About the stablizer of an element, $G_\alpha$ After reading some notes about permutation groups, I have tackled with this really simple question I hope it is not a ridiculous question. :)

From the first chapters of any book about these kinds of groups and knowing how a group $G$ acts on a set $\Omega$, we are faced to an important structure, named $G_\alpha$ where is $\alpha \in \Omega$. 

Why do we frequently regard $G_\alpha$ acting on set $\Omega-\{\alpha\}$. Why this set,$\Omega-\{\alpha\}$? Of course with this assumption a majority of theorems will be carried out properly and desirably. Thanks.
 A: Since $G_{\alpha}$ acts on $\Omega$ as a group of bijections, but every single one of those bijections maps $\alpha$ to $\alpha$, then the restriction of elements of $G_{\alpha}$ to $\Omega-\{\alpha\}$ will yield bijections from $\Omega-\{\alpha\}$ to itself. So $G_{\alpha}$ also acts on $\Omega-\{\alpha\}$. And given that we know exactly how it acts on $\{\alpha\}$, then understanding the action of $G_{\alpha}$ is equivalent to understanding the action on $\Omega-\{\alpha\}$. Moreover, for doubly transitive actions (which are of importance), having decided one of the actions (mapping $\alpha$ to $\alpha$), you now have a "leftover" transitive action on $\Omega-\{\alpha\}$. 
A: Consider two transitive groups on the same set: $G = \langle (1,2,3) \rangle$ and $H=\langle (1,2), (1,2,3) \rangle$ both acting on $\Omega = \{1,2,3\}$.
How well do $G$ and $H$ swirl the points of $\Omega$? Well they both do  a very thorough job! They are both transitive groups.  If the boss wants 1 moved to 3, no problem, $(1,2,3)^2$ will do the job in both $G$ and $H$.  Now move 3 back to 2? No problem! $(1,2,3)^2$ again will do the trick.
What if the boss goes mad with the power? What if he wants you to move 1 to 2 and at the same time leave 3 alone? In $G$ this cannot be done. Every element other than the identity moves all the points ($G$ is regular).  In $H$ this is easy, $(1,2)$ does the job.
What is the difference between these groups? They are both transitive, but once you start saying where one point goes, they react very differently about what they can do to the rest.
In other words, $G_3=\{()\}$, but $H_3=\langle (1,2) \rangle$. The first group cannot do anything, but the second is transitive on $\Omega \setminus \{3\}$.
To understand a group action, it is not enough to just ask if it is transitive. We often want to know how the group can move pairs of points.
Consider the actions of $G$ and $H$ on $\Omega \times \Omega$, the set of ordered pairs $\{ (i,j) : 1 \leq i,j \leq 3 \}$. Clearly these actions are not transitive, because $(i,i)^g = (i^g, i^g)$. One orbit is always the "diagonal orbit" $\{ (i,i) : i \in \Omega \}$.  What about the rest of $\Omega \times \Omega$?
Well for $G$ there are two orbits: $\{ (1,2), (2,3), (3,1) \}$ and $\{ (2,1), (3,2), (1,3) \}$. $G$ moves them in a circle, and does not change clockwise versus counter-clockwise.  We say $G$ is a rank three permutation group.
For $H$ there is only one orbit: any $i,j$ with $i \neq j$ can be sent to any other such pair. We say $H$ is a 2-transitive permutation group.
For groups acting on graphs, we restrict what part of $\Omega \times \Omega$ is allowed and of concern, and "transitive" becomes "vertex transitive" while "2-transitive" becomes "edge-transitive".
