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
Victor
Latex does not render on this page. In later writeups it renders correctly.
5 May 2009 at 12.29 pm
Henry Wilton
Many 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.