## Positive curvature versus negative curvature

A leitmotiv in geometric group theory is to make a group $G$ act on a geometric space $X$ in order to link algebraic properties of $G$ with geometric properties of $X$. Here, the expression geometric space is intentionally vague: it goes from rather combinatorial objects, like simplicial trees, to rather geometric objects, like Riemannian manifolds. However, it is worth noticing that the link between $G$ and $X$ turns out to be stronger when $X$ is non-positively curved in some sense. A possible definition of geodesic space of curvature bounded above is given by $CAT(\kappa)$ inequality:

Definition: Let $X$ be a geodesic space and $\kappa \in \mathbb{R}$. For convenience, let $M_{\kappa}$ be the only (up to isometry) simply connected Riemannian surface of constant curvature $\kappa$ and let $D_{\kappa}$ denote its diameter; therefore, $M_{\kappa}$ is just the hyperbolic plane $M_{-1}$, the euclidean plane $M_0$ or the sphere $M_1$ with a rescaling metric and $D_{\kappa}$ is finite only if $\kappa >0$.

For any geodesic triangle $\Delta=\Delta(x,y,z)$ – that is a union of three geodesics $[x,y]$, $[y,z]$ and $[x,z]$ – of diameter at most $2 D_{\kappa}$, there exists a comparison triangle $\overline{\Delta}= \overline{\Delta}(\overline{x},\overline{y},\overline{z})$ in $M_{\kappa}$ (unique up to isometry) such that

$d(\overline{x}, \overline{y})= d(x,y)$,  $d(\overline{y}, \overline{z})= d(y,z)$  and  $d(\overline{x}, \overline{z})=d(x,z)$.

We say that $\Delta$ satisfies $CAT(\kappa)$ inequality if for every $a,b \in \Delta$

$d(a,b) \leq d( \overline{a}, \overline{b})$.

We say that $X$ is a $CAT(\kappa)$ space if every geodesic triangle of diameter at most $2 D_{\kappa}$ satisfies $CAT(\kappa)$ inequality, and that $X$ is of curvature at most $\kappa$ if it is locally $CAT(\kappa)$.

This definition of curvature bounded above is good enough to agree with sectional curvature of Riemannian manifolds: A Riemannian manifold is of curvature at most $\kappa$ if and only if its sectional curvature is bounded above by $\kappa$. For more information, see Bridson and Haefliger’s book, Metric spaces of nonpositive curvature, theorem I.1.A6.

From now on, let us consider a nice kind of action on geodesic spaces:

Definition: A group acts geometrically on a metric space whenever the action is properly discontinuous and cocompact.

Such actions are fundamental in geometric group theory, notably because of Milnor-Svarc theorem: if a group $G$ acts geometrically on a metric space $X$, then $G$ is finitely-generated and there exists a quasi-isometry between $G$ and $X$. See [BH, proposition I.8.19].

Let us say that a group is $CAT(\kappa)$ if it acts geometrically on some $CAT(\kappa)$ space. Notice that, if $(X,d)$ is a $CAT(\kappa)$ space, then $(X, \sqrt{ | \kappa | } \cdot d )$ is a $CAT( \kappa / |\kappa|)$ space. Therefore, a $CAT(\kappa)$ group is either $CAT(-1)$ or $CAT(0)$ or $CAT(1)$.

According to the remark made at the beginning of this note, $CAT(-1)$ and $CAT(0)$ properties should give more information on our group than $CAT(1)$ property. In fact, we are able to prove the following result:

Theorem: Any finitely-presented group is $CAT(1)$.

Sketch of proof. Let $G$ be a finitely-presented group. If $X$ is a Cayley complex of $G$, it is know that the natural action $G \curvearrowright X$ is geometric – see Lyndon and Schupp’s book, Combinatorial group theory, section III.4. Now, the barycentric subdivision $X'$ of $X$ is a flag complex of dimension two. According to Berestovskii’s theorem mentionned in [BH, theorem II.5.18], the right-angled spherical complex $Y$ associated to $X'$ is a (complete) $CAT(1)$ space. Of course, the action $G \curvearrowright Y$ is again geometric, since the underlying CW complex is just $X'$. $\square$

Another important way to link a group $G$ with a geometric space is to write $G$ as a fundamental group. As above, we can prove the following result:

Theorem: Any group $G$ is the fundamental group of a $CAT(1)$ space $X$. Moreover, if $G$ is finitely-presented, $X$ can be supposed compact.

