The affine special linear group acts doubly transitive Let $F$ be a field and $d \ge 2$. Denote by $ASL_d(F)$ the affine special linear group, i.e. the group of all transformation on $F^d$ with $t_{A,v}(u) = Au + v$ and $\det A = 1$. I want to show that this group is $2$-transitive (also called doubly transitive) on the points from $F^d$. For this I have to show that for $x,y, x', y'$ with $x \ne y$ and $x' \ne y'$ I can find an element $t_{A,v} \in ASL_d(F)$ such that
$$
 x' = t_{A,v}(x) = Ax + v \qquad\mbox{and}\qquad
 y' = t_{A,v}(y) = Ay + v.
$$
Any ideas or hints how to solve this?
I have solved it in the case $d = 2$ by a cumbersome solution of the equations involved, with the restriction on the determinant for $d = 2$ with the above I have five equations and six unknows (the elements of the matrix and of the translation vector $v$), so this could be solved. But my solution is quite messy and involves a lot of rearrangemnt and equation handling, so any short solution would be preferred?
 A: It is easy to show the group is one-transitive.  Once you have done this, you can take any vector (the 0 vector is easy to work with), and show that the stabilizer of that vector is transitive on the other vectors (this is an equivalent way to show 2-transitivity).  (Hint: the stabilizer of the 0 vector is ${\rm SL}_{d}(F)$).
If you would rather prove this directly by showing any pair $(x,y)$ can be mapped to any other pair $(x^{\prime}, y^{\prime})$, you can think of transforming in several steps using several group elements.  Find a $g_{1}$ mapping $x$ to $0$ and $y$ to $y-x$, a $g_{2}$ mapping $0$ to $0$ and $y-x$ to $y^{\prime}-x^{\prime}$, and then a $g_{3}$ mapping $0$ to $x^{\prime}$ and $y^{\prime} - x^{\prime}$ to $y^{\prime}$.  Thinking of it in this sequence should let you guarantee the existence of a mapping $g_{3}(g_{2}(g_{1})))$ from $(x,y)$ to $(x^{\prime}, y^{\prime})$.
A: First you notice that you must have $$\vec{x^\prime y^\prime}  = A. \vec{xy}.$$ As $x^\prime \neq y^\prime$ and $x \neq y$, both vectors $e_1=\vec{xy}$ and $e_1^\prime=\vec{x^\prime y^\prime}$ are non zero.
Complete $\{e_1\}$ and $\{e_1^\prime\}$ into basis $\mathcal B=(e_1, \dots ,e_d)$ and $\mathcal B^\prime=(e_1^\prime, \dots ,e_d^\prime)$ of your associated vector space that I name $E$. You can find a linear map $A$ that transform $\mathcal B$ into $\mathcal B^\prime$, hence $\det A \neq 0$. Moreover, by changing $e_d^\prime$ into $\frac{e_d^\prime}{\det A}$ (which is possible as you supposed $d \ge 2$), you can even get $A$ such that $\det A=1$.
Finally, you compute $v=y-A.x$.
