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

Definition: Let S be a set. A free group on S is a group F_S with a set map i:S\to F_S such that, whenever G is a group and f:S\to G is a set map, there exists a unique group homomorphism \hat{f}:F_S\to G such that \hat{f}\circ i extends f.

This says that there’s a correspondence between group homomorphisms F_S\to G and set maps S\to G. If you like category theory, you can think of this as the assertion that the functor from \mathrm{Set} to \mathrm {Grp} given by S\mapsto F_S is adjoint to the forgetful functor from \mathrm{Grp} to \mathrm{Set}.

As usual with objects defined via a universal property, it’s immediate that if F_S 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 S is empty then F_S\cong 1.

Example 2: If S=\{s\} then F_S\cong\mathbb{Z}.

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 X be a path-connected topological space. Suppose that X=U\cup V where U and V are path-connected open subsets and U\cap V is also path-connected. For any x\in U\cap V, the commutative diagram

eqlatex

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 X be the rose with |S| petals – that is, the wedge of |S| copies of S^1 indexed by S. Then \pi_1(X)\cong F_S.

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

For the general case, in which X is a wedge of |S| circles, let U be (a small open neighbourhood of) the circle corresponding to some fixed element s_0\in S and let V be (a small open neighbourhood of) the union of the circles corresponding to T=S\smallsetminus{s_0}. Of course \pi_1(U)\cong\mathbb{Z}, and \pi_1(V)\cong F_T by induction. Choose an orientation on X (ie a choice of direction for each circle) and let i:S\to\pi_1(X) be the map that sends s to a path that goes round the circle corresponding to s.

Consider a set map f from S to a group G. As \pi_1(U)\cong\langle i(s_0)\rangle there is a unique group homomorphism f_1:\pi_1(U)\to G such that f_1\circ i(s_0)= f(s_0). As \pi_1(V)\cong F_T, there is a unique group homomorphism f_2:\pi_1(V)\to G such that f_1\circ i(t)= f(t) for all t\in T. It follows from the Seifert-van Kampen Theorem that there is a unique group homomorphism \hat{f}:\pi_1(X)\to G extending f_1 and f_2. QED

We will be interested in finitely generated groups. The free group on S is finitely generated if and only if S is finite. The cardinality of S is called the rank of F_S. 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 G be a graph, and let T be a maximal tree in G. Note that the quotient graph G'=G/T is a rose.

Consider the quotient map q: G\to G'. Because T is a tree, there is a map r:G'\to G, uniquely defined up to homotopy, such that q\circ r maps each edge of G' into itself and r\circ q maps each edge of G\smallsetminus T into itself. It is easy to check that q\circ r and r\circ q 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 F as the fundamental group of a graph X. Let H be a subgroup of F, and let X' be the covering space of X corresponding to H. Then X' is a graph and H=\pi_1(X') so H 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 H is a subgroup of F_r of finite index k then the rank of H is 1+k(r-1).

Proof: Again, let F_r=\pi_1(X) and let H=\pi_1(X'), where X is the rose with r petals and X' is a covering space of X. It is standard that

\chi(X')=k\chi(X).

If X is the rose with r petals then it’s clear that \chi(X)=1-r. Similarly, the rank of H is 1-\chi(X'). This completes the proof. QED

Exercise 1:

  1. Prove that F_2 has a subgroup isomorphic to F_k, for any countable k.
  2. Prove that F_2 has a subgroup of infinite index isomorphic to F_k, for any countable k.
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