Sketch of proof. Let $X$ be the CW complex associated to a presentation of $G$; then $G \simeq \pi_1(X)$ and $X$ is compact if the presentation were finite – see Lyndon and Schupp’s book, Combinatorial group theory, section III.4. As above, the right-angled spherical complex $Y$ associated to the barycentric subdivision $X'$ of $X$ is a $CAT(1)$ space, and its fundamental group is again isomorphic to $G$, since the underlying CW complex is just $X'$. $\square$

On the other hand, many properties are known for $CAT(0)$ groups; the usual reference on the subjet is Bridson and Haefliger’s book, Metric spaces of nonpositive curvature. Furthermore, is not difficult to prove that $CAT(-1)$ groups are Gromov-hyperbolic. However, it is an open question to know whether or not hyperbolic groups are $CAT(-1)$, or even $CAT(0)$.

Let us conclude this note by noticing that, for a fixed $CAT(1)$ space $X$, it is possible that only few groups are able to act on it. For example, we proved in our previous note Brouwer’s Topological Degree (IV): Jordan Curve Theorem that, for any even number $n$, $\{1 \}$ and $\mathbb{Z}_2$ are the only groups acting freely by homeomorphisms on the $n$-dimensional sphere $\mathbb{S}^n$.

## Cancellation property for groups I

A natural question in group theory is to know whether or not the following implication is true:

$G \times H \simeq G \times K \Rightarrow H \simeq K$.

Of course, such a cancellation property does not hold in general, for example

$\mathbb{Z} \times \mathbb{Z} \times ( \mathbb{Z} \times \cdots) \simeq \mathbb{Z} \times \mathbb{Z} \times \cdots \simeq \mathbb{Z} \times ( \mathbb{Z} \times \cdots)$

whereas $\mathbb{Z}$ and $\mathbb{Z} \times \mathbb{Z}$ are not isomorphic. However, cancellation property turns out to hold for some classes of groups. In his article On cancellation in groups, Hirshon proves that finite groups are cancellation groups, that is to say, for every groups $G,H,K$, if $G \times H \simeq G \times K$ with $G$ finite, then $H \simeq K$. Below, we present an elementary proof of a weaker statement due to Vipul Naik:

Theorem: Let $G,H,K$ be three finite groups. If $G \times H \simeq G \times K$ then $H \times K$.

Proof. For any finite groups $L$ and $G$, let $h(L,G)$ denote the number of homomorphisms from $L$ to $G$ and $i(L,G)$ denote the number of monomorphisms from $L$ to $G$. Notice that

$\displaystyle h(L,G)= \sum\limits_{N \lhd L} i(L/N,G) \hspace{1cm} (1)$

Let $G,H,K$ be three finite groups such that $G \times H \simeq G \times K$. Then

$\begin{array}{c} h(L,G \times H)=h(L,G \times K) \\ h(L,G) \cdot h(L,H)=h(L,G) \cdot h(L,K) \\ h(L,H)=h(L,K) \end{array}$

for any finite group $L$, since $h(L,G) \neq 0$. Using $(1)$, it is easy to deduce that $i(L,H)=i(L,K)$ for any finite group $L$ by induction on the cardinality of $L$. Hence

$i(H,K)=i(H,H) \neq 0.$

Therefore, there exists a monomorphism from $H$ to $K$. From $G \times H \simeq G \times K$, we deduce that $H$ and $K$ have the same cardinality, so we conclude that $H$ and $K$ are in fact isomorphic. $\square$

Because a classification of finitely-generated abelian groups or of divisible groups is known, it is easy to verify that cancellation property holds also for such groups (the classification of divisible groups was mentionned in a previous note). However, in his book Infinite abelian groups, Kaplansky mentions that the problem is still open for the class of all abelian groups.

We conclude our note by an example where the cancellation property does not hold even if the groups are all finitely-presented.

Let $\Pi$ be the following presentation

$\langle a,b,c \mid [a,b] = [a,c] = 1, b^{11}, cbc^{-1}=b^4 \rangle$.

Clearly, we have the following decomposition:

$\Pi \simeq \langle a \mid \ \rangle \times \underset{:=H}{ \underbrace{ \langle b,c \mid b^{11}=1, cbc^{-1}=b^4 \rangle }}$.

Now, we set $x=a^2c^5$ and $y= ac^2$. Because $a$ commutes with $b$ and $c$, we deduce that $[x,y]=1$. Then,

$y^2=a^4c^4=xc^{-1} \Rightarrow c=y^{-2}x$

and

$x=a(ac^2)c^3=ayc^3 \Rightarrow a=y^5x^{-2}$.

Now, let us notice that

$c^5bc^{-5} = c^4(cbc^{-1})c^{-4} = c^4b^4c^{-4} = c^3 (cbc^{-1})^4c^{-4}= \cdots = b$,

hence $[x,b] = [a^2c^5,b]=c^5bc^{-5}b^{-1}=1$. Therefore,

$\Pi \simeq \langle b,x,y \mid [x,b]=[x,y]=[y^5,b]=1, b^{11}=1, y^{-2}by^2=b^4 \rangle$.

Using $y^{-2}by^2=b^4$, it is not difficult to notice that the relation $[y^5,b]=1$ can be written as $yby^{-1}=b^5$. Then, we deduce that

$y^2b^4y^{-2}= y(yby^{-1})^4y^{-1} = (yby^{-1})^{-2}=b$

so that the relation $y^{-2}by^2=b^4$ can be removed from the presentation. Finally,

$\Pi \simeq \langle b,x,y \mid [x,y]=[x,b]=1, b^{11}=1, yby^{-1}=b^5 \rangle$.

Therefore, we find another decomposition of $\Pi$ as a direct product:

$\Pi \simeq \langle x \mid \ \rangle \times \underset{:=K}{\underbrace{ \langle b,y \mid b^{11}=1, yby^{-1}=b^5 \rangle }}$.

So $\mathbb{Z} \times H \simeq \Pi \simeq \mathbb{Z} \times K$. To conclude, it is sufficient to prove that $H$ and $K$ are not isomorphic. Notice that they are both a semi-direct product $\mathbb{Z}_{11} \rtimes \mathbb{Z}$.

Let $\varphi : H \to K$ be a morphism. It is not difficult to show that a torsion element of $K$ has to be conjugated to $b$, so, without loss of generality, we may suppose that $\varphi(b)= b^m$ for some $0 \leq m \leq 10$. Then, $\varphi(c)= b^ry^s$ for some $r,s \in \mathbb{Z}$. Let $\pi : K \to \mathbb{Z}$ be the canonical projection sending $b$ to $0$ and $y$ to $1$. If $\varphi$ is onto, then so is $\pi \circ \varphi$: because $\pi \circ \varphi(b)=0$ and $\pi \circ \varphi(c)=s$, we deduce that $s= \pm 1$. On the other hand,

$b^{4m} = \varphi (b^4) = \varphi (cbc^{-1} ) = b^r y^{\pm 1} b^m y^{ \mp 1} b^{-r} = y^{ \pm 1} b^m y^{\mp 1}$,

hence $b^{4m}= b^{5m}$ or $b^{20m}=b^m$. Therefore, $11$ has to divide $m$ and we deduce that $\varphi(b)=b^m=1$. We conclude that $\varphi$ cannot be an isomorphism.

Finally, we proved that

$\mathbb{Z} \times ( \mathbb{Z}_{11} \rtimes_{a} \mathbb{Z}) \simeq \mathbb{Z} \times ( \mathbb{Z}_{11} \rtimes_b \mathbb{Z})$,

where $a : n \mapsto 4n$ and $b : n \mapsto 5n$, but $\mathbb{Z}_{11} \rtimes_a \mathbb{Z}$ and $\mathbb{Z}_{11} \rtimes_b \mathbb{Z}$ are not isomorphic.

In particular, we deduce that:

Corollary: $\mathbb{Z}$ is not a cancellation group.

## Isomorphic groups from linear algebra

Although algebraic structures (such that groups, rings, fields, etc.) are generally difficult to classify, surprisingly linear algebra tells us that vector spaces (over a fixed field) are classified up to isomorphism by only one number: the dimension! But a vector field gives rise to a group and a vector space isomorphism gives rise to a group isomorphism. Therefore, it is not surprising that linear algebra can sometimes be used in order to verify that some groups are isomorphic. For instance,

Theorem: The additive groups $\mathbb{R}^n$, $\mathbb{C}^n$, $\mathbb{R}[X]$ and $\mathbb{C}[X]$ are all isomorphic.

Indeed, they are all $\mathbb{Q}$-vector spaces of dimension $2^{\aleph_0}$. Another less known example is:

Theorem: $T= \{ z \in \mathbb{C} \mid |z|=1 \}$ and $\mathbb{C}^{\times}$ are isomorphic.

Proof. Let $B$ be a basis of $\mathbb{R}$ as a $\mathbb{Q}$-vector space; without loss of generality, suppose $1 \in B$. We can write $B$ as a disjoint union $B_1 \coprod B_2$ where $|B_1|=|B_2|=|B|$ and $1 \in B_1$. Let $R_1$ and $R_2$ denote the subspaces generated by $B_1$ and $B_2$ respectively; as additive groups, they are both isomorphic to $\mathbb{R}$ since they are $\mathbb{Q}$-vector spaces of dimension $2^{\aleph_0}$. Finally, let $Z$ denote the subspace generated by $1$. Then

$\mathbb{R}/ \mathbb{Z} \simeq (R_1 \oplus R_2) / Z \simeq R_1/ Z \oplus R_2 \simeq \mathbb{R}/ \mathbb{Z} \oplus \mathbb{R}$.

Now, notice that the morphism

$\theta \in \mathbb{R}/ \mathbb{Z} \mapsto e^{i \theta}$

gives an isomorphism $( \mathbb{R} / \mathbb{Z} , +) \simeq ( T, \times )$, and the morphism

$x \in \mathbb{R} \mapsto e^x$

gives an isomorphism $(\mathbb{R},+) \simeq (\mathbb{R}_+, \times)$. Therefore, we deduce that

$T \simeq \mathbb{R} / \mathbb{Z} \simeq \mathbb{R}/ \mathbb{Z} \oplus \mathbb{R} \simeq T \oplus \mathbb{R}_+$.

Noticing that the polar decomposition

$(\theta,r) \in T \oplus \mathbb{R}_+ \mapsto r \cdot e^{i \theta}$

gives an isomorphism $T \oplus \mathbb{R}_+ \simeq \mathbb{C}^+$, we conclude the proof. $\square$

Nevertheless, the converse of our criterium does not hold, that is there exist two non-isomorphic vector spaces such that the associated groups are isomorphic: $\mathbb{R}$ and $\mathbb{R}^2$ give such an example as $\mathbb{R}$-vector spaces. In particular, it is worth noticing that the choice of the field is crucial.

Nota Bene: In this note, we used the axiom of choice dramatically, assuming that every vector space has a basis. In fact, the axiom of choice turns out to be necessary at least to prove the isomorphism $\mathbb{C} \simeq \mathbb{R}$, as noticed by C.J. Ash in his article A consequence of the axiom of choice, J. Austral. Math. Soc. 19 (series A) 1975 (pp. 306-308). More precisely, the author shows that the isomorphism $\mathbb{C} \simeq \mathbb{R}$ implies the existence of a set of reals not Lebesgue measurable, concluding thanks to a model for ZF due to Solovay in which every subset of reals is Lebesgue measurable.

## Representation theory of finite groups and commutativity degree

If $G$ is a finite group, we define its commutativity degree $P(G)$ as the probability that two elements of $G$ commute, that is

$\displaystyle P(G)= \frac{1}{| G \times G|} \cdot | \{ (g,h) \in G \times G \mid [g,h]=1 \} |$.

It is surprising that the value of $P(G)$ may give strong information about $G$. In particular, we prove here the two following results:

Theorem 1: Let $G$ be a finite group. If $P(G) > 5/8$ then $G$ is abelian.

Theorem 2: Let $G$ be a finite group. If $P(G)>1/2$ then $G$ is nilpotent.

In fact, the two theorems above will follow from the upper degree equation bound:

Theorem 3: Let $G$ be a finite group and let $G'$ denote its commutator subgroup. Then

$\displaystyle P(G) \leq \frac{1}{4} \left( 1+ \frac{3}{|G'|} \right)$.

