Stabilizer of a point and orbit of a point

You are not going to survive at all if you can't compute something after having its definition. Really spend time undestanding definitions, and do it a lot sooner than the night before the exam.

Doing the whole thing will not help you in the long run, but doing some samples is fair enough! I'm assuming the notation is doing composition this way:$ (fg)(x)=f(g(x))$.

Then $\mathrm{Orb}(1)=\{\phi(1)\mid \phi\in G\}=\underline{\{1,3,3,2,2,1\}}=\{1,2,3\}$. Each one of the numbers between the underlined braces is, in the order you listed them, the result of applying each element of $G$ to 1. For example $(132)(465)1=(132)1=3$.

If you cannot apply the permutations to a single number, then you indeed have a lot more studying to do.

For $\mathrm{Stab}_G(1)$, you just need to pick out all the elements of $G$ that don't move 1. Obviously $(1)$ and $(78)$ do not move 1. The first is just the identity permutation, and the latter does not move 1 at all, since 1 does not appear. Checking the others, you see that they move 1 either to 2 or to 3.

All of the others are like this: completely routine computation to see if you can read and understand the notation and definitions.

If $\Bbb{G}$ is a group of permutations of the set $\Bbb{S}$ then StabG={${\phi}\in\Bbb{G}$ | ${\phi}$(I)=I , for every $i\in\Bbb{S}$ } . And on other hand, OrbG={ ${\phi}$(s)| $s\in\Bbb{S}$ } .

These definitions mean that StabG is the collection of those mappings which maps an element of $\Bbb{S}$ to itself . And the OrbG is collection of images of elements of $\Bbb{S}$ . Here , $\Bbb{G}$ is the set of those mappings(permutations), which forms group under multiplication. If we go for an analogy, let's take A as United States , which is a collection of states . And f(A) be one of the ways joining the states of A. Let G be the collection of those f(A)s. Now if we consider StabG , then it will be a collection of those f(A)s which joins a state of A to itself i.e the collection of the internal transportation systems of that state . On the other hand OrbG is the collection of the state or states joined by an external transportation system, with a selected state of A. Though the analogy is not good enough but I think it will work. Now comes to the problem. StabG(x) is the set of permutations which maps x to x. So StabG(1) = {(1),(78)}. As f(7)=8 and f(8)=7. So with out 7 and 8 all other elements have been mapped to themselves. An easy way to find out the appropriate permutation for Stab(x) is , finding those permutations where x is missing ,as if there is x in that, then f(x)≠x( as far the construction of the given group). Now comes the OrbG. So, Orb(1) is the images of 1 by the permutations of G. So 1 has been mapped to 1,2 and 3. ( by, 2nd and 3rd permutation in G, 1 has been mapped to 3 and by 3rd and 4th it has been mapped to 2, and by 1st and 6th it has been mapped to itself. So the simple way of finding Orb(x) is seeing the permutations, finding where x has been mapped .

One more idea except the example that has been used in question, that may ensure the deep understanding of this topic is , “The Rotation Group Of a Cube”. A cube which have six faces can be used to understand it . If we label the six faces by numbers from 1 to 6. Then we will get a set S containing those numbers and also we can construct a permutation group G(under multiplication) , where different rotations induce different permutations on faces. Then as , by any rotation one face will come to another's place (including it's own). So, one labelled face ( from 1 to 6) can be mapped to any of those six faces . So Orb(1)={1,2,3,4,5,6}. Now we will seek for those rotations for which a face comes in it's own position. Likely, Stab(1) is the collection of those rotations i.e 0, 90, 180,270. So, Stab(1)={0,90,180,270}. And these are the basic ideas of Stab and Orb.