Some notations : $T=\begin{pmatrix} 1&1\\0&1 \end{pmatrix}$ , $D(N):=\langle gT^N g^{-1} : g \in SL_2(\mathbb{Z}) \rangle$
$\Gamma(N)$ denotes the principal congruence subgroup of level N .
(a) Using the Chinese remainder theorem , show that for given x, y, the following congruence $$x (\alpha y + \beta) \equiv\gamma \ (mod\ m) $$ has a solution in integers $\alpha,\beta,\gamma$ such that $(\alpha,\beta,m)=1 $.
(b) Using (a), show that given $g\in \Gamma(N)$ , there exists $$g'=\begin{pmatrix}a'&b' \\c' &d' \end{pmatrix} \in\Gamma(N)$$ such that $b'\equiv c' \equiv0 \ ( mod \ mN)$ and $g' \in D(N)gD(N)$ .
solution (b) . Let $$g=\begin{pmatrix} a&b\\c&d\end{pmatrix}$$be given and let L be the transpose of T. Consider the elements of the form $$g_1=L^{-y}T^{Nx}L^yg=\begin{pmatrix} a_1&b_1\\c_1&d_1\end{pmatrix}$$
where $b_1=b+Nx(by+d)$ . We choose x and y so that $$b_1=b+Nx(by+d)\equiv0 \ (mod \ mN) $$ and as b is divisible by N, this is equivalent to solving $b/N+x(by+d)\equiv0 \ ( mod \ m)$ which by (a) has a solution.
Now consider the elements of the form $$g_2=g_1L^{Nz}=\begin{pmatrix} a_2&b_2\\c_2&d_2\end{pmatrix} $$ where $c_2 = c_1 + d_1Nz.$ We observe that we may find z such that $$c_2=c_1+d_1NZ\equiv0 \ (mod \ mN) $$ which is equivalent to $$c_1/N+d_1z\equiv0 \ (mod \ mN) $$ . Thus $g_2 $ is the required modification .
I do not really understand this proof . Should $g_2$ be g' ?
I think the idea is not really clear to me .
Thanks for the help .