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.

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.

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.