Fortunately,
there is a drastic shortcut to the general tree-argument of last time, due to
Roger Alperin. Recall that the Moebius
transformations corresponding to u resp. v send z resp. to
and
whence the Moebius transformation
corresponding to send z to .
Consider
the set of all positive irrational real numbers and the
set of all negative irrational real numbers and observe
that
and
We have to show
that no alternating word in
u and can be the identity in .
If the
length of w is odd then either or depending on whether w starts with a u or with
a term. Either way, this proves that no odd-length word can
be the identity element in .
If the length of
the word w is even we can assume that (if necessary, after conjugating with u we get to this form).
There are two subcases, either in which case
and this latter set is contained in the set of all positive irrational
real numbers which are strictly larger than one .
Or, in which case
and this set is contained in
the set of all positive irrational real numbers strictly smaller than
one .
Either way, this shows that w cannot be the identity
morphism on so cannot be the identity element in
.
Hence we have proved that
A
description of in terms of generators and relations
follows
It is not true that is the free
product as there is the extra relation .
This relation says that the cyclic groups
and share a common subgroup and this extra condition is expressed by saying that
is the amalgamated free product of with
, amalgamated over the common subgroup and denoted
as
More
generally, if G and H are finite groups, then the free product consists of all words of the form (so alternating between non-identity
elements of G and H) and the group-law is induced by concatenation
of words (and group-laws in either G or H when end terms are
elements in the same group).
For example, take the dihedral groups and then the free product can be expressed
as
This almost fits in with
our obtained description of
except for the
extra relations and which express the fact that we
demand that and have the same subgroup
So, again we can express these relations by
saying that is the amalgamated free product of
the subgroups and , amalgamated over the common subgroup . We write
Similarly (but a bit easier) for
we have
- $PGL_2(\mathbb{Z}) = \langle u,v,R
- u^2=v^3=1=R^2 = (Ru)^2 = (Rv)^2 \rangle $
which can be seen as
the amalgamated free product of with , amalgamated over the common subgroup and therefore
Now let us turn to congruence subgroups of
the modular group.
With one denotes the kernel of the natural
surjection
that is all elements represented by a matrix
such that a=d=1 (mod n) and b=c=0
(mod n). On the other hand consists of elements
represented by matrices such that only c=0 (mod n). Both are finite
index subgroups of .
As we have seen that
it follows from general facts
on free products that any finite index subgroup is of the
form
that is the
free product of k copies of , l copies of and m copies
of where it should be noted that k,l and m are allowed
to be zero. There is an elegant way to calculate explicit generators of
these factors for congruence subgroups, due to Ravi S. Kulkarni (An
Arithmetic-Geometric Method in the Study of the Subgroups of the Modular
Group , American Journal of Mathematics, Vol. 113, No. 6. (Dec.,
1991), pp. 1053-1133) which deserves another (non-course) post.
Using this method one finds that is generated by
the Moebius transformations corresponding to the
matrices
and
and that
generators for are given by the
matrices
and
Next,
one has to write these generators in terms of the generating matrices
u and v of and as we know all relations between
u and v the relations of these congruence subgroups will follow.
We
will give the details for and leave you to figure out
that (that is that
there are no relations between the matrices A and
B).
Calculate that and that . Because the
only relations between u and v are we see that Y is an
element of order two as and that no power of
X can be the identity transformation.
But then also none of the
elements can be the identity
(write it out as a word in u and v) whence,
indeed
In fact,
the group is staring you in the face whenever you come to
this site. I fear I’ve never added proper acknowledgements for the
beautiful header-picture

so I’d better do it now. The picture is due to Helena
Verrill and she has a
page with
more pictures. The header-picture depicts a way to get a fundamental
domain for the action of on the upper half plane. Such a
fundamental domain consists of any choice of 6 tiles with different
colours (note that there are two shades of blue and green). Helena also
has a
Java-applet
to draw fundamental domains of more congruence subgroups.