The action of the group $SL_3(\mathbb{F}_p)$ on the group $\mathbb{F}_p^3/\{0\}$ Proof that the natural action of $SL_3(\mathbb{F}_p)$ on the group $\mathbb{F}_p^3/\{0\}$ is transitive action. ($\mathbb{F}_p$ is finite field of order p which is prime)
Any clue?
 A: Note if the matrix is 3x3, let the elements in its first column be $a_1$,$a_2$,$a_3$, and all other entries be 0 then it acts on (1,0,0) and sends it to ($a_1$,$a_2$,$a_3$). Now consider the matrix with the first elements in its first row as $(a_1)^{-1}$. And other entries being 0. It will send ($a_1$,$a_2$,$a_3$) in $F^3$\ {0} to (1,0,0). Hence using (1,0,0) as the intermediate, in $F^3$\ {0}, and suitable matrices in $M_3(F_p)$, we can send any vector ($a_1$,$a_2$,$a_3$) to any other vector ($b_1$,$b_2$,$b_3$). Hence the action is transitive. Now the matrix that sends (1,0,0) to ($a_1$,$a_2$,$a_3$) is with determinant 0 so it isn't in $SL_n(F_p)$. So to bring it in $SL_n(F_p)$, replace the zero entries in the above matrices with suitable elements from field to make the determinant 1, ensuring it still acts the same way.
A: *

*Justify that for any integer $n$ and any field $K$, $e_1:=(1,0,...,0)$ and any $v\in K^n-\{0\}$ you can find $M\in GL_n(K)$ such that $Me_1=v$ (this is a linear algebra property).

*Let $n\geq 2$. By multipliying $M$ by a well chosen matrix show that you can find $M'\in SL_n(K)$ such that $Me_1=v$ (your matrix will be diagonal with a lot of $1$'s). 
