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I really need help with this topic I have an exam tomorrow and am trying to get this stuff in my head. But the book is not explaining me these two topics properly.

It gives me the definition of a stabilizer at a point where $\mathrm {Stab}_G (i) = \{\phi \in G \mid \phi(i) = i\}$, and where $\mathrm{Orb}_G (i) = \{\phi(i) \mid \phi \in G\}$.

I do not know how to calculate the stabilizer nor the orbit for this. I am also given an example

Let $G = \{ (1), (132)(465)(78), (132)(465), (123)(456), (123)(456)(78), (78)\}$ and then

$\mathrm{Orb}_G (1) = \{1, 3, 2\}$,
$\mathrm{Orb}_G (2) = \{2, 1, 3\}$,
$\mathrm{Orb}_G (4) = \{4, 6, 5\}$, and
$\mathrm{Orb}_G (7) = \{7, 8\}$.

also

$\mathrm{Stab}_G (1) = \{(1), (78)\},\\ \mathrm{Stab}_G (2) = \{(1), (78)\},\\ \mathrm{Stab}_G (3) = \{(1), (78)\},\text {and}\\ \mathrm{Stab}_G (7) = \{(1), (132)(465), (123)(456)\}.$

If someone could PLEASE go step by step in how this example was solved it would be really helpful.

Thank you

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    Thank byron for editing it. and Tomasz even one computation from each with detailed steps would be helpful. And in order for me to learn I will do all the other computations2012-08-16

3 Answers 3

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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.

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    @user37012 Multiplication isn't a completely wrong way to think about this, but have you tried thinking of the cycles as functions? That's what they are! $(465)$ is notation for the function taking 4 to 6, 6 to 5 and 5 to 4, and implicitly that means 1 goes to 1, 2 goes to 2 and 3 goes to 3. that's why $(465)1=1$.2012-08-16
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In simple terms,

Stabilizer of a point is that permutation in the group which does not change the given point => for stab(1) = (1), (78)

Orbit of a point(say 1) are those points that follow given point(1) in the permutations of the group. =>orbit(1) = 1 for (1); 3 for (132)...; 2 for (123)...