Order of Permutation

• Nov 18th 2006, 04:52 PM
NYMFan
Order of Permutation
Hello. Could someone help me with this?

Find the order of the permutation, sigma =
(1 2 3 4 5 6 7 8
3 1 5 6 2 7 8 4)
and write it as a product of cycles.

It's for a graduate school project and honestly I have never seen a problem like this before. So I appreciate any guidance.
• Nov 18th 2006, 05:46 PM
ThePerfectHacker
Quote:

Originally Posted by NYMFan
Hello. Could someone help me with this?

Find the order of the permutation, sigma =
(1 2 3 4 5 6 7 8
3 1 5 6 2 7 8 4)
and write it as a product of cycles.

It's for a graduate school project and honestly I have never seen a problem like this before. So I appreciate any guidance.

So you have the premutation on $\{1,2,3,4,5,6,7,8\}$ defined as,
$\sigma = \left( \begin{array}{cccccccc}1&2&3&4&5&6&7&8\\ 3&1&5&6&2&7&8&4 \end{array}\right)$
This is an element of a finite group $S_8$ thus it has an order. We need to find the smallest positive integer $n$ such that $\sigma^n$ is the identity premutation.
Observe the following,
$3\to 5\to 2\to 1$
$1\to 3\to 5\to 2$
$5\to 2\to 1\to 3$
$6\to 7\to 8\to 4$
$2\to 1\to 3\to 5$
$7\to 8\to 4\to 6$
$8\to 4\to 6\to 7$
$4\to 6\to 7\to 8$
This show that $n=3$ is the order.

To express it as a product of disjoint cycles we find the orbits of $\sigma$ which are,
$\{1,3,5,2\}$
$\{4,6,7,8\}$
Thus,
$\sigma=(1,3,5,2)(4,6,7,8)$
(Note although the binary operation on the premutation group is not commutative here it is and I could have expressed the product the other way aroung because the cycles are disjoint).