Indeed, if $G$ is non-abelian then $|G'| \geq 2$, and the upper degree equation bound implies $P(G) \leq 5/8$; therefore, theorem 1 is proved. Then, if $P(G) >1/2$ then the upper degree equation bound implies $|G'| < 3$. If $|G'|=1$, $G$ is abelian and there is nothing to prove. Otherwise, $|G'|=2$ ie. $G' = \{1,c \}$ for some $c \neq 1$. For all $g \in G$, $[c,g] \in G'$, so either $[c,g]=1$ or $[c,g]=c$ which implies $c=1$; therefore, $c \in Z(G)$. We proved that $G' \subset Z(G)$, so $[G,G']= \{1\}$; in particular $G$ is nilpotent.

In order to prove Theorem 3, we first notice the following link between commutativity degree and the number of conjugacy classes of $G$:

Lemma 1: Let $G$ be a finite group. Then $\displaystyle P(G)= \frac{k(G)}{|G|}$ where $k(G)$ denotes the number of conjugacy classes of $G$.

Proof. If $x \in G$, let $C(x)$ denote the centralizer of $x$ and $C_1,\dots, C_{k(G)}$ denote the conjugacy classes of $G$; for convenience, let $x_i$ be an element of $C_i$. Then

Finally, the lemma follows. $\square$

Therefore, computing $P(G)$ is equivalent to computing $|G|$ and $k(G)$. But we know that the number of conjugacy classes is quite related to representation theory, and indeed we will prove Theorem 3 thanks to representation theory.

Definitions: Let $G$ be a finite group. A (complex) representation is a morphism $G \to \mathrm{Aut}(V)$ for some complex vector space $V$ (of finite dimension); the degree of the representation is the dimension of $V$. By extension, we often say that $V$ is a representation.

Any finite group admits a representation. For example, if $\mathbb{C}[G]$ denotes the complex vector space with $G$ as a basis, the left multiplication extends to a representation, namely the regular representation of $G$.

If $V$ is a representation, a subrepresentation is a subspace $W$ stable under the action of $G$. A representation is irreducible if it does not contain a nontrivial subrepresentation (ie. different from itself and $\{0\}$).

If $V,W$ are two representations, we define the direct sum $V \oplus W$ as the representation defined by the action $g \cdot (v \oplus w) = (g \cdot v) \oplus (g \cdot w)$ for all $g \in G$, $v \in V$ and $w \in W$.

Two representations $V$ and $W$ are isomorphic if there exists a $G$-invariant isomorphism $V \to W$.

Our first result is:

Property 1: (Maschke) Any representation is a direct sum of irreducible representations.

Proof. Let $G$ be a finite group and $V$ be a representation. If $V$ is irreducible, there is nothing to prove, so we suppose that $V$ contains a nontrivial representation $W$. Then, if $( \cdot , \cdot)$ is any scalar product on $V$, it is possible to define a scalar product $\langle \cdot, \cdot \rangle$ invariant under $G$, for example by

$\displaystyle \langle u,v \rangle = \sum\limits_{g \in G} ( g \cdot u, g \cdot v)$.

Therefore, because $W$ is stable under $G$, the orthogonal $W^{\perp}$ for $\langle \cdot , \cdot \rangle$ so is, and $V = W \oplus W^{\perp}$. Finally, Property 1 follows by induction on the dimension of $V$. $\square$

Of course, in the decomposition given by Property 1, the same representation may appear several times (up to isomorphism). Thanks to a theorem due to Frobenius, this number may be characterized.

Definitions: Let $G$ be a finite group. Let $R(G)$ denote the set of central functions $G \to \mathbb{C}$, that is the functions $f : G \to \mathbb{C}$ satisfying $f(ghg^{-1})=f(h)$ for all $g,h \in G$. We endow $R(G)$ with the scalar product

$\displaystyle \langle f , g \rangle = \frac{1}{ |G| } \sum\limits_{x \in G } \overline{ f(x) } g(x)$.

Clearly, the dimension of $R(G)$ is $k(G)$, the number of conjugacy classes of $G$ (a basis of $R(G)$ is given by the set of functions taking the value $1$ on a fixed conjugacy class and $0$ otherwise).

