Abstract Algebra - Orbit of a permutation

[vwishndaetr]

For this problem, I have to find all orbits of given permutation.

$$\sigma: \mathbb{Z} \rightarrow \mathbb{Z}$$

Where,

$$\sigma(n)=n-3$$

Now, the problem is I do not know how to approach this permutation in the given format.

All the permutations I dealt with were in the form:

$$\mu = \left( \begin{array}{cc} 1\ 2\ 3\ 4\ 5\ 6\\ 1\ 2\ 3\ 4\ 5\ 6 \end{array} \right)$$

Which I understand. But I do not understand the sigma permutation first mentioned. I tried another example where I had an answer to σ(n)=n+2, but I did not understand how that answer was achieved.

If someone can guide me with a start that'd be great.

 

[miqbal]

The sigma format makes it much harder to see what's going on with the permutation.

All it really is though is sending the nth element to the n - 3 element. So 4 goes to 1, 3 goes to 6, 2 goes to 5...


$$\mu = \left( \begin{array}{cc} 1\ 2\ 3\ 4\ 5\ 6\\ 4\ 5\ 6\ 1\ 2\ 3 \end{array} \right)$$

 

[Office_Shredder]

No that's not quite correct. Sigma is a map from the integers to the integers, so you can't express it like that. The map is what it says it is

$$\sigma(3) = 3-3=0$$
$$\sigma(5) = 5-3= 2$$ and so forth. Basically sigma is just shifting the integers three places over

 

[miqbal]

Sorry i thought it was an element of the symmetric group of degree 6.

 

[vwishndaetr]

The sigma format makes it much harder to see what's going on with the permutation.

All it really is though is sending the nth element to the n - 3 element. So 4 goes to 1, 3 goes to 6, 2 goes to 5...$$\mu = \left( \begin{array}{cc} 1\ 2\ 3\ 4\ 5\ 6\\ 4\ 5\ 6\ 1\ 2\ 3 \end{array} \right)$$

I tried thinking of it that way. I originally thought n=1 -> -2, n=2 -> -1, n=3 -> 0, n=4 -> 1, then 1 -> -2. So that would be an orbit. So looking at the structure so to say, looked like each orbit consisted of 4 elements. But then the -2, -1, 0 kinda disappeared? I was not comfortable with it. Let's just say that.

No that's not quite correct. Sigma is a map from the integers to the integers, so you can't express it like that. The map is what it says it is

$$\sigma(3) = 3-3=0$$
$$\sigma(5) = 5-3= 2$$ and so forth. Basically sigma is just shifting the integers three places over

Ok, but what about $$\sigma(1) = 1-3=-2$$

1 is an integer, -2 is an integer, why doesn't that work?

 

[Office_Shredder]

$$\sigma(1)=-2$$ is true.

What is the definition of an orbit here? It helps to find out what the question is actually asking in this context

 

[vwishndaetr]

$$\sigma(1)=-2$$ is true.

What is the definition of an orbit here? It helps to find out what the question is actually asking in this context

Taken from First course in Abstract Algebra:

Let $$\sigma$$ be a permutation of a set A. The equivalence classes in A determined by the equivalence relation (1) are the orbits of $$\sigma$$.

Definition is a bit high on vocabulary, but I think I understand it.

For example, in the following:

$$\sigma = \left( \begin{array}{cc} 1 \ 2 \ 3 \ 4 \ 5 \ 6 \ 7 \ 8\\ 4 \ 3 \ 2 \ 8 \ 5 \ 1 \ 7 \ 6 \end{array} \right)$$

The orbits are:

$$\{1, 4, 8, 6\} \ \ \{2, 3\} \ \ \{5\} \ \ \{7\}$$

But I am a bit confused how to determine an orbit with the form given in original post.

 

[Office_Shredder]

So look at the orbit of 1. Let's see what some elements in it are

$$\sigma(1)=-2, \sigma(-2)=-5, \sigma(-5)=-8,$$

and in the other direction

$$\sigma(4)=1, \sigma(7)=4$$

noticing a pattern?

 

[vwishndaetr]

So look at the orbit of 1. Let's see what some elements in it are

$$\sigma(1)=-2, \sigma(-2)=-5, \sigma(-5)=-8,$$

and in the other direction

$$\sigma(4)=1, \sigma(7)=4$$

noticing a pattern?

Omg. It totally clicked. I am so contributing to this forum. Member by tomorrow night that's for sure. The ultimate source for tutoring. Best thing is, never yet have I gotten just an answer. Always an explanation to help me solve it myself. A lot of nice people with patience on here. Thank you very much! I am so excited, high hopes for the next test! :)

Orbits will be:

$$\{3n\ |\ n \ \epsilon \ \mathbb{Z}\} \ \ \{3n-1\ |\ n \ \epsilon \ \mathbb{Z}\} \ \ \{3n-2\ |\ n \ \epsilon \ \mathbb{Z}\}$$