Family of generators for congruence subgroup $\Gamma_0(11)$ Consider the congruence subgroup
$$\Gamma_0(11)=\left\{\begin{pmatrix}a&b\\11c&d\end{pmatrix}\in M(2,\mathbb{Z}): ad-11bc=1\right\}$$
I want to prove that the family
$$\Omega=\{-\text{Id}\}\cup\left\{\begin{pmatrix}a&b\\11c&d\end{pmatrix}\in \Gamma_0(11): c>0,\  gcd(c,6)=1\right\}$$
generates the whole group.
Here is my attempt: let's start by a generic
$$\gamma=\begin{pmatrix}a&b\\11c&d\end{pmatrix}\in\Gamma_0(11)$$
My idea is to act with matrices in $\Omega$ until I find an element of this set. If $c\leq 0$
I can act with $-\text{Id}$ to get a matrix with $c\geq 0.$ If I have $c=0$ I also have $ad=1,$ hence $a=d=\pm1.$ Using $-\text{Id}$ I can suppose $a=d=1,$ and acting on the right, for instance, with the matrix
$$\begin{pmatrix}2&-1\\11&-5\end{pmatrix}\in\Omega$$
I get
$$\begin{pmatrix}1&b\\0&1\end{pmatrix}\begin{pmatrix}2&-1\\11&-5\end{pmatrix}=\begin{pmatrix}2+11b&-1-5b\\11&-5\end{pmatrix}$$
The issue is with the other condition. Suppose $gcd(c,6)\neq1:$ then $2$ or $3$ divide $c:$ how can I  manage these cases?
EDIT
Writing $c$ in the form $c=2^m\cdot 3^n\cdot C,$ if $m,n>0$ (i.e. $6\mid c$) I can write $\gamma$ using only elements of $\Omega$ in the following way: the determinant condition implies
$$(11\cdot2^m\cdot3^n c,a)=1$$
and in particular $a$ is odd and is not a multiple of $3.$ The matrix
$$\begin{pmatrix}
   d&2^m3^n b\\
   11c& a
  \end{pmatrix}\in\Omega,$$
and
$$\begin{pmatrix}
   d&2^m3^n b\\
   11c& a
  \end{pmatrix}\begin{pmatrix}
   a&b\\
   11\cdot 2^m\cdot 3^n c& d
  \end{pmatrix}=\begin{pmatrix}
   ad+2^{m}3^n(11c+b)&bd(1+2^m3^n)\\
   11ac(2^m3^n+1)&11bc+ad 
  \end{pmatrix}\in\Omega $$
Maybe I can generalize this approach even for the other cases

A calculation taking a tiny fraction of a second in Sage reveals that
$$\{ -\mathrm{Id}\} \cup \left\{\left(\begin{array}{rr}
18 & -5 \\
11 & -3
\end{array}\right), \left(\begin{array}{rr}
7 & -2 \\
11 & -3
\end{array}\right), \left(\begin{array}{rr}
8 & -3 \\
11 & -4
\end{array}\right)\right\}$$
is a generating set of $\Gamma_0(11)$. Since these all have $c = 1$ you are done.
