Let G be a group, and let it act on a set X. Then we also have an induced action of G on $X \times X$ that is given by $g\cdot (x,y)=(gx,gy) \ \forall \ (x,y) \in X\times X$. The action is called doubly-transitive if the induced action has two orbits.
(i) Show that $G_y$ acts on the set $X-\{y \}$, $y \in X$
(ii) Assume $|X|>2$ Show that the action of G on X is doubly-transitive if and only if
the action of $G_y$ on $X − \{y \}$ from (i) is transitive for all $y ∈ X.$
So I'm not quite sure what to do. For (i), $G_y=\{\forall g \in G \ | \ gy=y \}$, and if we take $g,h \in G_y$ and act it upon an element $x \in X-\{y\}$ we need to prove the group action axioms
(1) $(gh)\cdot x=g\cdot(hx)$ : Since $y$ is not in the set $X-\{y\}$ then for any $g,h \in G$, $gx \neq x$ and $hx\neq x$ and so $g\cdot (hx)=g(hx)=(gh)x$.
(2) Since the identity exists in the group $G_y$ we have $e\cdot x=x$
As you can see I have no idea what I'm doing, I need some help o'wise guys of MathStack
Thanks in advance.
Best Answer
For the first one:
You just have to show, that $gx \neq y$ for all $g \in G_y$ and $x \in X \setminus \{y\}$ since all group action axioms are already fullfilled from $G_y \subseteq G$. I hope you can do this (try it! (really!) If you get stuck, post a comment).
Towards the second one: First consider that you have always at least 2 orbits, since the orbit of $(x,x) \in X \times X$ has always(!) the same two elements in each component (make this yourself clear) and since you have two different elements $x,y \in X$ the element $(x,y)$ is not in this orbit.
For one direction you have to show two orbits are enough, if every $G_y$ is transitive on $X \setminus \{y\}$:
First show, that $G$ is transitive on $X$ (meaning there is only one orbit). Here it is sufficient, that $G$ does not have a fixpoint $y$ (because then there exists $g \in G$ with $gy\neq y$ and because $G_y$ is transitive on $X \setminus \{y\}$ the orbit of $y$ is already everything). Because you have at least 3 elements $x,y,z$ and each element is contained in $X \setminus \{p\}$ for some $p$ you can do this. But this shows, that $\{(x,x) \ | \ x \in X\}$ is already one orbit.
Now you have to show, that $\{(x,y) \ | \ x,y \in X, x \neq y \}$ is an orbit. For fixed $y$ and since $G_y$ acts transitive you can see, that $\{(z,y) \ | \ z \in X, z \neq y\}$ is in the orbit of $(x,y)$ for any $y \neq x \in X$. And samewise for fixed $x \in X \{(x,z) \ | \ z \in X z \neq x\}$is contained in the orbit of $(x,y)$ for any $x \neq y \in X$. Since $x,y$ where arbitrary you can deduce that $\{(x,y) \ | \ x,y \in X, x \neq y \}$ lies in an orbit of $(x,y)$ and since these are now every elements of $X \times X$ you are done.
For the direction backwards consider some $G_y$ which is not transitive and the orbit of $(x,y)$ for some $x \in X$. Now show, that you don't get all elements of this form and hence you need $\geq 3$ different orbits.