APPLICATIONS OF MODERN ALGEBRA: BURNSIDE’S LEMMA Anisha Dhalla 1

Motivating Problem  Claire is decorating her house for Christmas Eve dinner. She wants to decorate the top of a square table using 4 red and green tiles.  How many distinct ways can she arrange the tiles to get different tables? 2

Attempt #1  4 positions to place tiles  Each tile can be or  Hence, there are 2 4 =16 different arrangements.  What’s wrong with this solution? 3

Attempt #1  Problem :  Same arrangement can be obtained by rotating thetable.  2 4 is too many. 4

Attempt #2  The table can be rotated 4 times so there are 16/4=4 different arrangements. What is wrong with this solution? 5

Attempt #2  Some tables are represented by fewer than four pictures.  Example:  Looks the same no matter how you rotate it 6

Definition: Group Action on a Set  An action of a Group G on a set X is a map * : G × X  X such that: 1) g * x ∈ X  g in G, x in X 2) e * x = x  x in X 3) (g 1 g 2 ) * x = g 1 * ( g 2 * x)  g 1,g 2 in G, x in X  Under these conditions, X is a G-set. 7

Visualization of Group Action g g G= x g * x x g * x g = X 8

In our example:  G = { ρ 0, ρ 1, ρ 2, ρ 3 }  ρ 0 = identity  ρ 1 = 90 0 rotation CW  ρ 2 = rotation CW  ρ 3 = rotation CW  X = Set of 16 tiling arrangements ρ1ρ1 9

Group Action  Equivalence Relation  Let X be a G-set. For x 1, x 2 ∈ X, let x 1 ~x 2 iff there exists g ∈ G such that g 1 * x 1 =x 2. Then ~ is an equivalence relation on X.  Reflexive: x~x For each x ∈ X, e * x=x where e is the identity element of G. Hence x~x and ~ is reflexive.  Symmetric: x 1 ~x 2  x 2 ~x 1 Suppose x 1,x 2 ∈ X such that x 1 ~x 2 so that g * x 1 =x 2 for some g ∈ G. Then g -1 * x 2 = g -1 * (g * x 1 ) = (g -1 g) * x 1 = e * x 1 =x 1 Hence x 2 ~x 1 and ~ is symmetric.  Transitive: x 1 ~x 2, x 2 ~x 3  x 1 ~x 2 Suppose x 1,x 2,x 3 ∈ X such that x 1 ~x 2, x 2 ~x 3 so that g 1 * x 1 =x 2 and g 2 * x 2 =x 3 for some g 1,g 2 ∈ G. Then (g 2 g 1 ) * x 1 =g 2 * (g 1 * x 1 )=g 2 * x 2 =x 3. Hence x 1 ~x 3 and ~ is transitive. Therefore, ~ defines an equivalence relation 10

Orbits  Each cell in the partition of the equivalence relation is an orbit in X under G.  The cell containing x ∈ X is called the orbit of x denoted Gx  We are interested in counting the number of orbits in order to solve our motivating problem. 11

Relevant Terms  G x = {g ∈ G | g*x=x}  All elements in G that leave a specific x ∈ X undisturbed Examples: G = { ρ 0, ρ 1, ρ 2, ρ 3 } G = { ρ 0, ρ 2 } G = { ρ 0 } Stabilizer of x 12

Relevant Terms  X g ={x ∈ X | g*x=x}  All elements in X that stay the same when acted on by any g ∈ G. Examples: 13

Orbit Equation |Gx| = (G : G x ) The length of the orbit of an element is equal to the index of its stabilizer. If G is finite, then |Gx| is a divisor of |G|. 14

Burnside’s Lemma  Let G be a finite group that acts on a finite set X. If r is the number of orbits in X under G, then:  The number of orbits is equal to the average number of points fixed by an element of G. 15

Burnside’s Lemma: Proof 16

Burnside’s Lemma: Proof 17

Back to our Motivating Problem…  To use Burnside’s Lemma, we need to determine the cardinality of each X g. X = {all 2x2 red and green tiling designs} G = { ρ 0, ρ 1, ρ 2, ρ 3 } 18

Back to our Motivating Problem… X = {all 2x2 red and green tiling designs} G = { ρ 0, ρ 1, ρ 2, ρ 3 } Total ρ0ρ0 ✔✔✔✔✔✔✔✔✔✔✔✔✔✔✔✔ ρ1ρ1 ✔✔ ρ2ρ2 ✔✔✔✔ ρ3ρ3 ✔✔ Total |X ρ 0 |=16 |X ρ 1 |=2 |X ρ 2 |=4 |X ρ 3 |=2 19

Back to our Motivating Problem… |X ρ 0 |=16 |X ρ 1 |=2 |X ρ 2 |=4 |X ρ 3 |=2 20

Orbits The six orbits (equivalence classes) are shown below: { } {,,, } {, } {,,, } { } 21

Example #2: Wrapping Presents  Claire wants to wrap some cube-shaped presents using a different colour of wrapping paper on each face.  If she has six colours of wrapping paper available, how many distinctly coloured presents can she wrap? 22

Example #2: Wrapping Presents  Primary thoughts:  Set X is the total number of ways the 6 cube faces can be coloured. |X|=6!=720  Group consists of the set of possible rotations of the cube. When placed on a table, any one of six faces can be placed down and any one of four to the front: 6x4=24 possible positions. Achieve these different rotations by elements of G so |G|=24  Number of distinguishable presents is the number of orbits in X under G  Use Burnside’s Lemma! 23

Example #2: Wrapping Presents  X = {Set of 720 coloured presents}  G={Group of 24 rotations}  |G|=24  For g ∈ G where g≠e, |X g |=0  |X e |=720  X = {Set of 720 coloured presents}  G={Group of 24 rotations}  |G|=24  For g ∈ G where g≠e, |X g |=0  |X e |=720 (number of orbits) = 1/24 x 720 = 30 Using Burnside’s Lemma: Therefore, there are 30 distinguishable ways that Claire can wrap the presents. 24

Exam Question:  Claire is making bracelets which consist of 5 beads (with no clasp) equally spaced around a circle. Using five different coloured beads, how many bracelets does Claire need to make to have a complete set? 25

Exam Question Solution:  The bracelet can be turned over and rotated, so G = D 5 with 5 rotations and 5 reflections. |D 5 |=2x5=10  The set X consists of the 5!=120 possibilities for putting the beads on the bracelet.  Only the identity element leaves any arrangement fixed and it leaves all 5! elements of X fixed, so |X e |=120 and |X g |=0 for all g≠e.  If we consider the whole dihedral group D 5 acting on the set X, the number of bracelets to make a set is: 26