Associated to a representation $\rho : G \to V$, we define a character $\chi_V : G \to \mathbb{C}$ defined by $\chi_V(g)= \mathrm{tr}( \rho(g))$. A character is said irreducible if the associated representation so is.

Property 2: (Frobenius) Let $G$ be a finite group. The set of irreducible characters defines an orthonormal basis of $R(G)$.

Proof. We first prove that the family of irreducible characters is orthonormal. Of course, two isomorphic representations $\rho_i : G \to \mathrm{Aut}(V_i)$ are conjugated, so their characters are equal; therefore, it is sufficient to prove that two irreducible representations $V,W$ satisfy $\langle \chi_V, \chi_W \rangle =1$ if $V$ and $W$ are isomorphic and $\langle \chi_V, \chi_W \rangle=0$ otherwise.

We define the representation $\mathrm{Hom}(V,W)$ as the representation defined by the action $g \cdot f(v) = g \cdot f(g^{-1} \cdot v)$ for every $f \in \mathrm{Hom}(V,W)$, $g \in G$ and $v\in V$. Now, let $\mathrm{Hom}_G(V,W)$ denote the set of fixed points of the action of $G$ on $\mathrm{Hom}(V,W)$.

Claim 1: (Schur) $\mathrm{dim} ~ \mathrm{Hom}_G(V,W)=1$ if $V$ and $W$ are isomorphic, and $0$ otherwise.

A nontrivial element of $\mathrm{Hom}_G(V,W)$ defines a $G$-invariant isomorphism $V \to W$, hence $\mathrm{dim} ~ \mathrm{Hom}_G(V,W)=0$ if $V$ and $W$ are not isomorphic. From now on, suppose that $V$ and $W$ are isomorphic, and let $j : V \to W$ be a $G$-invariant isomorphism.

Let $f \in \mathrm{Hom}_G(V,W)$. Then $j^{-1} \circ f \in \mathrm{Hom}_G(V,V)$. Let $\lambda \in \mathbb{C}$ be an eigenvalue of $j^{-1} \circ f$, so that $g := j^{-1} \circ f - \lambda \mathrm{Id}_V$ is not invertible. Notice that $\mathrm{ker}(g)$ is a subrepresentation of $V$, so $\mathrm{ker}(g)=V$ because $V$ is irreducible and $g$ is not invertible. We deduce that $g \equiv 0$ that is $f = \lambda \cdot j$.

We just proved that $\mathrm{Hom}_G(V,W)$ is one-dimensional. $\square$

Therefore, it is sufficient to prove that $\langle \chi_V, \chi_W \rangle =\mathrm{dim} ~ \mathrm{Hom}_G(V,W)$. We left as an exercise that $\chi_{\mathrm{Hom}(V,W)} = \overline{\chi_V} \cdot \chi_W$. In particular, we deduce that $\langle \chi_V, \chi_W \rangle = \frac{1}{|G|} \sum\limits_{g \in G } \chi_{\mathrm{Hom}(V,W)} (g)$. Now,

Claim 2: For any representation $\rho : G \to \mathrm{Aut}(V)$, $\mathrm{dim} ~ V^G = \frac{1}{|G|} \sum\limits_{g \in G} \chi_V(g)$ where $V^G$ denotes the set of fixed points.

Let $f = \frac{1}{ |G| } \sum\limits_{g \in G} \rho(g) \in \mathrm{Aut}(V)$. It is easy to show that $\rho(h) \circ f= f$ for all $h \in G$, so that we deduce that $f \circ f = f$. Therefore, $V = \mathrm{ker}(f) \oplus f(V)$. Now we notice that $f(V)=V^G$. Indeed, if $u=f(v) \in f(V)$ then $g \cdot u = \rho(g) \circ f(v) = f(v)=u$ for all $g \in G$, hence $u \in V^G$; conversely, if $u \in V^G$, $u = \frac{1}{|G|} \sum\limits_{g \in G} \rho(g)(u) =f(u)$ hence $u \in f(V)$. We conclude that

$\displaystyle \mathrm{dim} ~ V^G = \mathrm{dim} ~ f(V) = \mathrm{tr}(f) = \frac{1}{|G|} \sum\limits_{g \in G} \chi_V(g)$. $\square$

Finally, from Claim 1 and Claim 2, we deduce that the family of irreducible characters is orthonormal. Now, if $E$ denotes the subspace of $R(G)$ spanning by the irreducible characters, we want to prove that $E= R(G)$ or equivalently that $E^{\perp}= \{ 0 \}$.

