Normal subgroups of matrices 
Let $G=\begin{bmatrix}1&a\\0&b\end{bmatrix}$ so that $a,b\in\mathbb C$ and $b\ne0$. I need to prove that $G$ has infinitely many normal subgroups. 

I attempt to do this by constructing some family of normal subgroups but I keep failing, as most of the things I try aren't even subgroups. 
 A: Let $g$ and $h$ be two elements in $G$. Calculate $ghg^{-1}$. 
You will immediately see under what conditions $ghg^{-1}$ is "of the same form" as $h$. That is, $ghg^{-1} \in H$ where $H$ is a subgroup of $G$.  
This condition $ghg^{-1} \in H$ means that $H$ is normal. This method will give you infinitely many subgroups. 
A: for each $p$ a prime number we get a sub group
$H_p=\{\left(\begin{array}{cc}
1 & d \\
0 & \frac mn
\end{array}\right)
,\mid d\in\Bbb{C}, (n,p)=1,(m,p)=1, n,m\in\Bbb{Z}^*\}$, $H_p$ is a subgroup because is not empty the
identity is an element, stable by product, and each element have
an inverse in $H_c$ , and is normal in $G$ because
$\left(\begin{array}{cc}
1 & a \\
0 & b
\end{array}\right)\left(\begin{array}{cc}
1 & d \\
0 & \frac mn
\end{array}\right)\left( 
\begin{array}{cc}
1 & -\frac a b \\
0 & \frac 1b
\end{array}\right)
=
\left(\begin{array}{cc}
1 & d' \\
0 & \frac mn
\end{array}\right)
$ is already an element of $H_p$, so there is infinity normal subgroup of $G$
