Finding another representation of a group represented by a set of matrices I want to find the representation of the group 
$$
\left\{ \left. \left( 
\begin{array}{cccc} a & 0 & 0 & 0 \\ 0 & b & 0 & 0 \\ c & 0 & ad & 0 \\ 0 & -\frac{b}{a}c & 0 & bd \end{array} 
\right) \in Gl_{4} \ \right| \ a,b,c,d \in \mathbb{C} \right\}
$$
currently I have 
$$
\left( 
\begin{array}{cccc} a_{1}a_{2} & 0 & 0 & 0 \\ 0 & b_{1}b_{2} & 0 & 0 \\ c_{1}a_{2} + a_{1}d_{1}c_{2} & 0 & a_{1}a_{2}d_{1}d_{2} & 0 \\ 0 & -\frac{b_{1}b_{2}}{a_{1}a_{2}}(c_{1}a_{2} + a_{1}d_{1}c_{2}) & 0 & b_{1}b_{2}d_{1}d_{2} \end{array} 
\right)
$$
(which is now closed under multiplication as you can take $a_{3}=a_{1}a_{2}$, $b_{3}=b_{1}b_{2}$, $c_{3}=c_{1}a_{2}+a_{1}d_{1}c_{2}$ and $d_{3}=d_{1}d_{2}$)
for the multiplication of 2 elements so I think it will look something like $\mathbb{C}^{*} \times \mathbb{C}^{*} \times \mathbb{C}^{*} \times$ with something on the end. I'm not sure how to find the other bit or if I'm right on the first bit.
The multiplication of two elements can be written as, 
$$(a_{1},b_{1},c_{1},d_{1}) \cdot (a_{2},b_{2},c_{2},d_{2}) = (a_{1}a_{2},b_{1}b_{2},c_{1}a_{2}+a_{1}d_{1}c_{2},d_{1}d_{2})$$
Any help would be appreciated with this specific case or just in general as I not really sure how to go about this.
 A: Note that for a matrix $A$ to be in this subset
$$
G = \left\{ \left. \left( 
\begin{array}{cccc} a & 0 & 0 & 0 \\ 0 & b & 0 & 0 \\ bc & 0 & ad & 0 \\ 0 & -ac & 0 & bd \end{array} 
\right) \in Gl_{4} \ \right| \ a,b,c,d \in \mathbb{C} \right\}
$$
it must satisfy $A_{11} A_{31} = - A_{22} A_{42}$. This allows to easily see $G$ is not closed under multiplication:
$\underbrace{\left(\begin{array}{rrrr}%
1&0&0&0\\%
0&1&0&0\\%
1&0&1&0\\%
0&-1&0&1\\%
\end{array}\right)}_{a=1,~b=1,~c=1,~d=1} \cdot 
\underbrace{\left(\begin{array}{rrrr}%
1&0&0&0\\%
0&2&0&0\\%
2&0&1&0\\%
0&-1&0&2\\%
\end{array}\right)}_{a=1,~b=2,~c=1,~d=1} =
\underbrace{\left(\begin{array}{rrrr}%
1&0&0&0\\%
0&2&0&0\\%
3&0&1&0\\%
0&-3&0&2\\%
\end{array}\right)}_{a=1,~b=2,~\color{red}{1.5=c=3},~d=1}
$
In particular, $G$ is not a group.
A: Consider a similar subset (as in the now edited question, but taking $c'=c/a$):
$$
H = \left\{ \begin{pmatrix} a & 0 & 0 & 0 \\ 0 & b & 0 & 0 \\ ac' & 0 & ad & 0 \\ 0 & -bc' & 0 & bd  \end{pmatrix} ~\middle|~ a,b,d \in \mathbb{C}^\times, c' \in \mathbb{C} \right\}
$$
This satisfies the multiplication property: $$(a_1,b_1,c_1',d_1) \cdot (a_2,b_2,c_2',d_2) = (a_1 a_2, b_1 b_2, c_1' + d_1 \cdot c_2', d_1 d_2).$$
Hence $$H \cong \mathbb{C}^\times \times \mathbb{C}^\times \times( \mathbb{C} \rtimes \mathbb{C}^\times ) \cong \operatorname{GL}_1 \times \operatorname{GL}_1 \times \operatorname{AGL}_1$$
Here $\operatorname{AGL}_1 = \left\{ \begin{pmatrix} d & c' \\ 0 & 1 \end{pmatrix} : d \in \mathbb{C}^\times, c' \in \mathbb{C} \right\}$
The field $\mathbb{C}$ can be replaced by any commutative, unital, associative ring.
