finitely generated group. Then $G$ is the quotient of a free group, that is there exists an epimorphism $\varphi : G \twoheadrightarrow F$ onto a free group of finite rank. Then, for all $n\geq 1$, $H \mapsto \varphi^{-1}(H)$ defines an injective map from to set of $n$-index subgroup of $G$ into the set of $n$-index subgroup of $F$. The theorem 7 follows. $\square$

Our aim is to prove Howson’s theorem for free groups followings Stallings’ article, Topology of finite graphs; this note is more or less a sequel of How Stallings proved finitely-generated free groups are subgroup separable. The construction introduced here, the fiber product, will be also useful for the generalization of ideas of Stallings to special cube complexes (due to Wise).

Another interesting proof, based on regular languages, can be found in Meier’s book untitled Graphs, groups and trees.

Theorem: (Howson) Let $F$ be a free group of finite rank and $H,K$ be two finitely-generated subgroups. Then $H \cap K$ is also finitely-generated.

Let $\alpha : X \to Z$ and $\beta : Y \to Z$ be two immersions (ie. locally injective cellular maps) between graphs. We define the fiber product $X \otimes_Z Y$ as the graph whose vertices are

$\{ (x,y) \in X \times Y \mid \alpha(x)= \beta(y) \}$

where two vertices $(x_1,y_1)$ and $(x_2,y_2)$ are linked by an edge if and only if $\alpha(x_1)=\beta(y_1)$ and $\alpha(x_2)=\beta(y_2)$ are linked by an edge in $Z$.

Notice that there are two obvious projections $p : X \otimes_Z Y \to X$, $q : X \otimes_Z Y \to Y$, and a natural map $\gamma : X \otimes_Z Y \to Z$ so that we have the following commutative diagram:

Claim 1: $\gamma : X \otimes_Z Y \to Z$ is an immersion.

Proof. Let $(x_1,y_1)$ be a vertex of $X \otimes_Z Y$ and $e,f$ two edges starting from $(x_1,y_1)$; let $(x_2,y_2)$ and $(x_3,y_3)$ denote the ending points of $e$ and $f$. If $\gamma(e)=\gamma(f)$, then $\gamma(x_3,y_3)=\gamma(x_2,y_2)$ hence $(x_2,y_2)=(x_3,y_3)$. Therefore, if $e \neq f$ then $e$ and $f$ are two loops based at $(x_1,y_1)$, sent via $\gamma$ on two loops based at $\gamma(x_1,y_1)$. So $\gamma$ is locally injective. $\square$

Claim 2: $\gamma_* \pi_1( X \otimes_Z Y) = \alpha_* \pi_1(X) \cap \beta_* \pi_1(Y)$.

First, we need the following easy lemma:

Lemma: Let $\Gamma$ be a graph and $c_1,c_2$ be two reduced (ie. without round-trip) loops. If $c_1=c_2$ in $\pi_1(\Gamma)$ then $c_1=c_2$.

Sketch of proof. The result is clear if $\Gamma$ is a bouquet of circles, using the usual normal form for free groups. In fact, it is the only case to consider, since the quotient of $\Gamma$ by a maximal subtree is a bouquet of circles. $\square$

Proof of claim 2. Because $\alpha \circ p = \gamma = \beta \circ q$, the inclusion

$\gamma_* \pi_1( X \otimes_Z Y) \subset \alpha_* \pi_1(X) \cap \beta_* \pi_1(Y)$

is clear. Conversely, let $c_0 \in \alpha_* \pi_1(X) \cap \beta_* \pi_1(Y)$, that is there exist $c_X \in \pi_1(X)$ and $c_Y \in \pi_1(Y)$ such that $\alpha(c_X)=c_0=\beta(c_Y)$ in $\pi_1(Z)$. Of course, $c_X$ and $c_Y$ may be chosen reduced so that $\alpha(c_X)$ and $\beta(c_Y)$ so are. We deduce that $\alpha(c_X)=\beta(c_Y)$ from our previous lemma. Therefore, by construction of $X \otimes_Z Y$, there exist a loop $c \in \pi_1(X \otimes_Z Y)$ such that $\gamma(c)=c_0$, hence the inclusion

$\alpha_* \pi_1(X) \cap \beta_* \pi_1(Y) \subset \gamma_* \pi_1 (X \otimes_Z Y)$. $\square$