So let $\varphi \in E^{\perp}$. We want to prove that, for any representation $\rho_V : G \to \mathrm{Aut}(V)$, the function $f = \sum\limits_{g \in G} \overline{ \varphi(g) } \rho_V(g)$ is zero. According to Property 1, $V$ is a direct sum of irreducible representations $V_1 \oplus \cdots \oplus V_n$; in particular, $\rho_V= \rho_{V_1} \oplus \cdots \oplus \rho_{V_n}$. Therefore, we may suppose without loss of generality that $V$ is irreducible.

Now, it is easy to show that $\rho(g) \circ f \circ \rho(g)^{-1} =f$ for every $g \in G$, so that $f \in \mathrm{Hom}_G(V,V)$. But $\mathrm{Id}_V \in \mathrm{Hom}_G(V,V)$ and $\mathrm{dim} ~ \mathrm{Hom}_G(V,V)=1$ according to Claim 1, so $f = \lambda \cdot \mathrm{Id}_V$ for some $\lambda \in \mathbb{C}$. Now,

$\lambda \cdot \dim (V)= \mathrm{tr}(\lambda \cdot \mathrm{Id}_V) = \mathrm{tr}(f)= \sum\limits_{g \in G} \overline{ \varphi(g) } \chi_V(g)= |G| \cdot \langle \varphi, \chi_V \rangle = 0$,

hence $f=0$. Now, we apply this result for the regular representation $\mathbb{C}[G]$: for every $h\in G$,

$\displaystyle 0= \sum\limits_{g \in G} \overline{ \varphi(g) } \rho_V(g) \cdot h = \sum\limits_{g \in G} \overline{ \varphi(g) } \cdot gh$,

hence $\varphi(g)=0$. We just prove that $\varphi=0$, so $E^{\perp} = \{ 0 \}$. $\square$

We now are able to precise Property 1:

Property 3: Let $G$ be a finite group and let $W_1,\dots, W_n$ denote its irreducible representations up to isomorphism. If $V$ is a representations of $G$, then

$\displaystyle V \simeq \bigoplus\limits_{i=1}^n W_i^{\langle \chi_V, \chi_{W_i} \rangle }$.

Proof. According to Property 1, $V$ may be written as a direct sum of $W_i$, say $W_{i_1} \oplus \cdots \oplus W_{i_r}$. Then $\langle \chi_V, \chi_{W_i} \rangle = \sum\limits_{k=1}^r \langle \chi_{W_k}, \chi_{W_i} \rangle$ because $\chi_V= \chi_{W_{i_1}} + \cdots + \chi_{W_{i_r}}$. But we saw during the proof of Property 2 that $\langle \chi_V, \chi_W \rangle=1$ if $V$ and $W$ are isomorphic and $0$ otherwise. Therefore, $\langle \chi_V, \chi_{W_i} \rangle$ is the number of representations in the decomposition isomorphic to $W_i$. So Property 3 follows. $\square$

Property 3 is our main result from representation theory, and now we are able to prove the two following corollaries:

Corollary 4: Let $G$ be a finite group. The number of conjugacy classes is equal to the number of irreducible representations up to isomorphism.

Proof. We already noticed that isomorphic representations lead to equal characters. Conversely, Property 3 implies that two representations with the same character are isomorphic. Therefore, $k(G)$ is equal to the dimension of $R(G)$, $\mathrm{dim} ~ R(G)$ is equal to the number of irreducible characters according to Property 2, and now the number of irreducible characters is equal to the number of irreducible representations. $\square$

Corollary 5: Let $G$ be a finite group. Then $\displaystyle |G|= \sum\limits_{W \ \mathrm{irr.}} \deg(W)^2$ where the sum is taken over the set of irreducible representations up to isomorphism.

Proof. Let $V= \mathbb{C}[G]$ be the regular representation. Let $V= \bigoplus\limits_{W \ \mathrm{irr.}} W^{ \langle \chi_V, \chi_W \rangle }$ be the decomposition given by Property 3. Then

$\displaystyle |G| = \dim(V)= \chi_V(1) = \sum\limits_{W \ \mathrm{irr.}} \langle \chi_V, \chi_W \rangle \cdot \chi_W(1)$.

