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

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 $\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.

## 2 Comments