Permutation question

Completionist

Given the permutation p below in S6, Calculate the orbit of p.

1 2 3 4 5 6
5 6 1 2 3 4

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What does it mean by S6, I've been told it is the name given to the group of permutations of the elements 1-6 however i still do not understand.

What is the orbit and how can it be gotten? Is it the cycle?

Spent the last few hours reading sources on the subject and none seem to have answered either of my questions.

LeixlipRed

http://en.wikipedia.org/wiki/Symmetric_group

That's the wiki page for symmetry groups. Here the elements are a permutation of the n whole numbers from 1 to n. The group operation is the composition of two permutations, i.e. performing one permutation followed by another. Review composition of functions if you can't recall how that works.

As for the orbit, I'm not sure to be honest, I'm not familiar with the term in this context. But I haven't done much group theory so someone else will sort you out with that I'm sure.

LeixlipRed

Sorry I should have given an example of what a permutation is. The one you have there can be written explicitly as 1 goes to 5, 2 goes to 6, 3 goes to 1, 4 goes to 2, 5 goes to 3, 6 goes to 4. That permutation can be written as (153)(264), look at the wiki page for an explanation of that. Let's call this one f.

If I then wanted to compose that with the operation g=(12)(34)(56) how would I do it? The way to think of this is applying the permutation, f of g or f after g. So we apply the permutation in g first and then follow it up with the permutation f asks us to apply. So 1 goes to 2 first (in g), then 2 goes to 6 (in f). So in the composition f of g, 1 goes to 6. 2 goes to 1 (in g), 1 goes to 5 (in f) so 2 goes to 5. Try it yourself and see can you get an answer of:

123456
652143

or in the alternative notation (163254).

equivariant

This permutation is a function p from {1,2,3,4,5,6} to {1,2,3,4,5,6}. It is the function defined by
p(1) = 5
p(2) = 6
p(3) = 1
p(4) = 2
p(5) = 3
p(6) = 4

The orbit of an element of {1,2,3,4,5,6} under the permutation p is the set of all elements that we obtain by repeatedly applying p starting with the given element. So the orbit of 2, for example, is {2,p(2),p(p(2)),p(p(p(2))), ........}. In other words it is the set {2,6,4}, Note that p(4) = 2 so we only have 3 elements in this orbit. On the other hand the orbit of 1 is {1,5,3}. You should be able to convince yourself that these are the only two orbits under this permutation. Thus the orbits of p are {1,3,5} and {2,4,6}.

Note that it doesn't make sense to ask for the orbit (singular) of this permutation (unless you specify a particular starting point).

Completionist

equivariant wrote: »

This permutation is a function p from {1,2,3,4,5,6} to {1,2,3,4,5,6}. It is the function defined by
p(1) = 5
p(2) = 6
p(3) = 1
p(4) = 2
p(5) = 3
p(6) = 4

The orbit of an element of {1,2,3,4,5,6} under the permutation p is the set of all elements that we obtain by repeatedly applying p starting with the given element. So the orbit of 2, for example, is {2,p(2),p(p(2)),p(p(p(2))), ........}. In other words it is the set {2,6,4}, Note that p(4) = 2 so we only have 3 elements in this orbit. On the other hand the orbit of 1 is {1,5,3}. You should be able to convince yourself that these are the only two orbits under this permutation. Thus the orbits of p are {1,3,5} and {2,4,6}.

Note that it doesn't make sense to ask for the orbit (singular) of this permutation (unless you specify a particular starting point).

Hi,

Some questions.

What is the purpose of this :
{2,p(2),p(p(2)),p(p(p(2)))................}

If the orbits of P = {1,3,5} {2,4,6} and the cycle of P = {1,5,3} {2,6,4} does that mean that orbit is just a different name for cycle. It would explain why orbit has not been mentioned in most of my sources.

Completionist

LeixlipRed wrote: »

So 1 goes to 2 first (in g), then 2 goes to 6 (in f).

Im not entirely sure what you mean. If 1 goes to 2(g), and 2 goes to 6(f), would 6 go to 5(g)?

LeixlipRed wrote: »

So in the composition f of g, 1 goes to 6. 2 goes to 1 (in g), 1 goes to 5 (in f) so 2 goes to 5.

If 1 goes to 6, How can 1 also go to 5?

Sorry it's late:o

LeixlipRed

I think you need to review composition of functions.

http://en.wikipedia.org/wiki/Function_composition

Look at the diagram on the right hand side of that link there, it should help. We're performing two permutations.

{1,2,3,4,5,6}-->g-->{1,2,3,4,5,6}-->f-->{1,2,3,4,5,6}

You send one element from the first set into the function g, out pops another element and then that element is sent into f and out pops the final output. So every element of the original set is permuted twice. That's where your confusion lies I think.

equivariant

Completionist wrote: »

Hi,

Some questions.

What is the purpose of this :
{2,p(2),p(p(2)),p(p(p(2)))................}

Not sure what you mean by "purpose". I am trying to indicate in that sentence what is meant by the orbit of 2 under the permutation p.

If the orbits of P = {1,3,5} {2,4,6} and the cycle of P = {1,5,3} {2,6,4} does that mean that orbit is just a different name for cycle. It would explain why orbit has not been mentioned in most of my sources.

the phrase "cycle of P = {1,5,3}{2,6,4}" doesn;t make sense. A cycle is a particular type of permutation - it is not a set or a collection of sets. For example Consider the permutation p given by

1 2 3 4 5 6
3 1 2 4 6 5

The cycle decomposition of p is (1 3 2)(5 6) - note the round brackets to denote cycles. However the orbits of p are {1,2,3}, {5,6} and {4}. Notice that (1 2 3) is not a cycle of p even though {1,2,3} is an orbit of p.

Completionist

equivariant wrote: »

Not sure what you mean by "purpose". I am trying to indicate in that sentence what is meant by the orbit of 2 under the permutation p.

Sorry, I didn't understand it at the time, Do now though:cool: