## Not all finitely-presented groups are fundamental groups of closed 3-manifolds

In our previous note Amalgamated products and HNN extensions (IV): Markov properties, we saw that for every $n \geq 4$ and for every finitely-presented group $G$, there exists a $n$-dimensional closed manifold whose fundamental group is $G$. However, such a result does not hold for $n\leq 3$; for the case $n=2$, see for example the note On subgroups of surface groups. The present note is devoted to the 3-dimensional case $n=3$.

In particular, we will prove that the groups $\mathbb{Z}^n$ for $n \geq 4$, $\mathbb{F}_m \times \mathbb{F}_n$ for $m \geq 3, \ n \geq 2$, and $\mathbb{F}_m \times \mathbb{Z}^n$ for $n ,m \geq 2$ are not 3-manifold groups.

Definitions: A handlebody of genus $g$ is an orientable 3-manifold (with boundary) obtained by attaching $g$ “handles” to a 3-ball. Roughly speaking, it is a full closed surface of genus $g$.

A way to construct 3-manifolds (without boundary) is to glue two handlebodies $H_1$ and $H_2$ of some genus $g$ along their boundaries by a homeomorphism $h : \partial H_1 \to \partial H_2$. We say that the triple $(H_1,H_2,h)$ is a Heegaard splitting of genus $g$ of the resulting manifold.

Theorem 1: Every closed 3-manifold admits a Heegaard splitting.

Sketch of proof. Let $M$ be a closed 3-manifold, which will be thought of as a three-dimensional simplicial complex. Let $M_1$ denote a small neighborhood of the 1-skeleton $\Gamma_1$ of $M$ and $M_2 = M \backslash M_1$ its complement. If $\Delta$ is a 3-simplex, then $M_1 \cap \Delta$ and $M_2 \cap \Delta$ are illustrated by the following picture:

Notice that, if $\Gamma_2 \subset M$ is the graph whose vertices are the centers of the 3-simplices of $M$ and whose edges link two vertices corresponding to adjacent 3-simplices, then $M_2$ may be thought of as a neighborhood of $\Gamma_2$.

For $i=1,2$, let $T_i$ denote a maximal subtree of $\Gamma_i$. Then, $M_i$ may be thought of as a neighborhood of $T_i$, which is homeomorphic to a 3-ball, with one handle for each edge of the complement $\Gamma_i \backslash T_i$. Therefore, $M_i$ is a handlebody of genus $1- \chi(\Gamma_i)$, since, if $v( \cdot)$ and $e(\cdot)$ denotes respectively the number of vertices and edges of a graph,

$\chi( \Gamma_i)= v(\Gamma_i)- e(\Gamma_i)$

and

$1= \chi(T_i)=v(T_i)-e(T_i)=v(\Gamma_i)-e(T_i)$

imply that

$e(\Gamma_i) - e(T_i) = 1- \chi(\Gamma_i)$.

Therefore, we have proved that $M$ is the union of two handlebodies which intersect along their boundaries. To conclude, we need to argue that $M_1$ and $M_2$ have the same genus, or equivalently, that $\chi(\Gamma_1)= \chi (\Gamma_2)$.

Let $v(M), \ e(M), \ f(M), \ c(M)$ denote respectively the number of vertices, edges, faces and 3-cells of $M$. By construction,

$\chi (\Gamma_1)= v( \Gamma_1)- e(\Gamma_1) = v( M) - e(M)$

and

$\chi( \Gamma_2)= v( \Gamma_2) - e( \Gamma_2) = c(M) - f(M)$.

However, by Poincaré’s duality, $\chi(M)=0$, hence

$\chi ( \Gamma_1) - \chi( \Gamma_2)= v(M)-e(M) + f(M)- c(M) = \chi (M)=0$.

The proof is complete. $\square$

From a Heegaard splitting $(H_1,H_2,h)$ of genus $g$, it is possible to find a presentation of the fundamental group of the associated 3-manifold $M$:

We apply van Kampen’s theorem to the decomposition $M = H_1 \cup H_2$. The handlebodies $H_1$ and $H_2$ have a free group of rank $g$ as fundamental groups, and they meet in $M$ along a closed surface $\Sigma$ of genus $g$. Let $\{ c_1, \ldots, c_{2g} \}$ be a generating set of $\pi_1(\Sigma) \leq \pi_1(H_1)$. Then,

$\pi_1(M)= \langle a_1,\ldots, a_g, b_1, \ldots, b_g \mid h_*(c_i)=c_i, \ 1 \leq i \leq 2g \rangle$,

where $\{ a_1, \ldots, a_g \}$ (resp. $\{ b_1, \ldots, b_g \}$) is a generating set of $\pi_1(H_1)$ (resp. $\pi_1(H_2)$).

It is worth noting that such a presentation has the same number of generators and relations; we say that the presentation is balanced. Therefore, we have proved:

Corollary 2: Every 3-manifold group admits a balanced presentation.

So, a natural question is: which groups admit a balanced presentation? In fact, this question is linked to (co)homology of groups, which we define now.

A space $X$ is aspherical whenever $\pi_n(X)= 0$ for all $n \geq 2$; according to Whitehead’s theorem, it amounts to require the universal covering $\tilde{X}$ to be contractible. An interesting property of such spaces is that they are uniquely determined, up to homotopy, by their fundamental groups, that is to say, two aspherical spaces are homotopy equivalent if and only if their fundamental groups are isomorphic. Given a group $G$, a $K(G,1)$ space or a classifying space for $G$ is an aspherical space whose fundamental group is isomorphic to $G$; according to our previous remark, such a space is uniquely determined by $G$, up to homotopy. In particular, this allows us to define the $n$-th (co)homology group of $G$ as the $n$-th (co)homology group of such a classifying space.

In order to justify that (co)homology groups are always defined, we need to prove that every group $G$ has a classifying space. First, to a given presentation $\langle X \mid \mathcal{R} \rangle$ of $G$ we associate a two-dimensional CW-complex $Y$ as follow: Define $Y^{(1)}$ as a bouquet of $| X |$ circles, each labelled by a generator of $X$. Then, for every relation $w \in \mathcal{R}$, glue a $2$-cell along the path labelling by $w$. Using van Kampen’s theorem, it is not difficult to notice that the fundamental group of our complex $Y$ is isomorphic to $G$. However, it may not be aspherical. Now, let us define an increasing sequence of CW-complexes $Y= Y_1 \subset Y_2 \subset \cdots$ such that $\pi_1(Y_n) \simeq G$ and $\pi_k(Y_n)= \{ 1 \}$ for all $k \leq n$ (such a sequence will define a Postnikov tower). The spaces $(Y_{n})$ are defined inductively in the following way: Let $(\varphi_{\alpha} : \mathbb{S}^n \to Y_{n-1})$ be a generating set for $\pi_n(Y_{n-1})$; then, define $Y_n$ by gluing $n$-cells to $Y_{n-1}$ using the $\varphi_{\alpha}$. By construction, $\pi_n(Y_n) = \{1 \}$, and we claim that the embedding $Y_{n-1} \hookrightarrow Y_n$ is $\pi_i$-injective for all $i< n$ by cellular approximation. Finally, let $Y_{\infty}= \bigcup\limits_{n \geq 1} Y_n$. By construction, $Y_{\infty}$ is aspherical, and because $Y_{\infty}$ have been constructed just by gluing $n$-cells to $Y$ with $n \geq 3$, we deduce that $\pi_1(Y_{\infty}) \simeq \pi_1(Y) \simeq G$ (since the fundamental group depends only on the $2$-skeleton of a CW-complex). Therefore, $Y_{\infty}$ is a classifying space of $G$.

For more information on Postnikov towers, see section 4.3 of Hatcher’s book, Algebraic Topology.

Now, we are ready to linked the deficiency of a presentation, that is the number of its generators minus the number of its relations, with the first and second homology groups of the underlying group. The result below is mentioned by Brown in the fifth exercice of chapter II.5 of his book Cohomology of groups. Here is the topological approach he suggests:

Theorem 3: Let $\langle X \mid R \rangle$ be a presentation of a group $G$ with $|X|=n$ and $|R|=m$. Let also $r$ denote $\mathrm{rk}_{\mathbb{Z}} G^{ab}$. Then $\mathrm{rk}_{\mathbb{Z}} H_2(G) \leq r-n+m$.

Proof. Let $Y$ be the CW-complex associated to the given presentation. According to what we have said above, we may add $n$-cells to $Y$, with $n \geq 3$, in order to “kill” higher homotopy groups, ie., our new CW-complex $Y_{\infty}$ is aspherical. Moreover, because we did not modify the $2$-squeleton of $Y$, $\pi_1(Y)$ is isomorphic to $\pi_1( Y_{\infty})$. Therefore, $Y_{\infty}$ is a $K(G,1)$, hence $H_{\ast}(Y_{\infty}) \simeq H_{\ast} (G)$.

Notice that $1-n+m = \chi(Y)= 1- \mathrm{rk}_{\mathbb{Z}} H_1(Y) + \mathrm{rk}_{\mathbb{Z}} H_2(Y)$, hence

$\mathrm{rk}_{\mathbb{Z}} H_2(Y)=r-n+m$.

On the other hand, we clearly have an epimorphism $H_2(Y) \twoheadrightarrow H_2(Y_{\infty})$, therefore:

$\mathrm{rk}_{\mathbb{Z}} H_2(G)= \mathrm{rk}_{\mathbb{Z}} H_2( Y_{\infty}) \leq \mathrm{rk}_{\mathbb{Z}} H_2(Y)=r-n+m.$ $\square$

Corollary 4: If a group $G$ admits a balanced presentation, then $\mathrm{rk}_{\mathbb{Z}} H_2(G) \leq \mathrm{rk}_{\mathbb{Z}} G^{ab}$.

Thus, now we have a necessary criterion to determine whether or not a group admits a balanced presentation. The Proposition below applies this criterion to the right-angled Artin groups (we defined them in our previous note Some SQ-universal groups).

Proposition 5: A right-angled Artin group $A(\Gamma)$ admits a balanced presentation if and only if $\Gamma$ has more vertices than edges.

Sketch of proof. To the canonical presentation of $A(\Gamma)$ is associated a cube complex $Y$, called Salvetti complex: $Y$ has only one point, one edge for each generator of $A(\Gamma)$, and $n+1$ generators span an $n$-cube if and only if they pairwise commute. Now, it can be proved that the Salvetti complex is nonpositively curved so that its universal cover is CAT(0); because a CAT(0) space is contractible, we deduce that $Y$ is a classifying space of $A(\Gamma)$. Moreover, it is not difficult to prove that the $n$-th homology group of $Y$ is free on the set of $n$-cubes. Therefore, $\mathrm{rk}_{\mathbb{Z}} H_1(A(\Gamma))$ corresponds to the number of vertices of $\Gamma$, and $\mathrm{rk}_{\mathbb{Z}} H_2(A(\Gamma))$ to its number of edges.

Thus, if $A(\Gamma)$ admits a balanced presentation, according to Theorem 3, $\Gamma$ must have more vertices than edges. The converse is obvious. $\square$

For more information on CAT(0) geometry, see Bridson and Haefliger’s book, Metric spaces of nonpositive curvature.

Now, we are ready to prove that the groups mentionned at the beginning of this note do not admit a balanced presentation, and thus are not the fundamental groups of closed 3-manifolds.

Corollary 6: Let $n \geq 1$. Then $\mathbb{Z}^n$ admits a balanced presentation if and only if $n \leq 3$.

Proof. $\mathbb{Z}^n$ is the right-angled Artin group associated to the complete graph $K_n$. Because $K_n$ has $n$ vertices and $(n-1)!$ edges, it is sufficient to apply Proposition 5 to conclude. $\square$

Corollary 7: Let $m,n \geq 1$. Then $\mathbb{F}_n \times \mathbb{F}_m$ admits a balanced presentation if and only if $m \leq 2$ and $n=1$ or $m=1$ and $n \leq 2$.

