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 (resp. ) denote the connected sum of tori (resp. projective spaces). Because , we easily deduce the Euler characteristics and ; the integer is called the *genus*.

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

,

or is the fundamental group of a non-orientable surface of genus and

.

(Just write or 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 be the fundamental group of a closed surface and be a subgroup. Either has finite index in and is isomorphic to the fundamental group of a closed surface whose Euler characteristic is , or is an infinite-index subgroup of 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 (resp. ) is isomorphic to (resp. ). Therefore, two closed surfaces and 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,

and, with ,

.

Consequently, the and define a family of pairwise non-isomorphic groups, and the conclusion follows from the classication of closed surfaces.

**Corollary 1:** Let be the fundamental group of a closed surface different from the projective space. Then 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 has no cyclic subgroup of finite index. Consequently, the cyclic subgroup generated by a non trivial element of is free, that its order is infinite.

**Corollary 2:** Let be the fundamental group of a closed surface satisfying and be a subgroup generated by elements. If then is free.

**Proof.** If , 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 . According to Property 3, the abelianization of has rank . We deduce that the smallest cardinality of a generating set of has cardinality ; in particular, because . Suppose that has finite index . Then is the fundamental group of a closed surface and . In the same way, necessarily . Therefore,

,

a contradiction.

**Corollary 3:** Let be the fundamental group of a closed surface satisfying . If commute, then there exist and such that and .

**Proof.** Let . Then it is easy to find two epimorphisms

and

.

Therefore, the groups and are not abelian for , and we deduce that the sphere, the projective plane and the torus are the only closed surfaces whose fundamental group is abelian.

Because satisfies , we conclude that has no finite-index abelian subgroup, and that the subgroup genereted by is necessarily free; since and commute, the subgroup turns out to be cyclic and the conclusion follows.

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

**Proof.** Let be the fundamental group of a closed surface . If is a projective space, then and its commutator subgroup is trivial (in particular free). Otherwise, thanks to Property 3 we know that the abelianization of is infinite; therefore, the commutator subgroup is an infinite-index subgroup and so is free.

**Proof of property 2.** First, we notice that the fundamental group of a connected compact surface with boundary is free. If is such a surface, by gluing a disk along each boundary component, we get a closed surface ; therefore, 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 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 be a non-compact surface. In order to prove that is free, we want to find a sequence of compact surfaces with boundary

such that and that the inclusions are -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 so that may be identified with a simplicial complex; let denote the 2-simplexes. Without loss of generality, we may suppose that is adjacent to one of the simplexes ; otherwise, number the simplexes by taking first the simplexes adjacent to a vertex cyclically, then the simplexes within from , then the simplexes within from , etc.

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

Let . Now suppose that is given. If , then set ; otherwise, define as the union of with and with the disks bounding a boundary component of .

Because is a union of , we know that it is a compact surface with boundary. To conclude, it is sufficent to prove that the inclusion is -injective. To do that, just notice that is attatched on the boundary of in six possible ways:

In the first three cases, because the boundary component bounds a disk in if and only if it bounds a disk in . So clearly retracts on so that the inclusion induces an isomorphism .

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 is obtained by adding one or two elements to a free basis of .

**Proof of theorem 1.** Let be the covering associated to the subgroup ; in particular, is a surface of fundamental group . If is a finite-index subgroup, is compact and the conclusion follows. Otherwise, is not compact, and the conclusion follows from Property 2.