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.
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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 $G$
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.