the generators of the modular group are S and T

Question

The claim is that the modular group $$\Gamma=\langle S,T\rangle$$ with $S=\begin{pmatrix} 0 &-1 \\ 1 & 0 \end{pmatrix}$ and $T=\begin{pmatrix} 1 &1 \\ 0 & 1 \end{pmatrix}$ .

I do not really understand the proof by induction .

Let $G$ be the group generated by $S$ and $T$. Now I want to show that $G =\Gamma$ . The one conclusion is trivial .

Let $A=\begin{pmatrix} a &b \\ c & d \end{pmatrix}\in \Gamma$ .

Then I want to prove by induction on $|c| $ that $A \in G$ .

First if $|c|=0$ then $A=\pm T^b \in G$ .

Now consider $|c|>0$

$ST^mA=\begin{pmatrix} -c &-d \\ a+mc & b+md \end{pmatrix}$ , m$\in \mathbb{Z}$

One can choose $m$ such that $|a+mc|<|c|$ .

Now I do not understand why

$ST^mA \in G$ saying that $A \in G$ .

Thanks for the help .

Answer 1

The argument is not complete. You have to keep on multiplying by $S$ and powers of $T$, so that the absolute value continues decreasing and eventually the lower left entry becomes zero. Now your back in the case $c =0$.

Answer 2

Using $T^m ,S$ you can transform the bottom row $(c,d)$ into $(c,d+mc)$ and $(d+mc,-c)$, this is what you need for the $\gcd$ algorithm, terminating with the bottom row $ = (0,\pm 1)$ which implies the obtained matrix is $\pm T^r$

Answer 3

As I see, the problem is the induction step. Consider this

Hint: $X \in G \implies T^{a}S^{b}X \in G$

Solution: $$ ST^mA \in G \implies T^{-m}S^{-1}(ST^mA) \in G \implies A \in G $$