Proof. $\mathbb{F}_n \times \mathbb{F}_m$ is the right-angled Artin group associated to the bipartite complete graph $K_{m,n}$. Because $K_{m,n}$ has $m+n$ vertices and $mn$ edges, it is sufficient to apply Proposition 5 to conclude. $\square$

Corollary 8: Let $m,n \geq 1$. Then $\mathbb{F}_m \times \mathbb{Z}^n$ admits a balanced presentation if and only if $m=1$ and $n \leq 2$ or $n \geq 2$ and $m \geq 2$.

Proof. Let $D_m$ denote the graph with $m$ vertices and no edges. Then $\mathbb{F}_m \times \mathbb{Z}^n$ is the right-angled Artin group associated to the join $K_n \ast D_m$. Because $D_n \ast D_m$ has $n+m$ vertices and $mn+ (n-1)!$ edges (here, we suppose $0!=0$), it is sufficient to apply Proposition 5 to conclude. $\square$

## An elementary application of ping-pong lemma

Let us denote by $\mathrm{SL}(2,\mathbb{Z})$ the set of $(2 \times 2)$-matrices whose determinant is $1$, and by $\mathrm{PSL}(2,\mathbb{Z})$ the quotient $\mathrm{SL}(2,\mathbb{Z}) / \{ \pm \mathrm{Id} \}$. Our aim here is to prove that $\mathrm{PSL}(2,\mathbb{Z})$ is a free product:

Theorem: $\mathrm{PSL}(2,\mathbb{Z}) \simeq \mathbb{Z}_2 \ast \mathbb{Z}_3$.

As noticed in the previous note Some SQ-universal groups, as a corollary we get that $\mathrm{SL}(2,\mathbb{Z})$ is SQ-universal, that is every countable group is embeddable into a quotient of $\mathrm{SL}(2, \mathbb{Z})$.

Usually, the theorem above is proved thanks to a natural action of $\mathrm{PSL}(2,\mathbb{Z})$ on the hyperbolic plane by Möbius transformations. However, as noticed by Roger Alperin in his article published in The American Mathematical Monthly, it is sufficient to make $\mathrm{PSL}(2,\mathbb{Z})$ act on the set of irrational numbers via:

$\displaystyle \left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \cdot r = \frac{a r + b }{ c r + d}$.

Notice that $cr+d \neq 0$ for all $c,d \in \mathbb{Z}$ precisely because $r$ is irrational. The argument below may be thought of as an application of ping-pong lemma (that we already met in the note Free groups acting on the circle in order to find free subgroups).

First, it is a classical exercice to prove that $\mathrm{SL}(2,\mathbb{Z})$ is generated by the two following matrices:

$A= \left( \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right)$  and  $B= \left( \begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix} \right)$.

For convenience, let $C= AB$. Then $B$ and $C$ also generate $\mathrm{SL}(2, \mathbb{Z})$, and consequently their images in $\mathrm{PSL}(2,\mathbb{Z})$ generate it; furthermore, $B$ (resp. $C$) has order two (resp. order three) in $\mathrm{PSL}(2, \mathbb{Z})$. Notice that

$\displaystyle B=B^{-1} : z \mapsto - \frac{1}{z}$,    $\displaystyle C : z \mapsto 1- \frac{1}{z}$    and    $\displaystyle C^{-1} : z \mapsto \frac{1}{1-z}$.

Now, let $\mathcal{P}$ and $\mathcal{N}$ denote the set of positive and negative irrationals respectively. Clearly,

$B( \mathcal{P} ) \subset \mathcal{N}$  and  $C^{ \pm 1} ( \mathcal{N} ) \subset \mathcal{P}$.

To conclude, we want to prove that, for any alternating word $w$ from $\langle B \rangle$ and $\langle C \rangle$, $w \neq 1$ in $\mathrm{PSL}(2,\mathbb{Z})$.

Case 1: $w$ has odd length. Then either $w$ begins and ends with a $B$, hence $w( \mathcal{P} ) \subset \mathcal{N}$, or $w$ begins and ends with a $C^{\pm 1}$, hence $w( \mathcal{N} ) \subset \mathcal{P}$. In particular, we deduce that $w \neq 1$ in $\mathrm{PSL} (2, \mathbb{Z})$.

