How to prove the equivalence of definitions of doubly transitive group Given a group $G$ which acts on a set $X$ we define that a group $G$ acts doubly transitive on $X$ as $G$ is transitive on $X$ and $G_x$ is transtive on $X \backslash \{x\}$ where $G_x$ the stabilizer is and $x \in X$. I want to prove that for $x\neq x'$ and $y \neq y'$ an element $g \in G$ exists with $(g(x),g(x')) = (y,y')$. How can I do that ?
 A: Let's just think about our definitions. We know that there is a $g_1 \in G$ with 
$$
g_1 \cdot x = y.
$$
 We also know that for any point $z \ne x$, there is a $g_2 \in G_x$ with
$$
g_2 \cdot x' = z. 
$$
Let's check that we can take 
$$
z = g_1^{-1}\cdot y';
$$ 
in other words, we need to check that $g_1^{-1} \cdot y' \ne x$. If this were the case, then we would have
$$
g_1\cdot x = y',
$$
but we already know that $g_1 \cdot x = y$ and that $y \ne y'$. Now I claim that $g_1g_2$ is the required element. We have
$$
(g_1g_2)\cdot x = g_1\cdot (g_2 \cdot x) = g_1 \cdot x = y,
$$
since $g_2 \in G_x$, and
$$
(g_1g_2)\cdot x' = g_1\cdot (g_2 \cdot x') = g_1 \cdot (g_1^{-1} \cdot y') = y'.
$$
So intuitively, what we're doing here is first acting by an element of the stabilizer so that $x'$ has moved to the right position for it to then be moved to $y'$ when we act by the element moving $x$ to $y$.
A: The idea will be to pick $g_1 \in G$ with $g_1 \cdot x = y$ and then use $g_2 \in G_x$ to move $x'$ to $y'$.  We'll have to use $g = g_1 g_2$ to insure that $g \cdot x = y$.  And in order to have $g \cdot x' = y'$, we have to pick $g_2 \cdot x' = g_1^{-1} \cdot y'$.