Proof of Howson’s theorem. As in the previous note, we see $F$ as the fundamental group of a finite bouquet of circles $B$, and we choose immersions of finite graphs $\alpha : X \to B$ and $\beta : Y \to B$ such that $\alpha_* \pi_1(X)=H$ and $\beta_* \pi_1(Y)= K$. Then the fiber product $\gamma : X \otimes_B Y \to B$ gives an immersion such that $\gamma_* \pi_1(X \otimes_B Y) = H \cap K$; because $X \otimes_B Y$ is a finite graph (since $X$ and $Y$ are itself finite), we deduce that $H \cap K$ is finitely-generated. $\square$

## Surfaces and commutators in free groups

Let $\mathbb{F}_n$ be a free group of rank $n \geq 2$ and $w \in \mathbb{F}_n$, viewed as a word on $n$ letters. Then $w$ is a product of commutators if and only if it belongs to the kernel of $\mathbb{F}_n \twoheadrightarrow \mathbb{F}_n^{\text{ab}} \simeq \mathbb{Z}^n$, that is if the sum of exponents on each letter is zero. From now on, we suppose that $w$ satisfies this property.

Question: How to find the least integer $k$ such that $w$ can be written as a product of $k$ commutators?

In their article Applications of topological graph theory to group theory, Gorenstein and Turner find a simple solution closely related to the theory of closed surfaces. More precisely, we first construct a polygon whose edges are labelled and oriented by the letters of $w$; then we we choose a pairing of edges, encoded in a circle graph $\Gamma$ (topologically, a circle with a collection of interior arcs whose endpoints are pairwise distinct), and we construct a closed surface $\Sigma_{\Gamma}$ by pairwise indentifying the edges of our polygon following the circle graph (see the figure below). Of course, the surface depends on the circle graph we chose; in particular, we define the genus $\gamma(w)$ of $w$ as the least genus of a surface obtained from $w$ in the previous way.

For example, $\gamma(a^2b^{-1}a^{-1}ba^{-1})=1$.

We then have the following surprising result:

Theorem 1: $w$ can be written as a product of $\gamma(w)$ commutators, and not fewer.

Moreover, Gorenstein and Turner give in their article a simple way to compute the genus $g(\Sigma_{\Gamma})$ from the circle graph $\Gamma$. First, we number the $n$ interior arcs of $\Gamma$, and we define the $n \times n$-matrix $A(\Gamma)$ by saying that the entry $(i,j)$ is equal to $1$ if the $i$-th and $j$-th interior arcs intersect; otherwise, it is equal to $0$. Notice that $A(\Gamma)$ is symmetric and that any diagonal element is zero, so, working in $\mathbb{Z}_2$, it can be written in some basis as a diagonal block matrix (see for example Kaplansky’s book, Linear algebra and geometry):

$A(\Gamma) = \left( \begin{array}{cccc} J & & & \\ & \ddots & & \\ & & J & \\ & & & 0 \end{array} \right)$ where $J = \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)$.

In particular, notice that $\mathrm{rank}(A(\Gamma))$ is even. Let us define the genus of $A(\Gamma)$ as

$g(A(\Gamma))= \frac{1}{2} \mathrm{rank}_{\mathbb{Z}_2} (A(\Gamma))$.

Finally, we have

Theorem 2: $g(\Sigma_{\Gamma})= \gamma(A(\Gamma))$.

Proof of theorem 1: We prove by induction on $\gamma(w)$ that $w$ can be written as a product of $\gamma(w)$ commutators. If $\gamma(w)=0$, then $w=1$ and there is nothing to prove.

From now on, suppose that our result holds for $\gamma(w) \leq n-1$ and suppose $\gamma(w)=n \geq 1$. Let $\Gamma$ be a circle graph such that $\gamma(w)= g( \Sigma_{\Gamma})$. First, we replace $w$ by a word $w'$ such that each pair of linked letters in $w$ are different in $w'$. For example,

Because $n \geq 1$, there exist two intersecting interior arcs; therefore, up to a cyclic permutation (that does not change a pairing), we may suppose that $w'$ can be written as

$w'= x_1Rx_2Sx_1^{-1} T x_2^{-1} U$.