Of course, $\chi_W(1)= \dim(W)$ so it is sufficient to prove that $\langle \chi_V, \chi_W \rangle = \dim(W)$. Notice that $\rho_V(g)$ is a permutation matrix, so that $\mathrm{tr}(\rho_V(g))$ is the number of fixed points of the function $h \mapsto gh$ from $G$ to itself; therefore, $\mathrm{tr}(\rho_V(g))=|G|$ if $g=1$ and $0$ otherwise. Hence

$\displaystyle \langle \chi_V, \chi_W \rangle = \frac{1}{ |G| } \sum\limits_{g \in G} \overline{ \chi_V (g) } \cdot \chi_W (g) = \chi_W (1) = \dim(W)$.

Corollary 5 follows. $\square$

Now we are ready to prove Theorem 3:

Proof of Theorem 3. Let $W_1, \dots, W_{k(G)}$ denote the irreducible representations of $G$ up to isomorphism. Representations of degree 1 are only morphisms from $G$ to $\mathbb{C}$, so the number of such representations is equal to the number of morphisms from the abelian group $G/G'$ to $\mathbb{C}$.

But this number turns out to be $[G:G']$ because the number of morphisms from a finite abelian group $A$ to $\mathbb{C}$ is precisely $|A|$; we leave this statement as an exercise (hint: argue by induction on the number of factors in the decomposition of $A$ as a direct sum of cyclic groups).

Therefore, the formula given by Corollary 5 may be written as

$\displaystyle |G| = [G:G'] + \sum\limits_{i=[G:G']+1}^{k(G)} \dim(W_i)^2$,

where $\dim(W_i) \geq 2$. Therefore,

$|G| \geq [G:G'] + 4 ( k(G) - [G:G'] )$

and Theorem 3 follows. $\square$

In fact, the relation

$\displaystyle |G| = [G:G'] + \sum\limits_{i=[G:G'] +1}^{k(G)} \dim(W_i)^2$

proved above may be quite useful to compute $P(G)$ for some small groups. For example, if $Q_8$ denotes the quaternion group, its commutator subgroup is $\{ \pm 1 \}$, so the relation above becomes $8= 1+1+1+1+2^2$; therefore, according to Corollary 4, $Q_8$ has five conjugacy classes, hence $P(Q_8)=5/8$. (Noticing that $Q_8$ is not abelian, we deduce that Theorem 1 is optimal.)

Another example is the symmetric group $S_3$: its commutator subgroup is $\{ \mathrm{Id}, \ (123), \ (132) \}$, so the relation above becomes $6= 1+1+2^2$; therefore, $S_3$ has three conjugacy classes, hence $P(S_3)=1/2$. (Noticing that $S_3$ is not nilpotent, since $[S_3, [S_3,S_3]]= [S_3,S_3]=A_3$, we deduce that Theorem 2 is optimal; however, a lot a nilpotent groups satisfies $P(G) \leq 1/2$ because we saw that a group satisfying $P(G) >1/2$ has nilpotent-length at most two.)

For more information on commutativity degree of finite groups, see Anna Castelaz’s thesis Commutativity degree of finite groups. In particular, an elementary proof of Theorem 1 based on conjugacy class equation can be found there, as well as a complete classification of groups satisfying $P(G) \geq 1/2$ (see Proposition 5.1.4):

Theorem: Let $G$ be a finite group. If $P(G) \geq 1/2$, then either

• $G$ is abelian,
• or $G \simeq P \times A$ for some abelian group $A$ and some $2$-group $P$ (in this case, $|G'|=2$),
• or $G \simeq G_m \times A$ where $A$ is an abelian group, $m \geq 1$ and

$G_m= \langle a,b \mid a^3=b^{2^m}=1, \ bab^{-1}=a^{-1} \rangle$.

## 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. $\square$

In the note Some SQ-universal groups, we prove also that almost all surface groups are SQ-universal. In particular, it implies that they have uncountably many norma subgroups.

## 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 $X \backslash K$ 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 $e(\mathbb{R}^n \backslash S)= |S|+1$ and similarly $e(\mathbb{R}^m \backslash R)= |R|+1$. $\square$

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. $\square$