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
−1z and 11−z
whence the Moebius transformation
corresponding to v−1 send z to 1−1z.
Consider
the set P of all positive irrational real numbers and the
set N of all negative irrational real numbers and observe
that
u(P)⊂N and
v±(N)⊂P
We have to show
that no alternating word w=(u)v±uv±u…v±(u) in
u and v± can be the identity in PSL2(Z).
If the
length of w is odd then either w(P)⊂N or w(N)⊂P depending on whether w starts with a u or with
a v± term. Either way, this proves that no odd-length word can
be the identity element in PSL2(Z).
If the length of
the word w is even we can assume that w=v±uv±u…v±u (if necessary, after conjugating with u we get to this form).
There are two subcases, either w=v−1uv±u…v±u in which case w(P)⊂v−1(N)
and this latter set is contained in the set of all positive irrational
real numbers which are strictly larger than one .
Or, w=vuv±u…v±u in which case
w(P)⊂v(N) 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 P so cannot be the identity element in
PSL2(Z).
Hence we have proved that
PSL2(Z)=C2∗C3=⟨u,v:u2=1=v3⟩
A
description of SL2(Z) in terms of generators and relations
follows
SL2(Z)=⟨U,V:U4=1=V6,U2=V3⟩
It is not true that SL2(Z) is the free
product C4∗C6 as there is the extra relation U2=V3.
This relation says that the cyclic groups C4=⟨U⟩
and C6=⟨V⟩ share a common subgroup C2=⟨U2=V3⟩ and this extra condition is expressed by saying that
SL2(Z) is the amalgamated free product of C4 with
C6, amalgamated over the common subgroup C2 and denoted
as
SL2(Z)=C4∗C2C6
More
generally, if G and H are finite groups, then the free product G∗H consists of all words of the form (g1)h1g2h2g3…gnhn(gn−1) (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 D4=⟨U,R:U4=1=R2,(RU)2=1⟩ and D6=⟨V,S:V6=1=S2,(SV)2=1⟩ then the free product can be expressed
as
D4∗D6=⟨U,V,R,S:U4=1=V6=R2=S2=(RV)2=(RU)2⟩
This almost fits in with
our obtained description of
GL2(Z)
GL2(Z)=⟨U,V,R:U4=1=V6=R2=(RU)2=(RV)2,U2=V3⟩
except for the
extra relations R=S and U2=V3 which express the fact that we
demand that D4 and D6 have the same subgroup
D2=⟨U2=V3,S=R⟩
So, again we can express these relations by
saying that GL2(Z) is the amalgamated free product of
the subgroups D4=⟨U,R⟩ and D6=⟨V,R⟩, amalgamated over the common subgroup D2=C2×C2=⟨U2=V3,R⟩. We write
GL2(Z)=D4∗D2D6
Similarly (but a bit easier) for
PGL2(Z) 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 D2=⟨u,R⟩ with D3=⟨v,R⟩, amalgamated over the common subgroup C2=⟨R⟩ and therefore
PGL2(Z)=D2∗C2D3
Now let us turn to congruence subgroups of
the modular group.
With Γ(n) one denotes the kernel of the natural
surjection
PSL2(Z)→PSL2(Z/nZ)
that is all elements represented by a matrix
[abcd]
such that a=d=1 (mod n) and b=c=0
(mod n). On the other hand Γ0(n) consists of elements
represented by matrices such that only c=0 (mod n). Both are finite
index subgroups of PSL2(Z).
As we have seen that
PSL2(Z)=C2∗C3 it follows from general facts
on free products that any finite index subgroup is of the
form
C2∗C2∗⋯∗C2∗C3∗C3∗⋯∗C3∗C∞∗C∞⋯∗C∞
that is the
free product of k copies of C2, l copies of C3 and m copies
of C∞ 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 Γ0(2) is generated by
the Moebius transformations corresponding to the
matrices
X=[1101] and
Y=[1−12−1]
and that
generators for Γ(2) are given by the
matrices
A=[10−21]
and B=[1−22−3]
Next,
one has to write these generators in terms of the generating matrices
u and v of PSL2(Z) and as we know all relations between
u and v the relations of these congruence subgroups will follow.
We
will give the details for Γ0(2) and leave you to figure out
that Γ(2)=C∞∗C∞ (that is that
there are no relations between the matrices A and
B).
Calculate that X=v2u and that Y=vuv2. Because the
only relations between u and v are v3=1=u2 we see that Y is an
element of order two as Y2=vuv3uv2=v3=1 and that no power of
X can be the identity transformation.
But then also none of the
elements (Y)Xi1YXi2Y…YXin(Y) can be the identity
(write it out as a word in u and v) whence,
indeed
Γ0(2)=C∞∗C2
In fact,
the group Γ0(2) 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 Γ0(2) 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.