Our first examples of infinite discrete groups will be (non-abelian) free groups. These are most elegantly defined via a universal property.

**Definition:** Let be a set. A *free group on * is a group with a set map such that, whenever is a group and is a set map, there exists a unique group homomorphism such that extends .

This says that there’s a correspondence between group homomorphisms and set maps . If you like category theory, you can think of this as the assertion that the functor from to given by is adjoint to the forgetful functor from to .

As usual with objects defined via a universal property, it’s immediate that if exists then it’s unique up to unique isomorphism. If we were to embrace the categorical point of view fully, then we could guarantee the existence of free groups by appealing to an adjoint functor theorem.

**Example 1:** If is empty then .

**Example 2:** If then .

To construct free groups on larger sets, we adopt a topological point of view. Our main tool will be the Seifert-van Kampen Theorem.

**Seifert-van Kampen Theorem:** Let be a path-connected topological space. Suppose that where and are path-connected open subsets and is also path-connected. For any , the commutative diagram

is a push out.

There are more sophisticated versions of this, which enable one to think about arbitrarily large open covers. But this will be sufficient for most of our purposes.

**Theorem 1:** Let be the rose with petals – that is, the wedge of copies of indexed by . Then .

**Proof for finite :** The proof is by induction on . In the base case where , we take the wedge of 0 circles to be a point and the proof is immediate.

For the general case, in which is a wedge of circles, let be (a small open neighbourhood of) the circle corresponding to some fixed element and let be (a small open neighbourhood of) the union of the circles corresponding to . Of course , and by induction. Choose an orientation on (ie a choice of direction for each circle) and let be the map that sends to a path that goes round the circle corresponding to .

Consider a set map from to a group . As there is a unique group homomorphism such that . As , there is a unique group homomorphism such that for all . It follows from the Seifert-van Kampen Theorem that there is a unique group homomorphism extending and . **QED**

We will be interested in finitely generated groups. The free group on is finitely generated if and only if is finite. The cardinality of is called the *rank* of . We will see a bit later that this is an invariant of the isomorphism type of a free group.

Theorem 1 implies that every free group is the fundamental group of a graph (ie a one-dimensional CW-complex). This has a strong converse.

**Theorem 2:** A group is free if and only if it is the fundamental group of a graph.

The idea of the proof is fairly straightforward: simply contract a maximal tree.

**Proof:** By Theorem 1, it is enough to show that every graph is homotopy equivalent to a rose. Let be a graph, and let be a maximal tree in . Note that the quotient graph is a rose.

Consider the quotient map . Because is a tree, there is a map , uniquely defined up to homotopy, such that maps each edge of into itself and maps each edge of into itself. It is easy to check that and are homotopic to the identity maps. **QED**

We are now in position to give a very easy topological proof of a fairly sophisticated group-theoretic result.

**Nielsen-Schreier Theorem:** Every subgroup of a free group is free.

**Proof:** Think of a free group as the fundamental group of a graph . Let be a subgroup of , and let be the covering space of corresponding to . Then is a graph and so is free. **QED**

Another theorem that we can now easily prove relates the rank of a finite-index subgroup of a free group to its index.

**Schreier Index Formula:** If is a subgroup of of finite index then the rank of is .

**Proof:** Again, let and let , where is the rose with petals and is a covering space of . It is standard that

.

If is the rose with petals then it’s clear that . Similarly, the rank of is . This completes the proof. **QED**

**Exercise 1:**

- Prove that has a subgroup isomorphic to , for any countable .
- Prove that has a subgroup
*of infinite index*isomorphic to , for any countable .

## 2 comments

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5 May 2009 at 10.25 am

VictorLatex does not render on this page. In later writeups it renders correctly.

5 May 2009 at 12.29 pm

Henry WiltonMany thanks for pointing that out, Victor. It seems to be a bug with WordPress – it spontaneously decides to remove every backslash! Anyway, it’s fixed now.