Let $\text{GL}(n,q)$ denote the group of all the invertible $n$ by $n$ matrices over finite field $\mathbb{F}_q$. $\text{GL}(n,q)$ be generated by two elements for all $n>2$. See here.
Now my first question is:
Given two elements $A$, $B$ in $\text{GL}(n,q)$, how to generate the group $G=\langle A, B \rangle$ ? Can we find out all the maximal subgroups of $\text{GL}(n,q)$ ?
It seems this question may be to hard. See here. Thanks for provide any information about the first question.
Then the second question is:
Let $n=2$, $q=5$. Denote $${\displaystyle A={\begin{pmatrix}1 & 3 \\ 3 & 0 \end{pmatrix}}}\in \text{GL}(n,q), {\displaystyle B={\begin{pmatrix}4 & 4 \\ 1 & 0 \end{pmatrix}}}\in \text{GL}(n,q).$$ Please compute all the elements in $\langle A, B \rangle$.
By the computation in magma software, $|\langle A, B \rangle|=|\langle A \rangle| \times |\langle B \rangle|$, where $|\langle A \rangle|=6$ and $|\langle B \rangle|=3$. It's special case.
So my third question is:
Let $n=2$, $q=7$. Denote $${\displaystyle C={\begin{pmatrix}6 & 6 \\ 1 & 0 \end{pmatrix}}}\in \text{GL}(n,q), {\displaystyle D={\begin{pmatrix}4 & 4 \\ 0 & 2 \end{pmatrix}}}\in \text{GL}(n,q).$$ Please compute all the elements in $\langle C, D \rangle$.
In the third question, it has $|\langle C, D \rangle| \ne|\langle C \rangle| \times |\langle D \rangle|$. $|\langle C \rangle|=3$ and $|\langle D \rangle|=3$ while $|\langle C, D \rangle|=24$.
Thanks for any replies.
Magma uses an algorithm known as Schreier-Sims using orbits of matrices on vectors and $1$-dimensional subspaces, which is not really suitable for hand calculation.
Here is a way of calculating the order by hand, but I have to admit that it makes it a lot easier if you know the order of the subgroup and something about the subgroups of ${\rm SL}(2,5)$. Knowing that that this subgroup has order $24$, I expect it to be isomorphic to ${\rm SL}(2,3)$, so I can look for generators that satisfy a known presentation of ${\rm SL}(2,3)$.
We observe that $A^3=-I$, which is central. Also $A^2B = \left(\begin{array}{cc}3&0\\1&2\end{array}\right)$, and $(A^2B)^2 = -I$. So putting $x=A^2$ and $y=B$, $z=-I$, we see that $G=\langle A,B \rangle$ satisfies the relations of the group $$\langle x,y,z \mid x^3=y^3=1, (xy)^2=z, zx=xz, zy=yz \rangle$$ which is a $2$-fold cover of $A_4$ and isomorphic to ${\rm SL}(2,3)$. Also, since $A^2B \ne \pm I$, $G$ cannot be a proper quotient of this group and hence $|G|=24$.
You can show similarly that $\langle C,D \rangle \cong {\rm SL}(2,3)$.