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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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Precisely showing all these seems to be so tedious and unrewarding that I doubt anyone's going to do that for you, especially since it's pretty much a homework (not that it's very hard; after all, the group has only 6 elements...). What exactly baffles you so much about this example? It seems to me that you just need to apply the definition. –  tomasz Aug 16 '12 at 12:43
    
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 computations –  user37012 Aug 16 '12 at 13:50
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1 Answer 1

up vote 5 down vote accepted

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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alright, I get how to do Stabilizers perfectly I just needed more practice on that. But for the orbit that still upsets me. i know we have permutation multiplied to a single number, we just cant do our regular multiplying technique and treat 1 as a permutation also. I tried doing that but it was obviously wrong. i noticed (465) dissapeared and then only 3 was left could you please explain why –  user37012 Aug 16 '12 at 13:48
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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 $(465)1=1$. –  rschwieb Aug 16 '12 at 13:55
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