Case 2: $w$ has even length. Without loss of generality, we may suppose that $w$ begins with a $C^{\pm 1}$; otherwise, just conjugate $w$ by $B$. Then either $w$ begins with a $C$ and ends with a $B$, hence

$w( \mathcal{P} ) \subset C(\mathcal{N}) \subset \{ r \ \text{irrationals} \mid r> 1 \}$,

or $w$ begins with a $C^{-1}$ and ends with a $B$, hence

$w( \mathcal{P} ) \subset C^{-1} ( \mathcal{N} ) \subset \{ r \ \text{irrationals} \mid r< 1 \}$.

In either case, we deduce that $w(1) \neq 1$, so $w \neq 1$ in $\mathrm{PSL}(2, \mathbb{Z})$. The proof is complete.

## Some SQ-universal groups

A group $G$ is said SQ-universal whenever every countable group can be embedded into a quotient of $G$. In some sense, SQ-universality is a “largeness property”. Motivating by this idea, we prove the two following properties:

Property 1: A SQ-universal group contain a non-abelian free group (of countable rank).

Proof. Let $G$ be a SQ-universal group. In particular, there exists a quotient $\overline{G}$ containing a free group of infinite rank $\langle \overline{x_1}, \overline{x_2}, \ldots \rangle$. Let $x_i$ be a lift of $\overline{ x_i }$ in $G$. If there existed a non-trivial relation between the $x_i$‘s then the same relation would hold between the $\overline{x_i}$‘s in $\overline{G}$. Therefore, $\langle x_1,x_2 , \ldots \rangle$ is a free subgroup of infinite rank in $G$. $\square$

Property 2: A countable SQ-universal group contains uncountably many normal subgroups.

Proof. Let $G$ be a countable SQ-universal group. Every quotient of $G$ contains only countably many two-generated subgroups. However, we know that there exist uncountably many two-generated groups up to isomorphism: see for instance the notes Amalgamated products and HNN extensions (I): A theorem of B.H. Neumann or Cantor-Bendixson rank in group theory: A theorem of B.H. Neumann. Therefore, $G$ must have uncountably many quotients, and a fortiori uncountably many normal subgroups. $\square$

Although SQ-universality seems to be a very strong property, many groups turn out to be SQ-universal. We already saw in previous notes that the free group of rank two $\mathbb{F}_2$ is SQ-universal; in fact, we gave two proofs, the one in Amalgamated products and HNN extensions (I): A theorem of B.H. Neumann, and the other, completely elementary, in A free group contains a free group of any rank. Other examples come from the following trivial observation: a group with a SQ-universal quotient is itself SQ-universal. Therefore, any group with a non-abelian free quotient is SQ-universal; in particular, we deduce that any non-abelian free group, of any rank, is SQ-universal. Together with the following result, we will be able to exhib a lot of other examples of SQ-universal groups.

Theorem 3: A group containing a SQ-universal subgroup of finite-index is itself SQ-universal.

For an (almost) elementary proof, see P. Neumann’s article, The SQ-universality of some finitely presented groups, J. Aust. Math. Soc. 16, 1 (1973) .

Example 1: Let $\Sigma$ be a closed surface. Then either $\pi_1(\Sigma)$ is SQ-universal or $\Sigma$ is a sphere $\mathbb{S}^2$, a projective plane $\mathbb{P}^2$, a torus $\mathbb{T}^2$ or a Klein bottle $\mathbb{K}$.

Notice that we already dealt with surface groups in the previous note On subgroups of surface groups. In particular, we know that $\pi_1(\mathbb{S}^2) \simeq \{ 1 \}$, $\pi_1(\mathbb{T}) \simeq \mathbb{Z}^2$, $\pi_1( \mathbb{P}^2) \simeq \mathbb{Z}_2$ are not SQ-universal. We also know that $\pi_1(\mathbb{K})$ contains $\pi_1(\mathbb{T}) \simeq \mathbb{Z}^2$ as a subgroup of finite-index two, so it cannot be SQ-universel.

