Is the regular group action of $G$ on itself doubly transitive? Is the action of $G$ on itself ($X=G$) via left multiplication doubly transitive?
(I know it's transitive but can't figure out how to prove or disprove $2$-transitivity)
Edit -
Doubly transitive - For any $x_1, x_2, y_1, y_2 \in X$ with $x_1\neq x_2$ and $y_1\neq y_2$, there exists some $g\in G$ such that $gx_1=y_1$ and $gx_2=y_2$. See: https://groupprops.subwiki.org/wiki/Doubly_transitive_group_action
 A: HINT: If $y_1 = g x_1$, $y_2 = g x_2$, then
$$y_1^{-1} y_2 = x_1^{-1} x_2$$
( like the "vectors" $x_1 x_2$ and $y_1 y_2$ are equal)
A: The regular action is not just transitive, it's sharply transitive. That means, given any $x,y\in G$ there exists a unique group element $g\in G$ for which $y=gx$. There is no wiggle room in the choice of $g$! And thus, if there is more than one choice of where we might want to simultaneously send a second element to, we will not be able to realize all these destinations since there is only one travel option, $g$, to work with! This occurs if $|G|>2$.
In particular, the only element of $G$ that sends the identity element $e\in G$ to $g\in G$ is $g$ itself. If we have a second element $x\in G$, the only place it could also be sent then is to $gx$, not to any other element. Since $x$ is distinct from $e$, we can say $gx$ is distinct from $x$, and if there were any third element $y\ne g,gx$ that would mean it's impossible to arrange for $e\mapsto g,x\mapsto y$ since $e\mapsto g$ automatically implies $x\mapsto gx$ in the regular representation.
