# Abstract Algebra Order of Permutation

1. Nov 3, 2012

### Fellowroot

1. The problem statement, all variables and given/known data

See image.

2. Relevant equations

3. The attempt at a solution

I am finding the orders of permutations.

I know that you first find the orbits or cycles I don't know the difference (but I should).

This is what my professor said:

If you have (1345)(897) the orders are 4 and 3 respectively. If 4 and 3 are relatively prime then you just multiply 4 and 3 to get 12. So the order of the both is 12.

Now if you have (1345)(6,7,8,9,10,11) the orders are 4 and 6 respectively. So now you look at the Least Common Multiple

4, 8, 12 and 6, 12 and so the order is the least common multiple which is 12.

But in my problem I have (135)(264)(7), 7 is fixed and I don't know what to do when I have 3 of them.

The orders should be 3, 3, 1, but I don't know how to combine them to get the order of the entire thing.

2. Nov 3, 2012

### tiny-tim

Hi Fellowroot!
Remember what "order" of a permutation P means …

it means the (smallest positive) number n for which Pn = the identity

and that means that Pnx = x, for all x

Sooo … you need Pnx = x, for all x in (135),

Pnx = x, for all x in (264),

Pnx = x, for all x in (7).​

For the first, that's n = 3 6 9 12 …

For the second, that's also n = 3 6 9 12 …

For the third, that's n = 1 2 3 4 …

You get the picture?

3. Nov 3, 2012

### Fellowroot

Thanks for your response.

So I think I get it now.

An order to find the order of a permutation you first identify all of its orbits and write them out as a product.

You look at each one and determine the order of each one.

Then for each order number you write out its multiples.

You look at all the multiples and you look for the smallest number they all have in common.

This smallest number is your n, and if you take your permutation and raise it to the n power you get the identity.

All the identity is, is a permutation where each element is fixed (has no cycles) and the orbits are one element subsets of the original permutation.

In my original problem I have n = 3 6 9 12 …, n = 3 6 9 12 …, and n = 1 2 3 4...

So the LCM is 3, and if I do P^3 you do indeed get the identity! Yea!

But may I ask, what is so important about the identity permutation?

Also many times I see the identity in the form of a^n = e, where e is the identity, but really what is e? Is e really just a^0?

If e is an identity, is it also a permutation identity? Thanks.

Last edited: Nov 3, 2012
4. Nov 4, 2012

### tiny-tim

Hi Fellowroot!
Which of course goes by the name of "least common multiple"!
That's like asking "what is so important about the number zero?"

The identity is fundamental to any algebraic structure.

It's the simplest possible element.
yes, e is the permutation (a)(b)(c)(d)…, made up of all the possible 1-cycles

as a function, it is defined by e:a -> a, e:b -> b, etc