The pairing $\Gamma$ clearly induces a pairing $\Gamma'$ whose underlying circle graph is the same; in particular, $g(\Sigma_{\Gamma})= g( \Sigma_{\Gamma'})$. Now, we cut and paste $\Sigma_{\Gamma'}$ in the following way:

From a purely algebraic viewpoint, we set $c=Rx_2S$ and we write $w'= x_1cx_1^{-1}TSc^{-1} RU$. Up to a cyclic permutation, we have

$w'= c^{-1}RUx_1cx_1^{-1}TS$.

Then, we set $d= x_1^{-1}U^{-1}R^{-1}$ and we write

$w' = c^{-1}d^{-1}c d URTS= [c^{-1},d^{-1}] \cdot URTS$.

Now, we want to apply our induction hypothesis to $\tilde{w} = URTS$. First, the pairing $\Gamma'$ induces a pairing $\Gamma''$ for $[c^{-1},d^{-1}] \cdot URTS$ just by adding some disjoint interior arcs (because the word is not reduced), but that does not change the surface $\Sigma_{\Gamma'}$ (in the polygon, we just add consecutive identified edges), hence $g(\Sigma_{\Gamma'})=g( \Sigma_{\Gamma''})$. Then, the pairing $\Gamma''$ induces a pairing $\tilde{\Gamma}$ for $URTS$; but the figure below justifies that we have the connected sum $\Sigma_{\Gamma''} \simeq \mathbb{T}^2 \sharp \Sigma_{\tilde{\Gamma}}$.

Therefore,

$\gamma(\tilde{w})= g( \Sigma_{\tilde{\Gamma}}) = g( \Sigma_{\Gamma''})-1 = g( \Sigma_{\Gamma'})-1 = g( \Sigma_{\Gamma})-1=n-1$.

Using our induction hypothesis, $\tilde{w}$ can be written as a product of $n-1$ commutators, so $w$ can be written as a product of $n$ commutators.

Conversely, we prove that if $w$ is a product of $n$ commutators then there exists a pairing $\Gamma$ satisfying $g(\Sigma_{\Gamma})=n$. We consider the case $n=3$, but the argument is completely general. So

$w = [a_1,a_2] \cdot [a_3,a_4] \cdot [a_5,a_6]$.

We use the pairing $\Gamma$:

The associated matrix is $A( \Gamma) = \left( \begin{array}{ccc} J & & 0 \\ & J & \\ 0 & & J \end{array} \right)$ where $J = \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)$, hence

$g(\Sigma_{\Gamma}) = \frac{1}{2} \mathrm{rank}_{\mathbb{Z}_2}(A(\Gamma)) = 3$. $\square$

Corollary: Let $w \in \mathbb{F}_n$ be a reduced word of length $m$ in the commutator subgroup. Then $w$ can be written as a product of at most $m/4$ commutators.

Proof. Let $\Gamma$ be a pairing of minimal genus, so that $w$ can be written as a product of $\frac{1}{2} \mathrm{rank}_{\mathbb{Z}_2}(A(\Gamma))$ commutators. The circle graph $\Gamma$ has $m/2$ interior arcs (of course, $m$ is even) and $A(\Gamma)$ is a $\frac{m}{2} \times \frac{m}{2}$-matrix, hence

$\gamma(w) = \frac{1}{2} \mathrm{rank}_{\mathbb{Z}_2}(A(\Gamma)) \leq \frac{1}{2} \cdot \frac{m}{2}= \frac{m}{4}$. $\square$

Notice that the proof of Theorem 1 is effective, that is it gives an algorithm to express a word as product with a minimal number of commutators. Let us apply this algorithm to the word $w = x^2yx^{-1}y^{-2}x^{-1}y$.

Step 1: Draw all the possible pairings.

Step 2: Compute the ranks of the associated matrices.

We have $\gamma(A_1)=\gamma(A_2)=\gamma(A_3)=2$ and $\gamma(A_4)=1$. So we are interesting in the fourth pairing.

Step 3: Cut and paste the surface of minimal genus.

Up to a cyclic permutation, $w$ is therefore

$cx^{-1}y^{-1}c^{-1}yx= [c, (yx)^{-1}]= [xy,x^{-1}y^{-1}]= xyx^{-1}y^{-2}x^{-1}yx$.

Finally,

$w=x[xy,x^{-1}y^{-1}]x^{-1} = [x^2yx^{-1}, y^{-1}x^{-1}]$.

Notice that this decomposition is not unique, since we also have

$w= [xy,y^{-1}xy^2]$.

