×

### Let's log you in.

or

Don't have a StudySoup account? Create one here!

×

or

18

0

3

# Class Note for MATH 796 at KU 6

Marketplace > Kansas > Class Note for MATH 796 at KU 6

No professor available

These notes were just uploaded, and will be ready to view shortly.

Either way, we'll remind you when they're ready :)

Get a free preview of these Notes, just enter your email below.

×
Unlock Preview

COURSE
PROF.
No professor available
TYPE
Class Notes
PAGES
3
WORDS
KARMA
25 ?

## Popular in Department

This 3 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at Kansas taught by a professor in Fall. Since its upload, it has received 18 views.

×

## Reviews for Class Note for MATH 796 at KU 6

×

×

### What is Karma?

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 02/06/15
Friday 44 Group Actions and Poly Theory How many different necklaces can you make with four blue two green and one red bead It depends what different means The second necklace can be obtained from the rst by rotation and the third by re ection but the fourth one is honestly different from the rst two 000 If we just wanted to count the number of ways to permute four blue two green and one red beads the answer would be the multinomial coef cient 7 7 7 105 421 41211 However what we are really trying to count is orbits under a group action Let G be a group and X a set An action of G on X is a group homomorphism a G A x the group of permutations of Xi Equivalently an action can also be regarded as a map G X X A X sending 91 to gr such that o 131 z for every I E X where lg denotes the identity element of G o ghz ghz for every gh E G and z E Xi The orbit of z E X is the set O gzlgEGCX and its stabilizer is S gEGlgzzCG which is a subgroup of Cl To go back to the necklace problem we now see that same really means in the same orbitlli In this case X is the set of all 105 necklaces and the group acting on them is the dihedral group D7 the group of symmetries of a regular heptagoni The number we are looking for is the number of orbits of D7i Lemma 1 For every I E X we have lellel Proof The element gr depends only on which coset of Sac contains 9 so lOgcl is the number of cosets which is Gllsxl 5 Proposition 2 Burnside s Theorem The number of orbits of the action of G on X equals the average number of xed points 1 Z16Xl911 gEG Proof For a sentence P let xP 1 if P is true or 0 if P is false the Garsia chi function Then 1 1 Number of orbits Z O E Z lel xeX l Tl l l zeX 1 Mar 7 1 zeX gEG 1 1 xgzz2IEXlgrI gEGrEX 9amp0 Typically it is easier to count xed points than to count orbits directlyi Example 1 We can apply this technique to the necklace example above 0 The identity of D7 has 105 xed points 0 Each of the seven re ections in D7 has three xed points the single bead lying on the re ection line must be red and then the two green beads must be equally distant from it one on each side 0 Each of the six nontrivial rotations has no xed points Therefore the number of orbits is lD7l 7 which is much more pleasant than trying to count them directly 10573 126 Example 2 Suppose we wanted to nd the number of orbits of 7bead necklaces with 3 colors without specifying how many times each color is to be used 0 The identity element of D7 has 37 2187 xed points 0 Each re ection xes one bead which can have any color There are then three pairs of beads ipped and we can specify the color of each pair Therefore there are 34 81 xed points 0 Each rotation acts by a 7cycle on the beads so it has only three xed points all the beads must have the same color Therefore the number of orbits is 218778163 198 14 More generally the number of inequivalent 7bead necklaces with h colors allowed is k7 7 6k 1 i 14 As this example indicates it is helpful to look at the cycle structure of the elements of G or more precisely on their images 19 6 Gxi Proposition 3 Let X be a nite set and let a G A 6X be a group action Color the elements ofX with h colors so that C also acts on the colorings 1 For g E G the number of xed points of the action ofg is 1629 where g is the number of cycles in the disjointcycle representation of 19 2 Therefore 2 equivalence classes of colorings 2 hag gEG Let7s rephrase Example 2 in this notation The identity has cycleshape 1111111 so Z 7 each of the six re ections has cycleshape 2221 so Z 4 and each of the seven rotations has cycleshape 7 so Z 1 Thus 1 is an example of the general formula Example 3 How many ways are there to kcolor the vertices of a tetrahedron7 up to moving the tetrahedron around in space Here X is the set of four vertices7 and the group G acting on X is the alternating group on four elements This is the subgroup of 64 that contains the identity7 of cycleshape 1111 the eight permutations of cycle shape 31 and the three permutations of cycleshape 22 Therefore7 the number of colorings is k4 11k 12 i

×

×

### BOOM! Enjoy Your Free Notes!

×

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

Bentley McCaw University of Florida

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

Anthony Lee UC Santa Barbara

#### "I bought an awesome study guide, which helped me get an A in my Math 34B class this quarter!"

Bentley McCaw University of Florida

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

Parker Thompson 500 Startups

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

Become an Elite Notetaker and start selling your notes online!
×

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com