From now on, let $\Sigma$ be a closed surface different from the surfaces mentioned above. If $\Sigma$ is non-orientable, there exist a finite-cover $\Sigma' \to \Sigma$ where $\Sigma'$ is orientable; in particular, $\pi_1(\Sigma')$ is a subgroup of finite-index in $\pi_1(\Sigma)$, so, according to Theorem 3, we may suppose that $\Sigma$ is orientable, that is

$\pi_1(\Sigma)= \langle a_1, b_1, \ldots, a_g, b_g \mid [a_1,b_1] \cdots [a_g, b_g] = 1 \rangle$

for some $g \geq 2$. Noticing that the following quotient is free,

$\pi_1( \Sigma) / \langle \langle a_1 b_1^{-1}, \ldots, a_g b_g^{-1} \rangle \rangle = \langle a_1, \ldots, a_g \mid \ \rangle \simeq \mathbb{F}_g$,

we deduce that $\pi_1(\Sigma)$ is SQ-universal.

Example 2: A right-angled Artin groups is either free abelian or SQ-universal.

The right-angled Artin group associated to a finite simplicial graph $\Gamma$ is defined by the presentation

$A(\Gamma)= \langle v \in V ( \Gamma ) \mid [v_1,v_2] = 1, \ (v_1,v_2) \in E(\Gamma) \rangle$,

where $V ( \Gamma)$ and $E(\Gamma)$ denote respectively the set of vertices and edges of $\Gamma$. In particular, if $\Gamma$ is a complete graph with $n$ vertices, then $A(\Gamma)$ is the free abelian group $\mathbb{Z}^n$; if $\Gamma$ is a graph with $n$ vertices and no edges, then $A(\Gamma)$ is the free group $\mathbb{F}_n$.

For another example, if $\Gamma$ is the graph given above, then

$A(\Gamma) = \langle a,b,c,d,e \mid [a,b]= [a,c] = [a,d] = [e,b]=[e,c]=[e,d]=1 \rangle$

is isomorphic to $\mathbb{F}_2 \times \mathbb{F}_3$.

We claim that, if $\Gamma$ is not a complete graph, then $A(\Gamma)$ has a non-abelian free quotient, and in particular is a SQ-universal group.

Because $\Gamma$ is not complete, there exist two vertices $v_1,v_2$ not linked by an edge. Let $\Gamma_1$ denote the subgraph induced by $v_1$ and its neighbors, and $\Gamma_2$ the subgraph induced by $V(\Gamma) \backslash \{ v_1 \}$. Then $\Gamma_1$ and $\Gamma_2$ are non-empty, since $v_1 \in \Gamma_1$ and $v_2 \in \Gamma_2$, they cover $\Gamma$, and $\Gamma \backslash (\Gamma_1 \cap \Gamma_2)$ is not connected, because $v_1$ and $v_2$ are separated by $\Gamma_1 \cap \Gamma_2$. Now, quotienting $A(\Gamma)$ by the subgroup

$\langle \langle v \in V(\Gamma_1) \cap V(\Gamma_2), \ uv^{-1} \ (u,v \in V(\Gamma_1)), \ uv^{-1} \ (u,v \in V( \Gamma_2 ) ) \rangle \rangle$,

we get the free group $\langle a , b \mid \ \rangle$. For an explicit example, let us consider the following graph:

So, the associated right-angled Artin group is

$\langle a, b , c, d \mid [a,b] = [a,c] = [b,c] = [b,d] = [c,e] = [d,e] = 1 \rangle$.

Quotienting by the subgroup $\langle \langle b,c, de^{-1} \rangle \rangle$, we get the free group $\langle a,d \mid \ \rangle$.

Example 3: If $A$ and $B$ are two finite groups with $|A| \geq 2$ and $|B| \geq 3$, then the free product $A \ast B$ is a SQ-universal group.

We proved in the previous note A free group contains a free group of any rank that

$\{ [a , b ] \mid a \in A \backslash \{ 1 \}, b \in B \backslash \{ 1 \} \}$

is a free basis of the derived subgroup $D$ of $A \ast B$. Furthermore, the quotient $(A \ast B) / D \simeq A \times B$ is finite, so that $D$ is a non-abelian free subgroup of finite index in $A \ast B$. From Theorem 3, we deduce that $A \ast B$ is SQ-universal.