Even more surprisingly, Stallings and Gersten noticed that such ideas may be applied in order to prove the fundamental theorem of algebra, following an intuition of Gauss! See their article On Gauss’s first proof of the fundamental theorem of algabra.

## How Stallings proved finitely generated groups are subgroup separable

In this note, we deal with a construction on finite graphs due to Stallings, exposed in his article Topology of finite graphs, admitting some exciting recent generalizations to some graphs of groups (see Henry Wilton’s articles Elementary free groups are subgroup separable and Hall’s theorem for limit groups) or cube complexes (see Haglund and Wise’s article A combination theorem for special cube complexes). A next note will be dedicated to Wise’s construction for Salvetti complexes, and more generally for special cube complexes, but our main result here is:

Main Theorem: Finitely generated groups are subgroup separable.

Definition: Let $G$ be a group and $H$ be a subgroup. We say that $H$ is separable if, for every $g \notin H$, there exists a finite-index subgroup $K$ containing $H$ and such that $g \notin K$.

Moreover, if the trivial subgroup is separable, $G$ is said residually finite or RF; if every subgroups of $G$ are separable, $G$ is said ERF (Extended Residually Finite); if every finitely-generated subgroups of $G$ are separable, $G$ is said subgroupe separable or LERF (Locally Extended Residually Finite).

Theses definitions have topological interpretations:

Definition: Let $G$ be a group. Defining the set of finite-index subgroups as a fundamental system of neighborhoods of the identity element, we get the profinite topology. Clearly, endowed with its profinite topology, $G$ is a topological group. Moreover, a morphism between two groups endowed with their profinite topologies is automatically continuous.

Nota Bene: Because a finite-index subgroup $H$ always contains a normal finite-index subgroup (namely $\bigcap\limits_{g \in G} gHg^{-1}$), finite-index subgroups in the definitions of profinite topology and separable subgroup may be supposed normal.

Lemma: Let $G$ be a group and $H$ be a subgroup. Then $H$ is separable if and only if it is closed for the profinite topology.

Proof. Suppose that $H$ is separable and let $g \notin H$. By hypothesis, there exists a finite-index subgroup $K$ containing $H$ but not containing $g$. Then $gK$ is an open set disjoint from $H$: if there exists $x \in H \cap gK$, then there exists $k \in K$ such that $x=gk$, hence $g=xk^{-1} \in K$, a contradiction. Therefore, $H$ is closed.

Conversely, suppose that $H$ is closed and let $g \notin H$. By hypothesis, there exists an open subset, say $pK$ for some normal finite-index subgroup $G$ and $p \in G$, containing $g$ and disjoint from $H$. Because we chose $K$ normal, $HK$ is a subgroup of $G$; clearly, it contains $H$ and it is a finite-index subgroup of $G$. Furthermore, it does not contain $g$: indeed, $g \in HK$ implies $pK \cap HK \neq \emptyset$, hence $pK \cap H \neq \emptyset$, a contradiction. Therefore, $H$ is separable. $\square$

Corollary: A group is residually finite if and only if its profinite topology is Hausdorff.

Genarally, it is not easy to prove that a given subgroup is separable. A useful tool for that is:

Definition: Let $G$ be a group and $H$ be a subgroup. We say that $H$ is a retract if there exists a morphism (a retraction) $r : G \to H$ such that $r_{|H}= \mathrm{Id}_H$. We say that $H$ is a virtual retract if it is a retract in some finite-index subgroup of $G$.

Lemma: Let $G$ be a residually finite group and $H$ be a subgroup. If $H$ is a virtual retract then it is separable.

Sketch of proof. It is sufficient to notice that the image of a (topological) retraction in a Hausdorff topological space is closed. So let $X$ be a Hausdorff space and $r : X \to Y \subset X$ be a retraction. Here, we only work with finitely-generated groups so that profinite topologies are second-countable, therefore it is sufficient to show that $Y$ is sequentially closed; for the more general case, the same argument works with ultrafilters.

Let $(x_n) \subset Y$ be a sequence converging to some $\in X$. Then, $(r(x_n))=(x_n)_n$ converges to $r(x)$. Because $X$ is Hausdorff, we deduce that $x=r(x) \in Y$. Therefore, $Y$ is sequentially closed. $\square$

The construction introduced by Stallings, is the following (recall that an immersion is a locally injective simplicial map):

Theorem: Let $B$ be a finite bouquet of circles, $X$ be a finite graph and $X \to B$ be an immersion. Just by adding edges to $X$, it is possible to construct a covering $C(X,B) \to B$ and a retraction $r : C(X,B) \to X$. Moreover, we get a commutative diagram:

We say that $C(X,B)$ is the canonical completion and $r$ the canonical retraction.

