Abstract Algebra: Groups of Permutations

[DEMJ]

Homework Statement

List the elements of the cyclic subgroup of S_6 generated by

f = \left(\begin{array}{llllll}<br /> 1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 &amp; 6\\<br /> 2 &amp; 3 &amp; 4 &amp; 1 &amp; 6 &amp; 5\\ <br /> \end{array}\right)

Homework Equations

The Attempt at a Solution

I really do not understand what the elements of a permutation really is. I know if I write this as the product of disjoint cycles I get (2341)(65) but other than that I have no idea what the elements could be. The only really element I can say I understand is the identity \epsilon = \left(\begin{array}{llllll}<br /> 1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 &amp; 6\\<br /> 1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 &amp; 6\\ <br /> \end{array}\right) should be an element since it is a subgroup. It also should have an inverse but don't know how to find that.
Please help me get jump started on this problem.

 

[Minimonster]

Homework Statement

List the elements of the cyclic subgroup of S_6 generated by

f = \left(\begin{array}{llllll}<br /> 1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 &amp; 6\\<br /> 2 &amp; 3 &amp; 4 &amp; 1 &amp; 6 &amp; 5\\ <br /> \end{array}\right)

Homework Equations

The Attempt at a Solution

I really do not understand what the elements of a permutation really is. I know if I write this as the product of disjoint cycles I get (2341)(65) but other than that I have no idea what the elements could be. The only really element I can say I understand is the identity \epsilon = \left(\begin{array}{llllll}<br /> 1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 &amp; 6\\<br /> 1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 &amp; 6\\ <br /> \end{array}\right) should be an element since it is a subgroup. It also should have an inverse but don't know how to find that.
Please help me get jump started on this problem.

You could just take powers of the element. Since it is cyclic you would eventually compute all the elements in the subgroup. This approach could take a while.

 

[DEMJ]

You could just take powers of the element. Since it is cyclic you would eventually compute all the elements in the subgroup. This approach could take a while.

What do you mean take the powers of the element? Could you give me an example of one? Thank you so much.

 

[Minimonster]

What do you mean take the powers of the element? Could you give me an example of one? Thank you so much.

With these type of symmetric groups, you take a power of an element by composing it with itself. Unfortunately I don't how to use latex else i could type up an example for you, you could google for such examples though.

Here's a start, but a bit lacking I think. The subsection titled "Elements" is where you wan to focus your attention.

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

 

[DEMJ]

you take a power of an element by composing it with itself

I don't understand because on the wiki page above it says that
(1 2 3 4 5 6)^2 = (1 3 5)(2 4 6) but when I compose (123456) with (123456) = (123456). I guess I am just royally confused.

 

[lanedance]

it may also help to write the permutation as
<br /> f = \left(\begin{array}{llllll}<br /> 1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 &amp; 6\\<br /> 2 &amp; 3 &amp; 4 &amp; 1 &amp; 6 &amp; 5\\ <br /> \end{array}\right) = (1234)(56)<br />

 

[lanedance]

I don't understand because on the wiki page above it says that
(1 2 3 4 5 6)^2 = (1 3 5)(2 4 6) but when I compose (123456) with (123456) = (123456). I guess I am just royally confused.

be careful on how you apply the composition

<br /> <br /> (1 2 3 4 5 6)^2 <br /> <br /> = \left(\begin{array}{llllll}<br /> 1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 &amp; 6\\<br /> 2 &amp; 3 &amp; 4 &amp; 5 &amp; 6 &amp; 1\\ <br /> 3 &amp; 4 &amp; 5 &amp; 6 &amp; 1 &amp; 2\\ <br /> \end{array}\right) <br /> <br /> = \left(\begin{array}{llllll}<br /> 1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 &amp; 6\\<br /> 3 &amp; 4 &amp; 5 &amp; 6 &amp; 1 &amp; 2\\ <br /> \end{array}\right)<br /> <br /> = (135)(246)<br /> <br />

 

[DEMJ]

it may also help to write the permutation as
<br /> f = \left(\begin{array}{llllll}<br /> 1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 &amp; 6\\<br /> 2 &amp; 3 &amp; 4 &amp; 1 &amp; 6 &amp; 5\\ <br /> \end{array}\right) = (1234)(56)<br />

So that is one element, right? But how do I go about finding the rest. And more important, how could I know exactly how many elements there will be?

 

[lanedance]

or you can read it off
(1 2 3 4 5 6) (1 2 3 4 5 6)
2 goes to 1 which goes to 6: 2->6
3 goes to 2 which goes to 1: 3->1

 

[lanedance]

So that is one element, right? But how do I go about finding the rest. And more important, how could I know exactly how many elements there will be?

the disjoint cycles will not affect each other. Consider re-applying each cycle, how long until it repeats a form?

what is (56)(56) equivalent to?

similarly for the other permutation (1234)?

 

[DEMJ]

Ok so I could write an element as (234165)^2 as \left(\begin{array}{llllll}<br /> 1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 &amp; 6\\<br /> 2 &amp; 3 &amp; 4 &amp; 1 &amp; 6 &amp; 5\\ <br /> \end{array}\right) \circ \left(\begin{array}{llllll}<br /> 1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 &amp; 6\\<br /> 2 &amp; 3 &amp; 4 &amp; 1 &amp; 6 &amp; 5\\ <br /> \end{array}\right) = (3412)(56)

 

[DEMJ]

Also, I still do not understand how to list all the elements because this is the only element I know how to find. Should I do (234165)^3, (234165)^4, ... on up to what exponent?

 

[lanedance]

i think your notation is a little out of whack, have a look at mine, and below
http://en.wikipedia.org/wiki/Permutation#Notation

that said it is just convention, though I've made a similar mistake before

when you ask up to what exponent, first have a think about what i wrote in post #10

 

[Minimonster]

Also, I still do not understand how to list all the elements because this is the only element I know how to find. Should I do (234165)^3, (234165)^4, ... on up to what exponent?

yes, just keep taking powers. The subgroup is cyclic so eventually you would end up with the identity element. As far as the size goes, by Lagrange's Theorem, it could be as big as S6 itself. Since that group has size 720, it could potentially be that large, although I doubt this particular exercise would torture you in that fashion.

 

[lanedance]

shouldn't the number of elements will be related to the order of the generator, in this case i think it will be 4

 

[Minimonster]

shouldn't the number of elements will be related to the order of the generator, in this case i think it will be 4

I guess it would, I didn't bother to calculate the order of the element given.

 

[lanedance]

and i think this group has no elements with order greater than 6

 

[jbunniii]

It's not too hard to see that the order of an element of S_n is the least common multiple of the orders of the cycles in the disjoint cycle decomposition.

In your case, you have f = (1 2 3 4)(5 6) so the order of f will be lcm(4,2) = 4. (Fortunately, it's not 720.)

You can easily use this fact to check all the decomposition types to conclude that indeed the maximum order of any element in S_6 is 6. There are two types of elements with this order: 6-cycles, such as (1 2 3 4 5 6), and those with a 3-cycle and a 2-cycle, such as (1 2 3)(4 5).