For example, $SL(2,\mathbb{Z})$ is SQ-universal, because it has the SQ-universal group

$PSL(2,\mathbb{Z}) \simeq \mathbb{Z}_2 \ast \mathbb{Z}_3$

as a quotient. (See An elementary application of ping-pong lemma for an elementary proof of this fact.)

## A theorem of Schur on commutator subgroup

The aim of this note is to prove the following theorem, due to Schur, and to exhibit some corollaries. The proof given here comes from J. Dixon’s book, Problems in Group Theory (problems 5.21 to 5.24).

Theorem: Let $G$ be a group. If the center $Z(G)$ is a finite-index subgroup of $G$, then the commutator subgroup $D(G)$ is finite.

We begin with two easy lemmae. The first one is left to the reader as an exercice: it is just a computation. In the sequel, we use the notation $[x,y] = x y x^{-1} y^{-1}$.

Lemma 1: For all $a,x,y \in G$, $a[x,y]a^{-1} = [axa^{-1},aya^{-1}]$.

Lemma 2: If $Z(G)$ is a subgroup of finite index $n$, then $[x,y]^{n+1} = [x,y^2] \cdot [yxy^{-1}, y ]^{n-1}$.

Proof. Notice that, because $Z(G)$ is a subgroup of finite index $n$, for all $a \in G$ we have $a^n \in Z(G)$. Therefore,

$[x,y]^{n+1} = [x,y]^n \cdot xyx^{-1} y^{-1} = xyx^{-1} \cdot [x,y]^n \cdot y^{-1}$.

Then,

$[x,y]^{n+1} = xyx^{-1} \cdot [x,y] \cdot [x,y]^{n-1} \cdot y^{-1} = xy^2 x^{-1} y^{-2} \cdot y [x,y]^{n-1} y^{-1}$.

Using Lemma 1, we conclude that

$[x,y]^{n+1} = [x,y^2] \cdot \left( y[x,y]y^{-1} \right)^{n-1} = [x,y^2] \cdot [yxy^{-1},y]^{n-1}$. $\square$

Proof of Theorem: Let us suppose that $Z(G)$ is a subgroup of finite index $n$.

It is not difficult to notice that $[x,y]=[a,b]$ in $G$ whenever $x=a$ and $y=b$ in $G/Z(G)$. Therefore, $G$ has at most $n^2$ commutators.

Now, any $c \in D(G)$ can be written as a product of commutators

$c = c_1 \cdot c_2 \cdots c_r$.

Suppose that $r$ is as small as possible, and assume by contradiction that $r>n^3$. Therefore, because there exist at most $n^2$ commutators, the product contains some commutator, say $k$, at least $(n+1)$ times. We claim that it is possible to write

$c = k^{n+1} \cdot c_1' \cdots c_{r-n-1}' \ \ \ \ \ \ (1)$

for some new commutators $c_j'$. For example, if $c_i=k$, notice that

$c = c_1 \cdots c_{i-1} \cdot k \cdot c_{i+1} \cdots c_r = k \cdot (k^{-1}c_1k) \cdots (k^{-1}c_{i-1}k) \cdot c_{i+1} \cdots c_r$.

Thanks to Lemma 1, we know that each $k^{-1}c_jk$ is again a commutator. We just proved our claim for $n=0$. The general case follows by induction.

From $(1)$ and Lemma 2, we deduce that $c$ can be written as a product of $r-1$ commutators, a contradiction with the minimality of $r$.

Therefore, we just proved that any element of $D(G)$ can be written as a product of at most $n^3$ commutators. Because there exist at most $n^2$ commutators, we deduce that the cardinality of $D(G)$ is bounded above by $n^{2n^3}$; in particular, $D(G)$ is finite. $\square$

Corollary 1: If $G$ has only finitely many commutators, then $D(G)$ is finite.

Proof. Let $[g_1,g_2], \ldots, [g_{2r-1},g_{2r}]$ denote the commutators of $G$ and $H$ be the subgroup generated by the $g_i$‘s. Notice that $D(G)=D(H)$, so it is sufficient to prove that $D(H)$ is finite to conclude. Furthermore, according to our previous theorem, it is sufficient to prove that $Z(H)$ is a finite-index subgroup of $H$.