Now, let us see  how to conclude thanks to Stallings’ construction:

Proof of the main theorem: Let $F$ be a finitely generated free groups. We first prove that $F$ is residually finite in order to apply the previous lemma.

Let $B$ be a bouquet of circles satisfying $\pi_1(B) \simeq F$; let $\widehat{B}$ denote its universal covering. Now, thinking of $g \in F \backslash \{1 \}$ as a non trivial loop in $B$, let $X \subset \widehat{B}$ be a finite subgraph containing a lift of $g$. Then, the fundamental group of the canonical completion $C(X,B)$ is a finite-index subgroup of $\pi_1(B) \simeq F$ not containing $g$. We just proved that $F$ is residually finite.

Now let $H$ be a finitely generated subgroup of $F$. Because $F$ is residually finite, it is sufficient to prove that $H$ is a virtual retract to conclude that $H$ is separable.

Again, let us see $F$ as the fundamental group of a bouquet of circles $B$ and let $Y \to B$ denote the covering associated to the subgroup $H$. If $T \subset Y$ is a maximal subtree, because $H$ is finitely generated, $Y \backslash T$ has only finitely many edges; let $X \subset Y$ be a ball containing all theses edges. Then the canonical retraction $r: C(X,B) \to X$ gives a retraction

$r_* : K:= \pi_1(C(X,B)) \to H \simeq \pi_1(X)$

so that $H$ be a retract in $K$, and therefore a virtual rectract in $F$. $\square$

During the proof above, the construction of an immersion $X \to B$ with $\pi_1(X) \simeq H$ can be made directly. For example, if $F = \langle a,b \mid \ \rangle$ and $H = \langle ab^{-2}a^{-1}, abab^2 \rangle$:

Notice that the main theorem cannot be extended to the separability of all subgroups, ie. every non-abelian (finitely-generated) free group is not ERF: Suppose by contradiction that there exists an ERF non-abelian free group. Because any quotient of an ERF group is RF, we deduce that any two-generator group is RF, and a fortiori Hopfian. However, we already proved in a previous note that the Baumslag-Solitar group $BS(1,2)$ is not Hopfian.

We now turn on the proof of Stallings’ construction. The figure below illustrates an example; the reader is encouraged to keep in mind this particular case in order to visualize how the argument is simple.

Proof of Stallings’ construction: Because $X \to B$ is an immersion, for any vertex $v \in V$ and any label $x_i$, there is at most one edge labelled $x_i$ starting from $v$ or ending at $v$. We want to add edges to $X$ so that exactly one edge of any label start from and end at any vertex; such a graph will be a covering of $B$.

Step 1: Let $e \in X$ be an edge labelled $x_i$ and let $p = (e_0, \dots, e_r)$ denote the path of oriented edges labelled $x_i$ containing $e$ of maximal length (such a path is uniquely defined). Two cases happens: either $p$ is a loop, and there is nothing to do, or $p$ is not a loop, and we add an edge labelled $x_i$ between the extremities of $p$.

Doing the same thing for every edges of $X$ and for every label, we get a graph $X'$ satisfying: for any vertex $v \in X'$ and any label $x_i$, either there is no edge labelled $x_i$ adjacent to $x_i$, or there are two such edges, one starting from $v$ and the other ending at $v$.

Step 2: Let $v \in X$ be a vertex and $x_i$ a label. If there is no edge labelled $x_i$ adjacent to $v$, then add a loop labelled $x_i$ based at $v$. Keep going for every every vertex and for every label.

Let $C(X,B)$ denote our new graph. It is a covering $C(X,B) \to B$ extending by construction the immersion $X \to B$, so we have the commutative diagram.

To conclude the proof, let us define a retraction $r : C(X,B) \to X$. Of course, we only have to define $r(e)$, where $e$ is an edge added to $X$. Two cases happen: either $e$ is a loop based at a vertex $v \in X$, and we define $r(e)=v$, or $e$ completes a path $p\subset X$ to a loop, and we send $e$ on $p$ (first choose a subdivision of $e$) so that $r(e)=p$. $\square$

Nota Bene: In general, the retraction defined above is not simplicial.

