How can I show that a subset of a group generates the group? In general, how can I show that a subset of a group generates the group?/ some subgroup is generated by some elements of the group?
In particular, I want to show that $SL(2,\Bbb F_3)$ is the subgroup of $GL(2,\Bbb F_3)$ generated by $\begin{pmatrix}1 & 1\\0 & 1\\\end{pmatrix}$ and $\begin{pmatrix}1 & 0\\1 & 1\\\end{pmatrix}$, assuming that I know that $SL(2,\Bbb F_3)$  has order $24$. I have absolutely no idea where to begin with. I dont want to list out $24$ distinct elements of $SL(2,\Bbb F_3)$ though because that is stupid.
 A: Let $A$ and $B$ be the two matrices given in the question.
$|A|=|B|=3$, Also it is easy to see $|AB|=4$ and $|BA|=4$, and observe that $AB$ and $BA$ generate different groups.
Call the group generated by $A$ and $B$ over the field $F_3$ be $G$
Clearly, from above , we have $12|$Order($G$)
Note that $G$ is a subgroup of $SL(2,F_3)$ as determinant of any element of $G$ is $1$ and so the order of $G$ can be $12$ or $24$.
Now if Order($G$)$=12$, then the number of $2$-sylow subgroups must leave a remainder $1$ when divided by $2$ and it also must divide $3$, hence the number of $2$-sylow subgroups would be $1$, but we already know that the number of $2$-sylow subgroups is at least $2$, hence a contradiction.
So the order($G$)$=24$ (Proved).
A: Understand what multiplication (from the left) with the given generators means: For the first matrix, it means we add the second row to the first, for the other matrix we add the first row to the second.
Thus the question can be reformulated as follows:

Given a matrix $M\in SL_2(\Bbb F_3)$, can we transform it into the identity matrix with row addition operations?

To this end, consider a given $M=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in SL_2(\Bbb F_3)$
If $a=0$, then necessarily $c\ne 0$. Thus by adding the second row to the first if necessary, we may assume that $a\ne 0$.
But then we can add the first row to the second row $-a^{-1}c$ times to obtain $\begin{pmatrix}a&b\\0&d'\end{pmatrix}$ with $d'=d-a^{-1}bc$. In fact, from the determinant $1$, we see that $d'=a^{-1}=a$. Now add the second row to the first row $ba$ times to arrive at $\begin{pmatrix}a&0\\0&a\end{pmatrix}$.
If $a=1$, we are done. 
Thus we are only have to deal with $\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$.
But then we can add the first row to the second row to obtain $\begin{pmatrix}-1&0\\-1&-1\end{pmatrix}$, then add the second to the first to obtain $\begin{pmatrix}1&-1\\-1&-1\end{pmatrix}$. By the method of the preceding paragraph, this will turn into the identity matrix as this time we start with $a=1$.

Remark: The above argument works without great changes for any $\Bbb F_p$ (but not $\Bbb F_q$).
