How to show congruence subgroup $\Gamma_0(4)$ is generated by $\begin{pmatrix} 1 &0\\ 4& 1 \end{pmatrix}$, $\begin{pmatrix} 1 &1\\ 0& 1 \end{pmatrix}$? This is a exercise in GTM228 P21 2.2.4.
Let $\Gamma_{u}=<\begin{pmatrix} 1 &0\\ 4& 1 \end{pmatrix}, \begin{pmatrix} 1 &1\\ 0& 1 \end{pmatrix}>$, I try to prove $\Gamma_{u}=\Gamma_0(4)$. It is obvious to show that $\Gamma_{u} \subset\Gamma_0(4)$. In the textbook, here is a hint but I do not know what is the meanings.
Let $\forall \alpha =\begin{pmatrix} a &b\\ c& d \end{pmatrix} \in \Gamma_0(4)$ and use the identity
$$\begin{pmatrix} a&b\\ c& d \end{pmatrix} \begin{pmatrix} 1 &n\\ 0& 1 \end{pmatrix}=\begin{pmatrix} a &b'\\ c& nc+d \end{pmatrix}$$ show that the bottom row $(c',d')$ with $\vert d'\vert <\vert c' \vert /2$.
Use the identity
$$\begin{pmatrix} a&b\\ c& d \end{pmatrix} \begin{pmatrix} 1 &0\\ 4n& 1 \end{pmatrix}=\begin{pmatrix} a' &b\\ c+4nd& d \end{pmatrix}$$ show that the bottom row $(c',d')$ with $\vert c'\vert <2\vert d' \vert$.
But why do the process stop with $c=0$?
You have two numbers $c$ and $2d$. Note that $c\equiv0\pmod4$ and $2d\equiv2\pmod4$.
The first identity shows that, unless $c=0$, you can reduce $|2d|$ until it is less than $|c|$. The second identity shows that you can reduce $|c|$ until it is less than $|2d|$.
This is a bit like Euclid's Algorithm to find the greatest common factor of two numbers.
Since both $|c|$ and $|2d|$ are positive, and reduce by a whole number each time, this must stop some time. $|2d|$ can't be zero because $2d=2\pmod4$, so it must stop when $c=0$