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

$-\frac{1}{z}$ and $\frac{1}{1-z}$

whence the Moebius transformation
corresponding to $v^{-1}$ send z to $1-\frac{1}{z}$.

Consider
the set $\mathcal{P}$ of all positive irrational real numbers and the
set $\mathcal{N}$ of all negative irrational real numbers and observe
that

$u(\mathcal{P}) \subset \mathcal{N}$ and
$v^{\pm}(\mathcal{N}) \subset \mathcal{P}$

We have to show
that no alternating word $w=(u)v^{\pm}uv^{\pm}u \dots v^{\pm}(u)$ in
u and $v^{\pm}$ can be the identity in $PSL_2(\mathbb{Z})$.

If the
length of w is odd then either $w(\mathcal{P}) \subset \mathcal{N}$ or $w(\mathcal{N}) \subset \mathcal{P}$ depending on whether w starts with a u or with
a $v^{\pm}$ term. Either way, this proves that no odd-length word can
be the identity element in $PSL_2(\mathbb{Z})$.

If the length of
the word w is even we can assume that $w = v^{\pm}uv^{\pm}u \dots v^{\pm}u$ (if necessary, after conjugating with u we get to this form).

There are two subcases, either $w=v^{-1}uv^{\pm}u \dots v^{\pm}u$ in which case $w(\mathcal{P}) \subset v^{-1}(\mathcal{N})$
and this latter set is contained in the set of all positive irrational
real numbers which are strictly larger than one .

Or, $w=vuv^{\pm}u \dots v^{\pm}u$ in which case
$w(\mathcal{P}) \subset v(\mathcal{N})$ 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 $\mathcal{P}$ so cannot be the identity element in
$PSL_2(\mathbb{Z})$.
Hence we have proved that

$PSL_2(\mathbb{Z}) = C_2 \ast C_3 = \langle u,v : u^2=1=v^3 \rangle$

A
description of $SL_2(\mathbb{Z})$ in terms of generators and relations
follows

$SL_2(\mathbb{Z}) = \langle U,V : U^4=1=V^6, U^2=V^3 \rangle$

It is not true that $SL_2(\mathbb{Z})$ is the free
product $C_4 \ast C_6$ as there is the extra relation $U^2=V^3$.

This relation says that the cyclic groups $C_4 = \langle U \rangle$
and $C_6 = \langle V \rangle$ share a common subgroup $C_2 = \langle U^2=V^3 \rangle$ and this extra condition is expressed by saying that
$SL_2(\mathbb{Z})$ is the amalgamated free product of $C_4$ with
$C_6$, amalgamated over the common subgroup $C_2$ and denoted
as

$SL_2(\mathbb{Z}) = C_4 \ast_{C_2} C_6$

More
generally, if G and H are finite groups, then the free product $G \ast H$ consists of all words of the form $~(g_1)h_1g_2h_2g_3 \dots g_nh_n(g_{n-1})$ (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 $D_4 = \langle U,R : U^4=1=R^2,(RU)^2=1 \rangle$ and $D_6 = \langle V,S : V^6=1=S^2,(SV)^2=1 \rangle$ then the free product can be expressed
as

$D_4 \ast D_6 = \langle U,V,R,S : U^4=1=V^6=R^2=S^2=(RV)^2=(RU)^2 \rangle$

This almost fits in with
our obtained description of
$GL_2(\mathbb{Z})$

$GL_2(\mathbb{Z}) = \langle U,V,R : U^4=1=V^6=R^2=(RU)^2=(RV)^2, U^2=V^3 \rangle$

except for the
extra relations $R=S$ and $U^2=V^3$ which express the fact that we
demand that $D_4$ and $D_6$ have the same subgroup

$D_2 = \langle U^2=V^3,S=R \rangle$

So, again we can express these relations by
saying that $GL_2(\mathbb{Z})$ is the amalgamated free product of
the subgroups $D_4 = \langle U,R \rangle$ and $D_6 = \langle V,R \rangle$, amalgamated over the common subgroup $D_2 = C_2 \times C_2 = \langle U^2=V^3,R \rangle$. We write

$GL_2(\mathbb{Z}) = D_4 \ast_{D_2} D_6$

Similarly (but a bit easier) for
$PGL_2(\mathbb{Z})$ 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 $D_2 = \langle u,R \rangle$ with $D_3 = \langle v,R \rangle$, amalgamated over the common subgroup $C_2 = \langle R \rangle$ and therefore

$PGL_2(\mathbb{Z}) = D_2 \ast_{C_2} D_3$

Now let us turn to congruence subgroups of
the modular group
.
With $\Gamma(n)$ one denotes the kernel of the natural
surjection

$PSL_2(\mathbb{Z}) \rightarrow PSL_2(\mathbb{Z}/n\mathbb{Z})$

that is all elements represented by a matrix

$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$

such that a=d=1 (mod n) and b=c=0
(mod n). On the other hand $\Gamma_0(n)$ consists of elements
represented by matrices such that only c=0 (mod n). Both are finite
index subgroups of $PSL_2(\mathbb{Z})$.

As we have seen that
$PSL_2(\mathbb{Z}) = C_2 \ast C_3$ it follows from general facts
on free products that any finite index subgroup is of the
form

$C_2 \ast C_2 \ast \dots \ast C_2 \ast C_3 \ast C_3 \ast \dots \ast C_3 \ast C_{\infty} \ast C_{\infty} \dots \ast C_{\infty}$

that is the
free product of k copies of $C_2$, l copies of $C_3$ and m copies
of $C_{\infty}$ 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 $\Gamma_0(2)$ is generated by
the Moebius transformations corresponding to the
matrices

$X=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ and
$Y=\begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix}$

and that
generators for $\Gamma(2)$ are given by the
matrices

$A=\begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix}$
and $B=\begin{bmatrix} 1 & -2 \\ 2 & -3 \end{bmatrix}$

Next,
one has to write these generators in terms of the generating matrices
u and v of $PSL_2(\mathbb{Z})$ and as we know all relations between
u and v the relations of these congruence subgroups will follow.

We
will give the details for $\Gamma_0(2)$ and leave you to figure out
that $\Gamma(2) = C_{\infty} \ast C_{\infty}$ (that is that
there are no relations between the matrices A and
B).

Calculate that $X=v^2u$ and that $Y=vuv^2$. Because the
only relations between u and v are $v^3=1=u^2$ we see that Y is an
element of order two as $Y^2 = vuv^3uv^2= v^3 = 1$ and that no power of
X can be the identity transformation.

But then also none of the
elements $~(Y)X^{i_1}YX^{i_2}Y \dots YX^{i_n}(Y)$ can be the identity
(write it out as a word in u and v) whence,
indeed

$\Gamma_0(2) = C_{\infty} \ast C_2$

In fact,
the group $\Gamma_0(2)$ is staring you in the face whenever you come to
this site. I fear I’ve never added proper acknowledgements for the 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 $\Gamma_0(2)$ 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.

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