A subgroup of $\textrm{GL}(3,q)$ of order $q^2(q-1)$

Question

Let $q$ be a prime power. Consider the multiplicative group $\textrm{GL}(3,q)$ of the $3 \times 3$ matrixes with coefficients in $\mathbb{F}_q$ which are invertible. The matrixes $$ M_{a,b,c} = \left( \begin{array}{ccc} a & 0 & 0 \\ b & 1 & 0 \\ c & 0 & 1 \end{array} \right) $$ with $a,b,c \in \mathbb{F}_q$, $a \neq 0$, form a subgroup $G$ of $\textrm{GL}(3,q)$. My objective is to represent this group $G$ as a known finite group. Note that $G$ has order $q^2(q-1)$. I studied the order of its elements and I found that $G$ has a normal elementary abelian $p$-subgroup $N$ of order $q^2$. $$ N = \{ M_{1,b,c} \ | \ b,c \in \mathbb{F}_q \}. $$ Any help to represent the group $G$ as a known finite group would be appreciated.

Answer 1

$M\cong(\mathbb{F}_q\oplus\mathbb{F}_q, +)\rtimes\mathbb{F}_q^\times$, where $a\in\mathbb{F}_q^\times$ acts on $\mathbb{F}_q\oplus\mathbb{F}_q$ by vector space scalar multiplication by $a^{-1}$.

Answer 2

Notice that all your matrices differ from the identity matrix only in the first column: So one can describe this as those non-singular transformations fixing the vectors $e_2$ and $e_3$. This means, $e_1$ should not go to a span of $e_2$ and $e_3$.

So the vector to which $e_1$ is sent, describes an element of the group completely: This is what Alex Fok's answer contains.