## Euler product formula from probability

From the discussion A simple way to obtain $\prod_{p \in \mathbb{P}} \frac{1}{1-p^{-s}} = \sum_{n =1}^{\infty} \frac{1}{n^s}$ on math.stackexchange, I learnt an elementary proof of Euler product formula; in fact, I found it so remarkable that I decided to write a post on it.

We only need the two following basic lemmas; only sketchs of proof are given, we refer to Jean Jacod’s book, Probability essentials, for more information.

Definition: $\{ E_i \mid i \in I \}$ is a family of independent events if for every finite subset $J \subset I$,

$\displaystyle P \left( \bigcap\limits_{j \in J} E_j \right)= \prod\limits_{j \in J} P(E_j)$.

Lemma: Let $P$ be a probability measure and $\{ E_i \mid i \geq 1 \}$ be a family of independent events. Then

$\displaystyle P \left( \bigcap\limits_{i \geq 1} E_i \right) = \prod\limits_{i \geq 1} P(E_i)$.

Sketch of proof. $\left( \bigcap\limits_{i=1}^n E_i \right)$ is a decreasing sequence of events, hence

$\displaystyle P \left( \bigcap\limits_{i \geq 1} E_i \right) = \lim\limits_{n \to + \infty} P \left( \bigcap\limits_{i=1}^n E_i \right)= \lim\limits_{n \to + \infty} \prod\limits_{i=1}^n P(E_i) = \prod\limits_{i \geq 1} P(E_i)$. $\square$

Lemma: If $\{ E_i \mid i \in I \}$ is a familiy of independent events, then $\{ E_i^c \mid i \in I \}$ so is.

Sketch of proof. Using inclusion-exclusion principle, prove by induction on $n$ that every subfamily $\{E_{i_1}, \dots E_{i_n} \}$ of cardinality $n$ satisfies

$P \left( \bigcap\limits_{k=1}^n E_{i_k}^c \right)= \prod\limits_{k=1}^n P \left( E_{i_k}^c \right)$.

For $n=2$, we have

$\begin{array}{lcl} P(E_1^c \cap E_2^c ) & = & 1- P(E_1 \cup E_2)= 1- P(E_1) - P(E_2 ) + P(E_1 \cap E_2) \\ \\ & = & 1- P(E_1) - P(E_2) + P(E_1) P(E_2) = ( 1 - P(E_1)) (1- P(E_2)) \\ \\ & = & P(E_1^c) P(E_2^c) \hspace{1cm} \square \end{array}$

Theorem: Let $s > 1$. Then $\displaystyle \zeta(s):= \sum\limits_{n \geq 1} \frac{1}{n^s} = \prod\limits_{p \in \mathbb{P}} \frac{1}{1-p^{-s}}$.

Proof. Let $X : \mathbb{N} \to \mathbb{N}$ be a random variable and $P$ be a probability measure such that

$\displaystyle P(X=n)= \frac{\zeta(s)^{-1}}{n^s}$;

for example, take $X= \mathrm{Id}$ and $P ( \{n \})= \frac{\zeta(s)^{-1}}{n^s}$. Let $E_k$ be the event “$X$ is divisible by $k$“. Notice that

$\displaystyle P(E_k)= \sum\limits_{i \geq 1} P(X=ik) = \sum\limits_{i \geq 1} \frac{\zeta(s)^{-1}}{n^s} = k^{-s}$.

Therefore, for any primes $k_1 , \dots, k_r$,

$\displaystyle P \left( \bigcap\limits_{j=1}^r E_{k_j} \right)= P \left( E_{\prod\limits_{j=1}^r k_j} \right) = \prod\limits_{j=1}^r k_j^{-s} = \prod\limits_{j=1}^r P(E_{k_j})$.

We deduce that $\{ E_k \mid k \ \text{prime} \}$ is a family of independent events. Because $X=1$ if and only if $X$ is not divisible by any prime, we have

$\displaystyle \frac{1}{\zeta(s)} = P(X=1)= P \left( \bigcap\limits_{p \in \mathbb{P}} E_p^c \right) =\prod\limits_{p \in \mathbb{P}} \left( 1- P(E_p) \right) = \prod\limits_{p \in \mathbb{P}} \left( 1-p^{-s} \right)$. $\square$