If $C(g_i)$ denotes the centralizer of $g_i$ in $H$, notice that

$Z(H)= \bigcap\limits_{i=1}^{2r} C(g_i)$.

Therefore, it is sufficient to prove that each $C(g_i)$ is a finite-index subgroup of $H$. But $H/C(g_i)$ is finite if and only if $g_i$ has only finitely-many conjugacy classes. Because $hg_ih^{-1}=[h,g_i] g_i$ for all $h \in H$, and because $H$ has only finitely-many commutators, it is clear that $g_i$ has finitely-many conjugacy classes, concluding our proof. $\square$

Corollary 2: The only infinite group all of whose non-trivial subgroups are finite-index is $\mathbb{Z}$.

Proof. Let $G$ be such a group. Let $g_0 \in G \backslash \{ 1 \}$ and $g_1, \ldots, g_r \in G \backslash \{ 1 \}$ be a set of representatives of the cosets of $\langle g_0 \rangle$. For each $0 \leq i \leq r$, $\langle g_i \rangle$ is a finite-index subgroup of $G$ by assumption; but it is also a subgroup of the centralizer $C(g_i)$, so $C(g_i)$ is a finite-index subgroup of $G$. Therefore, because $\{ g_0, \ldots, g_r \}$ generates $G$, the center

$Z(G)= \bigcap\limits_{i=0}^r C(g_i)$

is a finite-index subgroup of $G$. According to our previous theorem, $D(G)$ is finite, and by assumption, any subgroup of $G$ is either trivial, or finite-index and in particular infinite since $G$ so is. Therefore, $D(G)$ is trivial, that is $G$ is abelian.

From the classification of finitely-generated abelian groups, it is not difficult to prove that the only abelian group all of whose non-trivial subgroups are finite-index is $\mathbb{Z}$. $\square$

Notice however that there exist non-cyclic groups all of whose normal subgroups have finite-index: they are called just-infinite groups. The infinite dihedral group $D_{\infty}$ is such an example.

Corollary 3: A torsion-free virtually infinite cyclic group is infinite cyclic.

Proof. Let $G$ be a torsion-free group with an infinite cyclic subgroup $H= \langle g_0 \rangle$ of finite index $n$. As above, if $g_1, \ldots, g_r$ denotes a set of representatives of the cosets of $\langle g_0 \rangle$ then $\{ g_i \mid 0 \leq i \leq r \}$ generates $G$. Noticing that $g_i^n \in H$, we deduce that $C(g_i) \cap H$ is of finite index in $H$, so the centralizer $C(g_i)$ has a finite index in $G$. Again, we deduce that the center

$Z(G)= \bigcap\limits_{i=0}^r C(g_i)$

is a finite-index subgroup of $G$. According to our previous theorem, $D(G)$ is finite, and in fact trivial because $G$ is torsion-free, so $G$ is abelian. From the classification of finitely-generated abelian groups, it is not difficult to prove that the only abelian torsion-free virtually cyclic group is $\mathbb{Z}$. $\square$

In fact, using Stallings theorem, it is possible to prove more generally that any torsion-free virtually free group turns out to be free.

A group $G$ is abelian if and only if $Z(G)=G$ if and only if $D(G)= \{ 1 \}$. Therefore, a way to say that $G$ is almost abelian is to state that $Z(G)$ is big (eg. is a finite-index subgroup) or that $D(G)$ is small (eg. is finite). Essentially, our main theorem proves that the first idea implies the second. Although the converse does not hold in general, it holds for finitely-generated groups:

Theorem: Let $G$ be a finitely-generated group. If $D(G)$ is finite then $Z(G)$ is a finite-index subgroup.

Proof. Let $g_1, \ldots, g_r \in G$ be a finite generating set of $G$. Then

$Z(G)= \bigcap\limits_{i=1}^r C(g_i)$.

To conclude, it is sufficient to prove that each $C(g_i)$ is a finite-index subgroup, that is equivalent to say that each $g_i$ has finitely-many conjugacy classes. Noticing that for all $h \in G$, $hg_ih^{-1}=[h,g_i]g_i$, we deduce the latter assertion because $D(G)$ is finite. $\square$

## 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.